Chemistry student solutions manual ( 10e 2013 ISBN 9781133933526 ) kenneth w whitten, raymond e davis, larry peck, george g stanley

Chemistry student solutions manual ( 10e 2013 ISBN 9781133933526 ) kenneth w whitten, raymond e davis, larry peck, george g stanley Chemistry student solutions manual ( 10e 2013 ISBN 9781133933526 ) kenneth w whitten, raymond e davis, larry peck, george g stanley Chemistry student solutions manual ( 10e 2013 ISBN 9781133933526 ) kenneth w whitten, raymond e davis, larry peck, george g stanley Chemistry student solutions manual ( 10e 2013 ISBN 9781133933526 ) kenneth w whitten, raymond e davis, larry peck, george g stanley

Student Solutions Manual Chemistry TENTH EDITION Kenneth W Whitten University of Georgia, Athens Raymond E Davis University of Texas at Austin M Larry Peck Texas A&M University George G Stanley Louisiana State University Prepared by Wendy Keeney-Kennicutt Texas A&M University Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it © 2014 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to permissionrequest@cengage.com ISBN-13: 978-1-133-93352-6 ISBN-10: 1-133-93352-1 Brooks/Cole 20 Davis Drive Belmont, CA 94002-3098 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at: www.cengage.com/global Cengage Learning products are represented in Canada by Nelson Education, Ltd To learn more about Brooks/Cole, visit www.cengage.com/brookscole Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com Printed in the United States of America 17 16 15 14 13 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it This is an electronic version of the print textbook Due to electronic rights restrictions, some third party content may be suppressed Editorial review has deemed that any suppressed content does not materially affect the overall learning experience The publisher reserves the right to remove content from this title at any time if subsequent rights restrictions require it For valuable information on pricing, previous editions, changes to current editions, and alternate formats, please visit www.cengage.com/highered to search by ISBN#, author, title, or keyword for materials in your areas of interest Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it Foreword to the Students This Solutions Manual supplements the textbook, General Chemistry, tenth edition, by Kenneth W Whitten, Raymond E Davis, M Larry Peck and George Stanley The solutions of the 1441 even-numbered problems at the end of the chapters have been worked out in a detailed, step-by-step fashion Your learning of chemistry serves two purposes: (1) to accumulate fundamental knowledge in chemistry which you will use to understand the world around you, and (2) to enhance your ability to make logical deductions in science This ability comes when you know how to reason in a scientific way and how to perform the mathematical manipulations necessary for solving certain problems The excellent textbook by Whitten, Davis, Peck and Stanley provides you with a wealth of chemical knowledge, accompanied by good solid examples of logical scientific deductive reasoning The problems at the end of the chapters are a review, a practice and, in some cases, a challenge to your scientific problem-solving abilities It is the fundamental spirit of this Solutions Manual to help you to understand the scientific deductive process involved in each problem In this manual, I provide you with a solution and an answer to the numerical problems, but the emphasis lies on providing the step-by-step reasoning behind the mathematical manipulations In some cases, I present as many as three different approaches to solve the same problem, since we understand that each of you has your own unique learning style In stoichiometry as well as in many other types of calculations, the "unit factor" method is universally emphasized in general chemistry textbooks I think that the over-emphasis of this method may train you to regard chemistry problems as being simply mathematical manipulations in which the only objective is to cancel units and get the answer My goal is for you to understand the principles behind the calculations and hopefully to visualize with your mind's eye the chemical processes and the experimental techniques occurring as the problem is being worked out on paper And so I have dissected the "unit factor" method for you and introduced chemical meaning into each of the steps I gratefully acknowledge the tremendous help over the years provided by Frank Kolar in the preparation of this manuscript Wendy L Keeney-Kennicutt Department of Chemistry Texas A&M University iii Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it Table of Contents The Foundations of Chemistry Chemical Formulas and Composition Stoichiometry 13 Chemical Equations and Reaction Stoichiometry 29 The Structure of Atoms 49 Chemical Periodicity 69 Some Types of Chemical Reactions 81 Chemical Bonding 94 Molecular Structure and Covalent Bonding Theories 108 Molecular Orbitals in Chemical Bonding 126 10 Reactions in Aqueous Solutions I: Acids, Bases, and Salts 138 11 Reactions in Aqueous Solutions II: Calculations 150 12 Gases and the Kinetic-Molecular Theory 167 13 Liquids and Solids 188 14 Solutions 209 15 Chemical Thermodynamics 228 16 Chemical Kinetics 250 17 Chemical Equilibrium 270 18 Ionic Equilibria I: Acids and Bases 289 19 Ionic Equilibria II: Buffers and Titration Curves 306 20 Ionic Equilibria III: The Solubility Product Principle 328 21 Electrochemistry 343 22 