ACS General Chemistry Study Guide PDF

Preparing for Your ACS Examination. 1n General Chemistry Second Edition Examinations Institute® American Chemical Society Division of Chemical Education 2018 American Chemical Society Division of Chemical Education Examinations Institute. Printed in the United States of America 1st Printing 2018 2nd Printing 2019 All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without written permission from the publisher. American Chemical Society Division of Chemical Education Examinations Institute Chemistry and Biochemistry Department University of Wisconsin-Milwaukee 3210 N. Cramer St. Milwaukee, WI 53211 Foreword You have in your hand the second edition of Preparingfor Your A CS Examination in General Chemistry, The Official Guide designed to help students prepare for examinations produced under the auspices of the Division of Chemical Education (DivCHED) of the American Chemical Society (ACS). The first edition of this guide was published in 1998. In 2002, a study guide for organic chemistry students was published followed by the physical chemistry study guide. As is common for materials produced by ACS Exams, we called on colleagues in the chemistry education community to help us put this second edition together. The first edition of this guide was written and produced under previous Director, I. Dwaine Eubanks and previous Associate Director, Lucy T. Eubanks. Their time and effort was utilized by many, many students who used the first edition since 1998. In this second edition, we have utilized many of their underlying principles while updating a substantial portion of the guide to provide a guide aligned to current textbooks and curriculum as well as a better representation of what may be expected on an ACS Exam. This effort was in no way accomplished solely by the three authors, rather was a collective effort with special thanks to our expert editing team and student volunteers. Specific acknowledgments noting each contributor is listed in the Acknowledgments. As a discipline, chemistry is surely unique in the extent to which its practitioners provide beneficial volunteer service to the teaching community. ACS exams have been produced by volunteer teacher-experts for more than eighty-five years. Other projects of the Examinations Institute benefit from the abundance of donated time, effort, and talent. The result is that invariably high-quality chemistry assessment materials are made available to the teaching and learning community at a fraction of their real value. The three Official Guides that have been released so far are intended to be ancillary student materials, particularly in courses that use ACS exams. The care that goes into producing ACS exams may be lost on students who view the exams as foreign and unfamiliar. The purpose of this series of guides is to remove any barriers that might stand in the way of students demonstrating their knowledge of chemistry. The extent to which this goal is achieved in this new edition of the general chemistry study guide will become known only as future generations of chemistry students sit for an ACS exam in general chemistry. We wish them the best. Thomas Pentecost Jeffrey Raker Kristen Murphy Milwaukee, Wisconsin September, 2018 Acknowledgements The unselfish dedication of hundreds of volunteers who contribute their time and expertise make ACS Exams possible. It is from the reservoir of their work that we have drawn inspiration and examples to produce this book to help students who will be taking an ACS exam. We gratefully acknowledge the efforts of all past General Chemistry Committee members. This Official Guide also benefited from the careful proofreading by several colleagues. We extend our special thanks to these faculty members. William J. Donovan The University of Akron Christine Gaudinski Aims Community College Barbara L. Gonzalez California State University, Fullerton Daniel Groh Grand Valley State University Thomas A. Holme Iowa State University Cynthia J. Luxford Texas State University Eric Malina University ofNebraska Matthew Stoltzfus The Ohio State University Melonie A. Teichert United States Naval Academy Students also participated in the development of this Official Guide through using this guide in draft form and sharing their experiences back to us to aid in providing the student-user perspective. We extend our special thanks to these 45 students. General Chemistry I students: University of South Florida General Chemistry II students: Grand Valley State University The University of Akron University of Wisconsin-Milwaukee The personnel ofthe ACS Division of Chemical Education Examinations Institute played a central role in helping us to produce Preparing for Your ACS Examination in General Chemistry: The Official Guide. A very special thank you for all of the work involved is owed to our staff members. Julie Adams Cherie Mayes Jessica Reed Shalini Srinivasan Jaclyn Trate While all of these reviewers have been very helpful in finding problems large and small, any remaining errors are solely our responsibility. You can assist us in the preparation of an even better product by notifying ACS Exams of any errors you may find. Thomas Pentecost Jeffrey Raker Kristen Murphy Milwaukee, WI September, 2018 ii Table of Contents Foreword........................................................................................................................................ i Acknowledgements...................................................................................................................... ii Table of Contents........................................................................................................................ iii Toolbox (Foundational Concepts).............................................................................................. 1 How to Use This Book................................................................................................................. 9 Sample Instructions................................................................................................................... 10 Example General Chemistry Data Sheet................................................................................. 11 Big ldeas...................................................................................................................................... 12 First-Term Material: Chapter 1: Atomic Structure........................................................................................ 15 Chapter 2: Electronic Structure................................................................................... 25 Chapter 3: Formula Calculations and the Mole......................................................... 37 Chapter 4: Stoichiometry.............................................................................................. 47 Chapter 5: Solutions and Aqueous Reactions, Part 1................................................. 57 Chapter 6: Heat and Enthalpy..................................................................................... 69 Chapter 7: Structure and Bonding............................................................................... 81 Chapter 8: States of Matter.......................................................................................... 97 Second-Term Material: Chapter 9: Solutions and Aqueous Reactions, Part 2............................................... 113 Chapter 10: Kinetics.................................................................................................... 123 Chapter 11: Equilibrium............................................................................................. 139 Chapter 12: Acids and Bases...................................................................................... 151 Chapter 13: Solubility Equilibria............................................................................... 163 Chapter 14: Thermodynamics.................................................................................... 175 Chapter 15: Electrochemistry..................................................................................... 187 Chapter 16: Nuclear Chemistry................................................................................. 