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ANSWER KEYSection Review 1.1 Part A Completion 1. 2. 3. 4. 5. 6. mass space composition changes five carbon 7. 8. 9. 10. 11. 12. carbon organisms Analytical Chemistry physical chemistry energy transfer more than one Part D Questions and Problems 19. production of chemicals such as insulin; replacement of a gene that is not working properly (gene therapy) 20. Factors include poor soil quality, lack of water, weeds, plant diseases, and pests that eat crops. 21. Data collected by the robotic vehicle Opportunity indicated that the landing site was once drenched in water. Part B True-False 13. NT 14. ST 15. AT 16. ST 17. AT Section Review 1.3 Part A Completion 1. 2. 3. 4. 5. achemists tools techniques measurement systematic 6. 7. 8. 9. 10. scientific method hypothesis experiment theory scientific law Part C Matching 18. f 19. i 20. b 21. d 22. h 23. g 24. 25. 26. 27. c a e j Part D Questions and Problems 28. a. b. c. d. physical chemistry analytical chemistry biochemistry organic chemistry Part B True-False 11. NT 12. NT 13. ST © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Part C Matching 14. c 15. b 16. a 17. e 18. d 19. f Section Review 1.2 Part A Completion 1. 2. 3. 4. 5. specific microscopic energy conserve batteries 6. 7. 8. 9. 10. productivity crops specific space chemical composition Part D Questions and Problems 20. a. observation d. observation b. hypothesis e. scientific law c. experiment 21. collaboration and communication Section Review 1.4 Part A Completion 1. 2. 3. 4. 5. 6. plan implementing three analyze unknown plan 7. 8. 9. 10. 11. calculate evaluate sense unit significant figures Part B True-False 9. NT 10. NT 11. AT 12. AT Part C Matching 13. d 14. a 15. c 16. b 17. f 18. e Answer Key 759 Part B True-False 12. ST 13. AT 14. NT 15. NT 16. AT 17. AT Section 1.3 1. Examples include: The battery could be dead, the car could be out of gas, the spark plugs could be fouled, the wires could be loose. 2. Several experiments were performed to test hypotheses. Some of the experiments disproved some hypotheses, one experiment resulted in the car starting. Based on the experiment, you could hypothesize that a wire was loose. 3. Check students’ answers. 4. Theories are only as reliable as the knowledge on which they are based. Throughout the history of science, theories have been discarded or modified as scientific knowledge has increased. Part C Matching 18. b 19. e 20. d 21. a 22. c Part D Questions and Problems 23. Step 1: Knowns: length ϭ 10.0 inches. 1 inch ϭ 2.54 cm 2.54 cm Step 2: 10.0 in ϫ ᎏᎏ ϭ 25.4 cm i n Step 3: There are about two and a half centimeters per inch, so the answer 25.4 makes sense. 24. Step 1: Knowns: distance ϭ 5.0 km 1 km ϭ 0.62 mi 0.62 mi Step 2: 5.0 km ϫ ᎏᎏ ϭ 3.1 mi k m Step 3: There is a little more than half a mile per kilometer. 3.1 is a little more than half of 5.0. Section 1.4 1. a. cost of apples ϭ $1.50 a pound weight of an apple ϭ 0.50 pound dollars available ϭ $16 b. cost per apple ϭ $0.75 number of apples purchased ϭ 8 c. Two apples weigh a pound, and a pound costs $1.50. $6.00 is four times $1.50. 2. Figure out how many pounds of apples can be purchased with $6.00. Then figure out how many apples this represents. Practice Problems 1 Section 1.1 1. a. analytical b. biochemistry c. physical 2. a. applied b. pure c. pure d. organic e. inorganic d. applied e. pure © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Interpreting Graphics 1 1. five 2. Component D 3. 3.5 minutes 4. Component A 5. Component E 6. Component C Section 1.2 1. Answers may include development of new methods of energy conservation, such as new types of insulation; development of sources of energy other than fossil fuels, such as biodiesel; and the development of new methods of energy storage, such as improved batteries. 2. a. T f. T b. T c. F d. T e. F g. T h. T i. T Vocabulary Review 1 1. 2. 3. 4. 5. c l a d m 6. 7. 8. 9. 10. b f i n g 11. 12. 13. 14. 15. j o h e k 760 Core Teaching Resources Quiz for Chapter 1 1. 2. 3. 4. 5. 6. 7. T F F T T scientific method hypothesis 8. 9. 10. 11. observations hypothesis communicate a. The volume of the test tube is needed. b.There is enough information. figures. Express the answer in scientific notation, if appropriate. Chapter 1 Test B A. Multiple Choice 1. 2. 3. 4. c c c b 5. d 6. b 7. b 8. c 9. c 10. a Chapter 1 Test A A. Multiple Choice 1. d 2. c 3. b 4. b 5. b 6. c 7. a 8. b B. Problems 11. Organic chemistry is the study of essentially all chemicals containing carbon. Inorganic chemistry is the study of essentially all chemicals that do not contain carbon. Analytical chemistry is concerned with the composition of chemistry. Physical chemistry is concerned with mechanisms, rates, and energy transfer when matter undergoes a change. Biochemistry is the study of processes that take place in organisms. 12. Check students’ answers. 13. a. Hypothesis: The lawn needs water. Experiment: Water the lawn every day for one week. b. Hypothesis: The lawn needs fertilizer. Experiment: prescribed. Fertilize the lawn as B. Questions 9. Chemistry is the study of the composition of matter and the changes matter undergoes. 10. Making Observations: Use your senses to obtain information directly. Testing Hypotheses: A hypothesis is a proposed explanation for what you observed. Experiments are done to test a hypothesis. Developing Theories: A theory is a well-tested explanation for a broad set of observations. A hypothesis may become a theory after repeated experimentation. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. C. Essay 11. 1. Analyze. List the knowns and the unknown. A known may be a measurement of an equation that shows a relationship between measurements. Determine what unit, if any, the answer should have. Make a plan for getting from the knowns to the unknown. You might draw a diagram to help visualize the relationship between the knowns and the unknown; or use data from a table or graph; or select an equation. 2. Calculate. This step can involve converting a measurement from one unit to another or rearranging an equation to solve for an unknown. 3. Evaluate. Decide if the answer makes sense. Check your work. Make a quick estimate to see whether your answer is reasonable. Make sure the answer is given with the correct number of significant Chapter 1 Small-Scale Lab Safety goggles should be worn at all times when working in the laboratory. If glassware breaks, tell your teacher and nearby classmates. Dispose of the glass as instructed by the teacher. If you spill water near electrical equipment, stand back, notify your teacher, and warn other students in the area. When working near an open flame, tie back hair and loose clothing. Never reach across a lit burner. Keep flammable materials away from the flame. After cleaning up the work area, wash your hands thoroughly with soap and water. It is not always appropriate to dispose of chemicals by flushing them down the sink. You should follow your teacher’s instructions for disposal. Answer Key 761 substance shape or volume shape or volume shape gas Part D Questions and Problems 19. 7. solid e. d 18. extensive intensive mass volume amount type 7. compound b. 2. heterogeneous e. iron e. 17. mixture d. a.2 Part A Completion 1. ST 14. 22. 4. nitrogen Part B True-False 8. heterogeneous c. a. ST Part C Matching 12. All rights reserved.. Pb c. homogeneous 20. 20. compound e. 21. substance b. 8. a. AT 12. yes Part B True-False 9. b 17. a. f 24. 3. homogeneous b. 4. 5. e 16. 9. silver d. AT Part C Matching 13. 18. 11. 9. yes d. mixture Section Review 2. 6. e © Pearson Education. g 762 Core Teaching Resources . 6. copper b. NT 11. f 13. c 17. ST 10. 6. vapor b. 3.3 Part A Completion 1. AT 12. AT Part C Matching 15. ST 9. 8. d i c a 19. a. a 15. heterogeneous c. 7. j Part D Questions and Problems 26. 3. a 15. liquid 27. gas c. NT 13. K b. b 16. element d. Cl e. 10. 4. h e b g 23. AT 10. AT 11. yes b. 2. liquid c. Inc. 10. element c. 5. S d. publishing as Pearson Prentice Hall. Na 20. 16. hydrogen c. a. d 14. a.1 Part A Completion 1. 5. substance d.Section Review 2. element or compound compound or element elements ratio or proportions chemical substance mixture symbol C K Part B True-False 11. c 14. k 25. mixture heterogeneous or homogeneous homogeneous or heterogeneous solutions phase physical distillation Part D Questions and Problems 18. no d. compound 19. Section Review 2. 2. b 16. iron c. substance b. extensive Interpreting Graphics 2 1. 32. or a mixture of liquids. 2. a. 2. 5. c Part C Matching 14. carbonic acid b. ST 10. a. substance Section 2. phase 2. 6. Sn c. substance c. nitrogen. Inc. mixture c. C 5. chemical separation c. a. physical separation 4. a. compound 4. potassium 3. chemical physical chemical reactants chemical composition conservation of mass mass 4. Ag Part B True-False 9. 6. 4. intensive Section 2. a.4 1. a. distillation. mixture d. No. d 17. heterogeneous b. 3. 7.. solution 4. physical change b. chemical change c. a mixture with a uniform composition throughout 3. Chemical process. c 2. 8. publishing as Pearson Prentice Hall. Homogeneous mixture. substance no yes yes yes no no c. element d. water 4. compound b. It could be a liquid element.Section Review 2. e 15. 4. mixture b. elements Solution: substance 5. 5. which mix with the air. AT 11. 7. a Section 2. homogeneous c. b a b d gas no liquid no solid yes 6. 20. Na d. 6. physical change 3. The products of the reaction are gases.4 Part A Completion 1.3 1. hydrogen 2. 36 grams © Pearson Education. while pure iron is an element. it has a variable composition. 62 grams 5. ST 12. a. evaporation 2. Iron oxide is a compound. a. bromine. gold d. a. extensive d. lead b. mixture c. a chemical change 21. 3. a liquid compound. Since motor oil comes in grades. chemical change d. ethanol 8. NT b. 8. chlorine 7. carbon dioxide.1 1.4 grams 6. mass solid mixture gas Answer Key 763 . 2. c 18. 3. a. physical separation b. mixture d. NT 13. All rights reserved.2 1. heterogeneous Vocabulary Review 2 1. intensive b. chemical symbol 3. Practice Problems 2 Section 2. the law of conservation of mass Part D Questions and Problems 19. Quiz for Chapter 2 1. c 10. NT 28. A physical change alters a substance without changing its composition. 31. 12. c d d d a 21. ST 25. a c d c d 16. 35. Completion 26. i d h a 5. 2. 17. e 7. 2.. skim the sawdust directly off the surface of the water. AT B. To separate the mixture. 15. 4. All rights reserved. potassium b. 11. 9. g B. e 9. d 20. D. a Chapter 2 Test A A. 12. 3. 34. Essay 37. NT 26. 3. color change. © Pearson Education. sawdust floats in water. Matching 1. c C. 3. j 6. 13. True-False 24. Some possible clues to chemical change are a transfer of energy. b 6. Matching 1. h 8. b 23. 27. 20. i f g a 5. 35. 6. b 16. hydrogen Chapter 2 Test B A. 33. AT 24. NT 23. NT E. oxygen c. Salt dissolves in water. 5. reactants chemical physical substance C. 7. 10. 30. Inc. compounds iron homogeneous conserved products E. 34. Multiple Choice 11. 14. Boil off the water. Solid salt remains at the bottom of the flask. 13. and the burning of wood. b 22. physical blend element vary a. 764 Core Teaching Resources . f 8. the production of gas. Completion 29. j 10. 19. 14. Examples of chemical change include the rusting of iron. Multiple Choice 11. and the formation of a precipitate. iron filings are attracted to a magnet. AT 25. one or more substances change into one or more new substances. the reaction between iron and sulfur to produce iron sulfide. d b b b 15. 36. d 18. AT 22. 4. Melting or boiling are physical changes. b 7. 32. 30. recapturing the water through distillation. c 17. 32. 4. 2. d 9. 18. then expose the resulting mixture to a magnet to remove the iron filings. 28. a 19. 29. Essay 36. energy distillation vary chemical physical 31. a d d b a d d 8. publishing as Pearson Prentice Hall. True-False 21. In a chemical change. ST 27. d D. liquid vapor heterogeneous N 33. 24. 23. 18. NT 12. 7. Part B True-False 11.4 1 ϩ 2 ϩ 3 ϭ BLACK!. NT 13. f 18.2 Part A Completion 1. 5. 9. 4. which suggests the presence of starch. 2. A black color indicates the presence of Kl. 4. b 15.4 cm3 1. ST 12. NT Part C Matching 15. b Answer Key 765 . 4 21. A picture can be drawn with colored ink. 3. f 27. 3.1 ϫ 102 b.. page 56 Analysis NaClO KI KI Starch KI Paper KI Cereal yellow H2O2 yellow CuSO4 brown ppt black Section Review 3. A black color indicates the presence of starch. but different in other reactions. Section Review 3. 2. both turn blue-black.1 Part A Completion 1. 17. e 16. 8. Add CuSO4 or H2O2. 16. AT 14. a.35 ϫ 102 mL b. 5. 4. 8. 7. 5. All rights reserved. 3 c. If an antacid tablet contains starch.01% Kl. 21. They all turn a mixture of Kl and starch black. a 17. c black black Figure A black Part D Questions and Problems 20. the mixture turns a blue-black color. 2. Starch. 4. Wet only a portion of a small pile of salt with starch. AT 13. Certain areas can be treated with NaClO to bleach parts of the picture. d 19. AT black black Part C Matching black black black 14. The results may be the same in reactions that are simnilar to the one with Kl and starch. metric seven meter liter weight 6. j 26. publishing as Pearson Prentice Hall. 10. yellow 2. absolute value 100% scientific notation known estimated Part B True-False 11. 22. Inc. it will turn black whenn treated with Kl + NaClO. k e m g a 20. a. 19. l d c h i 25. 3. gram Celsius kelvin joule/calorie calorie/joule You’re The Chemist 1. 2. 2 c. 3. 10. 8. Most table salt contains 0. © Pearson Education. 4. The color in the ink becomes bleached. 9. Add Kl + NaClO to various foods. accuracy precision experimental value error accepted value 6.Chapter 2 Small-Scale Lab Section 2. 58 kg ϫ ᎏᎏ ϭ 1580 g kg mass 1580 g Density ϭ ᎏᎏ ϭ ᎏᎏ ϭ 1. c.99 ᎏᎏ 50.54 km/h h mi 4. 125 g ϫ ᎏᎏ ϭ 0. 3. 4. ST Part C Questions and Problems 1 cm3 6.58 g/cm3 volume 1000 cm3 Section Review 3.25 ϫ 10Ϫ1 kg 1000 mL b. 6. 9. Bruce’s 1 cm 5% 6. 8. density ϭ ᎏᎏ ϫ ᎏ ᎏ 49 m i2 640 acres ϭ 24 people/acre 1 can 454 g dollars 2. composition Part B True-False 4. 10Ϫ2 mass 127 g Density ϭ ᎏᎏ ϭ ᎏᎏ ϭ 3.8 cm ϫ 8.0 km/L gal 1L 1 mi 1 oz 1l b 9.96 g/cm3 volume 32.1 mL ϫ ᎏ ϭ 3. 6.1 cm3 1000 cm3 7.8 ϫ 102 cm3 29. The beef steak is less expensive per pound. The chicken costs $2. AT 14.3 750. NT Part C Questions and Problems 1 kg 16.2 15. p.93 ϫ 104 kg mi 0. 32. 6. 1.7 mg 1 lb 1 g © Pearson Education. the beef steak costs $1.0 ϫ 103 cm/s g 100 cm 3 1 kg 3 7. ᎏᎏ ϫ ᎏ 6ᎏ ϫ ᎏᎏ 1 s 1m ile 10 s 2. steps conversion factor denominator units unknown Practice Problems 3 Section 3..82/lb. publishing as Pearson Prentice Hall.1 1.2 ϫ 102 mL 1 L 2. 7. 1 day ϫ ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ ϭ 86. 1 m ϫ 19. 2. ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ 1 in 1m 1500 cells 100 cm ϭ 17 m/cell 186 000 miles 1. 20 cm 400 mm 24 liters 12. Volume ϭ 1. 40. 50 lb 6 ϫ 1010 293 K 4°C Part B True-False 11. ST 13. pop.21 cm3 1 mL 766 Core Teaching Resources . k.2 g 1 lb can dollars ϭ 27. 103 d. ST 5. cost ϭ ᎏᎏ ϫ ᎏᎏ ϫ 2.0 ᎏᎏ lb mi 1. density 2. 72 ᎏᎏ ϫ ᎏ h 1 km 1 m 60 min 60 s ϭ 2. 8.5 cm ϭ 4. 2. a. 10Ϫ12 c. 20 cm 21. . AT 1. Section Review 3.14/lb. 4. 10Ϫ6 b. 4.000 g 5.4 Part A Completion 1. 2.54 cm 106 m 1.61 km 1s 18. intensive 3.609 km 3.0 ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ ϭ 17.2 ϫ 102 cm2 Pete’s 5. a. known 1.99 ϫ 10Ϫ1 km ϭ ᎏᎏ 1 s Section 3. 3.0 in 1 m 17. 60 ᎏᎏ ϫ ᎏᎏ ϭ 96.000 people 1 mi2 1.264 gal 1.3 Part A Completion one (unity) conversion factor remains the same dimensional analysis 5. 1.00 dollar ϫ ᎏᎏ ϫ ᎏᎏ 375 d ollars 16 o z 454 g 1000 mg ϫ ᎏᎏ ϫ ᎏᎏ ϭ 75.0 m Section 3.61 km 8.00 L ϫ ᎏ ϭ 1000 cm3 1L 1000 g 1. All rights reserved.8 cm ϫ 30. NT 12. 24 h 60 s 60 min 5. °C ϭ K Ϫ 273 ϭ 20 Ϫ 273 ϭ Ϫ253°C 30. 7.0 cm 4.Part D Questions and Problems 28. 7. 8.125 kg 1000 g ϭ 1. 0. 10. Inc.3 ᎏᎏ ϫ ᎏᎏ 3 ϫ ᎏᎏ cm 1 m 1000 g ϭ 1.400 s 1 da y 1 min 1h km 1000 m 100 cm 1 h 1 min ᎏ ϫ ᎏᎏ ϫ ᎏ ᎏϫᎏ ᎏ 6.12 L ϫ ᎏᎏ ϭ 1. 3. 3.083 m b. 8.00 grams 2. NT 32.72 ϫ 10Ϫ4 m) ϭ 6. 15. 14. Density ϭ ᎏᎏ ϭ ᎏᎏ volume 18. ST 28. h Interpreting Graphics 3 1. 8.46 m c. 4. Air is less dense than carbon dioxide.0 cm3 ϭ 0. c 7. 20. i b j g 5. (1.9 ϫ 103 km b c K ϭ Ϫ55ЊC ϩ 273 ϭ 218 Percent Error 5.80 g/cm3 ϭ ᎏᎏᎏ ϫ 100 ϭ 5. 9. 0. 3. 19. 9. 7. not on the size of the sample.3 ϫ 10Ϫ9 m2 mass 14. d d b a c 16. 17. 2. ST 33. f 6. 2. Density is the ratio of the mass of an object to its volume. ST 27. All rights reserved. 23.15 cm ϫ 1. True-False 26. c.5 cm2 b.778 g/cm3 Quiz for Chapter 3 1. ST 29. d a d d b C. The density of water at a given temperature does not vary with the size of the sample. Two significant figures Three significant figures cylinder B either cylinder A or B the Celsius scale the Kelvin scale No. 13. 3. 8. 18. 7.7 ϫ 10Ϫ5 m) ϫ (3. 6. 2.4 1.Section 3.10 g/cm3 E. does vary with the size of the sample.09 liters 4. 8. 2. c 12.. a. Multiple Choice 11. 4. Essay 38.8 cm ϭ 7.88% 5. 6. 22.10 g/cm3 Ϫ 4. Problems mass 980 g 35. the volume of the water. 4. liquid B. Density is an intensive property that depends only on the composition of a substance. 13. AT 34.00 m Ϫ 0. 6900 km 5. 27 g Chapter 3 Test A A. Density ϭ ᎏᎏ ϭ ᎏᎏ ϭ 15 g/cm3. e 8. j l b a m 11. © Pearson Education. a.0 g 37. 25. 6.0 ϫ 101 ЊC 6. ST 31. 4. b a c b a 21. NT Vocabulary Review 3 1. Volume is a measure of the space occupied by an object.3 ϫ 10Ϫ2 m b. 3. a 9. 1. no 1 ϫ 10Ϫ3 kg/ 1 g 4 a. Answer Key 767 . Inc. 5. 7. 5. 3. however. d 13. e D.54 m ϭ 12. Another significant figure could be estimated. publishing as Pearson Prentice Hall. g i h f k 6. Matching 1. 24. no volume 64 cm3 36. 4. 12. d 10. Volume is an extensive property that depends on the amount of matter in a sample. 20 ЊC is known. The balloon will float. 10. 2. 20°C c. AT 30. 20. AT 36. 4. V ϭ ᎏᎏ ϭ ᎏᎏ ϭ 0. 14.0 L ϫ ᎏᎏ ϭ 3. NT 39. NT Chapter 3 Small-Scale Lab Section 3. 28.11 g penny ϫ ᎏᎏ 100 g penny g Zn ϭ 0. ᎏᎏ ϭ 2.019 g mass of pre-1982 penny ϭ 3. 2. V ϭ 0.0 ϫ 106 m2 8.019 g ϫ 1000 mg/g ϭ 19 mg m g 0. a.0 g Zn g Zn ϭ 3. 22. a 9. 19.0 ϫ 103 cm3 ϫ ᎏᎏ 3 ϭ 3. (5.1 cm ϭ 54. c d a c a d c 18. a.08 ϫ 1015 m 4. page 94 Analysis Answers are based on the following sample data: average mass of water drop ϭ 0.019 cm ϫ ᎏᎏ ϭ 0.721 cm ϩ 15.019 mL 1c m3 1000 L ϫ ᎏᎏ ϭ 19 L x L ϭ 0.576 g Ϫ 35.60 m ϫ ᎏ ϫ 10. 3. x g Cu ϭ 2. True-False 32. since V ϭ L ϫ W ϫ H.4 Now What Do I Do?.3 cm b. 30.0 ϫ 103 cm3 1L 3. e i g f 5. Essay 43. x mg ϭ 0. 12.50 g 1..6 g Zn x g Zn ϭ 2. 29.0 cm3 The metal is not pure silver.17 g c. ( ( ( ( ϭ 3.44 g Zn 768 Core Teaching Resources . you must first express the volume of the box in cubic units. based on the linear dimensions given. 21. NT 38. j 8. D ϭ ᎏᎏ ϭ ᎏᎏ ϭ 11.166 g ϭ 113.0 cm ϫ m 1cm ϭ 3000 cm3 50.019 cm3 d 1. 26.Chapter 3 Test B A. AT 33. you must select a conversion factor so that relates cm3 to liters. Thus.11 g penny ϫ ᎏ 100 g penny g Cu ϭ 2.2 ϫ 10Ϫ6 ϭ 2. ST 35. 23. Multiple Choice 11. x mg/cm ϭ 1. d d c d b a c 25. 31. 16. 27. 24.5 g/cm3 V 65.11 g mass of post-1982-penny ϭ 2. All rights reserved. g Cu ϭ 3.060 g Cu 97. 36.016 ϫ 106 m2 ϭ 2. AT 37. 148. c 6. 3 3 1000 mg/cm ϫ 1 cm /1 mL ϭ 1000 mg/mL 95.00 g/cm ϫ 1000 mg/g x mg/cm3 ϭ 1000 mg/cm3.50 g penny ϫ ᎏᎏ 100 g penny x g Cu ϭ 0.60 ϫ 10Ϫ2 m) ϭ 20.0 L 1000 cm 1 kg H2O b.0 g Cu ᎏ 4.50 g penny ϫ ᎏᎏ 100 g penny x g Zn ϭ 2. that the cm3 units cancel and the liter units remain. publishing as Pearson Prentice Hall. Problems 40. 17. b 7. a a b d c a d C. and then select a conversion factor that will allow you to convert the cubic units into liters. h 10. V ϭ 25 cm ϫ 10 cm ϫ 8 cm ϭ 2000 cm3 To convert the cm3 into liters.00 g /cm3 1 mL 3 x mL ϭ 0.41 g ϭ 113. To find the volume of the box in liters.47 cm ϩ 2. ST 34. Thus.019 mL 1m L 3 3 3.95 g Cu 5. Matching 1. since 1 L ϭ 1000 cm3 1L 2000 cm3 ϫ ᎏᎏ 3 ϭ2L 1000 cm B.16 g Zn D.0 kg H2O L M 750.74 ϫ 109 m d.16 ϫ 105 m2 ϭ 2. d E. 15. 3. Inc. 2.6 ϫ 107 m) ϫ (3.0 g 42. 13.0 mm ϫ ᎏ 10 mm © Pearson Education.291 cm ϭ 54.4 g Cu 5.1 ϫ 1015 m 100 cm 41.019 2. 3.2 lb ϫ 1 L/1.36 cm3 Volume of die: Part C Matching 13. 2.77 g. All rights reserved. neutrons Answer Key 769 .20 cm.36 cm3 Percent error ϭ (0.2 Part A Completion 1.000 L)(100%) = 0.2 lb) and a density of about 1. AT Part C Matching 8. simple whole-number ratios 100 cm 1 ϫ 108 atoms 13. The new penny is lighter because it is mostly zinc which has a lower density than copper.) (3) Read label: 12 oz = 355 mL 5.0242 g Pipets give different results.00 kg/L.57%. (2) Measure the height and radius and calculate volume. at 90°. separated. % error = (3600 L/630. 7. Sample answer: mass of one can: 14. a 10. Volume of hemisphere: 2/3πr3 V = 2/3πr3 = 0.36 cm3 = 3. e 16. Expel the air bubble so that the first drop will be the same size as the rest. 7.017 cm3 Volume of 21 hemispheres: 0. Volume of 30 people V = 30 ϫ 130 lb ϫ 1 kg/2.70 g/cm3. dA: 2.0 kg/L. 6. joined.47 cm3 4. 8.. NT 6. 15 tables. AT A die has 21 holes that are hemispheres with a radius of 0. d 17.0 kg = 1800 L The volume of 30 chairs. AT 11.72 cm3 Ϫ 0.55 cm on a side: V = (1. publishing as Pearson Prentice Hall.1 Part A Completion 1. Find weight in pounds and convert to kg.36 cm3 Error = 0. The volume of a cone is 1/3 πr2h. c 14. mass of 1 drop: 0. b Part D Questions and Problems 18. c 9.5 m ϫ 3. 6. subatomic negative protons neutrons 5. NT 12.55 cm) = 3. ᎏᎏ ϫ ᎏ ϭ 1010 atoms/m m cm Section Review 4. Use the mass and density of water to find the volume. Find mass of can and divide by density of aluminum. V = 3.8 m = 630 m3 ϫ 1000 L/m3 © Pearson Education. 3. b Part D Questions and Problems 12. V = weight ϫ 1 kg/2. Assume the density is about 1. mass of 1 drop: 0. = 630 000 L/ Assume 30 people with an average weight of 130 lb (1 kg = 2. NT 10. and 2 desks is about that of 30 people or 1800 L. 4. d 11. mass of 1 drop: 0.6. ST 7. nucleus Ernest Rutherford positive electrons Part B True-False 9. a 15. V = πr2h (Can is not a perfect cylinder. The volume of people and furniture is 3600 L.36 cm3)(100%) = 11% Note: The holes in some dice are cones. Sample answer: V = 16. atoms 2. The best angle is 90° because the pipet is easiest to control.0218 g at 0°. compounds 4. V = 5. If die measures 1. different 3.36 cm3/3. (1) Measure the mass before and after you fill the can with water.2 lb ϫ 1L/1. or rearranged Part B True-False 5.72 cm 3 3 Section Review 4.00 kg You’re The Chemist 1. 2.019 g at 45°.0 m ϫ 12. protons. Inc. 138 Atomic number: 25.9976 ϭ 15. a.. 46 b. b 23. 42. 10 b. 12. 89 Mass number: 55. 4. e 17. The few that come near the nucleus are deflected or bounce straight back. publishing as Pearson Prentice Hall. f 6.3 Part A Completion 1. simple whole-number chemical reaction Electrons neutrons proton atomic 8 mass 156 30 periods atomic number Part C Matching 16. isotopes atomic mass 1 no 1 Vocabulary Review 4 1. protons number electrons protons neutrons 6. f 18. 10.2 1. 10. a 21. 35. 2. 7. 4. b 9. d Part D Questions and Problems 24. 80 3.1 1 cm 10 mm 1. Atoms are mostly empty space. AT Quiz for Chapter 4 22. 2. 11. All rights reserved.0367 amu 16. 2.19. electrons 20. Number of protons: 11. e 7.67 ϫ 10Ϫ24 g) + 16(1. j g a c 5. 3 2. d 8. 9. f h g b 5. g 20. 5.3 1. AT 15. 75% XϪ100.96 amu oxygen-17:17 amu ϫ 0. 16(1.011 amu 5. i Part B True-False 11. 146 d.0063 amu oxygen-18:18 amu ϫ 0. Most alpha particles pass through the empty space without deflection. electrons are found outside the nucleus. 33. Inc. 12 c. 3. 33. 5 c. (1 ϫ 10 atoms ) ϫᎏᎏ ϫ ᎏ 8 1 ϫ 10 atoms 1 cm 7 Chapter 4 Test A A.00 amu ϭ 16 amu © Pearson Education. mass of eϪ is negligible. 2. protons. 89 Number of electrons: 25. AT 13. c 7. i ϭ 1 mm Section 4. 11. h 10.33 ϫ 10Ϫ23 g 2. 7. h 1. 25% XϪ104 Section Review 4. j 6. 5. 3. oxygen-16:16 amu ϫ 0. 6. Practice Problems 4 Section 4. 8. d 9.00204 ϭ 0. 8. 39. 12. NT 12. 9.00037 ϭ 0. 3. 4. a. Matching 1. 10 4. 39. 16 d. 3. b 770 Core Teaching Resources . a 10. approximately 6 ϫ 1023 neutrons 3. 4. Section 4. The relatively massive protons and neutrons are concentrated in a small region called the nucleus. c 19. 89 Number of neutrons: 50. 35. e 8. 23. ST 14.67 ϫ 10Ϫ24) ϭ 5. 19. 35. j 9.0062 ϭ 0.024 amu Ar-40 39. 6 10 4Be: 4. 16 27.. and thus. and thus.926 amu ϫ 0. 56.0411 ϭ 12. Essay © Pearson Education. 30 108 47Ag: 47.1857 ϭ 12. 13. 26 Mass number: 19. Multiple Choice 11. 79 56 197 Symbol: 39 19K. Multiple Choice 11. 79Au D. h 6. 14. Atomic number: 19.43 amu atomic mass ϭ 65. c d c c d 16. 80.00063 ϭ 0. c i b g 5. Inc. 13 Number of neutrons: 14.978 amu ϫ 0. c 22. 20. 26Fe. 13 40 18Ar: 18. 13. 22. d 23. 6. a b a b b 21. All rights reserved.00337 ϭ 0. 45 Number of electrons: 12. 7. 10. Problems 25. 7.947 amu 26. c C. e 10. 30 Number of electrons: 9. Matching 1. 19.929 amu ϫ 0. 19. 6C: 6.33 amu 66. 4 20 10Ne: 10. 197 Number of protons: 12. different mass numbers. 26.4889 ϭ 31. 18. 18. 4. 5 33 16S: 16. publishing as Pearson Prentice Hall. 18 65 30Zn: 30.925 amu ϫ 0. 26 13 26. b b a d Answer Key 771 . 26. 7. 23. 10.963 amu ϫ 0.37 amu 25. 61. D.61 amu 69. 14. 20. 24.99600 ϭ 39. c c b a c 21. 19 F: 9. 20. 15. d B. f 8. 13. Atoms of different elements have a unique number of protons and electrons. Ar-36 35. 35 Number of neutrons: 12. 6. 9 9 27 13Al: 13.75 amu 67. the same number of protons and electrons. Isotopes of the same element are alike in that they have the same atomic number.2781 ϭ 18. Problems 24. 35. 20. 3. 63.25 amu 65.962 amu ϫ 0. 17.925 amu ϫ 0. 27. Chapter 4 Test B A.B. 22. 14. 2. 79 Mass number 24. 35. 10 11 5B: 5. Isotopes of the same element are different in that they have different numbers of neutrons. 47 C.121 amu Ar-38 37.802 amu atomic mass ϭ 39. Isotopes of the same element differ in the number of neutrons in the nuclei. a c d d a 16. 17. 19.927 amu ϫ 0. Essay 28. 17. 15. 12. Atomic number: 7. 12. a 7. 41 Number of protons: 9. 7. the locations of electrons are not fixed. 2. NT 12.8833 g Figure A Part B True/False 9. Percent abundance is parts per hundred. 20.39 g You’re The Chemist 1.25 27. The individual relative abundances add up to 1.2883 g ϩ 0. which affect the relative abundances. See row 6 in Figure A. The total in row 3 is an average that ignores the relative abundances of particles. a. Inc.. AT Part C Matching 1. 8. Part D Questions and Problems 19.16 g B 13. Mass is likely to provide better results than volume.83 g C 15. b 17. Bohr proposed that electrons are arranged in concentric circular paths around the nucleus. AT 10. d © Pearson Education.2883 g 0.3125 0. which prevents them from falling into the nucleus.08 41. AT 13.064 g 0.4167 100. 3. page 120 Analysis Sample data provided.3 The Atomic Mass of Candium. Another student might not have had the same relative abundance of each candy. Dalton proposed that matter was made of indestructible particles called atoms. He proposed that electrons surround a dense nucleus.J. 5 orbitals d.2742 g ϩ 0.8773 g 1. See row 5 in Figure A. electrons John Dalton J. ST 11. 0. The individual percent abundances add up to 100. Rutherford discovered that atoms are mainly empty space. Thomson plum-pudding nucleus circular quantum mechanical probability 15 13 20 48 0. 7. In the modern atomic theory. Any differences are probably due to small variations in the numbers of each kind of candy in the samples. 2. The larger the samples. See row 3 in Figure A. a 18. Relative abundance tells you the decimal fraction of particles. 2.3208 g 0.3208 g ϭ 0. 5.0 31. 3 orbitals c. 1 orbital 772 Core Teaching Resources . they are described in terms of probability. Relative abundance is parts per one or the decimal form of percent. 15.8831 g 0.2742 g 0.40 g Totals 42. See row 4 in Figure A. 7 orbitals b. c 16. the better the results with any of the methods. 9. 4. 5.67 1. the electrons in a particular orbit have a fixed energy. The total in row 6 is a weighted average because it considers differences in mass and abundance among the particles. Section Review 5. 8. 3. 4. According to Bohr.Chapter 4 Small-Scale Lab Section 4.7700 g 0.2708 0. publishing as Pearson Prentice Hall. All rights reserved. 6.8833 g 6.000 0.1 Part A Completion 1. AT 14. A larger sample would provide a greater sampling of all isotopes. A Total mass (grams) Total Number Average mass (grams) Relative abundance Percent abundance Relative mass 13. Thomson proposed an atomic model in which negatively charged electrons were embedded in a positively charged mass. 3d. 7. c. 6.00 ϫ 1010 cm /s 18. e. ϭ ᎏᎏ ϭ ᎏᎏ Ϫ1 5.00 ϫ 1015 s ϭ 6. d. 16 7 h. b. 4p. c. d. 4 e. 3 orbitals. 1s22s22p63s23p63d104s24p3 i. c. 1 Part B True/False 11. The frequency of the light must be above the threshold frequency that will provide the necessary quanta of energy. f. NT 14. 1 orbital 2s. 10. ultraviolet 3. 5 orbitals. 4f. 2.1 orbital. b . 3 orbitals. 3. 2 c. a. b 20. a. 5 f. b.70 ϫ 10 m 14 Ϫ1 ϭ 4. 5s. 1 orbital. e 18. 5f. 3p.2 1. ϭ ᎏᎏ 2. 8.2 Part A Completion 1. 4. 5 orbitals. 2.40 ϫ 10Ϫ5 cm 15 Ϫ1 ϭ 1. AT 16.00 ϫ 108 m /s c 2. NT b.48 ϫ 10 s Answer Key 773 Part C Matching 13. All rights reserved. ST 12. Practice Problems 5 Section 5. d 19. ϭ ᎏᎏ ϭ ᎏᎏ Ϫ7 6. 1 orbital.1 1. d. electron configurations Aufbau principle equal Pauli exclusion two opposite a single electron superscripts electrons Chromium Part D Questions and Problems 3. 6 e. The photoelectric effect will not occur unless the frequency of the light striking a metal is high enough to cause an electron to be ejected from the metal. 