Nuclear Chemistry 366 23 Organic Chemistry I: Formulas, Names and Properties 378 24 Organic Chemistry II: Shapes, Selected Reactions and Biopolymers 394 25 Coordination Compounds 404 26 Metals I: Metallurgy 416 27 Metals II: Properties and Reactions 424 28 Some Nonmetals and Metalloids 432 v Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it The Foundations of Chemistry 1-2 Refer to the Introduction to Chapter and a dictionary (a) Organic chemistry is the study of the chemical compounds of carbon and hydrogen and a few other elements (b) Forensic chemistry deals with the chemistry involved in solving crimes, including chemical analyses of crime scene artifacts, such as paint chips, dirt, fluids, blood, and hair (c) Physical chemistry is the study of the part of chemistry that applies the mathematical theories and methods of physics to the properties of matter and to the study of chemical processes and the accompanying energy changes (d) Medicinal chemistry is the study of the chemistry and biochemistry dealing with all aspects of the medical field 1-4 Refer to the Sections 1-1, 1-4, 1-8, 1-13 and the Key Terms for Chapter (a) Weight is a measure of the gravitational attraction of the earth for a body Although the mass of an object remains constant, its weight will vary depending on its distance from the center of the earth One kilogram of mass at sea level weighs about 2.2 pounds (9.8 newtons), but that same one kilogram of mass weighs less at the top of Mt Everest In more general terms, it is a measure of the gravitational attraction of one body for another The weight of an object on the moon is about 1/7th that of the same object on the earth (b) Potential energy is the energy that matter possesses by virtue of its position, condition, or composition Your chemistry book lying on a table has potential energy due to its position Energy is released if it falls from the table (c) Temperature is a measurement of the intensity of heat, i.e the "hotness" or "coldness" of an object The temperature at which water freezes is 0qC or 32qF (d) An endothermic process is a process that absorbs heat energy The boiling of water is a physical process that requires heat and therefore is endothermic (e) An extensive property is a property that depends upon the amount of material in a sample Extensive properties include mass and volume 1-6 Refer to the Section 1-1 and the Key Terms for Chapter A reaction or process is exothermic, in general, if heat energy is released, but other energies may be released (a) The discharge of a flashlight battery in which chemical energy is converted to electrical energy is referred to as being exothermic the chemical reaction occurring in the battery releases heat (b) An activated light stick produces essentially no heat, but is considered to be exothermic because light is emitted 1-8 Refer to Sections 1-1 and 1-5, and the Key Terms for Chapter (a) Combustion is an exothermic process in which a chemical reaction releases heat (b) The freezing of water is an exothermic process Heat must be removed from the molecules in the liquid state to cause solidification Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it As the heats of formation of minerals become more exothermic, i.e., more negative, their thermodynamic stability increases And so the difficulty by which free metals can be extracted from the minerals also increases In other words, the more active is the metal, the easier it is to form compounds and the more difficult it is to retrieve the metal from its compounds This relationship is obvious in the methods by which the metals are removed from their mineral matrix as shown in the third column of the above table: heating is a less severe metallurgic process, whereas electrolysis is a more severe method 423 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it 27 Metals II: Properties and Reactions 27-2 Refer to the Introduction to Chapter 27 and Section 5-1 The representative elements have valence electrons in s or s and p orbitals in the outermost occupied energy level, whereas the d-transition metals must have a partially filled set of d orbitals 27-4 Refer to Chapter Metals are located at the left side of the periodic table and therefore, in comparison with nonmetals, have (a) fewer outer shell electrons, (b) lower electronegativities, (c) more negative standard reduction potentials and (d) less endothermic ionization energies 27-6 Refer to Section 4-10 and Table 4-6 Malleable refers to the ability of a substance to be shaped by being beaten with a hammer or by the pressure of rollers Ductile refers to the ability of a substance to be drawn out into wire or hammered into a thin sheet 27-8 Refer to Sections 27-1 and 27-4, and Tables 27-1 and 27-3 (a) Alkali metals are larger than alkaline earth metals in the same period due to the increased effective nuclear charge of the alkaline earth metals (b) Alkaline earth metals have higher densities since they are both heavier and smaller than alkali metals of the same period (c) Alkali metals have lower first ionization energies than alkaline earth metals of the same period due to both the increased effective nuclear charge and decreased size of the alkaline earth metals (d) Alkali metals have much higher second ionization energies than alkaline earth metals of the same period This is because removal of a second electron from an alkali metal ion involves destroying the very stable noble gas electronic configuration of the ion whereas removal of