Chem Section Reviews

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 10.0 inches. 1 inch 2.54 cm 2.54 cm Step 2: 10.0 in 25.4 cm in Step 3: There are about two and a half centimeters per inch, so the answer 25.4 makes sense. 24. Step 1: Knowns: distance 5.0 km 1 km 0.62 mi 0.62 mi 3.1 mi Step 2: 5.0 km km 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 $1.50 a pound weight of an apple 0.50 pound dollars available $16 b. cost per apple $0.75 number of apples purchased 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 f 24. substance shape or volume shape or volume shape gas Part D Questions and Problems 19. ST 10. d i c a 19.. f 13. All rights reserved. 3. a. 4. solid e. mixture Section Review 2. 9. substance b. AT 12. K b. homogeneous b. 4. 5. extensive intensive mass volume amount type 7. 18. c 17.3 Part A Completion 1. copper b. ST 14. AT 10. 9. Pb c. heterogeneous e. AT Part C Matching 15. 7. 22. h e b g 23. Cl e. Inc. 2. heterogeneous c. 17. e © Pearson Education. a. 5. g 762 Core Teaching Resources . ST Part C Matching 12. 3. 11. heterogeneous c. NT 11. 6. iron e. NT 13. 8. a. AT 12. liquid c. yes b. a. 2. hydrogen c. k 25. 6. nitrogen Part B True-False 8. substance d. 5. ST 9. 10. a. d 18. 4. mixture heterogeneous or homogeneous homogeneous or heterogeneous solutions phase physical distillation Part D Questions and Problems 18. yes Part B True-False 9. 16. a 15. liquid 27. b 17.2 Part A Completion 1. a. a. gas c. vapor b. S d. b 16. yes d. no d. j Part D Questions and Problems 26. 7. element or compound compound or element elements ratio or proportions chemical substance mixture symbol C K Part B True-False 11. a 15. element c.Section Review 2. Section Review 2. compound 19. e 16. compound e. 3. 10. compound b. d 14.1 Part A Completion 1. publishing as Pearson Prentice Hall. element d. 2. silver d. 20. AT Part C Matching 13. mixture d. Na 20. 8. c 14. homogeneous 20. 6. 21. AT 11. Ag Part B True-False 9. or a mixture of liquids. Sn c. 7. 6. mixture c. a Section 2. the law of conservation of mass Part D Questions and Problems 19. nitrogen. ST 10. extensive d. a. which mix with the air. a. bromine. a. substance Section 2. b 16. solution 4. chlorine 7. intensive Section 2. The products of the reaction are gases. carbon dioxide. heterogeneous Vocabulary Review 2 1. c 2. mixture c. gold d. intensive b. iron c. AT 11. extensive Interpreting Graphics 2 1. Inc.4 1. compound 4. ethanol 8. d 17. chemical physical chemical reactants chemical composition conservation of mass mass 4. heterogeneous b. physical change b. No. elements Solution: substance 5. c Part C Matching 14. phase 2. chemical change c. mass solid mixture gas Answer Key 763 . 4. All rights reserved. a. Iron oxide is a compound.4 Part A Completion 1.1 1.Section Review 2. 32. carbonic acid b. chemical symbol 3. 2. mixture b. Chemical process. e 15. c 18. physical separation 4. hydrogen 2. evaporation 2. a. substance no yes yes yes no no c. physical separation b. 3. substance c. Homogeneous mixture. mixture d. substance b. 8. physical change 3. chemical change d. b a b d gas no liquid no solid yes 6. publishing as Pearson Prentice Hall. Na d. 62 grams 5. 6. potassium 3. a chemical change 21. a. It could be a liquid element. homogeneous c. while pure iron is an element. a. lead b. compound b. 6. 20. 5. Practice Problems 2 Section 2.4 grams 6. a. distillation.3 1. 5. a. element d. water 4. ST 12. mixture d. 3.2 1. a. 4. NT b. 8. a liquid compound. it has a variable composition. 3. 2. C 5.. 36 grams © Pearson Education. Since motor oil comes in grades. chemical separation c. 7. a mixture with a uniform composition throughout 3. 2. NT 13. Examples of chemical change include the rusting of iron. h 8. Melting or boiling are physical changes. i f g a 5. 12. NT E. AT 22. AT B. 29. a 19. 15. b 16. 19. i d h a 5. liquid vapor heterogeneous N 33. 35. Matching 1. publishing as Pearson Prentice Hall. oxygen c. Solid salt remains at the bottom of the flask. NT 23. e 9. In a chemical change. c 17. 32. 33. 32. 2. 4. True-False 24. 4. 11. b 7. c C. recapturing the water through distillation. 4. g B. Multiple Choice 11. 764 Core Teaching Resources . a Chapter 2 Test A A. b 6. Essay 37. skim the sawdust directly off the surface of the water. hydrogen Chapter 2 Test B A. 31. j 10. ST 27. 3. A physical change alters a substance without changing its composition. 6. 2. and the burning of wood. 34. NT 26. iron filings are attracted to a magnet. 3. AT 24. 2. Completion 29. 17. 12. 18. 14. All rights reserved. 27. c 10. physical blend element vary a.Quiz for Chapter 2 1. Inc. 14. energy distillation vary chemical physical 31. j 6. the reaction between iron and sulfur to produce iron sulfide. sawdust floats in water. e 7. b 22. one or more substances change into one or more new substances. f 8. Boil off the water. compounds iron homogeneous conserved products E. Some possible clues to chemical change are a transfer of energy. Matching 1. d b b b 15. 34. Multiple Choice 11. d 20. potassium b. 9. To separate the mixture. 28. AT 25. c d d d a 21. then expose the resulting mixture to a magnet to remove the iron filings. 13. d 9. d D. 3. b 23. 36. True-False 21. the production of gas. Completion 26. 35. color change. 7. 13. d 18. a d d b a d d 8. 20. NT 28. 5. 10. reactants chemical physical substance C. 30. © Pearson Education. and the formation of a precipitate. 30. D. Essay 36. a c d c d 16. Salt dissolves in water. ST 25.. publishing as Pearson Prentice Hall. l d c h i 25. 4 21. a. 5. 5. yellow 2. 10. 5. ST 12. 2. AT 14.35 1. 4. NT 13. 4. 3. 21. 8. © Pearson Education. 17. f 18. c black black Figure A black Part D Questions and Problems 20. b Answer Key 765 . If an antacid tablet contains starch. 2. Add Kl + NaClO to various foods.2 Part A Completion 1. Most table salt contains 0. 4. 9. 19. metric seven meter liter weight 6. b 15. 4. Part B True-False 11. Certain areas can be treated with NaClO to bleach parts of the picture. but different in other reactions. j 26. b. Inc. 3. A picture can be drawn with colored ink. both turn blue-black. 24.. Starch. Part B True-False 11. absolute value 100% scientific notation known estimated 2 3 BLACK!. 22. NT Part C Matching 15. a 17. accuracy precision experimental value error accepted value 6. 8.Chapter 2 Small-Scale Lab Section 2. 3 c. The color in the ink becomes bleached. 3.4 1 page 56 Analysis NaClO KI KI Starch KI Paper KI Cereal yellow H2O2 yellow CuSO4 brown ppt black Section Review 3. All rights reserved. 10. 9. the mixture turns a blue-black color. Wet only a portion of a small pile of salt with starch. 23.4 cm3 Section Review 3. d 19. 2. gram Celsius kelvin joule/calorie calorie/joule You’re The Chemist 1. 3. The results may be the same in reactions that are simnilar to the one with Kl and starch. 18. f 27. They all turn a mixture of Kl and starch black. A black color indicates the presence of starch.01% Kl. 7. AT 13. 4.1 102 b. it will turn black whenn treated with Kl + NaClO. 8. NT 12. 2 102 mL c. e 16. Add CuSO4 or H2O2. A black color indicates the presence of Kl. k e m g a 20.1 Part A Completion 1. 2. 7. which suggests the presence of starch. 16. a. AT black black Part C Matching black black black 14. 3 Part A Completion one (unity) conversion factor remains the same dimensional analysis 5.8 cm 4. known 1. 1 in 1m 1500 cells 100 cm 17 m/cell 186 000 miles 1.5 cm 29.0 km/L gal 1L 1 mi 1 oz 1 lb 1.8 102 cm3 20 30. 7. pop. 3. 20 cm 21.2 15.82/lb.61 km 1s 18. © Pearson Education.54 cm 106 m 1. 3. the beef steak costs $1. Inc..58 kg 1580 g kg mass 1580 g Density volume 1000 cm3 Density 3. ST Part B True-False 4. 0. ST 13.2 g 1 lb can dollars 27.000 g 5. a.1 cm3 3 1000 cm 7. density 750.2 102 cm2 Pete’s 5. . 24 h 60 s 60 min 1 day 86. 3.00 L 1000 cm3 1L 1000 g 1. 6. 20 cm 400 mm 24 liters 12. AT 1.264 gal 1.Part D Questions and Problems 28.96 g/cm3 1. p. 2. 103 273 253°C c.1 1. 3. 10 2 mass 127 g volume 32. a. composition 5. 5.000 people 1 mi2 49 mi2 640 acres 24 people/acre 1 can 454 g dollars cost 2. 8.93 104 kg mi 0.0 m Section 3. 4. 1. 10 12 b.400 s 1 day 1 min 1h km 1000 m 100 cm 1h 1 min 72 h 1 km 1m 60 min 60 s 3 2. 4. 32.125 kg 1.0 cm 4. 8.99 10 1 km 1 s Section 3. Volume 1. 7. intensive 3.8 cm 8. 9.25 10 1 kg 1000 mL b.7 mg 1 lb 1g 2. c. Part C Questions and Problems 6. steps conversion factor denominator units unknown Practice Problems 3 Section 3. 3. All rights reserved.0 in 1m 17. density 2.12 L 1. ST 8. Section Review 3. 10 6 d.14/lb. NT 12. 8.4 Part A Completion 1. 7. 10.54 km/h h mi The chicken costs $2. Bruce’s 1 cm 5% 6. 6.3 3 cm 1m 1000 g 1.61 km 40. AT 14. 6. 6.3 1. 4. 125 g 1 kg 1000 g 0. NT Part C Questions and Problems 16.1 mL 1 cm3 1 mL 766 Core Teaching Resources .2 102 mL 1L 2.0 17.21 cm3 9. 7. 2. 4. 2.0 lb mi 1. publishing as Pearson Prentice Hall.00 dollar 375 dollars 16 oz 454 g 1000 mg 75.99 50. 50 lb 6 1010 293 K 4°C Part B True-False 11.0 10 cm/s g 100 cm 3 1 kg 1 m3 19.609 km 60 96. °C K 273 30. 6 1s 1 mile 10 s 2.58 g/cm3 Section Review 3. k. The beef steak is less expensive per pound. 3.5 cm2 35. 8. does vary with the size of the sample. 3. Another significant figure could be estimated. 4. j l b a m 11. 22. 8. 1 2 37. © Pearson Education.88% m 10 C c. 2. i b j g 5. however. 100 5. Inc. ST 31. d d b a c 16. 23. 5. 14. d 10. a.10 g/cm3 4. 2. ST 33.9 103 km b c K 55 C 273 218 Percent Error 5. Volume is a measure of the space occupied by an object. 24.54 m 5 12.7 6. True-False 26. c 12. NT 32.80 g/cm3 5. not on the size of the sample. Problems mass 980 g 15 g/cm3. 20 C is known.10 g/cm3 Answer Key 767 . 0.0 g volume 18. h Interpreting Graphics 3 1.0 6. 20°C c. AT 30. 7. Volume is an extensive property that depends on the amount of matter in a sample. 6. 17. a 9. 6900 km 5. 18.46 m 10 4 m) 9 2 (3. b a c b a 21. 13. d a d d b C. 7. 8. no 1 10 3 kg/ 1 g 4 a. The density of water at a given temperature does not vary with the size of the sample..3 10 0. 9. The balloon will float. 2. Multiple Choice 11.3 10 b.15 cm 1. Density 10 m mass 14. 20. Density is the ratio of the mass of an object to its volume. e D.0 cm3 0. 6. Essay 38. ST 28.083 m b.8 cm 7. Density b.09 liters 4. 4. 8. 15. 27 g Chapter 3 Test A A. publishing as Pearson Prentice Hall. g i h f k 6. liquid B. c 7. 4. NT Vocabulary Review 3 1. AT 34.72 m) Quiz for Chapter 3 1. 6.00 m c. Density is an intensive property that depends only on the composition of a substance. 13. a. 3. 2. 7. (1. 25. d 13. Matching 1. ST 29. 3. ST 27. 3.4 1.Section 3. the volume of the water. e 8. 10. f 6. Air is less dense than carbon dioxide. no volume 64 cm3 36. 12.778 g/cm3 E. Two significant figures Three significant figures cylinder B either cylinder A or B the Celsius scale the Kelvin scale No. 4. 2. 5. 1. 4. 19. All rights reserved.00 grams 2. 9. 27. Problems 40. (5.019 mL 1 cm3 1000 L 19 L x L 0.11 g penny 100 g penny 2. a a b d c a d C. you must first express the volume of the box in cubic units. that the cm3 units cancel and the liter units remain. a. 30. 28. 3. 20.019 g mass of pre-1982 penny 3.0 cm 3000 cm3 103 cm3 1L 1000 cm3 3.1 1015 m 41.019 mL 1 mL 3. 1000 mg/cm3 4.50 g penny 0. ST 34.019 g 1000 mg/g 19 mg m 0. b 7.. Thus.0 100 cm m 1cm 10 mm ( ( 10. and then select a conversion factor that will allow you to convert the cubic units into liters. Matching 1.016 106 m2 2. x mg/cm3 1.6 113. 19. ST 35. 16.0 cm3 The metal is not pure silver.4 Now What Do I Do?. AT 33.00 g/cm3 1000 mg/g x mg/cm3 1000 mg/cm3.019 g 2.74 109 m d.44 g Zn 97. you must select a conversion factor so that relates cm3 to liters.0 g Cu 3.0 L 3. Essay 43. AT 36.08 1015 m 4. NT 38. V 0. True-False 32. NT Chapter 3 Small-Scale Lab Section 3. 31. since V L W H.11 g penny 100 g penny 0. 148. 15. x mg 0.11 g mass of post-1982-penny 2. 29. d E. x g Cu x g Cu x g Zn x g Zn 1 cm3/1 mL 1000 mg/mL 95. 13.0 L g Cu g Zn g Zn 5. All rights reserved.0 g Zn 3. D. 768 Core Teaching Resources .0 106 m2 8.16 g Zn 2. since 1 L 1000 cm3 1L 2000 cm3 2L 1000 cm3 B. 2. based on the linear dimensions given.4 g Cu 100 g penny © Pearson Education. 17. AT 37. e i g f 5. g Cu 3. To find the volume of the box in liters. 12.0 103 cm3 1 kg H2O b. 24.0 kg H2O L M 750.41 g 113.060 g Cu 2.95 g Cu 5. 26. 36.0 g 42. D 11.16 105 m2 2.60 m 50. V ( ( 0.5 g/cm3 V 65.60 10 2 m) 20. 21. a.1 cm 54.721 cm 15.2 10 6 2.0 mm 3. Thus.Chapter 3 Test B A. V 25 cm 10 cm 8 cm 2000 cm3 To convert the cm3 into liters.019 cm3 d 1. 2. publishing as Pearson Prentice Hall. c 6. page 94 Analysis Answers are based on the following sample data: average mass of water drop 0. NT 39.50 g 1. 3. Inc.50 g penny 2.576 g 35. 23. Multiple Choice 11.3 cm b. c d a c a d c 18.17 g c.291 cm 54.6 g Zn 100 g penny 2. a 9.166 g 107 m) (3. d d c d b a c 25. h 10.00 g/cm3 1 mL x mL 0. 14. 22.019 cm3 0. j 8.47 cm 2. 4. 017 cm3 Volume of 21 hemispheres: 0. 15 tables.6.2 lb 1 L/1.00 kg Section Review 4.55 cm) = 3.5 m = 630 m © Pearson Education. (1) Measure the mass before and after you fill the can with water. and 2 desks is about that of 30 people or 1800 L.47 cm3 4.1 Part A Completion 1. NT 12. NT 6. b Part D Questions and Problems 18. 1 108 atoms cm 100 cm m 1010 atoms/m = 630 000 L/ Assume 30 people with an average weight of 130 lb (1 kg = 2. AT A die has 21 holes that are hemispheres with a radius of 0. 6. mass of 1 drop: 0.0242 g Pipets give different results. 7.0218 g at 0°. Volume of 30 people V = 30 = 1800 L The volume of 30 chairs.70 g/cm3.55 cm on a side: V = (1.019 g at 45°. d 11. b 3. neutrons Answer Key 769 . The volume of people and furniture is 3600 L.8 m 3 Part D Questions and Problems 12. 3.36 cm3)(100%) = 11% Note: The holes in some dice are cones. publishing as Pearson Prentice Hall.0 kg/L. or rearranged Part B True-False 5. V = 3. V = πr2h (Can is not a perfect cylinder. V = 5. 4. subatomic negative protons neutrons 5. The volume of a cone is 1/3 πr2h. The best angle is 90° because the pipet is easiest to control.57%. 8. compounds 4.77 g. 3. c 14. Find weight in pounds and convert to kg. mass of 1 drop: 0. 3 Error = 0.) (3) Read label: 12 oz = 355 mL 5. different 3. The new penny is lighter because it is mostly zinc which has a lower density than copper. % error = (3600 L/630. Use the mass and density of water to find the volume. c 9. atoms 2. V = weight 1 kg/2. mass of 1 drop: 0. Volume of hemisphere: 2/3πr3 V = 2/3πr3 = 0.000 L)(100%) = 0. AT Part C Matching 8.0 kg Part B True-False 9. 2.2 Part A Completion 1. nucleus Ernest Rutherford positive electrons 130 lb 1 kg/2. 2.00 kg/L. dA: 2. a 10.72 cm 3 3 Section Review 4. Assume the density is about 1. protons. at 90°. AT 11. 7. Inc. Find mass of can and divide by density of aluminum. d 17. Sample answer: mass of one can: 14.2 lb) and a density of about 1. NT 10.36 cm3 0. If die measures 1. 6. ST 7. a 15.2 lb 1L/1.36 cm3 Volume of die: Part C Matching 13.36 cm3 Percent error (0. e 16. Sample answer: V = 16.72 cm3 = 3.. simple whole-number ratios 13. separated. joined.0 m 1000 L/m 12. Expel the air bubble so that the first drop will be the same size as the rest. (2) Measure the height and radius and calculate volume.36 cm3 You’re The Chemist 1.36 cm3/3.20 cm. All rights reserved. approximately 6 1023 neutrons 3. 12. h 10. electrons are found outside the nucleus. d 9. AT 13. d Part D Questions and Problems 24. publishing as Pearson Prentice Hall. 39.67 10 5. Section 4. 138 Atomic number: 25. 3. 35. (1 10 atoms) 1 mm 7 1 1 cm 108 atoms 10 mm 1 cm Chapter 4 Test A A. c 7. d 8.3 Part A Completion 1. NT 12.19. 7.67 10 24 g) + 16(1. 146 d. electrons 20. b 23. 10 4. protons. 33. 2.00204 0. 3 2. Most alpha particles pass through the empty space without deflection. isotopes atomic mass 1 no 1 Vocabulary Review 4 1. e 8.96 amu oxygen-17:17 amu 0.00037 0. 89 Number of neutrons: 50. 16 d. c 19. 8. 33. 23. Matching 1.2 1. 3. 75% X 100. 10 b. f 6. e 17. ST 14. protons number electrons protons neutrons 6. 5.011 amu 5. 9. i Part B True-False 11. f h g b 5. 4. AT Quiz for Chapter 4 22. 11. Atoms are mostly empty space. 80 3.0063 amu oxygen-18:18 amu 0. a 10.. 2. a. 2. 42. 16(1. The few that come near the nucleus are deflected or bounce straight back. 4. 2. 6. g 20. e 7. 46 b. The relatively massive protons and neutrons are concentrated in a small region called the nucleus. 89 Number of electrons: 25. 4. h 1. Number of protons: 11.1 1. f 18. a. a 21. 8. 5. j g a c 5. 7. 3. 10. simple whole-number chemical reaction Electrons neutrons proton atomic 8 mass 156 30 periods atomic number Part C Matching 16. 12. b 9. 10. 4. 3. All rights reserved.3 1. 35.00 amu 16 amu © Pearson Education. j 6. i Section 4. oxygen-16:16 amu 0. 25% X 104 Section Review 4. Practice Problems 4 Section 4. 12 c. 9. 89 Mass number: 55. AT 15. 5 c. b 24 ) 770 Core Teaching Resources . 11. mass of e is negligible. Inc. 39.33 10 23 g 2.9976 15.0367 amu 16. All rights reserved. 22. publishing as Pearson Prentice Hall.4889 31. Multiple Choice 11. the same number of protons and electrons. e 10. 20. 17. 23.926 amu 0. 19F: 9. Problems 25. 35.75 amu 67. 19. Multiple Choice 11. 17. 15.947 amu C. 27. 6. 20. 47 0. 19. Atoms of different elements have a unique number of protons and electrons.925 amu 0. 79 Mass number 24. 19.802 amu 39. 4. 35. c d c c d 16. 61. d 23. and thus. b b a d Answer Key 771 . 3. 20.0062 atomic mass 65. 26 Mass number: 19.99600 atomic mass 26. different mass numbers.121 amu 0. Atomic number: 7. 35 Number of neutrons: 12. 19. c 22. 10 11 5B: 5.927 amu 0. c c b a c 21. Problems 24. 197 Number of protons: 12. h 6. 18. c C. and thus. 26. 17.00337 Ar-38 37.925 amu 0. Essay © Pearson Education. Essay 28. Isotopes of the same element differ in the number of neutrons in the nuclei. 13.25 amu 65.929 amu 0. 26.61 amu 0. 18 65 30Zn: 30.43 amu 69. 9 9 27 13Al: 13. 2.978 amu 0. a b a b b 21. 197Au 19 26 79 D. 14. Ar-36 35. 41 Number of protons: 9. a 7.963 amu 0.2781 18. 24. c i b g 5. 4 20 10Ne: 10. Isotopes of the same element are alike in that they have the same atomic number. d B. 13. D. 35. a c d d a 16. Inc..024 amu 39.33 amu 66.00063 Ar-40 39. 5 33 16S: 16. 16 27.B.37 amu 25. Matching 1. 15. 12. 13. Isotopes of the same element are different in that they have different numbers of neutrons. 13 40 18Ar: 18. 18. j 9. 22. 14. 10. 7. 6C: 6. 56. 80. Chapter 4 Test B A. 56Fe. 12. 26 13 26. 6 10 4Be: 4. 7. 13 Number of neutrons: 14. 10. 30 108 47Ag: 47. 6. 14.1857 12. 30 Number of electrons: 9. 63. Atomic number: 19. 45 Number of electrons: 12. f 8.962 amu 0. 79 Symbol: 39K. 7. 20.0411 12. which prevents them from falling into the nucleus. 20. He proposed that electrons surround a dense nucleus.2742 g 0. Relative abundance tells you the decimal fraction of particles. Part D Questions and Problems 19. 7. Thomson plum-pudding nucleus circular quantum mechanical probability 15 13 20 48 0. 3.. See row 5 in Figure A. All rights reserved.83 g C 15.8773 g 1. Inc. Another student might not have had the same relative abundance of each candy. Any differences are probably due to small variations in the numbers of each kind of candy in the samples.3208 g 0. According to Bohr.0 31. AT Part C Matching 1. A larger sample would provide a greater sampling of all isotopes. Relative abundance is parts per one or the decimal form of percent. they are described in terms of probability. The individual percent abundances add up to 100. electrons John Dalton J. See row 4 in Figure A. 4.08 41. AT 14.8833 g 6. AT 10. 2. which affect the relative abundances.8831 g 0.2742 g 0. 6. 5.3125 0. 15.J. 5 orbitals d. the locations of electrons are not fixed. 3 orbitals c. 7.16 g B 13.2883 g 0. NT 12. The larger the samples. See row 3 in Figure A. 7 orbitals b. A Total mass (grams) Total Number Average mass (grams) Relative abundance Percent abundance Relative mass 13. AT 13. 0.67 1.3 The Atomic Mass of Candium. See row 6 in Figure A.064 g 0. Rutherford discovered that atoms are mainly empty space. 3. Section Review 5. 8.39 g You’re The Chemist 1. Bohr proposed that electrons are arranged in concentric circular paths around the nucleus.2883 g 0. 8. the better the results with any of the methods. 1 orbital 772 Core Teaching Resources . c 16.2708 0. The total in row 3 is an average that ignores the relative abundances of particles.8833 g Figure A Part B True/False 9. 4. Dalton proposed that matter was made of indestructible particles called atoms. Percent abundance is parts per hundred. a.3208 g 0.1 Part A Completion 1.Chapter 4 Small-Scale Lab Section 4. 2. d © Pearson Education.40 g Totals 42. b 17.7700 g 0. publishing as Pearson Prentice Hall. ST 11. a 18. The individual relative abundances add up to 1. In the modern atomic theory. Thomson proposed an atomic model in which negatively charged electrons were embedded in a positively charged mass.4167 100. the electrons in a particular orbit have a fixed energy. page 120 Analysis Sample data provided. 2. Mass is likely to provide better results than volume.000 0.25 27. The total in row 6 is a weighted average because it considers differences in mass and abundance among the particles. 9. 5. 70 10 7 m 4. a. NT b. e. 1 f. 3. 3 orbitals. 5s. 6 e.00 10 6 cm. b 20. a. NT Part B True/False 8. 1 b. 2. AT 16. 1 orbital. a. 5f. 5p. electron configurations Aufbau principle equal Pauli exclusion two opposite a single electron superscripts electrons Chromium Part D Questions and Problems 3.1 orbital. NT 10. 16 7 h. 5 orbitals. B © Pearson Education.. 1 15. d 19. 10. 4. c Part D Questions and Problems 22. 3p. b 2.00 1010 cm/s 2. ultraviolet 3. NT 13. 7 orbitals e.1 1. 3d. 1s1 1s22s22p63s23p63d34s2 1s22s22p63s2 1s22s22p63s23p63d104s24p64d105s25p66s2 1s22s22p63s23p63d104s24p5 1s22s22p63s23p4 g. 4f. NT Section 5. b. NT 14. All rights reserved. 5 orbitals. 6.40 10 5 cm 1.2 1. d. 5d. c. 1s22s22p63s23p63d104s24p64d104f145s25p6 5d106s26p6 12. c 14. 1s22s22p63s23p63d104s24p6 h. Practice Problems 5 Section 5.00 108 m/s c 6. c. d.2 Part A Completion 1. 4 e. 6.3 Part A Completion 1. 8. 3 5 g. d. 1s22s22p63s23p63d104s24p3 i. 3. b. d. Inc. d. 7 orbitals Section 5. e 16. 18. a 15. 1s22s22p2 1s22s22p63s23p4 1s22s22p63s23p64s1 1s22s22p63s23p6 Ar b. The frequency of the light must be above the threshold frequency that will provide the necessary quanta of energy. 7. 3 orbitals. 4p. 3. c.Section 5. 5 orbitals 4s.48 1014 s 1 c Answer Key 773 Part C Matching 13. a 21. 5 f. 3. 5 1s. 1 orbital. 2p. 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. 1 orbital 2s. . 5. 3 orbitals 3s. NT 9. waves inversely light atomic emission spectrum light radiation photoelectric frequency Section 5. 2. d 17.00 1015 s 1 6. AT 11. 4.3 1.00 1010 cm/s 5. 3 2. e 18. f. 7.25 1015 s 1 19. 1 Part B True/False 11. b. publishing as Pearson Prentice Hall. 2 c. a. AT Part C Matching 17. c. 3 orbitals. a. 23. 1 orbital. ST 12. 9. a. 5. 4d. microwaves.47 10 19 J 6.21 10 7 m type of radiation: infrared. b g e c 5. h 6. Planck’s constant ground state photons photoelectric wavelike Interpreting Graphics 5 1. 5. ST NT AT ST AT NT 7. 9. 7. hertz Pauli exclusion principle quantum quantum mechanical model aufbau principle wavelength atomic emission spectrum photoelectrons Quiz for Chapter 5 1.28 10 6 m.60 1014. d 20. 1. infrared. infrared. 4. a 16. 1. 10. 4.14 1014. 7. Vocabulary Review 5 1. 4. 1.22 1014 s E 1.75 1014. 12.48 10 6 m.88 102 m 1 3.3. ultraviolet. Multiple Choice © Pearson Education.05 10 6 m. 4. 4.17 1014.. ultraviolet light.09 10 6 m.48 1015 : 7. 6. 6. infrared. visible light. ultraviolet light. The laser emits red light.52 10 8 m. Hund’s rule 2. 5.11 10 7 m. 11. 3. thus. d 9. Matching 1. 1. 4. 9. ultraviolet light.63 10 6 m.34 1014.00 108 m/s c 1600 103 s 1 1. 2.20 1015. Chapter 5 Test A A. b 18.30 1014.88 1014. 6. 8.74 10 8 m. ultraviolet light.02 10 7 m. 13. 9. 4. (s 1): 4. i 8.41 1013. 2.86 10 7 m.6262 10 34 J s 2. 6. a 3. 6.15 1015. infrared. 2. 7.6262 10 34 J s h 9.56 10 7 m.05 1018 s 1 radio waves. visible light. 9. 9. 2.38 10 8 m. 1. 2. B. j 7. 3. 6.00 10 15 J E 6. 8.08 1015. behave according to the Bohr model. 10. 2. publishing as Pearson Prentice Hall. c 17. 5. Li2 and He each have a single electron surrounding the nucleus and should. The Bohr model is adequate for explaining the emission spectra of atoms with a single electron. Inc. All rights reserved. 1. d 19. b c a d 15.93 1015. 11. 7. visible light. 5. 4. 3. infrared. 14. visible light. infrared.36 10 7 m. photons 774 Core Teaching Resources .88 10 6 m.01 1013.57 1014. cosmic rays 3. 1. E h 6. ultraviolet light 6→2 5→2 4→2 3→2 6 → 2 blue 5 → 2 blue 4 → 2 green 3 → 2 red All of the transitions end at the n 2 energy level. a 4. 6. 4. f 10. visible light. 2. 3. 3. 2. a. the 5s sublevel is lower in energy than the 4d sublevel. 2. Mn d. Sc g. 22. 14. c d c b C. Problems 1 © Pearson Education. phosphorus b. electrons travel around the nucleus along fixed paths much as planets orbit the sun. P c.2 1015 s 1) D.. b. but in the fourth level the energy ranges of the principal energy levels begin to overlap. a 29. Problems 21.25 1015 s 2. 19. Generally. d 6. d. Essay 26.25 10 5 cm 4. According to the Bohr model. Electrons occupy orbitals in a definite sequence. e 31. f 30. j 9. krypton 24. a. 20. 26.6262 10 3. 1s22s22p63s2 1s22s22p63s23p3 1s22s22p63s23p64s23d104p5 1s22s22p63s23p64s23d104p65s24d105p6 S e. 23. The quantum mechanical model. E E E h (6. NT 35. ST 37. As a result. Ne f. explains the positions of electrons in terms of probability clouds within which the electrons are most likely to Answer Key 775 . filling orbitals with lower energies first. a. a. c 32. b. a 10. g 7. aluminum c. AT 39. x w x 3p x w 3s x x x w w w 2p x w 2s x w 1s 1s22s22p63s23p4 b. neon 25. c c c b b 16. cobalt b. b 28. a c b d b 21. 13. 4.00 1010 cm s 6. d F. Fr 27.00 1010cm/s 22. 3. True/False 33. ST 41. 25. publishing as Pearson Prentice Hall. 24. 17.40 10 5cm 23. Energy level 1 Energy level 2 Energy level 3 Energy level 4 Energy level 5 2 8 18 32 50 c. 15. 18. i B. 1. AT 40. Br h. AT 36. Essay 29. Inc.4 10 18 J 34 Js) (5. h c f e 5. Sr D. NT 38.80 1014 s 1 28. x 3s x x x w w w 2p x w 2s x w 1s 1s22s22p63s1 3. Matching 1. Multiple Choice 11. Additional Matching 27. orbitals in a lower principal energy level have lower energies than those in a higher principal energy level. c. a. bromine x E. AT 34. AT Chapter 5 Test B A. b 8.C. All rights reserved. 12. c c/ 3. AT 39. 6. 7. AT Part B True-False Section Review 6. 11. 4. AT 37. NT Part C Matching 15. solid. 3s23p6 . bromine. 