199 1ll IV Toolbox (Foundational Concepts) Chapter Summary: This chapter will focus on foundational concepts and skills you will use throughout this study guide and likely on your exam. This should not be considered a comprehensive list of preparatory material, but does include important concepts that will be referenced in the content chapters. Specific topics covered in this chapter are: Unit conversations Significant figures Scientific notation Nomenclature Density Classification of matter Properties and representations of matter Additionally, this chapter will include supplemental information in sections at the end of this chapter for preparation for your exam including: Sample instructions Sample datasheet How to use this book Common representations used in questions related to this material: Name Example Used in questions related to ce i> Particulate representations Classification of matter ~ i Com_Q_ound units g·cm-3 Density or unit conversion Where to find this in your textbook: The material in this chapter typically aligns to an introductory chapter or chapters (could be labeled as "Matter and Measurement" or "Atoms, Molecules, and Compounds") in your textbook. The name of your chapter may vary. Practice exam: There are practice exam questions aligned to the material in this chapter. Because there are a limited number of questions on the practice exam, a review of the breadth of the material in this chapter is advised in preparation for your exam. How this fits into the big picture: The material in this chapter aligns to the Big Idea of Experiments, Measurement and Data (9) as listed on page 12 ofthis study guide. Study Questions (SQ) SQ-1. Plank's constant is 6.626xl0-34 J·s. What is this value in klJ.ts? (A) 6.626xl0-4°khts (B) 6.626x 1o-37 kJ- J.lS (C) 6.626xl0-34 klJ.lS (D) 6.626xl0-31 klJ.lS _T_o_o_Ib_o_x~(~F_o_un_d_a_t_io_n_a_I_C_o_n_ce~p_t~s)__________~,~-- 6-"-MP-·, ~~~ ------------------------------- Knowledge Required: (1) Definitions of metric prefixes. Thinking it Through: As you read this question, you see that the units for the value are given with no prefixes: "J" and "s". The units to which you are converting are prefixed: "kJ" and "J.Ls". Therefore, you could start this question by reviewing your metric prefixes (a good idea during studying as well): Prefix Name Value Prefix Name Value M me a 106 or I ,000,000 c centi 10-2 or 0.01 k kilo 103 or 1,000 m milli 10-3 or 0.00 I ).l micro 10-6 or 0.000001 n nano 10-9 or 0.000000001 j:)_ pi co I o- 12 or 0.000000000001 You can then see that you will use the equivalencies of 1 kJ = 1000 J and 1 J.LS = 1Q-6 s and then use this to convert the value: ( 6.626x 10- 34 J· s)(~)(~) 1000 J 10- s = 6.626x 10- 31 kJ · J.LS or choice (D) Choice (A) incorrectly defines micro as 1o-3 and inverts the conversion. Choice (B) inverts the conversion of seconds only and Choice (C) incorrectly defines micro as 10-3. Practice uestions Related to This: PQ-1 and PQ-2 SQ-2. What is the reading on this graduated cylinder to the correct number of significant figures? (A) 42 mL (B) 42.5 mL (C) 42.50 mL (D) 44mL Knowledge Required: (1) Significant figures. (2) Reading a meniscus. Thinking it Through: This question helps remind you that the rules for significant figures are in place because of measurements that are made. This example uses a graduated cylinder. Before you can determine the number of ~ 55; Read here significant figures you will have in your answer, you first recall that you will read the bottom of the meniscus as shown to the right: Now you can determine the value. You start with the largest increment (measured by I mL increments), and see that the meniscus is halfway between the 42"d and 43'd mL (so the value will be 42 mL as a start). Now you observe that the meniscus is halfway between 42 and 43, giving you 42.5 mL. Finally, because you estimated the last significant figure (5), the measurement of 42.5 mL or choice (B) is correct. Choices (A) and (C) do not use the correct number of significant figures based on the increments available to you and choice (D) is not correct as this uses the incorrect location of reading the meniscus. Practice Related to This: SQ-3. Which compound is named correctly? (A) FeC03, iron carbonate (B) K2S0 3, potassium sulfite (C) Sr(N03)2, strontium nitrite (D) Co2(S04) 3, cobalt(II) sulfate 2 ----------~/'"Jlli/J·-· A C S Toolbox (Foundational Concepts) ··Exams Knowledge Required: (1) Rules for nomenclature. (2) Formula and names ofpolyatomic ions. Thinking it Through: The question tells you that you are going to be using the rules of nomenclature (or naming) to answer this. When you skim the choices, you can see that this question is about ionic nomenclature. Therefore, you know the general rules for ionic nomenclature are: 1. Name the metal or polyatomic cation first (first word for simple compounds). a. For those metals with only one cationic charge (such as most of the main group metals), you use just the name of the metaL For example, Na+ would be "sodium" and Zn 2 + would be "zinc". b. For those metals with more than one cationic charge (such as most of the transition metals), you would use the name of the metal followed by the charge in Roman numerals in parentheses. For example, Fe 3+ would be "iron(III)" and Pb 2+ would be "lead(II)". c. Polyatomic cations are named with the name of the cation. 