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Chapter 3 Stoichiometry of Formulas and Equations This book was printed on recycled paper. Chemistry http://www.primisonline.com Copyright ©2008 by The McGraw−Hill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without prior written permission of the publisher. This McGraw−Hill Primis text may include materials submitted to McGraw−Hill for publication by the instructor of this course. The instructor is solely responsible for the editorial content of such materials. 111 CHEMGEN ISBN: 0−390−82332−5 Chemistry Contents Silberberg • Student Solutions Manual for use with Chemistry, Fourth Edition 1. Keys to the Study of Chemistry 1 Text 1 2. The Components of Matter 11 Text 11 3. Stoichiometry: Mole−Mass−Number Relationships In Chemical Systems 27 Text 27 4. The Major Classes of Chemical Reactions 56 Text 56 5. Gases and the Kinetic−Molecular Theory 73 Text 73 6. Thermochemistry: Energy Flow and Chemical Change 91 Text 91 7. Quantum Theory and Atomic Structure 103 Text 103 8. Electron Configuration and Chemical Periodicity 116 Text 116 9. Models of Chemical Bonding 127 Text 127 10. The Shapes of Molecules 137 Text 137 11. Theories of Covalent Bonding 162 Text 162 12. Intermolecular Forces: Liquids, Solids, and Phase Changes 171 Text 171 iii 13. The Properties of Mixtures: Solutions and Colloids 185 Text 185 14. Periodic Patterns in the Main−Group Elements: Bonding, Structure, and Reactivity 200 Text 200 15. Organic Compounds and the Atomic Properties of Carbon 214 Text 214 16. Kinetics: Rates and Mechanisms of Chemical Reactions 242 Text 242 17. Equilibrium: The Extent of Chemical Reactions 259 Text 259 18. Acid−Base Equilibria 279 Text 279 19. Ionic Equilibria in Aqueous Systems 309 Text 309 20. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 342 Text 342 21. Electrochemistry: Chemical Change and Electrical Work 357 Text 357 22. The Elements in Nature and Industry 385 Text 385 23. The Transition Elements and Their Coordination Compounds 396 Text 396 24. Nuclear Reactions and Their Applications 412 Text 412 iv Silberberg: Student 1. Keys to the Study of Text © The McGraw−Hill 1 Solutions Manual for use Chemistry Companies, 2005 with Chemistry, Fourth Edition Chapter 1 Keys to the Study of Chemistry Most numerical problems will have two answers, a “calculator” answer and a “true” answer. The calculator answer, as seen on a calculator, will have one or more additional significant figures. These extra digits are retained in all subsequent calculations to avoid intermediate rounding error. Rounding of a calculator answer to the correct number of significant figures gives the true (final) answer. FOLLOW-UP PROBLEMS 1.1 Plan: The real question is “Does the substance change composition or just change form?” Solution: a) Both the solid and the vapor are iodine, so this must be a physical change. b) The burning of the gasoline fumes produces energy and products that are different gases. This is a chemical change. c) The scab forms due to a chemical change. 1.2 Plan: We need to know the total area supplied the bolts of fabric, and the area required for each chair. These two areas need to be in the same units. The area units can be either ft2 or m2. The conversion of one to the other will use the conversion given in the problem: 1 m = 3.281 ft. Solution: 2 §·200.m2 §· 3.281 ft §· 1 chair 3bolts = 205.047 = 205 chairs ¨¸¨¸ ¨¸2 ©¹bolt©¹ 1 m ©¹31.5 ft Check: 200 chairs require about 200 x 32 / 10 = 640 m2 of fabric. (Round 31.5 to 32, and 1 / (3.281)2 to 1 / 10) Three bolts contain 3 x 200. = 600. m2. 1.3 Plan: The volume of in the ribosome must be determined using the equation given in the problem. This volume may then be converted to the other units requested in the problem. Solution: 3 43 4§· 21.4 nm 3 V S r 3.14159 ¨¸ = 5131.44 nm 33©¹ 2 3 3 §·109 m§· 1 dm 5131.44 nm3 = 5.13144 x 10-21 = 5.13 x 10-21 dm3 ¨¸¨¸ ©¹1nm©¹ 0.1m §·§·1L 1P L 5.13144 x 1021 dm 3 = 5.13144 x 10-15 = 5.13 x 10-15 PL ¨¸¨¸36 ©¹©¹1dm 10 L Check: The magnitudes of the answers are reasonable. The units also agree. 