2. a Part D Questions and Problems 23. AT 44. ST 13. AT 41. bismuth is a metal. Tellurium: metalloid. 10. 20. publishing as Pearson Prentice Hall. Cl.2 Part A Completion 1. NT 38. 4. 2. AT 13. Polonium: metal. b 21. True/False 36. NT 11. decrease increases energy levels charge ionization 6. nitrogen and phosphorus are nonmetals. 3s23p1. Part B True-False 10. AT 42. 5. ST 43. g 19. Additional Matching 30. Si. e 18. but describes the locations of greatest probability in terms of specified orbital shapes. Section Review 6.be found. b 776 Core Teaching Resources . c 22. f 34. 9. c Part A Completion 1. 3s23p4. 5. iodine Part B True-False 11. fluorine. AT 14. NT 15. 7. e 16. NT 40. 8. 9. 8. ST 12. Inc. The Bohr model places electrons at specific distances from the nucleus. e 20. d 32. a 19. solids at room temperature 21. b F. S. 9. f 18. ductile. 3s23p2. Mg. solid 12. c 17. 6. 3s1. 3.3 18. d Part C Matching 14. good conductors of heat and electric current.. 3. Sulfur: nonmetal. NT © Pearson Education. AT Part C Matching 16. Ar. c 31. Oxygen: nonmetal. 10. 10. AT 13. Selenium: nonmetal. malleable. NT 14. 3s23p5. 24. 3s2. All rights reserved. e 35. 5. b 15. 3. gas. 7. a 33. NT Section Review 6. 8. arsenic and antimony are metalloids. d 16. high luster. 2. d 20. solid. The quantum mechanical model allows electrons to be at virtually any distance from the nucleus. P 3s23p3. 4. f 17. solid. names atoms alkali metals alkaline earth metals representative elements halogens noble gases transition metals inner transition metals p not filled E. properties groups periods or rows atomic number group Metals gases metalloids less more 12.1 Part A Completion 1. Na. a 17. increases electrons smaller electronegativity increases Part D Questions and Problems 19. Al. 5. Lithium in Group 1A has only 1 electron in its highest occupied energy level. 2. metal 3. Its first four energy levels are full. a. this element is bromine. bromine Practice Problems Section 6. All rights reserved.1 1. it loses electrons from its highest occupied energy level. 1) The metals: good conductors of heat and electric current. It is Group 4A. A magnesium ion has filled first and second levels. 3. 6. Sulfur in Group 6A has 6 electrons in its highest occupied energy level. When chlorine reacts. Astatine is in Group 7A. tellurium is in Group 6A. it gains an electron in its highest occupied energy level. ductile. c. 2. publishing as Pearson Prentice Hall. When a magnesium atom reacts. Both Ne and Ar have a completely filled highest occupied energy level. Al b. Br 22. 8. Elements in Group 5A have 5 electrons in their highest energy level. The three classes are as follows. Astatine is in period 6. a. a. Na. . Rb. nonmetal b. The period 4 element with 2 electrons is calcium. the additional occupied energy level in astatine significantly increases the size of the astatine atom as compared to the tellurium atom. 4s24p5 represents the Group 7A element in period 4. c 2. The prediction is that atoms of astatine are larger than atoms of tellurium. The complete configuration is 1s22s22p63s23p63d104s24p64d105s25p5. so its electron configuration must end in 5s25p5. An ion with three Answer Key 777 © Pearson Education. 2) The nonmetals: poor conductors of heat and electric current. 3. The configuration s2p3 indicates 5 electrons in the highest occupied energy level. O c. Fr 5. and are nonlustrous. A magnesium atom is smaller than a calcium atom because there are fewer occupied energy levels. malleable.3 1. The third period element in Group 5A is phosphorus b. Its first and second energy levels are full (1s22s22p6). gallium b.2 1. The electron configurations for Ag and Fe are: Ag 1s22s22p63s23p64s24p64d105s1 Fe 1s22s22p63s23p63d64s2 Section 6. a. oxygen d. c. The period 2 element with six electrons is oxygen. Iodine is located in period 5. The chemical and physical properties are largely determined by their electron configurations. but the nuclear charge is greater in magnesium. chlorine d. Silicon is in the third period. which is a feature of Group 5A.. metal d. which is also known as the noble gases. Inc. So the electrons are drawn closer to the nucleus. high luster when clean. A magnesium atom is smaller than a sodium atom because the shielding effect is constant for elements in the same period. A chlorine atom is smaller than a magnesium atom because atomic size decreases from left to right across a period. 3) The metalloids: elements that have properties similar to those of metals and nonmetals depending on the conditions. nonmetal e. Ne: 1s22s22p6 Ar: 1s22s22p63s23p6 7. The element in period 4 with 2 electrons in the 4s sublevel and 10 electrons in the 3d sublevel is zinc. Magnesium and calcium have the same number of electrons in their highest occupied energy level. S c. selenium. Although atomic size decreases across a period. Tellurium is in period 5. Na e. Transition metals are elements whose highest occupied s sublevel and a nearby d sublevel contain electrons. Section 6. metalloid c. a. so its electron configuration must end in 3s23p2. a 4. Cs. It is the fourth element in the period. They are in Group 8A. Li. The complete configuration is 1s22s22p63s23p2.Part D Questions and Problems 21. 4. b. 53 g/cm3 5. whether an element is not found in nature 9. The correct order for increasing electronegativity is then sulfur oxygen fluorine. 2. Listing the elements. 8.. 4 8. physical state at room temperature. Because electronegativity increases from left to right across a period. Beryllium’s first ionization energy is greater because first ionization energy tends to increase from left to right across a period. 5. period 2 Mo: Group 6B (or Group 6). 6. No. © Pearson Education. n a m c k b 7. Interpreting Graphics 6 1. 9. Inc. Magnesium normally forms an ion with a 2 charge. but the nuclear charge increases. of the element (and the state if the square is not color coded for style). transition metal. Because magnesium has a relatively low first and second ionization energy. 14. 7. 6. because they are not located in the same group or family. Because the chloride ion has the greater nuclear charge. Li: Group 1A (or Group 1). fluorine is more electronegative than oxygen. occupied energy levels is larger than an ion with two occupied energy levels. the removal of two electrons from magnesium is likely. 11. is a transition metal. When a sulfur atom reacts to form an ion it adds two electrons while chlorine adds one electron. 9. Li. table B 7. Their keys need to include the color. The trend is less pronounced as the number of electrons increases because the inner electrons shield the electrons in the highest occupied energy level. This organization helps scientists predict and explain similarities and differences in the properties of elements based on their underlying atomic structure. rubidium. mp. it will be smaller than the sulfide ion. The increasing nuclear charge pulls the electrons closer to the nucleus. f h d p j 778 Core Teaching Resources .4. 5. 3. l g o i e 12. The relatively high third ionization energy indicates the difficulty of removing a third electron from the filled second energy level. Vocabulary Review 6 1. 2617 C 6. potassium. sulfur is less electronegative than oxygen. Answers may include sodium. period 5 b.g. 4. a. d. Because electronegativity decreases from top to bottom within a group. Sulfide and chloride ions have the same number of electrons. Molybdenum. Barium is less electronegative than struntium because electronegativity values tend to decrease from top to bottom within a group. cesium. 10. All rights reserved. 16. Sodium’s first ionization energy is higher than that of potassium because ionization energy tends to decrease from top to bottom within a group. 4. Check students’ work. c. 3. Atomic size increases as you move down a period because the electrons are added to higher principal energy levels. resulting in a smaller atomic radius. 13. Mo. and francium. in alphabetical order. In the periodic table elements with similar chemical and physical properties are grouped together in vertical columns. general class. 15. 11. is an alkali metal. 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. 10. 2. publishing as Pearson Prentice Hall. 42 table A atomic weight 0. Lithium. This enlarging effect is greater than the shrinking effect caused by increasing nuclear charge. 10. and bp. Across a period from left to right the principal energy level remains the same. e. 8. which shields the electrons in the highest occupied energy level from the attraction of protons in the nucleus. a. c c a b b c b. 22. Multiple Choice 11. Group 8A (or Group 18) f. a. Li. a. 26. 14. 6. K. 25. d f j b 5. 28. F F. I c. period 6. 2. a c d b b Chapter 6 Test B A. a 8. transition metal. Questions © Pearson Education. Mg b. a. Be. 4. c. e 10. period 5. j 8. i 7. There is an increase in the number of occupied energy levels. K. Li. 7A(17). 17. 2s2 2p4 28. b.. 20. Ca. 12. 2. d. 32. 29. Be 3. halogen. 13. P 3. b. 2. 8A(18). Ar. 22. g 9. period 2. c c d a d 21. c. Cl D. Si c. F Cs. Cl P. 15. c i b g 5. d 9. 4s2 4p6 e. 10. AT 30. a 7. From left to right across a period. There is an increase in nuclear charge. period 4. Matching 1. Li. 3. noble gas. alkali metal. 26. 13. transition metal. alkali metal. Group 6B (or Group 6) e. C. 12. Group 2A (or Group 2) b. C. which draws the electrons closer to the nucleus. 1A(1). All rights reserved. Cs b. NT Chapter 6 Test A A. noble gas. 27. 4B(4). Sc 4. 18. 5A(15). period 5. d c d a a 16. c 6. f. Group 2A (or Group 2) period 4. Ar 4. period 4. Group 7a (or Group 17) d. Ba 29. Group 6A (or Group 16) period 6. h 6. 7. K. Essay 31. 19. Group 1A (or Group 1) c. 2. C. 3s2 3p1 d. c. Matching 1. b. 4. transition metal. 30. electrons are being added to the same energy level. 24. a. 23. K. The increasing charge on the nucleus tends to pull the electrons closer and atomic radius decreases. 16. Two factors influence the size of an atom as the atomic number increases within a group. period 2. 15. a. 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. a. NT AT 9. e C. Multiple Choice 11. h B. 24. Questions 31. halogen. alkaline earth metal. Sr. 21. f 10. K 4. 4s1 c. 20. Group 4A (or Group 14) 1s2 B. e. 17. c d a d d b a 25. Inc.Quiz for Chapter 6 1. 14. 5. Li. 2A(2). 23. number seven group (column or family) six two less ST 8. Br Cs. Group 1B (or Group 11) C. 19. 3. alkaline earth metal. 4. 27. Cs Answer Key 779 . publishing as Pearson Prentice Hall. b d d a d c d 18. 3. K. Nonmetals attain stable noble gas configurations by gaining electrons and forming anions with 8 outer electrons in the existing energy level. Students divide the values of first ionization energies by 300 and measure the appropriate length of straws. You’re The Chemist 1. B e.. 8. F. 6. xenon appears to have the ability to attract electrons and form compounds. Metals. 5. Electronegativity generally increases from left to right along a period. Si c. 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. Metals attain noble gas configurations by losing electrons and forming cations with a complete octet in the next-lowest energy level. NT Chapter 6 Small-Scale Lab Analyze and Conclude 1. O. C Ca. e 20. c. Part B True/False 11. c Part D Questions and Problems 24. b 18. publishing as Pearson Prentice Hall. a. 1 electron lost. A high ionization energy indicates that an atom has a tight hold on its electrons. 5. Although hydrogen is placed in Group 1A based on its electron configuration. cation b. anion d. have lower electronegativity values than nonmetals.1 Part A Completion d. Mg. b. 780 Core Teaching Resources . Based on this value. 4. f 22. Li. Inc. S B C D Section Review 7. 3. bromide ion. a. which is consistent with the general trend. d 19. which are on the left side of the table. The value for xenon is similar to iodine. 34. a 23. ST 14. Ca S. 3. calcium ion. 3. © Pearson Education. 1 electron gained. NT 15. which are on the right. AT 13. Ionization energy and electronegativity are properties that reflect an atom’s ability to attract and retain electrons. All rights reserved. 2 electrons lost. NT 12. g 21. Students must determine their own scale before they begin. magnesium ion. 9. Electronegativity generally increases from bottom to top within a group. Fluorine 2. c. the rest of Group 3A shows a reverse in this trend. valence electrons group electron dot structures octet rule cations anions 1 Halide ions gain charges D. 2. 4. cation 26. 10. A high electronegativity indicates an ability to attract additional electrons. cation c. K C. Rb= 25 a. b. 2 electrons lost. Except for boron. Students often use two wells to represent both ionic and atomic radii. 2. d. AT 16. silver ion. Essay 35. =Ba= b. a. Part C Matching 17. B 1. hydrogen is classified as a nonmetal.33. 7. 1s22s22p6 4. AT 15. electrostatic forces oppositely ionic bonds neutral formula unit crystals high large stable molten 9. a Part D Questions and Problems 21. The superior properties of alloys result from the cumulative properties of all the constituents of the alloy. 7. Part B True-False 11. When ionic compounds dissolve in water. 3. c 19.1 1. b 17. 22. 1s22s22p6 d. publishing as Pearson Prentice Hall. When ionic compounds are melted. 1s22s22p63s23p6 e. Ionic bonds are the electrostatic forces of attraction that bind oppositely charged ions together. the orderly crystal structure breaks down. and anions will migrate to the other. which make metals good conductors of electric current. Section 7. AT 14. 7. 6. 6 b. e 18. b 19. 3 c. In an ionic compound. a. a Part D Questions and Problems 21. NT 14. 8. 2. 10. As electrons enter one end of a bar of metal. d 17. AT Part C Matching 16. a. 3. hexagonal close-packed 10. NT 12. ST 13. their ions are free to move. 8. This movement of ions means that there is a flow of electricity between the two electrodes. This behavior makes the metal malleable and ductile. an equal number leave the other end. Practice Problems 7 Section 7. 5. 6. Inc. 7 3.3 Part A Completion 1. Solid metals consist of closely packed cations surrounded by free-moving valence electrons. Each ion is then free to move throughout the molten mass.Section 7. the positive charges of the cations equal the negative charges of the ions. 4.. aqueous solutions of ionic compounds also conduct electricity. (i) 1 (ii) K = (iii) K 2. alloy Part B True-False 11. 22. NT Part C Matching 16. If a voltage is applied. cations electrons metallic electrical malleable/ductile ductile/malleable body-centered/face-centered face-centered/body-centered Answer Key 781 . When a metal is subjected to pressure. Thus. the metal cations easily slide past one another. 1s22s22p63s23p5 c. an alloy can be more durable than one constituent but more malleable than another. e 18. (i) 2 (ii) Ba 6 (iii) Ba2 b. 9. AT 12. (i) 7 (ii) I (iii) I c. 2. d 20. For example. a. Metal cations are insulated from one another by electrons.2 Part A Completion 1. 4. © Pearson Education. cations will migrate to one electrode. The number of valence electrons in an atom of a representative element is the same as the group number of the element. All rights reserved. c 20. 5. 1s22s22p63s23p64s2 b. ST 15. ST 13. BaCl2 c. sodium becomes positively charged and chlorine becomes negatively charged. The ions arrange themselves in an orderly.3 1. Cl potassium ion. loses 3 electrons. Vocabulary Review 7 1. In NaCl. cation gains 2 electrons. and nickel d. 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).5. each sodium atom gives up one valence electron to a chlorine atom. chlorine 1s22s22p63s23p5 Chlorine has 7 valence electrons. Ba2 2 lost 1 gained 1 lost 3 lost cation d. 4. An alloy is a mixture of two or more elements. All rights reserved. Bronze: copper and tin c. Sterling silver: silver and copper e. Although the electrons are attracted to the metal cations. Al2O3 b. cation gains 1 electron.. three-dimensional array characteristic of a crystalline solid. Spring steel: iron.2 1. The metal cations are arranged in a very compact and orderly structure or pattern. the electrons are free to move about the crystalline structure. In Step 1. Patterns are used to calculate the positions of ions in the crystal and to define the structure of the crystal. c. f. When electrical current is applied to a metal. The combinations in b and c will form ionic compounds 3. • Body-centered cubic: every atom (except those at the surface) has 8 neighbors. 4. Stainless steel: iron. these mobile electrons can carry charge from one end of the metal to the other. c. 2. O2 barium ion. a. 3. j i f g 6. at least one of which is a metal. d. • Hexagonal close-packed: every atom has 12 neighbors. NaCl is typical of many ionic compounds. CaI2 2. each ion is surrounded by six other ions of opposite charge. anion chloride ion. but in a different arrangement than face-centered cubic. a. a. loses 2 electrons. 4. To reverse the lattice formation through melting would require enough energy to overcome the multiple atttractions within the crystal lattice. 9. Section 7. m 782 Core Teaching Resources . The coordination number is determined by using x-ray diffraction crystallography. The metallic crystal is thought to consist of an array of metal cations in a “sea” of electrons. chromium. Ionic compounds are formed when metals react with nonmetals. 3. anion gains 3 electrons. 2. Each ion attains the electron configuration of the nearest noble gas. Inc. b. c. publishing as Pearson Prentice Hall. d. a. b. 6. l a c e 11. 2. Interpreting Graphics 7 1. Cast iron: iron and carbon f. K oxide ion. which results in a very stable ionic compound. and carbon Section 7. ionic bonds form between sodium cations and chlorine anions. a. cation 4. cation b. © Pearson Education. anion loses 1 electron. k 13. • Face-centered cubic: every atom has 12 neighbors. e. 7. b 12. In Step 2. The coordination number is the number of ions of the opposite charge that surround an ion in a crystal. In this process. A metallic bond is made up of cations that are surrounded by mobile valence electrons. carbon. Alloys have properties of metals. 8. 6. Metals are crystalline. rather. c. b. 8. d. 5. 7. anion cation e. chromium. cation anion f. Brass: copper and zinc b. a. sodium 1s22s22p63s1 Sodium has 1 valence electron. 3. Na2S e. no individual electron is confined to any specific cation. NaBr d. 1s22s22p63s23p6 b. 15. Thus metals are good conductors of electricity. a. c 11. True-False © Pearson Education. f g d e 5. 22. b 11. This makes the metal malleable. 9. AT 32. d c d c d c 24. 14. 5. Be cation 36. 1s22s22p6 B. so 3 atoms of Cl are needed to form the compound AlCl3. j g a i 9. 26. 27. 28. 3. 2. i 35. some of the free-floating electrons leave the other end. 20. d E. O2 anion c. 7. 10. 13.. Na cation d. 6. 3. 3. b k a j 9. All rights reserved. 13. Matching 1. AT NT NT ST 28. Cl Al Cl Cl F 1s2 2s2 2p5 → Na F 1s2 2s2 2p6 1s2 2s2 2p6 Both ions have the configuration of neon. An electric current is a flow of electrons. the cations slide past each other easily. Metallic bonds are the result of the attraction of free-floating valence electrons for positively charged metal ions. 24. c b c a a c 18. 14. =Ca = b. 8. e f h c 5. Br 2 Ca Br 2 c. AT 31. Na 1s2 2s2 2p6 3s1 y Al3 3 Cl Chapter 7 Test A A. Essay 39. 8. 4. a. a a a c C. 22. 25. ST 33. Al Al 1. Cl anion b. 25. 23. 18. 2. The electron dot formulas show that one atom of Al can give 3 electrons. 17. 6. 7. when metal is struck. h 10. 21. 11. 15. Thus. 17. 23. k 10. 1s2 c. Multiple Choice 12. c b d b 20.5. 16. Questions 34. d 37. ST NT AT AT AT valence octet 8. Matching 3 D. The cations in a piece of metal are insulated from each other by the free electrons. 27. c b c b 16. 19. 6. NT 30. 13. AT Chapter 7 Test B A. 12. c d a c d Answer Key 783 . 21. Multiple Choice 12. 2. gaining eight pseudo-noble gas metals anion formula unit 38. AT 29. 26. 7. Inc. 4. As electrons enter one end of a piece of metal. publishing as Pearson Prentice Hall. h 10. a. 4. B. 19. Al Cl Quiz for Chapter 7 1. 1s22s22p6 d. b. Thus. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 Na . 2s1. 41. All rights reserved. AT Chapter 7 Small-Scale Lab Section 7. NO3 and HCl Fe. b. 5. NaF MgCl2 CaS Al2O3 KSCN NVR NVR Figure B Cation Analysis © Pearson Education. 784 Core Teaching Resources . Each of the following pairs of ions produces a visible product that can be used to identify the ion in question: PO43 and Ag .2 Analysis of Anions and Cations. 7. a. Essay 44. Inc. observable products. The group number for Ca is 2A. c. d. N c. a. 32. which means that F has seven valence electrons and reacts by gaining one electron to attain the noble gas configuration. 4s24p5. No. 35. which means that two valence electrons will be lost. Li b. An intermediate compound. AT 38. Ca . An orangebrown color forms. 2s22p3. The formula of the compound formed is CaF2. 1s22s22p6 K . which reacts with the nitrate ion. 3. d. You’re the Chemist All designs should include tests that produce unique. 34.. 42. NVR: No Visible Reaction WP: White Precipitate 1. The group number for F is 7A. publishing as Pearson Prentice Hall. b. Bloodred soln d. a. Na Cl . Bubbles Bubbles WP WP KI (K ) NaOH NVR CaCl 2 (Ca 2 ) WP FeCl3 (Fe3 ) Rust ppt. Cl P . FeCl2. E. Si d. c. P 3 Na3PO4 (PO43 ) Light yellow ppt. Ca2 43.C. Br Na . 2. The formula for the anion produced is F . when Ca and F react. neither of the solutions produced a visible product. 36. Fe3 and SCN . forms. True-False 29. 3s23p2. AT AT AT NT 33. 1. 4. 30. Ca2 and OH . the cation Ca2 is produced. Questions 40. 1s2 2s2 2p6 3s2 3p6 Sr2 . two atoms of F are required to react with one atom of Ca. page 200 Analyze Sample data provided. 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. 31. a. 1s22s22p6 F . SO42 and Pb2 . c. AT ST ST AT 37. NT 39. molecule d. Inc. 8. e 17. H H 6. equally nonpolar unequally polar electronegativities dipole interactions hydrogen bond electronegative oxygen. a 17. atom b. d 18. AT 14. 5. 7. 1 fluorine atom 22. 4 carbon atoms.2 © Pearson Education. molecule Part B True-False 9. 10 hydrogen atoms b. publishing as Pearson Prentice Hall. 8. 8. e 18. Part A Completion 1. NT 14.Section Review 8. b 20. nitrogen. NT 13. 10. a. 4. 9. 7. 6 carbon atoms. NT 12. c Part D Questions and Problems 20. 6.N 6 H 9 H Part B True-False 10. d Answer Key 785 . 6-Br 6-Br 6 9 9 b. 7. AT 13. stable electron covalent shared single unshared pairs double/triple coordinate covalent bond Energy bond dissociation energy resonance structure 15. 9. 5. ST Section 8. compound molecular nonmetals diatomic molecular formula 6. AT 15. ST 11. 4. 3. low high atoms structure Part D Questions and Problems 22. NT 12.4 Part A Completion 1. 4. 8. e 16. b 19. 3. 4. 2. 3. 7. AT Part C Matching Section Review 8. 2. d 16. a 18. 6. molecular compound 21. AT 13. or fluorine Part B True-False 11. 5 hydrogen atoms. 2. AT 10. a.1 Part A Completion 1. 6. a. molecule e. atom c. c Part D Questions and Problems 20. 5. ST 16. All rights reserved. 9. b 19. Reading from left to right: sp3 sp2 sp2 sp sp sp3 Section 8. 2. ST 14. 3. NT Part C Matching 17. H 6 C 6 6 6 N 6 c. molecular orbitals bonding orbital lower sigma or pi or three-dimensional VSEPR theory orbital hybridization Part C Matching 15. ST 11. 5.. a 19. NT 12. c 21.3 Part A Completion 1. 3. Two additional electrons are added to account for the ion having a 2 charge. Inc. Each sp2 orbital overlaps with an Section 8. The unshared pair repels the shared pairs more strongly. AT 15. Each carbon atom shares one electron with one of the two hydrogen atoms. AT Part C Matching 16.3 1. polar covalent bonds c. The actual bond angle for NH3. which can shift to any one of the carbon-oxygen bonds giving rise to three resonance structures. The four valence electron pairs repel each other. not diatomic e. Cl N Cl Cl 4.1 1.5°. but the unshared pair is held closer to the phosphorus than the three bonding pairs. molecule 2. molecule e. 4. ST 13.5 . Phosphorous needs 3 more electrons to fill the 3p orbitals. ionic b. nonpolar covalent bonds Practice Problems 8 Section 8. a. a. ST 14. O C O O 2 O O C O O 2 O C O 2 Part D Questions and Problems 21. atom c. The remaining three electrons for each carbon atom form a triple covalent bond. a 3. a. the angle between bonds is expected to be slightly smaller than the tetrahedral bond angle of 109. The four shared pairs of electrons repel each other to the corners of a tetrahedron. e 19. 5. This is a clue that a carbon-carbon multiple bond exists in this compound. The four fluorine atoms are covalently bonded to the central carbon atom. Carbon has 4 valence electrons and each of the oxygens has 6 valence electrons. atom d. diatomic c. is 107 . Section 8. polar covalent bonds d.Part B True-False 10. The carbon and oxygen can satisfy the octet rule by having the oxygens bonded to a central carbon. Molecular compounds tend to have lower melting and boiling points that that of ionic compounds © Pearson Education. AT 12. a similar molecule. d 18. F P F F 786 Core Teaching Resources . Nitrogen needs 3 more electrons to fill its second energy level.. c 20. three chlorine atoms are required to bond with nitrogen. Since each fluorine atom only needs one electron and phosphorus needs 3 electrons. there is an apparent shortage of atoms with which to bond. H F 2. A molecular structure gives information about the kinds and numbers of atoms present in a molecule. 2. not diatomic b. There is one double covalent bond between a carbon and oxygen. All rights reserved. Fluorine needs one more electron to fill its second energy level. Molecular compounds are usually composed from two or more nonmetallic elements. Because each chlorine atom needs only one electron and nitrogen needs 3 electrons. Because carbon can form four single covalent bonds.2 1. dispersion forces. three fluorine atoms are required to bond with phosphorus. b 17. dipole interactions. Chlorine needs one more electron to achieve a noble gas configuration. Boron forms three sp2 orbitals by mixing one 2s orbital and two 2p orbitals. molecule b. The two atoms share a pair of electrons in order to form a single covalent bond. publishing as Pearson Prentice Hall. diatomic 3. NT 11. The three sp2 orbitals lie in the same plane. The electron dot structure is: H C C H 5. diatomic d. All four bond angles are 109. Thus. hydrogen bonds 22. 