a second electron from an alkaline earth metal ion involves creating a stable noble gas configuration 27-10 Refer to Sections 27-1 and 27-4, and Tables 27-1 and 27-3 When an alkali metal (Group 1A) with atomic configuration, ns1, reacts with a nonmetal, its outermost s electron is transferred to the nonmetal The atom becomes an ion with a +1 charge that is isoelectronic with a noble gas The size of the ion as set by the electron cloud becomes smaller than its parent atom 27-12 Refer to Sections 27-1 and 27-4, and Tables 27-1 and 27-3 (a) physical properties: Both the alkaline earth metals and the alkali metals are silvery-white, malleable and ductile, but the alkaline earths are somewhat harder than alkali metals Both are excellent electrical and thermal conductors The melting and boiling points of the 2A metals are higher than those for the 1A metals, which are relatively low chemical properties: Alkaline earth metals and alkali metals are easily oxidized and thus are strong reducing agents The 2A metals are not as reactive as 1A metals, but both groups are too reactive to occur as free elements in nature Alkali metals are characterized by the loss of electron per metal atom and form basic metal oxides, which react with water to produce hydroxides Alkaline earths are characterized by the loss of electrons per metal atom and form basic metal oxides (except BeO) which also react with water to produce hydroxides 424 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it (b) Alkali metals are larger than alkaline earth metals in the same period due to the increased effective nuclear charge of the alkaline earth metals Both increase in size going down their family Alkaline earth metals have higher densities since they are both heavier and smaller than alkali metals of the same period Alkali metals have lower first ionization energies than alkaline earth metals of the same period due to both the increased effective nuclear charge and decreased size of the alkaline earth metals Alkali metals have much higher second ionization energies than alkaline earth metals of the same period This is because removal of a second electron from an alkali metal ion involves destroying the very stable noble gas electronic configuration of the ion whereas removal of a second electron from an alkaline earth metal ion involves creating a stable noble gas configuration 27-14 Refer to Sections 4-18, 5-5, 7-2, 27-9 and Appendix B [Ar] np (b) Ca2 [Ar] 4s (e) Sn [Kr] np np np np np np n n 5s 5p np np np np np 4d np 4d 5s (g) Sn4 [Kr] np np np np np 4d (a) Ca (c) Mg (f) Sn2 (d) Mg2 [Ne] np 3s [Ne] [Kr] 27-16 Refer to Sections 27-1 and 27-4 The alkali metals (Group 1A) and the alkaline earth metals (Group 2A) are not found free in nature because they are so easily oxidized Their primary sources are seawater, brines of their soluble salts and deposits of sea salt The metals are obtained from the electrolysis of their molten salts 27-18 Refer to Sections 27-1, 27-2, 27-4 and 27-5, and Tables 27-1 and 27-3 The metals in Group 1A and Group 2A have standard reduction potentials that are more negative than that for H2 Therefore, the alkali metals and alkaline earth metals are stronger reducing agents than H2 They should be above H2 in the activity series Consequently, they all react with acids by reducing acidic H ions to H2, while they are oxidized to +1 (Group 1A) or +2 (Group 2A) ions All but Be reduce H2O to H2 gas 27-20 Refer to Section 27-6 (a) Calcium metal is used (1) as a reducing agent in the metallurgy of U, Th and other metals, (2) as a scavenger to remove dissolved impurities in molten metals and residual gases in vacuum tubes, and (3) as a component in many alloys Slaked lime, Ca(OH)2, is a cheap base used in industry and is also a major component of mortar and lime plaster Careful heating of gypsum, CaSO4˜2H2O, produces plaster of Paris, 2CaSO4˜H2O (b) Magnesium metal is used (1) in photographic flash accessories, fireworks and incendiary bombs, (2) as a component in alloys for structural purposes, and (3) as a reagent in organic syntheses Magnesia, MgO, is an excellent heat insulator used in furnaces, ovens and crucibles Milk of magnesia, an aqueous suspension of Mg(OH)2, is a stomach antacid and laxative Anhydrous MgSO4 and Mg(ClO4)2 are used as drying agents 27-22 Refer to Section 27-2 and Table 27-2 Let M = alkali metal, X = halogen (a) 2M + 2H2O o 2MOH + H2 (b) 12M + P4 o 4M3P (c) 2M + X2 o 2MX 425 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it 27-24 Refer to Section 27-5 and Table 27-4 Let M = alkaline earth metal (a) M + 2H2O o M(OH)2 + H2 (b) 6M + P4 o 2M3P2 (c) M + Cl2 o MCl2 27-26 Refer to Section 27-2 and the Key Terms for Chapter 27 Diagonal similarities refer to chemical similarities of Period elements of a certain group to Period elements, one group to the right This effect is particularly evident toward the left side of the periodic table One example is the pair, B and Si, which are both metalloids with similar properties Another example is the pair, Li and Mg They have similar ionic charge densities and electronegativities; their compounds are similar in many ways: (1) Li is the only 1A metal that combines with N2 to form a nitride, Li3N Mg readily forms the nitride, Mg3N2 (2) Li and Mg both form carbides (3) The solubilities of Li compounds are similar to those of Mg compounds (4) Li and Mg form normal oxides, Li2O and MgO, when oxidized in air at atm pressure, while the other members of Group 1A form peroxides