Schaum's Outlines - 3,000 Solved Problems in Chemistry

■ Concise explanations of all course concepts ■ Information on chemical formulas, structure of the atom, bonding, molecu

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■ Concise explanations of all course concepts ■ Information on chemical formulas, structure of the atom, bonding, molecules, stoichiometry, solids, liquids, gases, oxidation and reduction, solutions, and thermodynamics I

SEWITHLXHES 5

C0Hege Chemistry • General Chemistry • Beginning Chemistry • Chemical Principles • First Year Chemistry • Introduction to Chemistry • advanced High School Chemistry • AP High School Chemistry

David £■ Goldberg, Ph.D.

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3,000 Solved Problems in Chemistry

David E. Goldberg, Ph.D. Brooklyn College

Schaum’s Outline Series

New York

Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto

The McGraw-Hill Companies David E. Goldberg, Ph.D., who is the former department chairman at Brooklyn College, has taught chemistry for many years and has more recently entered the field of computer languages. He is the author of 8 textbooks, as well as Schaum's Outline of Beginning Chemistry. l

Copyright © 2011 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 the prior written permission of the publisher.

10 11 12 13 14 15 QVS/QVS 23 22 21 20 19 ISBN MHID

978-0-07-175500-9 0-07-175500-4

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Trademarks: McGraw-Hill, the McGraw-Hill Publishing logo, Schaum’s and related trade dress are trademarks or registered trademarks of The McGraw-Hill Companies and/or its affiliates in the United States and other countries and may not be used without written permission. All other trademarks are the property of their respective owners. The McGraw-Hill Companies is not associated with any product or vendor mentioned in this book.

McGraw-Hill books are available at special quantity discounts to use as premiums and sales promotions or for use in corporate training programs. To contact a representative, please e-mail us at [email protected].

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CONTENTS 1

Chapter 1

MEASUREMENT 1.1 Exponential Numbers / 1.2 Metric System / 1.3 Significant Figures / 1.4 Calculations with Metric Quantities / 1.5 English-Metric Conversions / 1.6 Temperature Scales /

Chapter 2

STRUCTURE OF MATTER 2.1 Elements, Compounds, Mixtures / 2.2 Elementary Atomic Structure / 2.3 Ionic and Covalent Bonding / 2.4 Electron Dot Structures and the Octet Rule /

20

Chapter 3

PERIODIC TABLE 3.1 Periodic Trends / 3.2 Inorganic Nomenclature /

32

Chapter

4

CHEMICAL FORMULAS 4.1 Percent Composition / 4.2 The Mole; Formula Calculations / 4.3 Empirical Formulas / 4.4 Molecular Formulas /

38

Chapter

5

MODERN STRUCTURE OF THE ATOM 5.1 Physical Background / 5.2 Light / 5.3 Photoelectric Effect / 5.4 Bohr Theory / 5.5 Electron Diffraction /

67

ELECTRONIC STRUCTURE OF THE ATOM

83

Chapter 6

6.1 Shells, Subshells, Orbitals / 6.2 Electronic Structures of Atoms and Ions / 6.3 Consequences of Electronic Structure / 95

Chapter 7

BONDING 7.1 Bond Lengths and Bond Energies / 7.2 Dipole Moment / 7.3 Other Intermolecular Forces / 7.4 Resonance / 7.5 Geometry of Molecules /

Chapter

BONDING THEORY 8.1 Valence Bond Theory / 8.2 Molecular Orbital Theory /

119

Chapter 9

ORGANIC MOLECULES 9.1 Organic Nomenclature and Classification / 9.2 Structural Isomerism / 9.3 Geometric and Optical Isomerism / 9.4 More Advanced Topics /

131

Chapter 10

CHEMICAL EQUATIONS 10.1 Balancing Chemical Equations / 10.2 Prediction of Products / 10.3 Net Ionic Equations /

151

Chapter 11

STOICHIOMETRY 11.1 Quantities in Chemical Reactions / 11.2 Limiting Quantities / 11.3 Solute Concentrations, Physical Units / 11.4 Molarity

165

Chapter 12

GASES 12.1 Units of Pressure and Temperature / 12.2 Boyle’s Law, Charles’Law, Combined Gas Law / 12.3 Moles of Gas and Ideal Gas Law / 12.4 Dalton’s Law / 12.5 Molecular Weights of Gases / 12.6 Reactions Involving Gases /

199

Chapter 13

ADVANCED GAS CONCEPTS 13.1 Van der Waals Equation / 13.2 Basics of Kinetic Molecular Theory / 13.3 Kinetic Energies of Gas Molecules / 13.4 Graham’s Law /

231

Chapter 14

SOLIDS AND LIQUIDS 14.1 Crystal Structure / 14.2 Crystal Energies / 14.3 Liquids /