2. Name the nonmetal or polyatomic anion second (second word for simple compounds). a. For monatomic anions, this is the stem of the element plus the suffix "-ide". For example, S2- is "sulfide" and 0 2- is "oxide". b. Polyatomic anions are named with the name ofthe anion. Additionally, for this question, you also need to know the names and formula ofpolyatomic anions: co;-= carbonate so;-= sulfite NO~= nitrate So~-= sulfate Finally, to give the charge for the metals that require them, you will also need to know the charge on the metaL You do this using charge balance ofthe compound. Algorithmically, this is commonly done through the "cross down" method or the subscript in the formula is the opposite charge on the ion (making sure to still check once you use this that your formula is represented correctly; i.e. it is an empirical formula with the correct ratio of cation:anion; also make sure you understand why this method works): Fr+ o\ ~ Therefore, combining all of these, choice (B) does not need a charge on the metal (potassium has only one charge, +I), sulfite is named correctly and the ratio for charge balance is correct Choice (A) is missing the charge (should be II). Choice (D) has a charge, but it is the wrong charge (should be III not II). Finally, choice (C) has the wrong name of the polyatomic anion. Practice Questions Related to This: PQ-5, PQ-6, and PQ-7 SQ-4. In the name arsenic trichloride, the tri- prefix specifies ~~.~o\ cof''e'9 (A) arsenic has an oxidation state of +3. (B) the geometry of the molecule is trigonal pyramidal. (C) arsenic is covalently bonded to three chlorine atoms. (D) arsenic(III) is ionically bonded to three chloride ions. Knowledge Required: (1) Rules for nomenclature. Thinking it Through: The question tells you that you are going to be using the rules of nomenclature again to answer this. From the question, you can see this is about molecular nomenclature (or naming molecules or covalent compounds). Therefore, you know the general rules for covalent nomenclature are (for simple compounds): 1. Name the first element shown in the formula first (first word in the name). This name is the name of the element with the prefix for the number of atoms of the element using Greek prefixes: number prefix number prefix 1 mono- 6 hexa- 2 di- 7 hepta- 3 _T_o_o_Ib_o_x_,_(F_o_u_n_d_a_ti_on_a_I_C_o_n_c_,ep'-t-'s)'---------~;vuw>y:.,·!W·' :x~~ number prefix number prefix 3 tri- 8 octa- 4 tetra- 9 nona- 5 penta- 10 deca- When there is only one atom of the element, the prefix of"mono-" is not used. For example, N204 is "dinitrogen" and !;.0 is simply "carbon". 2. Name the second element shown in the formula second (second word in the name). This name is the name of the stem of the element with the prefix for the number of atoms of the element using Greek prefixes for all numbers of atoms including one and the suffix "-ide". When two vowels are adjacent that are either o or a, the letter of the prefix is omitted. For example, Nz04 is "tetroxide" and CO is "monoxide". Choice (C) reflects the correct usage of the Greek prefix (and the formula of arsenic trichloride, AsCh, reflects this). Choice (A) incorrectly links the prefix of"tri" to oxidation state. Choice (D) implies this is an ionic compound, which is not correct. No structural information is provided with this name, so choice (B) is not correct either. Practice Questions Related to This: PQ-8 and PQ-9 SQ-5. What is the formula of iodic acid? (A) HIO (B) HI02 (C) HI03 (D) HI04 Knowledge Required: (1) Rules for nomenclature. (2) Formula and names ofpolyatomic ions. Thinking it Through: The question tells you that you are going to be using the rules of nomenclature (or naming) for acids to answer this. Therefore, you know the general rules for naming acids are based on the type of acid: Binary acids- those with hydrogen and a monatomic anion: First word: hydro+ the stem of the anion+ ic Second word: acid Ex: HI = hydro + iod + ic acid = hydroiodic acid Oxoacids- those with hydrogen and a polyatomic anion: First word: if the anion has the suffix "-ite" then remove the "-ite" and add "-ous" Ex: the nitrite