1s22s22p2 1s22s22p63s23p4 1s22s22p63s23p64s1 1s22s22p63s23p6 Ar b. publishing as Pearson Prentice Hall. NT 10. 1 b. 1 orbital. 5d. d. b. 4d. 3 orbitals 3s. a. 5 1s. 1 f. a. NT 13. AT 11. 3 2. 5. NT Section 5. Inc.3 Part A Completion 1. 2p.3 3. 23.00 ϫ 10Ϫ6 cm. NT 9. 5 orbitals 4s. d. c 14. 1s1 1s22s22p63s23p63d34s2 1s22s22p63s2 1s22s22p63s23p63d104s24p64d105s25p66s2 1s22s22p63s23p63d104s24p5 1s22s22p63s23p4 g. 5. 3 orbitals. 3. a 15. 1s22s22p63s23p63d104s24p64d104f145s25p6 5d106s26p6 12. e 16. waves inversely light atomic emission spectrum light radiation photoelectric frequency Section 5. c. 7 orbitals Section 5. 6. AT Part C Matching 17. a. c Part D Questions and Problems 22. 5p. 7. a. 4. 1s22s22p63s23p63d104s24p6 h.25 ϫ 10 s 19. d 17. NT Part B True/False 8. B © Pearson Education. 3. 9. 1 15.00 ϫ 1010 cm/s c 1. 3 5 g. a 21..Section 5. 7 orbitals e. Li2ϩ and Heϩ each have a single electron surrounding the nucleus and should. 6 → 2 5→2 4→2 3→2 5. The Bohr model is adequate for explaining the emission spectra of atoms with a single electron.74 ϫ 10Ϫ8 m. 4. ultraviolet light 4.28 ϫ 10Ϫ6 m. 4.30 ϫ 1014. 6.02 ϫ 10Ϫ7 m. 3. 1. infrared. 6. 2.60 ϫ 1014. 12. 1.88 ϫ 1014. 4. ST NT AT ST AT NT 7.6262ϫ10Ϫ34 J s ϫ 2. 3. 9.75 ϫ 1014. cosmic rays 3. 6. 4.14 ϫ 1014. 2.05 ϫ 10Ϫ6 m. 4. infrared.56 ϫ 10Ϫ7 m. E ϭ hϫ ϭ 6. 8.08 ϫ 1015. ultraviolet light. a B. ultraviolet light. d 19. photons 774 Core Teaching Resources .88 ϫ 10Ϫ6 m. 2. 9. 9.09 ϫ 10Ϫ6 m. visible light. 3. ultraviolet light. b 18.. visible light. thus. 2. 7. 10. Planck’s constant ground state photons photoelectric wavelike Interpreting Graphics 5 1.63 ϫ 10Ϫ6 m. infrared.05 ϫ 10 s radio waves. d 9. visible light. 2. The laser emits red light. 9. a 16.00 ϫ 10Ϫ15 J E ϭ ᎏᎏ ϭ ᎏᎏᎏ Ϫ34 6. i 8. Hund’s rule 2. b g e c 5. 7.01 ϫ 1013.57 ϫ 1014. 4. h 6. All of the transitions end at the n ϭ 2 energy level. 9. 14.34 ϫ 1014. 6. Inc. ultraviolet.93 ϫ 1015. 10. Matching 1. : 7. 3.41 ϫ 1013. 3. (s؊1): 4.52 ϫ 10Ϫ8 m. 5. 1. Chapter 5 Test A A.3. 11. f 10.20 ϫ 1015. infrared.6262 ϫ 10 s J h 18 Ϫ1 ϭ 9.17 ϫ 1014. ultraviolet light. 2. type of radiation: infrared. All rights reserved. 2.21 ϫ 10Ϫ7 m 3.15 ϫ 1015. 13. microwaves. d 20. 6 → 2 blue 5 → 2 blue 4 → 2 green 3 → 2 red 6.48 ϫ 10Ϫ6 m. 4. 1. 5.11 ϫ 10Ϫ7 m.36 ϫ 10Ϫ7 m. 6. 1. visible light.48 ϫ 1015 2. Multiple Choice © Pearson Education. 1. 8. 7. 5.47 ϫ 10Ϫ19 J 6. infrared. 7.86 ϫ 10Ϫ7 m. 4. behave according to the Bohr model. j 7.38 ϫ 10Ϫ8 m. 4. b c a d 15. c 17. hertz Pauli exclusion principle quantum quantum mechanical model aufbau principle wavelength atomic emission spectrum photoelectrons Quiz for Chapter 5 1. 1. visible light.88 ϫ 10 m 3.00 ϫ 108 m/s c ϭ ᎏᎏ ϭ ᎏᎏ 3 Ϫ1 1600 ϫ 10 s 2 ϭ 1. publishing as Pearson Prentice Hall. a Vocabulary Review 5 1.22 ϫ 1014 sϪ1 E ϭ 1. 11. 6. infrared. AT 34. c 32. 3. the 5s sublevel is lower in energy than the 4d sublevel. ϭ ᎏᎏ ϭ 1. Generally. 19. d F. 24. 15. 4.2 ϫ 1015 sϪ1) Ϫ18 E ϭ 3. f 30. P c. Br h. Essay 26. Additional Matching 27. phosphorus b. Sr © Pearson Education. i B. filling orbitals with lower energies first. True/False 33. a. b. b. NT 38. x w x 3p x w 3s x w x w x w 2p x w 2s x w 1s 1s22s22p63s23p4 b. h c f e 5. Electrons occupy orbitals in a definite sequence. 23.25 ϫ 1015 sϪ1 Ϫ5 2. 14. bromine x E. Matching 1. electrons travel around the nucleus along fixed paths much as planets orbit the sun. AT Chapter 5 Test B A. 22. a. a. The quantum mechanical model. NT 35. c d c b C. Essay 29.. Multiple Choice 11.00 ϫ 1010 cm րs ϭ ᎏᎏ 6. ST 41. Sc g.6262 ϫ 10Ϫ34 Js ) ϫ (5. AT 40. All rights reserved.4 ϫ 10 J D. a. Inc. 12. aluminum c. x 3s x w x w x w 2p x w 2s x w 1s 1s22s22p63s1 3.40 ϫ 10 cm 23. g 7.25 ϫ 10Ϫ5 cm 14 Ϫ1 ϭ4. Problems 25. a 29. krypton 24. orbitals in a lower principal energy level have lower energies than those in a higher principal energy level. d 6. 18. c ϭ ϭ c/ 3. AT 36.80 ϫ 10 s 28. Fr 27. a 10. a. 13. b 28. neon 25. 17. publishing as Pearson Prentice Hall. Ne f. Mn d. D. cobalt b. E ϭ h ϫ E ϭ(6. Problems 21. explains the positions of electrons in terms of probability clouds within which the electrons are most likely to Answer Key 775 . but in the fourth level the energy ranges of the principal energy levels begin to overlap.00 ϫ 1010 cm/s 22. 1s22s22p63s2 1s22s22p63s23p3 1s22s22p63s23p64s23d104p5 1s22s22p63s23p64s23d104p65s24d105p6 S e. c. j 9. Energy level 1 ϭ 2 Energy level 2 ϭ 8 Energy level 3 ϭ 18 Energy level 4 ϭ 32 Energy level 5 ϭ 50 c. a c b d b 21. d. 20. ST 37. As a result.C. 26. AT 39. c c c b b 16. According to the Bohr model. e 31. b 8. 2. solids at room temperature 21. 2. 2. Inc. solid. AT 13. Tellurium: metalloid. 7. ST 13. AT 44. AT Part B True-False Section Review 6. c 22. Ar. 8. bismuth is a metal. Sulfur: nonmetal. 7. NT Part C Matching 15. bromine. d Part C Matching 14. AT Part C Matching 16. a 33. fluorine. e 16. properties groups periods or rows atomic number group Metals gases metalloids less more 12. NT 11. f 34. 8. f 18. NT 38. ST 43. a Part D Questions and Problems 23. 6. decrease increases energy levels charge ionization 6.1 Part A Completion 1. 3s2. AT 39. 3s23p2. Si. 5. NT 14. ST 12. Al. b 21. The quantum mechanical model allows electrons to be at virtually any distance from the nucleus. AT 37. 9. 5. f 17.. 4. names atoms alkali metals alkaline earth metals representative elements halogens noble gases transition metals inner transition metals p not filled E. high luster.be found. Mg. Part B True-False 10. but describes the locations of greatest probability in terms of specified orbital shapes. solid. increases electrons smaller electronegativity increases Part D Questions and Problems 19. 3s23p4. NT 40. 5. 3. malleable.2 Part A Completion 1. g 19. 4. a 19. P . 3s23p1. 8. 3. Na. True/False 36. e 18. The Bohr model places electrons at specific distances from the nucleus. 3. NT Section Review 6. 10. d 16. ductile. 4. e 35. c Part A Completion 1. 3s23p6 24. AT 41. d 20. Selenium: nonmetal. 3s23p5. 3s23p3. b F. Section Review 6. S. iodine Part B True-False 11. 10. b 15. 2. publishing as Pearson Prentice Hall. Additional Matching 30. c 17. 10. NT © Pearson Education. c 31. good conductors of heat and electric current.3 18. 3s1. 7. 6. 11. 9. d 32. solid 12. Oxygen: nonmetal. AT 13. a 17. e 20. solid. nitrogen and phosphorus are nonmetals. gas. Cl. AT 14. NT 15. Polonium: metal. arsenic and antimony are metalloids. b 776 Core Teaching Resources . 9. AT 42. All rights reserved. 20. it gains an electron in its highest occupied energy level. nonmetal e. publishing as Pearson Prentice Hall. Astatine is in Group 7A. a. Silicon is in the third period.2 1. The electron configurations for Ag and Fe are: Ag 1s22s22p63s23p64s24p64d105s1 Fe 1s22s22p63s23p63d64s2 Section 6. c. 1) The metals: good conductors of heat and electric current. 2) The nonmetals: poor conductors of heat and electric current. 3. b. The third period element in Group 5A is phosphorus b. but the nuclear charge is greater in magnesium. it loses electrons from its highest occupied energy level. selenium. Br 22. metal 3. c. A magnesium atom is smaller than a sodium atom because the shielding effect is constant for elements in the same period. A magnesium atom is smaller than a calcium atom because there are fewer occupied energy levels. so its electron configuration must end in 5s25p5. Fr 5. The prediction is that atoms of astatine are larger than atoms of tellurium. and are nonlustrous. 6. which is also known as the noble gases. Na e. Although atomic size decreases across a period. Li. S c. 3) The metalloids: elements that have properties similar to those of metals and nonmetals depending on the conditions. The configuration s2p3 indicates 5 electrons in the highest occupied energy level. An ion with three Answer Key 777 © Pearson Education. It is the fourth element in the period. this element is bromine. 5. Lithium in Group 1A has only 1 electron in its highest occupied energy level. The complete configuration is 1s22s22p63s23p2. gallium b. Inc. O c. 4. a. Na. So the electrons are drawn closer to the nucleus. 2. When a magnesium atom reacts. Al b. malleable. which is a feature of Group 5A. a. It is Group 4A. Section 6. Magnesium and calcium have the same number of electrons in their highest occupied energy level. Tellurium is in period 5.3 1. Transition metals are elements whose highest occupied s sublevel and a nearby d sublevel contain electrons. oxygen d. The three classes are as follows. so its electron configuration must end in 3s23p2. Elements in Group 5A have 5 electrons in their highest energy level. high luster when clean. bromine Practice Problems Section 6. . Its first four energy levels are full.Part D Questions and Problems 21. The element in period 4 with 2 electrons in the 4s sublevel and 10 electrons in the 3d sublevel is zinc. c 2. A chlorine atom is smaller than a magnesium atom because atomic size decreases from left to right across a period. 4s24p5 represents the Group 7A element in period 4. Cs. Rb. a 4. Sulfur in Group 6A has 6 electrons in its highest occupied energy level. ductile. Both Ne and Ar have a completely filled highest occupied energy level. 2. 8. When chlorine reacts. chlorine d. a. a. The complete configuration is 1s22s22p63s23p63d104s24p64d105s25p5. nonmetal b. the additional occupied energy level in astatine significantly increases the size of the astatine atom as compared to the tellurium atom.. Its first and second energy levels are full (1s22s22p6). The chemical and physical properties are largely determined by their electron configurations. Astatine is in period 6. The period 2 element with six electrons is oxygen.1 1. They are in Group 8A. The period 4 element with 2 electrons is calcium. 3. metalloid c. A magnesium ion has filled first and second levels. Ne: 1s22s22p6 Ar: 1s22s22p63s23p6 7. All rights reserved. Iodine is located in period 5. metal d. tellurium is in Group 6A. This enlarging effect is greater than the shrinking effect caused by increasing nuclear charge. Sodium’s first ionization energy is higher than that of potassium because ionization energy tends to decrease from top to bottom within a group. Because electronegativity increases from left to right across a period. f h d p j 778 Core Teaching Resources . occupied energy levels is larger than an ion with two occupied energy levels. it will be smaller than the sulfide ion. 7. 4. 14. c. period 5 b. cesium. is an alkali metal. a. Because magnesium has a relatively low first and second ionization energy. All rights reserved. and bp. transition metal. Beryllium’s first ionization energy is greater because first ionization energy tends to increase from left to right across a period. 3. Across a period from left to right the principal energy level remains the same. No. and francium. The relatively high third ionization energy indicates the difficulty of removing a third electron from the filled second energy level. 42 table A atomic weight 0. n a m c k b 7. In the periodic table elements with similar chemical and physical properties are grouped together in vertical columns. sulfur is less electronegative than oxygen. 3. 6. Interpreting Graphics 6 1. 16. The increasing nuclear charge pulls the electrons closer to the nucleus.4. Listing the elements. fluorine is more electronegative than oxygen. 9. period 2 Mo: Group 6B (or Group 6). in alphabetical order.53 g/cm3 5. but the nuclear charge increases. Because electronegativity decreases from top to bottom within a group. e. When a sulfur atom reacts to form an ion it adds two electrons while chlorine adds one electron. 10. Vocabulary Review 6 1. rubidium. 5. general class. Atomic size increases as you move down a period because the electrons are added to higher principal energy levels. 4. 5. 9. 2617 ЊC 6. mp. Barium is less electronegative than struntium because electronegativity values tend to decrease from top to bottom within a group. 10. 4 8.g. table B 7. Because the chloride ion has the greater nuclear charge. Check students’ work. 2. potassium. 10. This organization helps scientists predict and explain similarities and differences in the properties of elements based on their underlying atomic structure. Inc. The correct order for increasing electronegativity is then sulfur Ͻ oxygen Ͻ fluorine. Mo. 2.. 8. 11. Answers may include sodium. resulting in a smaller atomic radius. l g o i e 12. Magnesium normally forms an ion with a 2ϩ charge. whether an element is not found in nature 9. is a transition metal. Sulfide and chloride ions have the same number of electrons. Li. Their keys need to include the color. makes it possible to quickly find information about the properties of a particular element without having to know the location of the element in the periodic table. d. Molybdenum. physical state at room temperature. 15. publishing as Pearson Prentice Hall. because they are not located in the same group or family. © Pearson Education. 6. 11. The trend is less pronounced as the number of electrons increases because the inner electrons shield the electrons in the highest occupied energy level. Lithium. 8. 13. of the element (and the state if the square is not color coded for style). the removal of two electrons from magnesium is likely. Li: Group 1A (or Group 1). Group 8A (or Group 18) f. 22. 4. K. Essay 31. 26. period 2. 26. 4s2 4p6 e. c d a d d b a 25. g 9. Group 1A (or Group 1) c. 5A(15). c c a b b c b. 19. K 4. b. K. i 7. Cs b. K. e. 3. 16. c c d a d 21. which shields the electrons in the highest occupied energy level from the attraction of protons in the nucleus. a. transition metal. a 8. 25. 7A(17). a. Group 7a (or Group 17) d. period 4. f 10. P 3. period 5. 24. c 6. noble gas. 3s2 3p1 d. I c. 23. 4s1 c. Questions 31. Li. c. AT 30. 17. b. number seven group (column or family) six two less ST 8. 20. 2. Cl D. alkaline earth metal. 29. 4. 2. There is an increase in nuclear charge. a 7.Quiz for Chapter 6 1. j 8. Matching 1. The increasing charge on the nucleus tends to pull the electrons closer and atomic radius decreases. d. Cs Answer Key 779 . Si c. 28. Two factors influence the size of an atom as the atomic number increases within a group. Li. Sc 4. transition metal. Inc. 21. Mg b. 27. period 2. halogen. h 6. Sr. 6. 3. 2. Matching 1. period 5. Questions © Pearson Education. 14. 2A(2). c. C. 15. NT Chapter 6 Test A A. noble gas. 13. transition metal. 20. 2s2 2p4 28. Ar. 22. alkali metal. 15. alkaline earth metal. Be 3. which draws the electrons closer to the nucleus. 8A(18). 4B(4). Ar 4. Group 6A (or Group 16) period 6. 18. There is an increase in the number of occupied energy levels. Be. 19. Group 4A (or Group 14) 1s2 B. Li. Group 2A (or Group 2) period 4. 2. F Cs. 12. 23. From left to right across a period. c. c i b g 5. C. period 6.. C. 30. 3. a c d b b Chapter 6 Test B A. 27. period 4. 5. b. 7. Cl P. NT AT 9. a. electrons are being added to the same energy level. e C. Li. d c d a a 16. f. 10. a. halogen. All rights reserved. 32. a. Multiple Choice 11. a. h B. a. alkali metal. 17. 24. Group 6B (or Group 6) e. b d d a d c d 18. 12. Ca. Ba 29. d f j b 5. Group 1B (or Group 11) C. F F. d 9. Br Cs. 14. The net effect is a decrease in the attraction of the nucleus on the electrons in the highest occupied energy level and an increase in atomic radius. K. publishing as Pearson Prentice Hall. 13. Multiple Choice 11. Group 2A (or Group 2) b. 1A(1). e 10. 4. f 22. F. 4. Inc. ST 14. A high electronegativity indicates an ability to attract additional electrons. which are on the right. O. 34. the rest of Group 3A shows a reverse in this trend. =Ba= b. Electronegativity generally increases from left to right along a period. 2 electrons lost. Metals attain noble gas configurations by losing electrons and forming cations with a complete octet in the next-lowest energy level. S B C D Section Review 7. Except for boron. AT 13. d 19. silver ion. Other students cut a straw to a length that represents the larger radius of an atom and mark the straw to show the smaller radius of the corresponding cation. 3. which are on the left side of the table.33. 6. C Ca. Electronegativity generally increases from bottom to top within a group. NT Chapter 6 Small-Scale Lab Analyze and Conclude 1. K C. Students divide the values of first ionization energies by 300 and measure the appropriate length of straws. Part C Matching 17. Essay 35. NT 12. a. Students often use two wells to represent both ionic and atomic radii. e 20. c Part D Questions and Problems 24. magnesium ion. AT 16. 9. 1 electron lost. b.1 Part A Completion d. xenon appears to have the ability to attract electrons and form compounds. which is consistent with the general trend. Nonmetals attain stable noble gas configurations by gaining electrons and forming anions with 8 outer electrons in the existing energy level. Ionization energy and electronegativity are properties that reflect an atom’s ability to attract and retain electrons. NT 15. K. b 18. 2 electrons lost. 3. have lower electronegativity values than nonmetals. a. Although hydrogen is placed in Group 1A based on its electron configuration. B e. 780 Core Teaching Resources .. 5. b. 5. a. publishing as Pearson Prentice Hall. a 23. Fluorine 2. Si c. 2. © Pearson Education. c. calcium ion. valence electrons group electron dot structures octet rule cations anions 1ϩ Halide ions gain charges D. g 21. You’re The Chemist 1. c. A high ionization energy indicates that an atom has a tight hold on its electrons. cation c. Rb= 25 a. Part B True/False 11. 2. 8. d. The value for xenon is similar to iodine. 4. bromide ion. 1 electron gained. 10. hydrogen is classified as a nonmetal. Li. anion d. B 1. cation b. Mg. 3. Metals. Students must determine their own scale before they begin. Ca S. cation 26. All rights reserved. 7. Based on this value. Metal cations are insulated from one another by electrons. aqueous solutions of ionic compounds also conduct electricity. All rights reserved. 2. 8. As electrons enter one end of a bar of metal. For example. a Part D Questions and Problems 21. hexagonal close-packed 10. © Pearson Education.3 Part A Completion 1.2 Part A Completion 1. When a metal is subjected to pressure. AT 14. 2. 1s22s22p63s23p6 e. 22. alloy Part B True-False 11. ST 15. 6. 6 b. (i) 1 (ii) K = (iii) Kϩ 2. 8. When ionic compounds dissolve in water. (i) 7 (ii) I (iii) IϪ c. 7. When ionic compounds are melted. 1s22s22p63s23p64s2 b. 10. and anions will migrate to the other. (i) 2 (ii) Ba 6 (iii) Ba2ϩ b. a. ST 13. a. AT 12.1 1. a Part D Questions and Problems 21. The superior properties of alloys result from the cumulative properties of all the constituents of the alloy. 1s22s22p6 d. 22. 7.. 3 c. electrostatic forces oppositely ionic bonds neutral formula unit crystals high large stable molten 9. a. AT Part C Matching 16. cations will migrate to one electrode. an equal number leave the other end. AT 15. the positive charges of the cations equal the negative charges of the ions. 5. 3. c 20. cations electrons metallic electrical malleable/ductile ductile/malleable body-centered/face-centered face-centered/body-centered Answer Key 781 . Section 7. Each ion is then free to move throughout the molten mass. publishing as Pearson Prentice Hall. d 20. NT 14. 1s22s22p6 4.Section 7. e 18. Part B True-False 11. Solid metals consist of closely packed cations surrounded by free-moving valence electrons. d 17. This movement of ions means that there is a flow of electricity between the two electrodes. 4. Inc. If a voltage is applied. the metal cations easily slide past one another. e 18. their ions are free to move. Ionic bonds are the electrostatic forces of attraction that bind oppositely charged ions together. ST 13. 1s22s22p63s23p5 c. The number of valence electrons in an atom of a representative element is the same as the group number of the element. In an ionic compound. NT Part C Matching 16. b 17. Practice Problems 7 Section 7. which make metals good conductors of electric current. This behavior makes the metal malleable and ductile. the orderly crystal structure breaks down. Thus. NT 12. 7 3. 6. b 19. 3. an alloy can be more durable than one constituent but more malleable than another. c 19. 9. 5. 4. 2. b. 4. • Body-centered cubic: every atom (except those at the surface) has 8 neighbors. but in a different arrangement than face-centered cubic. • Hexagonal close-packed: every atom has 12 neighbors. c. c. these mobile electrons can carry charge from one end of the metal to the other. Spring steel: iron. The metal cations are arranged in a very compact and orderly structure or pattern.2 1. The ions arrange themselves in an orderly. no individual electron is confined to any specific cation. f. 5. each ion is surrounded by six other ions of opposite charge. at least one of which is a metal. In Step 1. k 13. 7. a. b 12. loses 3 electrons. A metallic bond is made up of cations that are surrounded by mobile valence electrons. e. Stainless steel: iron. CaI2 2. cation anion f. the electrons are free to move about the crystalline structure. each sodium atom gives up one valence electron to a chlorine atom. chromium. a. sodium becomes positively charged and chlorine becomes negatively charged. cation 4. Although the electrons are attracted to the metal cations. loses 2 electrons. a. b. An alloy is a mixture of two or more elements. Vocabulary Review 7 1. anion chloride ion. Section 7. a. In this process. 2. cation gains 1 electron. The combinations in b and c will form ionic compounds 3. Kϩ oxide ion. The coordination number is the number of ions of the opposite charge that surround an ion in a crystal. 7. Interpreting Graphics 7 1. 3. a. All rights reserved. When electrical current is applied to a metal. The large amount of energy released when an ionic lattice is formed (Step 2) compensates for the endothermic nature of the electron transfer (Step 1). c. cation gains 2 electrons. Alloys have properties of metals. 4. Na2S e. l a c e 11.3 1. three-dimensional array characteristic of a crystalline solid. • Face-centered cubic: every atom has 12 neighbors. Inc. Al2O3 b. Each ion attains the electron configuration of the nearest noble gas. BaCl2 c. 8. Metals are crystalline. anion loses 1 electron. rather. d. Patterns are used to calculate the positions of ions in the crystal and to define the structure of the crystal. 4. 9. anion cation e. ionic bonds form between sodium cations and chlorine anions. 3. O2Ϫ barium ion. 8. cation b. The metallic crystal is thought to consist of an array of metal cations in a “sea” of electrons. To reverse the lattice formation through melting would require enough energy to overcome the multiple atttractions within the crystal lattice. Ionic compounds are formed when metals react with nonmetals. which results in a very stable ionic compound. The coordination number is determined by using x-ray diffraction crystallography. chromium. publishing as Pearson Prentice Hall. anion gains 3 electrons.5. j i f g 6. a. Ba2ϩ 2 lost 1 gained 1 lost 3 lost cation d. m 782 Core Teaching Resources . NaCl is typical of many ionic compounds. © Pearson Education. Sterling silver: silver and copper e. d. Cast iron: iron and carbon f. ClϪ potassium ion. 2. In Step 2. c. sodium 1s22s22p63s1 Sodium has 1 valence electron. b. NaBr d. chlorine 1s22s22p63s23p5 Chlorine has 7 valence electrons. Brass: copper and zinc b. Bronze: copper and tin c. 6. 6. In NaCl. and carbon Section 7. d. carbon. and nickel d. 3.. d c d c d c 24. j g a i 9. Metallic bonds are the result of the attraction of free-floating valence electrons for positively charged metal ions. 6. a. 9. AT 29. Be2ϩ cation 36. 5. i 35. Al Cl Quiz for Chapter 7 1. c b d b 20. 6. c 11. 8. when metal is struck. 27. 2. ST 33. Multiple Choice 12. 23. Inc. 12. a. 4. 1s22s22p6 d. Al Al 1. h 10. 13. 26. k 10. Thus metals are good conductors of electricity. 13. 21. 1s22s22p6 B. 11. True-False © Pearson Education. O2Ϫ anion c. c b c b 16. ClϪ anion b. Questions 34. 17. The electron dot formulas show that one atom of Al can give 3 electrons. some of the free-floating electrons leave the other end. 1s22s22p63s23p6 b. publishing as Pearson Prentice Hall. 4. 18. a. Thus. 16. 14. a a a c C. 2. As electrons enter one end of a piece of metal. AT 31. c b c a a c 18. 10. Naϩ cation d. f g d e 5. AT 32. 8. 17. NT 30. 28. This makes the metal malleable. The cations in a piece of metal are insulated from each other by the free electrons. d E. 14. Br Ca Br 2ϩ Ϫ c. 25. h 10. 3.. 15. Matching 3ϩ D. 22. so 3 atoms of Cl are needed to form the compound AlCl3. 3. 23. d 37. 3. y Al3ϩϩ 3 Cl Ϫ Chapter 7 Test A A. Essay 39. e f h c 5. 7. 22. 26. 6. 21. 2. 25. An electric current is a flow of electrons. 27. 15. ST NT AT AT AT valence octet 8. Multiple Choice 12. b 11.5. c d a c d Answer Key 783 . AT Chapter 7 Test B A. 20. AT NT NT ST 28. 19. Cl Al ϩ Cl Cl Na F 1s2 2s2 2p6 3s1 ϩ 1s2 2s2 2p5 → Naϩ FϪ 2 2 6 2 1s 2s 2p ϩ 1s 2s2 2p6 Both ions have the configuration of neon. 1s2 c. gaining eight pseudo-noble gas metals anion formula unit 38. 13. 4. Matching 1. 7. 24. All rights reserved. 19. the cations slide past each other easily. B. =Ca = b. b k a j 9. 7. page 200 Analyze Sample data provided. 3. 32. when Ca and F react. which means that F has seven valence electrons and reacts by gaining one electron to attain the noble gas configuration. AT Chapter 7 Small-Scale Lab Section 7. Essay 44. AT ST ST AT 37. observable products. FeCl2. E. AT 38. Br Naϩ . 1s22s22p6 Kϩ. 2. b. neither of the solutions produced a visible product. NO3Ϫ and HCl ϩ Fe. 1.. NaF MgCl2 CaS Al2O3 NaOH KI (K؉) NVR CaCl 2 (Ca 2؉) WP FeCl3 (Fe3؉) Rust ppt. forms. b. Ca2ϩ 43. Li b. 7. Si d. All rights reserved. SO42Ϫ and Pb2ϩ. 2s1. Bubbles Bubbles WP WP 3Ϫ d. NT 39. You’re the Chemist All designs should include tests that produce unique. 5. Each of the following pairs of ions produces a visible product that can be used to identify the ion in question: PO43Ϫ and Agϩ. a. c. 35. d. N c. the cation Ca2ϩ is produced. a. The group number for Ca is 2A. 31. which reacts with the nitrate ion. Cl P . The formula of the compound formed is CaF2. 42. The formula for the anion produced is FϪ. NVR: No Visible Reaction WP: White Precipitate 1. 36. An orangebrown color forms. which means that two valence electrons will be lost. 784 Core Teaching Resources . b.C. 4. No. 4s24p5. a. An intermediate compound. Inc. 41. Ca2ϩ and OHϪ. a. d. True-False 29. Bloodred soln KSCN NVR NVR Figure B Cation Analysis © Pearson Education. 3s23p2. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 Na . 30. P Ϫ Na3PO4 (PO43؊) Light yellow ppt. 34. 1s22s22p6 FϪ. The group number for F is 7A. Naϩ Cl . Ca . 1s2 2s2 2p6 3s2 3p6 Sr2ϩ. Thus. Na2SO4 (SO42؊) AgNO3 HCl plus 1 piece of Fe(s) Pb(NO3)2 NVR HNO3 (NO3؊) NVR Bubbles w/yellow solution NVR Figure A Anion Analysis D. 2s22p3. two atoms of F are required to react with one atom of Ca. Fe3ϩ and SCNϪ. AT AT AT NT 33. c. c.2 Analysis of Anions and Cations. Questions 40. publishing as Pearson Prentice Hall. 69 Br 69 Br 6 b. NT Part C Matching 17. NT 12. H 6 C 6 6 6 N 6 c. Reading from left to right: sp3 sp2 sp2 sp sp sp3 Section 8. low high atoms structure Part D Questions and Problems 22. 10. a. or fluorine Part B True-False 11. ST 16. NT 12. molecular orbitals bonding orbital lower sigma or pi or three-dimensional VSEPR theory orbital hybridization Part C Matching 15. d 16. 4. c Part D Questions and Problems 20. 2. stable electron covalent shared single unshared pairs double/triple coordinate covalent bond Energy bond dissociation energy resonance structure 15. 6. atom c. 5. a. d 18. 7. 4 carbon atoms.4 Part A Completion 1. AT 13. All rights reserved. 6. 3. 4. ST 11. nitrogen. NT 12. AT 14. H ϩ H 69N 6 H H Part B True-False 10. 7. 1 fluorine atom 22. molecule d. 9. e 18. ST 14. a 19. compound molecular nonmetals diatomic molecular formula 6. 7. 2. 2. b 20. 3. a 18.1 Part A Completion 1. NT 13. e 17. equally nonpolar unequally polar electronegativities dipole interactions hydrogen bond electronegative oxygen. c Part D Questions and Problems 20. d Answer Key 785 . publishing as Pearson Prentice Hall. AT Part C Matching Section Review 8. 7. 8. ST 11. 5. Inc. 9. b 19. c 21. AT 10. e 16. 3. AT 15. a 17. b 19. 10 hydrogen atoms b. AT 13. atom b. 9. molecular compound 21.Section Review 8. Part A Completion 1. 5 hydrogen atoms. 2. 4.3 Part A Completion 1. 3. 5. 6. 6 carbon atoms. a. 8. 5. NT 14. molecule e. ST Section 8.2 © Pearson Education. 8. molecule Part B True-False 9. 4. 8.. c 20. 120Њ apart from one another. ionic b. O C O O 2 O O C O O 2 O C O 2 Part D Questions and Problems 21. three fluorine atoms are required to bond with phosphorus. AT 15. Because carbon can form four single covalent bonds. The remaining three electrons for each carbon atom form a triple covalent bond. Molecular compounds are usually composed from two or more nonmetallic elements. e 19. there is an apparent shortage of atoms with which to bond. A molecular structure gives information about the kinds and numbers of atoms present in a molecule. atom c. Cl N Cl Cl 4. molecule e. the angle between bonds is expected to be slightly smaller than the tetrahedral bond angle of 109. Chlorine needs one more electron to achieve a noble gas configuration. All four bond angles are 109. This is a clue that a carbon-carbon multiple bond exists in this compound. Molecular compounds tend to have lower melting and boiling points that that of ionic compounds © Pearson Education. Thus. diatomic d.2 1. There is one double covalent bond between a carbon and oxygen. Two additional electrons are added to account for the ion having a 2Ϫ charge. diatomic 3. ST 13. The four valence electron pairs repel each other. The three sp2 orbitals lie in the same plane. publishing as Pearson Prentice Hall. a. Fluorine needs one more electron to fill its second energy level. a. The actual bond angle for NH3. d 18. Nitrogen needs 3 more electrons to fill its second energy level.Part B True-False 10. which can shift to any one of the carbon-oxygen bonds giving rise to three resonance structures. 