120 apart from one another. For a bond to be classified as nonpolar covalent. atomic orbital of chlorine to form three equivalent sigma bonds. Most bonds are between unlike atoms. The other 4 electrons are unshared pairs. therefore. there is a partial positive pole and a partial negative pole. Inc. 4. 4. The bond polarities cancel. Each carbon atom has two electrons left over. δ– F F δ– δ+ δ– C F F δ– 5.4 1. These 2 unshared pairs repel the two bonding pairs and prevent F2O from being linear. However. It has 6 valence electrons. The electronegativity difference between P and O is about 1. none Answer Key 787 . which form 4 sigma bonds. O S O O S O O O O S O O O S O 2. two of which are bonding electrons. The difference in electronegativity between Na and O is about 2. Both carbon dioxide and carbon monoxide contain polar bonds. they must be ionic or polar covalent. The non-hybridized p carbon orbital overlaps with an oxygen p orbital to form one pi bonding orbital. CaO is an ionic compound and CS2 is a polar covalent compound.4 and the bond is polar covalent. tetrahedral. like atoms must bond. F Be F . 5. This is a clue that CH2CF2 contains a carbon-carbon multiple bond. H δ+ δ– δ+ H N H b. none C H 3. Carbon 2 mixes one s and two p orbitals to form three sp2 hybrid orbitals. O 3 C 3 O .© Pearson Education. trigonal planar. The two hydrogen atoms bond with one carbon atom while the two fluorine atoms bond with the other carbon.4 and the bond is ionic. 4.5°. there exists in this compound an apparent shortage of atoms with which to bond. Because each carbon can form single covalent bonds with four other atoms. The more electronegative atom in a covalent bond will have the symbol and the less electronegative atom the symbol. 109. H C H C F F c. To form four equivalent bonds. linear. Oxygen is the central atom in this molecule. 2. 180°. ionic compounds have much higher melting points than molecular compounds. Carbon 1 mixes one s orbital and three p orbitals to form four sp3 hybrid orbitals. the effect of the polar bond on the polarity of the entire molecule depends on the shape of the molecule. δ+ a. In carbon monoxide. publishing as Pearson Prentice Hall. The hybridization in SiF4 is sp3. The molecule is a bent triatomic molecule. the molecule is a dipole. Interpreting Graphics 8 1.. silicon mixes one s orbital and all three of the p orbitals. With like atoms. All rights reserved. the carbon and oxygens lie along the same axis. H H O . Section 8.5°. with a bond angle of approximately 104. 7. Generally. b. 6. the difference is zero and the bond is nonpolar covalent. none H . the hybridization involved in the carbon-carbon bond is sp2. 3. These electrons form a carbon-carbon double covalent bond. H2C2H and F2C2F bond angles of 120°. The molecule looks very much like the ethene molecule (C2H4). a. producing a nonpolar molecule. 120°. In carbon dioxide. 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. as in diatomic molecules. Therefore. linear. which overlap with the hybrid orbitals of the carbon and oxygen atoms to form three equivalent sigma bonds. 180°. 6. The remaining 2p orbitals are found in regions above and below the axis. All rights reserved.5. 3. 6. difference 0.0). 10. a c b d 5. N N c. bent triatomic. a. Cl P Cl Cl D. 2. 96°.2. 4. h 33. F(4. i e j f 5.0). difference 1. c 9. trigonal bipyramidal.5. Cl Cl P Br 2 Br b. Questions 28. Essay 31. H(2. Matching 1. C—H 5 mol C—O 1 mol O—H 1 mol 393 kJ 1 mol 356 kJ 1 mol 464 kJ 1 mol 347 kJ 1 mol 1965 kJ 356 kJ 464 kJ 347 kJ Chapter 8 Test A A. 2. 13. Two examples are diamond and silicon carbide. . Vocabulary Review 8 1.7. b a d c The first 2p orbitals lie along the axis connecting the atoms.0. polar covalent bond d. C—C 1 mol Total 2px 3132 kJ sigma bond 2px pi bond 2py 2py B. Network solids are substances in which all of the atoms are covalently bonded to each other. 7. 14. difference 0. 4. ST E. d b d a 17. 4. c 16.8). O(3. Br(2. H Br b. c C. 3. Multiple Choice 11. 12. pyramidal. ST 11. b 8. d 6. 26. b 27. 24. 3. polar covalent bond 30. Quiz for Chapter 8 1. none 7. ST 10. 25.5). b 22. and form pi bonds.5). K(0. Samples of these solids are thought of as single molecules. 18. 8. 8.5). 2. S(2. 788 Core Teaching Resources . coordinate covalent bond bond dissociation energy bonding molecular orbital sigma bond VSEPR theory hybridization polyatomic ion van der Waals forces hydrogen bond molecule c. 5. ionic bond b. Cl Cl Cl 90° and 120°. c c b a 23. 105 . polar covalent bond c. O(3. C O 29. F P F F . 9. d 21. b c ST NT 9. H O H ..1). none 15.8). Additional Questions and Problems 32. difference 3. © Pearson Education. and so form a sigma bond. a 7. Check students’ work. O O H H 6. a. g 10. Inc. N(3. 20. publishing as Pearson Prentice Hall. 19. 7. none 8. a. . sp. f 8. a 10. 1 Yellow No. Differences of less than 2. e.0 result in covalent bonding. Differences in excess of 2. page 245 Analysis Blue No. c b a c b b b c b 29. Multiple Choice 11. 5 and Blue No. C!O Yellow Green Blue 1. nonpolar covalent. 18. polar covalent. 2. 35.34.1 e. 40 C. ionic bond.N 6 H 9 C H @ H 2 N2 H Red d. 3. Red. 31. c. H H 6. Additional Questions and Problems 41.4 c. sp3 35. 37. 30. 40. 25. b 6. 36.3 42. 0. 13. 15. See the above figure.0 result in the formation of ionic bonds.9 b. 24. 1. 4. 33. Green food color is usually a mixture of Yellow No. and blue is Blue No. H2H b. sp2. a. 3. because under the law food manufacturers are allowed to use up their current supplies. 28. 5 Red No. 27. 2. c a d d d a b d b 20. depends upon The electronegativity difference between two elements is used to predict which type of bonding will occur when specific atoms combine. publishing as Pearson Prentice Hall. 6 if it is orange in appearance). polar covalent. 1.5 d. sp3. and blue are pure compounds. g e h c 5. H 6 H . © Pearson Education. d c d a a c c a c Chapter 8 Small-Scale Lab Section 8.70 103 kJ/mol (2.25 mol) 675 kJ B. Essay 40. 26.4 Paper Chromatography of Food Dyes. Red is Red No. 1. d 7. N N . N4N O H. Chapter 8 Test B A. 0. 5 (or sometimes Yellow No. Matching 1. 34.) 4. 3. but sometimes appears. 6(C—H) 6 (393 kJ/mol) 2358 kJ/mol 1(C—C) 1 347 kJ/mol 347 kJ/mol Total 2705 kJ or 2. yellow. 0. 23. a. 21. An ionic bond is formed when one or more electrons are transferred from one atom to another. 22. Questions 38. 16. All rights reserved. Yellow is Yellow No. 17. polar covalent. polar covalent. 19. a. j E. 14. Inc. 39. From left to right: sp. (Red #3 has been banned. O O P O O 3 .70 103 kJ/mol) (0. Answer Key 789 H H 6-N 6 H 9 H b. O . Cl Cl P Cl Cl Cl F S F F F F F D. i 9.2 f. 32. 0. 12. Covalent bonds are formed when atoms share electrons. O 2 H @ H H . 1 Part A Completion 1. b 14. 3. monatomic lose 1 2 3 8 1 8. c. 5.3 Part A Completion 1. hydrogen carbonate Part C Matching 8. NT 7. e 21. ST 17. ST 12. publishing as Pearson Prentice Hall. sodium dichromate. Make a small spot of each colored marker pen on a piece of chromatography paper and develop in solvent. Inc. 5. d. NT 18. ionic with polyatomic ion c. 2. -ite or -ate 13. ammonium permanganate hydroxide loses 2 gains 2 c. 5 because of the water content of the paper. Wet a portion of a piece of candy and blot it with a paper towel to remove excess water. 3. Roman numeral anion oxygen zero Part B True-False 9. 1 is the most polar because it runs the fastest and appears at the top of the chromatogram. Stock 10. You’re the Chemist 1. 2 25. NaClO3 b. Different water content changes variations in the polarity of the stationary phase (the water molecules hydrogen bonded to the paper). 3.2 Part A Completion 1. transition (Group B) metals 9. NT Part D Questions and Problems 24. 8. 4. 2. Section Review 9. d 22. ST 15. c Part D Questions and Problems 17. 4. Red No. 2 d. a. b 20. AT Part C Matching 19. 2. a. 7. Repeat for other colors of candy. 2. ionic with polyatomic ion 18. Rubbing alcohol runs much more slowly and gives slightly better separation than 0. 1 b. Some papers cause a reversal of the positions of Blue No. cation anion -ide sodium iodide 5. potassium hydroxide. 4. c Part B True-False 5. a. -ite or -ate © Pearson Education.. a 23. Pb3(PO4)2 c.5. AT Part C Matching 13. polyatomic 12. binary ionic b. gains 1 d. loses 3 Section Review 9.1% NaCl. a. NT 10. ST 6. 26. 7. 4. 6. Mg(HCO3)2 Section Review 9. b. 1 and Yellow No. 6. nonmetallic -ide atoms diarsenic pentasulfide Part B True-False 14. Blue No. a 10. b. Press the wet side of the candy onto the chromatography paper so that it makes a colored spot. a.1% NaCl. 1 c. b 11. d 16. d 9. Use a toothpick to spot a solution of powdered drink on chromatography paper. 3. All rights reserved. Develop in 0. classical 11. a 15. AT 11. 40 is the least polar. c Part D Questions and Problems 790 Core Teaching Resources . AT 16. iron(III) bromide. a. b. 13. All rights reserved. b. CrO42 SO42 f. 3 1 e. © Pearson Education. 4.2 1. MgO d. 17. 3. Al2O3 CaS f. molecular 2. AT . e. c. b. 1 gained 1 lost f. c. b.12. a. publishing as Pearson Prentice Hall. a 14. a. a.1 1. b. 2 1 f. phosphoric acid 14. c 15. e. c. hydroxide ions ionic cation anion Practice Problems 9 Section 9. a. FeO manganese(II) oxide or manganous oxide lithium nitride calcium chloride Answer Key 791 Part B True-False 16. acid 13. 2. FeBr3 BaCl2 d. Inc. 4. 2 lost 3 gained e. lead(II) acetate 8. Mg(OH)2 c. c. c. H3PO4 HF d. 21. b.5 Part A Completion 1. LiOH potassium hydroxide hydroiodic acid sulfuric acid Section Review 9. anion 6. b. small whole 12. b. hydrogen hydrogen ions hydrobromic nitric ionic 6. NH4 cyanide anion hydrogen carbonate anion phosphate anion chloride anion calcium cation sulfite anion Part B True-False 10. OH CO32 NO22 e. AlCl3 SnF2 e. c. 5. c. multiple 11. a. 2 3 lost d. acid proportions 4. a. binary Section 9. a.. ST 17. NT 19. N2O4 Part C Questions and Problems 20. lead(IV) acetate hydrofluoric acid diphosphorus pentoxide lithium bromide PCl5 FeO HNO3 KCl Ca(NO3)2 Section Review 9. d. ST 18. a. 3. definite proportions 9. b. b. f. carbon tetrachloride 3. a. Na2S KI f. a. c. 2 gained tin(II) or stannous cation cobalt(III) or cobaltic cation bromide anion potassium cation hydride anion manganese(II) or manganous cation d. 9. NT 12. 2 d. c. NT Part C Matching 13. d. 7. b. 2. d. a. acid 15. b. f. 2. a. b Part D Questions and Problems 16. c. proportions 10. 4A 7. ST 11. c. phosphorus pentachloride sulfur dioxide tetraphosphorus decasulfide CBr4 b. 8. 5. c.4 Part A Completion 1. d. b. elements 5. 3. e. KBr AgI e. a. Na nickel cation. 2A two 12 10 7A one nine 8. 7. 6. 4. strontium bromide nickel chloride potassium sulfide copper(II) chloride or cupric chloride tin(IV) chloride or stannic chloride Na3PO4 d. d. e.3 1. dinitrogen pentoxide f. phosphorous triiodide h. a. c. Li2CO3 sodium cyanide iron(III) chloride or ferric chloride sodium sulfate potassium carbonate copper(II) hydroxide or cupric hydroxide lithium nitrate sodium cation. a. mercury(II) oxide or mercuric oxide e. c. NH4Cl NaOH f. f. a.4. d. Fe3 copper(I) cation. h.4 1. 6. a. f. e. d. c. e. potassium phosphate b. nitrous acid sulfuric acid Ca(OH)2 NH4OH c. d. 3. aluminum hydroxide c. a. N2O5 h. KCN MgSO4 e. 2. a. 2. The law of definite proportions states that samples of any compound will always contain the constituent elements in the same proportions. K2Cr2O7 (NH4)2SO4 c. j. N2F4 3. 5. k. b. b. a. d. b. g. i. a. sodium hydrogen sulfate d. 2. f. K iron(III) cation. 10 10 10 a. nitrogen tribromide g. Section 9. publishing as Pearson Prentice Hall. Cu Section 9. b. NO3 SnCl4 H2S CaO HBr AlF3 SO42 CaSO4 Fe2(CO3)3 SF6 MgCl2 H3PO4 HNO3 OH PO43 Ca(NO3)2 Ca(OH)2 Ca3(PO4)2 AlPO4 Na3PO4 Al2(SO4)3 Al(NO3)3 Al(OH)3 Na2SO4 NaNO3 NaOH Pb(SO4)2 Pb(NO3)4 Pb(OH)4 Pb3(PO4)4 Section 9. b. phosphorous pentachloride carbon tetrachloride nitrogen dioxide dinitrogen difluoride tetraphosphorous hexoxide xenon difluoride silicon dioxide dichlorine heptoxide NBr3 c. The law of multiple proportions states that in two compounds containing the same two elements.. a. d. Ni2 calcium cation. b. h. 7. 9. b. ammonium sulfate 4. hydrofluoric acid carbonic acid Al(OH)3 LiOH Interpreting Graphics 9 1. Ba(OH)2 KNO3 d. d. c. 2. f. a. b. © Pearson Education. l. 5. 10. All rights reserved. e. c. K2S b. d.5 1. SO2 Cl2O d. 11. f. g. Inc. c. e. Ca2 Al3 Na Pb4 g. the masses of one element that combines with a given mass of the other element will be in the ratio of small whole numbers. Ca2 potassium cation. 18 18 18 18 792 Core Teaching Resources . c. b. Common names do not describe the chemical composition of a compound. copper(I) chloride b.. a c c d 19. 5. b 8. Completion 22. a 7. b 7. 21. 7. i 2. 30. They may relate to a physical or chemical property. dinitrogen trioxide c. 2. e 9. 2 27. h 5. Inc. 23. e B. Completion 23.Vocabulary Review 9 1. f 7. 20. 6. d. 4. 28. Compounds exist in enormous numbers. 15. 17. nitric acid E. 10. c c a a a 15. 3. potassium acetate Quiz for Chapter 9 1. 20. Metals PCl5 anion definite proportions the group number lose hydroxide H2PO4 dinitrogen monoxide FeCl2 tin(IV) sulfide (or stannic sulfide) dinitrogen pentoxide sodium hydrogen carbonate (or sodium bicarbonate) 14. Matching 1. 11. sodium sulfate Fe2O3. 12. 28. 13. 11. 13. g 6. c 8. NH4 magnesium nitrate -ide Ca3(PO4)2 31. 29. Multiple Choice 10. f 2. 13. 24. d 6. b a d c B. d © Pearson Education. Chapter 9 Test A A. a. 16. 22. 25. c c c b C. 4. 16. 26. a 3. Ca(NO3)2. e 6. iron(III) oxide Al2(CO3)3. d d c c 18. Essay 33. h 9. 25. All rights reserved. publishing as Pearson Prentice Hall. 1 oxygen nonmetallic hydrogen D. 4 1 ions -ide 27. g 4. 14. 30. h 5. calcium nitrate Na2SO4. gives information on the ratio in which the atoms combined to form the compound and promotes efficient and effective communication between chemists. 24. 18. i 9. 2. a D. 19. 9. SF6 b. 17. c 4. more C. 11. a Mg(CN)2 c. c. 8. f 8. 12. HgBr2 32. a. aluminum carbonate Answer Key 793 . Chapter 9 Test B A. j 10. but usually do not reveal what elements are in the compound. i c g d 5. Cu(OH)2 15. gain lose cation definite proportions 26. The systemic method tells what atoms are in the compound. b 3. 12. 3. Matching 1. Problems 32. Problems 31. b. 21. Multiple Choice 10. c a a c 14. 29. Fe2(SO4)3 PBr3 h. a. Ca3(PO4)2 k. e.. no visible reaction e. c. Ionic compounds consist of a metallic and a nonmetallic ion. g. a. using prefixes to denote the numbers of atoms of each element present.2 Names and Formulas for Ionic Compounds. AgOH silver hydroxide Teacher’s note: This is Ag2O. f. CaCO3 j. The name of the second element always ends in -ide. Molecular compounds are named from the elements that comprise them. silver oxide d. whereas molecular compounds consist of nonmetallic elements. Name silver carbonate silver phosphate © Pearson Education. HNO3 Chapter 9 Small-Scale Lab Section 9. d. i CaCl2 (Ca2 ) grainy white ppt milky white ppt cloudy white ppt grainy white ppt CF4 f. Ionic compounds are named from the two ions that comprise them. PbCO3 f. 34. Essay 35. Na2SO4 2. SiO2 b. Ag3PO4 c.33. Pb3(PO4)2 g. b. Pb(OH)2 h. Inc. All rights reserved. Ag2CO3 b. Ca(OH)2 l. HgCl2 carbon disulfide ammonium carbonate diarsenic pentoxide carbon monoxide tin(IV) hydroxide sulfuric acid phosphorus pentiodide potassium permanganate E. AgNO3 did not form a precipitate. AgNO3 Zn(OH)2 g. publishing as Pearson Prentice Hall. f j g k h l Formula a. d. PbSO4 i. No prefix is used if only one atom of the first element is present. c. h. using a Roman numeral to distinguish between positive ions of the same element that have more than one charge. 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. 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 . e. Density 2.5 mol H2O2 6.You’re The Chemist 1.0 g C 1 mol C 1.4 L SO2/1.0 g C 24. b 12. 9.2 Part A Completion 1. Cu2+ 2PO43 → Cu3(PO4)2(s) 2OH → Cu(OH)2(s) e orange ppt f Section Review 10. d 14. Fe(OH)3 d.36 g K2SO4 molar mass of C2H6 17. FePO4 c. no visible reaction © Pearson Education. mole 3.02 1.3 1015 atoms Pb 1. Fe2(CO3)3 b. NT 9. no visible reaction 2.. publishing as Pearson Prentice Hall. ST 8.00 mol K2SO4 6. NT d Na2SO4 no visible reaction (SO42 ) Part C Matching 11. CuCO3 j. 2 mol K2SO4 10 2 g K2SO4 2.5 1023 representative particles 1 mol H2O2 1024 representative particles Section Review 10. MgCO3 f. Mg2+ i. Fe3+ e. molar volume 5. AT Answer Key 795 g. NT 9. 3. c Name iron(III) carbonate iron(III) phosphate iron(III) hydroxide magnesium carbonate magnesium phosphate magnesium hydroxide copper(II) carbonate copper(II) phosphate copper(II) hydroxide Part D Questions and Problems 15. ST 7.02 1023 Part B True-False 6. 5. mole Avogadro’s number atomic masses molar mass 6. Cu3(PO4)2 k. Cu2+ .0 mol Pb 6. Mg(OH)2 h. 22.1 Part A Completion 1. AT 10. 2Fe3+ b. ST 7.3 g K2SO4 1. e. 4.02 1023 atoms Pb 10 8 mol Pb 12. Inc. 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. c 13. 3Mg Part B True-False 6. Fe3+ c.0 g H 1 mol H 6.00 mol SO2 3CO32 → Fe2(CO3)3(s) PO43 → FePO4(s) 2OH → Fe(OH)2(s) CO32 → MgCO3(s) 2PO43 → Mg3(PO4)2(s) 2OH → Mg(OH)2(s) CO32 → CuCO3(s) f. All rights reserved.4 4. 2. no visible reaction i. 3. AT 8. 2 mol C 6 mol H 1. a.5 16.0 g H 30. Mg3(PO4)2 g. Mg 2+ 2+ j. Cu(OH)2 l. 22. 3Cu2+ MgSO4 (Mg2 ) CuSO4 (Cu2 ) i white ppt blue ppt j white ppt k blue ppt l no visible reaction blue ppt k.0 g 174.65 10 636 18. 90 mol O 16. c 12.0 g Density 1.27 mol S 32. AT 9.43 102 g 3. 0. 352. 84. 3. Molar mass O2 2(16.2 g/mol Part C Matching 11.0 g mass 44. a. 5. a.5 g/mol 2 mol 117 g 17. 153. 26 g c.0 g S 1.0 g O 1.4 L 16. 310.4 g O 1. c 12.3% Hg 233 g HgS 32. 342.1 g Ca 100 24.90 mol O/1. a 13. d 11.0 16.85 mol H2O 5 3.5% Ca 164 g Ca(NO3)2 28 g N 100 17. 112 g Fe %Fe g Fe g Fe2O3 100 48 g O 112 g 160 g 160 g Fe2O3 100 70.1 g S 100 13. 1.500 mol 11.80 e.5g/mol 58. percent composition 100 molar mass empirical whole-number molecular 40.5% O 164 g Ca(NO3)2 1. b 13. 100 38.0 g 35.0 kg Fe 100 kg Fe2O3 447 kg Fe Part B True-False © Pearson Education.480 g or 4.4% O 284 g Mn2P2O7 201 g Hg c. 3.27 1 2 2 1.6 1023 atoms 6.4% Cr 152 g Cr2O3 48 g O 100 31.6% O 152 g Cr2O3 110 g Mn b.0 g b. NT Practice Problems Section 10.0 44. 96. 208.0 g/mol 16.5 g S 1.0 g/mol) 32.00 mol Na 1.1% N 164 g Ca(NO3)2 96 g O 100 58. b 14. a.500 mol 18.5 2 3 Empirical formula Na2S2O3 16.1 g c.0 g/32. a Part D Questions and Problems 14.8 g Fe 16..4 L/mol 0. ST 8.2 1.1 g Na 23.0 g/mol 4.2 L 15. 204. All rights reserved. 4.2 g/mol 3. Molar mass NaCl 23.20 10 2 g d.2 g/mol b.2 g/mol 10 1 g 796 Core Teaching Resources .0 g O 2 mol Fe 3 mol O 1 mol Fe 1 mol O d. 5.0 g/mol 2.27 mol Na 15. a.5 g 58. Molar mass Section Review 10. 22.8 103 g b.8% S 233 g HgS Section 10.87 102 g 4.00 mol S 40.7% Mn 284 g Mn2P2O7 62 g P 100 21.1 1. 32. 158.2 g/mol d. 104 g Cr 100 68.27 mol Na/1. e Part D Questions and Problems 28.27 1 2 2 1.96 g/L volume 22. 2. 1.0 g/mol b. Inc. a.Part C Matching 10. 7. 1. 29.0 g Na 1. AT 10. 180.0 g/mol 0. 6.0% Fe 639 kg Fe2O3 70.27 1. publishing as Pearson Prentice Hall. molar mass Fe2O3 55.27 mol S/1. 100 86.8% P 284 g Mn2P2O7 112 g O 100 39.00 mol O 30.3 Part A Completion 1.3 g/mol 2. 61 g Cl 5. 8.3 g 19. 6. 1.00 g %O 100% 65. 9. 9.0 g O2 1 mol O2 AT AT ST NT AT 1. © Pearson Education.3 g H 21.1 g 2.69% 98. a.61 g Cl Percent of Cl 100 18. 6.97 c. Percent C 100 81.98 10 10 5 5 mol mol 6. 1. 1 atm) Quiz for Chapter 10 1.84 g cpd 0.79% H 89. 2. molar mass 5.56 g Au Answer Key 797 . H2SO3.08 g 32.03 mol NH3 4.781 g cpd 48.08 10 2 mol 5.10 10 2 mol e.08 g Cl 52. 6.08 g 64. 150.1% H 12. publishing as Pearson Prentice Hall. All rights reserved.874 g H Percent H 100 18. They are indistinguishable on the basis of percent composition alone. Percent C 100 158.5. 13.9 10 3 mol d. a.25% 98. 4. 2.35 g cpd 68.4 g ethene 100 g C2H4 3.1 g Ca(C2H3O2)2 30.42 g H 52.0 g C2H4 197 g Au 1.08 g Acid X represents sulfuric acid. 2.74 g Sn 12.3 1.781 g cpd 0.84 g cpd 100 100 100 10. 7. 9. Use the same approach to show that Acid Y represents sulfurous acid. 4. b. therefore.0 g C 4. 5.84 g cpd 47. CHCl3 Vocabulary Review 10 1. 8.35 g cpd 31.1% C 0.00 mol 5.06 g %S 100% 32.00 mol 3.35 g of compound 5.. 85. 3.1 g 59. 2.1% Cl 85. 3. 15.43 mol 15. Inc. 1.6 L CH4 6. 2. Mass of Cl total mass of compound mass of Sn 18. 7.3% Sn 12. H2SO4.2 g Al 6. 4.8 g 5.06% 98. Percent C Percent H Percent Cl 5.05 g Ag 108 g Ag Interpreting Graphics 10 1. Some compounds have the same empirical formula but different molecular formulas.7% C 14.8% C Section 10. percent composition empirical formula 22. 10.1% Br Bromine accounts for the largest percent of the mass of dibromoethane.7% C 4.4% C 65.7% Cl 3.74 g Sn Percent of Sn 100 18. molar mass of H2SO4 98.34 g C 52. 5.4% C Mass C 30. 11.4 L molar mass mole Avogadro’s number standard temperature and pressure (0 C. Cyclohexane and ethene have the same empirical formula and. the same percent composition.3% H 4. CCl4 b.907 g C 3. 7.02 g %H 100% 2.11 g Fe 7. a substance atomic mass molar mass atom 32. b 4. c b b c C. 30. Matching 1.3 g 28. d b a c b b 15. d Chapter 10 Test A A. 29. 2. Problems 25. c a d c c 16. 18. 4. A mass that has molar as a modifier must be the mass of a mole. 364 g Ar 9.1 mol H 1.3 g 28.3 g C.0 g 1.4 14. 0.25 mol 98. 8.0 g F 2 mol C 4 mol F 38.00 mol C 3.50 6.02 1023 atoms 197.39 1. 19.0 g C 1.02 1023 atoms mol © Pearson Education. 17. 31. Ar 39. Matching 1. 23.0 g O 1. 22.1 mol H/3.02 1023 molecules F2 31.25 mol 1. Inc. 15. 25.4 L g 1 mol 15.. Chapter 10 Test B A. 19. A mole is a unit that counts all kinds of representative particles. 23.39 mol C/3.0 g 7640 L g L 0. Multiple Choice 9. Multiple Choice 11. 40. j a g e 5. 20. 26. 22. 14. 11.0 g C 1 mol F 19. 26.0 g F 2 mol C . c 7.0 g H 5.00 mol O 54. 798 Core Teaching Resources .4% 148.39 mol O 16. 1 mol Ar 27. h 3.12. d a c b d c B. 24.9 g Ar The mole ratio of C to F is 32.3 g/mol Mg(NO3)2 24.5 mol H 3.6 g F2 1 mol C 12.39 1. 24. e 5.12 mol. 13.0 g 96.1 613 g mol 22. 7. 6. d 10. 20.9% 148. 4 mol F The lowest whole-number ratio of C to F is CF2.0 g C 76. 5.1 g H 5.0 g F2 1 mol F2 28. Essay 28. 18.00 mol H 5.6 g/mol L mol 1 mol 3.7 g C 1. 10. 12.5 mol H 2 3 mol H 1 mol C 2 2 mol C 1 mol O 2 2 mol O Empirical formula C2H3O2 D.0 g N: 100 18.0 g 148.39 1. a d a c c b 21. a 8. h B.39 mol C 12.0 g O: 100 64. 24. publishing as Pearson Prentice Hall. so molar mass can be used for the mass of Avogadro’s number of particles of any pure substance. 16. 1023 atoms g 6. Problems 27. g 6. c a d a c 21. 12. 3. f.0 kg 1000 7636 L mol kg 44. 13.00 1023 molecules F2 1 mol F2 6.7% 148.2 g O 3. 14.3 g 96.00 mol O 1.39 mol O/3. 2. b c i f 9. All rights reserved.23 10 6 g mol 24.00 mol C 3.75 1015 atoms 6. 17.3 g Mg: 100 16.650 22. but expressed in grams.68 D.0967 0.239 0.75 22 10 Ca 1022 C 1023 O © Pearson Education.5 100. expressed in atomic mass units. Essay 35.28 4.0967 mol CaCO3 2 mol H 3. or Avogadro’s number of atoms of that element has the same numerical value as the atomic mass.0 g or P4O6 33.0870 0.0 g C 1 mol O 72. 1 mol P 34. the empirical formula is CO2.54 mol O 16.30 5.38 g P) 1.239 mol H2O 1 mol H2O x mol O 0. Molar Mass (g/mol) Moles of each compound Moles of each element Atoms of each element 18.68 g CaCO3 100.. page 304 Analysis Student data may vary slightly.24 5.82 1.7 g O 4. The mass of one mole.1 g x mol CaCO3 0.0 g Empirical formula is P2O3 empirical formula mass 110.54 1.88 1.239 mol H2O 1 mol H2O x mol H 0. and 2 2. 2. Inc.30 g H2O 18.5 g x mol NaCl 0.3 g C Chapter 10 Small-Scale Lab Section 10.0 g x mol H2O 0.0 g O 2.478 H 0.28 Thus.0 58.09 g NaCl Answer Key 799 .0870 Na 0. x mol NaCl 5.28 mol C 12.82 mol P 31. The mass of a single atom of an element is the atomic mass given on the periodic table.62 g O) 2. 27.24 1022 Na 1022 Cl Figure A 1 mol 58.0870 mol NaCl 1 mol 2.82 5.44 1023 H 1023 O 5.290 O 5. x mol H 0. H2O(l ) Mass (grams) NaCl(s) CaCO3(s) 4. x mol H2O 4.0 g.2 Counting by Measuring Mass. so that 219.239 mol H2O 1 mol x mol CaCO3 9.0870 Cl 0. (56.1 mol C 2.73 mol O 16. publishing as Pearson Prentice Hall.239 mol O 1.0967 Ca 0.239 O 0. All rights reserved.9 g molecular formula 2 P2O3 110.28 2.0 g 1 mol O (43.0967 C 0.478 mol H 1 mol O x mol O 0.09 9.1 0. b 20. 2Al(s) 6HCl(aq) → 2AlCl3(aq) 3H2(g) b. Water has the greatest total number of atoms. subscripts (l) (s) (g) (aq) catalyst x atoms H x atoms O 0.. publishing as Pearson Prentice Hall. NT 14.83 1022 atoms C 0. 7. a.44 x atoms O x atoms Na 1023 atoms 1 mol O 1023 atoms O B. g 21. Inc. 10. 8. Completion 1. c 18.0967 mol CaCO3 1 mol Ca 0.02 1023 atoms 1 mol Cl 22 5. 2. d 23.290 mol O 6. 3. 12.0967 mol Ca 1 mol CaCO3 x mol C 0. Water has the greatest number of moles in one teaspoon.02 1023 atoms 1 mol Na 5. 6. Multiple Choice 17. 6.24 10 atoms Cl 0.09697 mol Ca 6.02 2. AT 15. 8. x atoms H 0. 12. 4. 11. Determine the mass of 100 drops of water and then calculate the mass in grams of one drop. 9. 10. Write your name with the chalk and determine the mass of the chalk again.02 1023 atoms 1 mol Ca 22 5.0870 mol NaCl 0. f © Pearson Education. D.02 1023 atoms 1 mol C 5. 3. 4. equation reactants products mass coefficients element 7. 11.0870 mol Na 0. x mol Ca 0.75 1023 atoms 1 mol O 1023 atoms O 5.02 x atoms O 1.0870 mol Cl 6. True-False 13. 9.2 A. a 22.x mol Na x mol Na x mol Cl x mol Cl 0. 5. Determine the mass of a piece of chalk.88 1023 atoms 1 mol H 1023 atoms H Section Review 11. 800 Core Teaching Resources .0870 mol Na 6. Completion 1. 6.83 10 atoms Ca C. e 19.0967 mol C 6. 2C2H2(g) 5O2(g) → 4CO2(g) 2H2O(g) x atoms Cl x atoms Ca x atoms Ca Section Review 11. Problems 24. All rights reserved.239 mol O 6. 2.0967 mol CaCO3 3 mol O 0.290 mol O 1 mol CaCO3 4. ST x atoms Na x atoms Cl 0.1 A.0967 mol C 1 mol CaCO3 x mol O 0.0870 mol Cl 1 mol Na 1 mol NaCl 1 mol Cl 1 mol NaCl You’re The Chemist 1. 5.478 mol H 6. Convert the mass difference to moles and atoms.24 1022 atoms Na 0. 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 x atoms C x atoms O 0.02 1.0870 mol NaCl 0. NT 16.0967 mol CaCO3 1 mol C 0. 2. or water.3 A. publishing as Pearson Prentice Hall. f 16. silver sulfur → silver sulfide Silver metal and sulfur react to produce solid silver sulfide. 4. a. H2O(l) SO3(g) → H2SO4(aq) 2AgNO3(aq) Cu(s)→2Ag(s) Cu(NO3)2(aq) 4P(s) 5O2(g) → P4O10(s) C. and the coefficients are in their lowest possible ratio. B. 8. 2. none d. b 19. Ca(s) 2HCl(aq) → H2(g) CaCl2(aq) 5. ST 11. NT C. Ca(s) Mg(NO3)2(aq) → Ca(NO3)2(aq) Mg(s) Answer Key 801 © Pearson Education. e D.2 1. ST 16.B.. or a gas. Thus. Magnesium is a Group 2A metal and forms cations with a 2 charge. Completion 1. combination 23. one of the products must be a solid (precipitate). Questions and Problems 22. C3H8(g) 5O2(g) → 3CΟ2(g) 4H2O(g) 6. KClO3 → KCl O2 Next. 9. 