244

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til

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U CONTENTS

Chapter 15

OXIDATION AND REDUCTION 15.1 Oxidation Number; Oxidizing and Reducing Agents / 15.2 Balancing Redox Equations / 15.3 Calculations Involving Redox Equations /

Chapter 16

fc-

268

OTHER CONCENTRATION UNITS 16.1 Normality in Acid-Base Reactions / 16.2 Normality in Redox Equations / 16.3 Mole Fraction and Molality /

295

PROPERTIES OF SOLUTIONS 17.1 Raoult’s Law and Vapor Pressure Lowering / 17.2 Freezing Point Depression and Boiling point Elevation / 17.3 Osmotic Pressure / 17.4 Other Properties ot' Solutions / 17.5 Solutions of Strong Electrolytes

312

t

Chapter 17

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Chapter 18

THERMODYNAMICS 18.1 Heat, Internal Energy, Enthalpy / 18.2 Heat Capacity and Calorimetry / 18.3 Law of Dulong and Petit / 18.4 Enthalpy Change / 18.5 Enthalpies of Ions in Solution / 18.6 Free Energy Change and Entropy /

335

I s 5

I Chapter 19 Chapter 20

CHEMICAL KINETICS 19.1 Rate Laws / 19.2 Half-Life / 19.3 Collision Theory / 19.4 Reaction Mechanisms

373

EQUILIBRIUM 20.1 Le Chatelier’s Principle / 20.2 Equilibrium Constants / 20.3 Kp / 20.4 Thermodynamics of Equilibrium /

399

ACIDS AND BASES 21.1 Acid-Base Theory / 21.2 Ionization Constants / 21.3 Ionization of Water / 21.4 Buffer Solutions / 21.5 Hydrolysis Equilibrium / 21.6 Polyprotic Acids and Bases / 21.7 Indicators and Titration /

427

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Chapter 21

Chapter 22 HETEROGENEOUS AND OTHER EQUILIBRIA 22.1 Solubility Product Equilibria / 22.2 Competitive Reactions / 22.3 Coordination Equilibria / 22.4 Miscellaneous Applications of Equilibria /

480

ELECTROCHEMISTRY 23.1 Electrical Units / 23.2 Electrolysis / 23.3 Galvanic Cells / 23.4 Nernst Equation / 23.5 Practical Applications / 23.6 Electrochemical Equilibrium and Thermodynamics /

510

Chapter 24 NUCLEAR AND RADIOCHEMISTRY 24.1 Nuclear Particles and Nuclear Reactions / 24.2 Half-Life / 24.3 Binding Energy / 24.4 Nuclear Cross Section / 24.5 Radiochemistry /

542

Chapter 25 NONMETALS 25.1 General / 25.2 Halogens / 25.3 Group VI Elements / 25.4 Group V Elements / 25.5 Groups IV and III / 25.6 Noble Gases /

563

Chapter 26

573

Chapter 23

METALS AND METALLURGY 26.1 Metallic Bonding / 26.2 Alloys / 26.3 Main Group Metals / 26.4 Transition and Inner Transition Metals / 26.5 Metallurgy /

Chapter 27 COORDINATION COMPOUNDS 27.1 Properties of the Coordinauon Sphere / 27.2 Nomenclature of Coordination Compounds / 27.3 Isomerism of Coordination Compounds / 27.4 Valence Bond Theory of Coordination Compounds / 27.5 Cry stal Field Theory / 27.6 Other Concepts /

584

Index

609

To the Student The best way to ensure that you understand the concepts of General Chemistry is to solve many problems on each topic. You should attempt a large number of different problems, rather than merely reworking the same problems again and again, since you might have a tendency to memorize the solution in the latter case. Be sure to read each problem carefully, since a small difference in the wording of a problem can make a large difference in its solution. Since there is no set order of topics in general chemistry texts, you will have to consult the Table of Contents in this book to find the problems you wish to do. The problems in each section start with the more basic ones and progress to those that are more difficult. In some chapters, you may find problems based on material which you have not yet covered in your course. Do not attempt to do these before you cover the material on which they are based. For example, some texts cover equilibrium before thermodynamics, while others cover it afterward. Do not attempt the section on the thermodynamics of equilibrium until you have been introduced to both topics in your course. There are numerous methods to solve most problems. The solution methods are usually related to each other but may seem very different. In this book, some related problems are solved using one method, and some with another. Many of the problems, especially in the early chapters, use several solution methods. You should attempt to do the problems yourself before looking at the solutions given. If you get the correct answer using a reasonable method, you need not worry that you did not use the method selected here. If the method presented is clearer that the one you used, however, you might consider adopting it for future similar problems.