4. 3. F P F F 786 Core Teaching Resources .3 1. The carbon and oxygen can satisfy the octet rule by having the oxygens bonded to a central carbon.5Њ. ST 14. atom d. b 17. 2. All rights reserved. polar covalent bonds c. AT Part C Matching 16.. The unshared pair repels the shared pairs more strongly. not diatomic b. not diatomic e. dipole interactions. but the unshared pair is held closer to the phosphorus than the three bonding pairs. hydrogen bonds 22. H F 2. The four fluorine atoms are covalently bonded to the central carbon atom. is 107Њ. NT 11. AT 12. molecule b. molecule 2. Each carbon atom shares one electron with one of the two hydrogen atoms.5°. polar covalent bonds d. Phosphorous needs 3 more electrons to fill the 3p orbitals. a. Since each fluorine atom only needs one electron and phosphorus needs 3 electrons. three chlorine atoms are required to bond with nitrogen. Inc. a 3. The two atoms share a pair of electrons in order to form a single covalent bond.1 1. Boron forms three sp2 orbitals by mixing one 2s orbital and two 2p orbitals. Each sp2 orbital overlaps with an Section 8. Because each chlorine atom needs only one electron and nitrogen needs 3 electrons. a similar molecule. Carbon has 4 valence electrons and each of the oxygens has 6 valence electrons. The four shared pairs of electrons repel each other to the corners of a tetrahedron. 5. nonpolar covalent bonds Practice Problems 8 Section 8. diatomic c. Section 8. The electron dot structure is: HϺCӇӇCϺH 5. dispersion forces. the hybridization involved in the carbon-carbon bond is sp2. For a bond to be classified as nonpolar covalent. the carbon and oxygens lie along the same axis. 3. H H O . Generally. 4. the effect of the polar bond on the polarity of the entire molecule depends on the shape of the molecule.5°. therefore. like atoms must bond. none Answer Key 787 . tetrahedral. 6. as in diatomic molecules. Inc. Because each carbon can form single covalent bonds with four other atoms. In carbon dioxide. which form 4 sigma bonds. These 2 unshared pairs repel the two bonding pairs and prevent F2O from being linear. silicon mixes one s orbital and all three of the p orbitals. 2. 180°. 7. δ+ a. atomic orbital of chlorine to form three equivalent sigma bonds. with a bond angle of approximately 104. Carbon 2 mixes one s and two p orbitals to form three sp2 hybrid orbitals. This is a clue that CH2CF2 contains a carbon-carbon multiple bond. publishing as Pearson Prentice Hall. The electronegativity difference between P and O is about 1. The other 4 electrons are unshared pairs. O S O O S O O O O S O O O S O 2. linear. trigonal planar. Each carbon atom has two electrons left over. The difference in electronegativity between Na and O is about 2. linear. which overlap with the hybrid orbitals of the carbon and oxygen atoms to form three equivalent sigma bonds.4 1.. two of which are bonding electrons. However. there exists in this compound an apparent shortage of atoms with which to bond. Most bonds are between unlike atoms. Oxygen is the central atom in this molecule. O 3 C 3 O .5°. producing a nonpolar molecule. Both carbon dioxide and carbon monoxide contain polar bonds. The molecule looks very much like the ethene molecule (C2H4). This angle is slightly smaller than the tetrahedral bond angle because the two unshared pairs repel each other more strongly than the two shared pairs. The molecule is a bent triatomic molecule.© Pearson Education. To form four equivalent bonds. 5. δ– F F δ– δ+ δ– C F F δ– 5. 4. All rights reserved. the difference is zero and the bond is nonpolar covalent. the molecule is a dipole. These electrons form a carbon-carbon double covalent bond. 120°. none H . 4. CaO is an ionic compound and CS2 is a polar covalent compound. The two hydrogen atoms bond with one carbon atom while the two fluorine atoms bond with the other carbon. Therefore. The more electronegative atom in a covalent bond will have the ␦Ϫ symbol and the less electronegative atom the ␦ϩ symbol. b. H C H C F F c. The hybridization in SiF4 is sp3. H2C2H and F2C2F bond angles of 120°. Carbon 1 mixes one s orbital and three p orbitals to form four sp3 hybrid orbitals. they must be ionic or polar covalent. Section 8. none C H 3. 109.4 and the bond is polar covalent. F Be F . Interpreting Graphics 8 1. there is a partial positive pole and a partial negative pole. The bond polarities cancel. In carbon monoxide. 180°. The non-hybridized p carbon orbital overlaps with an oxygen p orbital to form one pi bonding orbital. H δ+ δ– δ+ H N H b. ionic compounds have much higher melting points than molecular compounds. It has 6 valence electrons. a. With like atoms.4 and the bond is ionic. d 6. Vocabulary Review 8 1. and form pi bonds.8). 14. Cl P Cl Cl D. O O H H c. F(4. c C. 29. publishing as Pearson Prentice Hall. The remaining 2p orbitals are found in regions above and below the axis. 25.1). All rights reserved. 788 Core Teaching Resources . © Pearson Education. d 21. a. Check students’ work. trigonal bipyramidal. Br 2 Br ϺNϵNϺ ϺCϵOϺ K(0. 7. c 9. polar covalent bond 6.0. none 15. Inc. Cl Cl Cl 90° and 120°. 6. 10.5. Matching 1. difference 1. Samples of these solids are thought of as single molecules.7. 8. 4. Quiz for Chapter 8 1. g 10. H Br b. . h B. i e j f 5. Two examples are diamond and silicon carbide. c. 13.5). N(3. 6. 3. 4.2. c c b a The first 2p orbitals lie along the axis connecting the atoms. pyramidal. polar covalent bond c. 19. 105Њ. Cl Cl P b.5). difference 0. difference 0.. Additional Questions and Problems 393 kJ 32. F P F F . 12. none 7. none 8. b 22.0). 96°. 5. C—H 5 mol ϫ ᎏᎏ ϭ 1965 kJ 1m ol 356 kJ C—O 1 mol ϫ ᎏ ϭ 356 kJ 1 mol 464 kJ O—H 1 mol ϫ ᎏᎏ ϭ 464 kJ 1m ol 347 kJ ϫ ᎏᎏ ϭ 347 kJ C—C 1 mol 1m ol Total ϭ 3132 kJ 33. difference 3. 4. coordinate covalent bond bond dissociation energy bonding molecular orbital sigma bond VSEPR theory hybridization polyatomic ion van der Waals forces hydrogen bond molecule 30. 2. H O H . 7. 9. a c b d 5.5. Network solids are substances in which all of the atoms are covalently bonded to each other. b a d c 2py 2px ϭ pi bond Chapter 8 Test A A. Multiple Choice 11. 3. ionic bond S(2. bent triatomic. polar covalent bond d. 8. Essay 31. a. b. O(3. c 16. ST 11. ST E. H(2. Br(2. and so form a sigma bond. 24.8). b 27. ϩ ϭ sigma bond 2px ϩ 2py 23.0).5). a 7. Questions 28. 2. b c ST NT 9. O(3. 3. 18. 20. a. 2. 26. d b d a 17. ST 10. b 8. yellow. Multiple Choice 11. 2. Red is Red No. but sometimes appears. polar covalent. (Red #3 has been banned. page 245 Analysis Blue No. Cl Cl P Cl Cl Cl F S F F F F F D. e. 4. 25. 23. 1. sp. c b a c b b b c b 29. 5 (or sometimes Yellow No. Green food color is usually a mixture of Yellow No. 1. sp3. H 6 H . Differences of less than 2. 3. 0. 5 Red No. 40 C. O . i 9. a. 3. 18. depends upon The electronegativity difference between two elements is used to predict which type of bonding will occur when specific atoms combine.2 f. b 6. © Pearson Education. Matching 1. ionic bond. Inc. 0. 32. 12. 6(C—H) ϭ 6 ϫ (393 kJ/mol) ϭ 2358 kJ/mol 1(C—C) ϭ 1 ϫ 347 kJ/mol ϭ 347 kJ/mol Total ϭ 2705 kJ or 2.0 result in covalent bonding. a.70 ϫ 103 kJ/mol (2. All rights reserved.. H H 69N 6 H C H @ H 2 N2 H Red d. 34. 16. H2H b. 19. d c d a a c c a c Chapter 8 Small-Scale Lab Section 8. 13.9 b. c. sp2. 40. 24. Yellow is Yellow No.4 Paper Chromatography of Food Dyes. f 8. 3. Differences in excess of 2. c a d d d a b d b 20.70 ϫ 103 kJ/mol ) ϫ (0. O 2 H @ H H . d 7. polar covalent.1 e. 30. a 10. 0. Essay 40. polar covalent. N N . sp3 35.4 c. 1. 17. 14. From left to right: sp. Questions 38. 28. 2. 6 if it is orange in appearance). O O P O O 3Ϫ . N4N O H. a.0 result in the formation of ionic bonds. Covalent bonds are formed when atoms share electrons. Additional Questions and Problems 41. C!O ϩ Yellow Green Blue 1. publishing as Pearson Prentice Hall. 26.25 mol ) ϭ 675 kJ B. 0.34. g e h c 5. Red. polar covalent. and blue are pure compounds. 5 and Blue No. An ionic bond is formed when one or more electrons are transferred from one atom to another. 15. and blue is Blue No. 1 Yellow No.5 d. because under the law food manufacturers are allowed to use up their current supplies. 22. Chapter 8 Test B A. 37. Answer Key 789 H H 69N 6 H H b. See the above figure.) 4. 35. nonpolar covalent. 39. 27. 33. 36. j E.3 42. 21. 31. Develop in 0. 6. ionic with polyatomic ion c. cation anion -ide sodium iodide 5. NT 10. 2Ϫ d. 2. Pb3(PO4)2 c. All rights reserved. 1 is the most polar because it runs the fastest and appears at the top of the chromatogram. Mg(HCO3)2 Section Review 9. Wet a portion of a piece of candy and blot it with a paper towel to remove excess water. ST 12. 4. a. Roman numeral anion oxygen zero Part B True-False 9. ST 15. e 21. d 22. a 23. c. Use a toothpick to spot a solution of powdered drink on chromatography paper. 2. b 11. potassium hydroxide. gains 1 d. 3. Section Review 9. Press the wet side of the candy onto the chromatography paper so that it makes a colored spot. nonmetallic -ide atoms diarsenic pentasulfide Part B True-False 14. 3. 7. monatomic lose 1ϩ 2ϩ 3ϩ 8 1Ϫ 8. Repeat for other colors of candy. binary ionic b. NT 7. sodium dichromate. 7. NT 18. a 15.2 Part A Completion 1. a. NT Part D Questions and Problems 24.1% NaCl.5. 2. d 9. 4. AT 11. ammonium permanganate hydroxide loses 2 gains 2 c. -ite or -ate © Pearson Education. c Part D Questions and Problems 17. loses 3 Section Review 9. 5 because of the water content of the paper. hydrogen carbonate Part C Matching 8. b 14.1% NaCl. a. d 16. 1 and Yellow No. 26. ST 6.3 Part A Completion 1. publishing as Pearson Prentice Hall. polyatomic 12. c Part D Questions and Problems 790 Core Teaching Resources .1 Part A Completion 1. AT Part C Matching 19. 1Ϫ b. b. ionic with polyatomic ion 18. 40 is the least polar. Stock 10. 3. a. c Part B True-False 5. Different water content changes variations in the polarity of the stationary phase (the water molecules hydrogen bonded to the paper). a 10. classical 11. Rubbing alcohol runs much more slowly and gives slightly better separation than 0. Inc. Some papers cause a reversal of the positions of Blue No. 6. 2ϩ 25. AT Part C Matching 13. Make a small spot of each colored marker pen on a piece of chromatography paper and develop in solvent. a. d. 8.. -ite or -ate 13. 1ϩ c. 5. 2. Blue No. transition (Group B) metals 9. 4. ST 17. NaClO3 b. b. Red No. You’re the Chemist 1. 3. 5. 4. iron(III) bromide. AT 16. b 20. elements 5. small whole 12. 2. 21. e.12. NT Part C Matching 13. a.5 Part A Completion 1. a. a. b. a. H3PO4 HF d. b. b. FeBr3 BaCl2 d. acid 13. multiple 11. 4. d. 17. Mg(OH)2 c. a. a. c. NT 12. 7. d. a. b. c. a. 3. e. 2 lost 3 gained e. b. 8. 5. phosphorus pentachloride sulfur dioxide tetraphosphorus decasulfide CBr4 b. NH4ϩ cyanide anion hydrogen carbonate anion phosphate anion chloride anion calcium cation sulfite anion Part B True-False 10. b. lead(IV) acetate hydrofluoric acid diphosphorus pentoxide lithium bromide PCl5 FeO HNO3 KCl Ca(NO3)2 Section Review 9. NT 19. 3ϩ 1Ϫ e.. 2Ϫ d. Na2S KI f. MgO d. d. a 14. AlCl3 SnF2 e. hydrogen hydrogen ions hydrobromic nitric ionic 6. ST 18. a. anion 6. 2. hydroxide ions ionic cation anion Practice Problems 9 Section 9. 2 gained tin(II) or stannous cation cobalt(III) or cobaltic cation bromide anion potassium cation hydride anion manganese(II) or manganous cation d. b. b. d. acid proportions 4. LiOH potassium hydroxide hydroiodic acid sulfuric acid Section Review 9. c. publishing as Pearson Prentice Hall. 3. c. ST 11. b. 13. a. c. binary Section 9. lead(II) acetate 8.1 1. a. c. 2ϩ 1ϩ f.2 1. 2. © Pearson Education. 9. c 15. b. b Part D Questions and Problems 16. carbon tetrachloride 3. 3. b. f. acid 15. f. e. AT . Inc. phosphoric acid 14. proportions 10. c. molecular 2. CrO42Ϫ SO42Ϫ f. c. a. c. definite proportions 9. b. 4.4 Part A Completion 1. N2O4 Part C Questions and Problems 20. KBr AgI e. c. ST 17. b. a. OHϪ CO32Ϫ 2Ϫ NO2 e. 1 gained 1 lost f. FeO manganese(II) oxide or manganous oxide lithium nitride calcium chloride Answer Key 791 Part B True-False 16. Al2O3 CaS f. All rights reserved. 4A 7. 5. c. a. 2ϩ 3 lost d. c. 5. c. Kϩ iron(III) cation. a. d. 10 10 10 a. d. f. 6. g. Naϩ nickel cation. a. b. a. k. Ca2ϩ potassium cation. NH4Cl NaOH f. NO3Ϫ SnCl4 H2S CaO HBr AlF3 SO42Ϫ CaSO4 Fe2(CO3)3 SF6 MgCl2 H3PO4 HNO3 OHϪ PO43Ϫ Ca(NO3)2 Ca(OH)2 Ca3(PO4)2 AlPO4 Na3PO4 Al3ϩ Al2(SO4)3 Al(NO3)3 Al(OH)3 Naϩ Na2SO4 NaNO3 NaOH Pb4ϩ Pb(SO4)2 Pb(NO3)4 Pb(OH)4 Pb3(PO4)4 Section 9. © Pearson Education. 5. the masses of one element that combines with a given mass of the other element will be in the ratio of small whole numbers.5 1. b. phosphorous triiodide h. a. nitrogen tribromide g. 6. f. The law of definite proportions states that samples of any compound will always contain the constituent elements in the same proportions. e. d. c. 4. d. aluminum hydroxide c. publishing as Pearson Prentice Hall. c. N2O5 h. b. e. b. K2S b. mercury(II) oxide or mercuric oxide e. f. d. l. b. d. 2.4 1. 2A two 12 10 7A one nine 8. c. The law of multiple proportions states that in two compounds containing the same two elements. Section 9. Fe3ϩ copper(I) cation. a. N2F4 3. h. a. dinitrogen pentoxide f.4. 18 18 18 18 792 Core Teaching Resources . hydrofluoric acid carbonic acid Al(OH)3 LiOH Interpreting Graphics 9 1.3 1. sodium hydrogen sulfate d. 2. Ba(OH)2 KNO3 d. strontium bromide nickel chloride potassium sulfide copper(II) chloride or cupric chloride tin(IV) chloride or stannic chloride Na3PO4 d. a. j. b. Ca2ϩ g.. K2Cr2O7 (NH4)2SO4 c. b. Li2CO3 sodium cyanide iron(III) chloride or ferric chloride sodium sulfate potassium carbonate copper(II) hydroxide or cupric hydroxide lithium nitrate sodium cation. 11. f. nitrous acid sulfuric acid Ca(OH)2 NH4OH c. g. e. potassium phosphate b. 3. e. b. a. 2. e. ammonium sulfate 4. f. Inc. d. 10. Cuϩ Section 9. 9. a. a. All rights reserved. 2. c. h. phosphorous pentachloride carbon tetrachloride nitrogen dioxide dinitrogen difluoride tetraphosphorous hexoxide xenon difluoride silicon dioxide dichlorine heptoxide NBr3 c. b. i. Ni2ϩ calcium cation. KCN MgSO4 e. 7. c. SO2 Cl2O d. d. c. 7. 2. a. b 8. i 9. 13. j 10. Completion 22. publishing as Pearson Prentice Hall. Matching 1. e B.. a Mg(CN)2 c. Inc. f 2. Completion 23. g 4. i 2. 4. 9. 3. 18. g 6. sodium sulfate Fe2O3. a D. d 6. calcium nitrate Na2SO4. b. 2. Multiple Choice 10. Cu(OH)2 15. h 9. 10. 5. 20. d. Ca(NO3)2. Chapter 9 Test B A. 22. Common names do not describe the chemical composition of a compound. 20. 29. HgBr2 32. 29. 14. Problems 31. Compounds exist in enormous numbers. 11. Matching 1. 15. gain lose cation definite proportions 26. a 3. more C. 4. c 4. 23. 25. i c g d 5. 19. f 7. 26. 21. 24. 3. The systemic method tells what atoms are in the compound. copper(I) chloride b. f 8. 21. 2Ϫ 27. 12. 11. c c a a a 15. 24. but usually do not reveal what elements are in the compound. c. a 7. Essay 33. d d c c 18. 8. Problems 32. 4ϩ 1Ϫ ions -ide 27. iron(III) oxide Al2(CO3)3. 16. 28. potassium acetate Quiz for Chapter 9 1. 30. 7. a c c d 19. dinitrogen trioxide c. Metals PCl5 anion definite proportions the group number lose hydroxide H2PO4Ϫ dinitrogen monoxide FeCl2 tin(IV) sulfide (or stannic sulfide) dinitrogen pentoxide sodium hydrogen carbonate (or sodium bicarbonate) 14. a. 17. h 5. b 3. b 7. 25. e 9. 30. 11. b a d c B. d © Pearson Education. c a a c 14. 13. c c c b C. 12. They may relate to a physical or chemical property. h 5. gives information on the ratio in which the atoms combined to form the compound and promotes efficient and effective communication between chemists. Multiple Choice 10. 17. Chapter 9 Test A A. NH4ϩ magnesium nitrate -ide Ca3(PO4)2 31. 13. e 6. aluminum carbonate Answer Key 793 . 6. c 8. 16.Vocabulary Review 9 1. 28. 1Ϫ oxygen nonmetallic hydrogen D. All rights reserved. SF6 b. nitric acid E. 12. 2 Names and Formulas for Ionic Compounds. AgOH silver hydroxide Teacher’s note: This is Ag2O. publishing as Pearson Prentice Hall. f. Essay 35. 2. Molecular compounds are named from the elements that comprise them. PbSO4 i. c. Formula a. using a Roman numeral to distinguish between positive ions of the same element that have more than one charge. No prefix is used if only one atom of the first element is present. CaCO3 j. PbCO3 f. e. page 267 Analysis AgNO3 ( Ag؉) a Na2CO3 (CO32؊) b Na3PO4 (PO43؊) c NaOH (OH؊) d Na2SO4 (SO42؊) milky white ppt cloudy white ppt muddy brown ppt no visible reaction e Pb(NO3)2 (Pb2؉) cloudy tan ppt milky white ppt milky white ppt milky white ppt Figure A 1. Fe2(SO4)3 PBr3 h. g. Inc. Ionic compounds are named from the two ions that comprise them. no visible reaction e. All rights reserved. whereas molecular compounds consist of nonmetallic elements. AgNO3 Zn(OH)2 g. HgCl2 carbon disulfide ammonium carbonate diarsenic pentoxide carbon monoxide tin(IV) hydroxide sulfuric acid phosphorus pentiodide potassium permanganate E. e. Ca3(PO4)2 k. Ag3PO4 Name silver carbonate silver phosphate © Pearson Education. The name of the second element always ends in -ide. CaSO4 lead(II) carbonate lead(II) phosphate lead(II) hydroxide lead(II) sulfate calcium carbonate calcium phosphate calcium hydroxide calcium sulfate 794 Core Teaching Resources . Ionic compounds consist of a metallic and a nonmetallic ion. d. HNO3 Chapter 9 Small-Scale Lab Section 9. 34. Ag2CO3 b. Ca(OH)2 l. Pb3(PO4)2 g. b. c.33. Na2SO4 ϩ AgNO3 did not form a precipitate. d. a. Pb(OH)2 h. SiO2 b. f j g k h l c. a. i CaCl2 (Ca2؉) grainy white ppt milky white ppt cloudy white ppt grainy white ppt CF4 f. h.. silver oxide d. using prefixes to denote the numbers of atoms of each element present. Cu2+ ϩ CO32Ϫ → CuCO3(s) copper(II) carbonate copper(II) phosphate copper(II) hydroxide magnesium carbonate magnesium phosphate magnesium hydroxide Name iron(III) carbonate iron(III) phosphate iron(III) hydroxide l no visible reaction orange ppt f white ppt k blue ppt e white ppt j blue ppt MgSO4 (Mg2؉) i blue ppt CuSO4 (Cu2؉) j. Mg2+ ϩ 2OHϪ → Mg(OH)2(s) i. NT 9.2 Part A Completion 1. mole 3. Cu3(PO4)2 k. ST 7. d 14. c 13.3 g K2SO4 17. 22.65 ϫ 10Ϫ2 mol K2SO4 ϫ ᎏᎏ 1. Fe3+ ϩ 2OHϪ → Fe(OH)2(s) e. FePO4 c.4 4. 2Fe3+ ϩ 3CO32Ϫ → Fe2(CO3)3(s) b.0 g H 1 mol H molar mass of C2H6 ϭ 30.You’re The Chemist 1. AT 10. Cu2+ ϩ 2OHϪ → Cu(OH)2(s) Section Review 10. no visible reaction i. MgCO3 f.02 ϫ 1023 representative particles ᎏ 1 mol H2O2 ϭ 1.5 mol H2O2 ϫ 6. Mg(OH)2 h.0 g C 1 mol C 1. Cu(OH)2 l. 3. CuCO3 j.00 mol SO2 Part B True-False 6.1 Part A Completion 1.4 L SO2/1. Fe(OH)3 d. 2. NT d Na2SO4 no visible reaction (SO42؊) Part C Matching 11. Inc. AT Answer Key 795 . FeCl3 ( Fe3؉) a Na2CO3 (CO32؊) b Na3PO4 (PO43؊) c NaOH (OH؊) orange ppt orange ppt g white ppt h no visible reaction Figure B Formula a. 5. NT 9.0 g 174. Fe3+ ϩ PO43Ϫ → FePO4(s) c.02 ϫ 1023 atoms Pb ϭ 1. All rights reserved. e.5 ϫ 1024 representative particles Section Review 10. 9. Mg3(PO4)2 g. ST 8. c Part D Questions and Problems 1. 3Cu2+ ϩ 2PO43Ϫ → Cu3(PO4)2(s) k.36 g K2SO4 18. 3Mg2+ ϩ 2PO43Ϫ → Mg3(PO4)2(s) g. Density 2. a. 2. no visible reaction 2.00 mol K2SO4 ϭ 636 ϫ 10Ϫ2 g K2SO4 ϭ 6. b 12. publishing as Pearson Prentice Hall.0 g H 6 mol H ϫ ᎏᎏ ϭ 6. 2 mol C ϫ ᎏᎏ ϭ 24.02 ϫ 1023 Part B True-False 6. 22. 3. ST 7. mole Avogadro’s number atomic masses molar mass 6. molar volume 5. 4. AT 8. Mg2+ ϩ CO32Ϫ → MgCO3(s) f.3 ϫ 1015 atoms Pb ϫ ᎏᎏᎏ 6. Fe2(CO3)3 b.0 g C 16..5 ϫ 10Ϫ8 mol Pb 12. no visible reaction © Pearson Education.0 mol Pb 15. 158. Molar mass NaCl ϭ 23. a.8 ϫ 103 g c.0 g/mol 2. 153. 7.2 1. a.0 gS 1.0 g Density ϭ ᎏᎏ ϭ ᎏᎏ ϭ 1. 5. percent composition 100 molar mass empirical whole-number molecular 40.90 mol O/1.4% O 284 g Mn2P2O7 201 g Hg c.500 mol ϭ 11.43 ϫ 102 g 3.0 g/32.2 g/mol Part C Matching 11. 84.4 g O ϫ ᎏᎏ ϭ 1. 29.87 ϫ 102 g 4.27 mol Na/1.0 g O 2 mol Fe ϫ ᎏᎏ ϩ 3 mol O ϫ ᎏᎏ 1m ol Fe 1m ol O ϭ 112 g Fe ϩ 48 g O ϭ 160 g Fe2O3 112 g g Fe %Fe ϭ ᎏ ᎏ ϫ 100 ϭ ᎏᎏ ϫ 100 160 g g Fe2O3 ϭ 70.3% Hg 233 g HgS 32.4% Cr 152 g Cr2O3 48 g O ᎏᎏ ϫ 100 ϭ 31. Molar mass ϭ 28.27 ϭ 1 ϫ 2 ϭ 2 1.2 L Section Review 10. 22. 96. 32.5 g ϭ 58.5g/mol 58.00 mol Na 15.8% P 284 g Mn2P2O7 112 g O ᎏᎏ ϫ 100 ϭ 39.2 g/mol b. 2.1% N 164 g Ca(NO3)2 96 g O ᎏᎏ ϫ 100 ϭ 58. ᎏᎏ ϫ 100 ϭ 38. 3.480 g or 4.4 L/mol ϫ 0.00 mol O 30.20 ϫ 10Ϫ2 g d.5 g S ϫ ᎏᎏ ϭ 1. ᎏᎏ ϫ 100 ϭ 68.500 mol 18.8% S 233 g HgS Section 10.1 g Ca d. c 12. 1.90 mol O 16. 3.Part C Matching 10.0 ϭ 44.27 ϭ 1. 4. molar mass Fe2O3 ϭ 55.1 1.85 mol H2O 5 3.3 Part A Completion 1. 5. AT 10. 342.0 g/mol ϭ 0. c 12. ᎏᎏ ϫ 100 ϭ 24. 208.2 g/mol 3. 180. 204. Molar mass O2ϭ 2(16. a. b 13. Inc.27 mol S 32. ᎏᎏ ϫ 100 ϭ 86.27 mol Na 23. publishing as Pearson Prentice Hall.0 ϩ 16.4 L 16.00 mol S 40.0% Fe 70.27 mol S/1.8 g Fe 16.7% Mn 284 g Mn2P2O7 62 g P ᎏᎏ ϫ 100 ϭ 21.1 g S ᎏᎏ ϫ 100 ϭ 13.0 g/mol 4. a. a Part D Questions and Problems 104 g Cr 14. a 13.5% O 164 g Ca(NO3)2 1.6% O 152 g Cr2O3 110 g Mn b.6 ϫ 1023 atoms 6.80 ϫ 10Ϫ1 g e.0 gO 1.. 0. 26 g c.2 g/mol d. NT Practice Problems Section 10.1 g Na ϫ ᎏᎏ ϭ 1.0 g mass 44.0 g Na 1. 310. 352. 1. ST 8.2 g/mol b.0 g/mol) ϭ 32.96 g/L volume 22.3 g/mol 2.0 g ϩ 35. 1. 6. All rights reserved. e Part D Questions and Problems 15. b 14.0 g b.5 ϫ 2 ϭ 3 Empirical formula ϭ Na2S2O3 16.5 g/mol ϫ 2 mol ϭ 117 g 17. AT 9.5% Ca 164 g Ca(NO3)2 28 g N ᎏᎏ ϫ 100 ϭ 17.0 kg Fe 639 kg Fe2O3 ϫ ᎏᎏ ϭ 447 kg Fe 100 kg Fe2O3 Part B True-False © Pearson Education. a.0 g/mol b.1 g 796 Core Teaching Resources .27 ϭ 1 ϫ 2 ϭ 2 1. d 11.0 g/mol 16. 9.4% C ϫ 65.02 g %H ϭ ᎏᎏ ϫ 100% ϭ 2. 85. 150. 2.8% C Section 10. Use the same approach to show that Acid Y represents sulfurous acid.06% 98. 2. 9. 6. molar mass 5. 4.35 g of compound Ϫ 5.84 g cpd 47.8 g 5.7% Cl 3. percent composition empirical formula 22.98 ϫ 10Ϫ5 mol 6.907 g C 3.34 g C 1.35 g cpd ϭ 31.1% C 52.08 g Cl Percent Cl ϭ ᎏᎏ ϫ 100 ϭ 89. 1 atm) © Pearson Education. 9. Mass of Cl ϭ total mass of compound Ϫ mass of Sn ϭ 18.3 g H 2.3 5. 10. 4.4% C Mass C ϭ 30. 7.. publishing as Pearson Prentice Hall. CHCl3 85. Cyclohexane and ethene have the same empirical formula and. a. 3.00 mol Answer Key 797 Interpreting Graphics 10 1.08 g Acid X represents sulfuric acid.42 g H Percent H ϭ ᎏᎏ ϫ 100 ϭ 0.0 g O2 ᎏᎏ 1 mol O2 AT AT ST NT AT 1.7% C 14.06 g %S ϭ ᎏᎏ ϫ 100% ϭ 32.08 g 32.10 ϫ 10Ϫ2 mol e.25% 98.5. 4. Inc.00 mol 197 g Au 3. . H2SO3. Percent C ϭ ᎏᎏ ϫ 100 ϭ 81.4 g ethene 100 g C2H4 3. CCl4 b.08 g 64. Quiz for Chapter 10 1. 15. All rights reserved.6 L CH4 6. 7.69% 98. a substance atomic mass molar mass atom 32. 8.4 L molar mass mole Avogadro’s number standard temperature and pressure (0ЊC.1 g 2.00 g %O ϭ ᎏᎏ ϫ 100% ϭ 65. 5.781 g cpd 48.1% Br Bromine accounts for the largest percent of the mass of dibromoethane. Percent C ϭ ᎏᎏ ϫ 100 ϭ 10.3 g ϭ 19.3% H 4.97 ϫ 10Ϫ5 mol c. 6.0 g C 4.61 g Cl Percent of Cl ϭ ᎏᎏ ϫ 100 18.56 g Au 108 g Ag 1. 8. 1.9 ϫ 10Ϫ3 mol d. Vocabulary Review 10 1.2 g Al 6. 3. 5. 6.3% Sn 12. a.1 g 59. 2. H2SO4. 11.7% C 4.03 mol NH3 4. 7.1% H 12.84 g cpd 0. 1.05 g Ag ϫ ᎏᎏ ϫ ᎏᎏ ϭ 5.781 g cpd 0. therefore.35 g cpd ϭ 68.74 g Sn Percent of Sn ϭ ᎏᎏ ϫ 100 18.11 g Fe 7. 1. molar mass of H2SO4 ϭ 98.43 mol 15. They are indistinguishable on the basis of percent composition alone.84 g cpd 2.874 g H Percent H ϭ ᎏᎏ ϫ 100 ϭ 18.1 g Ca(C2H3O2)2 ϭ 30.61 g Cl 5.08 ϫ 10Ϫ2 mol 5.1% Cl 52.74 g Sn ϭ 12. 13. Some compounds have the same empirical formula but different molecular formulas. the same percent composition. Percent C ϭ ᎏᎏᎏ ϫ 100 158.79% H 52. b.0 g C2H4 ϫ ᎏᎏ ϭ 21. c b b c C.6 g/mol L mol 1 mol 31. 8.0 gF 2 mol C The mole ratio of C to F is ᎏᎏ. 17. Essay 28. 12. A mole is a unit that counts all kinds of representative particles. C. 11.0 gC 1.25 mol ϫ 98.39 mol O 16. 364 g Ar ϫ ᎏᎏ ϭ 9.0 g O: ᎏᎏ ϫ 100 ϭ 64. 14.0 g F ϫ ᎏᎏ ϭ 4 mol F 19.6 g F2 1 mol C 26. b c i f 9. 13.00 mol C 12. 4. so molar mass can be used for the mass of Avogadro’s number of particles of any pure substance.0 g N: ᎏᎏ ϫ 100 ϭ 18.650 ᎏᎏ ϫ 22. Multiple Choice 11.0 g ϫ ᎏᎏ ϭ 1.02 ϫ 1023 ᎏ mol 23 ϭ 1. h B. Inc.0 g ϭ 7640 L g L 30.3 g © Pearson Education.0 g F2 ϫ ᎏᎏᎏ ϫ ᎏᎏ 6. 24.0 g C ϫ ᎏᎏ ϭ 2 mol C 12.00 mol C 3. 20. 3. 3. d b a c b b 15. 4 mol F The lowest whole-number ratio of C to F is CF2. 2.39 mol O/3. 25. Multiple Choice 9.3 g 96. 10.1. 2. 24. 19. A mass that has molar as a modifier must be the mass of a mole. e 5.4% 148. 23. 19. 15. 6.00 ϫ 1023 molecules F2 1 mol F2 38. Matching 1. Problems 25.4 ᎏᎏ ϭ 14. 16.0 gO 1. 23. 18. Ar 39. d a c b d c B. g 6.7 g C ϫ ᎏᎏ ϭ 3. f. 26. 40. 24.0 g ϭ 148.1 ᎏᎏ ϭ 613 g mol 22. publishing as Pearson Prentice Hall. All rights reserved. j a g e 5.3 g Mg: ᎏᎏ ϫ 100 ϭ 16.50 ϫ 10 atoms g 28.1 mol H/3.39 ϭ 1.5 mol H ϫ 2 ϭ 3 mol H 1 mol C ϫ 2 ϭ 2 mol C 1 mol O ϫ 2 ϭ 2 mol O Empirical formula ϭ C2H3O2 D.00 mol O 1. 22. 7. d 10. 12.5 mol H 3. d Chapter 10 Test A A.39 mol C 12. 14.9 g Ar 798 Core Teaching Resources .1 mol H 1.0 gH 5. Problems atoms ᎏ 27.3 g 28.0 gC 1 mol F 76. 24. a d a c c b 21. h 3.02 ϫ 1023 atoms 197. 13.9% 148. 22.39 mol C/3.0 kg ϫ 1000 ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ ϭ 7636 L mol kg 44. 1 mol Ar 27. c 7.12 mol. 15. 6.3 g ϩ 28. c a d a c 21.00 mol O 54. Chapter 10 Test B A.25 mol ϫ 6. Matching 1.39 ϭ 1.75 ϫ 1015 atoms ϫ ᎏᎏ 6.02 ϫ 1023 molecules F2 1m ol F 2 ϭ 31..00 mol H 5. 17.3 g/mol Mg(NO3)2 24.1 g H ϫ ᎏᎏ ϭ 5.7% 148.0 g ϩ 96. 5.23 ϫ 10Ϫ6 g mol 32.39 ϭ 1.2 g O ϫ ᎏᎏ ϭ 3. 0. 18.4 L g 1 mol 29. 0. a 8. c a d c c 16. 20. b 4. 0967 C 0. x mol H ϭ 0.239 O 0. the empirical formula is CO2.68 g CaCO3 ϫ ᎏᎏ 100.0870 mol NaCl 1 mol 2.0 g 1 mol O (43.82 mol P 31.28 Thus.0 58. 1 mol P 34. The mass of one mole.0 g Empirical formula is P2O3 empirical formula mass ϭ 110.5 g x mol NaCl ϭ 0.0870 0. 27.0 g or P4O6 Chapter 10 Small-Scale Lab Section 10.2 Counting by Measuring Mass.82 ϫ 1022 C 1. (56.88 ϫ 1023 H 1.28 2..9 g molecular formula ϭ ᎏᎏ ϭ 2 ϫ P2O3 110. and ᎏᎏ ϭ 2 2.0 g.239 mol H2O ϫ ᎏᎏ 1 mol H2O x mol H ϭ 0.290 O 22 5.75 ϫ 1023 O © Pearson Education.24 ϫ 1022 Na 5.0870 Cl 0.0967 mol CaCO3 2 mol H 3.239 mol O Answer Key 799 .0967 Ca 0. publishing as Pearson Prentice Hall.68 D.30 g H2O ϫ ᎏᎏ 18.54 mol O 16. x mol H2O ϭ 4.1 0. Essay 35.09 9.0870 Na 0.38 g P) ϫ ᎏᎏ ϭ 1. but expressed in grams.239 0.0967 0.62 g O) ϫ ᎏᎏ ϭ 2. or Avogadro’s number of atoms of that element has the same numerical value as the atomic mass.478 mol H 1 mol O x mol O ϭ 0.73 mol O 16.0 gC 1 mol O 72.1 mol C 33. H2O(l ) Mass (grams) NaCl(s) CaCO3(s) 4. The mass of a single atom of an element is the atomic mass given on the periodic table.3 g C ϫ ᎏᎏ ϭ 2.0 g O 2. Inc.30 5.28 mol C 12.5 100. 2.7 g O ϫ ᎏᎏ ϭ 4.82 ϫ 10 Ca 5.28 4. page 304 Analysis Student data may vary slightly.239 mol H2O 1 mol x mol CaCO3 ϭ 9. so that 219. Molar Mass (g/mol) Moles of each compound Moles of each element Atoms of each element 18.1 g x mol CaCO3 ϭ 0. All rights reserved.24 ϫ 1022 Cl Figure A 1 mol NaCl ϫ ᎏᎏ 1. x mol NaCl ϭ 5.44 ϫ 1023 O 5.09 g 58.478 H 0.239 mol H2O ϫ ᎏᎏ 1 mol H2O x mol O ϭ 0. expressed in atomic mass units.0 g x mol H2O ϭ 0.54 ᎏᎏ ϭ 1. Write your name with the chalk and determine the mass of the chalk again. AT 15. 10.02 ϫ 1023 atoms ϫ ᎏᎏᎏ 1 mol O x atoms O ϭ 1. NT 16. True-False 13. 2. g 21. ST C. d 23.0870 mol Cl 6.02 ϫ 1023 atoms ϫ ᎏᎏᎏ 1 mol O x atoms O ϭ 1. 9.0870 mol NaCl ϫ ᎏᎏ 1 mol NaCl x mol Cl ϭ 0. Problems 24. D.239 mol O 6. Determine the mass of a piece of chalk. 6. 8. 8.0870 mol Na 6. 9. a 22.88 ϫ 1023 atoms H x atoms O ϭ 0. 12. 11. 4. subscripts (l) (s) (g) (aq) catalyst B. predict combination elements single decomposition single-replacement activity series of metals double-replacement aqueous oxygen carbon dioxide or water water or carbon dioxide x atoms C ϭ 0. 10.2 A. Multiple Choice 17.75 ϫ 1023 atoms O 5.290 mol O 1 mol CaCO3 4..02 ϫ 1023 atoms ϫ ᎏᎏᎏ 1 mol Ca 22 x atoms Ca ϭ 5. e 19. 4.0870 mol Cl x mol Ca ϭ 0. c 18. a. Completion 1.1 mol Na x mol Na ϭ 0. Convert the mass difference to moles and atoms.0967 mol CaCO3 1 mol Ca ϫ ᎏᎏ ϭ 0.44 ϫ 1023 atoms O x atoms Na ϭ 0. Determine the mass of 100 drops of water and then calculate the mass in grams of one drop. 6. Completion 1. 6.02 ϫ 1023 atoms ϫ ᎏᎏᎏ 1 mol Na x atoms Na ϭ 5. 2C2H2(g) ϩ 5O2(g) → 4CO2(g) ϩ 2H2O(g) Section Review 11. Inc. Section Review 11.1 A. NT 14. equation reactants products mass coefficients element 7.24 ϫ 1022 atoms Na x atoms Cl ϭ 0. 3.02 ϫ 1023 atoms ϫ ᎏᎏᎏ 1 mol H x atoms H ϭ 2.0967 mol Ca 1 mol CaCO3 x mol C ϭ 0. 2Al(s) ϩ 6HCl(aq) → 2AlCl3(aq) ϩ 3H2(g) b.0967 mol C 6. publishing as Pearson Prentice Hall. Water has the greatest number of moles in one teaspoon.83 ϫ 10 atoms Ca You’re The Chemist 1. 2.0967 mol CaCO3 1 mol C ϫ ᎏᎏ ϭ 0.02 ϫ 1023 atoms ϫ ᎏᎏᎏ 1 mol C x atoms C ϭ 5.478 mol H 6. Water has the greatest total number of atoms. 