9. True-False 13. First. Inc. balance the equation. 2KClO3(s) → 2KCl(s) 3O2(g) 4. ST Practice Problem Solutions Section 11. 8. d 15. H2(g) O2(g) → H2O(l) Fe(s) S(s) → FeS(s) MgCO3(s) y MgO(s) CO2(g) H2(g) Cl2(g) y 2HCl(g) Hydrochloric acid and solid calcium carbonate react to produce carbon dioxide gas. All rights reserved. a. True-False 10. and liquid water. 2. Cl2(g) Na (aq) Br (aq) → Br2(l) Na (aq) Cl (aq) The spectator ion is Na . 5.1 1. a 17. The balanced net ionic equation is Cl2(g) 2Br (aq) → Br2(l) 2Cl (aq) 21. NT 15. 4. a 21. ST 17. AgCl(s) b. 2Al(s) 3F2(g) → 2AlF3(s) 3. 5. water aqueous complete ionic equation spectator ions net ionic equation charge atoms precipitate solubility 7. aqueous calcium chloride. CaCO3(s) c. Oxygen is in Group 6A and forms anions with a 2 charge. b 18. ST 13. 3. FeCl3(aq) 3NaOH(aq) → Fe(OH)3(s) 3NaCl(aq) 7. 2Li3PO4(aq ) 3Zn(NO3)2(aq ) → Zn3(PO4)2(s) 6LiNO3(aq ) For any double-replacement reaction to occur. a. c 19. combination reactions: 1 and 2 decomposition reaction: 3 single-replacement reaction: 4 double-replacement reaction: 6 combustion reactions: 1 and 5 8. combustion b. AT 14. Matching 18. 6. Section Review 11. Matching 14. 6. PbCl2(s) . 7. the equation is balanced correctly. determine the formulas for the reactant and products and write them in their proper positions to form a skeleton equation. c D. Section 11. Mg O2 → MgO The balanced chemical equation is 2Mg(s) O2(g) → 2MgO(s) 2. There are 2 silver atoms and 1 sulfur atom on each side of the equation. Questions and Problems 20. They combine in a 1:1 ratio to form MgO. AT 12. 3. d 20. no reaction b. no reaction occurs in d. 9. 2. 5. AgCl(s) d. This reaction can be described as: Ba(NO3)2(aq) Na2SO4(aq) → BaSO4(s) 2NaNO3(aq) The net ionic equation is: Ba2 (aq) SO42 (aq) → BaSO4(s) 2. spectator single replacement balanced equation coefficients Solution: precipitate Section 11. f 9. 2. 2K(s) H2SO4(aq) → K2SO4(aq) H2(g) d. 15. 3. 4. (NH4)2CO3(aq) 2NaOH(aq) → Na2CO3(aq) 2NH3(g) 2H2O(l) 802 Core Teaching Resources . 3. d d b a c nitric acid HNO3 N2O O2 → NO 2N2O O2 → 4NO C6H6 O2 → CO2 H2O 2C6H6 15O2 → 12CO2 6H2O NH3 CO2 → CH4N2O H2O 2NH3 CO2 → CH4N2O H2O C6H6 HNO3 → C6H5NO2 H2O balanced as written C. c 8. 23. Vocabulary Review 11 1. Multiple Choice 11. 3. Matching 1. 2Na(s) Interpreting Graphics 11 nitrous oxide nitric oxide oxygen carbon dioxide water ammonia urea benzene nitrobenzene carbonic acid 1. right b.c. 7. 24. All rights reserved. 4. 2. 10. Al(OH)3(s) Quiz for Chapter 11 1. 17. a. d 10. a. 22. 7. 3. publishing as Pearson Prentice Hall. This reaction can be described as: Mg(s) 2HCl(aq) → Η2(g) MgCl2(aq) The net ionic equation is: Mg(s) 2H (aq) → Η2(g) Mg2 (aq) 3. 8. h 7. 4. a c b c a 16. i B. A. 14. skeleton equation formulas balanced a. left arrow 2HgO(s) → 2Ηg(l) O2(g) 2Ag (aq) 2Na (aq) 2NO3 (aq) CO2 (aq) 3 → 2Νa (aq) 2NO3 (aq) Ag2CO3(s) net: 2Ag (aq) CO32 (aq) → Ag2CO3(s) C4H8(g) 6O2(g) → 4CO2(g) 4H2O(g) Br2(l) → 2NaBr(s) 9. Problems 26. 25. a 6. 2. no precipitate c.. 6. 5. combustion decomposition net ionic equation catalyst reactants 6. Pb(NO3)2(aq) 2NH4Cl(aq) y PbCl2(s) 2NH4NO3(aq) 2 Pb 2Cl → PbCl2(s) 4. b b c d b 21. 13. no reaction In a bromine is less reactive than chlorine so no reaction occurs. N2O NO O2 CO2 H2O NH3 CH4N2O C6H6 C6H5NO2 H2CO3 Chapter 11 Test A © Pearson Education. 8. e g j b 5. 20. 19.3 1. 3Ca(s) 2H3PO4(aq) → Ca3(PO4)2(s) 3H2(g) b. 4. 2KBrO3(s) → 2KBr(s) 3O2(g) c. Because zinc is less reactive than sodium. CuS(s) b. Inc. In b calcium replaces a less reactive magnesium and in c potassium replaces the less reactive hydrogen. 12. 18. 15. B. Li2O(s) H2O(l ) → 2LiOH(aq) electricity b. AB CD → AD CB. 3. NaCl(aq) AgNO3(aq) → NaNO3(aq) AgCl(s) 27. Fe2O3(s) 3CO(g) → 2Fe(s) 3CO2(g) Answer Key 803 . 2H2O(l ) JJJJH 2H2(g ) O2(g ) c. C3H8(g ) 5O2(g ) → 3CO2(g ) 4H2O(l ) 29. double-replacement reaction. 1. 14. E FG → EG F.27. a. a. a. single-replacement decomposition double-replacement C5H10(g) 5O2(g) → 5CO(g) 5H2O(g) (incomplete) b. 2C3H7OH(l) 9O2(g)y6CO2(g) 8H2O(g) (complete) 29. the cations have exchanged positions such that two new compounds are formed. 22. The activity series of metals lists metals in order of decreasing reactivity. Whether one metal will replace another is determined by the relative reactivity of the two metals. KOH(aq) HCl(aq) → KCl(aq) H2O(l ) K (aq) OH (aq) H (aq) Cl (aq) → Κ (aq) Cl (aq) H2O(l) Spectator ions: K (aq) and Cl (aq) Net: H (aq) OH (aq) → H2O(l ) b. a. 23. 12. Essay 30. 2HNO3(aq) Ca(OH)2(aq) → Ca(NO3)2(aq) 2H2O(l ) e. Pb(NO3)2(aq) KI(aq) → PbI2(s) KNO3(aq) 2 Pb (aq) 2NO3 (aq) K (aq) I (aq) → PbI2(s) K (aq) NO3 (aq) Spectator ions: K (aq) and NO3 (aq) Net: Pb2 (aq) 2I (aq) → PbI2(s) c. 20. Multiple Choice 11. A metal will replace any metal found below it in the activity series. 4. d a c b d C. j 9. the metal E has replaced the metal F so that a new compound and a different element are produced. single-replacement reaction. a 7. 19. 28. publishing as Pearson Prentice Hall. b a d b d 21. 2Fe(NO3)3(aq) 3Na2CO3(aq) → Fe2(CO3)3(s) 6NaNO3(aq) net: 2Fe3 (aq) 3CO32 (aq) → Fe2(CO3)3(s) D. 24. d b c b a 16. Inc. 2. 13. ZnI2(aq) NaOH(aq) → NaI(aq) Zn(OH)2(s) Zn2 (aq) 2I (aq) Na (aq) OH (aq) → Na (aq) I (aq) Zn(OH)2(s) Spectator ions: Na (aq) and I (aq) Net: Zn2 (aq) 2OH (aq) → Zn(OH)2(s) D. CS2(s) 3O2(g) → CO2(g) 2SO2(g) b.. 25. 2HNO3(aq) Mg(OH)2(aq) → Mg(NO3)2(aq) 2H2O(l ) c. Problems 26. a. b 10. 2K3PO4(aq) 3MgCl2(aq) → Mg3(PO4)2(s) 6KCl(aq) net: 3Mg2 (aq) 2PO43 (aq) → Mg3(PO4)2(s) b. g h i f 5. d 8. 18. 2Al(s) 3Fe(NO3)2(aq) → 2Al(NO3)3(aq) 3Fe(s) d. Chapter 11 Test B A. a. 17. Essay 30. Matching © Pearson Education. b. All rights reserved. c 6. c. e 28. Pb2 i.. All rights reserved. Pb k. 3. Na2SO4 Pb(NO3)2 → 2NaNO3 PbSO4(s) j. moles/molecules balanced equation mass/atoms atoms/mass moles STP (standard temperature and pressure) 804 Core Teaching Resources . 4. 2NaOH Pb(NO3)2 → 2NaNO3 Pb(OH)2(s) Section Review 12. Ag f. a. 3Ag c. 3. 6. Ag I → AgI(s) © Pearson Education. 2NaCl k. KI AgNO3 → KNO3 AgI(s) Silver iodide is pale green. 2Na3PO4 3Pb(NO3)2 → 6NaNO3 Pb3(PO4)2(s) 2. Ca2 You’re the Chemist 1. publishing as Pearson Prentice Hall.3 Precipitation Reactions: Formation of Solids. Na3PO4 3AgNO3 → 3NaNO3 Ag3PO4(s) c. NaOH AgNO3 → NaNO3 AgOH(s) Teacher’s note: The students will write the above reaction but this is what really happens: 2NaOH 2AgNO3 → 2NaNO3 Ag2O(s) H2O e. 4.1 Part A Completion 1. 3. 2Na3PO4 3CaCl2 → 6NaCl Ca3(PO4)2(s) m. Place one drop of lead nitrate on a small pile of dry table salt. 3Ca2 m. 2. Adding one drop of lead nitrate to a few grains of table salt causes white crystals to grow on the salt. 5. a. Be sure to keep part of the pile dry and look carefully for signs of yellow lead iodide. Pb2 2I → PbI2(s) 2. Na2CO3 2AgNO3 → 2NaNO3 Ag2CO3(s) b. Ca 2 c h m white white ppt ppt j. Na2CO3 Pb(NO3)2 → 2NaNO3 PbCO3(s) h. 2NaOH 5. 3Pb2 h. Pb2 2 d no i n no white visible visible ppt reaction reaction e j o no white white visible ppt ppt reaction l. 2KI Pb(NO3)2 → 2KNO3 PbI2(s) Lead(II) iodide is bright yellow. Pb 2 CaCl2 → 2NaCl CO32 → Ag2CO3(s) PO43 → Ag3PO4(s) Ca(OH)2(s) a white ppt tan ppt brown ppt f white ppt white ppt k white ppt white ppt OH → AgOH(s) Cl → AgCl(s) CO32 → PbCO3(s) 2PO43 → Pb3(PO4)2(s) 2OH → Pb(OH)2(s) SO42 → PbSO4(s) 2Cl → PbCl2(s) CO32 → CaCO3(s) 2PO43 → Ca3(PO4)2(s) 2OH → Ca(OH)2(s) b g l g. Ag e. 2Ag b. 1. b. Na2CO3 Pb(NO3)2 → 2NaNO3 CaCl2 → 2NaCl PbCl2(s) CaCO3(s) l. n.Chapter 11 Small-Scale Lab Section 11. and o all gave no visible reaction so it is not necessary to write an equation. Silver nitrate produces a similar result. Sodium hydroxide reacts with calcium chloride to form sodium chloride and solid calcium hydroxide. page 345 Analysis AgNO3 Pb(NO3)2 CaCl2 ( Ag ) (Pb2 ) (Ca2 ) Na2CO3 (CO32 ) Na3PO4 (PO43 ) NaOH (OH ) Na2SO4 (SO42 ) NaCl (Cl ) i. NaCl AgNO3 → NaNO3 AgCl(s) f. Mixings d. Inc. AT 9. Inc. NT 14. 7. d 18. a. a 1. 2.4 L CO 6. representative particles volumes coefficients mole ratios product/reactant reactant/product moles Part D Questions and Problems 18. actual yield Part B True-False 7.0 g O2 1 mol O2 22. AT 10. a 17. AT 8. moles N2 2 moles O2 3 moles N2O3 2 molecules N2 2 molecules O2 3 molecules N2O3 2 volume N2 2 22. b 14. b 14.Part B True-False 7.4 L 67. NT 10. maximum 6. c 15.1 mol SO2 1 mol O2 2 mol SO2 1. 3. 4.02 1023 molecules O2 17. AT Practice Problems 12 Section 12.7 mol O2 1. e 16.8 L 2 mol N2 56 g 3 mol O2 96 g mass reactants 152 g 2 mol N2O3 152 g mass product 152 g 3 mol Cl2 21 mol Cl2 19. limiting reagent 2.0 g NH3 33.1 mol O2 in excess Part B True-False 8.1 mol SO2 2 mol SO2 mol SO3 can be formed 2. 4.6 mol O2 1. c 16. 3. publishing as Pearson Prentice Hall. 5. b 2.2 L volume N2O3 2 22. 3.2 © Pearson Education. e 16.6 mol O2 needed SO2 is the limiting reagent. 2 mol SO3 b.9 g NH3 1 mol NH3 4 mol NH3 7 mol O2 Section 12. ST 12.8 L volume O2 3 22. NT 10.1 Part D Questions and Problems 18. 6.. d 15. ST 9. AT 12. a 17. used up 3. All rights reserved. d Section Review 12. AT 12. product 4. AT 11. 2. 10A Ci y A10C2Ci 10A 25 A10C2Ci 250 apples A10C2Ci 2C 18 mol O2 17. e Part C Matching 15. AT Part C Matching 13. ST 11. 14 mol FeCl3 2 mol FeCl3 1024 molecules O2 1 mol O2 6. Part A Completion 1.4 L 44. NT 13.8 g O2 1 mol O2 2 mol CO 32.4 L 44. ST 11.7 L CO 1 mol CO Part C Matching 13. theoretical yield 5.3 Part A Completion 1. AT 8. 2KClO3(s) y 2KCl(s) 3O2(g) 3 mol O2 12 mol KClO3 2 mol KClO3 Answer Key 805 . c 21. ST Part D Questions and Problems 20.1 19. NT 9. 0 mol H2 needed Oxygen is the limiting reagent.0 g O2 32.0 3 mol O2 2 mol KClO3 21 mol O2 2.25g acetic anhydride 0. 100% = 54. Zn(s) 2HNO3 y H2(g) Zn(NO3)2 1 mol H2 1 mol Zn 7. publishing as Pearson Prentice Hall.0 g KClO3 1 mol KClO3 122.0 g O2 2 mol KClO3 1 mol O2 55.2% 2 mol Ag 2 mol AgCl 29.1 g/mol Student 1: 0.0 g CO2 4. 3. Inc.0 g C 1 mol C 44.1 g/mol 2.3 g Ag 71.4 g O2 5.0 g O2 1 mol O2 22. 2HCl(g) y H2(g) C12(g) 1 mol Cl2 1 mol HCl 25.3 g MgO 71..0 g C 12.8 g Mg 24.0 mol H2O 22.0155 moles SA Student 2: 0. 4 mol NO 2 mol HCl 1 mol H2 20 mol HCl 21 mol Cl2 3 mol Cl2 2 mol FeCl3 O2(g) y CO2(g) 1 mol C 1 mol CO2 18. 2Ag(s) Cl(g) y 2AgCl(s) 1 mol AgCl 2 mol Ag 84 g AgCl 43.5 L H2 22.0 g MgO 0.2 g Ag 1 mol Mg 2 mol MgO 6. 806 Core Teaching Resources .5 g AgCl 75.4 g Zn 22 g Zn 1 mol Zn mass of Ag(s) reclaimed = 0.946 75. 2 mol H2O 4 mol O2 8 mol H2O formed 1 mol O2 4. 2H2(s) 2.80 g O2 8.4 L H2 1 mol H2 65.25 g acetic anhydride 0.0 g AgCl 108 g Ag 1 mol Ag 13.3 1. 100. C(s) Section 12. All rights reserved. c.1 g/mol b.5 g AgCl 2 mol AgCl 108 g Ag 63 g Ag 1 mol Ag 1 mol O2 2 mol CO 32.3. 138. 5.1 g Cl2 1 mol Cl2 percent yield percent yield 5. a.0 g MgO 58.1 g Cl2 1 mol AgCl 143.0514 moles Student 2: 5.79 g Student 2: 2. 180.72 L CO 1 mol CO 3 L O2 22.3 g Ag © Pearson Education.817 Interpreting Graphics 12 1.6 g Cl2 25.0 g CO2 1 mol CO2 2 mol NO2 2 mol NO 184 g NO2 46 g NO2 1 mol NO2 4.0 g CO2 66. 42.00 mol O2 1 mol O2 1 mol H2O 180 g H2O 3. 2 mol H2O 18.0147 moles SA Student 1: 5.3 g Mg 2 mol Mg 40.0 L N2O3 2 L N2O3 6. 2H2(g) O2(g) y 2H2O(g) 1023 molecules O2 2 molecules H2O 1 molecule O2 4.65 g Section 12.3% 66. 2H2(g) O2(g) y 2H2O(g) 2 mol H2 8 mol H2 4 mol O2 1 mol O2 1 mol O2 16 mol H2 8 mol O2 2 mol H2 Oxygen is the limiting reagent.4 L CO 6.0 g O2 1 mol O2 10.0 g CO2 100% = 83. 4.0 1023 molecules H2O 2 mol H2O 1 mol O2 45.5 g HCl 2 mol HCl 71.0 g MgO 1 mol MgO actual yield 71. 102.2 1.6 g KClO3 3 mol O2 32.0 g Cl2 25.8 g HCl 36. 14 mol KClO3 4.5 mol O2 O2(g) y 2H2O(g) 1 mol O2 2 mol H2 160. 10 mol H2 2. 14 mol FeCl3 3. 75.0 g H2O 5.0514 moles salicylic acid Student 1: 2.5 L O2 7. 15. a. Percent yield: 28. 9. 2. 4.509 g 7.6 g KClO3 18.0 g H2 2. e 4. 5. 3. 10. a. The coefficients of a balanced chemical equation describe the relative number of moles of reactants and products. 4. volume. 5. 10. 8.0 L O2 1 mol Al2(SO4)3 342 g Al2(SO4)3 74.0 g CO 4.80 L 5. which is reflected by a higher percent yield than that obtained by Student 1. 4. 14.745 g Student 2: 2. Matching © Pearson Education.0 g H2 32. Quiz for Chapter 12 1.0% 27. 13. 558 g Fe 55. c Answer Key 807 4. 17.8 g Fe 2 mol Fe 28. Essay Vocabulary Review 12 1.6. 450 g 325 g 125 g excess Ca(OH)2 1 mol Ca(OH)2 125 g Ca(OH)2 74. 22. 3. All rights reserved. Problems 19. 1 mol Fe 3 mol CO 20.69 mol Ca(OH)2 remaining 25. Multiple Choice 7. the amounts of reactants and products can be calculated.1 g Ca(OH)2 1. a 6.20 102 g CO 1 mol CO 21. 12. c 3. i 9. or number of representative particles. c 23. There is no limiting reagent. The number of moles may be converted to mass. 10. d 4. Student 1: 62.00 102 g Al2(SO4)3 3 mol Ca(OH)2 1 mol Al2(SO4)3 5. b c b a 15. e 10. e 3.00 102 g Al2(SO4)3 3 mol CaSO4 1 mol Al2(SO4)3 596 g CaSO4 5.00 L Chapter 12 Test B A. Student 1: 1. 7. j 7. 2.0 g 1 mol CH3OH 2 mol H2 32. f B. g 6. Inc.7% 8. NT NT NT NT ST E. 16. d h f a 5. b. 9. a 5. From this information. Student 2 exhibited much better lab technique. c a c c C.0 g 32. Matching 1.0 g CH3OH 1 mol CH3OH b. coefficients reactant moles atoms 44. f 6. publishing as Pearson Prentice Hall. 18.0 g CH3OH 100% 87.0 g CO2 58 g C4H10 1 mol C4H10 1 mol CO2 44 g CO2 2 mol C4H10 8 mol CO2 26. b c b d 11. Theoretical yield: 1 mol H2 4.5% D. 5. b 8. b 2.94 g C4H10 .00 L O2 1 mol O2 2 mol KClO3 22.8 6. 8. d 325 g Ca(OH)2 Al2(SO4)3 is the limiting reagent.2 g KClO3 1 mol KClO3 100% 96. because the mole ratio of the reactants is 1 mol N2 to 3 mol H2.. b 2.4 L O2 3 mol O2 122. 1. 15. Student 2 should receive the higher grade.5% Student 2: 94. Additional Problems 24.0 L H2S 3 mol O2 2 mol H2S 15.1 g Ca(OH)2 1 mol Ca(OH)2 1 mol Al2(SO4)3 342 g Al2(SO4)3 136 g CaSO4 1 mol CaSO4 Chapter 12 Test A A. 1 g CaO 1 mol CaCO3 1 mol CaO 28. publishing as Pearson Prentice Hall. Multiple Choice 7. 19.4 g CaO % yield 100% 28. 15.0 g CuO 79. Problems 22.35 mol 0. CuO is the limiting reagent.8 g CaCO3 1 mol CaCO3 100.00 mol NH3 1 mol SnF2 1 mol HF 45.1 g N2 4 mol NO 1000 g 10. The maximum amount of A2B3 that can be produced (0.7 g SnF2 176 g SnF2 1 mol SnF2 1 mol CH4 1 mol CO2 2.80 14. 21.6% yield © Pearson Education. 10.35 mol NH3 present NH3 excess c. 4 mol CO2 5.50 mol H2O 6 mol H2O 180 g C6H12O6 135 g C6H12O6 1 mol C6H12O6 29.90 g Al2O3 31. Essay 27.1 g H2SO4 1 kg 1 kg 8. 20. D.1 kg NO 808 Core Teaching Resources . 3. 1 mol N2 3. Additional Problems 28.92 mol 2. 105 g CO2 2 mol H3PO4 1.0 g NH3 17.0 g H3PO4 1 mol H3PO4 98.4 25. 50.0 g CH4 1 mol CH4 24.0 g HF 2 mol HF 20.0 g C2H2 1 mol CO2 1. d d b d d 12. 8. 2 mol NH3 3. 23.5 g CaO actual 100% % yield theoretical 26.0 g NH3 2 mol NH3 17.0 g N2 1 mol N2 energy C.0 g NH3 1 mol NO 1000 g 0.50 mol of B remaining in excess.0 kg NH3 4 mol NH3 1 kg 1 mol NH3 30. 2 mol NH3 3 mol H2 12.1 g CaCO3 1 mol CaO 56. a.43 mol 28. 26. Inc. 13.65 mol CuO b.65 mol CuO 3 mol CuO 34.25 105 kg H2SO4 3 mol H2SO4 1 mol H2SO4 1000 g 98.B.0 g CO2 26.00 104 g C2H2 2 mol C2H2 1 mol C2H2 44.5 g CuO 3. 3 mol CuO 2. 150 mol CO2 103 g CH4 16. 16. 32. 9. c d a b c E.65 mol CuO present Thus.0 mol of A requires only 1. b c b a c 17.. 2 mol Al2O3 3 mol O2 1 mol O2 1L 22.5 g CaO % yield 92.5 mol of B in order to react completely.0 g NH3 3. All rights reserved.43 mol NH3 react 1 mol NH3 Since 57. 18.69 30.32 104 kg H3PO4 1000 g 3 mol CuO 1 mol NH3 57.50 mol) is thus limited by the amount of A that is available.03 mol CuO needed 1 mol CuO 290. Based on the 2:3 molar ratio between A and B.4 L O2 1000 mL 102 g Al2O3 625 mL O2 1 mol Al2O3 1. with 0. the 1.0 g HF 156.0 g NO 1 kg 17. 14.0 mol H2 8. 6CO2 6H2O uuy C6H12O6 6O2 1 mol C6H12O6 4. 11.0 g NH3 5. 2 % error (assuming baking soda is 100% NaHCO3). 6. AT Part C Matching 17. 4. ST 12.53 © Pearson Education.2 Analysis of Baking Soda. have students add just enough HCl from a third pipet to turn the mixture red. 10. AT 15. ST 16. 2.70 g D.78 g Section Review 13. 8. b Answer Key 809 .46 g 0.83 g B. Inc.41 mmol HCl 4. 10. 3. page 367 Procedure Sample answers are given. The % error is the % of baking soda in baking powder assuming no other errors.29) g HCl 1. NT 14.. 7. AT 15. 5. 22. 8. a Part D Questions and Problems 101.2 Part A Completion 1. surface vapor pressure manometer vapor pressure 101. (10. e 19.70 4. 8.83 g 0. 100% (0. Section 13.46 g NaHCO3 7.30 atm Analysis 1. c 20. A.29 g E.53 g G. 8. denser condensed vaporization boiling cooling 6. motion empty space far apart independently random or rapid 6. 7. ST 12. 9. F.3 kPa or 1 atm Part B True-False 11.45 g 2. 3. no attractive or repulsive forces exist between the particles. 4. publishing as Pearson Prentice Hall.3 kPa 436 kPa 1 atm 760 mm Hg 4. 10. Odors travel long distances from their sources. Part B True-False 11.Chapter 12 Small-Scale Lab Section 12.875 mmol NaOH (0.53 mmol NaHCO3 0. (10. (See Steps 2–7. They move independently of each other and travel in straight line paths until they collide with one another or other objects.53 mmol NaHCO3) 6.28 g C.28 g 2. According to the kinetic theory. If the mixture does not turn red when thymol blue is added. 3.3 kPa or 1 atm You’re the Chemist 1. 4. Repeat Steps A–G and 1–7 except use baking powder instead of baking soda. collisions kinetic energy atmospheric 0°C 101. 9.0840 g NaHCO3/mmol NaHCO3) 5. (0. 24. NT 14.00 mmol HCl/g HCl 6. AT 13. 3. 10. 2.875 mmol HCl unreacted 5.500 mmol NaOH/g NaOH 0.) 2.53 mmol HCl neutralized (5.30 atm 3.41 mmol HCl total 0. NT 16. the motion of the particles in a gas is constant and random. 5.27 103 mm Hg 1 atm 23. b 18. Students must measure the mass lost by the pipet to add to the total mass of HCl used. AT Part C Matching 17. d 21. c 19. d 20. e 21. Because the particles are relatively far apart. a 18. HCl NaHCO3(s) → CO2(g) H2O NaCl 2. 4. All rights reserved.45 g)/0.875 mmol HCl unreacted) 5. ST 13. 2.78) g NaOH 0.1 Part A Completion 1.45 g baking soda 3. 24. c 18. The temperature of the system remains constant while the change of state is occurring. 754. NT Part C Matching 15. NT 14. 100. 1 atm 3. compress fixed melts melting point freezing point 6. Inc. NT 14. there can be no collisions or pressure. 810 Core Teaching Resources . 3. sublimation vapor pressure carbon dioxide phase 5. d © Pearson Education. 8. d 18. Evaporation leads to cooling of a liquid because the particles with the highest kinetic energy tend to escape first. Section 13. 4. the disruptive vibrations of the particles are strong enough to overcome the attractions that hold them in fixed positions. a 20. 2. This process is called sublimation. equilibrium triple point 0. particles throughout the liquid have enough kinetic energy to vaporize. its particles vibrate more rapidly as their kinetic energy increases. 2. 22. but there are attractions between particles of a liquid. Eventually. high crystalline lattice unit cell amorphous Practice Problems 13 Section 13.3 mm Hg 100.2 1. e 19. The average kinetic energy of the particles of a substance is directly proportional to the Kelvin Temperature. Solids that have a vapor pressure that exceeds atmospheric pressure at or near room temperature can change directly to a vapor. Part B True-False 9. Liquid B would evaporate faster because it has a higher vapor pressure.4 Part A Completion 1. the beaker must be sealed so that the rate of condensation can equal the rate of evaporation. 10. 7. 8.3 Part A Completion 1. According to kinetic theory. Part B True-False 11. f 19. NT 10. a 20. 7. At the boiling point. there are no attractions between the particles in a gas.9925 atm 760 mm Hg 101. Because there are no particles of matter in a vacuum. the average kinetic energy increases by a factor of two.3 mm Hg 0. AT 13. All rights reserved.61 kPa Section 13. which indicates that it is more volatile. Setting aside fluctuations due to changes in the weather. 6.0 C 273 173 K 73 oC + 273 346 K Because the Kelvin temperature increases by a factor of two. b 21. NT 12. 2. When a solid is heated. 9.Part D Questions and Problems 22. publishing as Pearson Prentice Hall. 3.5 kPa 760 mm Hg 4. The organization of the particles within the solid breaks down and the solid becomes a liquid. The remaining particles have a lower average kinetic energy and a lower temperature. For a dynamic equilibrium to be established. Gas pressure is the result of collisions between between rapidly moving particles in a gas and an object. 4. g 22. f Part D Questions and Problems 21.016°C 0.3 kPa 754. Part D Questions and Problems 24. e 17. Section 13.. AT Part C Matching 16. AT 11. ST 15. b 16. c 17. 5. 2. ST 13. AT 12. 23. you would notice that the pressure reading on the barometer would decrease as you climbed in altitude.1 1. 7. boiling point Solution: barometer 5. condenses 9. Peanut brittle is an amorphous solid. 5. Allotropes are two or more different molecular forms of the same element in the same physical state. and buckminsterfullerene. The melting-point curve leans slightly to the right (has a positive slope) indicating that. and vapor phases of a substance are in equilibrium. In diamond. publishing as Pearson Prentice Hall.4 1. One example of a crystalline solid is sodium chloride. This line represents the set of all temperature–pressure values at which the solid and gas phases of water are in equilibrium. 4. 100 Normal Melting point Normal Boiling point B Solid 40 Triple point S C 0 30 10 0 20 40 Temperature (°C) 60 Liquid V Vapor 80 A M 60 Section 13. 20 © Pearson Education. 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. crystal allotropes melting point pascal Answer Key 811 . 2. 6. 101. 4. 4. Inc. repeating. 2. freezes Vocabulary Review 13 1. including diamond. 3. 7. Carbon has multiple allotropes. Interpreting Graphics 13 1. the melting point of bromine increases. The triple point is the temperature and pressure at which solid.Pressure (kPa) 3. See answer to 1. kinetic energy 2. Normal melting point 7.3 kPa (1 atm) 2. In general. 5. evaporation 4. 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. gas pressure 3.0°C Normal boiling point 59°C Triple point 8°C and 6 kPa 3. 50°C 6. A molecular solid. each carbon atom is strongly bonded to four other carbon atoms in a rigid three-dimensional array.3 1. threedimensional pattern called a crystal lattice. Section 13. as pressure is increased. The carbon atoms are arranged differently in each allotrope. the remaining particles have a lower average kinetic energy. Solid bromine is more dense than liquid bromine. In crystal atoms. graphite. 8. The fastest runner corresponds to the particles in a liquid with the greatest kinetic energy. 8. See answer to 1.. ions. Higher pressures favor the denser phase of a substance. liquid. The melting point of water decreases as the pressure increases. 3. This line represents the set of all temperature–pressure values at which the liquid and gas phases of water are in equilibrium. All rights reserved. The carbon atoms in graphite are arranged in widely-spaced sheets. 4. or molecules are arranged in an orderly. When these particles vaporize. All rights reserved. The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to atmospheric pressure. if atmospheric pressure is E. ST 33.3 kPa. 3. Only water at pressures of more than one atmosphere will boil at higher temperatures. 3. 25. The additional energy is being used to change the liquid water to water vapor. Because the Kelvin temperature is directly proportional to the average kinetic energy of the particles in a substance. 26. 7. c a b a 24. NT 30. 21. j n h e d 6. 12. Additional Questions 38.3 kPa. 39. 22. When water boils at standard atmospheric pressure. 24. d b d d c b 21.. 9. 4. 7. d b n e m 11. C. Multiple Choice 15. Matching 1. AT C. 18. 8.3 kPa 1 atm G. 22. Essay 38. 13. 9. then the boiling point of a liquid will be lower than its normal boiling point. 14.0 mm Hg 37. ST 32. 2. AT 35. 20. 0. 4. a d d c b 20. Inc. 29. 12. it cannot be heated above 100°C. kinetic decreases temperature kinetic elastic condensed minimum kinetic energy external (atmospheric) pressure unit cell amorphous ST AT AT F. 2. True-False 31. 17. 3. it doesn’t matter how many particles there are in the sample. AT 34. 12.3 kPa 81. i f g j c 6.Quiz for Chapter 13 1.803 atm 760 mm Hg 101.803 atm 1 atm 0.70 atm 101. 9. Additional Problems 36. 812 Core Teaching Resources . 10.70 atm 34. 27. 5. The normal boiling point is the boiling point of the liquid when the atmospheric pressure is 101. 10. 5. 10. k b g f l 11. 25. 28. b b c c B. AT 29. 8. 610.81 103 mm Hg 1 atm D. 8. Conversely. 13. AT 31. 17. 27. 16. publishing as Pearson Prentice Hall. a m c i B. 3. 2. AT D. NT 36. The pressure must be increased to the point at which water boils at 150°C or higher to kill bacteria. If atmospheric pressure is less than 101. 16.3 kPa 375 kPa 1 atm 760 mm Hg 2. 5. 19. 4. 7. Chapter 13 Test A A. Multiple Choice 15. 30. 13. k h l a Chapter 13 Test B A. 6. AT 32. a a a b b 26. True-False 28. 19. a a c d b © Pearson Education. The temperature of the water will not rise until all of the water is in the gaseous state. Essay 35. 18. 11. Problems 33. 23. 23. 14. ST 37. Matching 1. 3. page 400 Analyze and Conclude 1. the boiling point of the liquid will be higher than its normal boiling point. compressed spare volume temperature moles 6. ST 12. © Pearson Education. liquids are denser than gases. 2. Water beads up and alcohol spreads out due to stronger intermolecular attractions in water. The volume of the particles is insignificant and their collisions are perfectly elastic. a 19. The water drop increases in diameter over time as the alcohol evaporates and is captured by the water drop. 4. The volume of the particles in a gas is small compared the overall volume of a gas. 9. NT 13. Cover and observe. Part B True-False 10. more liquid freezes. In a liquid. The attractions in the resulting mixture are weaker overall. Answer Key 813 . 4. 5. 40. 5. a. AT 11. Place a drop of vinegar and a drop of BTB about 3 cm apart in a Petri dish. 3. b Part D Questions and Problems 20. 2. If the mixture is heated. If the mixture is cooled. kinetic doubles reducing pressure Chapter 13 Small-Scale Lab Section 13. The motion of particles in a gas is constant and random. The particles in a solid vibrate around fixed points. 2. All rights reserved. 