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CHAPTER 1

Measurement

2 3 «4

i

1.1

-

1.1

I -

EXPONENTIAL NUMBERS List the powers of ten, from 10~4 to 106, together with their explicit meanings. #

1 10"1 = — = 0.1 10

10° = 1 10' = 10

9

2

102 = 10 x 10 = 100

= _

10"2 =

1 102

1 — = 0.01 100

10"3 =

1 103

1 = 0.001 1000

10"4 =

1 104

1 = 0.0001 10000

103 = 10 x 10 x 10 = 1000

B

104 = 10 x 10 x 10 X 10 = 10 000 10s = 10 x 10 x 10 X 10 X 10 = 100000

106 = 10 X 10 X 10 X 10 X 10 X 10 = 1 000 000 In the expression 10s, the base is 10 and the exponent is 5.

1.2

Express the following numbers in standard exponential form: (a) (e)

22 400 0.0454

(b) 7 200 000 (/) 0.00006

(c) 454 (g) 0.003 06

(d) 0.454 (h) 0.000 000 5

# Any number may be expressed as an integral power of 10, or as the product of two numbers one of which is an integral power of 10 (e.g., 300 = 3 x 102). (a) 22 400 = 2.24 x 104 (d) 0.454 = 4.54 x 10 - i ($) 0.003 06 = 3.06 x 10"3

(6) 7 200 000 = 7.2 x 106 (e) 0.0454 = 4.54 x 10"2 (h) 0.000 0005 = 5 x 10"7

(c) 454 = 4.54 x 102 (/) 0.000 06 = 6 x 10"5

Moving the decimal point one place to the right is equivalent to multiplying a number by 10; moving the decimal point two places to the right is equivalent to multiplying by 100, and so on. Whenever the decimal point is moved to the right by n places, compensation can be achieved by dividing at the same time by 10"; the value of the number remains unchanged. Thus 3.25 0.0325 = -mr = 3.25 x 10"2 102 Moving the decimal point one place to the left is equivalent to dividing by 10. Whenever the decimal point is moved to the left n places, compensation can be achieved by multiplying at the same time by 10"; the value of the number remains unchanged. For example, 7296 = 72.96 x 102 = 7.296 x 103 1.3

Evaluate: (a) (d)

a3 x a5 107 x 10"3

(b) (e)

102 x 103 (4 x 104)(2 x 10"6)

(c) 10 x 10 (/) (2 x 105)(3 x 10"2)

# In multiplication, exponents of like bases are added. (a) a3 x a5 = a3 + s = a8 (c) 10 x 10= 101 + 1 = 102 («) (4 x 104)(2 x 10"6) = 8 x 104"6 = 8 x 10“2

(b) 102 x 103 = 102 + 3 = 103 (d) 107 x 10"3 = 107"3 = 104 (/) (2 x 105)(3 x 10"2) = 6 x 105"2 = 6 x 103 1

2

0

1.4

CHAPTER 1

Evaluate:

(,,) ?

w

w

8 x 102

m 56 x l0~2 ( } 1.6 x 10*

2 x 10“6

I

In division, exponents of like bases are subtracted. a5 102 (a) ~ = a 5-3 = a2 (b) ToJ=102-j = 10-3 a3 . , (c)

1.5

8 x 102 8 -z--- ——^ = - x 102 + 6 = 4 x 108 2 x 10 2

a0 = 1

(A)

Express as radicals: i (a)

1.7

10° =1

10-2-4 = 3 5 x 10-e

(c) (3 x 10)° = 1

(d) 7x10° = 7

(e) 8.2 x 10° = 8.2

(a) 102/3 (b) 103/2 (c) 10,/2 (d) 43/2

102/3 = VlO2

(b)

103/2 = v/lO3

(c)

101/2 = ^10

(d) 43/2 = yfi3 = >/64 = 8

Simplify: (a) (103)2 (A) (10-2)3 (c) (a3)-2 f (a) (103>2 = 103x2 = 106

1.8

5.6 x 10 2 _ 5^ 4 1.6 x 10 l.o

Evaluate the following expressions: (a) a0 (A) 10° (c) (3 x 10)° (d) 7 x 10° (e) 8.2 x 10° f (a)

1.6

(d)

(A)

(10"2)3 = 10"2x3 = 10"6

(c) (a3)-2 = a 6

Take the square root ofeach of the following numbers, using exponential notation as an aid: (a) 90 000 (A) 3.6 x 103 (c) 4.9 x 10-3 Take the cube roots of the following numbers: (d) 8 x 109 (e) 1.25 x 105. # To extract the square root of a power of 10, divide the exponent by 2. If the exponent is an odd number it should be increased or decreased by 1, and the coefficient adjusted accordingly. To extract the cube root of a power of 10, adjust so that the exponent is divisible by 3; then divide the exponent by 3. The coefficients are treated independently. (a) v/90 000 = V9 x 10* = 79 x JlO* = 3 x 102 or 300 (A) v/3^1F = v/36 x 102 = v/36 x VlO2 = 6x10* or 60 (c) V4.9 x 10”5 = n/49 x 10"6 = V49 x VUT® = 7 x 10-3 or 0.007 (d)