5. 2. 5.0870 mol Na 1 mol Cl x mol Cl ϭ 0.0967 mol C 1 mol CaCO3 x mol O ϭ 0. 3.09697 mol Ca 6.02 ϫ 1023 atoms ϫ ᎏᎏᎏ 1 mol Cl 22 x atoms Cl ϭ 5. b 20.24 ϫ 10 atoms Cl x atoms Ca ϭ 0.83 ϫ 1022 atoms C x atoms O ϭ 0. 11. x atoms H ϭ 0. All rights reserved.0870 mol NaCl ϫ ᎏᎏ 1 mol NaCl x mol Na ϭ 0. 7.0967 mol CaCO3 3 mol O ϫ ᎏᎏ ϭ 0. f © Pearson Education. 12.290 mol O 6. 800 Core Teaching Resources . balance the equation. Matching 14. 2Li3PO4(aq ) ϩ 3Zn(NO3)2(aq ) → Zn3(PO4)2(s) ϩ 6LiNO3(aq ) For any double-replacement reaction to occur.2 1. Inc. 6. one of the products must be a solid (precipitate). 9. 7. 2. AT 14. 5. or a gas. no reaction b. 3. the equation is balanced correctly. C3H8(g) ϩ 5O2(g) → 3CΟ2(g) ϩ 4H2O(g) 6.. 3. ST 11. ST 13. and the coefficients are in their lowest possible ratio. Oxygen is in Group 6A and forms anions with a 2Ϫ charge. Thus. 6. Section 11. combustion b. All rights reserved. Magnesium is a Group 2A metal and forms cations with a 2ϩ charge. CaCO3(s) c. H2O(l) ϩ SO3(g) → H2SO4(aq) 2AgNO3(aq) ϩ Cu(s)→2Ag(s) ϩ Cu(NO3)2(aq) 4P(s) ϩ 5O2(g) → P4O10(s) C. NT 15. water aqueous complete ionic equation spectator ions net ionic equation charge atoms precipitate solubility 7. Section Review 11. Completion 1. True-False 10. The balanced net ionic equation is Cl2(g) ϩ 2BrϪ(aq) → Br2(l) ϩ 2ClϪ(aq) 21. and liquid water. Cl2(g) ϩ Na (aq) ϩ Br (aq) → Br2(l) ϩ Naϩ(aq) ϩ ClϪ(aq) The spectator ion is Naϩ. Matching 18. a 21. 4. ST 17. or water. AT 12. 2. silver ϩ sulfur → silver sulfide Silver metal and sulfur react to produce solid silver sulfide. a 17. determine the formulas for the reactant and products and write them in their proper positions to form a skeleton equation. 2Al(s) ϩ 3F2(g) → 2AlF3(s) 3. d 15. They combine in a 1:1 ratio to form MgO. True-False 13. d 20. Mg ϩ O2 → MgO The balanced chemical equation is 2Mg(s) ϩ O2(g) → 2MgO(s) 2. B. none d. f 16. KClO3 → KCl ϩ O2 Next. H2(g) ϩ O2(g) → H2O(l) Fe(s) ϩ S(s) → FeS(s) ⌬ MgCO3(s) y MgO(s) ϩ CO2(g) H2(g) ϩ Cl2(g) y 2HCl(g) Hydrochloric acid and solid calcium carbonate react to produce carbon dioxide gas. a.1 1.3 A. Questions and Problems 20. Questions and Problems 22. 2KClO3(s) → 2KCl(s) ϩ 3O2(g) 4. Ca(s) ϩ 2HCl(aq) → H2(g) ϩ CaCl2(aq) 5. b 19. aqueous calcium chloride. NT C. b 18.B. FeCl3(aq) ϩ 3NaOH(aq) → Fe(OH)3(s) ϩ 3NaCl(aq) 7. PbCl2(s) ϩ Ϫ . AgCl(s) b. Ca(s) ϩ Mg(NO3)2(aq) → Ca(NO3)2(aq) ϩ Mg(s) Answer Key 801 © Pearson Education. 5. e D. 8. a. c D. combination reactions: 1 and 2 decomposition reaction: 3 single-replacement reaction: 4 double-replacement reaction: 6 combustion reactions: 1 and 5 8. 8. publishing as Pearson Prentice Hall. ST 16. First. c 19. a. 9. combination 23. ST Practice Problem Solutions Section 11. There are 2 silver atoms and 1 sulfur atom on each side of the equation. 4. Matching 1. 2K(s) ϩ H2SO4(aq) → K2SO4(aq) ϩ H2(g) d. 17. In b calcium replaces a less reactive magnesium and in c potassium replaces the less reactive hydrogen. 4. 10. d 10. 3. Al(OH)3(s) Quiz for Chapter 11 1. CuS(s) b.3 1. 4. a c b c a 16. 5. 4.c. i B. Pb(NO3)2(aq) ϩ 2NH4Cl(aq) y PbCl2(s) ϩ 2NH4NO3(aq) 2ϩ Pb ϩ 2ClϪ → PbCl2(s) 4. 2Na(s) ϩ Br2(l) → 2NaBr(s) Interpreting Graphics 11 nitrous oxide nitric oxide oxygen carbon dioxide water ammonia urea benzene nitrobenzene carbonic acid 1. 7. 7. no reaction In a bromine is less reactive than chlorine so no reaction occurs. 9. no reaction occurs in d. 15. 2. 4. 2KBrO3(s) → 2KBr(s) ϩ 3O2(g) c. 19. 8. publishing as Pearson Prentice Hall. d d b a c nitric acid HNO3 N2O ϩ O2 → NO 2N2O ϩ O2 → 4NO C6H6 ϩ O2 → CO2 ϩ H2O 2C6H6 ϩ 15O2 → 12CO2 ϩ 6H2O NH3 ϩ CO2 → CH4N2O ϩ H2O 2NH3 ϩ CO2 → CH4N2O ϩ H2O C6H6 ϩ HNO3 → C6H5NO2 ϩ H2O balanced as written C. 3. e g j b 5. skeleton equation formulas balanced a. This reaction can be described as: Ba(NO3)2(aq) ϩ Na2SO4(aq) → BaSO4(s) ϩ 2NaNO3(aq) The net ionic equation is: Ba2ϩ(aq) ϩ SO42Ϫ(aq) → BaSO4(s) 2. h 7. a.. b b c d b 21. combustion decomposition net ionic equation catalyst reactants 6. 18. 13. All rights reserved. 6. c 8. right b. 3. 22. AgCl(s) d. Because zinc is less reactive than sodium. a 6. N2O NO O2 CO2 H2O NH3 CH4N2O C6H6 C6H5NO2 H2CO3 Chapter 11 Test A © Pearson Education. 25. f 9. 14. no precipitate c. 12. Problems 26. 2. 3Ca(s) ϩ 2H3PO4(aq) → Ca3(PO4)2(s) ϩ 3H2(g) b. Vocabulary Review 11 1. 5. 23. 20. A. Inc. 8. This reaction can be described as: Mg(s) ϩ 2HCl(aq) → Η2(g) ϩ MgCl2(aq) The net ionic equation is: Mg(s) ϩ 2Hϩ(aq) → Η2(g) ϩ Mg2ϩ(aq) 3. spectator single replacement balanced equation coefficients Solution: precipitate Section 11. 24. 2. 2. a. Multiple Choice 11. left arrow 2HgO(s) → 2Ηg(l) ϩ O2(g) 2Ϫ 2Agϩ(aq) ϩ 2Naϩ(aq) ϩ 2NOϪ 3 (aq) ϩ CO3 (aq) → 2Νaϩ(aq) ϩ 2NO3Ϫ(aq) ϩ Ag2CO3(s) net: 2Agϩ(aq) ϩ CO32Ϫ(aq) → Ag2CO3(s) C4H8(g) ϩ 6O2(g) → 4CO2(g) ϩ 4H2O(g) 9. (NH4)2CO3(aq) ϩ 2NaOH(aq) → Na2CO3(aq) ϩ 2NH3(g) ϩ 2H2O(l) 802 Core Teaching Resources . 3. 2C3H7OH(l)ϩ9O2(g)y6CO2(g) ϩ 8H2O(g) (complete) 29. b a d b d 21. 2H2O(l ) JJJJH 2H2( g ) ϩ O2( g ) c. Fe2O3(s) ϩ 3CO(g) → 2Fe(s) ϩ 3CO2(g) Answer Key 803 . 2. Multiple Choice 11. C3H8( g ) ϩ 5O2( g ) → 3CO2( g ) ϩ 4H2O(l ) 29. the cations have exchanged positions such that two new compounds are formed. Essay 30. 2Fe(NO3)3(aq) ϩ 3Na2CO3(aq) → Fe2(CO3)3(s) ϩ 6NaNO3(aq) net: 2Fe3ϩ(aq) ϩ 3CO32Ϫ(aq) → Fe2(CO3)3(s) D. b 10. Problems 26. NaCl(aq) ϩ AgNO3(aq) → NaNO3(aq) ϩ AgCl(s) 27. B. 22. 13.. 4. Essay 30. 2HNO3(aq) ϩ Mg(OH)2(aq) → Mg(NO3)2(aq) ϩ 2H2O(l ) c. All rights reserved. 2K3PO4(aq) ϩ 3MgCl2(aq) → Mg3(PO4)2(s) ϩ 6KCl(aq) net: 3Mg2ϩ(aq) ϩ 2PO43Ϫ(aq) → Mg3(PO4)2(s) b. Inc. d a c b d C. E ϩ FG → EG ϩ F. the metal E has replaced the metal F so that a new compound and a different element are produced. Pb(NO3)2(aq) ϩ KI(aq) → PbI2(s) ϩ KNO3(aq) 2ϩ Pb (aq) ϩ 2NO3Ϫ(aq) ϩ Kϩ(aq) ϩ IϪ(aq) → PbI2(s) ϩ Kϩ(aq) ϩ NO3Ϫ(aq) Spectator ions: Kϩ(aq) and NO3Ϫ(aq) Net: Pb2ϩ(aq) ϩ 2IϪ(aq) → PbI2(s) c. d b c b a 16. single-replacement decomposition double-replacement C5H10(g) ϩ 5O2(g) → 5CO(g) ϩ 5H2O(g) (incomplete) b. Matching © Pearson Education. j 9. 18. 19. 28. Whether one metal will replace another is determined by the relative reactivity of the two metals. a. publishing as Pearson Prentice Hall. a. a. 2Al(s) ϩ 3Fe(NO3)2(aq) → 2Al(NO3)3(aq) ϩ 3Fe(s) d. 17. Chapter 11 Test B A. d 8. 20. b. 1. 25. g h i f 5. KOH(aq) ϩ HCl(aq) → KCl(aq) ϩ H2O(l ) Kϩ(aq) ϩ OHϪ(aq) ϩ Hϩ(aq) ϩ ClϪ(aq) → Κϩ(aq) ϩ ClϪ(aq) ϩ H2O(l) Spectator ions: Kϩ(aq) and ClϪ(aq) Net: Hϩ(aq) ϩ OHϪ(aq) → H2O(l ) b. a. Li2O(s) ϩ H2O(l ) → 2LiOH(aq) electricity b. c 6. 23. 24. A metal will replace any metal found below it in the activity series. a. e 28. 12. single-replacement reaction. double-replacement reaction. a. 15. c. 3. The activity series of metals lists metals in order of decreasing reactivity. a 7. CS2(s) ϩ 3O2(g) → CO2(g) ϩ 2SO2(g) b. AB ϩ CD → AD ϩ CB. 2HNO3(aq) ϩ Ca(OH)2(aq) → Ca(NO3)2(aq) ϩ 2H2O(l ) e. ZnI2(aq) ϩ NaOH(aq) → NaI(aq) ϩ Zn(OH)2(s) Zn2ϩ(aq) ϩ 2IϪ(aq) ϩ Naϩ(aq) ϩ OHϪ(aq) → Naϩ(aq) ϩ IϪ(aq) ϩ Zn(OH)2(s) Spectator ions: Naϩ(aq) and IϪ(aq) Net: Zn2ϩ(aq) ϩ 2OHϪ(aq) → Zn(OH)2(s) D. 14.27. Agϩ ϩ ClϪ → AgCl(s) f. 2Na3PO4 ϩ 3Pb(NO3)2 → 6NaNO3 ϩ Pb3(PO4)2(s) 2. and o all gave no visible reaction so it is not necessary to write an equation. Agϩ ϩ OHϪ → AgOH(s) e. Ca2ϩ ϩ CO32Ϫ → CaCO3(s) l. 3. 2Na3PO4 ϩ 3CaCl2 → 6NaCl ϩ Ca3(PO4)2(s) m. 3Ca2ϩ ϩ 2PO43Ϫ → Ca3(PO4)2(s) m. 2Agϩ ϩ CO32Ϫ → Ag2CO3(s) b. Na3PO4 ϩ 3AgNO3 → 3NaNO3 ϩ Ag3PO4(s) c. 3Pb2ϩ ϩ 2PO43Ϫ → Pb3(PO4)2(s) h. publishing as Pearson Prentice Hall.3 Precipitation Reactions: Formation of Solids. 2. Na2SO4 ϩ Pb(NO3)2 → 2NaNO3 ϩ PbSO4(s) j. 5. n. Na2CO3 ϩ 2AgNO3 → 2NaNO3 ϩ Ag2CO3(s) b. 1. Adding one drop of lead nitrate to a few grains of table salt causes white crystals to grow on the salt. Pb2ϩ ϩ 2OHϪ → Pb(OH)2(s) i. All rights reserved. KI ϩ AgNO3 → KNO3 ϩ AgI(s) Silver iodide is pale green. Inc. Sodium hydroxide reacts with calcium chloride to form sodium chloride and solid calcium hydroxide.Chapter 11 Small-Scale Lab Section 11. Place one drop of lead nitrate on a small pile of dry table salt. Agϩ ϩ IϪ → AgI(s) 2KI ϩ Pb(NO3)2 → 2KNO3 ϩ PbI2(s) Lead(II) iodide is bright yellow. 3Agϩ ϩ PO43Ϫ → Ag3PO4(s) c. Pb2ϩ ϩ SO42Ϫ → PbSO4(s) j. 4. 2NaOH ϩ CaCl2 → 2NaCl ϩ Ca(OH)2(s) 5. Na2CO3 ϩ CaCl2 → 2NaCl ϩ CaCO3(s) l. Mixings d. 2NaOH ϩ Pb(NO3)2 → 2NaNO3 ϩ Pb(OH)2(s) Section Review 12. Silver nitrate produces a similar result. NaCl ϩ AgNO3 → NaNO3 ϩ AgCl(s) f. 4. moles/molecules balanced equation mass/atoms atoms/mass moles STP (standard temperature and pressure) 804 Core Teaching Resources . Be sure to keep part of the pile dry and look carefully for signs of yellow lead iodide. Pb2ϩ ϩ CO32Ϫ → PbCO3(s) g. 6. 3. 2NaCl ϩ Pb(NO3)2 → 2NaNO3 ϩ PbCl2(s) k. Na2CO3 ϩ Pb(NO3)2 → 2NaNO3 ϩ PbCO3(s) h.. © Pearson Education. page 345 Analysis AgNO3 Pb(NO3)2 CaCl2 ( Ag؉) (Pb2؉) (Ca2؉) Na2CO3 (CO32؊) Na3PO4 (PO43؊) NaOH (OH؊) Na2SO4 (SO42؊) NaCl (Cl؊) i. Pb2ϩ ϩ 2ClϪ → PbCl2(s) k. NaOH ϩ AgNO3 → NaNO3 ϩ AgOH(s) Teacher’s note: The students will write the above reaction but this is what really happens: 2NaOH ϩ 2AgNO3 → 2NaNO3 ϩ Ag2O(s) ϩ H2O e. b. a. Pb2ϩ ϩ 2IϪ → PbI2(s) 2. 3.1 Part A Completion 1. a. Ca2ϩ ϩ 2OHϪ → Ca(OH)2(s) a white ppt tan ppt brown ppt f white ppt white ppt k white ppt white ppt b g l c h m white white ppt ppt d no i n no white visible visible ppt reaction reaction e j o no white white visible ppt ppt reaction You’re the Chemist 1. NT 9. a. AT Practice Problems 12 Section 12. 3. e Part C Matching 15.1 mol SO2 ϫ ᎏᎏ 2 mol SO2 ϭ 1. NT 13. c Part D Questions and Problems 18. 5. b .4 L ϭ 67. 4. e 16. 2. 14 mol FeCl3 ϫ ᎏᎏ 2 mol FeCl3 Section 12. AT 11.1 mol O2 in excess Part B True-False 8. ST 11.8 L 2 mol N2 ϭ 56 g 3 mol O2 ϭ 96 g mass reactants ϭ 152 g 2 mol N2O3 ϭ 152 g mass product ϭ 152 g 3 mol Cl2 ϭ 21 mol Cl2 19. maximum 6. 2. a 1. e 16. d Section Review 12. AT 12. c 16. AT 8. NT 14.4 L CO ϫ ᎏᎏ ϭ 6. theoretical yield 5.4 L ϭ 44.2 L volume N2O3 ϭ 2 ϫ 22. AT Part C Matching 13. d 18.7 mol O2 Ϫ 1.6 mol O2 needed SO2 is the limiting reagent. AT 9.0 g O2 1 mol O2 22. 3.2 © Pearson Education. NT 10.6 mol O2 ϭ 1. moles N2 ϭ 2 moles O2 ϭ 3 moles N2O3 ϭ 2 molecules N2 ϭ 2 molecules O2 ϭ 3 molecules N2O3 ϭ 2 volume N2 ϭ 2 ϫ 22.Part B True-False 7.1 19.1 ϫ 1024 molecules O2 1 mol O2 4 mol NH3 ϫ ᎏᎏᎏ ϫ ᎏᎏ 23 6. AT 8. d 15. ST 12. b 14. 2KClO3(s) y 2KCl(s) ϩ 3O2(g) 3 mol O2 12 mol KClO3 ϫ ᎏᎏ ϭ 18 mol O2 2 mol KClO3 Answer Key 805 17. ST 9.02 ϫ 10 molecules O2 7 mol O2 17. AT 10.7 L CO 1 mol CO 21. limiting reagent 2. All rights reserved. publishing as Pearson Prentice Hall. 6. AT 12. a 17. representative particles volumes coefficients mole ratios product/reactant reactant/product moles Part D Questions and Problems 1 mol O2 18. ST Part D Questions and Problems 1 mol O2 2 mol CO O2 ϫ ᎏᎏ ϫ ᎏᎏ 20.8 L volume O2 ϭ 3 ϫ 22. 10A ϩ 2C ϩ Ci y A10C2Ci 10A 25 A10C2Ci ϫ ᎏᎏ ϭ 250 apples A10C2Ci 2. NT 10.0 g NH3 ϫ ᎏᎏ ϭ 33. 3. 2 mol SO3 b.8 g 32.1 mol SO2 ϫ ᎏᎏ 2 mol SO2 mol SO3 can be formed 2. Inc.3 Part A Completion 1. product 4. used up 3.9 g NH3 1 mol NH3 Part C Matching 13. b 14. Part A Completion 1. a 17. actual yield Part B True-False 7. ST 11. c 15. 4.4 L ϭ 44.. 7. 5.5 g AgCl 2 mol AgCl 108 g Ag ϫ ᎏᎏ ϭ 63 g Ag 1 mol Ag 1 mol O2 2 mol CO 6. 4.3 g Ag ϭ 71.3 mol O2 ϭ 21 mol O2 3.0147 moles SA Student 1: 5.0 g CO2 1 mol CO2 55.0 g CO2 ϫ ᎏᎏ ϭ 66.4 g O2 2 mol KClO3 1 mol O2 5. 10 mol H2 ϫ ᎏᎏ ϭ 20 mol HCl 1 mol H2 3 mol Cl2 ϭ 21 mol Cl2 2. 180. 15.1 g/mol Student 1: 0. 2H2(s) ϩ O2(g) y 2H2O(g) 2 molecules H2O 2. 4 mol NO ϫ ᎏᎏ ϫ ᎏᎏ 2 mol NO 1 mol NO2 ϭ 184 g NO2 1 mol KClO3 4. 42. 14 mol FeCl3 ϫ ᎏᎏ 2 mol FeCl3 2 mol NO2 46 g NO2 3. 2Ag(s) ϩ Cl(g) y 2AgCl(s) 1 mol AgCl 2 mol Ag 84 g AgCl ϫ ᎏᎏ ϫ ᎏᎏ 43.25 g acetic anhydride 0.817 ϭ 58. All rights reserved.5 g AgCl 2 mol AgCl 108 g Ag ϫ ᎏᎏ ϭ 75.0 L N2O3 ϫ ᎏ 2 L N2O3 8.0 g O2 1 mol O2 ϭ 10. a.0 ϫ 1023 molecules O2 ϫ ᎏᎏ 1 molecule O2 ϭ 4. 102. 3.3 g Mg 2 mol Mg 40. publishing as Pearson Prentice Hall.0 g O2 ϫ ᎏᎏ ϫ ᎏᎏ 32.79 g Student 2: 2. 14 mol KClO3 ϫ ᎏᎏ 2 mol KClO3 4.0 g MgO © Pearson Education.2% 25.0 g KClO3 ϫ ᎏᎏ 122.65 g Section 12.0 g MgO 1 mol MgO actual yield ϭ 71.0 ϫ 1023 molecules H2O 2 mol H2O 22.0 g CO2 percent yield ϭ ᎏᎏ ϫ 100% = 83.1 g/mol 2.3 g MgO ϫ ᎏᎏ ϭ 71. 2H2(g) ϩ O2(g) y 2H2O(g) 1 mol O2 2 mol H2 160. 2HCl(g) y H2(g) ϩ C12(g) 1 mol Cl2 1 mol HCl 25..5 mol O2 ϫ ᎏᎏ ϭ 45. 138. 2H2(g) ϩ O2(g) y 2H2O(g) 2 mol H2 ϭ 8 mol H2 4 mol O2 ϫ ᎏᎏ 1 mol O2 1 mol O2 16 mol H2 ϫ ᎏᎏ ϭ 8 mol O2 2 mol H2 Oxygen is the limiting reagent.4 L CO ϫ ᎏᎏ ϭ 6.5 L H2 ϫ ᎏᎏ ϫ ᎏᎏ 22.00 mol O2 ϫ ᎏᎏ ϫ ᎏᎏ 1 mol O2 1 mol H2O ϭ 180 g H2O 3.0 g C 1 mol C 44.0 g C ϫ ᎏ ϫ ᎏᎏ 12.1 g/mol b.1 g Cl2 1 mol AgCl 2 mol Ag 5. 100.0 g O2 ϫ ᎏᎏ ϫ ᎏᎏ ϭ 29.1 g Cl2 1 mol Cl2 13. Zn(s) ϩ 2HNO3 y H2(g) ϩ Zn(NO3)2 1 mol H2 1 mol Zn 7.0 g Cl2 ϫ ᎏᎏ ϭ 25.8 g HCl ϫ ᎏᎏ ϫ ᎏᎏ 36.6 g Cl2 percent yield ϭ ᎏᎏ ϫ 100% = 54.3 g Ag 1 mol Ag mass of Ag(s) reclaimed = 0. Inc.8 g Mg ϫ ᎏᎏ ϫ ᎏᎏ 24.4 L H2 1 mol H2 65.0 g MgO ϫ 0.80 g O2 ϫ ᎏᎏ ϫ ᎏᎏ 32.2 2 mol HCl 1.2 g Ag 1 mol Mg 2 mol MgO 6. 806 Core Teaching Resources . c.25g acetic anhydride 0.5 L O2 7. 2 mol H2O ϭ 8 mol H2O formed 4 mol O2 ϫ ᎏᎏ 1 mol O2 4.0 mol H2O 1 mol O2 2.6 g KClO3 3 mol O2 32.0 g O2 1 mol O2 22.0514 moles Student 2: 5.4 g Zn ϫ ᎏᎏ ϭ 22 g Zn 1 mol Zn Interpreting Graphics 12 1.5 g HCl 2 mol HCl 71. C(s) ϩ O2(g) y CO2(g) 1 mol C 1 mol CO2 18. Section 12.3% 66.0 mol H2 needed Oxygen is the limiting reagent.0 g AgCl ϫ ᎏᎏ ϫ ᎏᎏ 143.0 g H2O 5.946 ϫ 75. 75.0514 moles salicylic acid Student 1: 2.0 g CO2 4.72 L CO 1 mol CO 3 L O2 ϭ 22. 2 mol H2O 18.3 1.0155 moles SA Student 2: 0. 0 g CO2 ϫ ᎏᎏ ϫ ᎏᎏ ϫ 44 g CO2 8 mol CO2 58 g C4H10 ᎏᎏ ϭ 4. b 8. Additional Problems 1 mol Al2(SO4)3 24.69 mol Ca(OH)2 remaining 3 mol O2 ϭ 15. Inc. coefficients reactant moles atoms 44. 4.5% Student 2: 94. the amounts of reactants and products can be calculated. ᎏᎏ ϫ 100% ϭ 96.7% 8. 5.0 g CO ϫ ᎏᎏ ϭ 4. 9.509 g 7. j 7. The number of moles may be converted to mass. b 2. Quiz for Chapter 12 1. f B.6 g KClO3 ϫ ᎏᎏ ϭ 18. 3. Student 1: 1. which is reflected by a higher percent yield than that obtained by Student 1. Essay Vocabulary Review 12 1.94 g C4H10 1 mol C4H10 1 mol O2 2 mol KClO3 O2 ϫ ᎏᎏ ϫ ᎏᎏ 26. e 4. 450 g Ϫ 325 g ϭ 125 g excess Ca(OH)2 1 mol Ca(OH)2 125 g Ca(OH)2 ϫ ᎏᎏ 74.745 g Student 2: 2.0 g H2 ϫ ᎏᎏ 2.6. 10. Student 1: 62. 9. 1 mol Fe 3 mol CO 20. a. 18.80 L 27. 14. c 3.2 g KClO3 1 mol KClO3 4. All rights reserved.8 6. e 10.0 L O2 25. a 5.20 ϫ 102 g CO 1 mol CO 1 mol CO2 2 mol C4H10 21. 17. volume.0% 5. There is no limiting reagent.00 L Chapter 12 Test B A. g 6. 5. Problems 19. From this information. b c b d 11.0 L H2S ϫ ᎏᎏ 2 mol H2S Chapter 12 Test A A. c 23. 16. Student 2 exhibited much better lab technique. a. Multiple Choice 7. 10. f 6.4 L O2 3 mol O2 122. Theoretical yield: 1 mol H2 1 mol CH3OH ϫ ᎏᎏ 4. b. 2. because the mole ratio of the reactants is 1 mol N2 to 3 mol H2. 2. 8. 15.0 g CH3OH ϭ 32.00 ϫ 102 g Al2(SO4)3 ϫ ᎏᎏ 342 g Al2(SO4)3 3 mol Ca(OH)2 74.0 g D.1 g Ca(OH)2 ϫ ᎏᎏ ϫ ᎏᎏ 1 mol Al2(SO4)3 1 mol Ca(OH)2 5. i 9. 13. 8. d ϭ 325 g Ca(OH)2 Al2(SO4)3 is the limiting reagent.0 g CH3OH ϫ ᎏᎏ 1 mol CH3OH 28.8 g Fe 2 mol Fe 28. publishing as Pearson Prentice Hall. b 2.0 g H2 2 mol H2 32. a 6. 3. 5. d h f a 5. Matching 1. 12. 1.5% 32. 7. d 4. b c b a 15.0 g b. The coefficients of a balanced chemical equation describe the relative number of moles of reactants and products. 558 g Fe ϫ ᎏᎏ ϫ ᎏᎏ 55. c a c c C. c Answer Key 807 . 4. or number of representative particles. NT NT NT NT ST E. 10. e 3.. Matching © Pearson Education.1 g Ca(OH)2 ϭ 1.00 L 22. 22.00 ϫ 102 g Al2(SO4)3 ϫ ᎏᎏ 342 g Al2(SO4)3 3 mol CaSO4 136 g CaSO4 ϫ ᎏᎏ ϫ ᎏᎏ 1 mol Al2(SO4)3 1 mol CaSO4 ϭ 596 g CaSO4 1 mol Al2(SO4)3 5. Student 2 should receive the higher grade. Percent yield: ᎏᎏ ϫ 100% ϭ 87. 0 g NH3 2 mol NH3 17.0 mol H2 ϭ 8. ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ 3 mol O2 22. ᎏᎏ 2 mol C2H2 1 mol C2H2 44.50 mol H2O 6 mol H2O 180 g C6H12O6 ϫ ᎏᎏ ϭ 135 g C6H12O6 1 mol C6H12O6 4 mol CO2 ϫ 5.1 g N2 4 mol NO 1000 g 32. 6CO2 ϩ 6H2O uuy C6H12O6 ϩ 6O2 1 mol C6H12O6 ᎏᎏ ϫ 4..32 ϫ 104 kg H3PO4 1000 g 3 mol CuO 1 mol NH3 31. 808 Core Teaching Resources .1 g H2SO4 1 kg 1 kg ϫ ᎏ ϭ 8.1 g CaCO3 1 mol CaO 56. 14.1 kg NO energy C.1 g CaO ϫ ᎏᎏ ϫ ᎏᎏ 1 mol CaCO3 1 mol CaO ϭ 28. ᎏᎏ 1 mol CO2 1 mol CH4 ϭ 2. ᎏᎏ ϫ 3. ᎏᎏ 3 mol H2SO4 1 mol H2SO4 1000 g 98.0 g C2H2 1 mol CO2 ϭ 1. Multiple Choice 7.00 mol NH3 22. ᎏᎏ ϫ ᎏᎏ ϫ 57. Additional Problems 28.0 kg NH3 ϫ ᎏ 4 mol NH3 1 kg 1 mol NH3 30.0 g CH4 ϫ 150 mol CO2 ϫ ᎏᎏ 24. publishing as Pearson Prentice Hall. 50. CuO is the limiting reagent.69 ϫ 105 g CO2 2 mol H3PO4 ϫ 1. ᎏᎏ ϫ 3. ᎏᎏ ϫ 10. 20.65 mol CuO present Thus. Problems 2 mol NH3 ϫ 12.6% yield © Pearson Education.0 g H3PO4 ϫ ᎏᎏ ϫ ᎏ ϫ ᎏᎏ 1 mol H3PO4 98.5 mol of B in order to react completely. 18.25 ϫ 105 kg H2SO4 30. b c b a c 17. All rights reserved. Essay 27.00 ϫ 104 g C2H2 29.0 g NH3 ϭ 3. 13.65 mol CuO ϫ ᎏᎏ 3 mol CuO 1 mol N2 ϭ 34.7 g SnF2 ϫ ᎏᎏ ϭ 176 g SnF2 1 mol SnF2 1 mol CH4 16.0 g HF 156.03 mol CuO needed 1 mol CuO 290. 10. a. Inc. Based on the 2:3 molar ratio between A and B. ᎏᎏ 3 mol H2 1 mol SnF2 1 mol HF HF 23.5 g CuO ϭ 3.0 g NH3 ϭ 5.4 g CaO % yield ϭ ᎏᎏ ϫ 100% 28. c d a b c E.65 mol CuO 3 mol CuO ϭ 2.0 mol of A requires only 1. The maximum amount of A2B3 that can be produced (0. 2 mol NH3 b. ᎏᎏ ϫ ᎏᎏ ϫ 45.0 g NH3 ϫ ᎏᎏ 17. 8. D.0 g NH3 1 mol NO 1000 g ϫ 0.50 mol) is thus limited by the amount of A that is available.92 mol 1 mol N2 28. 11.B.35 mol NH3 present NH3 excess ϭ 3.5 g CaO actual % yield ϭ ᎏᎏ ϫ 100% theoretical 26.0 g CuO ϫ ᎏᎏ 79.0 g CO2 ϫ ᎏᎏ ϫ ᎏᎏ 26. the 1.8 g CaCO3 ϫ ᎏᎏ 100. with 0.4 ϫ 103 g CH4 2 mol Al2O3 1 mol O2 1L 25.5 g CaO % yield ϭ 92.50 mol of B remaining in excess.43 mol ϭ 0.90 g Al2O3 1 mol CaCO3 26.0 g N2 c.4 L O2 1000 mL 102 g Al2O3 ϫ 625 mL O2 ϫ ᎏᎏ 1 mol Al2O3 ϭ 1.35 mol Ϫ 2. 16. 19. d d b d d 12. 15.80 ϭ 14. 21.0 g 2 mol HF 20. 9.43 mol NH3 react 1 mol NH3 Since 57.0 g NO 1 kg ϫ ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ 17. 83 g ϭ 0.46 g NaHCO3 7. 3. 2.875 mmol HCl unreacted) 5.Chapter 12 Small-Scale Lab Section 12. d 21. Because the particles are relatively far apart. a Part D Questions and Problems 101.500 mmol NaOH/g NaOH ϭ 0. NT 14. 10. 5. a 18. e 21.2 Analysis of Baking Soda. AT Part C Matching 17. AT Part C Matching 17. surface vapor pressure manometer vapor pressure 101. have students add just enough HCl from a third pipet to turn the mixture red. c 19. 4. 10.78 g Section Review 13.2 Part A Completion 1.3 kPa or 1 atm Part B True-False 11. 7.29) g HCl ϫ 1.) 2.45 g ϭ 2. Inc. If the mixture does not turn red when thymol blue is added. AT 13.70 Ϫ 4. 10.30 atm 1 at m 23. ST 16. F.70 g D.2 % error (assuming baking soda is 100% NaHCO3). page 367 Procedure Sample answers are given. (0. c 20. 7. They move independently of each other and travel in straight line paths until they collide with one another or other objects.30 atm ϫ ᎏᎏ ϭ 436 kPa 1a tm 760 mm Hg ϫ ᎏᎏ ϭ 3. no attractive or repulsive forces exist between the particles.1 Part A Completion 1. denser condensed vaporization boiling cooling 6. The % error is the % of baking soda in baking powder assuming no other errors. Section 13. 8.53 mmol HCl neutralized (5.3 kPa 22. 10.29 g E. motion empty space far apart independently random or rapid 6.875 mmol HCl unreacted ϭ 5.53 g G. 100% ϫ (0. d 20.27 ϫ 103 mm Hg 4.3 kPa or 1 atm You’re the Chemist 1. Repeat Steps A–G and 1–7 except use baking powder instead of baking soda.28 g C.46 g Ϫ 0. 9. publishing as Pearson Prentice Hall. 4.53 mmol NaHCO3) 6.78) g NaOH ϫ 0. 2.875 mmol NaOH (0.53 mmol NaHCO3 ϭ 0. Odors travel long distances from their sources. NT 14.83 g B. All rights reserved. the motion of the particles in a gas is constant and random. ST 13. 2. According to the kinetic theory. 8.41 mmol HCl total Ϫ 0. ST 12. HCl ϩ NaHCO3(s) → CO2(g) ϩ H2O ϩ NaCl 2.00 mmol HCl/g HCl ϭ 6. AT 15. 24. 8. Part B True-False 11. ST 12. 4. b Answer Key 809 . © Pearson Education. (10. collisions kinetic energy atmospheric 0°C 101. e 19.53 Ϫ 8. 3.45 g)/0. 3. A.0840 g NaHCO3/mmol NaHCO3) ϫ 5. 5. b 18. NT 16. 9. (10.. (See Steps 2–7.41 mmol HCl 4. 6. Students must measure the mass lost by the pipet to add to the total mass of HCl used. Analysis 1. 3.28 g Ϫ 2.45 g baking soda 3. 4. AT 15. 2. ST 15. This process is called sublimation.0 ЊC ϩ 273 ϭ 173 K 73 oC + 273 ϭ 346 K Because the Kelvin temperature increases by a factor of two. d © Pearson Education. At the boiling point. publishing as Pearson Prentice Hall. NT 14. 23. a 20. All rights reserved. c 17. 1 atm 3. g 22.3 mm Hg ϫ ᎏᎏ ϭ 0. 6. Solids that have a vapor pressure that exceeds atmospheric pressure at or near room temperature can change directly to a vapor. NT 12. compress fixed melts melting point freezing point 6. equilibrium triple point 0. Evaporation leads to cooling of a liquid because the particles with the highest kinetic energy tend to escape first.2 1.9925 atm 760 mm Hg 101. 2. f 19. 8. b 16. 2. AT 13. The temperature of the system remains constant while the change of state is occurring. f Part D Questions and Problems 21. 754.3 Part A Completion 1. Part B True-False 11. When a solid is heated. Section 13. 7. ST 13. Section 13.Part D Questions and Problems 22. 7. 10. Eventually. The average kinetic energy of the particles of a substance is directly proportional to the Kelvin Temperature. 4. AT Part C Matching 16. NT Part C Matching 15. its particles vibrate more rapidly as their kinetic energy increases. e 17. Gas pressure is the result of collisions between between rapidly moving particles in a gas and an object. the disruptive vibrations of the particles are strong enough to overcome the attractions that hold them in fixed positions. 4.3 kPa 754. The remaining particles have a lower average kinetic energy and a lower temperature. For a dynamic equilibrium to be established. 5.3 mm Hg ϫ ᎏᎏ ϭ 100. Liquid B would evaporate faster because it has a higher vapor pressure. Part D Questions and Problems 24. 2. but there are attractions between particles of a liquid. 24. Because there are no particles of matter in a vacuum. c 18. there can be no collisions or pressure. which indicates that it is more volatile.1 1. particles throughout the liquid have enough kinetic energy to vaporize. you would notice that the pressure reading on the barometer would decrease as you climbed in altitude.4 Part A Completion 1. Ϫ100.5 kPa 760 mm Hg 4. Inc. sublimation vapor pressure carbon dioxide phase 5. The organization of the particles within the solid breaks down and the solid becomes a liquid. 3. Setting aside fluctuations due to changes in the weather. 9. the average kinetic energy increases by a factor of two. high crystalline lattice unit cell amorphous Practice Problems 13 Section 13. a 20.61 kPa Section 13. d 18. 3. According to kinetic theory. NT 10. AT 11. NT 14. 22..016°C 0. 810 Core Teaching Resources . e 19. b 21. there are no attractions between the particles in a gas. 8. AT 12. Part B True-False 9. the beaker must be sealed so that the rate of condensation can equal the rate of evaporation. graphite. as pressure is increased. Inc. The melting-point curve leans slightly to the right (has a positive slope) indicating that. The fastest runner corresponds to the particles in a liquid with the greatest kinetic energy. 8. publishing as Pearson Prentice Hall. evaporation 4.4 1. The carbon atoms in graphite are arranged in widely-spaced sheets. 4.. boiling point Solution: barometer 5. ionic solids have higher melting points because the forces that hold particles together in an ionic solid are usually stronger than the forces that hold particles together in a molecular solid. gas pressure 3. Allotropes are two or more different molecular forms of the same element in the same physical state. 4. 20 © Pearson Education. 101. 8. condenses 9. 2. or molecules are arranged in an orderly. 2. 5. Ethanol must have the greater vapor pressure because 75°C is very close to the boiling point of ethanol and the vapor pressure is equal to the external pressure at a liquid’s boiling point. the remaining particles have a lower average kinetic energy. and vapor phases of a substance are in equilibrium. 4. 7. 4. All rights reserved. Solid bromine is more dense than liquid bromine. In general. kinetic energy 2. Normal melting point ϭ Ϫ7.3 kPa (1 atm) 2. The carbon atoms are arranged differently in each allotrope. repeating. The triple point is the temperature and pressure at which solid. The melting point of water decreases as the pressure increases. Interpreting Graphics 13 1. 100 Normal Melting point Normal Boiling point B Solid 40 Triple point S C 0 Ϫ30 Ϫ10 0 20 40 Temperature (°C) 60 Liquid V Vapor 80 A M 60 Section 13. One example of a crystalline solid is sodium chloride.3 1. Section 13.0°C Normal boiling point ϭ 59°C Triple point ϭ Ϫ8°C and 6 kPa 3. In crystal atoms. including diamond. threedimensional pattern called a crystal lattice. each carbon atom is strongly bonded to four other carbon atoms in a rigid three-dimensional array. Higher pressures favor the denser phase of a substance. When these particles vaporize. This line represents the set of all temperature–pressure values at which the liquid and gas phases of water are in equilibrium. 50°C 6. and buckminsterfullerene. Peanut brittle is an amorphous solid. 7. See answer to 1. liquid. 5. This line represents the set of all temperature–pressure values at which the solid and gas phases of water are in equilibrium. A molecular solid.Pressure (kPa) 3. 3. freezes Vocabulary Review 13 1. See answer to 1. Carbon has multiple allotropes. 6. 3. the melting point of bromine increases. ions. crystal allotropes melting point pascal Answer Key 811 . In diamond. 16.3 kPa. a a c d b © Pearson Education. 8. 14. True-False 31. 8. 23. 27. 610. 7. 28. 17. 4. 12. The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to atmospheric pressure. All rights reserved. 16. 25. 12. 3. 10. 26. 14. then the boiling point of a liquid will be lower than its normal boiling point. AT 32. Essay 35. C. Multiple Choice 15. Matching 1.3 kPa 1 atm G. ST 32. Problems 101. If atmospheric pressure is less than 101. 9.3 kPa. When water boils at standard atmospheric pressure. i f g j c 6. if atmospheric pressure is E. kinetic decreases temperature kinetic elastic condensed minimum kinetic energy external (atmospheric) pressure unit cell amorphous ST AT AT F. 2.70 atm 1 atm D. 0. a a a b b 26. 5. 23. 22. 39. j n h e d 6. AT 31. 30. 812 Core Teaching Resources . 18. NT 30. The normal boiling point is the boiling point of the liquid when the atmospheric pressure is 101. 29. 22. Conversely. Only water at pressures of more than one atmosphere will boil at higher temperatures. 9. AT 35.803 atm ϫ ᎏᎏ ϭ 81.81 ϫ 103 mm Hg 34. 24. d b n e m 11. 2. 6.0 mm Hg ϫ ᎏᎏ ϭ 0. 5. a d d c b 20. k b g f l 11. it doesn’t matter how many particles there are in the sample. 25. AT 29. 10. The temperature of the water will not rise until all of the water is in the gaseous state. a m c i B. 11.803 atm 760 m m Hg 101. Matching 1. b b c c B.3 kPa 37. 2. ST 33. it cannot be heated above 100°C. NT 36. Multiple Choice 15.3 kPa 33. 27. 3. The additional energy is being used to change the liquid water to water vapor. publishing as Pearson Prentice Hall.. AT D. 20. ST 37. Additional Problems 1 atm 36. 8. 13. 4. 9. k h l a Chapter 13 Test B A. 4. 13. True-False 28. 19. 12. The pressure must be increased to the point at which water boils at 150°C or higher to kill bacteria. 10. c a b a 24. AT 34. 3. Chapter 13 Test A A. Additional Questions 38. 18. 7.70 atm ϫ ᎏᎏ ϭ 375 kPa 1a tm 760 mm Hg ϫ ᎏᎏ ϭ 2.Quiz for Chapter 13 1. 3. 17. AT C. 7. Essay 38. 3. Inc. d b d d c b 21. 5. 19. Because the Kelvin temperature is directly proportional to the average kinetic energy of the particles in a substance. 21. 13. ST 14. The water drop increases in diameter over time as the alcohol evaporates and is captured by the water drop. The BTB turns from green to yellow in the presence of vinegar. The BTB slowly changes from green to yellow even though there is no mixing of the drops. In solids. The particles travel in straight paths until they collide with other particles or the walls of their container. 2. Water beads up and alcohol spreads out due to stronger intermolecular attractions in water. 8. e 18. The BTB slowly changes from green to blue even though there is no mixing of the drops. You’re the Chemist 1. the particles are closely packed together. The volume of the particles in a gas is small compared the overall volume of a gas. In a liquid. which is why liquids have a definite volume. but the temperature of the mixture remains the same as long as liquid water is present. E. 9. page 400 Analyze and Conclude 1. The volume of the particles is insignificant and their collisions are perfectly elastic. AT 11. Inc. ST 12. If the mixture is heated. a 19. 7. 2.greater than 101. b Part D Questions and Problems 20. The particles of a gas are relatively far apart and there are no attractive or repulsive forces among them. As a result. 4. Calcium chloride absorbs water from the environment in the dish. 4. c 16. 2. 3. more particles of the liquid acquire enough kinetic energy to escape. The motion of particles in a gas is constant and random. © Pearson Education. The many pieces of calcium chloride effectively dry the atmosphere leaving no water vapor in the dish. more ice melts but the temperature of the mixture remains the same as long as ice is present. The particles in a solid vibrate around fixed points. 