4. The particles of a gas are relatively far apart and there are no attractive or repulsive forces among them. Ammonia thatt evaporates is captured by the BTB.. but the temperature of the mixture remains the same as long as liquid water is present. Additional Questions 39. the particles are closely packed together. Inc. the particles are attracted to each other. As a result. b. Cover and observe.greater than 101. The BTB slowly changes from green to yellow even though there is no mixing of the drops. c 16. more spread out pools.3 The Behavior of Liquids and Solids. 8. Water in the dish evaporates and condenses into a cloud when it contacts the cold surface under the ice. e 18.3 kPa. ST 14.1 Part A Completion 1. 3. Solids are dense and difficult to compress. Because the particles are closer together. the alcohol cloud is made up of larger. publishing as Pearson Prentice Hall. You’re the Chemist 1. The particles travel in straight paths until they collide with other particles or the walls of their container. AT Part C Matching 15. more ice melts but the temperature of the mixture remains the same as long as ice is present. The BTB slowly changes from green to blue even though there is no mixing of the drops. There are no significant attractive or repulsive forces between particles in a gas. 3. The many pieces of calcium chloride effectively dry the atmosphere leaving no water vapor in the dish. E. As more energy is added. more particles of the liquid acquire enough kinetic energy to escape. The drop of water on top of the dish provides enough cooling to cause cloud formation. usually in an organized array. which is why a gas can expand to take the shape and volume of its container. d 17. Calcium chloride absorbs water from the environment in the dish. Section 14. which is why a gas can be compressed. which is why liquids have a definite volume. Place a drop of ammonia and a drop of BTB about 3 cm apart in a Petri dish. 7. The water “cloud” consists of tiny individual beads. In solids. all the liquid boils away at a constant temperature. The BTB turns from green to yellow in the presence of vinegar. Ethanoic acid that evaporates is “captured” by the BTB. 8.31 (K • mol) 1. inversely increases Boyle’s mass Kelvin 6. 8.2 Part A Completion 1. 2. At a given temperature. AT n 2. 4. a 16.0 kPa 0. 2. a 814 Core Teaching Resources . 2.Section 14. b 19. AT Part C Matching 15. NT 12. AT Part C Matching 15. Inc.4 Part A Completion 1.3 kPa 0. molecules all have the same average kinetic energy. d Section 14. All rights reserved. Charles’s Gay-Lussac’s directly combined amount Part D Questions and Problems 19. 3. NT 14. If two molecules with different masses have the same kinetic energy. publishing as Pearson Prentice Hall. ST 14.21 g NO2 Part C Matching 17. V2 V2 V2 T2 55 kPa 473 K 173 K T1 2 1. 9. b 18. 6.64 10 mass of NO2 2. AT 13. 1 2 Part B True-False 10. 4.3 Part A Completion number of moles PV nRT n ideal gas constant (L • kPa) 5. 5. 7.64 10 mol NO2 2 mol NO2 46. NT 13. a 21.31 2 12. Part D Questions and Problems 22. the less massive molecule must have a higher velocity.061 L 61 mL P1 Section 14. 5. 3. 8. 3. 7. AT 12. ST 14.0 g NO2 1 mol NO2 1. 9. n P R V T 25. 10.29 10 mol O2(g) P V 240. ideal real attractions volume Part B True-False 11. NT 13. 6.325 kPa 8. AT 11. e 20. 7.5 10 kPa P1 V1 T2 T1 P2 91 kPa 0. 8. ST 12. 4.31 K mol 301 K 2 Part B True-False 11. The kinetic energy of a molecule is equal to where m isthe mass and v is the mv2 velocity of the molecule. NT 15.5 L 295 K L K kPa mol 1.075 L 273 K 303 K 101. P2 P2 23. 10. c 18. total sum lower uniform diffusion effusion hole inversely molar mass Graham’s law © Pearson Education. n n 20. 9. d 16.. c 17. NT 16. b 17.275 L L kPa R T 8. c Part D Questions and Problems 18. The motion of the tires causes the air in the tires to heat up. So.43 g b. n n P R V T 102. the particles inside the tire have a greater average kinetic energy.568 mol V V 12.0 L 10.24 8. n n 2. 103 kPa b.0 L 297 K 4.5 kPa (22.5 kPa P V 26.31 L K kPa mol 25. V2 6.875 1. 4.0 g 5. which in this case is about half the initial pressure. 1.31 25 L 297 K n 3. 3.1 1.31 L K kPa mol mass of He 4.00 L P2 V2 T1 P1 T2 V1 105. Interpreting Graphics 14 1. the balloon will only expand until internal pressure is equal to the external pressure. the pressure inside the mattress decreases. P2 2. V1 T2 10.31 K mol 273 K 101. Thus. 3. P1 V1 V2 P2 341 kPa No.00 g He 1 mol He 1. they collide less frequently and less forcefully with the walls of the mattress.25 mol 273 K 101.355 mol 8. temperatures are higher in the summer than in the winter. V V 10.0 g n R P 1 mol\44. 0.0 L V 7. Overnight the air in the mattress cools down. a.Practice Problems Section 14.. 372 K 4. n V 7. Section 14.267 L 3.41 g 2.0 kPa 8. a.3 1.4 kPa 55.31 K mol 3.0 L T1 248 K P1 V1 T2 P2 T1 V2 501 kPa 5. publishing as Pearson Prentice Hall.4 1. as does the volume.0 kPa 22.9 kPa 248 K 105 L 104 kPa L K kPa mol 155.0 g 32. 0.2 1. 108 kPa Answer Key 815 .935 28.03 mol He 1. On average. T T P R V T 2 2.12 g He T 8. resulting in a greater pressure inside the tire.03 mol He 4.4 L V1 5.3 kPa Section 14. 2. Inc.0 kPa 35.31 L K kPa mol 0. a.75 RateO2 RateN2 0.3 kPa © Pearson Education.00 mol 8. V V n R P 2. At higher temperatures.3 10 mol argon 500. All rights reserved. the frequency and force of the collisions between the particles and the walls of the tire are greater.0 kPa) 26.01 102 K Oxygen effuses slightly slower than nitrogen.3 kPa V 50.95 L 25.2 L 373 K P2 466 kPa 298 K 7.0 L 323 K P2 71.7 L 273 K 101.5 L 300 K T2 6. Section 14. PO2 PO2 PO2 n Ptotal (PN2 PAr) 98.0 L P V L kPa n R 7. 0.66 L 473 K T1 n R T P L kPa 0. 372 K b.0 g T 8.5 kPa 3.5 L L kPa R T 8.568 mol 0.31 K mol 298 K 10 2 mol O2(g) molar massO2 molar massN2 0. the average kinetic energy of the particles in the air decreases.0 L 373 K V2 15.0 kPa 50. Consequently. V2 T1 T2 V2 © Pearson Education. T1 T2 V2 V2 24. b c b c c 19. g i j f 5.5. Problems 21. e 7. d 6.6 105 m3 Ptotal (PN2 2PCO2) 145. 12. 5.0°C 273 273 K P1 V1 T2 T1 P2 180 mL 273 K 95. a. The behavior of a real gas deviates from the behavior of an ideal gas. 6. Multiple Choice 9. 18. 46 g/mol 8. molar mass molar mass 7.33 L K L K kPa mol 0. 2.5 kPa 388 kPa T1 T2 22.0°C 273 303 K 303 K 550 mL 764 mL 218 K P1 Vocabulary Review 14 1. 2. 15. publishing as Pearson Prentice Hall. P2 7.33 10 3 mol 46 g/mol 3 Chapter 14 Test A A. PO2 PO2 PO2 151 mL D. Matching 1.9 kPa 308 K 101. Inc.0 L 215 kPa 10.41 g 8. 10. 11. 13. collisions doubles small real diffusion P1 273 K 149 kPa 713 K 27°C 273 300 K P1 115 kPa 10°C 273 263 K P2 99 kPa P1 T2 V1 P2 T1 115 kPa 263 K 3. h 10. An ideal gas is one that follows the gas laws at all conditions of pressure and temperature. a 7. b Quiz for Chapter 14 1. d 20.43 g 9. c 8. especially at low 816 Core Teaching Resources . P1 P2 V2 23. c 2.5 105 m3 99 kPa 300 K 3.90 10 3 mol 46 g/mol 0. n n b.267 L 372 K mol 0. Essay 25. f 6. b 6. All rights reserved. a.5 kPa 76. T1 T2 V2 V2 V2 8.31 9. C2H6O (ethanol) C. b 4. P1 T1 P2 T1 T2 P2 P2 T2 T2 T1 227°C 273 500 K 27°C 273 300 K 300 K 655 kPa 500 K 393 kPa V1 P2 V2 P1 V1 V2 156 kPa 15. n n 103 kPa 8. 17.31 8. molar mass molar mass b. d b a c d 14. h 3. d 10 3 108 kPa kPa mol 10 B. a 9.267 L 372 K mol 0. 3.0°C 273 218 K 30. g 5.3 kPa T2 V1 T1 55.0 kPa) 40. 4.0°C 273 308 K 0.9 L 35. 16. e 8.0 kPa (28.90 8.. 3. 4. AT V2 P 29.0 kPa)(310 K) (85.49 mol 44. Matching © Pearson Education. True-False 26.29 L 31. AT 35.0 g 44.2 g/mol formula mass 2. kinetic theory assumes that the particles of an ideal gas have no volume and are not attracted to each other. g i j f 5. 1 T1 T2 T2 F.31 L kPa 300 K K mol n 0. 25.004 times faster. B. PV nRT P V 152 kPa 8. e 7.0 L 300 K 2.829 L V1 F. 20.50 L)(273 K) (330 K)(101.925 L)(250 K) (1.0 UF6 containing U-235 diffuses 1. AT 36. Essay 17. Multiple Choice 11. n n P R V T 216 kPa 8. V1 T1 V2 V2 V2 T2 V1 T1 (0.17 mol 96.0 g CO2 1 mol CO2 22 g CO2 0. 4. d c b d d b E. 28.temperatures and high pressures. AT 28.4°C PV P2V2 31. True-False 33. 1 1 T1 T2 V2 V2 P1 T1 V1 V2 T2 Chapter 14 Test B A. NT C.009 1. 22..650 L)(313 K) (293 K) 0.49 mol CO2 D. If all gases behaved ideally. 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. a 9. This is not true for real gases. P1V1 V2 V2 V2 P2V2 P1 P2 (425 kPa)(1. 26. c b a b 32. ST 34. Inc.0 kPa)(3.694 L P2 T2 P2 P1 (98.3 kPa) V2 2. 24. publishing as Pearson Prentice Hall. 15. h 10. d 6.0 kPa) T1 T2 E. 13. 18. Problems 27. V1 T1 T2 T2 T2 V2 T2 V2 V1 (0.31 L K kPa mol 25. 2. 14. Also. All rights reserved. 19.0 L n R T 8. 21. b (80. Additional Problems 30. 1. which can be liquefied and sometimes solidified by cooling and applying pressure. ST 29.0 1. T2 357 K or 84. 16.17 mol Rate235 252.25 L) 185 K or 88°C T1 Answer Key 817 .20 L) 615 kPa 0. Rate238 349. 3. NT 27. c 8. a c a b b d 23.004 31. 12. Additional Problems 37. 11. 12. ST 818 Core Teaching Resources .0 L 200 K 19.38. homogeneous 7. c 22. f 20. a 20. larger filtration Colloids Tyndall effect Brownian 6. publishing as Pearson Prentice Hall. c 21. 5. 2. AT Part C Matching 15. solvent 6. 2. Vary the size of the BTB drops from “pinheads” to “puddles. a. 4. partially conduct nonelectrolyte hydrates efflorescence Part B True-False 11. 10.1 Part A Completion 1. NT 16. 9. strong 10. 13. AT 17. AT 12. 2. electrolytes 9. 4. a 17. “like dissolves like” 8. ST 14. All rights reserved.2 Part A Completion 1. b Part D Question 22. 23. 3. 3. Succeeding pictures should show the yellow area gradually increasing until the entire dot is yellow. 9. As ammonia diffuses.4 Diffusion. 3NaNO2 + 2HCl → 2NO + H2O + NaNO3 + 2NaCl Part D Questions and Problems © Pearson Education. 7. lower b. As the particles diffuse from the center.” Tiny drops are better able to detect small quantities of gas. The Kl turned orange in the same manner as the BTB turned yellow.3 Part A Completion 1. NT 15.31 L K kPa mol 10. 3. The drops near the center change immediately. As the gas diffuses.. 4. AT 18. The color change begins at the outer edge of each drop. 5. d 16. higher Chapter 14 Small-Scale Lab Section 14.0 g 100% 55.6 kPa 8. 8.0 g 180. 3. high surface spherical surfactant Ice dense hydrogen bonding Section Review 15. Inc. polar negative positive polar hydrogen low 7. The first picture should show one edge turning yellow. BTB changes from yellow to blue.9% 322 g 24. 5. molar mass Na2SO4 10H2O Mass of 10H2O 180. page 437 Analyze and Conclude 1. 2. 2. molecules/ions ions/molecules Emulsions stability emulsions Part B True-False 14. 3. PV n n nRT P V R T 0. AT 13. 10. The particles of gas produced are in motion. NaHSO3 + HCl → SO2 + H2O + NaCl Section Review 15.304 mol Part C Matching 50. 6. b 19. h 18. all the drops change color. e You’re the Chemist 1. 4. b and c 322 g Section Review 15. 8. g 21. they collide and react with molecules of BTB. colloid e. insoluble b.1 1. water of hydration 9. colloid b.. aqueous solutions 8. solvent 6. 4.7 g/mol Section 15. The structure of ice is a regular. Possible answers include glucose (C6H12O6) and ethyl alcohol (C2H6O). b Answer Key 819 . CoCl2 6H2O d.0 g 100% 45. aqueous solution 5. soluble NO3 (aq) OH (aq) 6. a.Part B True-False 11.0 g mass of water Percent H2O 100% mass of hydrate 108.44% 237. a 5. suspension 4. strong electrolyte 8.3 1.2 1. NH3(g) H2O(l) 1 NH4 (aq) OH (aq) 3. colloid c. framework like a honeycomb. soluble c. KOH(s) → K (aq) 7. e 17. Molar mass of NiCl2 6H2O 237. CaSO4 2H2O b. 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. The particles in a suspension are retained on a filter and will settle out slowly upon standing. ST 15. a Practice Problems 15 Section 15. solution d. d 19. soluble b. surfactant 7. Hydrogen bonds hold the water molecules in place in the solid phase. c 4. H 2. Brownian motion refers to the chaotic movement of colloidal particles caused by the collisions of water molecules with the small. a. hydrogen bonding 3. 2. like dissolves like 3. solution Part D Questions and Problems 22. The solute is potassium chloride (KCl).7 g/mol Mass of 6H2O 108. open. AT 14. 5. Brownian motion SOLUTION: 1. H H Vocabulary Review 15 1. 3. LIQUID WATER Section 15. f 21. c 8. a 2. colloid g. a. 2. H O H O H O © Pearson Education. solvent 7. NT Part C Matching 16. WATER VAPOR 2. a 20. suspension f. Hygroscopic compounds are those compounds that remove moisture from air. The solvent is water. NT 12. a. All rights reserved. NH4NO3(s) → NH4 (aq) Quiz for Chapter 15 1. b 18. Colloids and suspensions exhibit the Tyndall effect and have larger particles than solutions. b 23. desiccants 2. Inc. solute 6. publishing as Pearson Prentice Hall. ICE 3. dispersed colloidal particles. 3. ST 13. . True-False 23. aqueous 8. effloresce 6. solvation 3. a C. nonelectrolyte 2. c 19. Essay 31. AT D. surfactants 5. the ions are surrounded by molecules of solvent Chapter 15 Small-Scale Lab Section 15. d 15. AT 23. AT 27. Because polar water molecules can attract charged particles. hydrogen 9. AT 29. True-False 22. b 22. the beads of water that would normally have formed collapse.0 g 100% 55. a 21. hydrate 9. d 17. ST 26.1 g B. allowing the water to spread out to cover and penetrate the fabric. Completion 1. c 19. dessicants 6.0 g 6 mol H2O 108 g 108 g 100% 45. d 13. Essay 28.4% H2O 238 g © Pearson Education. Soaps and detergents also are emulsifying agents that allow oils and greases to form colloidal dispersions. Tyndall effect 10. d 21. NT 28. c E. nonelectrolytes 10. a 17. b 14. With surface tension reduced. As the solute dissolves. Soaps and detergents are surfactants that reduce the surface tension of water by interfering with the hydrogen bonding between water molecules. c 18. c 20. Emulsions 8. C. a 18. publishing as Pearson Prentice Hall. NT D. solvent 5. a 16. are removed from the surface of the fabric. c 12. effloresce NaHCO3(s) Aqueous conducts KCl(s) Aqueous conducts 820 Core Teaching Resources . Surface tension B. which are normally insoluble in water.Chapter 15 Test A A. The oil and grease particles. Problems 27. Multiple Choice 11. c 16. colloids 7. suspension 4. Multiple Choice 11. solute 4. ST 24. Problem 30. ST 24. they cause solute ions to break away from the surface of the solid. c 15. deliquescent 3. Molar Mass of hydrate 238 g 1 mol H2O 18. a 13. AT 25. AT 26. a 12. ST 25. All rights reserved. %H2O mass of H2O 100% mass of hydrate 180. Brownian motion 7.2 Electrolytes. Inc. a 14.9% H2O 322. E. Completion 1. c 20. hygroscopic 2. 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. 5. HNO3. NaOH Weak electrolytes (dim light) CH3COOH. These are electrolytes: NaCl.2 Part A Completion 1. KCl.3 g 0. c 17. 6. corn starch.15 mol 51 g 1 mol © Pearson Education. NT Cl (aq) 2. c 15. H2SO4. a 17. NaCl. solubility 9. d 22. a 18. NT 15. AT Answer Key 821 . AT 11. Henry’s 5.6 g/L 2. None of the electrolytes conduct electricity in the solid form because the ions are locked in a crystal lattice and cannot move. NT 10. Section Review 16. miscible 8. MgSO4. NaHCO3. NaHCO3.1 Part A Completion 1.15 mol 1L 342. Part D Problem 23. supersaturated Section Review 16. rate 3. MgSO4(s) → Mg2 (aq) NaHCO3(s) → Na (aq) KCl(s) → K (aq) KI(s) → K (aq) I (aq) SO42 (aq) HCO3 (aq) Part B True-False 9.6 g/L S2 1. coffee Part C Matching 13. pickle juice Weak electrolytes: orange juice. 8. e 16. pressure 4. Na2CO3. NT 12. NT 12. distilled water 3. 3.5 atm 1. solvent You’re the Chemist 1. These are nonelectrolytes: sugar. increases 10. diluting 7. solvent 2. NT 14. KI. KI are ionic compounds. ST 13. b 14. MgSO4. 7. 1. Strong electrolytes: soft drinks. liter 6. KCl. 2. f 19. solution 3. particle size 2. NH3 Nonelectrolytes (no light): rubbing alcohol. b Part B True/False 11. dilute 4.. AT 12. g 21. 10. Strong electrolytes (bright light): HCl. AT Part C Matching 16. Inc. molar mass C12H22O11 342. Na2CO3.5 atm S2 1. e 20. temperature 6. All rights reserved. 4. 9. d Part D Problem 18.3 M V mol solute needed 0. to be an electrolyte a compound must dissociate into ions in solution.0 atm 4. Test a drop of each solution with a conductivity device. 2. saturated 7. solute colligative properties freezing lowering/depression elevation directly solution particles twice twice Part B True-False 11.3 Part A Completion 1. AT 14. publishing as Pearson Prentice Hall. 3. ST 13.50 mol 0.1.3000 L 0.0 g/L Section Review 16. In general. solute 8. Table sugar and corn starch are covalent compounds. moles 5.0 atm 2. 5m 7.0% methanol (v/v) 25.0 g/100 g H2O Solubility of AgNO3 at 50 C 455.25M 1.0 g 222.0 mL 2. a.86 C Tf Kf m 7. a. moles HNO3 2. M1V1 M2V2 0.0 mol HNO3 63 g HNO3 1 mol HNO3 2.Part C Matching 15. b 14. publishing as Pearson Prentice Hall.0 mol HNO3 b. moles KOH 1L 1000 mL 3. AT 4. 2.0 L M1 V1 5. S2 S1 P2 0.0 g/100 g H2O 455. 500.83M 3.0 mL solution % (v/v) b.4 Part A Completion 1. b.5m 1000 g 1 kg molality of total particles 3 2.0 L L moles HNO3 4.5m 1. Therefore.0 mol KOH 7.0 mL methanol 6. 250.5m 14 C m freezing point of solution 0 C 14 C freezing point of solution 14 C 2. All rights reserved. a. b. add 11 L of H2O.5% C3H7OH (v/v) Practice Problems 16 Section 16.2 102 g KOH 1 mol KOH 2.5 g Part D Problem 18.0 mL 1000 mL 1L Section Review 16. c 3. d 16.0 mL C3H7OH 100% 200. d 16. 3. 20.50 mol 0.6 L 20.00M 1.0 mol HNO3 mass of HNO3 11.0 mL 101.0 g © Pearson Education. solute solvent kilogram mole fraction molal boiling point depression molal elevation 2.7 822 Core Teaching Resources .500M M2 The final volume should be 12 L.0 g H2O 2. two one K2CO3 NaCl c.40 mol NaCl 0.0 g Add 233.0 mol CaCl2 800. moles NaNO3 molarity Part C Matching 13. % (v/v) 100% 500. V2 12 L 0.0 g H2O 100 g H2O 1.0 mL V2 113 mL 2.. AT 12. four d.5 mol KOH 1L mass of KOH 7.82 g/L 34.22 atm 102 g KCl % (v/v) 1.50 mol Part B True-False 9. e 15. 0.0 mL solution 12. c 18. 2.1 P1 S2 0. NaCl 3.1 g KOH 4. 7.0 g to maintain saturation at 50 C.799M 2500. b Part D Questions and Problems 19.54 g/L 1. 60. molality molality 2.2 1.00M V2 0. Section 16. a 17.0 L 4.75M 300.0 g 233.1 g KNO3 0. a. 4. two c. Inc. a 212. 6.2 g KNO3 1 mol KNO3 b.0 g KCl 2. NT mass of HNO3 17. 8.750M 300. Solubility of AgNO3 at 20 C 222. % (v/v) 12.5 102 g HNO3 1 mol 85. ST 10.86 atm 1.0M 6. 5.5 mol KOH 56. a. 1 g 1 mol 101.0 g H2O 1000 g H2O 45.00 g 1L 100 mL 103 mL 5.0 g C12H22O11 1 mol C12H22O11 342.00m 2 Tb 2. a.512 C 3.8 C/m 0. a.6m glucose b.00 L 1L 100 mL 1.3 g C12H22O11 0.05 C The boiling point of this solution is 100 C 2. The solution containing calcium chloride has a lower freezing point.4 g NaCl 22 g NaCl 1 mol NaCl 2.0 g H2O 103 g 1 kg 4. a.35 mol NaCl 2 1 kg 900. a. molality of solute particles 103 g 0.3 1.501 mol Ba(NO3)2 1 kg 750.0 g NaCl Section 16. produce 9 mol of particles because each formula unit of Na2SO4 dissociates into three ions. The boiling point of water increases by 0. 3.3861 mol C12H22O11 2500. b.45 C 1. 2.78m NaCl Tf Kf m Tf 1.512 C/m 2. 6.45 C The freezing point of this solution is 0 C 1.0 mol Li2S 3. 1600.00 g b. 6. a.0 g molality of solute particles 0. XLiBr XLiBr b. b.3681 mol C12H22O11 103 g 1 kg © Pearson Education.78m Tf 1. All rights reserved. publishing as Pearson Prentice Hall. The boiling point of the solution is 100 C 3.3 mol glucose 500.274 C Answer Key 823 . and freezing-point depression.0 mol 6. Section 16. Three moles of Na2SO4. 2.1472m Tf 0. 750.512 C for every mole of particles that the solute forms when dissolved in 1000 g of water. Tb Kb m Tb 0. moles Ba(NO3)2 moles Ba(NO3)2 131 g Ba(NO3)2 1 mol Ba(NO3)2 261.668m Ba(NO3)2 5. the boiling point of the solution increases by 6 0.30 C The boiling point of this solution is 100 C 2. 0.501 mol Ba(NO3)2 103 g 0.512 C/m 1. The solution containing sodium chloride has a lower boiling point.0 g H2O 0.0 g H2O 0.4 1. The solution containing calcium chloride has a lower vapor pressure.30 C 102. 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. 2. when dissolved in water. boiling-point elevation.0271 1 mol 101. When 2 mol of MgCl2 dissolve in water.0 g H2O 0.50 mol NaCl 1000 g H2O 58.0 mol 3.0 g H2O 4. Thus. 5. Each formula unit of K2CO3 produces three particles in solution.7.50m 3 Tb 2. 6 mol of particles are produced because each formula unit of MgCl2 dissociates into three ions. Vapor-pressure lowering is a colligative property.86 C/m 0.05 C 102.0 mol moles C12H22O11 800..07 C. 1.33 nKNO3 nKNO3 nH2O 125 g 125 g 0.0 g 1 mol 18. c.45 C.1 g 3. Inc. b. Three important colligative properties are vapor-pressure lowering. 4.07 C 103. a.5 102 g Li2S 1 mol Li2S 2.07 C. Tb Kb m Tb 0. XKNO3 nLiBr nLiBr nH2O 0.30 C. moles C12H22O11 126.3 g Ba(NO3)2 0.05 C.00 102 g KNO3 30.00 L 103 mL 3.9 g Li2S 1.1472m C12H22O11 Tf Kf m Tf 1. V2 250 mL. 5. Multiple Choice 11.6 g 3.0 g H2O) (230. publishing as Pearson Prentice Hall. 2. All others are associated with the preparation of solutions.0 48. ST 31. Na2SO4 Vocabulary Review 16 1. Solubility KCl at 20 C 34.60M 250 mL V1 75 mL 2. j 8. Solubility of sucrose at 100 C 487 g/100 g H2O Solubility of sucrose at 20 C 230. 16. 33. M 2 V2 M1 0.0 g 8.945 mol molarity 1. b 21. c 2. 25. Concentration. 12.. a 4. M2 M1V1 M2V2 V1 0. All others are units of concentration. All others are associated with colligative properites of solutions. a 9. Molarity. 3.1 g/mol 0.0M Al2(SO4)3 to enough distilled water to make 250 mL of solution. 23. 4. c 8.0 g/100 g H2O Solubility KCl at 50 C 42.945 mol 101. c b a b 19.0 g O: 3 16.1 g N: 1 14. c 7.0 g H2O) 4870 g 2309 g 2561 g 4. 2. Saturated solution. 6. 17.6 g/100 g H2O 42.60M. 24. d LiBr C. 18. a 5.0M © Pearson Education. f 6. All rights reserved. b i g c 5. ST AT AT AT 26.0M Add 75 mL of 2. Boiling-point elevation. Inc. The other terms are used to describe mixtures of liquids. NT 29. 8. b. All others are associated with the solubility of gases. h 10.274 C. Colligative properties. All others are related to freezingpoint depression 7.3 g/L 16. Molar mass KNO3: K: 1 39. LiBr b.0 g 101. b 6. Quiz for Chapter 16 1. NT 27. b Interpreting Graphics 16 1. a 20. d 7. AT 30. Matching 1. 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. b 2. Problems 32.9 g/100 g H2O 1000. All others are factors affecting the rate at which a substance dissolves. e B. 13.0 14. Henry’s Law.1 39.1 g 95. All others are qualitative terms used to describe solutions.5 g mol KNO3 0.26M 0. NT 28.750 L 34. Molarity. a.274 C 0.The freezing point of this solution is 0 C 0. M1 2. S1 P1 S2 S2 P2 S1 P2 P1 20. 14. c d a c 15.6 g 34. d 3.9 g/100 g H2O (487 g/100 g H2O 1000. 4. Mole fraction. 3. NT Temperature °C D.9 g/L 606 kPa 505 kPa 824 Core Teaching Resources . True-False 22. ST 32.327m K3PO4 → 3K PO43 4 particles 4 particles 0.9 g AgNO3 5. 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. 37. 13. f 8. 23.671 C The boiling point of the solution is 100°C 0. Boiling-point elevation. Molar mass KNO3: K: 1 39. 19.31m Tb Kb m Tb © Pearson Education.91 atm 50. j 10. Matching 1. 100.1 g N: 1 14. Problems 37. 12. mol K3PO4 1000 g H2O 0.1 g 175 g KNO3 1 mol 101. All rights reserved.0 g O: 3 16.671 C.671°C 100.31m Tb 0. 4. a 7. AT 35. 2. 30. d 6. 1000 225 mL L 1 mol Mg(NO3)2 148. 16. 24. b B.327m 1. NT 33. Essay 35.50M 40.E.900 mol K3PO4 1000 g H2O 2750 g H2O 1 kg H2O 0. NT 34. d a b b c a 17.0 g 101. a b d b d 22. 18.3 g Mg(NO3)2 1. They depend solely on the number of particles in the solution. Chapter 16 Test B A. 28. 25. 38.0 14. 14. Boiling-point elevation: Additional attractive forces exist between solute and solvent that must be overcome for the solution to boil.0 48.25 g/L 0.0 g H2O 216 g AgNO3 S1 P1 (725 g AgNO3) 336 g H2O S2 P2 S2 P1 P2 S1 6. 21. g c h e 5. 0. and freezing-point and vapor-pressure lowering are colligative properties.0 g Mg(NO3)2 mL 39.. 29. NT 36.1 g KNO3 1. 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.45 g/L P2 1.73 mol KNO3 molality mol solute 1000 g solvent 1. 20.1 39.750 atm P2 2. Additional Problems 36.73 mol KNO3 1250 g H2O 1. Multiple Choice 11.31 g AgNO3 1 mol AgNO3 1. AT D. 26. publishing as Pearson Prentice Hall.512 m C C. a c a b a F. i 9. True-False 27. Inc. 3.250 mol AgNO3 1 kg 125 g 1 kg 1000 g 169. 15. 0. AT NT NT NT 31.38m 103 g H2O 1 kg H2O Answer Key 825 . 88 g 46. The salt causes the freezing point of water to be depressed because it interferes with the crystallization process. Additional Problems 42. mass percent and mole fraction. You’re the Chemist Sample answers are given..98 g 6. 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.0 g/mol 0.118 mol NaCl b.050 L 6.21 g 5. page 497 Analysis Sample answers are given.90 g (6.88 g water 6.490 C boiling point 100. a.118 mol (0.1 g/mL 50 mL Notice that because the flask measures less accurately than the balance. and prevents additional ice from forming down to temperatures below0 C. All rights reserved. Inc.98 g 22. a.15 mol H2SO4 1L 750 mL 1L 1000 mL 98.958m Tb Kb m 0.118 mol NaCl 3.118 mol b.4 Making a Solution.0 g 1 mol CH3OH 1000 g 32.21) g 2.57) mol 100 13. Density 1. 6.98 g 22. Molarity 2.21 g 2. Chapter 16 Small-Scale Lab Section 16. publishing as Pearson Prentice Hall. molality of total particles 62.04621 kg H2O 4.57 mol 18.21 g c.1 g H2SO4 11 g H2SO4 1 mol H2SO4 Tf Tf Tf Tf Kf m 27.55m 1L 1000 mL © Pearson Education.09 g 22.0 g 261 g Ba(NO3)2 1000 g 3 moles particles 1 kg 1 mol Ba(NO3)2 0.E. This causes any ice that was initially present to melt. moles of NaCl solute 0. 0. mole fraction 46.21 g 0.0439. a.39 C 6.958m Tb 0. mass of dry flask mass of flask mass of flask 15.39 C 44. 0. mass of the solvent (water) 69.88 g 15. molarity and density have fewer significant figures than molality. Essay 41.4M 0. Sample data: dry flask flask flask 15. 1.490 C 0.512 C/m 0.90 g b.90 g 46.88 g water 69.5 g CH3OH 250.90 g 69. liters of solution 50 mL 2.09 g NaCl NaCl F.90 g 58.90 46.490 C boiling point 100 C 0. mass of the solute (NaCl) 22. Molality 0.0% NaCl 43.39 C 1. % by mass of NaCl 6.050 L 0.5 g/mol freezing point 0 C 6. moles of water c.897 g mass of NaCl solution mass of water mass of NaCl 826 Core Teaching Resources .86 C/m 1.5 g Ba(NO3)2 1 mol Ba(NO3)2 750.0 g CH3OH 1 kg 6.118 2. liters of solution 0.09 g NaCl solution NaCl solution 46. H 4.85 g 4.68 g % by mass of sugar % by mass of sugar 4. Heat is energy that is transfered because of a temperature difference. NT 12.0121 mol NaCl Molarity 0.0121 2.54 10 46. NT 10. C 0. Chemical potential energy is energy stored within the structural units of chemical substances.57 mol 5. e 14.68 g density 1.2 Part A Completion 1.16 kJ/mol molar heat of vaporization Condensation Answer Key 827 . d Part D Questions and Problems 18. heat potential energy thermochemistry calorie joule specific heat or specific heat capacity metals water Section 17. 