and superoxides 27-28 Refer to Sections 27-4 and 27-5, Table 27-3, and the Key Terms for Chapter 14 Hydration energy is the energy released when a mole of ions in the gaseous phase forms a mole of ions in the aqueous phase The higher the charge to size ratio of a cation, the stronger is its interaction with polar water molecules and the more exothermic is its hydration energy Therefore, hydration energies of the alkaline earth metals become less exothermic from top to bottom within a group because the size of the ions increases, whereas the charge of the ions remains 2+ 27-30 Refer to Section 27-4 and Table 27-3 Standard reduction potentials of the alkaline earth metals are, in general, very negative, indicating that alkaline earth metals are easily oxidized and hence are good reducing agents Progressing down Group 2A, the atoms are larger, the outer electrons are more easily lost, the metals become better reducing agents and standard reduction potentials become more negative 27-32 Refer to Section 27-2 and 27-5, and Appendix K Note: The metal hydroxide product is in the solid phase because only stoichiometric amounts of water are added (a) Balanced equation: Li(s) + H2O(A) o LiOH(s) + 1/2 H2(g) 'Hqr xn = ['Hqf LiOH(s) + 1/2 'Hqf H2(g)] - ['Hqf Li(s) + 'Hqf H2O(A)] = [(1 mol)(487.23 kJ/mol) + (1/2 mol)(0 kJ/mol)] - [(1 mol)(0 kJ/mol) + (1 mol)(285.8 kJ/mol)] = 201.4 kJ/mol rxn (b) Balanced equation: K(s) + H2O(A) o KOH(s) + 1/2 H2(g) 'Hqr xn = ['Hqf KOH(s) + 1/2 'Hqf H2(g)] - ['Hqf K(s) + 'Hqf H2O(A)] = [(1 mol)(424.7 kJ/mol) + (1/2 mol)(0 kJ/mol)] - [(1 mol)(0 kJ/mol) + (1 mol)(285.8 kJ/mol)] = 138.9 kJ/mol rxn 426 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it (c) Balanced equation: Ca(s) + 2H2O(A) o Ca(OH)2(s) + H2(g) 'Hqr xn = ['Hqf Ca(OH)2(s) + 'Hqf H2(g)] - ['Hqf Ca(s) + 2'Hqf H2O(A)] = [(1 mol)(986.6 kJ/mol) + (1 mol)(0 kJ/mol)] - [(1 mol)(0 kJ/mol) + (2 mol)(285.8 kJ/mol)] = 415.0 kJ/mol rxn In explanation, refer to the data table: Cation Li K Ca2 'Hqr xn from above (kJ) 201.4 138.9 415.0 Eq (V) 3.045 2.925 2.87 From the Eq values, we see that the relative strengths of reducing agents are : Li > K > Ca It is expected that for the reactions of these metals with water, 'Hqr xn for Li would be more negative than 'Hqr xn for K, which is in turn more negative than 'Hqr xn for Ca This trend is true only for Li and K 'Hqr xn for Ca is much more negative than predicted since it is a 2A metal (Li and K are 1A metals) and reacts with twice as much water to produce twice as much H2 gas 27-34 Refer to Section 27-8 The following are properties of most d-transition elements: (1) All are metals (2) Most are harder, more brittle and have higher melting points and boiling points and higher heats of vaporization than nontransition metals (3) Their ions and compounds are usually colored (4) They form many complex ions (5) With few exceptions, they exhibit multiple oxidation states (6) Many of the metals and their compounds are paramagnetic (7) Many of the metals and their compounds are effective catalysts 27-36 Refer to Section 27-9 and Appendix B (a) Sc [Ar] 3d1 4s2 (e) Cr3 [Ar] 3d3 (b) Fe [Ar] 3d6 4s2 (f) Ni2 [Ar] 3d8 (c) Cu [Ar] 3d10 4s1 (g) Ag [Kr] 4d10 5s1 (h) Ag [Kr] 4d10 (d) Zn2 [Ar] 3d10 27-38 Refer to Chapters 27 and 24, and Tables 27-1, 27-3, 27-5, 27-6, 27-7, 28-4, 28-6, 28-7, and others The Group A or representative elements have their last electrons filling the outer ns and np orbitals, whereas most Group B elements have their last electrons filling the (n-1)d orbitals Look to the tables to compare their properties For example, Group 3A elements are representative elements and have the ns2np1 outer electron configuration, with valance electrons Boron is a metalloid and crystallizes as a covalent solid (oxidation states range from +3 to –3), while the rest are metals forming ions with +1 or +3 oxidation states Aluminum ions only have +3 oxidation state Group 3B elements are transition metals, with the ns2(n-1)d1 outer electron configuration and tend to have a +3 oxidation state 427 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it 27-40 Refer to Sections 27-2and Table 27-1 The two properties listed in Table 27-1 that suggest that Group 1A metals are unlikely to exist as free metals are (1) the low ionization energies, which show how easily the outermost electron can be removed and (2) very negative standard reduction potentials, which indicate that the aqueous ions are not easily reduced to metals and that the free metals are easily oxidized to 1+ cations 27-42 Refer to Sections 27-7, 27-9 and 27-10, and Table 27-8 The acidity and the covalent nature of transition metal oxides generally increases with increasing oxidation state of the metals This is shown by the oxides of chromium Cr Oxide CrO Cr2O3 CrO3 Ox No of Cr +2 +3 +6 Character basic amphoteric weakly acidic/acidic 27-44 Refer to Section 27-10 and Table 27-8 Chromium(VI) oxide, CrO3, is the acid anhydride of chromic acid, H2CrO4, and dichromic acid, H2Cr2O7 Recall that there is no change in oxidation state when an acid anhydride is converted to the corresponding acid and so the oxidation state of Cr is +6 in both acids CrO3 + H2O o H2CrO4 2CrO3 + H2O o H2Cr2O7 27-46 Refer to Chapter 20, Section 6-1, Table 6-3 and Appendix H (1) Plan: Calculate the molar solubility of the metal hydroxides, then determine [OH] and pOH (i) Balanced equation: Be(OH)2(s) o m Be2(aq) + 2OH(aq) Let x Ksp = x 1022 = molar solubility of Be(OH)2 Then x = moles/L of