3. The water “cloud” consists of tiny individual beads. The drop of water on top of the dish provides enough cooling to cause cloud formation. more spread out pools. Additional Questions 39. Ethanoic acid that evaporates is “captured” by the BTB. a. Section 14. the alcohol cloud is made up of larger. kinetic doubles reducing pressure Chapter 13 Small-Scale Lab Section 13. which is why a gas can expand to take the shape and volume of its container.3 The Behavior of Liquids and Solids. Because the particles are closer together. Part B True-False 10. publishing as Pearson Prentice Hall.1 Part A Completion 1. b.3 kPa. Place a drop of vinegar and a drop of BTB about 3 cm apart in a Petri dish. Ammonia thatt evaporates is captured by the BTB. Water in the dish evaporates and condenses into a cloud when it contacts the cold surface under the ice. All rights reserved. NT 13. Cover and observe. There are no significant attractive or repulsive forces between particles in a gas. more liquid freezes. As more energy is added. Place a drop of ammonia and a drop of BTB about 3 cm apart in a Petri dish. 5. 4. 40. usually in an organized array. liquids are denser than gases. 5. The attractions in the resulting mixture are weaker overall. which is why a gas can be compressed. Answer Key 813 . Cover and observe. the boiling point of the liquid will be higher than its normal boiling point.. d 17. If the mixture is cooled. the particles are attracted to each other. Solids are dense and difficult to compress. AT Part C Matching 15. compressed spare volume temperature moles 6. all the liquid boils away at a constant temperature. 3. 10.5 ϫ 102 kPa P1 ϫ V1 ϫ T2 23. 9. inversely increases Boyle’s mass Kelvin 6. a 21.3 Part A Completion number of moles PV ϭ nRT n ideal gas constant (L • kPa) 5. AT 13. 3. NT 12.325 kPa ϫ 12. 9. 10. 8. 8. All rights reserved. the less massive molecule must have a higher velocity. total sum lower uniform diffusion effusion hole inversely molar mass Graham’s law © Pearson Education. 4. n ϭ ᎏᎏ ϭ ᎏᎏᎏ ϫ L kPa RϫT 8. b 17. publishing as Pearson Prentice Hall. NT 16. Part D Questions and Problems P1 ϫ T2 55 kPa ϫ 473 K 22. 5. AT 11.64 ϫ 10Ϫ2 mol NO2 mass of NO2 ϫ mol K Part B True-False 11. n ϭ ᎏᎏ ϭ ᎏᎏᎏ ϫ L kPa RϫT ᎏ 8. AT Part C Matching 15.5 L 19. ideal real attractions volume Part B True-False 11. 9. c 17. 8. 8. 3. At a given temperature. 4.2 Part A Completion 1.0 g NO2 ϭ 2.21 g NO2 Part C Matching 17. b 18.64 ϫ 10Ϫ2 mol NO2 ϫ ᎏᎏ 1 mol NO2 ϭ 1.29 ϫ 102 mol O2(g) PϫV 240. 7. 7. d Section 14. ST 12. d 16. 2.31 ᎏ ϫ mol ϫ 295 K K n ϭ 1. Inc.275 L 20. AT Part C Matching 15. NT 14. Charles’s Gay-Lussac’s directly combined amount Part D Questions and Problems PϫV 25. a 16. molecules all have the same average kinetic energy. 7. a 814 Core Teaching Resources . P2 ϭ ᎏ ᎏ ϭ ᎏᎏ 173 K T1 P2 ϭ 1. 3.31 ᎏᎏ ϫ 301 K n ϭ 2. 2.Section 14. ST 14. 6.. AT 12. c 18. V2 ϭ ᎏ ᎏ T1 ϫ P2 91 kPa ϫ 0.0 kPa ϫ 0. Part B True-False 10.4 Part A Completion 1.075 L ϫ 273 K V2 ϭ ᎏᎏᎏ 303 K ϫ 101. c Part D Questions and Problems 18. AT 46.061 L ϭ 61 mL Section 14. ST 14.3 kPa V2 ϭ 0. If two molecules with different masses have the same kinetic energy. 6. The kinetic energy of a molecule is equal to 1 ᎏ 2 where m isthe mass and v is the ᎏ ᎏ mv2 velocity of the molecule. NT 13. NT 15. NT 13.31 ϫ ᎏ ᎏ (K • mol) 1. e 20. 4. 2. 5. b 19. 5 kPa ϫ 3. resulting in a greater pressure inside the tire. 0. 372 K 4.0 kPa) PO2 ϭ 26. P2 ϭ ᎏ ᎏ ϭ ᎏᎏᎏ 10.3 kPa Section 14.31 ᎏ ϫ K mol V ϭ ᎏᎏᎏᎏ 101. No.0 L ϫ 323 K P2 ϭ ᎏᎏᎏᎏ ϭ 71. publishing as Pearson Prentice Hall.31 ᎏ ϫ mol ϫ 297 K K n ϭ 1.43 g b.75 ϫ 10 mol O2(g) ᎏᎏ RateO2 molar massO2 ϭ 2. they collide less frequently and less forcefully with the walls of the mattress.03 mol He 4. Overnight the air in the mattress cools down. The motion of the tires causes the air in the tires to heat up.00 mol ϫ 8.4 kPa ϫ 55.0 g ᎏᎏ 32. n ϭ 25.0 g ϭ 0. 103 kPa b.0 L P1 ϫ V1 1. the average kinetic energy of the particles in the air decreases. T ϭ 3.31 ᎏᎏ Kϫ mol n ϭ 3. T ϭ ᎏᎏ ϭ ᎏᎏᎏ L ϫ kPa nϫR 7.2 L ϫ 373 K P2 ϭ ᎏᎏᎏ ϭ 466 kPa 298 K ϫ 7. the balloon will only expand until internal pressure is equal to the external pressure.0 g ϫ 1 mol\44.31 ᎏ ϫ K mol V ϭ ᎏᎏᎏᎏ 101. n ϭ ᎏᎏ ϭ ᎏᎏᎏ ϫ L kPa RϫT ᎏ 8.31 ᎏ ϫ K mol V ϭ ᎏᎏᎏᎏ 101.01 ϫ 10 K 2 Interpreting Graphics 14 1.0 kPa ϩ 50. PϫV 102.4 1. 108 kPa Answer Key 815 .0 kPa ϫ 35.568 mol ϫ 8.3 kPa V ϭ 7. a. At higher temperatures.3 kPa Vϭ 50. a.0 L 3.5 kPa Ϫ (22. temperatures are higher in the summer than in the winter. V ϭ ᎏᎏ P Lϫ kPa ᎏ ϫ 273 K 2. V1 ϫ T2 10. V2 ϭ ᎏ ᎏ ϭ ᎏᎏ ϭ 15. P1 ϭ ᎏᎏ T2 ϫ V1 105.12 g He nϫRϫT 4.31 ᎏ ϫ mol ϫ 297 K K n ϭ 1.0 kPa ϫ 22. PO2 ϭ Ptotal Ϫ (PN2 ϩ PAr) PO2 ϭ 98. So.31 ᎏᎏ ϫ 298 K ϫ mol K Section 14. ᎏᎏ ϭ ᎏᎏ molar massN2 RateN2 Ϫ2 ͙ ͙ ᎏᎏ 28.0 L PϫV 2.95 L 7.5 L ϫ 300 K V1 ϫ T2 5.1 1. n ϭ ᎏᎏ ϭ ᎏᎏᎏ L ϫ kP a RϫT ᎏ 8.Practice Problems Section 14. which in this case is about half the initial pressure. the particles inside the tire have a greater average kinetic energy.875 ϭ 0.0 kPa ϫ 25.0 L V2 P2 ϭ 341 kPa 2. 0. All rights reserved.4 L 10. Section 14.00 L P2 ϫ V2 ϫ T1 5.00 g He mass of He ϭ 1.3 1. P2 ϭ ᎏ T1 ϫ V2 501 kPa ϫ 5.5 L n ϭ ᎏᎏ ϭ ᎏᎏᎏ L ϫ kP a RϫT 8. a.03 mol He ϫ ᎏᎏ 1 mol He ϭ 4.24 ϫ 104 kPa ϫ 25 L PϫV 1.3 ϫ 102 mol argon 500. the frequency and force of the collisions between the particles and the walls of the tire are greater. Thus. 0.2 155. V ϭ ᎏᎏ P Lϫ kPa ᎏ ϫ 273 K 0. the pressure inside the mattress decreases.0 g ᎏᎏ ϭ ͙0. V2 ϭ ᎏ ᎏ ϭ ᎏᎏ ϭ 6.5 kPa PϫV 26. 372 K b.66 L 473 K T1 nϫRϫT 6.0 L T1 248 K P1 ϫ V1 ϫ T2 ᎏ 4. as does the volume.355 mol ϫ 8.0 L ϫ 373 K 3.267 L 3.41 g 2. On average.25 mol ϫ 8.7 L © Pearson Education. Inc.9 kPa 248 K ϫ 105 L V ϭ 12. Consequently.568 mol nϫRϫT V ϭ ᎏᎏ P Lϫ kPa ᎏ ϫ 273 K 0..935 Oxygen effuses slightly slower than nitrogen. 2. e 8. All rights reserved. 5. a. n ϭ ᎏᎏᎏ ϫ L kPa ᎏ 8.6 ϫ 105 m3 8. e 7.3 kPa T2 24. Multiple Choice 9.31 ᎏ ϫ mol ϫ 372 K K n ϭ 8.0°C ϩ 273 ϭ 303 K 303 K V2 ϭ 550 mL ϫ ᎏᎏ ϭ 764 mL 218 K Vocabulary Review 14 1. V2 ϭ V1ϫ ᎏᎏ T1 T1 ϭ Ϫ55.90 ϫ 10Ϫ3 mol molar mass ϭ 46 g/mol 0. c 8. PO2 ϭ Ptotal Ϫ (PN2 ϩ 2PCO2) PO2 ϭ 145.5 kPa ϩ 76.33 ϫ 10Ϫ3 mol 0.0°C ϩ 273 ϭ 273 K P1 ϫ V1 ϫ T2 V2 ϭ ᎏ ᎏ T1 ϫ P2 95. d b a c d 14. publishing as Pearson Prentice Hall. b 4. a 7.9 kPa ϫ 180 mL ϫ 273 K V2 ϭ ᎏᎏᎏᎏ ϭ 151 mL 308 K ϫ 101.33 ϫ 10Ϫ3 mol molar mass ϭ 46 g/mol 7.5 kPa D.267 L b. g i j f 5. f 6. d 20. 11. 2. An ideal gas is one that follows the gas laws at all conditions of pressure and temperature. 18. g 5. 17.5 ϫ 105 m3 ϫ ᎏᎏ ϫ ᎏᎏ 99 kPa 300 K V2 ϭ 3. T1 273 K 6. Problems P1 P2 21.103 kPa ϫ 0. T1 ϭ 35. P2ϭ P1 ϫ ᎏᎏ ϭ 388 kPa ϫ ᎏᎏ ϭ 149 kPa T2 713 K P1 ϭ 115 kPa 7. b C. Inc. a 9. Matching 1.0 kPa Ϫ (28. 4.90 ϫ 10 Ϫ3 Chapter 14 Test A A. a. d 6.9 L 23. P1 ϫ V1 ϭ P2 ϫ V2 P1 ϫ V1 P2 ϭ ᎏᎏ V2 156 kPa ϫ 15. 2.0 kPa) PO2 ϭ 40.43 g b. T1 ϭ 27°C ϩ 273 ϭ 300 K T2 ϭ Ϫ10°C ϩ 273 ϭ 263 K P2 ϭ 99 kPa P1 T2 V2 ϭ V1 ϫ ᎏᎏ ϫ ᎏᎏ P2 T1 115 kPa 263 K V2 ϭ 3. 4.267 L 5.31 ᎏ ϫ mol ϫ 372 K K B. b Quiz for Chapter 14 1. especially at low 816 Core Teaching Resources . b c b c c 19. h 3. d mol n ϭ 9. molar mass ϭ ᎏᎏ 9. ᎏᎏ ϭ ᎏᎏ T1 T2 T2 ϫ P1 P2 ϭ ᎏᎏ T1 T1 ϭ 227°C ϩ 273 ϭ 500 K T2 ϭ 27°C ϩ 273 ϭ 300 K 300 K ϫ 655 kPa P2 ϭ ᎏᎏ 500 K ϭ 393 kPa 22. The behavior of a real gas deviates from the behavior of an ideal gas.0°C ϩ 273 ϭ 308 K T2 ϭ 0. 46 g/mol 8. h 10.0 L V2 ϭ ᎏᎏ 215 kPa ϭ 10. c 2. molar mass ϭ ᎏᎏ 8. 15. n ϭ ᎏᎏᎏ ϫ L kPa ᎏ 8. 16. Essay 25.0°C ϩ 273 ϭ 218 K T2 ϭ 30. 13.. C2H6O (ethanol) 108 kPa ϫ 0. 3. 3.41 g 6. 10. collisions doubles small real diffusion © Pearson Education. 12. If all gases behaved ideally.3 kPa) V2 ϭ 2.31 ᎏ K ϫ mol ϫ 300 K n ϭ 2. n ϭ ᎏᎏ ϭ ᎏᎏᎏ ϫ L kPa RϫT ᎏ 8.49 mol CO2 ϫ ᎏᎏ ϭ 22 g CO2 1 mol CO2 D. 12. c 8. Essay 17.50 L)(273 K ) V2 ϭ ᎏᎏᎏ (330 K)(101. This is not true for real gases.0 ᎏᎏ 31. All rights reserved.925 L )(250 K) T2 ϭ ᎏᎏ (1.0 g CO2 0. Additional Problems PϫV 216 kPa ϫ 25. 24. 14. Matching © Pearson Education.20 L) V2 ϭ ᎏᎏ 615 kPa V2 ϭ 0. Additional Problems V1 V2 37. e 7. 20. AT 36. P1V1 ϭ P2V2 P1 ϫ V1 V2 ϭ ᎏᎏ P2 (425 kPa)(1. AT 35. True-False 26. the individual particles that make up each gas could never exert the attractive forces on each other that are necessary for them to condense to liquids and solids. d 6.694 L P1 P2 29.0 kPa)(310 K) T2 ϭ ᎏᎏ (85.49 mol 44. Also. ᎏᎏ T1 T2 V1 ϫ T2 V2 ϭ ᎏᎏ T1 (0. a c a b b d 23. 21. kinetic theory assumes that the particles of an ideal gas have no volume and are not attracted to each other.0 L 30.004 times faster. 13. 1. d c b d d b E. True-False 33. NT C. AT 28. 16.31 ᎏᎏ ϫ 300 K ϫ mol K E. ST 34. a 9.29 L 31.004 Rate238 349. V1 V2 ϭ ᎏᎏ 28. 18.4°C P1V1 P2V2 31.temperatures and high pressures.17 mol 96.25 L ) T2 ϭ 185 K or Ϫ88°C Answer Key 817 .009 ϭ 1.. ᎏᎏ ϭ ᎏᎏ T1 T2 V2 ϫ T1 T2 ϭ ᎏᎏ V1 (0.2 g/mol formula mass 2. 22. 25.0 g ᎏᎏ ϭ 44. B. 26. PV ϭ nRT PϫV 152 kPa ϫ 8.650 L)(313 K ) V2 ϭ ᎏᎏ (293 K ) V2 ϭ 0. 4.17 mol ᎏᎏ ᎏᎏ Rate235 252. publishing as Pearson Prentice Hall. ᎏᎏ ϭ ᎏᎏ T1 T2 P2 ϫ T1 T2 ϭ ᎏᎏ P1 (98.0 ͙ ͙ UF6 containing U-235 diffuses 1.829 L F. Problems 27.0 L n ϭ ᎏᎏ ϭ ᎏᎏᎏ L ϫ kP a RϫT 8.0 kPa)(3. 15. Chapter 14 Test B A. g i j f 5. c b a b 32. Multiple Choice 11.0 kPa) T2 ϭ 357 K or 84. ᎏ ᎏϭᎏ ᎏ T1 T2 P1 ϫ V1 ϫ T2 V2 ϭ ᎏᎏ T1 ϫ V2 (80. ᎏᎏ ϭ ᎏᎏ ϭ ͙1. AT F. which can be liquefied and sometimes solidified by cooling and applying pressure. 2. Inc. 3. 19. NT 27. ST 29. h 10. b n ϭ 0. AT 13. f 20. The first picture should show one edge turning yellow.9% 322 g 24. 2.31 ᎏ ϫ mol ϫ 200 K K n ϭ 0. AT Part C Matching 15.1 Part A Completion 1.2 Part A Completion 1. Vary the size of the BTB drops from “pinheads” to “puddles. publishing as Pearson Prentice Hall. 3. The color change begins at the outer edge of each drop. c 21. electrolytes 9. all the drops change color. As the gas diffuses. 8. molecules/ions ions/molecules Emulsions stability emulsions Part B True-False 14. 9. NT 15. solvent 6. 6. b and c © Pearson Education. b 19.0 g 180. 2. lower b. strong 10. Section Review 15.38. 2. b Part D Question 22. As ammonia diffuses. 5. BTB changes from yellow to blue. 8. 2. 10. NaHSO3 + HCl → SO2 + H2O + NaCl Section Review 15. AT 12. molar mass Na2SO4 ؒ 10H2O ϭ 322 g Mass of 10H2O ϭ 180. 11. Inc. 3. Succeeding pictures should show the yellow area gradually increasing until the entire dot is yellow. 4.6 kPa ϫ 10. 3.0 g ᎏᎏ ϫ 100% ϭ 55. The drops near the center change immediately. 3.3 Part A Completion 1. e You’re the Chemist 1. 3NaNO2 + 2HCl → 2NO + H2O + NaNO3 + 2NaCl Part D Questions and Problems 23. PV ϭ nRT PϫV 50. All rights reserved. AT 18. 4.304 mol Part C Matching 19. larger filtration Colloids Tyndall effect Brownian 6. 2. NT 16. a 17. partially conduct nonelectrolyte hydrates efflorescence Part B True-False 11. “like dissolves like” 8. higher Chapter 14 Small-Scale Lab Section 14. ST 14. 9. g 21. As the particles diffuse from the center. The Kl turned orange in the same manner as the BTB turned yellow. homogeneous 7. 5.0 L n ϭ ᎏᎏ ϭ ᎏᎏᎏ ϫ L kPa RϫT ᎏ 8. d 16. c 22. polar negative positive polar hydrogen low 7. 13. The particles of gas produced are in motion.4 Diffusion. 7. 5.” Tiny drops are better able to detect small quantities of gas. 4. a 20. 12. page 437 Analyze and Conclude 1. AT 17.. 10. 4. 3. a. ST 818 Core Teaching Resources . they collide and react with molecules of BTB. high surface spherical surfactant Ice dense hydrogen bonding Section Review 15. h 18. The solute is potassium chloride (KCl). surfactant 7. soluble 6. desiccants 2.Part B True-False 11. a 20. Hydrogen bonds hold the water molecules in place in the solid phase. soluble b. NT Part C Matching 16. CaSO4 ؒ 2H2O b. open. H O H O H O © Pearson Education. 3.3 1. solvent 7. All rights reserved.44% 237. colloid c. KOH(s) → Kϩ(aq) ϩ OHϪ(aq) 7.2 1. c 8. Colloids and suspensions exhibit the Tyndall effect and have larger particles than solutions. dispersed colloidal particles. hydrogen bonding 3. 2. solvent 6. CoCl2 ؒ 6H2O Quiz for Chapter 15 1. The solvent is water. H 2. 4. solution Part D Questions and Problems 22.7 g/mol Mass of 6H2O ϭ 108. framework like a honeycomb. NH4NO3(s) → NH4ϩ(aq) ϩ NO3Ϫ(aq) b. a. colloid b. Inc. Brownian motion refers to the chaotic movement of colloidal particles caused by the collisions of water molecules with the small. water of hydration 9. suspension 4. e 17. b Answer Key 819 .0 g ϭ ᎏᎏ ϫ 100% ϭ 45. aqueous solution 5. insoluble d. NH3(g) ϩ H2O(l) 1 NH4ϩ(aq) ϩ OHϪ(aq) 3. a Practice Problems 15 Section 15. NT 12. The particles in a suspension are retained on a filter and will settle out slowly upon standing. colloid g.1 1. publishing as Pearson Prentice Hall. Hygroscopic compounds are those compounds that remove moisture from air. 5. a. d 19. aqueous solutions 8. solute 6.7 g/mol Section 15. b 18. 2. Hydrogen bonds are attractive forces in which a hydrogen atom that is covalently bonded to a very electronegative atom is also weakly bonded to an unshared electron pair of an electronegative atom in the same molecule or in a nearby molecule. like dissolves like 3. suspension f. b 23. strong electrolyte 8. The structure of ice is a regular. a. H H Vocabulary Review 15 1. soluble c. ST 13. Molar mass of NiCl2 ؒ 6H2O ϭ 237. a 5. AT 14. 3. colloid e. Brownian motion SOLUTION: 1. c 4. a. LIQUID WATER Section 15. Possible answers include glucose (C6H12O6) and ethyl alcohol (C2H6O).. ST 15. f 21. solution d. ICE 3. a 2.0 g mass of water Percent H2O ϭ ᎏᎏ ϫ 100% mass of hydrate 108. WATER VAPOR 2. c 20. c 18. AT 27. colloids 7. a 12. a 17. solute 4. NT D. Because polar water molecules can attract charged particles. d 15. With surface tension reduced. ST 25. c 19. allowing the water to spread out to cover and penetrate the fabric. Soaps and detergents also are emulsifying agents that allow oils and greases to form colloidal dispersions. ST 26.. b 22. Soaps and detergents are surfactants that reduce the surface tension of water by interfering with the hydrogen bonding between water molecules. page 458 Analysis NaCl(s) Aqueous conducts Na2CO3(s) Aqueous conducts MgSO4(s) Aqueous conducts Sugar Aqueous does not conduct Corn Starch Aqueous does not conduct Kl(s) Aqueous conducts Chapter 15 Test B A. nonelectrolytes 10. True-False 23. Essay 31. c 15. d 21. nonelectrolyte 2. All rights reserved.0 g ϭ ᎏᎏ ϫ 100% ϭ 55. Completion 1. the ions are surrounded by molecules of solvent Chapter 15 Small-Scale Lab Section 15. d 13. ST 24. E.4% H2O 238 g © Pearson Education. NT 28. hygroscopic 2. publishing as Pearson Prentice Hall. Problems 27.9% H2O 322. Brownian motion 7. %H2O ϭ ᎏᎏ ϫ 100% mass of hydrate 180. hydrogen 9. Multiple Choice 11. As the solute dissolves. they cause solute ions to break away from the surface of the solid. a 13. AT 25.Chapter 15 Test A A. Multiple Choice 11. True-False 22. Essay 28. solvent 5. The oil and grease particles. b 14. a 18. Completion 1. Inc. d 17. the beads of water that would normally have formed collapse. surfactants 5. aqueous 8. C. c 19. Surface tension B. c 20. AT 26. ST 24. deliquescent 3. a 16. which are normally insoluble in water. dessicants 6. solvation 3. a C. c 16. are removed from the surface of the fabric. effloresce 6. suspension 4. c E. a 14.2 Electrolytes. AT D.0 g 6 mol H2O ϭ 108 g 108 g ᎏᎏ ϫ 100% ϭ 45. Emulsions 8. effloresce NaHCO3(s) Aqueous conducts KCl(s) Aqueous conducts 820 Core Teaching Resources . a 21.1 g B. AT 29. hydrate 9. c 12. Molar Mass of hydrate ϭ 238 g 1 mol H2O ϭ 18. AT 23. Tyndall effect 10. Problem mass of H2O 30. corn starch. NH3 Nonelectrolytes (no light): rubbing alcohol. a 18. NT 10. These are electrolytes: NaCl.3 M ϫ V ϭ mol solute needed 0. Table sugar and corn starch are covalent compounds. increases 10. KCl. MgSO4.6 g/L ϫ 2. dilute 4. particle size 2. solvent 2. ST 13. pressure 4. moles 5. These are nonelectrolytes: sugar. pickle juice Weak electrolytes: orange juice. ᎏᎏ ϭ ᎏᎏ 1. Inc. b 14. KI are ionic compounds. 4. None of the electrolytes conduct electricity in the solid form because the ions are locked in a crystal lattice and cannot move.5 atm S2 ϭ ᎏᎏᎏ ϭ 4. publishing as Pearson Prentice Hall.3 g ᎏᎏ ϫ 0. 9..15 mol ϭ 51 g 1 mol © Pearson Education. c 17. d Part D Problem 18. 2. c 15. In general. liter 6. MgSO4. Part D Problem 1. 5. Na2CO3.1 Part A Completion 1. HNO3. solvent You’re the Chemist 1. temperature 6. g 21. NaOH Weak electrolytes (dim light) CH3COOH. NT Part C Matching 13.5 atm 1. AT Part C Matching 16. solute colligative properties freezing lowering/depression elevation directly solution particles twice twice Part B True-False 11. solute 8.0 g/L 1.50 mol ᎏᎏ ϫ 0. molar mass C12H22O11 ϭ 342. Test a drop of each solution with a conductivity device. 6. H2SO4. NT 14. d 22.0 atm 2.0 atm Section Review 16. 2. All rights reserved. NaHCO3. distilled water 3. miscible 8.1. 3. e 16.6 g/L S2 23. solution 3. saturated 7. 10. KI. ST 13. NT 12. coffee ϩ Ϫ Part B True-False 9. a 17. diluting 7. 7. e 20. MgSO4(s) → Mg2ϩ(aq) ϩ SO42Ϫ(aq) NaHCO3(s) → Naϩ(aq) ϩ HCO3Ϫ(aq) KCl(s) → K (aq) ϩ Cl (aq) KI(s) → Kϩ(aq) ϩ IϪ(aq) 2. Henry’s 5. 8. Strong electrolytes: soft drinks. b Part B True/False 11.15 mol 1L 342. to be an electrolyte a compound must dissociate into ions in solution.2 Part A Completion 1. AT 11. supersaturated Section Review 16. NaHCO3. KCl. rate 3. AT 14. Section Review 16. Strong electrolytes (bright light): HCl. NT 15. NaCl. 3. f 19.3 Part A Completion 1. solubility 9.3000 L ϭ 0. AT Answer Key 821 . Na2CO3. NT 12. AT 12. 00 M ϫ 1. d 16.5 mol KOH 56. a. V2 ϭ ᎏ ϭ ᎏᎏ ϭ 12 L 0. ᎏᎏ ϫ 500.0 M 6.0 g Add 233.6 L 1000 mL 20.5m ϭ 7.00M ϫ V2 0.54 g/L ϫ 1.0 L 4. Practice Problems 16 Section 16.1 g KOH ϫ ᎏᎏ ϭ 4. a 17. e 15.0 g/100 g H2O 455. moles KOH ϭ 2500.50 mol molarity ϭ ᎏᎏ ϭ 0. 60.2 g KNO3 1 mol KNO3 b.0 g Ϫ 222.0 mL C3H7OH b. a 17.0% methanol (v/v) 25. c 18.22 atm P1 S2 ϭ 0. 3. M1V1 ϭ M2V2 0.0 g H2O ϭ 1. Solubility of AgNO3 at 20ЊC ϭ 222. d 16. 5.799M 1 L 2. Inc.0 g/100 g H2O Solubility of AgNO3 at 50ЊC ϭ 455.1 0.4 Part A Completion 1.5m 1. solute solvent kilogram mole fraction molal boiling point depression molal elevation Part B True-False 9.5 mol KOH 1 L mass of KOH ϭ 7.50 mol 85..0 mol KOH ϫ ᎏᎏ ϭ 7. four d.0 mol HNO3 b. % (v/v) ϭ ᎏᎏᎏ 500.0 mol CaCl2 1000 g 18.5 g ϫ ᎏᎏ ϭ 2.0 mL ϫ ᎏᎏ 1000 mL 3. S2 ϭ ᎏ ᎏ ϭ ᎏᎏᎏ 1. molality ϭ ᎏᎏ ϫ ᎏᎏ 800.Part C Matching 15.0 L ϫ ᎏᎏ L moles HNO3 ϭ 4.2 ϫ 102 g KOH 1 mol KOH 2. publishing as Pearson Prentice Hall.0 L M1 ϫ V1 5.86ЊC ⌬Tf ϭ Kf ϫ m ϭ ᎏ ᎏ ϫ 7. two one K2CO3 NaCl c. ᎏᎏ ϭ 0. a.0 mL ϭ 2. b Part D Questions and Problems 19.5% C3H7OH (v/v) © Pearson Education. moles HNO3 ϭ 2.0 g KCl 2. NT 11.82 g/L 34.5 m ϭ 14ЊC m freezing point of solution ϭ 0ЊC Ϫ 14ЊC freezing point of solution ϭ Ϫ14ЊC 1 mol 3.0 mL methanol 6.0 g 2. 20.75 M ϫ 300. 4.0 mL 101.5m molality of total particles ϭ 3 ϫ 2. 6.500 M M2 The final volume should be 12 L. c Part D Problem 2. two c.0 mL V2 ϭ ᎏᎏᎏ ϭ 113 mL 2.40 mol NaCl 1.0 g to maintain saturation at 50ЊC. 8.5 ϫ 102 g HNO3 Section Review 16. AT 12.0 mL solution % (v/v) ϭ 12.0 mol HNO3 63 g HNO3 ϫ ᎏᎏ 1 mol HNO3 mass of HNO3 ϭ 2. All rights reserved. b 14. moles NaNO3 ϭ 212.0 g H2O 1 kg molality ϭ 2. a. b.2 0.0 mol HNO3 mass of HNO3 ϭ 4. % (v/v) ϭ ᎏᎏᎏ ϫ 100% 200. add 11 L of H2O. NaCl 3. a. 7. Section 16.0 mL solution ϫ 100% % (v/v) ϭ 12.86 atm S1 ϫ P2 1. Therefore.0 g ϭ 233.83M 3. 2. AT Part C Matching 13.1 g KNO3 1L ϭ 0. ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ 250.7 ϫ 102 g KCl 100 g H2O 822 Core Teaching Resources .25M 1.750M ϫ 300. b. ST 10. a. 00 L ϫ ᎏ ϫ ᎏ ϭ 1 L 100 mL 2 ϭ 1. a. Section 16.0271 . a. 6 mol of particles are produced because each formula unit of MgCl2 dissociates into three ions.4 g NaCl ϫ ᎏᎏ ϭ 22 g NaCl 1 mol NaCl 2.103 mL 3. moles C12H22O11 ϭ 126.00 mϫ2 ⌬Tb ϭ 2.1472m ⌬Tf ϭ 0. The solution containing calcium chloride has a lower freezing point.00 L ϫ ᎏ ϫ ᎏ ϭ 30. 1.50 mol NaCl 2. 6. 3.0 g H2O b.78m NaCl ⌬Tf ϭ Kf ϫ m ⌬Tf ϭ 1. and freezing-point depression. b.9 g Li2S ϫ ᎏᎏ ϭ 1. moles Ba(NO3)2 ϭ 131 g Ba(NO3)2 1 mol Ba(NO3)2 ϫ ᎏᎏ 261. Inc. a. a.0 mol ϩ 6.0 g H2O ϫ ᎏᎏ 1000 g H2O 45.0 g H2O ϫ ᎏᎏ 1000 g H2O 58.501 mol Ba(NO3)2 ᎏᎏᎏ ϫᎏ 1 kg 750. produce 9 mol of particles because each formula unit of Na2SO4 dissociates into three ions.0 g ϭ 0. when dissolved in water.0 mol nLiBr ϩ nH2O XLiBr ϭ 0.0 mol Li2S 3.05ЊC The boiling point of this solution is 100 ЊC ϩ 2. Each formula unit of K2CO3 produces three particles in solution. 0.50 mϫ3 ⌬Tb ϭ 2.6m glucose 1 kg 500.30ЊC The boiling point of this solution is 100ЊC ϩ 2.512 ЊC for every mole of particles that the solute forms when dissolved in 1000 g of water. boiling-point elevation. The boiling point of the solution is 100 ЊC ϩ 3. the boiling point of the solution increases by 6 ϫ 0. All rights reserved. 750.07ЊC.05ЊC ϭ 102. Thus. molality of solute particles 103 g 0.274 ЊC Answer Key 823 © Pearson Education.512ЊC/m ϫ 1. XLiBr ϭ ᎏ ᎏ ϭ ᎏᎏᎏ 3.45ЊC.512ЊC/m ϫ 2. Three important colligative properties are vapor-pressure lowering.07ЊC.0 g H2O ϭ 0.3 mol glucose 4.86 ЊC/ m ϫ 0.3 g Ba(NO3)2 moles Ba(NO3)2 ϭ 0. c.1 g 1 mol ᎏᎏ 101.05 ЊC. 2.45ЊC The freezing point of this solution is 0 ЊC Ϫ 1. ⌬Tb ϭ Kb ϫ m ⌬Tb ϭ 0. b.3681 mol C12H22O11 molality of solute particles 103 g 0. 2. Three moles of Na2SO4. The solution containing sodium chloride has a lower boiling point. ⌬Tb ϭ Kb ϫ m ⌬Tb ϭ 0. The solution containing calcium chloride has a lower vapor pressure.45ЊC ϭ Ϫ1.0 mol nLiBr 1.00 g 7. Colligative properties of solutions are the physical properties of solutions that depend on the concentration of solute particles in solution but not on the chemical identity of the solute.07ЊC ϭ 103.0 g ϫ 1 mol ᎏᎏ 101.5 ϫ 102 g Li2S 1 mol Li2S 103 g 2.8ЊC/ m ϫ 0.0 g H2O ϭ 0. XKNO3 ϭ ᎏᎏ nKNO3 ϩ nH2O 125 g ϫ ϭ ᎏᎏᎏᎏᎏ 125 g ϫ ϩ 800.0 g H2O ϭ 0.501 mol Ba(NO3)2 103 g 0.78 m ⌬Tf ϭ 1. 6.4 3. Vapor-pressure lowering is a colligative property.30ЊC ϭ 102. b.0 g C12H22O11 1 mol C12H22O11 ϫ ᎏᎏᎏ 342.00 g b. a.0 g NaCl 1 L 100 mL 3 10 mL 5.33 nKNO3 b. The boiling point of water increases by 0. 4.35 mol NaCl ϭ 2 ϫ ᎏᎏ ϫ ᎏ 1 kg 900. ᎏᎏ ϫ ᎏ ϭ 4.00 ϫ 10 g KNO3 Section 16. 1600.1472m C12H22O11 ⌬Tf ϭ Kf ϫ m ⌬Tf ϭ 1. publishing as Pearson Prentice Hall. 5.668m Ba(NO3)2 5. When 2 mol of MgCl2 dissolve in water.512ЊC ϭ 3.3861 mol C12H22O11 ϭ ᎏᎏᎏ ϫᎏ 1 kg 2500.30ЊC.. a.1 g 1 mol ᎏᎏ 18.3 1.3 g C12H22O11 moles C12H22O11 ϭ 0. 6. NT Temperature °C D. a 4. V2 ϭ 250 mL. Molarity. Quiz for Chapter 16 1.6 g Ϫ 34. Boiling-point elevation.0M M1V1 ϭ M2V2 M 2 ϫ V2 V1 ϭ ᎏᎏ M1 0. 24.60M. 25. c d a c 15. Colligative properties. c 2. 8.1 g 95. 4. All others are factors affecting the rate at which a substance dissolves.0 g H2O) Ϫ (230.945 mol 101. Concentration (g/100 g H2O) 350 300 250 200 150 100 Na2SO4 50 KCl 0 10 20 30 40 50 60 70 80 90 100 Chapter 16 Test A A. NT 27.60M ϫ 250 mL V1 ϭ ᎏᎏ ϭ 75 mL 2.9 g/L ϫ 606 kPa S1 ϫ P2 S2 ϭ ᎏᎏ ϭ ᎏᎏᎏ 505 kPa P1 ϭ 20. AT 30. d LiBr C.6 g/100 g H2O ϭ 42. d 7.1 g /mol 0. LiBr b.750 L S1 S2 34. 17. 5.274ЊC. 18. True-False 22.0 g/100 g H2O Solubility KCl at 50ЊC ϭ 42.The freezing point of this solution is 0ЊC Ϫ 0. 824 Core Teaching Resources . M2 ϭ 0. Solubility of sucrose at 100ЊC ϭ 487 g/100 g H2O Solubility of sucrose at 20ЊC ϭ 230. e B.1 ϭ 39. b 6. 16. 12. c 8.0 M Add 75 mL of 2. c 7. c b a b 19. All rights reserved. 3. Na2SO4 Vocabulary Review 16 1. a.9 g/100 g H2O (487 g/100 g H2O ϫ 1000.26M 0. All others are associated with the solubility of gases. d 3. b.945 mol molarity ϭ ᎏᎏ ϭ 1. Molarity. Mole fraction. Henry’s Law. All others are related to freezingpoint depression 7. Solubility KCl at 20ЊC ϭ 34.1 g N: 1 ϫ 14. a 5. 23.3 g/L © Pearson Education.0 ϭ 48. f 6.0 g O: 3 ϫ 16. b Interpreting Graphics 16 1. 33. NT 28. ST 31.0M Al2(SO4)3 to enough distilled water to make 250 mL of solution. b i g c 5. M1 ϭ 2. a 20. The other terms are used to describe mixtures of liquids.9 g/100 g H2O ϫ 1000. Saturated solution.0 ϭ 14. Molar mass KNO3: K: 1 ϫ 39. 2. Concentration. 14.6 g 3. b 2. All others are associated with the preparation of solutions. 4.274ЊC ϭ Ϫ0. 3. b 21. ᎏᎏ ϭ ᎏᎏ P1 P2 16. Multiple Choice 11.0 g ϭ 8. ST AT AT AT 26. Matching 1.0 g 101. Problems 32. h 10.. All others are associated with colligative properites of solutions. NT 29.0 g H2O) ϭ 4870 g Ϫ 2309 g ϭ 2561 g 4. publishing as Pearson Prentice Hall. All others are qualitative terms used to describe solutions.5 g mol KNO3 ϭ ᎏᎏ ϭ 0. 2. 13. Inc. All others are units of concentration. a 9. j 8. 37.91 atm 50. NT 36.31m ЊC C.0 ϭ 14. ᎏᎏ ϫ 1000 ᎏᎏ 225 mL L 1 mol Mg(NO3)2 ᎏᎏᎏ ϭ 1. 3.45 g/L P2 ϭ 1.31m ⌬Tb ϭ Kb ϫ m ᎏ ⌬Tb ϭ 0.0 g O: 3 ϫ 16. Vapor-pressure lowering: The formation of solvent shells around the solute particles reduces the number of solvent particles that have sufficient kinetic energy to vaporize.0 g 101. Multiple Choice 11.750 atm P2 ϭ ᎏᎏᎏ 2. 26.E. 19. j 10.1 g N: 1 ϫ 14. Additional Problems mol K3PO4 36. i 9.31 g AgNO3 1 mol AgNO3 © Pearson Education.512ᎏ m ϫ 1.0 g Mg(NO3)2 mL 39.0 ϭ 48. 14.1 g KNO3 ϭ 1. NT 33. 16. 12. NT 34. d a b b c a 17.73 mol KNO3 mol solute molality ϭ ᎏᎏ 1000 g solvent 103 g H2O 1. AT D. g c h e 5. Chapter 16 Test B A. 4. 13. Matching 1. 24. Boiling-point elevation.671°C ϭ 100.671ЊC The boiling point of the solution is 100°C ϩ 0. They depend solely on the number of particles in the solution.327m ϭ 1. 18. All rights reserved. publishing as Pearson Prentice Hall. Freezing-point depression: The solute particles interfere with the formation of the orderly pattern that the solvent particles assume as the solvent changes from liquid to solid.73 mol KNO3 ϭ ᎏᎏ ϫ ᎏᎏ 1 kg H 1250 g H2O 2O ϭ 1. a 7. 20.250 mol AgNO3 1k g 40. a c a b a F.38m Answer Key 825 . 29. 23.50M ϫ 148. 28. Molar mass KNO3: K: 1 ϫ 39. ⌬Tb ϭ 0. True-False 27.671ЊC. Boiling-point elevation: Additional attractive forces exist between solute and solvent that must be overcome for the solution to boil. and freezing-point and vapor-pressure lowering are colligative properties. 30. AT 35.0 g H2O ϫ (725 g AgNO3) ϭ 336 g H2O 37. AT NT NT NT 31. ST 32. Inc. f 8. 15.1 ϭ 39. 25. Problems 100.1 g 1 mol 175 g KNO3 ϫ ᎏᎏ 101.327m K3PO4 → 3Kϩ ϩ PO43Ϫ ϭ 4 particles 4 particles ϫ 0. ᎏᎏ 216 g AgNO3 S1 S2 38. ᎏᎏ ϫ ᎏᎏ ϫ 125 g 1 kg 1000 g 169.900 mol K3PO4 1000 g H2O ϭ ᎏᎏ ϫ ᎏᎏ 2750 g H2O 1 kg H2O ϭ 0..3 g Mg(NO3)2 0.25 g/L ϫ 0. 21. d 6. b B. ᎏᎏ 1000 g H2O 0.9 g AgNO3 ϫ ᎏᎏ ϭ 5. Essay 35. 2. ᎏᎏ ϭ ᎏᎏ P1 P2 S2 ϫ P1 P2 ϭ ᎏᎏ S1 6. a b d b d 22. F.90 g mass of water ϭ 46.90ϩ46. and prevents additional ice from forming down to temperatures below0 ЊC.39 ЊC freezing point ϭ 0ЊC Ϫ 6.4M 0.958m m ϫ 0.09 g mass of NaCl solution ϭ 6.0 g 1 mol CH3OH 1000 g ⌬Tf ϭ ϫ ᎏᎏ ϫᎏ 32. % by mass of NaCl 6.21 g c.897 g 826 Core Teaching Resources .86 ЊC/ m ϫ ᎏᎏ 250.490ЊC © Pearson Education.21 g mass of NaCl ϭ 0.98 g flask ϩ NaCl solution ϭ 22.88 g ϭ 46.55m 0.0 g 261 g Ba(NO3)2 1000 g 3 moles particles ϫ ᎏᎏ ϫ ᎏ ᎏ 1 kg 1 mol Ba(NO3)2 ϭ 0.490ЊC boiling point ϭ 100.0% NaCl (6.98 g mass of flask ϩ NaCl ϭ 22.0439. Density ϭ ᎏᎏ ϭ 1. ⌬Tf ϭ Kf ϫ m 27. mass of the solute (NaCl) ϭ 22.21 g b.490 ЊC boiling point ϭ 100ЊC ϩ 0. mole fraction ϭ ᎏᎏᎏ (0.5 g/mol ϭ 0. Inc.050 L 0.21 g 5.90 g ϩ 46. mass of dry flask ϭ 15. Chapter 16 Small-Scale Lab Section 16.5 g CH3OH ⌬Tf ϭ 1.88 g mass of flask ϩ NaCl ϩ water ϭ 69.512 ЊC/ ⌬Tb ϭ 0.118 mol NaCl 3. a.09 g 1. Additional Problems 0.39ЊC 44. This causes any ice that was initially present to melt. a.1 g/mL 50 mL Notice that because the flask measures less accurately than the balance.1 g H2SO4 ϫ ᎏᎏ ϭ 11 g H2SO4 1 mol H2SO4 43.0 g CH3OH 1 kg ⌬Tf ϭ 6.118ϩ2. Sample data: dry flask ϭ 15.0 g/mol 0.90 g b. mass percent and mole fraction.90 g ϭ ᎏᎏ ϫ 100 ϭ 13. 0. a. 1. mass of the solvent (water) ϭ 69. Salt is often used on bridges and sidewalks because it dissolves in an ice/ice water mixture to produce a solution with a lower freezing point than that of water alone. ᎏᎏ ϫ ᎏᎏ ϫ 750 mL 1L 1000 mL 98.90 g 2. Essay 41.15 mol H2SO4 1L 42.09 g Ϫ22.04621 kg H2O 1L 4.050 L 6. liters of solution ϭ 50 mL ϫ ᎏᎏ 1000 mL liters of solution ϭ 0.88 g flask ϩ NaCl solution ϩ water ϭ 69. publishing as Pearson Prentice Hall. All rights reserved.958m ⌬Tb ϭ Kb ϫ m ϭ 0. moles of NaCl solute ϭ ᎏᎏ 58.4 Making a Solution.88 g Ϫ15.57 mol 18.118 mol 46. molarity and density have fewer significant figures than molality. page 497 Analysis Sample answers are given. moles of water ϭ ᎏᎏ ϭ 2.118 mol c.. The salt causes the freezing point of water to be depressed because it interferes with the crystallization process.57) mol ϭ 0. Molarity ϭ ᎏᎏ ϭ 2. You’re the Chemist Sample answers are given. Molality ϭ ᎏᎏ ϭ 2. molality of total particles 62.21) g 6.5 g Ba(NO3)2 1 mol Ba(NO3)2 ϭ ᎏᎏ ϫ ᎏᎏ 750.118 mol NaCl b.98 g ϭ 6.E.39ЊC ϭ Ϫ6. 3.0 g 1 mol moles of water ϭ 4.16 kJ/mol molar heat of vaporization Condensation Answer Key 827 . 4.13 g ϫ ᎏᎏ 342 g ϭ 2. 2.050 L 4. a 15.53 g mass of the sugar ϭ 20.242M 0. NT 10. Section 17.72 g mass of the sugar ϭ 4.13 ϩ 48.333m density ϭ 1.0121 mol NaCl Molarity ϭ ᎏᎏ ϭ 0.460 J/(g • °C) 18.82% moles of sugar (C12H22O11) ϭ 1 mol 48. 8. 5. 6.21 g moles of water ϭ 2.0121 mol sugar Molality ϭ ᎏᎏ ϭ 0.68 g 4. 2. mass Part B True-False 7. AT 9.68 g density ϭ ᎏᎏ ϭ 1. 6. 4.85 g mass of flask ϩ sugar ϩ water ϭ 69.2 kJ 19.3 Part A Completion 1. e 16.249m 0. 