34.53 g 16. 5.68) g 7.13 g mass of the solvent (water) 69.68 g 1 mol 18. 7. 2. AT 9.13 g 2.0121 mol 4. calorimeter 2.0 g CH4 103 kJ Section Review 17.85 g 48.0121 mol sugar Molality 0.04 kJ 890.percent mass of NaCl moles NaCl mass of water moles of water mole fraction molality density 1.69% 2 Part B True-False 9.72 g sugar sugar 2.0 g 15.0 C The unknown metal is iron.82% 100 moles of sugar (C12H22O11) 48. molar heat of fusion molar heat of solidification equal 3.1 Part A Completion 1.333m 1. Work is done when a force moves an object. AT 11.72 g 20. enthalpy 3. NT 11. mass of the sugar mass of the sugar 20.0 g 0.242M 0. mass of dry flask mass of flask mass of flask Part D Questions and Problems 19.0121 mol (0.94 1 mol CH4 16.1 g/mL 50 mL Mole fraction 1 mol 342 g Section 17.13 g 48. 124. initial or final 5.70 mol © Pearson Education. AT Part C Matching 13. 2NO O2 → 2NO2 H 113. 3. AT 12. e 16.21 g 1. publishing as Pearson Prentice Hall. AT mol 2. 8. mass moles of water Part B True-False 7.460 J/(g • °C) 18. b 17. 4.050 L 4. AT 13. ST 8. All rights reserved.53 g 20. 6..04868 kg H2O 0. AT 10. c 17.2 kJ 1 mol CH4 19.85 g water 69.13 g (4. d 18.1 g/mL 16. b 16. 3. 6.70) mol Mole fraction 0.2 J 20.96 10 3 Part C Matching 14. 4.13 48.3 Part A Completion 1.249m 0.00446 0. 2. final or initial 6. Inc.8 g CH4 1. a 0. c 15. a 15. 5. 0. 1 1.7.0 g 1.8 kJ 525.0 cal 2.0 g H2O(s) 6. 7. ST 11. b 17. 155 kJ 242 kJ 87 kJ 828 Core Teaching Resources . NT 12. c 18. NT H0 (products) f 3 mol CO2(g) H0 f 1181 kJ (reactants) 3 mol CO(g) 110. 6.3 2. d 20. 3.2 1.1 3 20. exothermic 19.0 C 1255. 10.0 g 15. endothermic b. exothermic f.1 g Ca(OH)2(s) 13.0 g 2. d.0 Cal 1000 cal 4.01 kJ 9. NT 10.18 J g C 10 C 6. Section 17.2 kJ 1 mol CH4(g) 16. 5.4 g CH4(g) 890.0 C 0. sum enthalpy indirectly changed (reversed) standard heat of formation change one H0 f zero subtracting 120. exothermic 1 mol H2O(s) H 28. NT 15.40 cal/(g C) 25.45 kJ 1 mol H2O(s) H 5. e 15.0 C 100. 2. publishing as Pearson Prentice Hall.1 kJ) 393. C 1. H 103 J 4. Section 17. NT 13.0 g Ca(OH)2(s) 65. c 17.3 kJ 1 mol Cu(OH)2(s) 74.4 Part A Completion 1.0 g CH4(g) 2.5 kJ 1 mol CO(g) ( 822. H 52. endothermic c. 8.2 kJ 1 mol CH4(g) Part C Matching 16. 9.0 J 3.. e 4.184 J 1 kJ 1 Cal 1 cal 103 J 836.0 g 6. a 19.2 kJ 1 mol Ca(OH)2(s) Part B True-False 11. 4.3 g H2O(s) 18. NT Part C Matching 13. T 6. b 16. 4NH3 H 5O2(g) y 4NO(g) 6H2O(g) 226 kJ 4 mol NH3(g) 1 mol NH3(g) 904 kJ Part D Questions and Problems CuO(s) y Cu(s) 1 O2 H 2 H2(g) 1 O2(g) y H2O(g) H 2 CuO(s) H2(g) y Cu(s) H2O(g) Hrxn 21. All rights reserved.5 kJ 1 mol CO2(g) Part B True-False 8. q 100. a.7 kJ 142 kJ 1 mol NH4NO3(s) Practice Problems Section 17. d H 0 1154 kJ H0 (products) H0 (reactants) f f 1181 kJ ( 1154 kJ) 27 kJ Part D Questions and Problems 18. 200. AT 9.1 J/(g C) 4. Inc.93 103 kJ 15.53 mol NH4NO3(s) 25. endothermic e. AT 14. NT 3. H 150.90 J g C 10 J © Pearson Education. a 14. 12. molar heat of condensation 22. see Table 17.184)(39.6 kJ) H0 178. H 2.0 g 4.1 kJ ( 393. 2.1 kJ 1 mol H2S(g) Answer Key 829 . specific heat 9.25 J/(g °C) 7.35)( 73.7 kJ 2.512 kJ 429. 9.6 kJ 1077kJ H0(reactants) f 2 mol H2S(g) 40. 27. 8.8 kJ 1 mol SO2(g) 241. 8. 3. 5 °C b.0) 102 J b. a.33)( 73. a.6 kJ 2 2H2(g) N2(g) y N2H4(l) H 50.0 g H2O(s) 1 mol H2O(s) 18.22 J/(g °C) b.6 kJ H0(reactants) f 0 H0(reactants) f Interpreting Graphics 17 1.0 g H 429.01 kJ 1 mol H2O(s) 2.184)(39.01 kJ 1 mol H 96 g 32 kJ 1 mol 18. 3. 9. Inc. 4.543 mol NaOH(s) 1 mol NaOH(s) H 1.7 kJ 1 mol H 190.18 28 C g C H 1.8 kJ 32 kJ 11 kJ 47.512 103 J 6. H (4.6 kJ Htotal 6.8 kJ 1 mol H2O(g) Quiz for Chapter 17 1. Step 1: H2O(s) at 24 C y H2O(s) at 0 C J H 96 g 2. specific heat (50. a is reversed zero minus ( 483. 2.4 1. N2(g) 2H2O(l) y N2H4(l) O2(g) H 622.04 103 kJ 296.6 kJ H 1207.1 kJ 3. 26.6 kJ 436.0 g Step 3:H2O(l) at 0 C y H2O(l) at 28 C J H 96 g 4.1 kJ 445. a. 73. b 6. H 35.6 kJ) 20.2 kJ) 1. 11.Section 17. publishing as Pearson Prentice Hall. 5.132 103 kJ 4. © Pearson Education.1 104 J 11 kJ Htotal 4.8 kJ 635. H0 (products) f 2 mol SO2(g) 2 mol H2O(g) 593.2 kJ 2[H2(g) 1 O2(g) y H2O(l)] H 571.5) specific heat 0.0 g 1 mol 18.3 1.8 103 J g C Step 2: H2O(s) at 0 C y H2O(l) at 0 C 6.5 kJ) 1028. 4. 73. b. 7.2 102 J b. b c a d b 6.0 g H2O(s) 6.5 °C (4. Vocabulary Review 17 1. 5.1 102 J 5. 4.4 kJ The reaction is exothermic.2 102 6. a h f g 5.1 24 C 4. a.1 102 J 8. All rights reserved.2 kJ H0 f 1077 kJ ( 40.2 b.0 kJ ( 1028.452)(5. j 10. d 9.0 °C H 3. i 8. heat gained by H2O heat lost by metal a.100)(5. e 7.5 °C b.512 kJ Step 2: H2O(l) at 100 C y H2O(g) at 100 C 40.1.5 °C 8.5) 9. c Section 17.1 102\(50. 3.0 °C 2..0) specific heat 0. a.18 82 C g C H 6. Step 1: H2O(l) at 18 C y H2O(l) at 100 C J H 190. Endothermic examples include the melting of ice. Specific heat 0. 3. 14.01 kJ 31. b d b c a 21. 13.7 kJ 3 mol CO 18.7 kJ B.0 g CO 55. Endothermic processes absorb heat.5 kJ) 565. Problems 27. 20. H H m C 96 J 12 g 20°C °C) T 1.Chapter 17 Test A A.5 kJ 24. 18. 25. H H m C 18 g °C) T 1. H H g mL (35. 2[C(s) O2(g) → CO2(g)] 2( H 2CO2(g) → 2CO(g) O2(g) H 2C(s) O2(g) → 2CO(g) H 787. e i g f 5. 14. g f h c 5. 75. 8. D. and the freezing of water. Matching 1.00 4. 13. the sublimation of mothballs. 2[Mg(s) Cl2(g) → MgCl2(s)] 2( H 641 kJ) SiCl4(l) → Si(s) 2Cl2(g) H 687 kJ 2Mg(s) SiCl4(l) → Si(s) 2MgCl2(s) H 1282 kJ 687 kJ 595 kJ 29.0 mL 1 mol CO 28.7 kJ 221. 12.0°C J g °C 24. h 7..0 g H2O 1 mol 30. 2. j 6.0 103 J or 2.18 103 J 9. C2H6 7 O2(g) → 2CO2(g) 3H2O(l) 2 H0 ? H0 H0 (products) H0 (reactants) f f 0 H [2( 393. 19.0 kJ)] H0 1559.0°C) 60. c b a d b 16. c b c a d 16. In vaporizing.18 J g °C D.40 J/(g 28. 24. Essay 26. 15.0 kJ 165 g 1 mol H2O 6.68 J) (0. c 8. 20. b 9. 2.0 kJ 565.0°C g °C 2. while exothermic processes release heat. a C. Thus.8 kJ)] [( 84. 19. 23. e 8. d 7. 23. Matching 1. Essay 26. and the heat used to cook food.7 kJ/mol). 3. a 10.0 kJ 60. b d c b d C. Specific heat 0. 12. Problems 27. a c b d b 21. j 6.0°C H H 55. d 32. the cooling of skin as perspiration evaporates. b c a d b Chapter 17 Test B A.5 kJ) 3( 285. Inc.0 mL 29.17 J/(g 28. i 9. 24. 25. the evaporation of a puddle. 22.0 g H2O 1 mol 1 mol H2O 6.00 g mL J g °C 2. steam absorbs the heat required for vaporization (40. 22.1 kJ 1411 kJ 3 mol O2 118 kJ 225 g 393.01 kJ 31. steam at 100 C contains more energy than boiling water at the same temperature.0 g CO 16.0 g 4.18 J 8. 17.7 kJ 830 Core Teaching Resources .0 kJ 1 mol O2 32. 4.1 4. Multiple Choice 11. 18. 4. 55. Multiple Choice 11. publishing as Pearson Prentice Hall.0°C 27. All rights reserved.3 kJ 4. 56. 17.0°C) 75 J 25°C © Pearson Education. Exothermic examples include the combustion of fossil fuels such as gasoline.00 g O2 30.18 (33.0 g 2. 15. b 10. B.0 g O2 18. 8. AT 14. 7. page 533 Analysis 1. Those who think the wick burns may suggest that the wax slows the rate of burning. The candle flame might be round in zero gravity. 8. which is drawn up into the wick. 4. e 20. The wick draws melted wax to the flame.1 Part A Completion 1. The mass loss is consistent with the wax burning.8 kJ)] [(48. 120 kg/24 h 22. 2.700 kJ/mol 21( 242) ( 2230) 10. © Pearson Education. The wax burns but many students will say the wick. in this case CO2 and H2O. hot gases rise. f 18.50 g 61 2 O2 13.0 kJ)] H0 3266. 4. 11. d 5. 3. 5. b 16. Depending on the candle and the time burned.9 kJ 10. d 17. means to draw a liquid from one place to another by capillary action. ST 12. evaporated. ∆H 20( 394) ∆H comb 11. C: 20 H: 42 0. Rates react kinetic energy activation minimum 6. AT Part C Matching 15. 10. 7..4 Heat of Combustion of a Candle. 2. Wick. Chapter 17 Small-Scale Lab Section 17. Black soot will appear on a glass Petri dish held over the flame. 10. 9. a 19. 10.5 kJ) 3( 285.32. 8. Heat from the combustion melts the wax. All rights reserved. 3.2 Part A Completion 1. Liquid water will form on the underside of a glass Petri dish filled with ice held over the flame. and burned. C20H42 9. 6.0018 mol 1 mol/282 g Answer Key 831 . products slower temperature catalyst increasing Part B True-False 11. Because of gravity. 5.50 kJ) (0. Section Review 18. b. as a verb. 5. c Part D Questions and Problems 21. publishing as Pearson Prentice Hall.700 kJ/mol 0. the wick loses a few millimeters and the candle a few tenths of a gram. C6H6 15/2O2(g) → 6CO2(g) 3H2O(l) H ? H0 H0 (products) H0 (reactants) f f 0 H [6( 393. 4.0018 mol 61/2(0) 19 kJ You’re the Chemist 1. 6. 2. Inc. 9.0 kg/h Section Review 18. NT 2. reversible products forward reactants reverse reactants products equilibrium equilibrium constant ratio Le Châtelier’s → 20CO2 21H2O 12 1 240 42 282 0. 7. 3. 3.5 10 3M) 1. 6. solubility product constant common ion effect addition precipitate Part B True-False 12. 3. 8. 7.072)2 7. publishing as Pearson Prentice Hall. 24. AT Section Review 18. Inc. All rights reserved. b 25. e 20.Part B True-False 12. 8.42)2 (0.0 10 9). 5. rate concentration rate law specific rate constant order first-order second order experiment elementary reaction mechanism 832 Core Teaching Resources . b Part D Questions 23. 6. NT 15. 10. b 19. ST Part C Matching 17. d. c Part D Questions and Problems 21. 4. a 21. c. spontaneous nonspontaneous energy work free energy energy greater entropy disorder law of disorder maximum Part C Matching 16. c 18. Section 18.. c 7. NT 13. 4. [CO32 ] 0. AT 15. a 18.145 7. 4. AT 16. 11. 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.0 10 4M) (1.1 10 6 Precipitation occurs because the ion product (1.4 Part A Completion 20.21) (0. 5. f 19. b. 10.5 Part A Completion 1. Part C Problem 8. e 1.0015M [CO32 ] [Ba2 ] (7. 9. 2. Keq Keq [SO2]2 [O2] [SO3]2 (0. a.1 Section Review 18. 7. d 17.3 Part A Completion 1. ST 14. ST 13. 9. d 22. 3. b Part B Matching 5.00070M [Ba2 ] 0.1 10 6) is greater than the Ksp of BaCO3 (5. 2. 2. AT 14. a 6. 20) 8. a. 2 mol/4 h 0.34 . a. Rates of chemical reactions can usually be increased by (1) increasing the temperature. This diagram represents a reaction that takes place in two elementary steps.1 10 4 Practice Problems 18 Section 18.6 [N2O4] 5. a. NT 12. Keq 2.20] [0. d 16.18)2 2.0 10 11 x 3.2 10 5)2 Ksp 8. AT 5. Keq [NO2]4 [O2] [N2O5]2 [0.6 0.0 10 5M 5. Ag2CO3(s) 1 2Ag (aq) CO32 (aq) Ksp [Ag ]2 [CO32 ] Let x [CO32 ]. Keq [CO] [H2]2 [SO3] [NO] d.6 10 16 4. [CO32 ] 0. shift right 4. 2.33 3.1 10 12 4x3 8. no shift 7.014)2 (0. increase the rate Section 18.0 10 6)(1.1 1.035) Keq (0.5[OH ] 6. Inc. All rights reserved. shift right b. [CO32 ] x 0. b 18. publishing as Pearson Prentice Hall. e 20. shift right c.2 1. Section 18. f 17. (2) increasing the concentration of the reactants. (3) decreasing the reactant particle size. [N2O4]2 [NO]4 [O2]2 [NOBr]2 b..20 Ksp (2x)2(0. c Part D Questions 21. [N O ] 2 4 [NO2]2 [NO2]2 [NO2]2 5. Ca(OH)2(s) 1 Ca2 (aq) 2OH (aq) Ksp [Ca2 ] [OH ]2 b.6 0. Keq [NO]2 [Br2] [CH3OH] c.3 1.0 10 12 x 1.0 10 6 Ksp (6. a 19.80]4 [0. Ksp [Ag ]2 [CO32 ] Let [Ag ] 2x. a. Keq [H2S]2 (0. a.3 10 10 x [Sr2 ] 3. Point D represents the energy level of the final product. Keq 1 1012 [H2]2 [S2] 8. Fe(OH)2(s) 1 Fe2 (aq) 2OH (aq) Ksp [Fe2 ] [OH ]2 [Fe2 ] 0. ST 13.5 mol/h 3.4 10 6M Answer Key 833 © Pearson Education.66 5. Point B represents the energy level of the intermediate product.3 10 4M [CO32 ] [Ag ] 2x 2.1 10 12 x3 2. The reaction is exothermic. Ag2CO3(s) 1 2Ag (aq) CO32 (aq) Ksp [Ag ]2 [CO32 ] 2. shift right d. and (4) using of a catalyst. AT 14. Points A and C represent the energy level of the activated complexes.20 assume x << 0. decrease the rate b. Keq [SO2] [NO2] 6.50]2 Keq 0.1 10 12 x2 1. SrCO3 1 Sr2 (aq) CO32 (aq) Ksp [Sr2 ] [CO32 ] Let x [Sr2 ] [CO32 ] Ksp x2 9.20 mol.Part B True-False 11.6 10 4M 3.2 10 6M [CO32 ] [Ag ] 6. shift left b. Keq Part C Matching 15. increase the rate 4. 2x Ag Ksp (2x)2(x) 8. 14.4 10 6 Because this product exceeds the Ksp value. Ksp(PbSO4) [SO4 ] [Pb2 ] 2 6. d 3. less 4. 4.0 mL 1000.2 102 2. rate k[J][K] The reaction is first order in both J and K. d 6. b 7. positive 0.2 102 [CO] [H2]2 24.020 7. a. Matching 1. 3. h Quiz for Chapter 18 1.60)2 1.0012M Interpreting Graphics 18 1. 8.0168 s 1 5.0021M [CO32 ] 0. rate k[C2H4O2][CH4O] B. precipitation will occur. 3. b 6.30)2 [NOCl]2 [CH3OH] 25.0020M 400.4 mol/(L • s) 8 0. b 5. negative 2. d 9. increasing b. 7.6 (0. Keq [CH3OH] 2. negative c. negative b. a. 19.00842 mol/(L • s) 0.0 mL 600.58 mol/L [CH3OH] 834 Core Teaching Resources .0020M 0. 2. rate k[H2O2]. 22. 2. c 10.2)2 (0.7 10 6 Because this product exceeds the Ksp value. Problems (1. e g i a 5. AT 9. NT Section 18. 1. Section 18. 15.5 1. 13. publishing as Pearson Prentice Hall. positive f.6. 12. e 7.30 mol/(L • s) 2. NT 8. All rights reserved.0021)(0. a 4. 18.5 10 9 [Ca2 ] [CO32 ] The total volume is 1000 mL. h 8. 4. k rate/[H2O2] k 0. D b.4 1. c d b c 20.0020) [Pb2 ] [SO42 ] 2. Example a 5. Ksp(CaCO3) 4. decrease 3.2 102 [CO] [H2]2 9. a. greater 3. A Chapter 18 Test A A. f 10.3 10 7 [Pb2 ] [SO42 ] 0. b b c d d 16. D d.0013M [Ca2 ] [CO32 ] (0. yes d. increasing c. NaCl has no ion in common with Mg(OH)2. g 6. 2. j 8.. 17. d 9. 0.020)(0. N c. C3H8O3 c.0050M 0. so [Ca2 ] 0. rate k[HgCl2][Na2C2O4]2 3. 21.0 mL [Pb2 ] [SO42 ] (0. increasing 4. 2. precipitation will occur. f 7.500 mol/L k 0. Vocabulary Review 18 1.0021 mol/L 0.0013) [Ca2 ] [CO32 ] 2. positive e. Multiple Choice 11. 4.0013 mol/L 0.60) [NO]2 [Cl2] (0. increase 2. b 2. 23. a b a d C. b © Pearson Education. Inc.0012)(0. a. two elementary reactions b. i a j c 5.0. three elementary reactions 6.0 mL 1000. b 10. All of the solids dissolved rapidly. PbF2 10 2)2 10 3)4 1010 (5. 14. c 7.59 [N2] ? 1. 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. 23. j g h i 5. decreases shifts left. b. ∆S is positive in each case. 22. a 1011 mol/L Chapter 18 Small-Scale Lab Section 18. 19. Mixture a.23 10 3) 10 4)2 (2. Essay 28. 20. c. d. f. Keq [HCl]4 [O2] (5. c d b d a B. NaCl H2O(l) did not change much in temperature. Chapter 18 Test B A. Matching 1. Doubling A doubles the rate first order in A.8 (1. under the conditions specified. NaCl H2O(l ) T1 21°C T2 21°C T 0°C 16. e. 12.8 10 2)2 10 4) b. increases shifts right. ∆H is close to 0.10 (1. which consumes that reactant. decreases shifts right. 3. shifts left. decreases [H2O]2 [Cl2]2 26. 4. Keq [N2] [NH3]2 [N2] [H2]3 [NH3]2 Keq [H2]3 (6.2 Keq 1. D. Some spontaneous reactions apprear to be nonspontaneous because their rates are slow. 18. 17. d. d a d b c C. increases shifts right. a. 15. 24. heat NH4Cl(s) → NH4 (aq) 2+ Cl (aq) heat CaCl2(s) → Ca (aq) 2Cl (aq) 5. Problems 25. f 6. Doubling B increases the rate 8 times (23 8) third order in B. An increase in temperature causes the endothermic reaction to speed up in an effort to consume the additional heat. E. The addition of more reactant causes an increase in the rate of the forward reaction. Entropy usually increases in the dissolving process. ∆H is negative.8 (3. a. NH4Cl H2O(l ) 21°C 5°C 16°C c..26.4 27. page 574 Analyze Sample data are provided. c. 2. are known to favor the formation of products. Inc.4 Enthalpy and Entropy. favors products favors reactants favors products favors reactants D. Nonspontaneous reactions do not favor the formation of products under the specified conditions. © Pearson Education. 3. Additional Problem 28. b. 4. Spontaneous reactions are reactions that. 13. Additional Problem 29. Essay 27. d 8. publishing as Pearson Prentice Hall. e 9. b. c. a. Answer Key 835 . All rights reserved. increases shifts left. First order third order fourth order overall. 21.75 10 6)3 E. c b c b 1. ∆H is positive. Multiple Choice 11. NH4Cl H2O is endothermic. CaCl2 H2O(l ) 21°C 53°C 32°C 2. CaCl2 H2O is exothermic. NT 14. Dimethyl ether is a Lewis base because it donates an electron pair to form a bond. NT Part C Matching 17. ST 14. Endothermic reactions absorb heat. 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.. f 25. Kw [H ][OH ] 1 10 Kw [OH ] 1 10 [H ] The solution is basic. 8. publishing as Pearson Prentice Hall. 4. b Part D Problems 26. c 18. Na2CO3 and Na3PO4 dissolve endothermically or with little or no change in temperature. 3. 2. a 21. ( ) or ( ) . 11. You’re the Chemist 1. 9. T( ) for NaCl(s) T( ) for NH4Cl(s) . ∆G G G is G G is G ∆G is ∆H (0) . pH log[H ] [H ] 1 10 6M c. The temperature of the NaCl and ice dropped dramatically. 9. a. g 18. f 19. AT 13. All rights reserved. Many salts such as KCl. cooling the environment. i 23. Inc. ST 16. T( ) for CaCl2(s) Part C Matching 17. d 19. 7. 10. Both CaCl2 and NH4Cl depress the freezing point of ice and cause a drop in temperature. 6. Sample data: Mixture NaCl H2O(s ) T1 0°C T2 15°C T 15°C Section Review 19. 8. AT 15. 10. 4. b 22. 5. e 20. NaHCO3. 5. Boron trifluoride is a Lewis acid because it accepts an electron pair from dimethyl ether. AT 15. 11.1 Part A Completion 1. 4. Section Review 19. 3. 2. c 24. e 22. T∆S. pH log[H ] [H ] 1 10 3M b. g 23. Melting ice is endothermic. h 21. a 20. 25. 2. 6.6. 7. NT 13. AT 836 Core Teaching Resources . This explains the drop in temperature of the NaCl and ice mixture. 3. NT 16. d Part D Problems 24. pH log[H ] [H ] 1 10 10 14 10 1 10 4 Part B True-False 12. self-ionize 1 10 7 0 to 14 hydrogen ion acidic basic neutral 7 ion-product hydronium/hydroxide hydroxide/hydronium © Pearson Education. triprotic 4. basic Part D Problem 22. ST 13. lithium reacts violently with water to produce hydrogen and the base lithium hydroxide. a 15.1 10 2 0. 5.1 10 2) [H ][X ] Ka 0. a. Inc. H3PO4 H2O 1 H3O H2PO4 H2PO4 H2O 1 H3O HPO42 2 HPO4 H2O 1 H3O PO43 3. d 20. degree of ionization Ka larger pH completely strong weak bases water acid strong 17.309 [HX] Ka 5. 2. H2SO4 H2O 1 H3O HSO4 2.309 (4. 3.1 10 2 [HX] 0. 7. H3PO4 Al(OH)3 y AlPO4 3H2O b. 2HI Ca(OH)2 y CaI2 2H2O Section Review 19. 4. 5. All ions formed are shown in the following chemical equations. publishing as Pearson Prentice Hall. monoprotic c. salt acidic basic neutral hydrolyze 6. AT 15. a. 7.1 10 2) (4. Answer Key 837 © Pearson Education. strong weak buffer capacity Part B True-False 10. f 16. 3. b 16. b Part D Questions and Problems 18.041 [HX] 0. All rights reserved. NT 11. NT Part C Matching 12. AT 11. end point 7.. 9. equivalence Part B True-False 8. a 19. H2SO4 and H3O are proton donors and H2O and HSO4 are the proton acceptors. 4. c 13. 10. The conjugate acid–base pairs are H2SO4/HSO4 and H3O /H2O. ST 12.3 Part A Completion 1. e 14. 2. a. d Part D Problem . Section Review 19. titration 6. a 15. e 18. c Part C Matching 16.5 Part A Completion 1. AT 10.35 0. HX 1 H X [H ] [X ] 4. 9. H3PO4 is a triprotic acid able to ionize three hydrogens. acidic b. b 17. AT Part B True-False 12. NT 14.4 10 3 Practice Problems Section 19.Section Review 19. 8. 4. NT Part C Matching 14.4 Part A Completion 1.35 4. 6. 8. 3. 11. AT 9. ST 13.1 1. neutral c. Only hydrogens bonded to highly electronegative atoms are ionizable. 2. acid hydroxide water neutralization 5. Like other alkali metals. diprotic d. c 17. monoprotic b. d 21. The weakest acid has the smallest Ka. At equilibrium. 7. weak acid.46 4. acidic c.2Li(s) 2H2O → H2(g) 2LiOH(aq) 5.6 5.00) pOH 2.6 b. (4) d.5 10 2) pOH (0. C6H5COOH(aq)1 C6H5COO (aq) H (aq) [C6H5COO ][H ] Ka [C6H5COOH] 6. Ka (4. pOH log[OH ] pOH log(3.36 d. acidic c.5 10 3M [H ] 10. 10.3 1. neutral b. They react with acids to produce a salt and water. Aqueous solutions of bases taste bitter and feel slippery. publishing as Pearson Prentice Hall. a. Acids have a tart or sour taste and cause indicators to change color.0 c.0 ( 3)) 3. Acids react with compounds containing hydroxide ions to produce a salt and water.20M x 0. Most acidic solutions of interest have a hydrogen ion concentration of less than 1M. 6. 838 Core Teaching Resources . 5. H2S(aq) 1 H (aq) HS (aq) Ka [H ][HS ] [H2S] Section 19. basic b.3 10 6M)2 Ka 4.00 0.34 b. pH pH log[H ] log(1 10 3) (0.6 7.0 12.2 1.. Inc.8 10 M 9. Kb [C6H5NH2] 10 3M)(4.20M (since x 0. (2) e. basic 8. pOH 14. All rights reserved. HF(aq) 1 H (aq) F (aq) [H ][F ] Ka [HF] 3.096M 4 Ka 1.20M x 3. pH log[H ] pH log(1 10 6) reminder: the log(a b) log a log b pH (0.0 2. Taking the negative log (minus sign in the pH definition) ensures that the pH values will usually be positive.00 0.0 10 10M 0. pH pOH 14.0 pH 14.0 ( 6)) reminder: the log 1 0.4 1.0 pOH 14.86) ( 9.0 pH pOH 14. (3) c. BF3 can accept a pair of electrons to form a covalent bond and is therefore a Lewis acid.20M) x2 Ka 6. (1) [C6H5NH3 ][OH ] 7.54 pOH 1. 7. a. [H ] [CN ] 6. a. At equilibrium.00) Use log tables or your calculator to find the log of 7. 12. basic d.54) ( 2. basic d.3 10 6M [H ][CN ] Ka [HCN] (6. 10. a. strong acid 2. N2H4(aq) H2O(l) 1 N2H5 (aq) OH (aq) [N2H5 ][OH ] Kb [N2H4] 4. pH log[H ] pH log(7. Bases cause indicators to change colors. The log of this concentration would always be a negative number. weak base.2.2 10 3M) 0. a. acidic 6. 9. NH4 (aq) 1 NH3(aq) H (aq) [NH3][H ] Ka [NH4 ] c.2 © Pearson Education. HCO3 H2PO4 HCOOH H2C2O4 5.0 3.4 pOH 10.10M 8.0 pH 6. strong base. it is a Lewis base.2 10 9) pH (0. (5) b.86 pH 8.0 Section 19. Since F donates the pair of electrons.3 10 5M 0. pH 9. acidic e. [H ] x [C6H5COO ] [C6H5COOH] 0.14 3. 0.0142 L HC2H3O2 2 mol HC2H3O2 9.014 L H2SO4 0.0198 L Ca(OH)2 © Pearson Education.10 mol/L NaOH 0.00212 mol HCl moles 0. a.75 mol H2SO4 1 L H2SO4 2 mol NaOH 0.021 mol NaOH liters 0. at the equivalence point. Benzoic acid is a weak acid. The equivalence point occurs when the number of moles of NaOH added equals the number of moles of C6H5COOH originally present.050 L 6.5. The neutralization of a weak acid with a strong base. Based on the answers to questions 4 and 5. publishing as Pearson Prentice Hall.025 L 0.00404M 3.025 L Answer Key 839 9.0246 L Ca(OH)2 Molarity 0. produces a basic solution at the equivalence point. neutral solution b.00212 mol HCl moles 0. Ca(OH)2(aq) 2HC2H3O2(aq) y Ca(C2H3O2)2(aq) 2H2O(l) Molarity 0. find the area of the titration curve where the pH changes abruptly when a small volume of NaOH is added.038 L NaOH Molarity 0.0160M H2SO4 4. such as NaOH. 4.0106 L HCl 10.25 mol HCl 0. Al(OH)3(aq) 3HCl(aq) y AlCl3(aq) 1 mol Al(OH)3 78.000198 mol H2SO4 0. Thus. basic solution H2O mol Ca(OH)2 Interpreting Graphics 19 1.12M Ba(OH)2 molarity 0.0100 mol Ca(OH)2 1 L Ca(OH)2 0.0025 mol Thus.0015 mol Ba(OH)2 liters liters 0.0 g Al(OH)3 3H2O(l) 0.55M NaOH 2.0142 L HC2H3O2 9.94 Molarity 10 5 0.4 1. The pH at the equivalence point is approximately 8.0025 mol C6H5COOH were originally present in a volume of 25 mL.0140 mol HC2H3O2 1 L HC2H3O2 1 mol Ca(OH)2 0.000198 mol H2SO4 moles Molarity 0. [C6H5COOH] [NaOH] 0M and 0. C6H5COOH NaOH y C6H5COONa H2O One mole of sodium hydroxide will neutralize one mole of benzoic acid. Locate the point on this steep portion of the curve equidistant between the two plateaus.5 1. All rights reserved. Because NaOH is a strong base. 0.10M 0. CHO2 H 1 HCHO2 HCHO2 OH 1 CHO2 2. 0.94 10 5 mol Ca(OH)2 0. 0.0122 L HCl 2 mol HC 0. Ba(OH)2(aq) 2HCl(aq) y BaCl2(aq) 0. 3.050M. acidic solution c. Ca(OH)2(aq) H2SO4(aq) y CaSO4(aq) 0. 2NaOH(aq) H2SO4(aq) y Na2SO4(aq) 2H2O(l) 0.021 mol NaOH 1 mol H2SO4 5.0125 L Ba(OH)2 13 mL Ba(OH)2 2H2O(l) . To determine the equivalence point.0550 g Al(OH)3 3 mol HCl 1 mol Al(OH)3 liters liters 0.94 10 5 mol Ca(OH)2 0.0124 L H2SO4 liters Molarity 0.Section 19. Inc.200M HCl molarity 0.0025 mol NaOH 5.6 mL HCl Section 19. each mole of NaOH added reacts with each mole of C6H5COOH present.0122 L HCl 1 L HCl 1 mol Ba(OH)2 0.0015 mol Ba(OH)2 moles 0. 2.000198 mol H2SO4 0.0025 mol [C6H5COONa] 0.. 1 mol H2SO4 1 mol Ca(OH)2 2H2O(l) 0. the solution is slightly basic. [C6H5COOH] 0. Matching 1. Thymol blue might also be a good candidate. h 6. hydrolyzing salt 8. acidic c. 4. Multiple Choice 11. phenolphthalein would be a good choice.7. 4. a 4. All rights reserved. Amphoteric.2 10 6M [C6H5COO ] 0. 2.5 [OH ] 3. a.. c 6.2 10 6M At the equivalence point [OH ] [C6H5COOH] 3. 27. d 7. 28. basic 840 Core Teaching Resources . Strong acids. 2Al(OH)3 y Al2(SO4)3 c. neutral 31. The dissociation constant reflects the fraction of a weak base or weak acid that is in ionized form. 17.050) Quiz for Chapter 19 1. a. 21. d a b d d d a 25. 3. 22. 2HBr b. 24. b Chapter 19 Test A A. Kb [C6H5COO ] 10. b 3. a. g 7. b d d a c d c 18. 8. Problems 30. basic b. i a j f 5. 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. buffer © Pearson Education. acidic d.5 pOH 14 pOH 5. Acidic solution. 14. 23. Kw [H ] [OH ] 1 10 14 Kw [H ] 1 10 12 [OH ] 2 [H ] 1 10 acidic 32. a d a c b C. 13. 12. Lewis acid. The other terms describe aqueous solutions based on their pH.0 10 10 (0. c 5. basic b. the benzoate ion establishes the equilibrium shown. pH 7. b 2. c 8. C6H5COO H2O 1 C6H5COOH OH At the equivalence point. pH log[H ] 9. 19. The resulting solution is slightly basic because [OH ] > [H ]. equivalence point 7. H2SO3 1 H HSO3 [H ] [HSO3 ] Ka [H2SO3] b. 2. neutral Mg(OH)2 y MgBr2 2H2O 6H2O Vocabulary Review 19 1. 3. a. 6. 26. 16. The other terms refer to ways of describing acids and bases according to the Brønsted-Lowry theory. 15. publishing as Pearson Prentice Hall. The other terms are theories used to classify acids and bases. H NO3 31H [H ] [NO3 ] Ka [HNO3] 33. 5. 3H2SO4 34. 29. b B.050M (3.2 10 6)2 Kb 2. pH 4. Hydronium ion. e 9. Inc. d 8. pH pOH 14 8. A faint pink color should be detected at the equivalence point. Weak acids and bases are only partially ionized in aqueous solution. neutral 9. 20. [C6H5COOH][OH ] 9. Because the equivalence point occurs between pH 6 and pH 11. [H ] 1 10 4. Alkaline is another name for a basic solution and basic solutions would have a high hydroxide ion concentration. c 10. 31.4 10 4] pH (log 3.09775M [H ][C2H3O2 ] Ka [HC2H3O2] [0. acidic c. acids are electron-pair acceptors. C2H3O2 (aq) 36. [OH ] 1 10 11.35 mol KOH 2 mol KOH 0. 23. 19.0 10 9 mol/L © Pearson Education. e D. c a b a a 34. Essay 33. H3O (aq).D. I b. a.7 10 4M [Cu ] Chapter 19 Test B A. Answer Key 841 . j 11. publishing as Pearson Prentice Hall. 4. 3. 2. Both acids and base. 30. 29.53) ( 4)] pH [ 3. H2SO4(aq) 2LiOH(aq) y Li2SO4(aq) 2H2O(l) H2SO4(aq) 2KOH(aq) y K2SO4(aq) 2H2O(l) 1 mol 2 mol 1 mol 2 mol 1 mol H2SO4 0. Kw [H ][OH ] 1. Acids taste sour. CuCl(s) y Cu (aq) Cl (aq) Ksp [Cu ][Cl ] 3.25 10 3M [HC2H3O2] 0. pH log [H ] pH log [3. b d a b c 17. HF(aq). H3O (aq).00225M [HC2H3O2] 0. F (aq) b. Cl (aq) c. H3O (aq). [OH ] 1. a. Multiple Choice 12. Ka [H2SO4] a. 6. acidic b. 35.00225][0. 