Be2 and 2x = moles/L of OH Ksp = [Be2][OH]2 = (x)(2x)2 = 4x3 = x 1022 Solving, x = x 108 However, because x < x 107 (the concentration of [OH] in pure water), we must include ionization of water when determining the [OH] Ksp = [Be2][OH]2 = (x)(2x + x 107)2 = x 1022 Rather than solve this mathematically, one can solve this by successive approximation Simply pick a value for x, plug into the above expression and see if you get the value of the Ksp If your value is different (which it probably is), pick a higher or lower number Keep going until you find x, in this case, to one significant figure Start with x = x 108 Value of x x 108 x 108 x 108 x 108 Calculated Ksp 2.9 x 1021 1.4 x 1022 7.7 x 1022 x 1022 x is too large x is too small x is slightly too large x is too small Therefore, the best answer to significant figure is x = x 108 Therefore, [OH] = (2x + 1.0 x 107) = 1.6 x 107 M; pOH = 6.80 (ii) Balanced equation: Mg(OH)2(s) o m Mg2(aq) + 2OH(aq) Ksp = 1.5 x 1011 Let x = molar solubility of Mg(OH)2 Then x = moles/L of Mg2 and 2x = moles/L of OH Solving, x = 1.6 x 104 Ksp = [Mg2][OH]2 = (x)(2x)2 = 4x3 = 1.5 x 10141 Therefore, [OH] = 2x = 3.1 x 104 M; pOH = 3.51 428 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it (iii) Balanced equation: Ca(OH)2(s) o m Ca2(aq) + 2OH(aq) Ksp = 7.9 x 106 Let x = molar solubility of Ca(OH)2 Then x = moles/L of Ca2 and 2x = moles/L of OH Solving, x = 0.013 Ksp = [Ca2][OH]2 = (x)(2x)2 = 4x3 = 7.9 x 106 Therefore, [OH] = 2x = 0.025 M; pOH = 1.60 (iv) Balanced equation: Sr(OH)2(s) o m Sr2(aq) + 2OH(aq) Let x Ksp = 3.2 x 104 = molar solubility of Sr(OH)2 Then x = moles/L of Sr2 and 2x = moles/L of OH Ksp = [Sr2][OH]2 = (x)(2x)2 = 4x3 = 3.2 x 104 Therefore, [OH] Solving, x = 0.043 = 2x = 0.086 M; pOH = 1.06 (v) Balanced equation: Ba(OH)2(s) o m Ba2(aq) + 2OH(aq) Ksp = 5.0 x 103 Although Ba(OH)2 is considered to be a strong soluble base, it does have a solubility product Let x = molar solubility of Ba(OH)2 Then x = moles/L of Ba2 and 2x = moles/L of OH Ksp = [Ba2][OH]2 = (x)(2x)2 = 4x3 = 5.0 x 103 Therefore, [OH] = 2x = 0.22 M; pOH = 0.67 Solving, x = 0.11 (2) A strong base is defined as an inorganic metal hydroxide that is soluble in water and dissociates completely into its ions Providing one works at concentrations less than 0.11 M for Ba(OH)2, 0.043 M for Sr(OH)2, and less than 0.013 M for Ca(OH)2, these bases are considered strong Mg(OH)2 is considered an “insoluble” base (3) Be(OH)2 produces less hydroxide ion than the autoionization of water (4) The 2A metal hydroxides become stronger bases as one goes down Group 2A on the periodic table, largely because they become more soluble 27-48 Refer to Section 2-7 and Example 2-12 mass of mol [Cr(H2O)5(OH)]Cl2 x Cr = x 52.00 g = 52.00 g 11 x H = 11 x 1.008 g = 11.09 g x O = x 16.00 g = 96.00 g x Cl = x 35.45 g = 70.90 g mass of mol = 229.99 g percent Cr by mass % Cr = (52.00/229.99) x 100% = 22.61% 27-50 Refer to Section 27-6 and the Internet Calcium carbonate is the primary component of seashells, antacids, marble and limestone (e.g stalactites and stalagmites in caves), blackboard chalk, scale in water pipes, and calcium supplements for people and animals It is also used to capture SO2 gas in fossil fuel burning boilers, thereby helping to prevent acid rain, and as a soil additive to provide pH adjustment and calcium to farmers’ soil 27-52 Refer to Section 27-3 Lithium compounds, not lithium metal, are used in the treatment of some types of mental disorders The chemical properties of lithium metal are very different from lithium compounds containing the ion, Li Li metal is very reactive with water, forming the strong base, LiOH, and hydrogen gas and releasing much heat, none of which are good for the human body 429 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it 27-54 Refer to Section 27-9 and Table 27-7 The maximum oxidation states for the metals in Groups 3B through 7B are given by the element’s group number However, this is unlikely to be the element’s most stable oxidation state 27-56 Refer to Section 3-4 Balanced equation: 3Co3O4 + 8Al o 9Co + 4Al2O3 Plan: (1) Calculate the theoretical yield of Co metal (2) Calculate the mass of Co3O4 required to produce the theoretical yield of Co actual yield (1) % yield = theoretical yield x 100% Substituting, 190 g 67.5% = theoretical yield x 100% Solving, theoretical yield = 281 g Co mol Co mol Co3O4 241 g Co3O4 x mol Co O = 384 g Co3O4 (2) ? g Co3O4 = 281 g Co x 58.9 g Co x mol Co 27-58 Refer to Section 27-2, Exercise 27-49, and Appendix K Balanced equation: Rb(s) + H2O(A) o RbOH(aq) + 1/2 H2(g) 'Hqr xn = ['Hqf RbOH(aq) + 1/2 'Hqf H2(g)] - ['Hqf Rb(s) + 'Hqf H2O(A)] = [(1 mol)(481.16 kJ/mol) + (1/2 mol)(0 kJ/mol)] - [(1 mol)(0 kJ/mol) + (1 mol)(285.8 kJ/mol)] = 195.4 kJ/mol Rb(s) 'Sqr xn = [Sqf RbOH(aq) + 1/2 Sqf H2(g)] - [Sqf Rb(s) + Sqf H2O(A)] = [(1 mol)(110.75 J/mol˜K) + (1/2 mol)(130.60 J/mol˜K)] - [(1 mol)(76.78 J/mol˜K) + (1 mol)(69.91 J/mol˜K)] = 29.4 J/K per mol Rb(s) 'Gqr xn = ['Gqf RbOH(aq) + 1/2 'Gqf H2(g)] - ['Gqf Rb(s) + 'Gqf H2O(A)] = [(1 mol)(441.24 kJ/mol) + (1/2 mol)(0 kJ/mol)] - [(1 mol)(0 kJ/mol) + (1 mol)(237.2 kJ/mol)] = 204.0 kJ/mol Rb(s) In Exercise 27-49, the 'Grxn was calculated for the following reaction: Na(s) + H2O(A) o NaOH(aq) + 1/2 H2(g) 'Gqr xn = ['Gqf NaOH(aq) + 1/2 'Gqf H2(g)] - ['Gqf Na(s) + 'Gqf H2O(A)] = [(1 mol)(419.2 kJ/mol) + (1/2 mol)(0 kJ/mol)] - [(1 mol)(0 kJ/mol) + (1 mol)(237.2 kJ/mol)] = 182.0 kJ/mol Na(s) Therefore, the reaction between