34. All rights reserved. b 16.57 mol mole fraction ϭ 5. publishing as Pearson Prentice Hall.1 Part A Completion 1. d 18. ST 8. a Part D Questions and Problems 19. calorimeter 2.0121ϩ2.0 g ϫ 15.13 g ϩ 48. 0. d Part D Questions and Problems 18.68) g % by mass of sugar ϭ 7. AT 12. AT mol Part C Matching 14.04868 kg H2O 0.0121 mol 18.percent mass of NaCl ϭ 1.1 g/mL 50 mL Ϫ2 Part B True-False 9.1 g/mL 2. NT 11. enthalpy 3.72 g mass of flask ϩ sugar ϭ 20. molar heat of fusion molar heat of solidification equal 3.0 g CH4 1 mol CH4 ϭ 1. AT 11. C ϭ ᎏᎏ ϭ 0. AT Part C Matching 13. 3.53 g Ϫ 20. AT 10.2 J 20.69% moles NaCl ϭ 1. 124.94 ϫ 103 kJ Section Review 17.13 g % by mass of sugar ϭ ᎏᎏ ϫ 100 (4.0 ЊC The unknown metal is iron.0121 mol Mole fraction ϭ ᎏᎏᎏ (0. Chemical potential energy is energy stored within the structural units of chemical substances.68 g ϫ ᎏᎏϭ 0. final or initial 6.04 kJ 1 mol CH4 890. Heat is energy that is transfered because of a temperature difference. heat potential energy thermochemistry calorie joule specific heat or specific heat capacity metals water Section 17.00446 0. e 14.54 ϫ 10 mass of water ϭ 46. b 17. AT 13.96 ϫ 10Ϫ3 molality ϭ 0. mass of dry flask ϭ 16. 5. Work is done when a force moves an object. Inc.13 g mass of the solvent (water) ϭ 69. c 17. c 15.2 Part A Completion 1.70) mol Mole fraction ϭ 0. ⌬H 4..85 g Ϫ16. NT 12.8 g CH4 ϫ ᎏᎏ ϫ ᎏᎏ 16. initial or final 5. 2NO ϩ O2 → 2NO2 ⌬H ϭ Ϫ113.85 g ϭ 48.70 mol © Pearson Education. 7. 2 kJ 1 mol Ca(OH)2(s) 1 mol CH4(g) 3.0 Cal ϫ ᎏᎏ ϫ ᎏᎏ ϫ ᎏᎏ 1C al 1c al 103 J ϭ 836. Inc. 4NH3 ϩ 5O2(g) y 4NO(g) ϩ 6H2O(g) Ϫ226 kJ ⌬H ϭ ᎏᎏ ϫ 4 mol NH3(g) 1 mol NH3(g) ϭ Ϫ904 kJ © Pearson Education. e 15. sum enthalpy indirectly changed (reversed) standard heat of formation change one ⌬H0 f zero subtracting Section 17.0 cal 2.0 J 3. AT 14.0 g Ca(OH)2(s) ϫ ᎏᎏᎏ 74. d Part D Questions and Problems 18.18 ᎏᎏ ϫ 10Њ C ϫ Њ g C ϭ 6.1 ϫ 10 J Section 17.4 g CH4(g) ϫ ᎏᎏ 16. NT Part C Matching 13. 10.8 kJ 525. exothermic d. 4. NT Ϫ393.7 kJ ϫ ᎏᎏᎏ ϭ 142 kJ 1 mol NH4NO3(s) Practice Problems Section 17. 5.3 g H2O(s) ϫ ᎏᎏ 18. 2. endothermic b.0 g ϫ 2. ⌬H ϭ 150.0 g H2O(s) 6. ⌬H ϭ 15.4 Part A Completion 1.0ЊC 1255. NT 13. All rights reserved.01 kJ ϫ ᎏᎏ ϭ 9. NT Part C Matching 16. b 17. ⌬H ϭ 28.93 ϫ 103 kJ 1 mol CH4(g) 4. d 20.53 mol NH4NO3(s) 25. 6.3 ϫ 103 J ϭ 6. 8.0 g ϫ 4.1 kJ) 1 mol CO(g) ϭ Ϫ1154 kJ 0 ⌬H0 ϭ ⌬H0 f (products) Ϫ ⌬Hf (reactants) ϭ Ϫ1181 kJ Ϫ (Ϫ1154 kJ) ϭ Ϫ27 kJ Part B True-False 8.5 kJ ϭ 3 mol CO2(g) ϫ ᎏᎏ 1 mol CO2(g) ϭ Ϫ1181 kJ ⌬H0 f (reactants) Ϫ110. ⌬H ϭ 52. 828 Core Teaching Resources . c 18.0 g ϫ 120.2 kJ ϫ ᎏᎏ ϭ Ϫ13. publishing as Pearson Prentice Hall. c 17. molar heat of condensation 22.0 g CH4(g) Ϫ890.5 kJ ϭ 3 mol CO(g) ϫ ᎏᎏ ϩ (Ϫ822. 3. Part B True-False 11. a 14. NT 10.45 kJ 1 mol H2O(s) 20.1 J/(g ϫ ЊC) J 4. ⌬H0 f (products) 12. ST 11.1 g Ca(OH)2(s) Ϫ65. NT 12. 7..2 kJ ϫ ᎏᎏ ϭ 2. AT 9. e Part D Questions and Problems ᎏ CuO(s) y Cu(s) ϩ ᎏ1 2 O2 ⌬H ϭ ϩ155 kJ 1 ᎏ ᎏ H2(g) ϩ 2 O2(g) y ϩH2O(g) ⌬H ϭ Ϫ242 kJ CuO(s) ϩ H2(g) y Cu(s) ϩ H2O(g) ⌬Hrxn ϭ Ϫ87 kJ 21. exothermic 1 mol H2O(s) 19. 200. NT 15. q ϭ 100.184 J 1 kJ 1. exothermic f.2 J 1.0ЊC 100.7.40 cal/(g ϫ ЊC) 25. a 19.0 g ϫ 15. C ϭ ᎏ ᎏ ϭ 1. endothermic c.90 ᎏᎏ ϫ Њ g C 3 ϭ 1. b 16.0ЊC ϫ 0. ⌬H ϭ 5.1 1000 c al 4. endothermic e. ⌬T ϭ ᎏᎏᎏ ϭ 6. 9.3 kJ 1 mol Cu(OH)2(s) 2. a. 0) ϭ 8.5 °C 4.512 kJ ϩ 429. c Section 17. 2.6 kJ 2H2(g) ϩ N2(g) y N2H4(l) ⌬H ϭ 50.8 kJ 2. heat gained by H2O ϭ heat lost by metal a.5 kJ) ⌬H0 f (reactants) ϭ Ϫ1028. a. 7.4 1.452)(5.33)(Ϫ73.0 g ⌬H ϭ 429.8 kJ ϭ 2 mol SO2(g) ϫ ᎏᎏ 1 mol SO2(g) ΄ ΅ ΅ Ϫ241. see Table 17.2 kJ ᎏ 2[H2(g) ϩ ᎏ1 2 O2(g) y H2O(l)] ⌬H ϭ Ϫ571.2 ϫ 102 J b.2 kJ 3 ⌬H0 f ϭ Ϫ1077 kJ Ϫ (Ϫ40.7 kJ 1 mol ⌬H ϭ 190. a. a h f g 5.512 kJ Step 2: H2O(l) at 100 ЊC y H2O(g) at 100 ЊC 40.184)(39. Vocabulary Review 17 1.5 °C b.Section 17.0 °C 3. ⌬H0 f (products) Ϫ296. 3. specific heat ϭ Ϫ9. j 10.7 kJ 2.18 ᎏᎏ ϫ 28 ЊC ϫ Њ g C 4 ⌬H ϭ 1.100)(5. ⌬H ϭ (4.1 ϫ 102 J Ϫ8.512 ϫ 103 J ϭ 6.1 kJ ϭ 2 mol H2S(g) ϫ ᎏᎏ 1 mol H2S(g) ϭ Ϫ40. a.2 ϫ 102 6.0 °C 2. a. ⌬H0 f (reactants) ϭ Ϫ635. b. N2(g) ϩ 2H2O(l) y N2H4(l) ϩ O2(g) ⌬H ϭ 622. © Pearson Education.8 ϫ 103 J ϫ Њ g C Step 2: H2O(s) at 0 ЊC y H2O(l) at 0 ЊC 6. Ϫ73.18 ᎏᎏ ϫ 82 Њ C ϫ Њ g C ⌬H ϭ 6.6 kJ Interpreting Graphics 17 1.4 kJ The reaction is exothermic. ⌬H ϭ 35.1 ᎏᎏ ϫ 24 ЊC ϭ 4.1 ϫ 10 J ϭ 11 kJ ⌬Htotal ϭ 4.22 J/(g ϫ °C) b. 9.04 ϫ 10 kJ Answer Key 829 .35)(Ϫ73.0 g H2O(s) 6.5) ϭ 9.6 kJ ϭ 436.184)(39. publishing as Pearson Prentice Hall.0 g H2O(s) ϫ ᎏᎏ 18.0 kJ Ϫ (Ϫ1028.0 g ϫ ᎏᎏ ϫ ᎏᎏ 1 mol 18.5 °C b.1.1 kJ Ϫ445. a is reversed zero minus ϭ Ϫ593. e 7.543 mol NaOH(s) ϫ ᎏᎏ 1 mol NaOH(s) ⌬H ϭ 1.1 kJ 3. Step 1: H2O(s) at Ϫ24 ЊC y H2O(s) at 0 ЊC J ⌬H ϭ 96 g ϫ 2. 8. d 9.1 kJ ϩ (Ϫ393. 4.2 kJ) ϭ 1. Ϫ8.0) specific heat ϭ 0. specific heat ϭᎏᎏ (50. ⌬H ϭ (4.01 kJ 1m ol ⌬H ϭ 96 g ϫ ᎏᎏ ϫ ᎏᎏ ϭ 32 kJ 1m ol 18.0 g ϫ 4. 5 °C b.6 kJ ϩ (Ϫ483. 3..3 1 mol H2O(s) 1. 26. 2.5) specific heat ϭ 0.6 kJ) ⌬H0 ϭ Ϫ178.1 ϫ 102 J 5. i 8.0 g Step 3:H2O(l) at 0 ЊC y H2O(l) at 28 ЊC J ⌬H ϭ 96 g ϫ 4. Ϫ9.25 J/(g ϫ °C) 7.132 ϫ 103 kJ 4. 3. b 6.6 kJ) ϭ Ϫ1077kJ ⌬H 0 f (reactants) Ϫ20. 5.6 kJ ⌬H0 ϭ Ϫ1207.01 kJ ϫ ᎏᎏ 1 mol H2O(s) ϭ 11. 27. ⌬H ϭ 2. 5.8 kJ ϩ 2 mol H2O(g) ϫ ᎏᎏ 1 mol H2O(g) ΄ Quiz for Chapter 17 1. All rights reserved. Inc. Step 1: H2O(l) at 18 ЊC y H2O(l) at 100 ЊC J ⌬H ϭ 190.2 ϫ 102 J b. a.6 kJ ⌬Htotal ϭ 6.8 kJ ϩ 32 kJ ϩ 11 kJ ϭ 47.1 ϫ 102\(50. Ϫ73. b c a d b 6. 4. e i g f 5. 13. 19.0 kJ ϩ 565. b d b c a 21.0 kJ D. C2H6 ϩ ᎏ7 2 O2(g) → 2CO2(g) ϩ 3H2O(l) ⌬H0 ϭ ? 0 ⌬H0 ϭ ⌬H0 f (products) Ϫ ⌬H f (reactants) 0 ⌬H ϭ [2(Ϫ393. the sublimation of mothballs.01 kJ H2O 31. 2.0°C) J ⌬H ϭ 55. 23.0 g 28. 22.40 J/(g ϫ °C) 28. b c a d b Chapter 17 Test B A.0°C Ϫ 24. steam absorbs the heat required for vaporization (40. Endothermic processes absorb heat. a c b d b 21.0 g H 1m ol 2O 30. Problems 96 J 27. B. g f h c 5. 13. j 6.0 kJ ϫ ᎏᎏ ϫ ᎏᎏ ϭ 225 g 1m ol 6.0 kJ ϫ ᎏᎏ ϫ ᎏᎏ ϭ 165 g 1m ol 6. c b c a d 16. Specific heat ϭ ᎏ 18 g ϫ 25°C ϭ 0.0°C gϫ° C ⌬H ϭ 2.18 ᎏᎏ mL g ϫ °C ϫ (35.00 32.7 kJ ϭ Ϫ221. 8.1 ϫ 103 J ϭ 2.68 J) ϩ (0. 25. 14. 2. the evaporation of a puddle. ⌬H ϭ m ϫ C ϫ ⌬T g J ⌬H ϭ 55. steam at 100 ЊC contains more energy than boiling water at the same temperature. d 7. 19. 24. Essay 26.0 g H 1m ol 2O 30.0°C ϫ °C g © Pearson Education.7 kJ CO ϫ ᎏᎏ ϫ ᎏᎏ 29. 2[Mg(s) ϩ Cl2(g) → MgCl2(s)] 2(⌬H ϭ Ϫ641 kJ) SiCl4(l) → Si(s) ϩ 2Cl2(g) ⌬H ϭ ϩ687 kJ 2Mg(s) ϩ SiCl4(l) → Si(s) ϩ 2MgCl2(s) ⌬H ϭ Ϫ1282 kJ ϩ 687 kJ ϭ Ϫ595 kJ ⌬H ϭ 2.00 ᎏᎏ ϫ 4. b 9. Problems 75 J 27. 3.01 kJ H2O 31. D. ⌬H ϭ m ϫ C ϫ ⌬T g J ⌬H ϭ 60. Endothermic examples include the melting of ice.7 kJ B.7 kJ 2C(s) ϩ O2(g) → 2CO(g) ⌬Hϭ Ϫ787. c 8. Multiple Choice 11. 3.Chapter 17 Test A A. 20. j 6.0 g O2 3 mol O2 18.1 kJ 1 mol O2 1411 kJ g O2 ϫ ᎏᎏ ϫ ᎏᎏ ϭ 118 kJ 29. the cooling of skin as perspiration evaporates. Specific heat ϭ ᎏ 12 g ϫ 20°C ϭ 0. 15. publishing as Pearson Prentice Hall. 24. Multiple Choice 11. a 10.18 J ⌬H ϭ ᎏᎏ ϫ 8. All rights reserved. while exothermic processes release heat. In vaporizing. Matching 1.. a C. Inc.0 g ϫ 4. b d c b d C. 17. Thus.17 J/(g ϫ °C) 28. 4.18 ᎏᎏ mL g ϫ °C ϫ (33. 20.0°C Ϫ 27. b 10. 23. and the freezing of water. 1 mol CO 24.0 g CO 3 mol CO ϭ 16. h 7.0 ϫ 103 J or 2.8 kJ)] Ϫ [(Ϫ84. 55. 12. 2[C(s) ϩ O2(g) → CO2(g)] 2(⌬H ϭ Ϫ393. and the heat used to cook food. 17. 4.18 ᎏᎏ ϫ 9.0 mL ϫ 1.00 ᎏᎏ ϫ 4. i 9. Matching 1. 25.7 kJ/mol). c b a d b 16.0 g ϫ 4.0 kJ)] ⌬H0 ϭ Ϫ1559. 18. 56. 18.0 mL ϫ 1. 12.5 kJ 18. d ᎏ 32. 22.5 kJ) ϩ 3(Ϫ285.5 kJ) 2CO2(g) → 2CO(g) ϩ O2(g) ⌬H ϭ ϩ565. e 8. 15. Exothermic examples include the combustion of fossil fuels such as gasoline. Essay 26. 75.3 kJ 830 Core Teaching Resources .0°C) 60. 14. 5. a 19. 3. Inc. d 17. and burned. The mass loss is consistent with the wax burning. 4.32.0018 mol Answer Key 831 .0 kJ)] ⌬H0 ϭ Ϫ3266. © Pearson Education. Section Review 18. in this case CO2 and H2O. 7.8 kJ)] Ϫ [(48. publishing as Pearson Prentice Hall. Chapter 17 Small-Scale Lab Section 17. Depending on the candle and the time burned. 2. evaporated. AT 14. 3. All rights reserved. 6. 11. Because of gravity. the wick loses a few millimeters and the candle a few tenths of a gram. NT 2. Wick. C: 20 ϫ 12 ϭ 240 H: 42 ϫ 1 ϭ 42 282 0. C6H6 ϩ 15/2O2(g) → 6CO2(g) ϩ 3H2O(l) ⌬Hϭ ? 0 ⌬H0 ϭ ⌬H0 f (products) Ϫ ⌬Hf (reactants) 0 ⌬H ϭ [6(Ϫ393. 2. 7. Liquid water will form on the underside of a glass Petri dish filled with ice held over the flame. The wax burns but many students will say the wick.1 Part A Completion 1. C20H42 ϩ ᎏ 2 O2 → 20CO2 ϩ 21H2O 13. c Part D Questions and Problems 21. 8. 10. 2.. hot gases rise. 7. ST 12. reversible products forward reactants reverse reactants products equilibrium equilibrium constant ratio Le Châtelier’s 9. 9. page 533 Analysis 1. 61 ᎏ 8. 120 kg/24 h ϭ 5. Black soot will appear on a glass Petri dish held over the flame. The wick draws melted wax to the flame. means to draw a liquid from one place to another by capillary action. products slower temperature catalyst increasing Part B True-False 11. e 20. f 18. ∆H ϭ 20(Ϫ394) ϩ 21(Ϫ242) Ϫ(Ϫ2230) Ϫ61/2(0) ∆H comb ϭ Ϫ10.0018 mol ϭ 19 kJ You’re the Chemist 1. b.50 g ϫ 1 mol/282 g ϭ 0. which is drawn up into the wick.50 kJ) ϩ (0. as a verb.5 kJ) ϩ 3(Ϫ285. b 16. 3. 5. The candle flame might be round in zero gravity.4 Heat of Combustion of a Candle. AT Part C Matching 15. 10.700 kJ/mol ϫ 0. Rates react kinetic energy activation minimum 6.2 Part A Completion 1. 4. 8. 9.9 kJ 10.0 kg/h 22. Heat from the combustion melts the wax. 6. Those who think the wick burns may suggest that the wax slows the rate of burning.700 kJ/mol 11. 4. 10. d Section Review 18. 5. 10. 3. a.145 ϭ 7. 8. 4. 24. b 25.. spontaneous nonspontaneous energy work free energy energy greater entropy disorder law of disorder maximum Part C Matching 16. AT 15. d.42)2 (0. 5. 8. 3.1 ϫ 10Ϫ6 Precipitation occurs because the ion product (1.1 ϫ 10Ϫ6) is greater than the Ksp of BaCO3 (5. d a heap of loose stamps ice cubes in a bucket 10 mL of steam at 100°C the people watching the parade © Pearson Education. 10. c. a 18. 6. b. d 17. 11. b Part D Questions 23.21) ᎏᎏ ϭ 7.5 Part A Completion 1. NT 13. AT 16.5 ϫ 10Ϫ3M) ϭ 1. 4. 7. b 19. 2. AT 14. rate concentration rate law specific rate constant order first-order second order experiment elementary reaction mechanism 832 Core Teaching Resources .3 Part A Completion 1. Keq ϭ ᎏ ᎏ [SO3]2 Keq ϭ (0. ST 14. publishing as Pearson Prentice Hall. 6. [CO32Ϫ] ϭ 0. d 22. e 1. ST 13. c Part D Questions and Problems [SO2]2 [O2] 21. a 21.072)2 Section Review 18.1 (0.0 ϫ 10Ϫ9). solubility product constant common ion effect addition precipitate Part B True-False 12. Section 18. All rights reserved. 9. 7.4 Part A Completion 20. Inc. 5. b Part B Matching 5. NT 15.00070M [Ba2ϩ] ϭ 0.Part B True-False 12. a 6. ST Part C Matching 17. 3. c 18. 2. AT Section Review 18. Part C Problem 8. e 20. f 19. c 7.0 ϫ 10Ϫ4M) ϫ (1. 4. 9. 2.0015M [CO32Ϫ] ϫ [Ba2ϩ] ϭ (7. a. f 17. Keq ϭ ᎏ ᎏ [CO] ϫ [H2]2 [SO3] ϫ [NO] d. AT Part C Matching 15. SrCO3 1 Sr2ϩ(aq) ϩ CO32Ϫ(aq) Ksp ϭ [Sr2ϩ] ϫ [CO32Ϫ] Let x ϭ [Sr2ϩ] ϭ [CO32Ϫ] Ksp ϭ x2 ϭ 9. decrease the rate b.2 ϫ 10Ϫ6M ϭ [CO32Ϫ] [Agϩ] ϭ 6.6 ϫ 10Ϫ4M 3.5 mol/h 3.0 ϫ 10Ϫ12 x ϭ 1. (2) increasing the concentration of the reactants. Point B represents the energy level of the intermediate product. Keq ϭ ᎏ ᎏ [NO]2 ϫ [Br2] [CH3OH] c. Keq ϭ ᎏ ᎏ [NO]4 ϫ [O2]2 [NOBr]2 b. Points A and C represent the energy level of the activated complexes.014)2 ϫ (0.Part B True-False 11.0 ϫ 10Ϫ5M 5. [N O ] 2 4 ϭ 5.66 [NO2]2 ϭ ᎏᎏ ϭ 0. b 18. shift right d. This diagram represents a reaction that takes place in two elementary steps. The reaction is exothermic. ST 13.1 ϫ 10Ϫ12 4x3 ϭ 8. shift right b.1 ϫ 10Ϫ4 (0. Point D represents the energy level of the final product.2 ϫ 10Ϫ5)2 Ksp ϭ 8. Keq ϭ ᎏ ᎏ [N2O5]2 [0. Inc.20 Ksp ϭ (2x)2(0.3 ϫ 10Ϫ4M ϭ [CO32Ϫ] [Agϩ] ϭ 2x ϭ 2. [CO32Ϫ] ϭ x ϩ 0.20) ϭ 8.20] 2. increase the rate 4.. All rights reserved. Ag2CO3(s) 1 2Agϩ(aq) ϩ CO32Ϫ(aq) Ksp ϭ [Agϩ]2 ϫ [CO32Ϫ] Let x ϭ [CO32Ϫ]. a 19. Fe(OH)2(s) 1 Fe2ϩ(aq) ϩ 2OHϪ(aq) Ksp ϭ [Fe2ϩ] ϫ [OHϪ]2 [Fe2ϩ] ϭ 0.2 [NO2]4 ϫ [O2] 1.6 ᎏᎏ [NO2]2 [N2O4] [NO2]2 ϭ ᎏᎏ 5. 2.6 ͙ . a. [CO32Ϫ] ϭ 0. Section 18.80]4 ϫ [0. c Part D Questions 21. shift right 4.20 assume x << 0. increase the rate Section 18.0 ϫ 10Ϫ6 Ksp ϭ (6. e 20. NT 12.4 ϫ 10Ϫ6M Answer Key 833 © Pearson Education. Ca(OH)2(s) 1 Ca2ϩ(aq) ϩ 2OHϪ(aq) Ksp ϭ [Ca2ϩ] ϫ [OHϪ]2 b. shift left b.1 1.3 ϫ 10Ϫ10 x ϭ [Sr2ϩ] ϭ 3. a.5[OHϪ] ϭ 6.6 ᎏᎏ 0.3 1. AT 14. d 16.1 ϫ 10Ϫ12 x2 ϭ 1. Keq ϭ ᎏ ᎏ [H2S]2 (0.6 ϫ 10Ϫ16 4.0 ϫ 10Ϫ11 x ϭ 3. Keq ϭ 1 ϫ 1012 [H2]2 ϫ [S2] 8. [N2O4]2 5. 2x ϭ Agϩ Ksp ϭ (2x)2(x) ϭ 8. Keq ϭ ᎏᎏᎏᎏ [0. (3) decreasing the reactant particle size. no shift 7. Ksp ϭ [Agϩ]2 ϫ [CO32Ϫ] Let [Agϩ] ϭ 2x.50]2 Keq ϭ 0. Keq ϭ ᎏ ᎏ [SO2] ϫ [NO2] 6. Rates of chemical reactions can usually be increased by (1) increasing the temperature. a.0 ϫ 10Ϫ6)(1. shift right c.34 5.20 mol.1 ϫ 10Ϫ12 x3 ϭ 2. and (4) using of a catalyst.33 3. 2 mol/4 h ϭ 0. Ag2CO3(s) 1 2Agϩ(aq) ϩ CO32Ϫ(aq) Ksp ϭ [Agϩ]2 ϫ [CO32Ϫ] 2.18)2 Practice Problems 18 Section 18. a. publishing as Pearson Prentice Hall.035) Keq ϭ ᎏᎏᎏ ϭ 2. 30 mol/(L • s) 2.30)2 [NOCl]2 [CH3OH] ᎏ ϭ 2.4 1. greater 3. c 10.6 24. c d b c 20. 13.00842 mol/(L • s) Ϭ 0.0 mL 600. Vocabulary Review 18 1.5 ϫ 10Ϫ9 ϭ [Ca2ϩ] ϫ [CO32Ϫ] The total volume is 1000 mL. rate ϭ k[C2H4O2][CH4O] B. 2. negative 2.7 ϫ 10Ϫ6 Because this product exceeds the Ksp value. Keq ϭ ᎏᎏ ϭ (0.4 mol/(L • s) Ϭ 8 ϭ 0. rate ϭ k[J][K] The reaction is first order in both J and K. 8.0012M 1000. 3. d 9. increase 2. C3H8O3 c.0 mL ] ϭ 0. D d. Example a 5. ᎏ [CO] ϫ [H2]2 [CH3OH] ϭ 2. yes d. two elementary reactions b. h 8. Problems (1.2 ϫ 102 ϫ [CO] ϫ [H2]2 ϭ 2.2 ϫ 102 25. j 8.0013 mol/L ϭ 0. publishing as Pearson Prentice Hall. rate ϭ k[HgCl2][Na2C2O4]2 3.020)(0.3 ϫ 10Ϫ7 ϭ [Pb2ϩ] ϫ [SO42Ϫ] [SO4 2Ϫ Interpreting Graphics 18 1. NT 8.0168 sϪ1 5.0. b © Pearson Education. b b c d d 16.0021)(0.2)2 ϫ (0. increasing 4. 2. 4.020 7.60)2 [CH3OH] ϭ 1.. b 7. All rights reserved. 3. 0. g 6. 4.0 mL [Pb2ϩ] ϫ [SO42Ϫ] ϭ (0. 14. d 3. b 2. positive 400.0 mL [Pb2ϩ] ϭ 0. a. 4. precipitation will occur. Ksp(CaCO3) ϭ 4.0013M [Ca2ϩ] ϫ [CO32Ϫ] ϭ (0. positive e. 12. less 4. 15. positive f.0021 mol/L ϭ 0. Section 18. N c. 1. 19. three elementary reactions 6. a. a 4.0020M 1000.5 1.0012)(0. f 10. Matching 1.0020) [Pb2ϩ] ϫ [SO42Ϫ] ϭ 2. negative b. A Chapter 18 Test A A. 2. a b a d C. NT Section 18.0013) [Ca2ϩ] ϫ [CO32Ϫ] ϭ 2. increasing c. k ϭ rate/[H2O2] k ϭ 0. so [Ca2ϩ] ϭ 0. h Quiz for Chapter 18 1. 22. 23. 17.500 mol/L k ϭ 0. Ksp(PbSO4) ϭ 6.0050M ϭ ᎏᎏ ϭ 0. b 5. d 6. e g i a 5.58 mol/L 834 Core Teaching Resources . i a j c 5. d 9. 7. AT 9. NaCl has no ion in common with Mg(OH)2. rate ϭ k[H2O2]. Multiple Choice 11. e 7.60) [NO]2 ϫ [Cl2] ᎏ ϭ 9. a.0021M [CO32Ϫ] ϭ 0. 18. f 7. a. 21.2 ϫ 102 ϫ (0.0020M ϭ ᎏᎏ ϭ 0.4 ϫ 10Ϫ6 Because this product exceeds the Ksp value. negative c. increasing b. Inc. b 6.6. decrease 3. D b. precipitation will occur. j g h i 5. Some spontaneous reactions apprear to be nonspontaneous because their rates are slow.75 ϫ 10Ϫ6)3 [N2] ?ϭ 1. 3. Keq ϭ ᎏ [HCl]4 ϫ [O2] (5. An increase in temperature causes the endothermic reaction to speed up in an effort to consume the additional heat. a. publishing as Pearson Prentice Hall. 19. a Chapter 18 Small-Scale Lab Section 18.2 ϫ 10Ϫ3)4 ϫ (3. 23. Multiple Choice 11. Essay 27. ∆H is negative. 24. All of the solids dissolved rapidly. c. First order ϩ third order ϭ fourth order overall. d 8. Doubling A doubles the rateᎏfirst order in A. NH4Cl ϩ H2O(l ) c. c 7. decreases shifts right. CaCl2 ϩ H2O is exothermic.. NaCl ϩ H2O(l ) b. a. Mixture a. heat ϩ NH4Cl(s) → NH4ϩ(aq) ϩ ClϪ(aq) CaCl2(s) → Ca2+(aq) ϩ 2ClϪ(aq) ϩ heat 5. © Pearson Education. [NH3]2 Keqϭ ᎏ [N2] ϫ [H2]3 [NH3]2 [N2]ϭ ᎏ Keq ϫ [H2]3 (1.8 ϫ 10Ϫ4) Keq ϭ 1. 2. c. decreases [H2O]2 ϫ [Cl2]2 26. ∆H is close to 0. favors products favors reactants favors products favors reactants D. b. 4. 15. NH4Cl ϩ H2O is endothermic. c b c b 1.10 ϫ 1011 mol/L E. d a d b c C. e. Nonspontaneous reactions do not favor the formation of products under the specified conditions. e 9. Essay 28. c d b d a B. b. f 6. d. increases shifts left. Answer Key 835 .8 ϫ 10Ϫ2)2 ϫ (5. ∆H is positive. 3. under the conditions specified. Doubling B increases the rate 8 times (23 ϭ 8) ᎏthird order in B. 14. d. b 10. ∆S is positive in each case. Chapter 18 Test B A. 13.26. Additional Problem 28. 18. All rights reserved. Additional Problem 29.4 ϫ 1010 27. Spontaneous reactions are reactions that.8 ϫ 10Ϫ2)2 ϭᎏ (1.59 ϫ 10Ϫ3) ϫ (2. shifts left.4 Enthalpy and Entropy. c. 22. increases shifts right. PbF2 2. An increase in pressure (for a gaseous system with an unequal number of molecules) causes the reaction that produces the fewest number of molecules to speed up. 12.23 ϫ 10Ϫ4)2 ϭᎏ (6. decreases shifts left. Entropy usually increases in the dissolving process. are known to favor the formation of products. b. E. a. D. The addition of more reactant causes an increase in the rate of the forward reaction. 17. increases shifts right. 20. which consumes that reactant. Inc. NaCl ϩ H2O(l) did not change much in temperature. Problems 25. 21. page 574 Analyze Sample data are provided. Matching 1. CaCl2 ϩ H2O(l ) T1 21°C 21°C 21°C T2 21°C 5°C 53°C ⌬T 0°C Ϫ16°C ϩ32°C 16. 4. f. ⌬G ϭ (Ϫ) ϪT(ϩ) for CaCl2(s) ∆G is Ϫ. 8. 11. 3. i 23. NT 13. Endothermic reactions absorb heat.. 4. 5. 4. 10. e 22. Dimethyl ether is a Lewis base because it donates an electron pair to form a bond. AT 836 Core Teaching Resources . a 20. cooling the environment. d 19. 25. three Arrhenius hydroxide ions proton acceptor electron-pair donor monoprotic diprotic conjugate acid–base pair amphoteric Part B True-False 12.2 Part A Completion 1. d Part D Problems 24. 6. b Part D Problems 26.6. Many salts such as KCl. NaHCO3. pH ϭ Ϫlog[Hϩ] [Hϩ] ϭ 1 ϫ 10Ϫ10 Part B True-False 12. 5. ⌬T You’re the Chemist 1.1 Part A Completion 1. Sample data: Mixture NaCl ϩ H2O(s ) T1 0°C T2 Section Review 19. a. g 23. AT 15. ⌬G ϭ (ϩ) ϪT(ϩ) for NH4Cl(s) ⌬G is ϩ or Ϫ. 11. 2. Boron trifluoride is a Lewis acid because it accepts an electron pair from dimethyl ether. Kw ϭ [Hϩ][OHϪ] 1 ϫ 10Ϫ14 Kw [OHϪ] ϭ ᎏᎏ ϭ 1 ϫ 10Ϫ4 ϩ ϭ ᎏᎏ 1 ϫ 10Ϫ10 [H ] The solution is basic. a 21. This explains the drop in temperature of the NaCl and ice mixture. pH ϭ Ϫlog[Hϩ] [Hϩ] ϭ 1 ϫ 10Ϫ3M b. h 21. The temperature of the NaCl and ice dropped dramatically. Part C Matching 17. 8. 10. b 22. Inc. 3. f 19. publishing as Pearson Prentice Hall. g 18. ∆G ϭ ∆H Ϫ T∆S. NT Part C Matching 17. AT 13. 6. c 18. ST 16. Na2CO3 and Na3PO4 dissolve endothermically or with little or no change in temperature. c 24. 4. f 25. pH ϭ Ϫlog[Hϩ] [Hϩ] ϭ 1 ϫ 10Ϫ6M c. All rights reserved. 9. ST 14. 9. 2. Both CaCl2 and NH4Cl depress the freezing point of ice and cause a drop in temperature. Section Review 19. e 20. NT 14. Melting ice is endothermic. Ϫ15°C Ϫ15°C 2. AT 15. 3. 7. 7. self-ionize 1 ϫ 10Ϫ7 0 to 14 hydrogen ion acidic basic neutral 7 ion-product hydronium/hydroxide hydroxide/hydronium © Pearson Education. NT 16. ⌬G ϭ (0) ϪT(ϩ) for NaCl(s) ⌬G is Ϫ. 2. 6. H3PO4 ϩ H2O 1 H3Oϩ ϩ H2PO4Ϫ H2PO4Ϫ ϩ H2O 1 H3Oϩ ϩ HPO42Ϫ HPO42Ϫ ϩ H2O 1 H3Oϩ ϩ PO43Ϫ 3. 2. f 16.35 Ϫ 4. a 19. c 17. c 13. diprotic d. Answer Key 837 © Pearson Education. b Part D Questions and Problems 18.4 Part A Completion 1. 2HI ϩ Ca(OH)2 y CaI2 ϩ 2H2O Section Review 19.35 Ϫ 0. 2.309 (4. monoprotic c. 4. triprotic 4. a 15. All rights reserved. 9. publishing as Pearson Prentice Hall.. c Part C Matching 16. 9.1 ϫ 10Ϫ2 [HX] ϭ 0. monoprotic b. 8. Section Review 19. a. 8. Only hydrogens bonded to highly electronegative atoms are ionizable. a. 10. H2SO4 and H3Oϩ are proton donors and H2O Ϫ and HSO4 are the proton acceptors.1 ϫ 10Ϫ2) (4.041 [HX] ϭ 0. 7. d 21. AT 10. strong weak buffer capacity Part B True-False 10. ST 13. ST 13. 7. a.Section Review 19. AT 11. HX 1 Hϩ ϩ XϪ [Hϩ] ϭ [XϪ] ϭ 4.1 1. e 18. d Part D Problem . 3. 4. titration 6. H2SO4 ϩ H2O 1 H3Oϩϩ HSO4Ϫ 2.5 Part A Completion 1. acidic b. H3PO4 is a triprotic acid able to ionize three hydrogens. AT Part B True-False 12. H3PO4 ϩ Al(OH)3 y AlPO4 ϩ 3H2O b. The conjugate acid–base pairs are H2SO4/HSOϪ 4 and H3Oϩ/H2O. 3. 11. e 14. 5. salt acidic basic neutral hydrolyze 6. d 20. basic Part D Problem 22. ST 12.4 ϫ 10Ϫ3 Practice Problems Section 19. a 15. b 17. NT 11. Like other alkali metals. acid hydroxide water neutralization 5. All ions formed are shown in the following chemical equations. Inc. NT Part C Matching 12. b 16. 5. AT 15. end point 7. lithium reacts violently with water to produce hydrogen and the base lithium hydroxide.309 [HX] Ka ϭ 5.3 Part A Completion 1. NT Part C Matching 14. degree of ionization Ka larger pH completely strong weak bases water acid strong 17. 3. neutral c.1 ϫ 10Ϫ2) [Hϩ][XϪ] Ka ϭ ᎏᎏ ϭ ᎏᎏᎏ 0. 4. AT 9. NT 14.1 ϫ 10Ϫ2 ϭ 0. equivalence Part B True-False 8. 4 pOH ϭ 10. 12. Bases cause indicators to change colors. acidic c.00) Use log tables or your calculator to find the log of 7. At equilibrium.20M x ϭ 3.00 Ϫ 0.2 ϫ 10Ϫ3M) 8.20M) x2 Ka ϭ ᎏᎏ ϭ 6. 10. Most acidic solutions of interest have a hydrogen ion concentration of less than 1M.00 Ϫ 0. C6H5COOH(aq)1 C6H5COOϪ(aq) ϩ Hϩ(aq) [C6H5COOϪ][Hϩ] Ka ϭ ᎏ ᎏ [C6H5COOH] 6.2 1.20M Ϫ x ϭ 0.0 pH ϭ 6. (4. it is a Lewis base. basic b. pH ϭ Ϫlog[Hϩ] pH ϭ Ϫlog(1 ϫ 10Ϫ3) ϭ Ϫ(0.0 Section 19. basic d. 6.46 4.8 ϫ 10 M 9. pH ϩ pOH ϭ 14. H2S(aq) 1 Hϩ(aq) ϩ HSϪ(aq) ᎏ Ka ϭ ᎏ [H S] 2 Section 19. Acids react with compounds containing hydroxide ions to produce a salt and water. [Hϩ] ϭ [CNϪ] ϭ 6. Ka ϭ ᎏᎏᎏᎏ 0. Aqueous solutions of bases taste bitter and feel slippery.3 1. a.0 pH ϭ 14.4 ϭ 1. 5. acidic 6. (2) e. pOH ϭ 14. pOH ϭ Ϫlog[OHϪ] pOH ϭ Ϫlog(3. Kb ϭ ᎏᎏᎏ [C6H5NH2] b. 9. The weakest acid has the smallest Ka.6 7.86) Ϫ (Ϫ9.0 Ϫ pOH ϭ 14.6 5.2. They react with acids to produce a salt and water. 10. HCO3Ϫ Ͻ H2PO4Ϫ Ͻ HCOOH Ͻ H2C2O4 5. Since FϪ donates the pair of electrons. strong base.10M NH4ϩ(aq) © Pearson Education.36 d. (3) c.0 2.3 ϫ 10Ϫ6M [Hϩ][CNϪ] Ka ϭ ᎏᎏ [HCN] (6. a. Acids have a tart or sour taste and cause indicators to change color. 7.5 ϫ 10Ϫ3M ϭ [Hϩ] 10.34 b. At equilibrium.54 pOH ϭ 1. pH ϭ Ϫlog[HϪ] pH ϭ Ϫlog(1 ϫ 10Ϫ6) reminder: the log(a ϫ b) ϭ log a ϩ log b pH ϭ Ϫ(0. (4) d. pH ϭ Ϫlog[HϪ] pH ϭ Ϫlog(7.5 ϫ10Ϫ2) pOH ϭ Ϫ (0. pH ϭ 9.3 ϫ 10Ϫ5M 0.0 c. (1) [C6H5NH3ϩ][OHϪ] 7. acidic c.3 ϫ 10Ϫ6M)2 Ka ϭ ᎏᎏ ϭ 4. All rights reserved. [Hϩ] ϭ x ϭ [C6H5COOϪ] [C6H5COOH] ϭ 0.096M Ϫ4 Ka ϭ 1.0 ϩ (Ϫ3)) ϭ 3.. basic 8.2 ϫ 10Ϫ9) pH ϭ Ϫ(0. BF3 can accept a pair of electrons to form a covalent bond and is therefore a Lewis acid.0 Ϫ pH pOH ϭ 14.0 ϫ 10Ϫ10M 0. The log of this concentration would always be a negative number.0 ϩ (Ϫ6)) reminder: the log 1 ϭ 0. 838 Core Teaching Resources . HF(aq) 1 Hϩ(aq) ϩ FϪ(aq) [Hϩ][FϪ] Ka ϭ ᎏᎏ [HF] 3.54) Ϫ (Ϫ2. publishing as Pearson Prentice Hall. (5) b.86 pH ϭ 8. neutral b. basic d. a. strong acid 2.20M (since x ϽϽ 0. Taking the negative log (minus sign in the pH definition) ensures that the pH values will usually be positive.2Li(s)ϩ 2H2O → H2(g) ϩ 2LiOH(aq) 5. weak acid.0 Ϫ 3. Inc.0 Ϫ 12.2 ϫ 10Ϫ3M)(4.14 3.6 [Hϩ][HSϪ] 1 NH3(aq) ϩ Hϩ(aq) [NH3][Hϩ] Ka ϭ ᎏ ᎏ [NH4ϩ] c. weak base.00) pOH ϭ 2. N2H4(aq)ϩH2O(l) 1 N2H5ϩ(aq) ϩ OHϪ(aq) [N2H5ϩ][OHϪ] Kb ϭ ᎏ ᎏ [N2H4] 4. a. acidic e. 7. a. 5. basic solution Interpreting Graphics 19 1. 0.021 mol NaOH Molarity ϭ ᎏᎏ ϭ ᎏᎏ liters 0. neutral solution b.0198 L Ca(OH)2 ϫ ᎏᎏᎏ 1 L Ca(OH)2 1 mol H2SO4 ϫ ᎏᎏ ϭ 0. Because NaOH is a strong base. such as NaOH.0140 mol HC2H3O2 0. To determine the equivalence point. Al(OH)3(aq) ϩ 3HCl(aq) y AlCl3(aq) ϩ 3H2O(l) 1 mol Al(OH)3 0. 2.4 1. Ca(OH)2(aq) ϩ H2SO4(aq) y CaSO4(aq) ϩ 2H2O(l) 0.0100 mol Ca(OH)2 0..75 mol H2SO4 0.55M NaOH 2. 2NaOH(aq) ϩ H2SO4(aq) y Na2SO4(aq) ϩ 2H2O(l) 0. Locate the point on this steep portion of the curve equidistant between the two plateaus.038 L NaOH Molarity ϭ 0. find the area of the titration curve where the pH changes abruptly when a small volume of NaOH is added.0124 L H2SO4 liters Molarity ϭ 0.Section 19.12M Ba(OH)2 molarity liters ϭ 0.10M 0. 0.0025 mol C6H5COOH were originally present in a volume of 25 mL. 3.0160M H2SO4 4. Ba(OH)2(aq) ϩ 2HCl(aq) y BaCl2(aq) ϩ 2H2O(l) 0. [C6H5COOH] ϭ [NaOH] ϭ 0M and 0. 0.94 ϫ 10Ϫ5 mol Ca(OH)2 ϭ 9.00404M 3. .200M HCl molarity liters ϭ 0.94 ϫ 10Ϫ5 mol Ca(OH)2 9.0 g Al(OH)3 3 mol HCl ϫ ᎏᎏ ϭ 0. [C6H5COOH] ϭ ᎏᎏ ϭ 0.014 L H2SO4 ϫ ᎏᎏ 1 L H2SO4 2 mol NaOH ϫ ᎏᎏ ϭ 0. Ca(OH)2(aq) ϩ 2HC2H3O2(aq) y Ca(C2H3O2)2(aq) ϩ 2H2O(l) 0.050M.0122 L HCl ϫ ᎏᎏ 2 mol HC ϭ 0. Thus. publishing as Pearson Prentice Hall. each mole of NaOH added reacts with each mole of C6H5COOH present. a.0125 L Ba(OH)2 ϭ 13 mL Ba(OH)2 5.025 L Answer Key 839 © Pearson Education. Inc.025 L ϫ 0. acidic solution c. produces a basic solution at the equivalence point.0025 mol NaOH 5. 4. The neutralization of a weak acid with a strong base.000198 mol H2SO4 1 mol Ca(OH)2 ϭ 0.0142 L HC2H3O2 ϫ ᎏᎏᎏ 1 L HC2H3O2 1 mol Ca(OH)2 0. Based on the answers to questions 4 and 5. CHO2Ϫ ϩ Hϩ 1 HCHO2 HCHO2 ϩ OHϪ 1 CHO2Ϫ ϩ H2O 2.0122 L HCl ϫ ᎏᎏ 1L HCl 1 mol Ba(OH)2 0.021 mol NaOH 1 mol H2SO4 moles 0. at the equivalence point.00212 mol HCl 1 mol Al(OH)3 0.0106 L HCl ϭ 10.0015 mol Ba(OH)2 moles liters ϭ ᎏᎏ ϭ ᎏᎏᎏ 0.5 1. 0.000198 mol H2SO4 moles Molarity ϭ ᎏᎏ ϭ ᎏᎏᎏ 0.050 L 6.0025 mol Thus.10 mol/L NaOH ϭ 0.25 mol HCl 0.0550 g Al(OH)3 ϫ ᎏᎏ 78. The pH at the equivalence point is approximately 8.0015 mol Ba(OH)2 0. C6H5COOH ϩ NaOH y C6H5COONa ϩ H2O One mole of sodium hydroxide will neutralize one mole of benzoic acid. All rights reserved. Benzoic acid is a weak acid.0246 L Ca(OH)2 Molarity ϭ 0.94 ϫ 10Ϫ5 mol Ca(OH)2 Molarity ϭ ᎏᎏᎏ 0.000198 mol H2SO4 0.6 mL HCl Section 19.00212 mol HCl moles liters ϭ ᎏᎏ ϭ ᎏᎏ 0.0142 L HC2H3O2 ϫ ᎏᎏ 2 mol HC2H3O2 ϭ 9. the solution is slightly basic.0025 mol [C6H5COONa] ϭ ᎏᎏ ϭ 0. The equivalence point occurs when the number of moles of NaOH added equals the number of moles of C6H5COOH originally present. 3H2SO4 ϩ 2Al(OH)3 y Al2(SO4)3 ϩ 6H2O 34. Multiple Choice 11.050M (3. c 8. The resulting solution is slightly basic because [OHϪ] > [Hϩ]. 4. 17. i a j f 5. d 8. pH ϩ pOH ϭ 14 8. 16. phenolphthalein would be a good choice. neutral c. Problems 30. Students should draw a horizontal band on the graph encompassing the pH range 8-10 to show the region of the curve where phenolphthalein would be an effective indicator of neutralization. d a b d d d a 25. basic b.050) Quiz for Chapter 19 1. c 6. [C6H5COOH][OHϪ] 9. 2. buffer © Pearson Education. publishing as Pearson Prentice Hall.5 [OHϪ] ϭ 3. basic b. A faint pink color should be detected at the equivalence point. The dissociation constant reflects the fraction of a weak base or weak acid that is in ionized form. All rights reserved. d 7. 3. Matching 1. b d d a c d c 18. 29.7. Acidic solution. Hydronium ion. 26. H2SO3 1 Hϩ ϩ HSO3Ϫ [Hϩ] ϫ [HSO3Ϫ] Ka ϭ ᎏᎏ [H2SO3] b. 8. 2. 27. Inc. the benzoate ion establishes the equilibrium shown. 4. 840 Core Teaching Resources . Thymol blue might also be a good candidate. Kb ϭ ᎏᎏᎏ [C6H5COOϪ] 10. 21. e 9. a 4. hydrolyzing salt 8.5 ϩ pOH ϭ 14 pOH ϭ 5. pH ϭ 7. Weak acids and bases are only partially ionized in aqueous solution. 23. Because the equivalence point occurs between pH 6 and pH 11. Amphoteric. 19. [Hϩ] ϭ 1 ϫ 10Ϫ4. acidic d. g 7. Lewis acid. 22. Kw ϭ [Hϩ] ϫ [OHϪ] 1 ϫ 10Ϫ14 Kw [Hϩ] ϭ ᎏᎏ ϭ ᎏ ᎏ 1 ϫ10Ϫ12 [OHϪ] [Hϩ] ϭ 1 ϫ 10Ϫ2 acidic 32. 3.. The other terms are theories used to classify acids and bases. basic Vocabulary Review 19 1. a.2 ϫ 10Ϫ6M [C6H5COOϪ] ϭ 0. 20. b B. acidic c. a. equivalence point 7. b 3. c 5. b 2. neutral 31. C6H5COOϪ ϩ H2O 1 C6H5COOH ϩ OHϪ At the equivalence point. pH ϭ 4. c 10. 28. h 6. Strong acids. a. 12. 6. b Chapter 19 Test A A. a d a c b C. 5.0 ϫ 10Ϫ10 (0. H⌵⌷3 1 Hϩ ϩ NO3Ϫ [Hϩ] ϫ [NO3Ϫ] Ka ϭ ᎏᎏ [HNO3] 33. 2HBr ϩ Mg(OH)2 y MgBr2 ϩ 2H2O b.2 ϫ 10Ϫ6)2 Kb ϭ ᎏᎏ ϭ 2. The other terms describe aqueous solutions based on their pH. pH ϭ Ϫlog[Hϩ] ϭ 9. The other terms refer to ways of describing acids and bases according to the Brønsted-Lowry theory. 15. neutral 9.2 ϫ 10Ϫ6M At the equivalence point [OHϪ] ϭ [C6H5COOH] ϭ 3. Alkaline is another name for a basic solution and basic solutions would have a high hydroxide ion concentration. 14. a. 24. 13. 3. H2O(l). 7. Ka ϭ ᎏᎏ [H2SO4] a. cause indicators to change colors and react with each other to form water and a salt. 20. H2O(l).18 ϫ 10Ϫ5 B. H2O(l). f c h k 9.4 ϫ 10Ϫ4] pH ϭ Ϫ(log 3. a. publishing as Pearson Prentice Hall. Matching 1.00225][0. HCl(g). H . acids are electron-pair acceptors. I b.0 ϫ 10Ϫ14 (mol/L)2 [OHϪ] ϭ Kw/[Hϩ] 1.7 ϫ 10Ϫ4M ϭ [Cuϩ] Chapter 19 Test B A. H3Oϩ(aq). CuCl(s) y Cuϩ(aq) ϩ ClϪ(aq) Ksp ϭ [Cuϩ][ClϪ] 3. C2H3OϪ 2 (aq) ϩ Ϫ 36.47] pH ϭ 3. 14. Problems 27.35 mol KOH 2 mol KOH ϭ 0. 2. E. HC2H3O2(aq) 1 Hϩ(aq) ϩ C2H3O2Ϫ(aq) [Hϩ] ϭ [C2H3O2Ϫ] ϭ 2. i 10. H2SO4(aq) ϩ 2LiOH(aq) y Li2SO4(aq) ϩ 2H2O(l) H2SO4(aq) ϩ 2KOH(aq) y K2SO4(aq) ϩ 2H2O(l) 1 mol 2 mol 1 mol 2 mol 1 mol H2SO4 ᎏᎏ ϫ 0. All rights reserved. [Hϩ] ϭ 1 ϫ10Ϫ6 . According to the Lewis theory. whereas bases are electron-pair donors. b d a b c 17. ClϪ(aq) c. pH ϭ Ϫlog [Hϩ] pH ϭ Ϫlog [3. a. j 11. e D.0 ϫ 10Ϫ5 mol/L The solution is basic. Essay 35. c a b a a 34.53) ϩ (Ϫ4)] pH ϭ Ϫ[Ϫ3. Bases feel slippery.4 ϩ log 10Ϫ4) pH ϭ Ϫ[(0. [Hϩ] ϭ 1 ϫ10Ϫ7 .09775] Ka ϭ 5. 4. BCl3. acidic c. Multiple Choice 12. Kw ϭ [Hϩ][OHϪ] ϭ 1. Answer Key 841 . 23. H3Oϩ(aq).1000M Ϫ 0. FϪ(aq) b.2 ϫ 10Ϫ7 ϭ [Cuϩ]2 5. 25.2 ϫ 10Ϫ7 ϭ [Cuϩ][Cuϩ] 3. 7.47 The solution is acidic.. The Brønsted-Lowry theory defines acids as proton donors and bases as proton acceptors.00225M [HC2H3O2] ϭ 0. H3Oϩ(aq). 