13. E. 20. HCl(g). H .2 10 7 [Cu ]2 5.4 log 10 4) pH [(0. 14. H2O(l). H2O(l). [H ] 1 10 6 . BCl3. HC2H3O2(aq) 1 H (aq) C2H3O2 (aq) [H ] [C2H3O2 ] 2. HF(aq) KOH(aq) y KF(aq) H2O(l) b. Additional Problems 22. 16. b g d a 5. Ka [HI] [H ][HSO4 ] b.47] pH 3. 26. [H ] 1 10 7 . Inc.18 mol H2SO4 28. 8.18 10 5 B. Bases feel slippery.. 7. H2O(l). NH3 37.0 10 5 mol/L The solution is basic. neutral [H ][I ] a.00225] Ka [0. Problems 27. According to the Lewis theory. 6. HC2H3O2(aq). i 10. cause indicators to change colors and react with each other to form water and a salt. Essay 35. 7.0 10 14 (mol/L)2 [OH ] Kw/[H ] 1. 18. 3. whereas bases are electron-pair donors. The Brønsted-Lowry theory defines acids as proton donors and bases as proton acceptors. 15. Matching 1. a. 25. 24.0 10 14 (mol/L)2 [OH ] 1. 32. 21. f c h k 9. bases taste bitter.09775] Ka 5.2 10 7 [Cu ][Cu ] 3. b d a d a C. All rights reserved.47 The solution is acidic.1000M 0. 2. The copper ion. 2. NT Part C Matching 13. g 22. AT 11.2 Part A Completion 1. NT 15. f 16.1 Part A Completion 1. AT 11. AT Part C Matching 17. 8. To measure the Ka of a colored weak acid. This pH is the Ka of the acid. The pH solutions 5-12 are blue. Oxidation is the complete or partial loss of electrons. 3. 5. 4. Figure A 1. 3. h 19. HBCG is yellow. was reduced and is the oxidizing agent. c 17. Inc. 7. Part B True-False 9. c 23. 2. Results will vary depending on the indicator chosen. The conjugate acid. Reduction is the complete or partial gain of electrons. Zn. NT 12. 7. e 18. The pH solution 4 is green. NT 14. © Pearson Education. a 21. redox away toward reduction 5. 6. 4. Section Review 20. AT 10. 8. they immediately transfer electrons to the iron ions. mix one drop of the weak acid with one drop of each pH 1-12 buffer solution. b 842 Core Teaching Resources . d 20. a 18. e 14. b 15. the iron atoms lose electrons as the iron begins to be oxidized. an intermediate between yellow and blue. BCG is blue. 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. zero sign charge zero 5. The zinc metal. When oxygen and water attack iron. 4.Chapter 19 Small-Scale Lab Section 19.4 Ionization Constants of Weak Acids. 6. An equal mixture of HBCG and BCG is green at pH 4. reducing them back to neutral iron atoms. Look for the pH of the color change. d Part D Questions and Problems 19.. charge on the ion electron oxidation decrease You’re the Chemist 1. The pH solutions 1-3 are yellow. 6. 21. was oxidized and is the reducing agent. Cu2 . i 25. The conjugate base. AT 16. oxidizing reduced reducing oxidized Part B True-False 9. 3. NT 12. f 24. All rights reserved. 20. Since aluminum and zinc are better reducing agents than iron and are more easily oxidized. AT 10. NT 13. 2. publishing as Pearson Prentice Hall. Cl is oxidized ( 1 to 0) N is reduced ( 5 to 2). a. 3. 2.1 1. thus the oxidation number is 1. 27. The ionic charge on iron is 3 . The oxidation number is 0 f. NT Part C Matching © Pearson Education. b 18. a loss of 10e for I2 10HNO3 I2 y 2HIO3 10NO2 4H2O 22. The ionic charge on magnesium is 2 . Fe2 → Fe3 e and 6e 14H Cr2O72 → 2Cr3 7H2O. Mg: N2: 5. Sn is tin in an uncombined state. AT 11. Multiply the oxidation reaction by 3 and the reduction reaction by 2.2 1. thus the oxidation number is 2. a gain of 3e . NT 13. Fe: O2: 6. g. oxidation number half-reaction balanced ionic 5. 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. 3H2S 2HNO3 → 3S 2NO 4H2O Section 20. a 16. e 19. c. The oxidation number is 0. The ionic charge on sulfur is 2 . A decrease in the oxidation number indicates reduction.. Sb is oxidized (0 to 5) S is reduced ( 6 to 4). ionic Part B True-False 8. An increase in the oxidation number of an atom indicates oxidation.3 Part A Completion 1. AT 9. I changes from 0 to 5. 4. thus the oxidation number is 3. d. a. Mg: H : 10. multiply by 3 2HNO3 6HI y 2NO 3I2 4H2O b. thus the oxidation number is 4. N changes 5 to 2. two 6. NT 12. All rights reserved. N changes 5 to 4. g Part D Questions and Problems 21. N is reduced ( 5 to 2). Sr: O2: 2. b. The ionic charge on tin is 4 . Br : Cl2: 7. C is oxidized (0 to 4) b. thus the oxidation number is 2. a loss of 2e for I2. d 20. Answer Key 843 . multiply by 10 I2 y 2HIO3. c 15. a gain of 1e . Si: F2 8. multiply by 2 2HI y I2. I changes from 1 to 0. S2 → S 2e and 3e 4H NO3 → NO 2H2O. Inc. AT 10. e. HNO3 y NO. publishing as Pearson Prentice Hall. added 7.Part D Questions and Problems 26. 14. HNO3 y NO2. Ca: O2 9. Cs: Br2: 4. The ionic charge on potassium is 1 . f 17. Se is selenium in an uncombined state. Li: S: 3. multiply the oxidation reaction by 6 14H 6Fe2 Cr2O72 → 6Fe2 2Cr3 7H2O Practice Problems Section 20. a. Br is oxidized ( 1 to 0) Mn is reduced ( 7 to 2). 2KMnO4 16HCl y 2MnCl2 5Cl2 8H2O 2KCl Interpreting Graphics 20 1. Reducing agent is chlorine. C 2H2SO4 y CO2 2SO2 2H2O b. KMnO4 HClyMnCl2 Cl2 H2O Manganese is reduced ( 7 y 2). Oxidizing agent is nitrogen. a. 4OH Mn → MnO2 2H2O 2e 2e H2O H2O2 → H2O 2OH 3.30 mL 0. Sulfur is reduced ( 6 y 4). 3 2. publishing as Pearson Prentice Hall. 1 8H © Pearson Education. 4 f. a.3 1. 6 3. 2OH CrO2 ClO y CrO42 Cl H2O c.. Reducing agent is iodine. a. Sb HNO3 y Sb2O5 NO H2O Antimony is oxidized (0 y 5). Volume KMnO4 Initial Volume Final Volume 48. Increase in oxidation number of chlorine ion 1. Iodide ion is oxidized ( 1 y 0). 2 c. a. 0 1 5 2 5 2 2 2 1 2 KCl Section 20. decrease in oxidation number of nitrogen 3. b. HNO3 HI y NO I2 H2O Nitrogen is reduced ( 5 y 2). 3Zn 2Cr2O72 28H y 3Zn2 4Cr3 14H2O 2 3 2. Increase in oxidation number of iodine ion 1. Inc. thus the oxidation number is 1. 2HNO3 6HI y 2NO 3I2 4H2O d.02530 L L 4 Moles MnO4 5. decrease in oxidation number of chromium 3. Increase in oxidation number of carbon 4. Reducing agent is antimony. 5Fe2 y Mn2 4H2O 5Fe3 3. and the solution in the flask turns light purple.0200 mol Moles MnO4 0. a. c. d. 6H 3Sn IO3 y 3Sn I 3H2O 2 d. a. The end point occurs when the number of equivalents of MnO4 added equals the number of equivalents of Fe2 originally present in the reaction flask. Increase in oxidation number of iodine ion 1. 3 c. S2 → S 2e 3e 4H NO3 y NO 2H2O 2 d. All rights reserved. Sn y Sn 2e 6H 6e IO3 → I 3H2O c. Oxidizing agent is nitrogen. 8H 5Fe MnO4 y 5Fe3 Mn2 4H2O 2 4 c. decrease in oxidation number of manganese 5. decrease in oxidation number of nitrogen 3. 1 7 2 1 1 2 1 0 1 2 1 1 1 5 2 1 1 2 2 0 1 2 0 1 6 2 4 2 4 2 1 2 f. decrease in oxidation number of sulfur 2. Increase in oxidation number of zinc 2. Chloride ion is oxidized ( 1 y 0). Reducing agent is carbon. MnO4 d. 6 d. 8H 3S 2NO3 y 3S 2NO 4H2O e. 2. signaling the end point of the titration. b. Oxidizing agent is sulfur. 4. 6Sb 10HNO3 y 3Sb2O5 10NO 5H2O e. 2 e. Increase in oxidation number of antimony 5.35 mL 25.h. 4. The ionic charge on bromine is 1 . KIO4 7KI 8HCl y 8KCl 4I2 4H2O g. Nitrogen is reduced ( 5 y 2).06 10 mol MnO4 844 Core Teaching Resources . 2OH Mn2 H2O2 → MnO2 2H2O f. decrease in oxidation number of iodine 7. Fe y Fe e 5e 8H MnO4 y Mn2 4H2O 2 4 b. Oxidizing agent is manganese. Increase in oxidation number of sulfur 2. When all the Fe2 in the flask is oxidized.65 mL 23. a. 2 b. d. the next drop of MnO4 remains unreacted. One equivalent is the amount of reducing agent (or oxidizing agent) that can give (or accept) one mole of electrons. C H2SO4 y CO2 SO2 H2O Carbon is oxidized (0 y 4). 3 b. 3H2S 2HNO3 y 3S 2NO 4H2O c. decrease in oxidation number of nitrogen 3. 2OH Zn HgO y ZnO22 Hg H2O 2 b. 7. An oxidation number is assigned to an element in a compound according to a set of arbitrary rules. Al 3..80% 2. The sum of the oxidation numbers of the elements in a neutral compound is zero.85 g Fe2 0. Oxidation-number change method: 2 3 2 ( 3) 2 2 6 0 4 2 Vocabulary Review 20 1. a 6. Multiple Choice 11. 13. 4.141 g % Fe in ore 100% 4. 8. Na oxidized. a 7. S reduced. 9. a. 6. e f d g 5. d j f a 5. i a c h 9. reducing agent. F 1 b. Chapter 20 Test A A. 8. 7. 2. publishing as Pearson Prentice Hall. oxidizing agent oxidation-number-change method reduction half-reaction method oxidation number reducing agent half-reaction oxidation-reduction reaction oxidation redox reaction Fe2O3 CO y Fe 3 ( 2) 6 CO2 Fe2O3 3CO y 2Fe 3CO2 Half-reaction method: 6H Fe2O3 6e y 2Fe 3H2O 3(H2O CO y CO2 2H 2e ) 6H Fe2O3 6e 3H2O 3CO y 2Fe 3H2O 3CO2 6H 6e Fe2O3 3CO y 2Fe 3CO2 Quiz for Chapter 20 1. 15. 4. 25. Oxidation numbers help keep track of electrons in redox reactions. 10. S 0 (element) 30. b d c b a b 17. d c c c Chapter 20 Test B A. c 10. 8. K oxidized. b 2. 2. In a polyatomic ion. b 5. 26. 18. 3. 6. b 10. Matching 1. 6. a 3. 19. 5. h B. however. oxidizing agent b.53 10 3 mol Fe2 1 mol MnO4 Mass Fe 2. Li 1. 3. 24. The oxidation number of a monatomic ion is the same in magnitude and sign as the ionic charge. 14. a. i g e b 9. Br2 reduced. 4. 12. Essay 31. A decrease is reduction. b © Pearson Education. j Answer Key 845 . 2. All rights reserved. Na 1.53 10 3 mol Fe2 55. 22. Questions 27. 3. 7. oxidizing agent.141 g 1 mol Fe2 0. c 8. 4. 2Cr 3Br2 y 2Cr3 6Br 29. Matching 1.Moles iron(II) 5. reducing agent 28. Inc.06 10 4 mol MnO4 5 mol Fe2 2. c c a a c b 23. d D.938 g C. 21. 20. O 2 c. An oxidation-number increase is oxidation. the sum is equal to the charge on the ion. 16. The oxidation number of an element in an uncombined state is zero. 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 1, 6, 2 Cu(NO3)2 2, 5, 2 HAsO3 1, 5, 2 MnO4 7, 2 3 ( 1) 3 Chapter 20 Small-Scale Lab Section 20.3 Half Reactions, page 655 Analysis HCl HNO3 H2SO4 Zn Bubbles Bubbles Bubbles 31. a. 4HNO3 3Ag y 3AgNO3 1 1 ( 3) ( 2) 2 3 NO 2H2O Mg Bubbles Bubbles Bubbles b. Br2 SO2 2H2O y H2SO4 2 ( 1) 2 2HBr Cu No visible reaction No visible reaction No visible reaction 32. a. HNO2 HI y I2 NO H2O H (aq) NO2 (aq) H (aq) I (aq) y I2(aq) NO(g) H2O(l) Oxidation: 2I (aq) y I2 2e Reduction: 2[2H (aq) NO2 (aq) 1e y NO H2O] 4H 2NO2 2e y 2NO 2H2O 4H 2I 2NO2 y I2 2NO H2O Final: 2HNO2 2HI y I2 2NO 2H2O b. K2Cr2O7 FeCl2 HCl y CrCl3 KCl FeCl3 H2O 2K (aq) Cr2O72 (aq) Fe2 (aq) 2Cl 2K (aq) H (aq) Cl (aq) y Cr3 (aq) 3Cl (aq) K (aq) Cl (aq) y Cr3 (aq) Fe3 (aq) 3Cl (aq) H2O Oxidation: 6[Fe2 y Fe2 1e ] 6 Reduction: 2Cr 6e y 2Cr3 6Fe2 2Cr6 y 6Fe3 2Cr3 Final: K2Cr2O7 6FeCl2 14HCl y 2CrCl3 2KCl 6FeCl3 7H2O © 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 Zn Fe Cu. 2. H2(g) is the gas produced all the reactions. 3. Mg(s) Mg(s) Fe(s) Fe(s) 2HCl(aq) → H2(g) 2H (aq) → H2(g) 2HCl(aq) → H2(g) 2H (aq) → H2(g) MgCl2(aq) Mg2 (aq) FeCl2(aq) Fe2 (aq) All are redox reactions because the oxidation number of reactants change. 4. Mg(s) → Mg2 (aq) Fe(s) → Fe (aq) 2 2e 2e 2e 5. 2H 2e → H2(g). Mg(s) → Mg2 (aq) 846 Core Teaching Resources Mg(s) 2H (aq) → H2(g) Mg2 (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 Mg2 2e Reduction: 2e Cl2 y 2Cl Redox: Mg Cl2 y Mg2 2Cl 0 0 0 Ecell Ered Eoxid E0 E0 2 E0 cell Cl Mg E0 1.36 V ( 2.37 V) cell E0 3.73 V cell 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 Mg2 at the anode (the negative electrode) and Pb2 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 (SO42 ) 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 e e 2. a. E0 0.14 V ( 2.90 V) 2.76 V; cell spontaneous b. E0 0.80 V 1.36 V 0.56 V; cell nonspontaneous c. E0 2.87 V ( 0.76 V) 3.63 V; cell spontaneous d. E0 0.28 V ( 3.05 V) 2.77 V; cell spontaneous e. E0 2.93 V 0.54 V 3.47 V; cell nonspontaneous Interpreting Graphics 21 1. 2. 3. 4. anode( ) cathode( ) electrorefining a. The anode( ) of the electrolytic cell should be connected to the positive( ) terminal of the battery. The cathode( ) of the electrolytic cell should be connected to the negative( ) 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( ) and into the cathode( ). 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. Zn2 and Fe2 c. copper Cathode Anode Ag Solution of silver ions Anode (oxidation): Ag(s) y Ag (aq) e Cathode (reduction): Ag (aq) e y Ag(s) Practice Problems Section 21.2 1. a. Cl2(g) + Mg(s) y 2Cl (aq) + Mg2 (aq) E0 1.36 V ( 2.37 V) 3.73 V cell cathode: Cl2(g) 2e y 2Cl (aq) b. 2Ag (aq) Ni(s) y Ni2 (aq) 2Ag(s) E0 0.80 V ( 0.25 V) 1.05 V cell cathode: Ag (aq) e y Ag(s) c. 2MnO4 (aq) 16H (aq) 5Cd(s) y 5Cd2 (aq) 2Mn2 (aq) 8H2O(l) E0 1.51 V ( 0.40 V) 1.91 V cell cathode: MnO4 (aq) 8H (aq) 5e y Mn2 (aq) 4H2O(l) d. Br2 2Na(s) y 2Na (aq) 2Br (aq) E0 1.07 V ( 2.71 V) 3.78 V cell cathode: Br2(l) 2e y 2Br (aq) e. MnO2(s) 4H (aq) H2(g) y 2H (aq) Mn2 (aq) 2H2O(l) E0 1.28 V 0.00 V 1.28 V cell cathode: MnO2(s) + 4H (aq) 2e y Mn2 (aq) 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 25. In both voltaic and electrolytic cells.66 V 2 3[Co (aq) 2e y Co(s)] E0 0. 19. 20. NT AT NT AT AT NT F. 4. 14. 8. 13. a C. 6. 11. 22. Matching 1. NT 29. NT 30. d b c c c 16.25 V Zn2 (aq) 2e y Zn(s) E0 0. oxidation occurs at the anode and reduction occurs at the cathode.28 V 0 0 0 Ecell Ered Eoxid E0 1.Quiz for Chapter 21 1. b 10. 17. Multiple Choice 11. 2. h d f i 5. 25. 18. 5. Question 31. the anode is positive and the cathode is negative. 24. NT D. 22. 28. a d c a a 21. b. the cathode is positive and the anode is negative. NT © Pearson Education. 24. AT 33. 35.. 8. 16. 6. 12. 13. 7. In the voltaic cell. 9. 20. 15. Matching 1. 6. 34. ST Answer Key 849 . 15. Inc.51 V This reaction is not spontaneous. 27. Ni2 (aq) 2e y Ni(s) E0 0. 3.76 V 0 0 0 Ecell Ered Eoxid E0 0. 8. 19. 18. j g i e 9. 12. 17. f 10. All rights reserved. e e Salt bridge B. AT 32. Essay 32.76 V ( 0. 21.28 V) cell 0 Ecell 1. 26. 14. AT 31. 2. 16. AT 29.25 V) cell 0 Ecell 0. 10. True-False 26. a B. 15. 23. 7. 12. 3. 4. NT AT NT AT ST NT 7. 4. True-False 28. NT 27. c g j e 9. c d b c d a 17. Chapter 21 Test A A. b h d c 5. NT 30. NT AT AT ST NT AT 13. 2. a. In the electrolytic cell. 14. Multiple Choice 11. c c a a c Chapter 21 Test B A.66 V ( 0. 18. The reduction potential of a half-cell is a measure of the tendency of a given halfreaction to occur as a reduction. publishing as Pearson Prentice Hall. The negative value means that the tendency for zinc ions to be reduced is less than that of hydrogen ions to be reduced. d a d c b c 23. 3. Additional Questions 33.38 V This reaction is not spontaneous. 2[Al3 (aq) 3e y Al(s)] E0 1. b a b c d Anode Cathode E. C. so zinc metal is oxidized when paired with the standard hydrogen half-cell. 80 V Reduction: Mg2 (aq) 2e y Mg(s) E0 2.34 V) cell 0. Y appears above Z.46 V E0 cell 2Ag(s) E. the two half-cell reactions are as follows: 2[Al(s) y Al3 (aq) 3e ] 3[Pb2 (aq) 2e y Pb(s)] Net: 2Al(s) 3Pb2 (aq) y 2Al3 (aq) 3Pb(s) E0 E0 E0 cell red oxid E0 0. reduction occurs at the cathode. a. publishing as Pearson Prentice Hall. F. In both voltaic and electrolytic cells.76 V 0 Ecell E0 E0 red oxid E0 0. Cu(s) y Cu2 (aq) 2e 2Ag (aq) 2e y 2Ag(s) Net: Cu(s) 2Ag (aq) y Cu2 (aq) E0 E0 E0 cell red oxid E0 0. Reaction 3: since Y is oxidized and Z2 is reduced. e e Salt bridge Cu Anode Cu(NO3)2 (aq) Ag Cathode AgNO3 (aq) 35. the redox reaction will be spontaneous. the redox reaction will be nonspontaneous. The half-reactions are: Oxidation: Na(s) y Na (aq) e E0 2.37 V 0 Ecell E0 E0 red oxid E0 2. The half-reactions are: Oxidation: Ag(s) y Ag (aq) e E0 0.53 V cell 38. Reaction 2: since W is not oxidized in the presence of X2 .66 V Reduction: Zn2 (aq) 2e y Zn(s) E0 0. in an electrolytic cell the flow of electrons is being pushed by an outside source such as a battery.17 V cell Since the standard cell potential is negative.66 V) cell E0 1.34 V ( 2. Additional Questions 37. X. 39. Al is oxidized and Pb is reduced. electrons flow from the anode to the cathode through the external circuit. Question 34. Thus.34 V 0 Ecell E0 E0 red oxid E0 0. while the anode is the negative electrode and the cathode is the positive electrode in a voltaic cell.80 V) cell E0 3. 850 Core Teaching Resources . Z appears above X in the activity series. the reverse is true in an electrolytic cell—the anode in an electrolytic cell is the positive electrode and the cathode is negative.80 V ( 0. However. Essay 36. W should appear below X. Z.37 V ( 0.66 V) cell E0 0. The half-reactions are: Oxidation: Al(s) y Al3 (aq) 3e E0 1.71 V Reduction: Cu2 (aq) 2e y Cu(s) E0 0. © Pearson Education.. All rights reserved. and oxidation occurs at the anode. Since Al is above Pb in the reduction potential table. and W.71 V) cell E0 3. Additionally. the redox reaction will be spontaneous. b. Inc.76 V ( 1.05 V cell Since the standard cell potential is positive. c. Reaction 1: since Z is oxidized and X2 is reduced.90 V cell Since the standard cell potential is positive. while the flow of electrons in a voltaic cell is caused by a spontaneous chemical reaction. The elements should be listed as follows: Y.D.13 V ( 1. The yellow solution is I2(aq). Figure B The bubbles are H2(g) and the blue solution indicates of the presence of OH ions. KBr: Cathode: 2H2O 2e → H2(g) Bubbles. a 18. 8.Chapter 21 Small-Scale Lab Section 21. publishing as Pearson Prentice Hall. 2. NaCl: Cathode: 2H2O 2e → H2(g) Bubbles. CuSO4: Cathode: Cu2 2e → H2(g) Copper plates out. 3. The bubbles are H2(g) and the blue BTB solution indicates the presence of OH ions. f 20. 2. which is black in the presence of starch. Inc. yellow BTB 3. 9. page 684 Analysis H2O No visible reaction Na2SO4 Bubbles at both the anode and cathode Figure A 1.1 Part A Completion 1. blue BTB Anode: 2Br → Br2(aq) 2e Yellow solution. Yellow soln at the anode. NT 12. Pure water has too few ions to carry an electric current. e 19. KI KI starch KI BTB Bubbles at Bubbles at Bubbles and the cathode. Anode: H2O → 1 O2(g) 2H 2 Bubbles. 4. 7. at the anode. Sodium sulfate is an electrolyte. CH3 @ CH3 CH3 CH2 @ @ @ CH3 2 C 2 CH 2 CH2 2 CH 2 CH2 2 CH2 2 CH3 @ CH3 Answer Key 851 .00 V 0. AT Part C Matching 16. All rights reserved.54 V H2O → O2(g) 2I → I2(aq) I is more likely to oxidize (lose electrons) than H2O because it has a more favorable (more positive) Eo value. 2H2O 2e → H2(g) H2O → O2(g) 1 2 1 2 1 2 1 2 2. b 21.2-dimethylbutane 23. c Part D Questions and Problems 22. 16 24. straight branches alkyl longest parent 3H2O → H2(g) © Pearson Education. 5. the cathode. 2. It dissociates into ions in solution. ST 13. 2H2O 1 2 2OH Na2SO4 BTB 2OH Anode area turns yellow. 16 b.3 Electrolysis of Water. blue soln at Yellow soln Black solution the cathode. 10.. AT 14. 2OH 2H 2OH 2H2O 2e 2H Section Review 22. The bubbles are O2(g) and H ions in solution impart the yellow color to the BTB solution. 3H2O → H2(g) H2O → H2(g) O2(g) O2(g) O2(g) You’re the Chemist 1. d 17. carbon organic hydrocarbons four single 6. blue BTB Anode: 2Cl → Cl2(aq) 2e Yellow solution. ST 15. cathode area turns blue 2OH 2e Eo 2e → H2(g) 2H 2e 2OH 2e 0.82 V 0. at the anode. Part B True-False 11. 3. a. 4. which carry an electric current through the solution. 5. Part B True-False 12. b Part D Questions and Problems 21. c 20. cyclic aromatic six hydrogen double resonance 7. c 15.Section 22. p Section 22. 3. AT 16. 5. 11. CH2 3 CH 2 CH 2 CH3 @ 852 Core Teaching Resources . 6. f 22. 5. a 23. ST 12. 3-methyl-2-hexene 19. 10. c 21. 3.5-tetramethylnonane 20. 7. ST Part C Matching 14. alkane -ene double bond -yne Part D Problems 25.4. 8. Carbon 2 is the asymmetric carbon. trans stereoisomer asymmetric optical mirror superimposed © Pearson Education. 5. NT 16. 12.. d 16. molecular structures butane properties Geometric cis 7. unsaturated double triple longest double 6. 11. 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. 10. 4. AT 13. d 18. 4-methyl-1-hexene CH3 21.4 Part A Completion 1. 8. 2 CH2CH2CH3 22.3 Part A Completion 1. a 24. h 18. d 20. g 19. NT 11. 2. ST 15. 3. a 17.2 Part A Completion 1. b Part D Questions and Problems 18. Methylbenzene xylenes ortho. 4. ST Part B True-False 13. 27. 9. ST 15. 8. CH3 CH2 H % ^ C3C ^ % CH2 CH3 H trans-3-hexene Part B True-False 10. publishing as Pearson Prentice Hall. AT 14. 9. o meta. 6. 2. 4. All rights reserved. AT Part C Matching 17. 2. 2. NT 14. ST 13. e 19.3. b Part C Matching 17. Inc. 9. m para. 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. 23. a.5-tetramethylheptane 3.3 1. Carbon 3 is the asymmetric carbon. 6. b 16. 3. 5. ST Part C Matching 15. 5-ethyl-3.4-dimethyl-2-hexene 2. 5-phenyl-2-hexene CH3 3. 1.4 1. a. NT 12. 2C5H12(l) 21. CH3 @ CH 4 C 2 C 2 CH3 @ CH3 Section 22. d 5. heptane: CH3—CH2—CH2—CH2—CH2—CH2—CH3 octane: CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH3 5. CH2 3 CH 2 CH 2 CH2 2 CH 2 CH3 @ @ CH3 CH3 b. publishing as Pearson Prentice Hall. c Practice Problems Section 22..1 © Pearson Education. All rights reserved. boiling point lignite bituminous anthracite aromatic Part B True-False 11. 2. a. natural gas coal methane aliphatic distillation 6.3-diethylbenzene or m-diethylbenzene Section 22. 7.3. 3-ethyl-2. e 18. 3. CH2 3 CH 2 CH2 2 CH 2 CH3 @ CH3 c. CH3 H % ^ C3C ^ % CH2 2 CH2 2 CH2 2 CH3 H 4. cis-2-pentene 2. 9. CH3 2 CH2 2 CH 2 CH 2 CH2 2 CH3 @ @ CH2 CH2 @ @ CH3 CH3 c. AT 14.5 Part A Completion 1.2 1. d 17.5-trimethyloctane 2. cyclooctane b. a. @ CH % ^ CH2 CH2 @ @ CH2 CH % ^ % CH2 CH3 b. 2. 8. CH3 2 CH2 2 2 CH2 2 CH3 Answer Key 853 . 2C6H6 11O2(g) → 10CO(g) 6H2O 12H2O(g) 15O2 → 12CO2 Section 22. H % ^ C3C ^ % H b.4-dimethyl-1-pentyne 3. 1. CH3 2 CH 2 CH 2 CH 2 CH 2 CH2 2 CH2 2 CH2 2 CH3 @ @ @ @ CH3 CH3 CH3 CH3 4. a. NT 13. 19 c. Inc. 4. c 19. 1-pentyne: CH C—CH2—CH2—CH3 2-pentyne: CH3—C C—CH2—CH3 3-methyl-1-butyne: CH 4 C 2 CH 2 CH3 @ CH3 4. 1-ethyl-3-methylbenzene 2. 10. a Part D Problems 20. CH3 @ CH3 2 CH 2 C 2 CH2 2 CH3 @ @ CH3 CH2 @ CH3 Section 22. a.5.3. trans-6-methyl-3-heptene 3. j 10.3 C and 36.5. Compounds D and F are geometric isomers. 5. 2. Vocabulary Review 22 1. AT Chapter 22 Quiz 1. AT 9. Number the carbons in the chain. a c c b C. 2C6H14(l) D. (In fact. 26. e 9.. 3. 9. 8.3-dimethylpentane D and F. 4. 2. Multiple Choice 11. NT 6. 22. 3. f 8. c 7.7-trimethyl-3-octene 12. b Interpreting Graphics 22 1. they are different compounds with different properties.) B. Attach the substituents to the numbered chain at the proper positions. 2-phenylbutane. 1. 23. ST NT NT AT 5. 8. 4.Section 22. c c d a 20. homologous series 5. 4. ethane. substituent 3. NT 7.5 1. 15. 2. All rights reserved. NT E. Natural gas contains mainly low molar mass. cis-2-pentene and trans-2-pentene E. Like all isomers. C-2 is asymmetric 22 10 Compounds A (2-phenylpropane) and E (2-phenylbutane) are aromatic compounds. Find the root word (ending in -ane) in the hydrocarbon name. Gasoline is composed of alkanes with five to twelve carbon atoms. 21. 854 Core Teaching Resources . Kerosene is composed of alkanes with twelve to fifteen carbon atoms. cis configuration 4. saturated compounds Solution: hydrocarbons CH3 @ CH3 CH2 CH3 @ @ @ CH3 2 C 3 CH 2 C 2 CH2 2 CH 2 CH3 19O2(g) → 12CO2(g) 14H2O(l) 27. and butane. Additional Problems 29. 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. Carbon 3 is asymmetric. the boiling points of trans-2-pentene and cis-2pentene are 36. 18. 17. cracking 6. 4. Identify the substituent groups. 12. arene 7. Problems 24. stereoisomers 8.9 C respectively. Inc. 3. 7. alkynes 2. publishing as Pearson Prentice Hall. straight-chain alkanes— methane. the boiling points of compounds D and F are not expected to be the same. 3. 2C8H18 25O2 y 16CO2 18H2O 2. 2. The refining process yields fractions that differ with respect to the length of the carbon chains. d 6. 19. Matching 1. g i a h 5. 2-methyl-2-phenylbutane 25. H H @ H @ @ @ @ @ @ 2C2C2C2C2C2H @ @ @ @ @ H H H H H Chapter 22 Test A A. a small but measurable difference. 14. Thus. Then draw the longest carbon chain to form the parent structure. 2. propane. Add hydrogens as needed so that each carbon added has four bonds. 13. 11. A C 3. c c a b d 16. 6. Essay 28. ST 10. 14. Carbon 3 is asymmetric because there are four different groups attached to it—a methyl. 22.7-diethyl-2.30.4. Inc. 32. Essay 43.3-dimethyl-3-phenylhexane b. j 10. Multiple Choice 11. d c b b c d 17. 2. AT 32. 25. 45. 3. AT 35. AT 34. 28. 27. 12.5-diethyl-4-methylheptane b. 30. an ethyl. 18. hexane C6H14 CH3 —CH2—CH2—CH2—CH2—CH3 b. d 8. CH3 2 CH 2 CH2 2 CH3 @ CH3 Chapter 22 Test B A. All rights reserved. 3. 19.. C. AT AT NT AT 33. b d d d b c 23. 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. Matching 1. CH3—CH2—CH2—CH2—CH3 pentane b.4-trimethyl-2-pentene c. 4. a. 15. AT 34. CH3 H CH3 H % ^ C3C ^ % % ^ C3C ^ % CH3 @ CH 2 CH2 2 CH3 H H CH 2 CH2 2 CH3 @ CH3 c. @ CH3 2 C 2 CH3 @ CH3 2. a 7. a. plus a 3carbon branched chain. 2. NT 38. 13. 2-hexyne C6H10 CH3—C C—CH2—CH2—CH3 The number of hydrogen atoms decreases when carbon atoms form double or triple bonds in the alkene and alkyne. 26. © Pearson Education. 3-ethyl-2. 2-hexene C6H12 CH3—CH CH—CH2—CH2—CH3 c. Problems 39. CH3 @ CH2 CH3 CH3 @ @ @ CH3 2 C 4 C 2 CH 2 C 2 CH 2 CH 2 CH3 @ @ CH2 CH2 @ CH3 F. Additional Problems 44.2-dimethylpropane H CH3 42. 16. AT 33. 24. a. ST D. propyl groups. ethylbenzene b. 20. CH3 CH ^ ^ ^ ^ 3 C3C C3C % % % % H CH3 H H trans-2-butene cis-2-butene B. 6. 21. e 2-methylbutane CH3 c. True-False 31. c 9. 31.8-dimethyl-5-propyl-3decyne 41. AT 36. h 6. a. True-False 29. NT 40. NT 37. respectively. c d a d b d E. Answer Key 855 . publishing as Pearson Prentice Hall. a. The number of hydrogen atoms is at a maximum in the unsaturated alkane. b i f g 5. NT 13. CH CH CH CH 3 2 2 3 butane CH3CHCH3 CH3 methylpropane Section Review 23. 9.. 2-dimethylpropane © Pearson Education. Inc. H H H H H @ @ @ @ @ H2C2C2C2C2C2H @ @ @ @ @ H H H H H pentane 2. count the number of lines drawn to any point and subtract from four. 4. hexane functional reactive/functional alkenes alkynes Halocarbons 6. ST 14. To find the number of hydrogen atoms on any carbon of a line-angle formula.1 Part A Completion 1. 2-methylpentane You’re the Chemist 1. 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. 3. publishing as Pearson Prentice Hall. AT 3-methylpentane 2. 8.3-dimethylbutane 856 Core Teaching Resources . 2. AT 12.2-dimethylbutane 2. 7. All rights reserved. substitution hydrogen bromine alcohol salt Part B True-False 11. 5. 3. 10.Chapter 22 Small-Scale Labs Section 22. 2. page 708 Analyze 1.3 Hydrocarbon Isomers. Cl @ CH3 C CH2 CH2 CH3 @ Cl Section Review 23. hydroxyl carbonyl carbonyl carboxyl CH3 @ ^ Cl b. Addition 3.. publishing as Pearson Prentice Hall. Polyethylene 5. 11. d 20. NT 15.Part C Matching 15. AT 16. 5. 4. b 22. d 16. 3–hexanone Section Review 23. 