Rb(s) and water is more favored with a greater degree of spontaneity than the reaction between Na(s) and water, since the 'Gq for the reaction between Rb(s) and water is more negative 27-60 Refer to Chapters 2, and 12 Balanced equations: Q2CO3 o CO2 + Q2O QCO3 o CO2 + QO if Q is a 1A element if Q is a 2A element 430 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it Plan: (1) Determine the moles of CO2 formed from the ideal gas law, PV=nRT (2) Determine the moles of the unknown compound and its formula weight (3) Determine the atomic weight of the unknown element for each case, identify the element and compound PV (1) n = RT = (1.00 atm)(4.48 L) = 0.200 mol CO2 (0.0821 L˜atm/mol˜K)(0qC + 273q) (2) ? mol compound = 0.200 mol CO2 x mol compound = 0.200 mol compound mol CO2 14.78 g compound g compound FW (g/mol) = mol compound = 0.200 mol compound = 74.0 g/mol (3) Since the carbonate ion, CO32, has a formula weight of 60.0 g/mol, then the mass of the metal, Q, in one mole of compound must be ? g Q = g compound - g CO32 = 74.0 g - 60.0 g = 14.0 g If the formula is Q2CO3, then the atomic weight of Q must be 14.0/2 = 7.00 g/mol If the formula is QCO3, then the atomic weight of Q must be 14.0 g/mol Since there is no metal with an atomic weight of 14.0 g/mol (nitrogen is not a metal), then the metal Q must be lithium, a IA element with an atomic weight of 6.9 g/mol), and the compound must be Li2CO3 27-62 Refer to Chapter 6, and Sections 27-2, 27-5 and 27-10 (a) formula unit: total ionic: net ionic: Mg(s) + H2O(g) o MgO(s) + H2(g) Mg(s) + H2O(g) o MgO(s) + H2(g) Mg(s) + H2O(g) o MgO(s) + H2(g) (b) formula unit: 2Rb(s) + 2H2O(A) o 2RbOH(aq) + H2(g) total ionic: 2Rb(s) + 2H2O(A) o 2Rb+(aq) + 2OH(aq) + H2(g) net ionic: 2Rb(s) + 2H2O(A) o 2Rb+(aq) + 2OH(aq) + H2(g) (c) formula unit: total ionic: net ionic: Cr(OH)3(s)+ NaOH(aq) o Na[Cr(OH)4](aq) Cr(OH)3(s)+ Na+(aq) + OH(aq) o Na+(aq) + [Cr(OH)4] (aq) Cr(OH)3(s)+ OH(aq) o [Cr(OH)4] (aq) 431 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it 28 Some Nonmetals and Metalloids 28-2 Refer to Section 28-1 The noble gases’ inert nature and low atmospheric abundances were two factors causing their late discovery 28-4 Refer to Section 28-1 The noble gases in order of increasing radii are: He < Ne < Ar < Kr < Xe < Rn The size increases as one goes down Group 8A, because the outer electrons which set their size exist in electron clouds that become farther from the nucleus 28-6 Refer to Section 28-2 The accidental preparation of O2PtF6 by the reaction of O2 with PtF6 led Bartlett to reason that xenon should also be oxidized by PtF6, since the first ionization energy of molecular oxygen is actually slightly larger than that of xenon He obtained a red crystalline solid initially believed to be XePtF6, but now known to be more complex At present, Xe and Kr are the only noble gases known to form compounds, mostly combining with F and O Our textbook discusses the compounds of Xe 28-8 Refer to Section 28-2 Balanced equation: XeF4(s) + F2(g) o XeF6(s) mol XeF mol XeF 245 g XeF ? g XeF6 = 1.85 g XeF4 x 207 g XeF4 x mol XeF6 x mol XeF6 = 2.19 g XeF6 4 28-10 Refer to Sections 28-4 and 28-1, and Exercise 28-5 To the right is a graph of the melting point (mp) and boiling point (bp) trends for the diatomic halogens (X2) and the noble gases (Y) as a function of the number of electrons in the species It is difficult to see the differences in mp and bp for the noble gases because they are very similar, e.g mp and bp for He are 25 K and 27 K, respectively The graphs are alike in that the boiling points and melting points increase with increasing size as set by the number of electrons in the species Melting points and boiling points of these non-polar molecules increase with increasing size because London forces increase with molecular size The graphs differ in that the melting and boiling points of the diatomic halogens are greater than they are for the same sized monatomic noble gases The London forces must be stronger between the halogens than between the noble gases Another disparity is that there is almost no difference between the melting and boiling points of the noble gases, whereas there is a significant difference between the melting and boiling points of the halogen molecules 432 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it 28-12 Refer to Section 28-3 and Table 28-4 (a) (b) (c) (d) In order of increasing atomic radii: F < Cl < Br < I < At In order of increasing ionic radii: F < Cl < Br < I < At In order of increasing electronegativity: At < I < Br < Cl < F In order of increasing melting points and boiling points: F2 < Cl2 < Br2 < I2 < At2 In nature, the halogens exist as nonpolar diatomic molecules London dispersion forces are the only forces of attraction acting between the molecules These forces increase with increasing molecular size (e) See (d) (f) In order of increasing standard reduction potentials: At2 < I2 < Br2 < Cl2 < F2 F2 has the most positive standard reduction potential and therefore is the strongest of all common oxidizing agents Oxidizing strengths of the diatomic halogen molecules decrease down Group 7A 28-14 Refer to Section 28-4 Christe's preparation of F2 is not a direct chemical oxidation, but rather it involves the formation of unstable MnF4, which spontaneously decomposes into MnF3 and F2 28-16 Refer to Sections 28-5 and 6-8, and Example 6-8 (a) Cl2(g) + 2Br–(aq) o Br2(l) + 2Cl–(aq) (c) I2(s) + Br–(aq) o no reaction (b) I2(s) + Cl–(aq) o no reaction (d) Br2(l) + Cl–(aq) o no reaction 28-18 Refer to Section 28-5 To the right are the electrostatic charge potential plots for FCl and ClBr Initially you can identify which is which simply based on the relative sizes of the atoms: F < Cl < Br However, from the greater range of colors across FCl, we can see that FCl is a more polar molecule than ClBr This is verified by the electronegativity differences In FCl, 'EN = [4.0 (for F) - 3.0 (for Cl)] = 1.0, whereas in ClBr, 'EN = [3.0 (for Cl) - 2.8 (for Br)] = 0.2 Note that Cl has a G+ charge in FCl, but has a G in ClBr G G+ FCl G G+ ClBr 28-20 Refer to Section 28-6 Hydrogen bromide, HBr(g), is a colorless gas which dissolves in water to give hydrobromic acid, HBr(aq) The latter is a strong acid which completely dissociates in aqueous solutions giving H3O(aq) and Br(aq) 28-22 Refer to Section 28-6 Hydrofluoric acid is used to etch glass by reacting with the silicates in glass to produce a very volatile and thermodynamically stable compound, silicon tetrafluoride, SiF4 For example, CaSiO3(s) + 6HF(aq) o CaF2(s) + SiF4(g) + 3H2O(A) 28-24 Refer to Sections 28-7 and 4-6, and Table 28-5 (a) KBrO3 potassium bromate (d) NaBrO2 sodium bromite (b) KOBr potassium hypobromite (e) HOBr hypobromous acid (c) NaClO4 sodium perchlorate (f) HBrO3 bromic acid 433 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it (g) HIO3 iodic acid (h) HClO4 perchloric acid 28-26 Refer to Sections 28-6 and 28-7 (a) X2 + H2O o HX + HOX (X = Cl, Br, I) (b) X2 + 2NaOH o NaX + NaOX + H2O (X = Cl, Br, I) (c) Ba(ClO2)2 + H2SO4 o BaSO4 + 2HClO2 (d) KClO4 + H2SO4 o KHSO4 + HClO4 (explosive) 28-28 Refer to Sections 28-7 and 10-7 The ranking of the acids from strongest to weakest is: (a) HOCl > HOBr > HOI, since for most ternary acids containing different elements in the same oxidation state from the same group in the periodic table, acid strengths increase with increasing electronegativity of the central element (b) HClO4 > HClO3 > HClO2 > HOCl, since the acid strengths of most ternary acids containing the same central element increase with increasing oxidation state of the central element and with increasing numbers of oxygen atoms (c) HClO4 > HBrO3 > HOI, since we know HClO4 is one of our strong acids and the others are not The acid with the most electronegative central element and the most oxygen atoms, have an H-O bond that is the weakest 28-30 Refer to Section 28-8 and Table 28-6 oxide, O2 sulfide, S2 selenide, Se2 [Ne] [Ar] [Kr] or or or 1s2 2s2 2p6 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 28-32 Refer to Section 28-8 Every Group 6A element has six valence electrons in the highest energy level, and is therefore two electrons away from achieving an octet of electrons This is why all of them exhibit an oxidation state of –2 In addition, Group 6A elements below oxygen, since they have empty d orbitals available for containing electron pairs, can share their six valence electrons to various degrees with other elements to give different positive oxidation states up to +6 An oxidation state of –3 is impossible because it would require placing an electron into the next higher energy d or s orbitals An oxidation state of +7 is impossible because the Group 6A elements only possess six valence electrons to share or to lose 28-34 Refer to Chapter (a) H2S (b) SF6 (c) SF4 434 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it (d) SO2 (e) SO3 28-36 Refer to Section 28-9 (a) E + 3F2(excess) o EF6 (E = S, Se, Te) (b) O2 + 2H2 o 2H2O E + H2 o H2E (E = S, Se, Te) (c) E + O2 o EO2 (E = S, Te; with Se - use O2 and NO2) 28-38 Refer to Section 28-10 Group 6A hydrides dissociate in two stages Their acid ionization constants, K1 and K2, are shown below: H2 S H2Se H2Te 7 4  + HE 1.0 x 10 1.9 x 10 2.3 x 103 K : H H2E o m a1 o 19   2 11 1.0 x 10 K : HE m H + E | 1.6 x 1011 | 10 a2 Acid strength increases upon descending the group: H2O < H2S < H2Se < H2Te This results from the corresponding decrease in the average E-H bond energy 28-40 Refer to Sections 28-11 and 28-12 (a) NaOH + H2SO4 o NaHSO4 + H2O (e) NaOH + H2SeO4 o NaHSeO4 + H2O (b) 2NaOH + H2SO4 o Na2SO4 + 2H2O (f) 2NaOH + H2SeO4 o Na2SeO4 + 2H2O (c) NaOH + H2SO3 o NaHSO3 + H2O (g) NaOH + TeO2 o NaHTeO3 (d) 2NaOH + H2SO3 o Na2SO3 + 2H2O (h) 2NaOH + TeO2 o Na2TeO3 + H2O 28-42 Refer to Sections 28-12 and 3-5 4FeS2 + 11 O2 o 2Fe2O3 + 8SO2 2SO2 + O2 o 2SO3 SO3 + H2SO4 o H2S2O7 H2S2O7 + H2O o 2H2SO4 2000 lb FeS 454 g FeS mol FeS mol SO mol SO ? ton H2SO4 = 1.25 ton FeS2 x ton FeS x lb FeS x 120.0 g FeS2 x mol FeS2 x mol SO3 2 2 mol H2S2O7 mol H2SO4 98.1 g H2SO4 lb H2SO4 ton H2SO4 x mol SO x mol H S O x mol H SO x 454 g H SO x 2000 g H SO 2 4 = 4.09 tons H2SO4 total However, half of the H2SO4 was a reactant in Step (3) Therefore, the net mass of H2SO4 produced = 4.09 ton/2 = 2.04 tons H2SO4 Balanced equations: (1) (2) (3) (4) 28-44 Refer to Sections 28-11, 17-2 and 17-5 Balanced equation: 2SO2(g) + O2(g) o m 2SO3(g) Let x 1.00 - 2x 5.00 - x 2x = [O2]reacted Then, = [SO2] = [O2] = [SO3] 435 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it o 2SO2 + O2 2SO3 m initial 1.00 M 5.00 M 0M change - 2x M - xM + 2x M at equilibrium (1.00 - 2x) M (5.00 - x) M 2x M However, [SO3] = 2x = 77.8% of [SO2]initial = 0.778 x 1.00 M = 0.778 M [SO ]2 (2x)2 (0.778)2 Therefore, Kc = [SO ]23[O ] = (1.00 - 2x)2(5.00 - x) = (1.00 - 0.778)2(5.00 - 0.778/2) = 2.7 2 28-46 Refer to the Introduction to Section 28-13, Table 28-7 and Appendix B N P As Sb Bi N3 P3 1s2 2s2 2p3 1s2 2s2 2p6 3s2 