32. Both acids and base. Inc. b d a d a C.09775M [Hϩ][C2H3O2Ϫ] Ka ϭ ᎏ ᎏ [HC2H3O2] [0. 6.0 ϫ 10Ϫ14 (mol/L)2 [OHϪ] ϭ ᎏᎏᎏ 1. [OHϪ] ϭ 1 ϫ 10Ϫ11. NH3 37. 30.0 ϫ 10Ϫ9 mol/L © Pearson Education. b g d a 5. HF(aq). 31. neutral [Hϩ][IϪ] a. [OHϪ] ϭ 1. 8. Additional Problems 22. HF(aq) ϩ KOH(aq) y KF(aq) ϩ H2O(l) b. 16. 6. bases taste bitter. Essay 33. 21. 15. 29. 24.00225] Ka ϭ ᎏᎏᎏ [0. 19. 26. a. Ka ϭ ᎏᎏ [HI] [Hϩ][HSO4Ϫ] b.18 mol H2SO4 28. 3. 35. 18. HC2H3O2(aq). 13.25 ϫ 10Ϫ3M [HC2H3O2] ϭ 0. Acids taste sour.D. acidic b. 2 Part A Completion 1. The pH solutions 1-3 are yellow. AT 11. d 20. b 15. NT Part C Matching 13. NT 12. 8. NT 14. 2. 4. 2. Results will vary depending on the indicator chosen. AT 16. redox away toward reduction 5. was oxidized and is the reducing agent. BCGϪ is blue. d Part D Questions and Problems 19. g 22. AT 11. NT 12. they immediately transfer electrons to the iron ions. 2. Section Review 20. the iron atoms lose electrons as the iron begins to be oxidized. When oxygen and water attack iron. 4. 4. This pH is the Ka of the acid. a 18. AT 10. Since aluminum and zinc are better reducing agents than iron and are more easily oxidized. 3. page 617 Analysis 1 Yellow 4 Green 7 Blue 10 Blue 11 Blue 8 Blue 12 Blue 5 Blue 9 Blue 2 Yellow 6 Blue 3 Yellow Section Review 20. oxidizing reduced reducing oxidized Part B True-False 9. was reduced and is the oxidizing agent. Figure A 1. 6. The zinc metal. reducing them back to neutral iron atoms. 6. publishing as Pearson Prentice Hall. HBCG is yellow. The copper ion. 3. The conjugate acid. 3. © Pearson Education.. 21. The conjugate base. e 18. a 21. Zn. AT Part C Matching 17. 7. The pH solutions 5-12 are blue. zero sign charge zero 5. Look for the pH of the color change. Inc. 8. Cu2ϩ. mix one drop of the weak acid with one drop of each pH 1-12 buffer solution. An equal mixture of HBCG and BCGϪ is green at pH ϭ 4. Reduction is the complete or partial gain of electrons.1 Part A Completion 1. i 25. e 14. f 24. Part B True-False 9. AT 10. c 17. The pH solution 4 is green. f 16. 6. 7. 5. To measure the Ka of a colored weak acid. 2.Chapter 19 Small-Scale Lab Section 19. Oxidation is the complete or partial loss of electrons.4 Ionization Constants of Weak Acids. All rights reserved. b 842 Core Teaching Resources . an intermediate between yellow and blue. charge on the ion electron oxidation decrease You’re the Chemist 1. 20. c 23. h 19. NT 13. NT 15. A decrease in the oxidation number indicates reduction. Mg: Hϩ: 10. Sr: O2: 2. d. The ionic charge on tin is 4ϩ. c 15.2 1. 4. The oxidation number is 0 f. N changes ϩ5 to ϩ4. Answer Key 843 . Fe: O2: 6. Sn is tin in an uncombined state. Ca: O2 9. N changes ϩ5 to ϩ2. Se is selenium in an uncombined state. thus the oxidation number is ϩ3. g Part D Questions and Problems 21. Br is oxidized (Ϫ1 to 0) Mn is reduced (ϩ7 to ϩ2). The ionic charge on magnesium is 2ϩ. BrϪ: Cl2: 7. Na: H2O: oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) oxidized (reducing agent) reduced (oxidizing agent) Section Review 20. e 19. f 17. All rights reserved. 3H2S ϩ 2HNO3 → 3S ϩ 2NO ϩ 4H2O Section 20. multiply by 3 2HNO3 ϩ 6HI y 2NO ϩ 3I2 ϩ 4H2O b. An increase in the oxidation number of an atom indicates oxidation. NT 12. I changes from 0 to ϩ5. Cl is oxidized (Ϫ1 to 0) N is reduced (ϩ5 to ϩ2). b. oxidation number half-reaction balanced ionic 5.1 1. a loss of 10eϪ for I2 10HNO3 ϩ I2 y 2HIO3 ϩ 10NO2 ϩ 4H2O 22. d 20. The ionic charge on potassium is 1ϩ. two 6. Multiply the oxidation reaction by 3 and the reduction reaction by 2.. 3. a. a 16. The oxidation number is 0. multiply by 2 2HI y I2. c. AT 10. 2. publishing as Pearson Prentice Hall. multiply by 10 I2 y 2HIO3. added 7. HNO3 y NO. ionic Part B True-False 8. Li: S: 3.Part D Questions and Problems 26. b 18. a. thus the oxidation number is ϩ1. I changes from Ϫ1 to 0. e. Mg: N2: 5. The ionic charge on sulfur is 2Ϫ. 14. multiply the oxidation reaction by 6 14Hϩ ϩ 6Fe2ϩ ϩ Cr2O72Ϫ → 6Fe2ϩ ϩ 2Cr3ϩ ϩ 7H2O Practice Problems Section 20. NT 13. Cs: Br2: 4. a gain of 3eϪ. Sb is oxidized (0 to ϩ5) S is reduced (ϩ6 to ϩ4). thus the oxidation number is ϩ2. a loss of 2eϪ for I2. Inc. a. thus the oxidation number is Ϫ2. The ionic charge on iron is 3ϩ. Fe2ϩ → Fe3ϩ ϩ eϪ and 6eϪ ϩ 14Hϩ ϩ Cr2O72Ϫ → 2Cr3ϩ ϩ 7H2O. a gain of 1eϪ.3 Part A Completion 1. AT 9. thus the oxidation number is ϩ4. N is reduced (ϩ5 to ϩ2). Si: F2 8. HNO3 y NO2. g. AT 11. 27. C is oxidized (0 to ϩ4) b. S2Ϫ → S ϩ 2eϪ and 3eϪ ϩ 4Hϩ ϩ NOϪ 3 → NO ϩ 2H2O. NT Part C Matching © Pearson Education. Increase in oxidation number of antimony ϭ ϩ5. C ϩ 2H2SO4 y CO2 ϩ 2SO2 ϩ 2H2O b. Inc. publishing as Pearson Prentice Hall. Increase in oxidation number of sulfur ϭ ϩ2. 4. 4OHϪ ϩ Mn2ϩ → MnO2 ϩ 2H2O ϩ 2eϪ 2eϪ ϩ H2O ϩ H2O2 → H2O ϩ 2OHϪ 3. Reducing agent is chlorine. 2. a. Oxidizing agent is manganese. 8Hϩ ϩ 5Fe2ϩ ϩ MnO4Ϫ y 5Fe3ϩ ϩ Mn2ϩ ϩ 4H2O c. decrease in oxidation number of manganese ϭ Ϫ5. Increase in oxidation number of chlorine ion ϭ ϩ1. decrease in oxidation number of nitrogen ϭ Ϫ3. Sulfur is reduced (ϩ6 y ϩ4). a. decrease in oxidation number of iodine ϭ Ϫ7. 2 e. HNO3 ϩ HI y NO ϩ I2 ϩ H2O Nitrogen is reduced (ϩ5 y ϩ2).06 ϫ 10 mol MnO4Ϫ © Pearson Education. Iodide ion is oxidized (Ϫ1 y 0). All rights reserved. a. b.3 1. Fe y Fe ϩ eϪ 5eϪ ϩ 8Hϩ ϩ MnO4Ϫ y Mn2ϩ ϩ 4H2O b. 2OHϪ ϩ CrO2 ϩ ClOϪ y CrO42Ϫ ϩ ClϪ ϩ H2O Section 20. ϩ2 b. Volume KMnO4 ϭ Initial Volume Ϫ Final Volume ϭ 48. Nitrogen is reduced (ϩ5 y ϩ2). c. thus the oxidation number is Ϫ1. 3H2S ϩ 2HNO3 y 3S ϩ 2NO ϩ 4H2O c. 1 2. Reducing agent is antimony. d. a. d.30 mL 0. The ionic charge on bromine is 1Ϫ. Chloride ion is oxidized (Ϫ1 y 0). 2 b. MnO4Ϫ ϩ 8Hϩ ϩ 5Fe2ϩ y Mn2ϩ ϩ 4H2O ϩ 5Fe3ϩ 3. 2KMnO4 ϩ 16HCl y 2MnCl2 ϩ 5Cl2 ϩ 8H2O ϩ 2KCl Interpreting Graphics 20 1. 2HNO3 ϩ 6HI y 2NO ϩ 3I2 ϩ 4H2O d.. decrease in oxidation number of sulfur ϭ Ϫ2. 2OHϪ ϩ Mn2ϩ ϩ H2O2 → MnO2 ϩ 2H2O f. KMnO4ϩ HClyMnCl2ϩ Cl2ϩ H2O ϩ KCl Manganese is reduced (ϩ7 y ϩ2). 4. 844 Core Teaching Resources . decrease in oxidation number of nitrogen ϭ Ϫ3. ϩ6 d.65 mL Ϫ 23. a. a. S2Ϫ → S ϩ 2eϪ 3eϪ ϩ 4Hϩ ϩ NO3Ϫ y NO ϩ 2H2O d. The end point occurs when the number of equivalents of MnO4Ϫ added equals the number of equivalents of Fe2ϩ originally present in the reaction flask. 0 ϩ1 ϩ5 Ϫ2 ϩ5 Ϫ2 ϩ2 Ϫ2 ϩ1 Ϫ2 ϩ1 ϩ7 Ϫ2 ϩ1 Ϫ1 ϩ2 Ϫ1 0 ϩ1 Ϫ2 ϩ1 Ϫ1 ϩ1 ϩ5 Ϫ2 ϩ1 Ϫ1 ϩ2 Ϫ2 0 ϩ1 Ϫ2 0 ϩ1 ϩ6 Ϫ2 ϩ4 Ϫ2 ϩ4 Ϫ2 ϩ1 Ϫ2 f. Sb ϩ HNO3 y Sb2O5 ϩ NO ϩ H2O Antimony is oxidized (0 y ϩ5). One equivalent is the amount of reducing agent (or oxidizing agent) that can give (or accept) one mole of electrons. 4 c. ϩ3 c. Oxidizing agent is nitrogen.02530 L ϫ ᎏᎏ L Ϫ Ϫ4 Moles MnO4 ϭ 5. Sn2ϩ y Sn4ϩ ϩ 2eϪ 6Hϩ ϩ 6eϪ ϩ IO3Ϫ → IϪ ϩ 3H2O c. signaling the end point of the titration. 6Sb ϩ 10HNO3 y 3Sb2O5 ϩ 10NO ϩ 5H2O e. Oxidizing agent is nitrogen. ϩ6 3. and the solution in the flask turns light purple. When all the Fe2ϩ in the flask is oxidized. a. b. Oxidizing agent is sulfur. 3Zn ϩ 2Cr2O72Ϫ ϩ 28Hϩ y 3Zn2ϩ ϩ 4Cr3ϩ ϩ 14H2O 2ϩ 3ϩ 2. decrease in oxidation number of chromium ϭ Ϫ3. KIO4 ϩ 7KI ϩ 8HCl y 8KCl ϩ 4I2 ϩ 4H2O g. decrease in oxidation number of nitrogen ϭ Ϫ3. Reducing agent is carbon. the next drop of MnO4Ϫ remains unreacted. 6Hϩ ϩ 3Sn2ϩ ϩ IO3Ϫ y 3Sn4Ϫ ϩ IϪ ϩ 3H2O d. c.h.0200 mol Moles MnO4Ϫ ϭ 0. 3 d. Increase in oxidation number of iodine ion ϭ ϩ1. 8Hϩ ϩ 3S2Ϫ ϩ 2NO3Ϫ y 3S ϩ 2NO ϩ 4H2O e.35 mL ϭ 25. Increase in oxidation number of zinc ϭ ϩ2. C ϩ H2SO4 y CO2 ϩ SO2 ϩ H2O Carbon is oxidized (0 y ϩ4). 2OHϪ ϩ Zn ϩ HgO y ZnO22Ϫ ϩ Hg ϩ H2O b. Reducing agent is iodine. 3 f. Increase in oxidation number of iodine ion ϭ ϩ1. Increase in oxidation number of carbon ϭ ϩ4. 13. A decrease is reduction. reducing agent. 2. 4. 20. d D.141 g % Fe in ore ϭ ᎏᎏ ϫ 100% ϭ 4. Matching 1. 10. Matching 1. b 2. 21. 6. Na ϩ1. however.53 ϫ 10Ϫ3 mol Fe2ϩ 1 mol MnO4Ϫ Mass Fe ϭ 2. oxidizing agent b. the sum is equal to the charge on the ion. i g e b 9. An oxidation-number increase is oxidation.80% 2.141 g 1 mol Fe2ϩ 0. The sum of the oxidation numbers of the elements in a neutral compound is zero. Oxidation numbers help keep track of electrons in redox reactions. a. In a polyatomic ion. 15. S reduced. 6. K oxidized. 14.85 g Fe2ϩ ϫ ᎏᎏ ϭ 0.53 ϫ 10Ϫ3 mol Fe2ϩ 55. reducing agent 28. Br2 reduced. 3. c 10. oxidizing agent oxidation-number-change method reduction half-reaction method oxidation number reducing agent half-reaction oxidation-reduction reaction oxidation redox reaction Fe2O3 ϩ CO y Fe ϩ CO2 3 ϫ (ϩ2) ϭ ϩ6 Fe2O3 ϩ 3CO y 2Fe ϩ 3CO2 Half-reaction method: 6Hϩ ϩ Fe2O3 ϩ 6eϪ y 2Fe ϩ 3H2O 3(H2O ϩ CO y CO2 ϩ 2Hϩ ϩ 2eϪ) 6Hϩ ϩ Fe2O3 ϩ 6eϪ ϩ 3H2O ϩ 3CO y 2Fe ϩ 3H2O ϩ 3CO2 ϩ 6Hϩ ϩ 6eϪ Fe2O3 ϩ 3CO y 2Fe ϩ 3CO2 Quiz for Chapter 20 1. e f d g 5. a. Li ϩ1. 4. 18. a 7. Na oxidized. oxidizing agent. 3. S 0 (element) 30. The oxidation number of an element in an uncombined state is zero. Essay 31. 7. b d c b a b 17. 8. 6. 8. d c c c Chapter 20 Test B A. 4. 12. c c a a c b 23. 2Cr ϩ 3Br2 y 2Cr3ϩ ϩ 6BrϪ 29. 9.06 ϫ 10Ϫ4 mol MnO4Ϫ 5 mol Fe2ϩ ϫ ᎏᎏ ϭ 2. a 6. i a c h 9.938 g C. h B. 8. Multiple Choice 11. b 5. c 8. 19. All rights reserved. 3. 2. 25. 7. An oxidation number is assigned to an element in a compound according to a set of arbitrary rules. 16. Al ϩ3. 4. b © Pearson Education. j Answer Key 845 . a 3. Oxidation-number change method: 2 ϫ (Ϫ3) ϭ Ϫ6 ϩ3 Ϫ2 ϩ2 Ϫ2 0 ϩ4 Ϫ2 Vocabulary Review 20 1.. publishing as Pearson Prentice Hall. 22. F Ϫ1 b. d j f a 5. Chapter 20 Test A A. Questions 27. O Ϫ2 c. 26.Moles iron(II) ϭ 5. b 10. 5. The oxidation number of a monatomic ion is the same in magnitude and sign as the ionic charge. 7. 24. Inc. 2. B. Multiple Choice 11. 12. 13. 14. 15. 16. d a a d d b 17. 18. 19. 20. 21. 22. a b c c b b 23. 24. 25. 26. 27. 28. d c d b b a D. Essay 33. Since oxidation is the loss of electrons, it can only occur in the presence of another substance that will accept the lost electrons. The accepting substance gains electrons, and thus, undergoes reduction. In other words, a loss of electrons can only occur if a gain takes place concurrently. C. Questions 29. a. b. c. d. 30. a. b. c. d. K; I; I2; K Na; H; H2O; Na H; Cu; CuO; H2 Mg; Cu; Cu(NO3)2; Mg K2SO4 ϭ ϩ1, ϩ6, Ϫ2 Cu(NO3)2 ϭ ϩ2, ϩ5, Ϫ2 HAsO3 ϭ ϩ1, ϩ5, Ϫ2 MnO4Ϫ ϭ ϩ7, Ϫ2 3 ϫ (ϩ1) ϭ ϩ3 Chapter 20 Small-Scale Lab Section 20.3 Half Reactions, page 655 Analysis HCl HNO3 H2SO4 Zn Bubbles Bubbles Bubbles 31. a. 4HNO3 ϩ 3Ag y 3AgNO3 ϩ NO ϩ 2H2O 1 ϫ (Ϫ3) ϭ Ϫ3 1 ϫ (ϩ2) ϭ ϩ2 Mg Bubbles Bubbles Bubbles b. Br2 ϩ SO2 ϩ 2H2O y H2SO4 ϩ 2HBr 2 ϫ (Ϫ1) ϭ Ϫ2 32. a. HNO2 ϩ HI y I2 ϩ NO ϩ H2O Hϩ(aq) ϩ NO2Ϫ(aq) ϩ Hϩ(aq) ϩ IϪ(aq) y I2(aq) ϩ NO(g) ϩ H2O(l) Ϫ Oxidation: 2I (aq) y I2 ϩ 2eϪ Reduction: 2[2Hϩ(aq) ϩ NO2Ϫ(aq) ϩ 1eϪ y NO ϩ H2O] Ϫ ϩ Ϫ 4H ϩ 2NO2 ϩ 2e y 2NO ϩ 2H2O 4Hϩ ϩ 2IϪ ϩ 2NOϪ 2 y I2 ϩ 2NO ϩ H2O Final: 2HNO2 ϩ 2HI y I2 ϩ 2NO ϩ 2H2O b. K2Cr2O7 ϩ FeCl2 ϩ HCl y CrCl3 ϩ KCl ϩ FeCl3 ϩ H2O ϩ 2K (aq) ϩ Cr2O72Ϫ(aq) ϩ Fe2ϩ(aq) ϩ 2ClϪ 2Kϩ(aq) ϩ Hϩ(aq) ϩ ClϪ(aq) y Cr3ϩ(aq) ϩ 3ClϪ(aq) ϩ Kϩ(aq) ϩ ClϪ(aq) y Cr3ϩ(aq) ϩ Fe3ϩ(aq) ϩ 3ClϪ(aq) ϩ H2O Oxidation: 6[Fe2ϩ y Fe2ϩ ϩ 1eϪ] Reduction: 2Cr6ϩ ϩ 6eϪ y 2Cr3ϩ 6Fe2ϩ ϩ 2Cr6ϩ y 6Fe3ϩ ϩ 2Cr3ϩ Final: K2Cr2O7 ϩ 6FeCl2 ϩ 14HCl y 2CrCl3 ϩ 2KCl ϩ 6FeCl3 ϩ 7H2O Cu No visible reaction No visible reaction No visible reaction © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Fe Bubbles Bubbles Bubbles Figure A 1. Mg is most reactive because it bubbles most vigorously. Cu did not react. The order of reactivity is Mg Ͼ Zn Ͼ Fe Ͼ Cu. 2. H2(g) is the gas produced all the reactions. 3. Mg(s) ϩ 2HCl(aq) → H2(g) ϩ MgCl2(aq) Mg(s) ϩ 2Hϩ(aq) → H2(g) ϩ Mg2ϩ(aq) Fe(s) ϩ 2HCl(aq) → H2(g) ϩ FeCl2(aq) Fe(s) ϩ 2Hϩ(aq) → H2(g) ϩ Fe2ϩ(aq) All are redox reactions because the oxidation number of reactants change. 4. Mg(s) → Mg2ϩ(aq) ϩ 2eϪ Fe(s) → Fe2ϩ(aq) ϩ 2eϪ 5. 2Hϩ ϩ 2eϪ→ H2(g). Mg(s) → Mg2ϩ(aq) ϩ 2eϪ 846 Core Teaching Resources Mg(s) ϩ 2Hϩ(aq) → H2(g) ϩ Mg2ϩ(aq) Section Review 21.2 Part A Completion 1. 2. 3. 4. 5. 6. 7. electric potential electrons cell potential standard hydrogen electrode 0.00 V less spontaneous You’re the Chemist 1. Add a drop of any acid to the damaged part of the penny and notice that only the zinc interior reacts. 2. Many toilet-bowl cleaners and vinegar dissolve metals. Keep products containing acids away from metal pipes and fixtures. Section Review 21.1 Part A Completion 1. 2. 3. 4. 5. 6. 7. electrochemical process electrons voltaic cells salt bridge ions anode cathode Part B True-False 8. NT 9. ST 10. NT 11. NT Part C Matching 12. b 13. d 14. f 15. c 16. a 17. e Part D Problem 18. Oxidation: Mg y Mg2ϩ ϩ 2eϪ Reduction: 2eϪ ϩ Cl2 y 2ClϪ Redox: Mg ϩ Cl2 y Mg2ϩ ϩ 2ClϪ 0 0 E0 cell ϭ Ered Ϫ Eoxid 0 0 0 Ecell ϭ ECl2 Ϫ EMg E0 cell ϭ ϩ1.36 V Ϫ (Ϫ2.37 V) E0 cell ϭ ϩ3.73 V Part B True-False 8. NT 9. AT 10. NT 11. ST Part C Matching © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. 12. g 13. f 14. d 15. b 16. c 17. e 18. a Part D Problem 19. The shorthand notation Mg(s) | MgSO4(aq) || PbSO4(aq) | Pb(s) represents a magnesiumlead voltaic cell. The single vertical lines indicate boundaries of phases that are in contact, and the double vertical lines represent the salt bridge that separates the anode compartment from the cathode compartment. In this electrochemical cell, Mg is oxidized to Mg2ϩ at the anode (the negative electrode) and Pb2ϩ is reduced to Pb at the cathode (the positive electrode). Electrons flow from the anode, through an external circuit (connected to a light bulb or voltmeter), to the cathode. To complete the circuit, sulfate (SO42Ϫ) anions move from the cathode compartment to the anode compartment, and magnesium and sodium cations move from the anode compartment to the cathode compartment. Check students’ diagrams. Section Review 21.3 Part A Completion 1. 2. 3. 4. electrolysis electrolytic cell electrons battery 5. 6. 7. 8. electrolyte hydrogen/oxygen oxygen/hydrogen hydrogen gas Part B True-False 9. NT 10. ST 11. ST 12. AT Part C Matching 13. b 14. d 15. e 16. c 17. a Answer Key 847 Part D Questions and Problems 18. In electrolytic cells, electrical energy is used to bring about a normally nonspontaneous chemical reaction. In a voltaic cell, chemical energy is converted to electrical energy by a spontaneous redox reaction. Electrolytic cells are used in electroplating, in refining metals, and in the production of substances such as sodium hydroxide, aluminum, sodium, and chlorine. Voltaic cells are used in pacemakers, hearing aids, and cameras. 19. DC eϪ eϪ 2. a. E0 cell ϭ Ϫ0.14 V Ϫ (Ϫ2.90 V) ϭ ϩ2.76 V; spontaneous b. E0 cell ϭ ϩ0.80 V Ϫ 1.36 V ϭϪ0.56 V; nonspontaneous c. E0 cell ϭ ϩ2.87 V Ϫ (Ϫ0.76 V) ϭϩ3.63 V; spontaneous d. E0 cell ϭ Ϫ0.28 V Ϫ (Ϫ3.05 V) ϭ ϩ2.77 V; spontaneous e. E0 cell ϭ Ϫ2.93 V Ϫ 0.54 V ϭ Ϫ3.47 V; nonspontaneous Interpreting Graphics 21 1. 2. 3. 4. anode(ϩ) cathode(Ϫ) electrorefining a. The anode(ϩ) of the electrolytic cell should be connected to the positive(ϩ) terminal of the battery. The cathode(Ϫ) of the electrolytic cell should be connected to the negative(Ϫ) terminal of the battery. b. The anode of the electrolytic cell is connected to the cathode of the battery. The cathode of the electrolytic cell is connected to the anode of the battery. Oxidation occurs at the anode, labeled number 1 in the diagram. Reduction occurs at the cathode, labeled number 2 in the diagram. Students should indicate the flow of electrons out of the anode(ϩ) and into the cathode(Ϫ). The voltage should be great enough to oxidize copper metal at the anode and reduce copper(II) ions at the cathode, but not high enough to oxidize other metals at the anode and reduce them at the cathode. The voltage should be greater than 0.34 V but less than 0.44 V. a. gold, silver, and platinum b. Zn2ϩ and Fe2ϩ c. copper Cathode Anode Ag Solution of silver ions Anode (oxidation): Ag(s) y Agϩ(aq) ϩ eϪ Cathode (reduction): Agϩ(aq) ϩ eϪ y Ag(s) Practice Problems Section 21.2 1. a. Cl2(g) + Mg(s) y 2ClϪ(aq) + Mg2ϩ(aq) E0 cell ϭ 1.36 V Ϫ (Ϫ2.37 V) ϭ 3.73 V cathode: Cl2(g) ϩ 2eϪ y 2ClϪ(aq) b. 2Agϩ(aq) ϩ Ni(s) y Ni2ϩ(aq) ϩ 2Ag(s) E0 cell ϭ 0.80 V Ϫ (Ϫ0.25 V) ϭ 1.05 V cathode: Agϩ(aq) ϩ eϪ y Ag(s) c. 2MnO4Ϫ(aq) ϩ 16Hϩ(aq) ϩ 5Cd(s) y 5Cd2ϩ(aq) ϩ 2Mn2ϩ(aq) ϩ 8H2O(l) E0 cell ϭ 1.51 V Ϫ (Ϫ0.40 V) ϭ 1.91 V cathode: MnO4Ϫ(aq) ϩ 8Hϩ(aq) ϩ 5eϪ y Mn2ϩ(aq) ϩ 4H2O(l) d. Br2 ϩ 2Na(s) y 2Naϩ(aq) ϩ 2BrϪ(aq) E0 cell ϭ 1.07 V Ϫ (Ϫ2.71 V) ϭ 3.78 V cathode: Br2(l) ϩ 2eϪ y 2BrϪ(aq) e. MnO2(s) ϩ 4Hϩ(aq) ϩ H2(g) y 2Hϩ(aq) ϩ Mn2ϩ(aq) ϩ 2H2O(l) E0 cell ϭ 1.28 V Ϫ 0.00 V ϭ 1.28 V cathode: MnO2(s) + 4Hϩ(aq) ϩ 2eϪ y Mn2ϩ(aq) ϩ 2H2O(l) 5. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. 6. 7. 8. Vocabulary Review 21 1. voltaic cell 2. fuel cell 3. electrochemical cell 4. electrochemical process 5. cathode 6. reduction potential Solution: aluminum 848 Core Teaching Resources 9. 3. Matching 1. 19. 8. 4. 23. In the electrolytic cell. f 10. 15. 3. 21.. 24. 26. d b c c c 16. 6. 15. AT 29. c g j e 9. NT 27. 11. True-False 28. 28. a. 12. The reduction potential of a half-cell is a measure of the tendency of a given halfreaction to occur as a reduction. 35.28 V) E0 cell ϭ Ϫ1. 4. 22. 18.76 V 0 0 E0 cell ϭ Ered Ϫ Eoxid 0 Ecell ϭ Ϫ0. b 10. 14. 34. 8. Additional Questions 33. the cathode is positive and the anode is negative. 12. 2[Al3ϩ(aq) ϩ 3eϪ y Al(s)] E0 ϭ Ϫ1. publishing as Pearson Prentice Hall. 27. 14. 12. 17. so zinc metal is oxidized when paired with the standard hydrogen half-cell. 22. b h d c 5. 8. h d f i 5. d a d c b c 23.66 V 2ϩ Ϫ 3[Co (aq) ϩ 2e y Co(s)] E0 ϭ Ϫ0. 2. All rights reserved.38 V This reaction is not spontaneous. 5. The negative value means that the tendency for zinc ions to be reduced is less than that of hydrogen ions to be reduced. Chapter 21 Test A A. In the voltaic cell. c d b c d a 17. a d c a a 21. Inc. 16. a B.25 V Zn2ϩ(aq) ϩ 2eϪ y Zn(s) E0 ϭ Ϫ0. 6. In both voltaic and electrolytic cells.28 V 0 0 E0 cell ϭ Ered Ϫ Eoxid E0 cell ϭ Ϫ1.51 V This reaction is not spontaneous. the anode is positive and the cathode is negative. eϪ eϪ Salt bridge B. ST Answer Key 849 . Essay 32. NT 30.25 V) E0 cell ϭ Ϫ0. NT 30. 13. 17. 18. 15. b. 16. Multiple Choice 11. 25. True-False 26. NT AT AT ST NT AT 13. C. oxidation occurs at the anode and reduction occurs at the cathode. Multiple Choice 11. NT AT NT AT AT NT F. 20. AT 33. 4. 7. 3. 6. 2. NT D. 25. Question 31. Matching 1. 20.76 V Ϫ (Ϫ0. AT 32. b a b c d Anode Cathode E. AT 31. a C. 24.66 V Ϫ (Ϫ0. c c a a c Chapter 21 Test B A. 7. 10. NT AT NT AT ST NT 7. 19. NT 29. Ni2ϩ(aq) ϩ 2eϪ y Ni(s) E0 ϭ Ϫ0. 2. 13.Quiz for Chapter 21 1. 18. NT © Pearson Education. j g i e 9. 14. 34 V Ϫ (Ϫ2. while the flow of electrons in a voltaic cell is caused by a spontaneous chemical reaction. The half-reactions are: Oxidation: Al(s) y Al3ϩ(aq) ϩ 3eϪ E0 ϭ Ϫ1.17 V Since the standard cell potential is negative.71 V) ϭ ϩ3. Additionally. © Pearson Education. Y appears above Z.53 V 38. Z appears above X in the activity series.71 V 850 Core Teaching Resources .80 V Reduction: Mg2ϩ(aq) ϩ 2eϪ y Mg(s) E0 ϭ Ϫ2. X.46 V E. the redox reaction will be spontaneous. Reaction 3: since Y is oxidized and Z2ϩ is reduced. The elements should be listed as follows: Y.66 V Reduction: Zn2ϩ(aq) ϩ 2eϪ y Zn(s) E0 ϭ Ϫ0. in an electrolytic cell the flow of electrons is being pushed by an outside source such as a battery. and oxidation occurs at the anode.80 V) E0 cell ϭ Ϫ3. the reverse is true in an electrolytic cell—the anode in an electrolytic cell is the positive electrode and the cathode is negative. In both voltaic and electrolytic cells. publishing as Pearson Prentice Hall. while the anode is the negative electrode and the cathode is the positive electrode in a voltaic cell. the two half-cell reactions are as follows: 2[Al(s) y Al3ϩ(aq) ϩ 3eϪ] 3[Pb2ϩ(aq) ϩ 2eϪ y Pb(s)] Net: 2Al(s) ϩ 3Pb2ϩ(aq) y 2Al3ϩ(aq) ϩ 3Pb(s) 0 0 E0 cell ϭ Ered Ϫ Eoxid E0 cell ϭ Ϫ0. Additional Questions 37.80 V Ϫ (ϩ0. Al is oxidized and Pb is reduced. The half-reactions are: Oxidation: Na(s) y Naϩ(aq) ϩ eϪ E0 ϭ Ϫ2.34 V) E0 cell ϭ ϩ0.34 V 0 ϭ E0 red Ϫ Eoxid ϭ ϩ0. Reaction 1: since Z is oxidized and X2ϩ is reduced. reduction occurs at the cathode.37 V Ϫ (ϩ0. Thus. a.76 V Ϫ (Ϫ1. Essay 36. Reduction: Cu2ϩ(aq) ϩ 2eϪ y Cu(s) E0 ϭ ϩ0. and W.37 V 0 0 Ecell ϭ E0 red Ϫ Eoxid E0 cell ϭ Ϫ2. The half-reactions are: Oxidation: Ag(s) y Agϩ(aq) ϩ eϪ E0 ϭ ϩ0. W should appear below X.90 V Since the standard cell potential is positive. Since Al is above Pb in the reduction potential table. Question 34.. However. F.D. Inc. Cu(s) y Cu2ϩ(aq) ϩ 2eϪ 2Agϩ(aq) ϩ 2eϪ y 2Ag(s) Net: Cu(s) ϩ 2Agϩ(aq) y Cu2ϩ(aq) ϩ 2Ag(s) 0 0 E0 cell ϭ Ered Ϫ Eoxid E0 cell ϭ ϩ0. the redox reaction will be spontaneous. the redox reaction will be nonspontaneous. c. All rights reserved. 39.66 V) E0 cell ϭ ϩ0.05 V Since the standard cell potential is positive.66 V) E0 cell ϭ ϩ1.76 V 0 0 Ecell ϭ E0 red Ϫ Eoxid E0 cell ϭ Ϫ0. Reaction 2: since W is not oxidized in the presence of X2ϩ. b. electrons flow from the anode to the cathode through the external circuit. Z. E0 cell E0 cell E0 cell eϪ eϪ Salt bridge Cu Anode Cu(NO3)2 (aq) Ag Cathode AgNO3 (aq) 35.13 V Ϫ (Ϫ1. AT 14. ϩ ᎏ Anode: H2O → ᎏ1 ϩ 2eϪ 2 O2(g) ϩ 2H Bubbles. 2H2O ϩ 2eϪ → H2(g) ϩ 2OHϪ H2O → O2(g) ϩ 2H ϩ 2e 1 ᎏᎏ 2 Na2SO4 ϩ BTB Anode area turns yellow. ST 15. ST 13.. 8. publishing as Pearson Prentice Hall. page 684 Analysis H2O No visible reaction Na2SO4 Bubbles at both the anode and cathode Figure A 1. a 18. 4. straight branches alkyl longest parent © Pearson Education. 3. at the anode. blue BTB Anode: 2ClϪ → Cl2(aq) ϩ 2eϪ Yellow solution. 2. Pure water has too few ions to carry an electric current. cathode area turns blue Eo 0. All rights reserved. Inc.82 V Ϫ0. 5. KI KI ϩ starch KI ϩ BTB Bubbles at Bubbles at Bubbles and the cathode. 16 b. The bubbles are H2(g) and the blue BTB solution indicates the presence of OHϪ ions.Chapter 21 Small-Scale Lab Section 21.3 Electrolysis of Water. a. which carry an electric current through the solution. You’re the Chemist 1.00 V Ϫ0. NaCl: Cathode: 2H2O ϩ 2eϪ → H2(g) ϩ 2OHϪ Bubbles.2-dimethylbutane 23. CuSO4: Cathode: Cu2ϩ ϩ 2eϪ → H2(g) ϩ 2OHϪ Copper plates out. at the anode. Yellow soln at the anode. The yellow solution is I2(aq). which is black in the presence of starch. the cathode. 2. e 19. blue BTB Anode: 2BrϪ → Br2(aq) ϩ 2eϪ Yellow solution.54 V ϩ Ϫ 2I → I2(aq) ϩ 2e Ϫ Ϫ IϪ is more likely to oxidize (lose electrons) than H2O because it has a more favorable (more positive) Eo value. 9. 3.1 Ϫ H2O → O2(g) ϩ 2H ϩ 2e 1 ᎏᎏ 2 ᎏ H2O → H2(g) ϩ ᎏ1 3 2 O2(g) ϩ 2H 2O ᎏ H2O → H2(g) ϩ ᎏ1 2 O2(g) ϩ Ϫ ᎏ 3H2O → H2(g) ϩ ᎏ1 ϩ 2Hϩ 2 O2(g) ϩ2OH Part A Completion 1. Sodium sulfate is an electrolyte. 10. The bubbles are O2(g) and Hϩ ions in solution impart the yellow color to the BTB solution. Ϫ 2H2O ϩ 2e → H2(g) ϩ 2OHϪ 2. 7. yellow BTB 3. blue soln at Yellow soln Black solution the cathode. carbon organic hydrocarbons four single 6. 5. b 21. AT Part C Matching 16. It dissociates into ions in solution. NT 12. f 20. KBr: Cathode: 2H2O ϩ 2eϪ → H2(g) ϩ 2OHϪ Bubbles. 4. Part B True-False 11. 2. Section Review 22. d 17. 16 24. Figure B The bubbles are H2(g) and the blue solution indicates of the presence of OHϪ ions. c Part D Questions and Problems 22. CH3 @ CH3 CH3 CH2 @ @ @ CH3 2 C 2 CH 2 CH2 2 CH 2 CH2 2 CH2 2 CH3 @ CH3 Answer Key 851 . trans stereoisomer asymmetric optical mirror superimposed © Pearson Education. ST 12. ST 15. 9. 12. p Section 22. 6. 11. ST 13. CH3 2 CH2 2 CH2 2 CH2 2 CH3 CH3 2 CH 2 CH2 2 CH3 @ CH3 CH3 @ CH3 2 C 2 CH3 @ CH3 26. NT 16. 2. ST Part B True-False 13. f 22. Carbon 2 is the asymmetric carbon. 3. molecular structures butane properties Geometric cis 7.. AT 16. 7. 2. 9. 27. ST 15. 2. unsaturated double triple longest double 6. b Part D Questions and Problems 18. CH3 CH2 H % ^ C3C ^ % CH2 CH3 H cis-3-hexene @ CH3 CH3 CH2 CH3 @ @ @ @ CH3 2 CH 2 CH2 2 CH 2 C 4 C 2 CH 2 CH 2 CH 2 CH3 Section 22. h 18. m para. 8. 4-methyl-1-hexene CH3 21. 5. CH2 3 CH 2 CH 2 CH3 @ 852 Core Teaching Resources . 3.2 Part A Completion 1. c 21. 5. AT 14. 6. ST Part C Matching 14.4. 2. Methylbenzene xylenes ortho. Inc.4 Part A Completion 1. 5. CH3 CH2 H % ^ C3C ^ % CH2 CH3 H trans-3-hexene Part B True-False 10. 3. 3-methyl-2-hexene 19. b Part C Matching 17. a 23. d 16. b Part D Questions and Problems 21. 4.3 Part A Completion 1. e 19. alkane -ene double bond -yne Part D Problems 25.3.Section 22. 4. 10. AT Part C Matching 17. Part B True-False 12. d 18. NT 14. All rights reserved. o meta. a 17. 11. 4. c 15. a 24. NT 11. 8. AT 13. publishing as Pearson Prentice Hall. 10. c 20.5-tetramethylnonane 20. cyclic aromatic six hydrogen double resonance 7. 8. 2 CH2CH2CH3 22. g 19. d 20. 9. 5. c Practice Problems Section 22. 2.3 1. 8. Carbon 3 is the asymmetric carbon. 3. CH3 2 CH 2 CH 2 CH 2 CH 2 CH2 2 CH2 2 CH2 2 CH3 @ @ @ @ CH3 CH3 CH3 CH3 4. AT 14. cis-2-pentene 2.4-dimethyl-2-hexene 2. 2. NT 13. 1-ethyl-3-methylbenzene 2. c 19. a.2 1. 4.3. a. a. CH2 3 CH 2 CH2 2 CH 2 CH3 @ CH3 c. CH3 @ CH 4 C 2 C 2 CH3 @ CH3 Section 22. publishing as Pearson Prentice Hall. 19 c. 6. 9. 1. 10.3. 1-pentyne: CHϵC—CH2—CH2—CH3 2-pentyne: CH3—CϵC—CH2—CH3 3-methyl-1-butyne: CH 4 C 2 CH 2 CH3 @ CH3 4.1 © Pearson Education. d 17. a. a Part D Problems 20. CH3 2 CH2 2 2 CH2 2 CH3 Answer Key 853 . ST Part C Matching 15. d 5. CH3 @ CH3 2 CH 2 C 2 CH2 2 CH3 @ @ CH3 CH2 @ CH3 Section 22. @ CH ^ % CH2 CH2 @ @ CH2 CH % ^ % CH2 CH3 b.4 1. 1. 5-phenyl-2-hexene CH3 3. a. boiling point lignite bituminous anthracite aromatic Part B True-False 11.5-trimethyloctane 2. Inc. CH3 H % ^ C3C ^ % CH2 2 CH2 2 CH2 2 CH3 H 4. 3-ethyl-2. CH3 2 CH2 2 CH 2 CH 2 CH2 2 CH3 @ @ CH2 CH2 @ @ CH3 CH3 c.4-dimethyl-1-pentyne 3. cyclooctane b. trans-6-methyl-3-heptene 3.5.5 Part A Completion 1. H % ^ C3C ^ % H b. 3. 5-ethyl-3. 2C5H12(l) ϩ 11O2(g) → 10CO(g) ϩ 12H2O(g) 21. b 16.3-diethylbenzene or m-diethylbenzene Section 22.. 2C6H6 ϩ 15O2 → 12CO2 ϩ 6H2O Section 22. 7. heptane: CH3—CH2—CH2—CH2—CH2—CH2—CH3 octane: CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH3 5. e 18. CH2 3 CH 2 CH 2 CH2 2 CH 2 CH3 @ @ CH3 CH3 b. natural gas coal methane aliphatic distillation 6.5-tetramethylheptane 3. NT 12. a. All rights reserved.23. 1. substituent 3. 2.9ЊC respectively. NT 6. Gasoline is composed of alkanes with five to twelve carbon atoms. the boiling points of trans -2-pentene and cis -2pentene are 36. 3. 3. Essay 28.Section 22. Inc. ethane. 13. j 10. Like all isomers. a c c b C. homologous series 5. f 8. Number the carbons in the chain. d 6.. c 7. NT E.) B. c c d a 20. Then draw the longest carbon chain to form the parent structure. 5. the boiling points of compounds D and F are not expected to be the same. All rights reserved. 2. Carbon 3 is asymmetric. b Interpreting Graphics 22 1. Attach the substituents to the numbered chain at the proper positions. a small but measurable difference. 19. 2C8H18 ϩ 25O2 y 16CO2 ϩ 18H2O 2. Multiple Choice 11. Identify the substituent groups. CH3 @ H H H H H H @ @ @ @ @ @ @ @ H 2 C 2 C 2 C 2 C 2 H and H 2 C 2 C 2 C 2 H @ @ @ @ @ @ @ H H H H H H H © Pearson Education. 3. 2-methyl-2-phenylbutane 25. 6. 3. AT Chapter 22 Quiz 1. stereoisomers 8. 11. 8. 4. Vocabulary Review 22 1. cracking 6.7-trimethyl-3-octene 12. cis-2-pentene and trans-2-pentene E. saturated compounds Solution: hydrocarbons CH3 @ CH3 CH2 CH3 @ @ @ CH3 2 C 3 CH 2 C 2 CH2 2 CH 2 CH3 27. 2-phenylbutane. 23. Matching 1.3-dimethylpentane D and F. 2. ST 10. 2. 21. Kerosene is composed of alkanes with twelve to fifteen carbon atoms. Additional Problems 29. Find the root word (ending in -ane) in the hydrocarbon name. 4. cis configuration 4. (In fact. 15. they are different compounds with different properties. NT 7. g i a h 5. C-2 is asymmetric 22 10 Compounds A (2-phenylpropane) and E (2-phenylbutane) are aromatic compounds. 12. and butane. Add hydrogens as needed so that each carbon added has four bonds. 2C6H14(l) ϩ 19O2(g) → 12CO2(g) ϩ 14H2O(l) D. c c a b d 16. straight-chain alkanes— methane. Thus. H H @ H @ @ @ @ @ @ 2C2C2C2C2C2H @ @ @ @ @ H H H H H Chapter 22 Test A A. Natural gas contains mainly low molar mass.3ЊC and 36. Problems 24. The refining process yields fractions that differ with respect to the length of the carbon chains. ST NT NT AT 5. arene 7.5 1. 22. 14. 7. alkynes 2. 18. Compounds D and F are geometric isomers. 8. 854 Core Teaching Resources . propane. 4. 26. 17. e 9. A C 3. 2.5. 9. 4. publishing as Pearson Prentice Hall. AT 9. propyl groups. h 6. 19. Inc.3-dimethyl-3-phenylhexane b. e 2-methylbutane CH3 c. 27. 2. 32. 2. d 8. 45. Essay 43. CH3 CH3 @ @ CH3 2 CH 2 CH2 2 CH2 2 CH3 CH3 CH3 @ @ CH3 2 C 3 C 2 C 2 CH2 2 CH3 @ CH2 @ CH3 F. AT 34. 24. True-False 29. publishing as Pearson Prentice Hall. respectively. NT 40. j 10. 3. 12. CH3 @ CH2 CH3 CH3 @ @ @ CH3 2 C 4 C 2 CH 2 C 2 CH 2 CH 2 CH3 @ @ CH2 CH2 @ CH3 F. ethylbenzene b. AT 36. a. 31.5-diethyl-4-methylheptane b. an ethyl. True-False 31. ST D. Answer Key 855 . 6.30. b d d d b c 23. CH3 2 CH 2 CH2 2 CH3 @ CH3 Chapter 22 Test B A. 28. CH3 CH ^ ^ ^ ^ 3 C3C C3C % % % % H CH3 H H trans -2-butene cis-2-butene B. Carbon 3 is asymmetric because there are four different groups attached to it—a methyl. 3-ethyl-2. c 9. 30.7-diethyl-2. 3. a. 2-hexyne C6H10 CH3—CϵC—CH2—CH2—CH3 The number of hydrogen atoms decreases when carbon atoms form double or triple bonds in the alkene and alkyne. CH3 H CH3 H % ^ C3C ^ % % ^ C3C ^ % CH3 @ CH 2 CH2 2 CH3 H H CH 2 CH2 2 CH3 @ CH3 c. Multiple Choice 11.4. 22.2-dimethylpropane H CH3 42. b i f g 5. a. 14. 