7. NT Part D Questions and Problems 23. 2. polyesters 6. length Part B True-False 7. d 20. 5. NT 17. a 18. NT 16. a. hydrogenation tertiary 10. 3. ST 8. 6. Condensation 4.2 Part A Completion 1. a. c Part D Problems 22. 13. Alcohols 7. 3. alkane secondary 11. AT 9. AT 14. b. 20. 2. AT Answer Key 857 . 6. AT 15. a. 4. primary 23. b 21. AT 11. e 21. 12. H H @ @ CH3 2 C 2 C 2 CH3 @ @ H H b. polymer 2. a 19. c Section Review 23. a Part B True-False © Pearson Education. tertiary b. a.4 Part A Completion 1. OH H @ @ CH3 2 C 2 C 2 H @ @ H H 24. 9. b 17. e 18. Inc. water secondary 9. NT 10. c. c 19. OH O O @ # # K2 Cr2 O7 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 R 2 C 2 R –2H @ H alcohol ketone Part C Matching 17. ethers hydrogen bonding 12. All rights reserved.3 Part A Completion 1. 8. AT Part C Matching 18. d. hydration primary 8. lower Part B True-False 14. 10. oxygen double ketones/carboxylic acids ketones/carboxylic acids aldehyde carboxylic acid formaldehyde carboxylic acids esters propanol oxidation–reduction potassium dichromate Part D Problems 19. a. Dacron™ is one example of a polyester. 3.3-butanediol is expected to be most soluble due to its two 2OH groups. Section 23. 858 Core Teaching Resources . x H H % ^ C3C uuy H ^ % H H CH2 2 CH2 H x CH3 CH3 CH2 C CH2 OH H 2. c. c 14. a. esterification b. e. a. CH3 CH2 CH2 2 CH 2 CH 2 CH2 CH2 Br @ @ CH3 CH3 © Pearson Education. addition b. 3. chloroethene (vinyl chloride) CH3 3. d.3-butanediol: CH3 2 CH 2 CH 2 CH3 @ @ OH OH 2. carboxyl d. a. c. ether c. publishing as Pearson Prentice Hall. ^ Br @ CH2 CH3 b. hydroxyl 2. CH3 CH2 CH 3 CH2 HCl uy CH3 CH2 CH CH3 @ Cl @ @ 18. a 2. halogen b. primary 1-pentanol. All rights reserved. Br2 uuuy catalyst ^ Br HBr 4. 2. substitution Section 23. oxidation-reduction. d. c. m-bromobenzene b. a. Inc. b.. a. a. a. dipropyl ether: CH3 CH2 CH2 2 O 2 CH2 CH2 CH3 2-methyl-1-butanol: Part D Questions and Problems 17. b 13. a. secondary ethylphenyl ether 3-methyl-1-butanol. b. 1-bromo-1-chloroethane c. 1–butanol is oxidized to butanoic acid.1 1. They are all halocarbons. 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 % CH 2 OH ^ O # CH3 CH2 CH2 C % OH b. 2-butanol. e 16. CH3 CH2 CH2 CH2 OH K2Cr2 O7 uuuy H2 SO4 O # CH3 CH2 CH2 C % OH 5. CCl4 4HCl 4. c. CH3 % CH 2 CH2 2 CH2 2 OH ^ NaBr b. primary 4. Practice Problems Section 23.3 1. b. Polyesters are polymers consisting of many repeating units of dicarboxylic acids and dihydroxy alcohols joined by ester bonds. a. a.Part C Matching 12. a. which can form hydrogen bonds with water. b. d 15. CH3 CH3 O # uy CH3 CH2 CH2 C % H CH ^ 3 O 2 CH % CH3 H2O b.2 1. PET is a polyester. All rights reserved. a primary alcohol.304. 11. 15. 14. e i f g 5. 4. functional group 2.362 min 1 Polypropylene is used extrensively in utensils and containers. 0.000. which indicates a reaction between ethanol. hydration reactions 6.600 0. 2. publishing as Pearson Prentice Hall.4 1. Answers will vary slightly.800 0. NT AT ST AT AT 7.157 min 1 2. ketones Solution: aspirin © Pearson Education. Interpreting Graphics 23 1. CH3CH2OH uuuuy CH3CHO H2 SO4 Quiz for Chapter 23 1. such as 2-methyl-2-propanol. 0. H @ I2 y 2. 3.000 c. d 6.465. 4. a. 3.140. Polyethylene terephthalate (PET) is formed from the condensation of terephthalic acid and ethylene glycol. Because the repeating units are joined by ester bonds.157 min 1 k 0.Section 23. This tertiary alcohol serves as a negative control Investigators use negative and positive controls to check that a chemical assay is functioning properly. Only primary and secondary alcohols are oxidized by dichromate ion. Matching 1. alcohols 5. 2 2 7 3. also known as Teflon™. Polytetrafluoroethene. b K Cr O [CH3CH2OH] k [CH3CH2OH] t 4. 0. 16. and the oxidizing agent. 0. 8. h 7. j 9. are not expected to react.627.303 k 0. a 10. propene (propylene) CH3 @ xCH2 3 CH polypropylene CH3 @ CH2 2 CH x b. slope 0. 5. 2. aryl halides 3. Table 1 shows a change in absorbance values with time. 9. NT NT AT NT ST I @ HI 12. c 8. substitution reaction 4. Inc.200 Time (min) 0 1 2 3 4 5 tetrafluoroethene xCF2 3 CF2 polytetrafluoroethene (PTFE) CF2 2 CF2 x 0. Tertiary alcohols. 10. (The slight fluctuation is due to random electronic noise in the instrument. is used to coat nonstick cookware and to make bearings and bushings in chemical reactors. One molecule of water is lost for each bond formed. AT AT AT AT ST Chapter 23 Test A A. 6.. hydrogenation 7. 13.) 2. a. Log (absorbance) 1. Rate Answer Key 859 . 0. 0.000 0.766 b.400 0. O # C2 O # 2 C 2 O 2 CH2 CH2 2 O x Vocabulary Review 23 1. The data in Table 2 show no change even after five minutes. carboxylic acid ester e. such as alkenes. a. 13. b. 20. a. 13. 21. 15. Multiple Choice 11. Problems 29. In addition polymerization. d. 22. Thus.B. d. are joined to one another. 12. 4. 19. 19. b. such as dicarboxylic acids and dihydroxy alcohols. 3–bromohexane c. b c a b d a 17. and durability. ketone CH3CH2I KOH → CH3CH2OH KI ethanol CH3 — CH2—CH CH—CH2—CH2—CH3 HBr → CH3—CH2—CHBr—CH2—CH2—CH2—CH3 Cl @ HCl 2-butanol chlorobenzene/phenyl chloride R—X R—O—R O # R2C2O2R O # R 2 C 2 OH © Pearson Education. 3. c a b c c c 17. called monomers. Problems 28. d 9. react in a head-to-tail fashion. I2 y ^I HI D. 30. d d c b c b 23. 16. chain-like molecules formed by the covalent bonding of repeating smaller molecules. and freezes at a temperature lower than water alone. 21. f 8. 24. Polymers are large. 27. Inc. 15. e 10. 26. 30. b 7.. e. i a g h 5. b. 14. 25. b. monomers with two functional groups. a. 18. 16. c. Cl2 uuy 29. Essay 31. c d d c c a 23. and synthetic fibers. All rights reserved. When ethylene glycol is added to the water in a car radiator. H H H H @ @ @ @ H 2 C 2 C 3 C 2 C2 H @ @ H H HOH B. 12. unsaturated monomers. Because of their malleability. 14. iodobenzene D. 860 Core Teaching Resources . Essay 32. b. ethylene glycol protects against boiling in summer and freezing in winter. catalyst C. 18. c C. 22. 24. a. insulation. 26. Matching 1. c. d c c c c Chapter 23 Test B A. 28. 20. 3–chloro–2–methylpentane 2. Ethylene glycol is soluble in water. Ethylene glycol is an alcohol with both a high boiling point and a low freezing point due to intermolecular hydrogen bonding. 27. publishing as Pearson Prentice Hall. c. polymers have many commercial uses such as packaging. a. high strength-to-weight ratio. Multiple Choice 11. In condensation polymerization. d a c c c b H H H H @ @ @ @ y H 2 C 2 C 2 C 2 C 2 H` @ @ @ @ H H OH H b. 31.3–dimethyl–2–butanol butanal 2–hexanone propyl ethanoate aldehyde d. ether alcohol f. j 6. the resulting mixture boils at a temperature higher than water alone. 25. 2. a. 9. 10. prokaryotic/eukaryotic prokaryotic/eukaryotic bacteria green plants organelles Mitochondria lysosomes Part C Matching 16. 3. 2. NT Part C Matching OH OH OH RO B OR OR 4HOH 17. 10. O B O O O O O B O O Section Review 24. page 753 Analysis 1. ST Section Review 24. d Part D Questions 21. nucleus Sunlight Photosynthesis oxygen Part B True-False 12. The hydroxyl group.Chapter 23 Small-Scale Lab Section 23. b 18. Like animals. AT 14. 5. The chain is like a polymer because it contains many repeating units linked end to end. Through experimentation. 11. squirms. 8. The polymer is a gel-like liquid which is very viscous. 9. HO B HO OH OH 4R OH RO 8. It will not hold its shape like a solid and will flow slowly if left to stand.4 Polymers.2 Part A Completion You’re the Chemist © Pearson Education. Chloroplasts contain the biological molecules necessary for the conversion of solar energy into chemical energy. 1. AT 15. and oozes. 3. AT 16. Inc. AT 14. glucose and fructose 23. 2–3. 22. b 18. 6. Starches are a source of energy for plants. Answer Key 861 . 7. students are able to produce an amazing variety of polymers with different properties. 5.. c 19. OH 4. NT 12. polysaccharide starch glucose Glycogen liver Part B True-False 11. c 19. 1. Cellulose is used by plants to construct cell walls that are hard and rigid. 5. they meet their energy demands by breaking down these stored compounds. publishing as Pearson Prentice Hall. 7. 4. 2. OH. NT 13. 4. AT 15. Carbohydrates energy cellulose monosaccharides disaccharides 6.1 Part A Completion 1. a 20. 2. The rings that link two chains together are like the borate ion that cross links polymer chains. Plants store the excess chemical energy in carbon compounds. AT 13. It wiggles. e 17. a 20. All rights reserved. d Part D Question 21. These oxidation reactions take place in mitochondria. Part B True-False 10. 3. 3. Part B True-False 15. e 21. 8. 12. or cytosine adenine. lipid not soluble/insoluble triglycerides Triglycerides Saponification glycerol Phospholipids hydrophilic/polar hydrophobic/nonpolar lipid bilayer Cell membranes © Pearson Education. AT Part C Matching 19. amino acid side-chain group side-chain group peptide peptide 6. The molecules of both types of lipids have hydrophilic and hydrophobic ends. 8. 11. 4. ST 11. AT 13. AT Part C Matching 14. In animals. thymine. All rights reserved. d 18. guanine. 9. c 20. 5. 9. water protein catalysts enzymes Part D Questions 21. or cytosine adenine. nucleotide deoxyribonucleic acid ribonucleic acid proteins nitrogen base adenine. Wax coats on the surface of plant leaves protect against water loss and attack by microorganisms. 10. c 22. AT 16. b 19. 14. or cytosine uracil double helix hydrogen bonds thymine cytosine Part D Problem 17. When a mutation occurs within a gene. 2 COOH @ @ 5 CH2 CH2 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. the ability of the Part B True-False 12. b 15. 4. d Part D Question 24. or cytosine adenine. AT 13. NT 14. ST 18. and feathers. The cleansing action of soaps relies on this physical property. 4. hair. guanine. guanine. a 23. 6. AT 17. 10. thymine. it may stop production of the specified protein or cause production of a protein with an altered amiono acid sequence. both types of lipids can interact with polar and nonpolar phases simultaneously. thymine. or substitutions of one or more of the nucleotides. ST Part C Matching 17. AT 16. 22. 7. 2. NT 12. thymine. publishing as Pearson Prentice Hall.Section Review 24. AT 15. 2. which keep these structures pliable and waterproof. guanine. 8. 2. 7. 7.5 Part A Completion 1. deletions. a 16.. b 20. waxes coat the skin. 11.3 Part A Completion 1. Thus. Mutations are random changes in the sequence of nucleotides in a DNA molecule. Mutations may arise from additions. 6. c Section Review 24. 3. a 862 Core Teaching Resources . 9. 5.4 Part A Completion 1. Inc. 5. Sometimes the change is beneficial. more often. 13. structures that enable plants to produce carbohydrates through photosynthesis. Skin cells are less active. Figure 1a: 0. The free energy of ATP hydrolysis is used to drive many nonspontaneous biological reactions. People with this mutation have a molecular disease called sickle cell anemia. which provide structure and rigidity to plant cells. b 18. c 19. cell wall 4. 7. 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. is found primarily in the nucleus of eukaryotic cells and in the cytoplasm of prokaryotic cells. 4. RNA. Mitochondria produce energy needed for cellular activities. Figure 1b: 0. The cells of all other organisms are eukaryotic. Plant cells have cell walls. DNA. cell membrane 3. is found in the cytoplasm of all cells.0001 to 0. AT 12. Muscle cells are highly active cells. They contain significantly fewer mitochondria. Part B 1. 3. 1000 to 10. AT 13. For example. All of the organelle types labeled in Figure 1b are found in a typical plant cell. They comprise the cell walls. 7.protein to function is seriously impaired. 16.001 to 0. Proteins embedded in cell membranes help to transport molecules and ions across this barrier. 6. 2. 10.000 nm.0 m. Interpreting Graphics 24 Part A 1.0 to 10. 6. Plant cells are eukaryotic. 5.6 Part A Completion 1. Because proteins catalyze metabolic reactions.01 mm. 4.1 to 1. 9. Inc. cytoplasm 2. a molecule that participates in the transfer of information between DNA and protein. they are found throughout the cell.0 m. which are present in eukaryotic cells such as the one depicted in panel b. Prokaryotic cells are the cells of bacteria. 8. 2.001 mm. publishing as Pearson Prentice Hall. 5. 100 to 1000 nm. where it is used to fuel nonspontaneous processes. adenosine triphosphate adenosine diphosphate -ADP oxidation 30. a molecule that stores the information needed to make proteins. The cell in panel a represents a prokaryotic cell. which is named for the distorted shape of the defective red blood cells. Section Review 24.. All rights reserved. d 17. a Part D Question 21. cytoplasm 6. It lacks a nucleus and organelles. 0. ATP captures energy from catabolism reactions to drive anabolism reactions. NT 14. 1. Carbohydrates are found in the cytoplasm and are attached to the extracellular surfaces of membrane-bound proteins. NT Part C Matching © Pearson Education. nucleus 5. a mutation in the peptide chain of hemoglobin reduces its ability to transport oxygen. AT 15. 3. ATP is produced in the mitochondrion and transported out to the cytoplasm. Plant cells contain chloroplasts. ATP hydrolysis provides the extra energy needed to shift the equilibrium of a nonspontaneous reaction in favor of the products. cell membrane Answer Key 863 .5 kJ nonspontaneous catabolic or anabolic catabolic or anabolic metabolism catabolism anabolism Part B True-False 11. which require many mitochondria to meet their energy demands. 15. DNA governs the reproduction and growth of cells. All rights reserved. b 9. NT 4. 14. Multiple Choice 11. 28. d b c c d 864 Core Teaching Resources . 17. They are composed of nucleotides that contain a phosphate group. ST 31. d Chapter 24 Test A A. Matching 1. DNA stores the information needed to make proteins. each phospholipid molecule must be approximately 2. 13. Nitrogen is required for the synthesis of amino acids and nitrogen-containing bases of nucleic acids. 14. Because the lipid bilayer is composed of two sheets of phospholipid molecules arranged tail to tail. B. d a c c b 21. which together form an important buffer.5 to 5 nm long from head to tail. f 7. NT 34. AT Chapter 24 Test B A. ATP H2O → ADP Pi [ADP ][Pi] b. © Pearson Education. b. AT 3. 2. Multiple Choice 11. b 6. a Quiz for Chapter 24 1. AT 5. 2. 25. c d b a d 16. Phosphorus is found in the bloodstream as HPO42 and H2PO4 . 2. 6. spontaneous 38. 24. RNA has a key role in the transmission of the information stored in DNA. a fivecarbon sugar. Phosphorus is essential for the synthesis of phospholipids. a. 20. j D. and a nitrogen-containing base. 17. Inc. c 6. A membrane protein that acts as a channel must have contacts inside and outside the cell membrane. Questions and Problems 36. 29. C. True-False 26. 22. NT 33. 7. Prokaryotic cells do not contain a nucleus or organelles. i 10. AT 35. To span the entire bilayer. h 10. and energy-rich molecules such as ATP. e b g i 5. 40. Test the aqueous solubility of the substance. e 9. 23. h 10. d b b d b 16. Trp-Arg-Ala-Leu-Asn-end 37.Vocabulary Review 24 1. 18. c d g i 5. Keq 1 [AT P] c. 18. 8. a 8. Eukaryotic cells contain a nucleus and other membrane-enclosed structures called organelles. f 8. 19. 24. 3. Matching 1. Essay 42. j 39. NT 2. AT 32. 3. There are two types: DNA and RNA. 12. d 9. B. Eukaryotic cells are typically much larger than prokaryotic cells. membrane proteins must have dimensions similar to the observed thickness of the lipid bilayer. 25. d d c a c E. 23. d d c c b 21. k h e j 9. 15. 4. d 7. AT AT NT ST 30. Nucleic acids are polymers found primarily in cell nuclei. Many carbohydrates are soluble in water. a. nucleic acids. publishing as Pearson Prentice Hall. The length along the transmembrane axis of the protein must be approximately 5 to 10 nm. 3. 12. 41. 20. whereas lipids are not. 4. 4. 27. 13. d 11. 19. 22. ST 6.. c f b g 5. b 8. a 7. 28.2 g ATP 1 mol ATP 1 kg 103 g called substrates. when all other conditions are kept the same. The molecules on which an enzyme acts are 3. 38.0 cm3 0. NT D.5632lw2 68.5236 lw2 (0. Two dipeptides are possible. Thus. but do not change the normal position of the equilibrium.90 cm)(4. CH3 O @ # H3N 2 C 2 C 2 OH @ H H O @ # H2N 2 C 2 C 2 OH @ H width = 4. 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. nonpolar substances have the greatest chance of becoming incorporated into this protion of the cell membrane. 2000 kJ 1 mol ATP 30. 27.55 cm)2 64.1 5.4 g. 4. the products dissociate from the enzyme leaving the enzyme free to bind new substrate and begin a second reaction cycle. All rights reserved.55 cm)2]2/3 77. whereas lipids are not. The measured mass of 62.90 cm)(4. 29.5236)(5. the rate at which product is formed will increase with the number of enzyme molecules present in the reaction system. They reduce the time required for a chemical reaction to reach equilibrium. Essay 42. Inc. An enzyme catalyzes the conversion of a substrate to product.55 cm 100 77.3 The Egg: A Biochemical Storehouse. True-False 26.90 cm 0. Test the aqueous solubility of the substance. 4. NT AT AT ST 30. Shape Index 2. Finally. 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. page 774 Analysis Sample answers are given.3 kg ATP 507. like act as biological catalysts.42 g 0. 39.0 cm3 This is less than the density of a freshly laid egg. AT 32.42 g 1. One possibility: © Pearson Education.C. Most proteins are soluble in water. AT 34. publishing as Pearson Prentice Hall. AT 35.8 g by 6. Enzymes are not changed by the reactions they promote. Answer Key 865 .55 cm)2 3. ST 31. Enzymes are proteins that. bond-breaking and bondmaking occur at the active site to produce the products of the reaction. AT 33. The egg may have lost water over time.42 g is less than the calculated mass of 68. One possibility: ACAGTTGGTACT 37. d m 62. Chapter 24 Small-Scale Lab Section 24.8 g (0.5 kJ 33.90 cm 40. In an enzyme-catalyzed reaction the substrate binds to the active site on the enzyme form an enzyme-substrate complex. Questions and Problems 36. V V M M A A 41. Next. length = 5.5632)(5.138 [(5.975 g/cm3 v 64. More than one answer is possible due to the redundancy of the genetic code.3 cm2 E.. Beacuase the interior of the lipid bilayer is a hydrophobic environment.90 cm)(4.55 cm JJJH H O CH3 O @ @ # # H3N 2 C 2 C 2 N 2 C 2 C 2 OH @ @ @ H H H mass H2O 62. Inc. 2. 4. a positive test for protein. NT 13. medium eggs are less than 50 grams.1 Part A Completion 1. 6. d 866 Core Teaching Resources . band of stability beta positron rate half-life radioactive 7. 2. 5. 8. radioactive radioisotopes nuclei stable energy beta Alpha helium 9. Weigh it once a day for three more days and store it at room temperature. AT 19. The larger the egg. 15. Extra large eggs are usually more than 70 grams. e 19.20 g per day. NT 20. the smaller the shape index. 9. 14. 10. 9. 13. a 17. 12. b. d Part D Questions 22. a 23. a 21. NT 18. After 4 half-lives 1/16 of the original 18. Powdered milk NaOH CuSO4 produces a violet color. Measure the volume of water displaced by the egg. c 24.13 g © Pearson Education. c 18. electrons metal foil Gamma mass Lead concrete stop 0. NT 14. d Part C Matching 16. Assume Step 1 reveals the egg loses 0.0 g 1 1 2 2 42 days 3 1 2 Section Review 25. 2. e 19 b 20. 6.You’re the Chemist 1. which are rounder. 3. 5. b 20. 3. 6. fission neutrons fissionable atom energy moderation absorption 7. 4. 4. NT 15. 5. Extra large eggs tend to be more oblong than small eggs. 8. The warmer the temperature the greater the mass loss.3 Part A Completion 1. a positive test for protein.20 g Age of egg 32 days old 3. fusion mass energy hydrogen helium Part B True-False 16. 11.2 Part A Completion 1. AT Part B True-False 12. 1/2 1/2 1/2 1/2 mass will remain. 60 hr 15 hr 4 half-lives. ST Part C Matching 17. 2.5 grams per day depending on the temperature at which they are stored. ST 17. publishing as Pearson Prentice Hall. Egg shell NaOH CuSO4 produces a violet color. 218 4 84 Po → 2He 210 210 82 Pb → 83Bi 214 82Pb 0 1e Section Review 25. AT 15. Part D Problems 26. Weigh an egg once a day for three days and store in a refrigerator between weighings.2 to 0. 10. 10. All rights reserved. NT Part C Matching 21.0 g 1.42 g) 0. 8. a. ST 16. billions transmutation radioactive decay atomic numbers synthesized Part B True-False 12. e 25. HCl produces bubbles at the surface of the egg shell. c 18.25 g or 3 half-lives 14 days Section Review 25. 5. 11. 2HCl CaCO3 → CO2 H2O CaCl2 7. Weigh the egg once each day for two or three days. 1 day Age of egg (68. ST 13. NT 14. 1/16 23. 4. Typical eggs will loose 0. 8.. 6. 11.8 g 62. 3. b 22. 7. 2. 25 g 4. slow fast-moving neutrons so they can be absorbed by the fuel atoms b.. a.0 g→ 4. 79 protons and 116 neutrons 4.0 g→ 1.2 1. 4. Teletherapy is the use of gamma radiation to destroy cancerous tissue.50 g→ 0. 28 protons and 36 neutrons b. 2. the mass number remains the same. 2.1 min decrease by a factor of more than 1000. a. scintillation all iodine-131 phosphorus-32 neutron activation Part B True-False 11. 20 1 1 5. or 820 s. Tracing the pathways of radioactive isotopes allows scientists to study reaction mechanisms and reaction rates. e 17. 1 10 1 2 1024 6. 4Be 1e → 3Li 0 c. 1. All rights reserved. 3. 6. The half-life of polonium-214 is insignificant compared to the half-life of bismuth-214. Radioisotopes are used to study chemical reactions and molecular structures. 9F → 8O 1e 3 2. 3 b. 53 protons and 83 neutrons c.1 1. The mass would 5. or three half-lives. The atomic number increases by one. 4He 2 ([1 235] [90 144] 2) ([1 235] [87 146] 3) ([1 235] [72 160] 4) 107 kcal 8. a.0 g→ 0.0 g→ 2. 94Pu c. Radioisotopes replace non-radioactive isotopes in the structure of a compound without changing its chemical properties. 36Kr e. 3. d Section 25. 1H d. The half-life is 5. lead-210. 0e 7 1 b. Section 25. 8. 144Ce 58 239 92 b. 30P 15 3. a. b 21.0 1023 1 3. ionizing electrons senses Geiger gas 6. AT Part C Matching 15.5 106 g Part D Questions 20. Three half-lives 15 days. 16 g y 8 g y 4 g y 2.4 Part A Completion 1. 37Ar → 37K 1e 18 19 17 0 17 d. The atomic number decreases by two. Practice Problems 25 Section 25. AT 12. 2 b. the mass number decreases by four.0 g → 1. © Pearson Education.0 3.Part D Questions and Problems 21. a 19. 238U y 206Pb 8 4He 6 0e 92 82 2 1 8 days 2 half-lives. 8.0 g 2 2 20 minutes 1 half-life. c 18. 4 2. It takes five half-lives. 3 c.4 1.0 g. 2. Inc. 3. 9. ST 14. decrease the number of slow-moving neutrons and slow the chain reaction Section 25. b 16. Neutron activation analysis is used to detect trace amounts of elements in samples. 4 22.0 g Four half-lives 4 17 days 68 days 51 min 10 half-lives. 4. 5. 7. NT 13.0 kcal/g 2.0 mol 6.0 1023 atoms 2 5.49/3 183 s Section Review 25. 2. 8 6. 16 g → 8 g → 4 g → 2. a. a. 5. publishing as Pearson Prentice Hall. 10. a. The mass decreases by a factor of 1/8. Radioisotopes are used to diagnose and treat diseases such as cancer. Answer Key 867 .3 1. 208Fr → 4He 204At 87 2 85 7 7 0 b. 14N c. 2. 237Np 93 Interpreting Graphics 25 1. a 6. Problems 26.6 10 4 s). 20. 18. e 6. 12. publishing as Pearson Prentice Hall. 25. a. 13. D. 7. 22. 1 of the original sample remains 32 0.7. b l g a 5. B. 4. d b c b 5.0 g 1 1 1 1 0. a d c b a 16. Chemical reactions occur in an effort to attain stable electron configurations. b C. 8. 20. Matching 1. Nuclear reactions release far more energy than typical exothermic chemical reactions. 2. or 5 half-lives. d 7. f 9. h 9. 90Th → 91Pa 1e 27. 3.750 g 2 2 2 2 28. or thermonuclear reaction. 15. 23. 4. All rights reserved. polonium-210 has the shortest half-life (1. 226Ra → 222Rn 4He 88 86 2 0 234 234 b. b d b c a 22. a b a d b C. a B.. a. 41H 2 0e → 4He energy 1 1 2 Vocabulary Review 25 1. For heavier isotopes. 29. 11. b 10. h 10. Problems 27. 3. the isotope decayed through 3 half-lives. 14. d 8. 13. Essay 30. 17.00 g © Pearson Education. 40 days 5 half-lives. Three half-lives is 252 days. 21.5 8. e 7.125 g 4. Fusion occurs when two light nuclei combine to produce a nucleus of heavier mass. 19. c 8. Essay 29. Uranium-238 has the longest half-life (4. 26. f Quiz for Chapter 25 1. 24. a.13 gram remaining. a i g c 5. 24. j 6. Multiple Choice 11. so one half-life period 84 days. 2. k f h j Chapter 25 Test B A.5 109 yr). 5 half-lives 0.5 years 14.3 years/half-life. d b a a c 21. i d e c 9. Inc. 4. such as lead-206. 2. 6. 10. b 9. 18. hdrogen nuclei (protons) are fused to make helium nuclei. 19. b. The energy released from the sun is the result of a nuclear fusion.00 half-lives Thus. Nuclear reactions occur in an effort to obtain stable nuclear configurations. 868 Core Teaching Resources . j i c d 5. 3.0 h/6.5 neutrons to 1 proton. Multiple Choice 11. 4. 12. the stability ratio is about 1. 2 32 71. 27. a 8. 2. 15. 12. 16. Matching 1. a 10. 25. The half-life of scandium-42 is 84 days. D. 3. d d d a d Chapter 25 Test A A. b c b b a c 17. 92U → 2He 231Th 90 28. 124 n 82 p 1. In solar fusion. 14. The reaction requires two beta particles. 12. 1 of the sample remains 32 1 5 1 Since represents .75 h 4. g 7. 42K → 0e 42Ca 1 19 20 235 4 b. 91Pa → 92U 1e 234 234 0 c. If one-eighth of the sample remains. 23. . For each flip the probability of a head is 0. After 3. Do trials until the number of events equals zero. 4. Each trial represents one half-life because the number of heads approximately halves for each trial.4 days. Answer Key 869 .50. Unlike chemical reactions. 3. pressure. one-eighth remains. 100 80 Flips 60 40 20 0 0 1 2 3 4 5 6 7 Trial 2. nuclear reactions are unaffected by changes in temperature. Trial # 1 2 3 4 5 Number of flips 100 42 20 9 5 Number of heads 42 20 9 5 3 © Pearson Education.8 days.460 years/5730 2) of carbon-14. 6 7 3 1 Figure A 1 0 1. You’re the Chemist 1. Chapter 25 Small-Scale Lab Section 25. publishing as Pearson Prentice Hall. after 11. 2. trial.6 days. one-fourth of the sample remains. After 7. page 809 Analysis Sample data are provided. 3.2 Radioactivity and Half Lives. Inc. and. The rate decreases over time. or the presence of a catalyst. half the sample remains. After two halflives. All rights reserved. Roll the die again a number of times equal to the number obtained in the first trial. The rate of disappearance of heads is nonlinear. This time period is two half-lives (11. Count the total number of even numbers that result in 100 rolls of the die. one-fourth remains. Plot number of evens vs.c.