3p3 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p3 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f 14 5s2 5p6 5d10 6s2 6p3 1s2 2s2 2p6 1s2 2s2 2p6 3s2 3p6 28-48 Refer to Section 28-13 The nitrogen cycle is the complex series of reactions by which nitrogen is slowly but continually recycled in the atmosphere, lithosphere (earth) and hydrosphere (water) Atmospheric nitrogen is made accessible to us and other life-forms in mainly two ways (1) A class of plants, called legumes, has bacteria which extract N2 directly, converting it to NH3 This nitrogen fixation process, catalyzed by an enzyme produced by the bacteria, is highly efficient at usual temperatures and pressures (2) N2 and O2 react in the atmosphere near lightning, forming NO and NO2, which dissolve in rainwater and fall to earth These nitrogen compounds are absorbed and incorporated into plants forming amino acids and proteins The plants are eaten by animals or die and decay, releasing their nitrogen to the environment The animals, in turn, excrete waste and/or die, releasing their nitrogen to the environment 28-50 Refer to Sections 28-14 and 17-7, and Examples 17-9 and 17-10 The Haber process is the economically important industrial process for making ammonia, NH3, from atmospheric N2, according to: 'Hq = 92 kJ/mol N2(g) + 3H2(g) o m 2NH3(g) (1) effect of temperature: The reaction is exothermic ('H < 0), so one might expect that to increase the amount of NH3, one would need to lower the temperature This action would increase the relative amount of NH3 present, however, the reaction rates are lowered as well So, Haber investigated other ways to increase the yield (2) effect of pressure: In Chapter 17, we learned that increasing the pressure favors the reaction that produces the smaller number of moles of gas (forward in this case) This reaction is run under pressures ranging from 200 to 1000 atmospheres to increase the yield of NH3 (3) effect of catalyst: The addition of finely divided iron and small amounts of selected oxides speeds up both the forward and reverse reactions This allows NH3 to be produced not only faster but at a lower temperature, which increases the yield of NH3 and extends the life of the equipment 436 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it 28-52 Refer to Section 28-15 Oxidation number of N: (a) N2 (b) N2O +1 (c) N2O4 +4 (d) HNO3 +5 (e) HNO2 +3 28-54 Refer to Chapter and the Sections as stated (a) NH2Br The Lewis dot formula predicts regions of high electron density around the central N atom, a tetrahedral electronic geometry and a pyramidal molecular geometry The N atom has sp3 hybridization (Sections 8-8 and 28-14) The three-dimensional structure is shown below (b) HN3 Around the outer two N atoms, the Lewis dot formula predicts regions of high electron density, a trigonal planar electronic geometry, sp2 hybridization, and the N atom bonded to the H has a bent molecular geometry The Lewis dot formula also predicts regions of high electron density around the central N atom, a linear electronic and molecular geometry and sp hybridization for the central N atom (Sections 8-5, 8-13 and 28-14) The three-dimensional structure is shown below (c) N2O2 The Lewis dot formula predicts regions of high electron density, trigonal planar electronic geometry and angular molecular geometry around each N atom The N atoms have sp2 hybridization (Sections 8-13 and 28-15) The three-dimensional structure is shown below (d) NO2 The Lewis dot formula predicts regions of high electron density, a linear electronic and ionic geometry around the N atom and sp hybridization for the N atom (Section 28-15) The three-dimensional structure is shown on the next page NO3 The Lewis dot formulas for the three resonance structures (one is shown) predicts regions of high electron density around the central N atom and a trigonal planar electronic and ionic geometry The N atom has sp2 hybridization (Section 28-16) The three-dimensional structure is shown on the next page (e) HNO3 The Lewis dot formulas for the two resonance structures (one is shown) predicts regions of high electron density for the N atom, and a trigonal planar electronic and molecular geometry about the N atom The N atom has sp2 hybridization (Section 28-16) The three-dimensional structure is shown on the next page (f) NO2 The Lewis dot formulas for the two resonance structures (one is shown) predicts regions of high electron density for the central N atom, and a trigonal planar electronic geometry and a bent ionic geometry The N atom has sp2 hybridization (Section 28-16) The three-dimensional structure is shown below (a) NH2Br (b) HN3 437 Copyright 201 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s) Editorial review has deemed that any suppressed content does not materially affect the overall learning experience Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it ... Substance 1~ = ~amount of heat gained by Substance 2~ ~(mass)(Sp Ht . )( temp change)~1 = ~(mass)(Sp Ht . )( temp change)~2 In this exercise, ~(mass)(Sp Ht . )( temp change)~limestone = ~(mass)(Sp Ht . )( temp... Figure 1-7 (a) Box (i) represents the very ordered, dense solid state (b) Box (iii) represents the less ordered, slightly less dense liquid state (c) Box (ii) represents the disordered, much less... Re: Re2O3 (Re=+ 3) < ReO2 (Re=+ 4) < ReO3 (Re=+ 6) < Re2O7 (Re=+ 7) (4 ) As the charge on Re increased, the percentage of Re in the rhenium oxide decreased 2-104 Refer to Section 2-6 When organic