21. a.8-dimethyl-5-propyl-3decyne 41. NT 37. All rights reserved. 13. a. c d a d b d E. Problems 39. a 7. plus a 3carbon branched chain. C.4-trimethyl-2-pentene c. 2-hexene C6H12 CH3—CHCH—CH2—CH2—CH3 c. The number of hydrogen atoms is at a maximum in the unsaturated alkane. 25. 15. 26.. CH3—CH2—CH2—CH2—CH3 pentane b. 16. Matching 1. hexane C6H14 CH3 —CH2—CH2—CH2—CH2—CH3 b. AT 32. AT 34. © Pearson Education. 20. AT AT NT AT 33. 4. NT 38. d c b b c d 17. AT 33. AT 35. 18. Additional Problems 44. @ CH3 2 C 2 CH3 @ CH3 2. 2-dimethylbutane 2.Chapter 22 Small-Scale Labs Section 22. CH CH CH CH 3 2 2 3 butane CH3CHCH3 CH3 methylpropane Section Review 23. Inc. To find the number of hydrogen atoms on any carbon of a line-angle formula.3 Hydrocarbon Isomers. 7.3-dimethylbutane 856 Core Teaching Resources . 2. publishing as Pearson Prentice Hall. page 708 Analyze 1. 2-dimethylpropane © Pearson Education. substitution hydrogen bromine alcohol salt Part B True-False 11. count the number of lines drawn to any point and subtract from four.1 Part A Completion 1. NT 13. AT 12. 8. 2. 3. 10. AT 3-methylpentane 2. ST 14. H H H H H @ @ @ @ @ H2C2C2C2C2C2H @ @ @ @ @ H H H H H pentane 2. 2-methylpentane You’re the Chemist 1. All rights reserved. 9. 5. 3. 4.. hexane functional reactive/functional alkenes alkynes Halocarbons 6. H H CH2CH2CH3 1-pentene CH3 ^ ^ C3C % % H H @ ^ C3C % % H H H H @ @ @ @ H2C2C2C2C2H @ @ @ @ @ H H H @ H2C2H @ H 2-methylbutane H CH2CH3 trans-2-pentene CH3 ^ ^ C3C % % CH2CH3 H H cis-2-pentene H ^ ^ C3C % % CH3CH2 H 2-methyl-1-butene CH3 H ^ ^ H C ^ % ^ C3C CH3 % % H H 3-methyl-1-butene H ^ C 3 C 2 CH3 % @ H CH2CH3 2-ethyl-1-propene CH3 H @ H2C2H @ H2C2C2C2H @ H2C2H @ H 2. d 20. AT Answer Key 857 . 7. b 21. polyesters 6. AT 11. tertiary b. 6. a. b 22. AT 9. oxygen double ketones/carboxylic acids ketones/carboxylic acids aldehyde carboxylic acid formaldehyde carboxylic acids esters propanol oxidation–reduction potassium dichromate Part D Problems 19. NT 17. length Part B True-False 7. 3–hexanone Section Review 23. AT 16. 6. H H @ @ CH3 2 C 2 C 2 CH3 @ @ H H b. OH O O @ # # Cr O K 2 2 7 oxidation R 2 C 2 H uuuy R 2 C 2 H uuuy R 2 C 2 OH –2H H2 SO4 @ H alcohol aldehyde carboxylic acid OH O @ # oxidation R 2 C 2 R uuuy R2C2R –2H @ H alcohol ketone Part C Matching 17. 4. NT 16. a. d 16. a. 20.4 Part A Completion 1. Condensation 4. c Section Review 23.3 Part A Completion 1. NT 10. d. polymer 2. 11. c. b 17. 5. NT Part D Questions and Problems 23. 3. a 19. 4. Inc. All rights reserved. 8. primary 23.. AT 15. 9. c 19. a 18. ST 8. 3. NT 15. a Part B True-False © Pearson Education. d 20. 5. hydrogenation tertiary 10. hydration primary 8. Alcohols 7. lower Part B True-False 14. publishing as Pearson Prentice Hall. hydroxyl carbonyl carbonyl carboxyl CH3 @ ^ Cl b. c Part D Problems 22.Part C Matching 15. ethers hydrogen bonding 12. 13.2 Part A Completion 1. AT Part C Matching 18. Polyethylene 5. 12. b. alkane secondary 11. water secondary 9. e 21. 2. a. 2. Addition 3. OH H @ @ CH3 2 C 2 C 2 H @ @ H H 24. Cl @ CH3 C CH2 CH2 CH3 @ Cl Section Review 23. AT 14. 10. e 18. a. Dacron™ is one example of a polyester. a.3-butanediol is expected to be most soluble due to its two 2 OH groups. chloroethene (vinyl chloride) CH3 3. a. b. b. dipropyl ether: CH3 CH2 CH2 2 O 2 CH2 CH2 CH3 2-methyl-1-butanol: Part D Questions and Problems 17. e. Inc. secondary ethylphenyl ether 3-methyl-1-butanol. c. 2. e 16. a. CH3 CH2 CH 3 CH2 ϩ HCl uy CH3 CH2 CH CH3 @ Cl @ @ 18. a.Part C Matching 12. substitution Section 23. CH3 CH2 CH2 CH2 OH K2Cr2 O7 uuuy H2 SO4 5. 3. which can form hydrogen bonds with water. d 15. CH3 CH ^ 3 O 2 CH ϩ H2O % CH3 O # CH3 CH2 CH2 C % OH b. c. ϩ Br2 uuuy catalyst ^ Br ϩ HBr 4.1 1. ether c. 3. Polyesters are polymers consisting of many repeating units of dicarboxylic acids and dihydroxy alcohols joined by ester bonds.. x H H % ^ C3C uuy H ^ % H H CH2 2 CH2 H x CH3 CH3 CH2 C CH2 OH H 2. All rights reserved. a 2. benzaldehyde 2-butanone 3-methylpentanoic acid ethyl butanoate 3-phenyl-2-propenal ethanal (acetaldehyde) propane 1-butanol 2-pentanone octanoic acid 1-butene or 2-butene O # % CH 2 OH ϩ CH3 CH2 CH2 C % ^ OH CH3 O # uy CH3 CH2 CH2 C % Hϩ b. a. oxidation-reduction. primary 4. 1-bromo-1-chloroethane c. a. ^ Br @ CH2 CH3 b. a.3 1. publishing as Pearson Prentice Hall. addition b. c 14. CH3 CH2 CH2 2 CH 2 CH 2 CH2 CH2 Br @ @ CH3 CH3 © Pearson Education. b. c. 1–butanol is oxidized to butanoic acid. hydroxyl 2. halogen b. c.3-butanediol: CH3 2 CH 2 CH 2 CH3 @ @ OH OH 2. carboxyl d. d. 2-butanol. esterification b. CCl4 ϩ 4HCl 4. primary 1-pentanol. CH3 % CH 2 CH2 2 CH2 2 OH ϩ NaBr ^ b.2 1. a. 858 Core Teaching Resources . a. They are all halocarbons. Section 23. d. m-bromobenzene b. a. Practice Problems Section 23. b 13. b. a. 0. publishing as Pearson Prentice Hall. Tertiary alcohols. O # C2 O # 2 C 2 O 2 CH2 CH2 2 O x Vocabulary Review 23 1.766 Answer Key 859 .000. 5. 10. a. which indicates a reaction between ethanol.157 minϪ1 k 0. propene (propylene) CH3 @ xCH2 3 CH polypropylene CH3 @ CH2 2 CH x b.157 minϪ1 ϭ ᎏᎏ 2.465. Polyethylene terephthalate (PET) is formed from the condensation of terephthalic acid and ethylene glycol. 2. 6.4 1.627. c 8. 4. 14. 9.304. Because the repeating units are joined by ester bonds.) 2. 3. d 6. CH3CH2OH uuuuy CH3CHO H2 SO4 Quiz for Chapter 23 1.Section 23.600 0. H @ ϩ I2 y 2. j 9. PET is a polyester. 13. 8. NT AT ST AT AT 7. Rate ϭ ᎏ ᎏ ϭ k ϫ [CH3CH2OH] ⌬t 4. is used to coat nonstick cookware and to make bearings and bushings in chemical reactors. hydrogenation 7. 15. e i f g 5. Interpreting Graphics 23 1. b K Cr O ⌬[CH3CH2OH] b. The data in Table 2 show no change even after five minutes. such as 2-methyl-2-propanol. also known as Teflon™.800 0. Answers will vary slightly.000 c. 4.000 0. slope ϭ 0. Log (absorbance) 1. Only primary and secondary alcohols are oxidized by dichromate ion.140. functional group 2. One molecule of water is lost for each bond formed. h 7. This tertiary alcohol serves as a negative control Investigators use negative and positive controls to check that a chemical assay is functioning properly. 0. Table 1 shows a change in absorbance values with time. 0. 16. hydration reactions 6. Matching 1. 0. ketones Solution: aspirin © Pearson Education..400 0. and the oxidizing agent. 3.362 minϪ1 Polypropylene is used extrensively in utensils and containers. Inc. 2 2 7 3. Polytetrafluoroethene. 0. are not expected to react. alcohols 5. a.303 k ϭ 0. a 10. 2. aryl halides 3. I @ ϩ HI NT NT AT NT ST 12. AT AT AT AT ST Chapter 23 Test A A. a primary alcohol. (The slight fluctuation is due to random electronic noise in the instrument. substitution reaction 4. 0. All rights reserved.200 Time (min) 0 1 2 3 4 5 tetrafluoroethene xCF2 3 CF2 polytetrafluoroethene (PTFE) CF2 2 CF2 x 0. 11. 25. 3. 18. called monomers. a. iodobenzene D. d d c b c b 23. Ethylene glycol is an alcohol with both a high boiling point and a low freezing point due to intermolecular hydrogen bonding. a. d a c c c b C. Ethylene glycol is soluble in water. 24. 3–bromohexane c. c C. react in a head-to-tail fashion. high strength-to-weight ratio. and freezes at a temperature lower than water alone. j 6. 21. the resulting mixture boils at a temperature higher than water alone. c.B. 27. 860 Core Teaching Resources . 30. and synthetic fibers. In addition polymerization. 20. 2. b. 27. Polymers are large. 16. d c c c c Chapter 23 Test B A. c. 19. 3–chloro–2–methylpentane 2. In condensation polymerization. Problems 29. i a g h 5. 4. 30.. Inc. 22. b. ether alcohol f. monomers with two functional groups. H H H H @ @ @ @ H 2 C 2 C 3 C 2 C 2 H ϩ HOH @ @ H H H H H H @ @ @ @ y H 2 C 2 C 2 C 2 C 2 H` @ @ @ @ H H OH H b. 13. 28. 20. 19. b c a b d a 17. 22. When ethylene glycol is added to the water in a car radiator. polymers have many commercial uses such as packaging. c. 18. Multiple Choice 11. Essay 32. 21. b 7. ϩ I2 y ^I ϩ HI D. 15. 31. 26. ϩ Cl2 uuy 29. ketone CH3CH2I ϩ KOH → CH3CH2OH ϩ KI ethanol CH3 — CH2—CH CH—CH2—CH2—CH3 ϩ HBr → CH3—CH2—CHBr—CH2—CH2—CH2—CH3 Cl @ ϩ HCl 2-butanol chlorobenzene/phenyl chloride R—X R—O—R O # R2C2O2R O # R 2 C 2 OH © Pearson Education. ethylene glycol protects against boiling in summer and freezing in winter. publishing as Pearson Prentice Hall. c d d c c a 23. carboxylic acid ester e. e 10. 16. unsaturated monomers. are joined to one another. d 9. c a b c c c 17. such as alkenes. catalyst B. a. 12. such as dicarboxylic acids and dihydroxy alcohols. a. 13. 12. 26. a. b. chain-like molecules formed by the covalent bonding of repeating smaller molecules. b. Problems 28. Thus. Essay 31.3–dimethyl–2–butanol butanal 2–hexanone propyl ethanoate aldehyde d. Multiple Choice 11. Because of their malleability. a. b. Matching 1. d. insulation. 25. 14. e. f 8. and durability. 24. d. 15. All rights reserved. 14. 10. Carbohydrates energy cellulose monosaccharides disaccharides 6. AT 16. Starches are a source of energy for plants. 4. The polymer is a gel-like liquid which is very viscous. It wiggles. OH 4. 9. 11. b 18. 5. ϪOH. 9. 22. NT Part C Matching OH OH OH RO OH RO B Ϫ 17. NT 13. AT 13. Inc. 6. 1. AT 14. Plants store the excess chemical energy in carbon compounds. 4. squirms. nucleus Sunlight Photosynthesis oxygen Part B True-False 12. All rights reserved. Through experimentation. d Part D Questions 21. The chain is like a polymer because it contains many repeating units linked end to end. c OR ϩ 4HOH OR 19. NT 12. 10. Chloroplasts contain the biological molecules necessary for the conversion of solar energy into chemical energy. These oxidation reactions take place in mitochondria. 7.2 Part A Completion You’re the Chemist © Pearson Education. glucose and fructose 23. O B O O O O O B O O Section Review 24. students are able to produce an amazing variety of polymers with different properties. 2. 3. 8.4 Polymers. publishing as Pearson Prentice Hall. Like animals. The hydroxyl group. a 20. page 753 Analysis 1. Cellulose is used by plants to construct cell walls that are hard and rigid. they meet their energy demands by breaking down these stored compounds. 5. 2–3.1 Part A Completion 1. c 19.. Answer Key 861 . a 20. AT 14. and oozes. AT 15. It will not hold its shape like a solid and will flow slowly if left to stand. 3.Chapter 23 Small-Scale Lab Section 23. e 17. ST Section Review 24. 7. b 18. 2. polysaccharide starch glucose Glycogen liver Part B True-False 11. d OH ϩ 4R OH Part D Question 21. 2. AT 15. prokaryotic/eukaryotic prokaryotic/eukaryotic bacteria green plants organelles Mitochondria lysosomes Part C Matching 16. 5. 1. The rings that link two chains together are like the borate ion that cross links polymer chains. HO B HO Ϫ 8. c Section Review 24. 2. water protein catalysts enzymes Part D Questions 21. 8. or cytosine adenine. a 16. 4. 12. guanine. d Part D Question 24. Thus. thymine. NT 14. b 20. 10.. 3. 5. 7. c 20. The cleansing action of soaps relies on this physical property. 6. 10. waxes coat the skin. 11. AT Part C Matching 19.3 Part A Completion 1. 4. Mutations are random changes in the sequence of nucleotides in a DNA molecule. 2 COOH @ @ 5 CH CH 2 2 4 1 @ 1 @ H2N 2 C 2 C 2 NH 2 C 2 C 2 O 2 CH3 @ @ # 3 # H O H O Section Review 24. AT 15. AT 17. it may stop production of the specified protein or cause production of a protein with an altered amiono acid sequence. 14. nucleotide deoxyribonucleic acid ribonucleic acid proteins nitrogen base adenine. ST 11. 8. 11. e 21. both types of lipids can interact with polar and nonpolar phases simultaneously. the ability of the Part B True-False 12. c 22. more often. Part B True-False 10. All rights reserved. a 23. amino acid side-chain group side-chain group peptide peptide 6. Inc. guanine. which keep these structures pliable and waterproof. or substitutions of one or more of the nucleotides. 4.4 Part A Completion 1. guanine. or cytosine adenine. lipid not soluble/insoluble triglycerides Triglycerides Saponification glycerol Phospholipids hydrophilic/polar hydrophobic/nonpolar lipid bilayer Cell membranes © Pearson Education. 5. Sometimes the change is beneficial. thymine. 3. 13. AT 16. AT 13. hair. b 19. 3. ST 18. 5. thymine. 8. 7. 7. ST Part C Matching 17. a 862 Core Teaching Resources . d 18. 6. and feathers. 9. 9.5 Part A Completion 1. b 15. AT Part C Matching 14. publishing as Pearson Prentice Hall. Wax coats on the surface of plant leaves protect against water loss and attack by microorganisms. or cytosine adenine. Mutations may arise from additions. or cytosine uracil double helix hydrogen bonds thymine cytosine Part D Problem 17. AT 13. In animals. Part B True-False 15. thymine. guanine. deletions. 22. When a mutation occurs within a gene. NT 12. 2.Section Review 24. The molecules of both types of lipids have hydrophilic and hydrophobic ends. 2. 9. AT 16. 7. a mutation in the peptide chain of hemoglobin reduces its ability to transport oxygen. 1. Carbohydrates are found in the cytoplasm and are attached to the extracellular surfaces of membrane-bound proteins. The cells of all other organisms are eukaryotic. c 19. cytoplasm 6. The free energy of ATP hydrolysis is used to drive many nonspontaneous biological reactions. Interpreting Graphics 24 Part A 1. Prokaryotic cells are the cells of bacteria. 2. which are present in eukaryotic cells such as the one depicted in panel b. is found primarily in the nucleus of eukaryotic cells and in the cytoplasm of prokaryotic cells. a Part D Question 21. Skin cells are less active. Inc. People with this mutation have a molecular disease called sickle cell anemia. is found in the cytoplasm of all cells. Plant cells contain chloroplasts.001 to 0. Figure 1b: 0. 4. a molecule that participates in the transfer of information between DNA and protein. All of the organelle types labeled in Figure 1b are found in a typical plant cell. ATP hydrolysis provides the extra energy needed to shift the equilibrium of a nonspontaneous reaction in favor of the products. Mitochondria produce energy needed for cellular activities. 8. they are found throughout the cell. AT 12. 10.0001 to 0. 3. NT 14. DNA. 3. NT Part C Matching © Pearson Education. AT 15. 1000 to 10. which is named for the distorted shape of the defective red blood cells. ATP is produced in the mitochondrion and transported out to the cytoplasm. 100 to 1000 nm. 7.0 m.0 m. 16.1 to 1. Plant cells have cell walls. 6. 5. Section Review 24.000 nm. All rights reserved. They contain significantly fewer mitochondria. Figure 1a: 0. For example. ATP captures energy from catabolism reactions to drive anabolism reactions. 2. They comprise the cell walls. cell wall 4. 0. Muscle cells are highly active cells. cell membrane 3.0 to 10. adenosine triphosphate adenosine diphosphate -ADP oxidation 30. cytoplasm 2. 9.01 mm. 5.. d 17. which require many mitochondria to meet their energy demands. Proteins embedded in cell membranes help to transport molecules and ions across this barrier. where it is used to fuel nonspontaneous processes. It lacks a nucleus and organelles. cell membrane Answer Key 863 . publishing as Pearson Prentice Hall. 4. a molecule that stores the information needed to make proteins. AT 13.001 mm. Part B 1.5 kJ nonspontaneous catabolic or anabolic catabolic or anabolic metabolism catabolism anabolism Part B True-False 11. structures that enable plants to produce carbohydrates through photosynthesis. b 18. Because proteins catalyze metabolic reactions. nucleus 5. 6. Lipids are found mainly in cell and organelle membranes where they form a barrier to the free flow of ions and molecules into and out of the membrane-enclosed compartments. The cell in panel a represents a prokaryotic cell. which provide structure and rigidity to plant cells.6 Part A Completion 1. RNA. Plant cells are eukaryotic.protein to function is seriously impaired. 5 to 5 nm long from head to tail. 2. i 10. b 9. AT 5. NT 34. Essay 42. nucleic acids. 3. B. DNA stores the information needed to make proteins. Many carbohydrates are soluble in water. publishing as Pearson Prentice Hall. 4. 4. d d c c b 21. j 39. There are two types: DNA and RNA. 15. 15. 23. 6. d b c c d 864 Core Teaching Resources . AT AT NT ST 30. Test the aqueous solubility of the substance. 22. a. 40. 25. h 10. ST 31. and energy-rich molecules such as ATP. They are composed of nucleotides that contain a phosphate group. a fivecarbon sugar. 13. 24. b 8. 12. AT Chapter 24 Test B A. f 7. Phosphorus is found in the bloodstream as HPO42Ϫ and H2PO4Ϫ. NT 4. RNA has a key role in the transmission of the information stored in DNA. 20. 3. f 8. 4. The length along the transmembrane axis of the protein must be approximately 5 to 10 nm. e 9. 14. AT 32. membrane proteins must have dimensions similar to the observed thickness of the lipid bilayer. 25. 18. Nucleic acids are polymers found primarily in cell nuclei. d b b d b 16. d 7. A membrane protein that acts as a channel must have contacts inside and outside the cell membrane. AT 3. 14. Matching 1. Matching 1. 23. c 6. 2. e b g i 5. ATP ϩ H2O → ADP ϩ Pi [ADP ][Pi] b. Multiple Choice 11. which together form an important buffer. Multiple Choice 11. c f b g 5. 41. c d b a d 16. 8. AT 35. b 6. a Quiz for Chapter 24 1. 12.. 13. Inc. NT 33. Questions and Problems 36. d 11. © Pearson Education. d a c c b 21. each phospholipid molecule must be approximately 2. All rights reserved. 24. 29. a 8. Eukaryotic cells contain a nucleus and other membrane-enclosed structures called organelles. Trp-Arg-Ala-Leu-Asn-end 37. 18. 17. True-False 26. B. d 9. j D. Prokaryotic cells do not contain a nucleus or organelles. and a nitrogen-containing base.Vocabulary Review 24 1. whereas lipids are not. 7. NT 2. 27. d d c a c E. b. Because the lipid bilayer is composed of two sheets of phospholipid molecules arranged tail to tail. 20. a 7. 22. DNA governs the reproduction and growth of cells. d Chapter 24 Test A A. spontaneous 38. Keq ϭ ᎏ ᎏ Ͼ1 [AT P] c. 17. 19. c d g i 5. Nitrogen is required for the synthesis of amino acids and nitrogen-containing bases of nucleic acids. 28. 19. C. a. h 10. Phosphorus is essential for the synthesis of phospholipids. k h e j 9. 3. To span the entire bilayer. ST 6. 2. Eukaryotic cells are typically much larger than prokaryotic cells. page 774 Analysis Sample answers are given.4 g. The extent to which the physical properties of a cell membrane are altered by a substance may depend on the solubility of the substance in the lipid bilayer. Two dipeptides are possible. All rights reserved. Finally. AT 32.42 g ϩ H2O 41.2 g ATP 1 kg kJ ϫ ᎏᎏ ϫ ᎏᎏ ϫ ᎏ3 ᎏ 39.55 cm)2 M ϭ 68. m 62.8 g A ϭ 3. d ϭ ᎏᎏ ϭ ᎏᎏ ϭ 0. when all other conditions are kept the same. V ϭ 0. bond-breaking and bondmaking occur at the active site to produce the products of the reaction.90 cm width = 4. Thus. True-False 26. NT D.55 cm)2 V ϭ 64. Beacuase the interior of the lipid bilayer is a hydrophobic environment. Questions and Problems 36. 4.55 cm)2]2/3 A ϭ 77. E. Shape Index ϭ ᎏᎏ ϫ 100 ϭ 77. 28. length = 5. 29.. Doubling and tripling the number of enzyme molecules in the reaction mixture is equivalent to doubling and tripling the number of active sites to which substrate can bind.90 cm)(4. 38.1 5.90 cm)(4. but do not change the normal position of the equilibrium.90 cm)(4. nonpolar substances have the greatest chance of becoming incorporated into this protion of the cell membrane. NT AT AT ST 30. Chapter 24 Small-Scale Lab Section 24.55 cm 1. The molecules on which an enzyme acts are Answer Key 865 . Next.3 The Egg: A Biochemical Storehouse. Essay 42. ST 31.138 [(5. One possibility: © Pearson Education. CH3 O @ # H3N 2 C 2 C 2 OH @ H ϩ H O @ # H2N 2 C 2 C 2 OH @ H called substrates. 2000 30.0 cm3 This is less than the density of a freshly laid egg. Most proteins are soluble in water. 1 mol ATP 507.5632lw2 ϭ (0.5236 lw2 ϭ (0. One possibility: ACAGTTGGTACT 37.5236)(5.8 g by 6. like act as biological catalysts. publishing as Pearson Prentice Hall. Inc. Enzymes are proteins that.42 g 4. the rate at which product is formed will increase with the number of enzyme molecules present in the reaction system. In an enzyme-catalyzed reaction the substrate binds to the active site on the enzyme form an enzyme-substrate complex. AT 33.3 cm2 3. More than one answer is possible due to the redundancy of the genetic code.C. AT 35. They reduce the time required for a chemical reaction to reach equilibrium.975 g/cm3 v 64. AT 34. The measured mass of 62.3 kg ATP 40. Enzymes are not changed by the reactions they promote. whereas lipids are not.5632)(5. An enzyme catalyzes the conversion of a substrate to product.55 cm JJJH H O CH3 O @ @ # # H3N 2 C 2 C 2 N 2 C 2 C 2 OH @ @ @ H H H mass ϭ 62.0 cm3 M ϭ 0. The egg may have lost water over time. the products dissociate from the enzyme leaving the enzyme free to bind new substrate and begin a second reaction cycle.5 kJ 1 mol ATP g 10 ϭ 33. Test the aqueous solubility of the substance.90 cm 2. 27.42 g is less than the calculated mass of 68. 10. 218 4 214 84 Po → 2He ϩ 82Pb 210 210 0 82 Pb → 83Bi ϩ Ϫ1e Section Review 25. 4.0 g ϫ ᎏ1 2 ϫ 2 ϫ 2 ϭ 0. 2. ST 17.1 Part A Completion 1. a positive test for protein. 2HCl ϩ CaCO3 → CO2 ϩ H2O ϩ CaCl2 7. b. 5. The warmer the temperature the greater the mass loss. Assume Step 1 reveals the egg loses 0. 4. 10. a positive test for protein.5 grams per day depending on the temperature at which they are stored. 11. 13. NT 14. e 19 b 20. 3. d Part D Questions 60 hr 22. Part D Problems 26. 11. Weigh the egg once each day for two or three days. 5. electrons metal foil Gamma mass Lead concrete stop Section Review 25. 2.0 g ϭ 1. 2. 12. NT 18. which are rounder. d 866 Core Teaching Resources . c 24. band of stability beta positron rate half-life radioactive 7. After 4 half-lives 15 hr 1/2 ϫ 1/2 ϫ 1/2 ϫ 1/2 ϭ 1/16 of the original mass will remain. Weigh it once a day for three more days and store it at room temperature.42 g) ϫ ᎏᎏ 0. Extra large eggs are usually more than 70 grams. Powdered milk ϩ NaOH ϩ CuSO4 produces a violet color. billions transmutation radioactive decay atomic numbers synthesized Part B True-False 12. Weigh an egg once a day for three days and store in a refrigerator between weighings. publishing as Pearson Prentice Hall. a 23.8 g Ϫ 62. ST 13. radioactive radioisotopes nuclei stable energy beta Alpha helium 9. 14.2 to 0. ST Part C Matching 17. 5. 1 day Age of egg ϭ (68. 11. 8.13 g © Pearson Education. 9. d Part C Matching 16. Typical eggs will loose 0. 8. c 18. Egg shell ϩ NaOH ϩ CuSO4 produces a violet color. HCl produces bubbles at the surface of the egg shell. NT 15. ᎏᎏ ϭ 4 half-lives. b 20.20 g Age of egg ϭ 32 days old 3. 2. 8.You’re the Chemist 1.2 Part A Completion 1. NT 20. 5. The larger the egg. 3. 4. AT 19. fusion mass energy hydrogen helium Part B True-False 16. b 22. e 19. fission neutrons fissionable atom energy moderation absorption 7. NT Part C Matching 21. 1 1 ᎏ ᎏᎏ ᎏᎏ 23. a 21. 3. AT Part B True-False 12.25 g or 3 half-lives 42 days Ϭ 3 ϭ 14 days Section Review 25. 9. 6.. a 17. AT 15.3 Part A Completion 1. medium eggs are less than 50 grams. Extra large eggs tend to be more oblong than small eggs. 8. Measure the volume of water displaced by the egg. All rights reserved. a. 6. NT 13. 2. e 25. 1/16 ϫ 18. c 18. NT 14. 4. 7. Inc. 6. ST 16.20 g per day. 15. the smaller the shape index. 10. 6. a. publishing as Pearson Prentice Hall.Part D Questions and Problems 21. decrease the number of slow-moving neutrons and slow the chain reaction Section 25.1 min decrease by a factor of more than 1000.25 g 4. 2. 3.0 g 20 minutes ϭ 1 half-life. All rights reserved.0 g→ 1. or 820 s. a.4 Part A Completion 1. 36Kr e.0 g→ 0. 9. 2. lead-210.2 4 204 1. scintillation all iodine-131 phosphorus-32 neutron activation Part B True-False 11. 2.3 1.49/3 ϭ 183 s Section Review 25. 8. a. 28 protons and 36 neutrons b. ionizing electrons senses Geiger gas 6. 79 protons and 116 neutrons 4. Section 25. Teletherapy is the use of gamma radiation to destroy cancerous tissue. 4.0 ϫ 1023 ϫ ᎏ1 atoms 2 ϭ 3. 2. The atomic number decreases by two. 238 92U y 82Pb ϩ 8 2He ϩ 6 Ϫ1e 1 ᎏ ᎏᎏ 8 days ϭ 2 half-lives. Practice Problems 25 Section 25. © Pearson Education.0 g → 1.0 g.0 g→ 4. Radioisotopes are used to diagnose and treat diseases such as cancer. 1. c 18. 7. e 17. Ϫ0 7N 1e 237 b. 5. 30 P 15 3. Radioisotopes are used to study chemical reactions and molecular structures. 3. 7 4Be ϩ Ϫ1e → 3Li 37 37 c. a. Three half-lives ϭ 15 days.0 ϫ 10 5. slow fast-moving neutrons so they can be absorbed by the fuel atoms b. 3 ([1 ϩ 235] Ϫ [87 ϩ 146] ϭ 3) c. The mass would 5. the mass number remains the same. or three half-lives. 16 g → 8 g → 4 g → 2. 8 206 4 0 6. b 16. b 21. a. 208 87 Fr → 2He ϩ 85At 7 0 b. Answer Key 867 .4 1. a 19. 10. AT 12. 6.50 g→ 0. Inc. It takes five half-lives. 3 b. 8.0 mol ϭ 23 ᎏ 6. 93Np Interpreting Graphics 25 1.0 g Four half-lives ϭ 4 ϫ 17 days ϭ 68 days 51 min 5. Neutron activation analysis is used to detect trace amounts of elements in samples. 1 10 1 ᎏᎏ ϭ ᎏᎏ 2 1024 6. 3 d. 4 22. The half-life is 5. 9F → 8O ϩ ϩ1e 2.. 144 58 Ce 1H 239 92 b. a. 4 ([1 ϩ 235] Ϫ [72 ϩ 160] ϭ 4) 2. 16 g y 8 g y 4 g y 2. The mass decreases by a factor of 1/8. Radioisotopes replace non-radioactive isotopes in the structure of a compound without changing its chemical properties. 14 c. 2 ([1 ϩ 235] Ϫ [90 ϩ 144] ϭ 2) b. The half-life of polonium-214 is insignificant compared to the half-life of bismuth-214. The atomic number increases by one. 2. 20 ϫ ᎏ1 2 ϫ 2 ϭ 5. a. 4 2He Part D Questions 20.1 1.0 kcal/g ϭ 2. 4. 94Pu c.0 g→ 2. 53 protons and 83 neutrons c. ST 14. AT Part C Matching 15. d Section 25. 3. 18Ar → 19K ϩ Ϫ0 1e 17 0 17 d. the mass number decreases by four. NT 13. Tracing the pathways of radioactive isotopes allows scientists to study reaction mechanisms and reaction rates.5 ϫ 106 g 3.0 ϫ 107 kcal Ϭ 8. ᎏᎏ ϭ 10 half-lives. The energy released from the sun is the result of a nuclear fusion. 24. 2. 13. a 6. 234 91Pa → 92U ϩ Ϫ1e 234 234 c. 2 32 71. a i g c 5. i d e c 9. 12. b c b b a c 17.7. 18. 90Th → 91Pa ϩ Ϫ0 1e 27. publishing as Pearson Prentice Hall. B. The half-life of scandium-42 is 84 days. 6.125 g 1 28. 2. 13. All rights reserved. d 7. 25. b 9. In solar fusion. e 6. Problems 222 4 26. hdrogen nuclei (protons) are fused to make helium nuclei. a. 8. 11. 20. ᎏϭ ᎏᎏ of the sample remains 4. 18. 12.00 g 32 1 5 1 Since ᎏᎏ represents ᎏᎏ . h 10. b d b c a 22. 3. 12. b l g a 5. b 10. 19.. 5 half-lives C. b C.00 half-lives 1 1 1 ᎏ ᎏᎏ ᎏᎏ ᎏᎏ Thus. the isotope decayed through 3 half-lives. 26. 20. D. Nuclear reactions release far more energy than typical exothermic chemical reactions. Problems 0 42 27. 15. 4. 4. 14. so one half-life period ϭ 84 days. Essay 29. a 8. e 7. Essay 30. a 10. 0 4 41 1H ϩ 2Ϫ1e → 2He ϩ energy Vocabulary Review 25 1. 40 days ϭ 5 half-lives. 235 92 U → 2He ϩ 90Th 28.13 gram remaining. 3.0 h/6.6 ϫ 10Ϫ4 s). The reaction requires two beta particles. Fusion occurs when two light nuclei combine to produce a nucleus of heavier mass.5 years ᎏᎏ ϭ 14. d b a a c 21. 23. such as lead-206. k f h j Chapter 25 Test B A.5 neutrons to 1 proton. 1 ᎏᎏ of the original sample remains 32 ϭ 0. 25. b. a. Nuclear reactions occur in an effort to obtain stable nuclear configurations.75 h ϭ 4. Multiple Choice 11. the stability ratio is about 1. 124 n Ϭ 82 p ϭ 1. a b a d b 0. Multiple Choice 11. 16. j i c d 5. 3. a d c b a 16. 2. or 5 half-lives. 17. polonium-210 has the shortest half-life (1. a. Inc. j 6. 23. h 9. g 7.0 g ϫ ᎏ1 2 ϫ 2 ϫ 2 ϫ 2 ϭ 0. 12.3 years/half-life.5 ϫ 109 yr). d 8. 7. 42 19K → Ϫ1e ϩ 20Ca 4 231 b. 226 88 Ra → 86Rn ϩ 2He 0 234 b.750 g © Pearson Education. 21. 19. Chemical reactions occur in an effort to attain stable electron configurations. 868 Core Teaching Resources . c 8. Matching 1. 27.5 8. 24. Matching 1. 4. If one-eighth of the sample remains. f 9. 3. 29. 14. Uranium-238 has the longest half-life (4. or thermonuclear reaction. 10. Three half-lives is 252 days. 22. d b c b 5. For heavier isotopes. 4. a B. d d d a d Chapter 25 Test A A. 15. D. 2. f Quiz for Chapter 25 1. 6 days. This time period is two half-lives (11. After 7. All rights reserved. 3. Inc. Each trial represents one half-life because the number of heads approximately halves for each trial. The rate of disappearance of heads is nonlinear. Plot number of evens vs. page 809 Analysis Sample data are provided. 4. Unlike chemical reactions. Do trials until the number of events equals zero.2 Radioactivity and Half Lives. Roll the die again a number of times equal to the number obtained in the first trial. 3. Trial # 1 2 3 4 5 Number of flips 100 42 20 9 5 Number of heads 42 20 9 5 3 © Pearson Education. After 3. publishing as Pearson Prentice Hall. one-eighth remains. trial. 100 80 Flips 60 40 20 0 0 1 2 3 4 5 6 7 Trial 2. one-fourth of the sample remains.c. one-fourth remains. Count the total number of even numbers that result in 100 rolls of the die. You’re the Chemist 1. After two halflives. The rate decreases over time. Chapter 25 Small-Scale Lab Section 25. 2. 6 7 3 1 Figure A 1 0 1. nuclear reactions are unaffected by changes in temperature.8 days. and.50. For each flip the probability of a head is 0. half the sample remains. pressure. after 11. or the presence of a catalyst.4 days.. Answer Key 869 .460 years/5730 ϭ 2) of carbon-14.
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