1.4 Plan: The time is given in hours and the rate of delivery is in drops per second. A conversion(s) relating seconds to hours is needed. This will give the total number of drops, which may be combined with their mass to get the total mass. The mg of drops will then be changed to kilograms. The other conversions are given in the inside back cover of the book. Solution: §·§·§·§·60 min 60s 1.5 drops 65 mg§· 103 g§· 1 kg 8.0 h = 2.808 = 2.8 kg ¨¸¨¸¨¸¨¸¨¸¨¸3 ©¹©¹©¹©¹1 h 1 min 1 s 1 drop©¹ 1 mg ©¹10 g Check: Estimating the answer — (8 h) (3600 s / h) (2 drops / s) (0.070 kg/103 drop) = 4 kg. The final units in the calculation are the desired units. 1.5 Plan: The volume unit may be factored away by multiplying by the density. Then it is simply a matter of changing grams to kilograms. 1 2 Silberberg: Student 1. Keys to the Study of Text © The McGraw−Hill Solutions Manual for use Chemistry Companies, 2005 with Chemistry, Fourth Edition Solution: §·7.5 g§· 1 kg 4.6 cm3 = 0.0345 = 0.034 kg ¨¸3 ¨¸ ©¹cm ©¹1000 g Check: 5 x 7 / 1000 = 0.035, and the calculated units are correct. 1.6 Plan: Using the relationship between the Kelvin and Celsius scales, change the Kelvin temperature to the Celsius temperature. Then convert the Celsius temperature to the Fahrenheit value using the relationship between these two scales. T (in°C) = -273.15 + T (in K) T (in°F) = 9/5 T (in°C) +32 Solution: T (in°C = -273.15 + 234) K = -39.15 = -39°C T (in°F) = 9/5(-39.15) + 32 = -38.47 = -38°F Check: Since the Kelvin temperature is below 273, the Celsius temperature must be negative. The low Celsius value gives a negative Fahrenheit value. 1.7 Plan: Determine the significant figures by determining the digits present, and accounting for the zeros. Only zeros between non-zero digits and zeros on the right, if there is a decimal point, are significant. The units make no difference. a) 31.070 mg; five significant figures b) 0.06060 g; four significant figures c) 850.°C; three significant figures — note the decimal point that makes the zero significant. d) 2.000 x 102 mL; four significant figures e) 3.9 x 10-6 m; two significant figures — note that none of the zeros are significant. f) 4.01 x 10-4 L; three significant figures Check: All significant zeros must come after a significant digit. 1.8 Plan: Use the rules presented in the text. Add the two values in the numerator before dividing. The time conversion is an exact conversion, and, therefore, does not affect the significant figures in the answer. The addition of 25.65 and 37.4 gives an answer where the last significant figure is the one after the decimal point (giving three significant figures total). When a four significant figure number divides a three significant figure number, the answer must round to three significant figures. An exact number (1min / 60 s) will have no bearing on the number of significant figures. 25.65 mL 37.4 mL = 51.434 = 51.4 mL/min §·1min 73.55 s¨¸ ©¹60 s Check: (25 + 35) / (5 / 4) = (60)4/5 = 48. The approximated value checks well. END-OF-CHAPTER PROBLEMS 1.2 Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples. Solution: Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. a) The helium fills the volume of the entire balloon. The addition or removal of helium will change the volume of a balloon. Helium is a gas. b) At room temperature, the mercury does no completely fill the thermometer. The surface of the liquid mercury indicates the temperature. c) The soup completely fills the bottom of the bowl, and it has a definite surface. The soup is a liquid, though it is possible that solid particles of food will be present. 2 Silberberg: Student 1. Keys to the Study of Text © The McGraw−Hill 3 Solutions Manual for use Chemistry Companies, 2005 with Chemistry, Fourth Edition 1.4 Plan: Define the terms and apply these definitions to the examples. Solution: Physical property – A characteristic shown by a substance itself, without interacting with or changing into other substances. Chemical property – A characteristic of a substance that appears as it interacts with, or transforms into, other substances. a) The change in color (yellow-green and silvery to white), and the change in physical state (gas and metal to crystals) are examples of physical properties. The change in the physical properties indicates that a chemical change occurred. Thus, the interaction between chlorine gas and sodium metal producing sodium chloride is an example of a chemical property. b) The sand and the iron are still present.
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