ion, N0 2-, becomes nitrous if the anion has the suffix "-ate" then remove the "-ate" and add "-ic" Ex: the nitrate ion, N03-, becomes nitri£ Second word: acid Ex: HN02 = nitrous acid HN0 3 = nitric acid Working backwards, iodic acid means this is the acid with iodate ion or I03-. Therefore, the acid formula is HI0 3 or choice (C). Choice (A) is hypoiodous acid; choice (B) is iodous acid; and choice (D) is periodic acid. Practice Questions Related to This: PQ-10 and PQ-11 SQ-6. Twenty irregular pieces of an unknown metal have a collective mass of 28.225 g. When carefully placed in a graduated cylinder that originally contained 7.75 mL of water, the final volume read 11.70 mL. What is the density of the metal (in g·cm-3)? (A) 2.41 g·cm-3 (B) 3.64 g·cm-3 (C) 7.15g·cm-3 (D) 112 g·cm-3 4 , ACS Toolbox (Foundational Concepts) ----------------------1~~-~-- Exams Knowledge Required: ( 1) Definition of density. (2) Displacement. Thinking it Through: As you read this question, you recognize that you will need to calculate the density of the metal which you know is the mass of the substance divided by the volume it occupies. You are given the mass in the problem, so you will need to use the information provided to determine the volume. The volume of the metal is determined by displacement of water: Vr. water +metal - Vi. water= Vmctal or 11.70 mL- 7. 75 mL = 3.95 mL Therefore, you can calculate the density of the metal (and convert from mL to cm3 ): gJ( m~ J = 7.15 g · cm- 3 which is choice (C) 28 225 1 · ( 3.95 mL 1 em' Choice (A) uses the final volume of water and metal and choice (B) uses the initial volume of only the water. Choice (D) uses the correct volume of the metal but then calculates the density incorrectly by multiplying the mass and volume. Practice uestions Related to This: PQ-12 and PQ-13 SQ-7. What are the correct classifications for 0 """'\)o\ I, II, and III? ~,e.,. (,0 0 0 0 I II III I II III (A) element compound mixture (B) element mixture compound (C) compound mixture element (D) mixture element compound..................................... Knowledge Required: ( 1) Classification of matter/definitions. (2) Particulate diagrams. Thinking it Through: As you read through this and evaluate the diagrams, you realize that you need to use your definitions of element, compound and mixtures. First, you know that an element is a group of atoms with the same or similar properties. Merging this with the diagrams you are provided, you see that you are looking for one atom type only which would be sample I. Second, you know that a compound is a pure substance which contains two or more elements chemically combined (or bonded together). Bonded atoms are shown in diagrams I and III, where the bonded atoms in diagram I are all the same type (so showing a molecular element) where diagram III is showing bonded atoms of different types (so showing a molecular compound); III is the compound. Finally, a mixture is two or more pure substances physically combined. Sample II shows two different atom types mixed together and not bonded together, so this is the mixture of two elements. These combine to the correct choice of (B). Practice Questions Related to This: PQ-14 and PQ-15 5 _T_o_o_Ib_o_x_,(~F_o_un_d_a_t_io_n_a_I_C_o_n_ce_.p_t--"s)_ _ _ _ _--IIIII&'S:., AC S.Exams Practice Questions (PQ) PQ-1. The speed oflight is 3.0x 108 m·s- 1. What is this speed in nm·ms- 1? (A) 3.0x 10-4 nm·ms- 1 (B) 3.0x 102 nm·ms- 1 (C) 3.0x10 14 nm·ms- 1 (D) 3.0x1020 nm·ms- 1 PQ-2. The density of helium is 0.164 kg·m-3 What is this density in lb·ft-3? I kg= 2.20 lb and 1 m = 3.28 ft (A) 0.0102lb·ft-3 (B) 0.110 lb·ft-3 (C) 1.18 lb·ft-3 (D) 12.7 lb·ft-3 PQ-3. Based on the figure, what volume should be reported for the liquid? (A) 20.8 mL (B) 20.68 mL (C) 20.6 mL (D) 20.57 mL PQ-4. What is the correctly reported mass of water Mass of beaker and water based on the data in the table? Mass of beaker on! (A) 1.3 g (B) 1.30 g (C) 1.298 g (D) 1.2980 g PQ-5. What is the name ofTh(P04)4? (A) titanium phosphate (B) titanium(III) phosphate (C) titanium(IV) phosphate (D) titanium tetraphosphate PQ-6. The formula of strontium hexafluorosilicate is SrSiF6. What is the formula of aluminum hexafluorosilicate? (A) AISiF6 (B) Ah(SiF 6)3 (C) AbSiF6 PQ-7. Which is NOT named correctly? (A) Mn02 manganese(II) oxide (B) CuS04 copper(II) sulfate (C) Na3P04 sodium phosphate (D) CaF2 calcium fluoride PQ-8. What is the correct name for N20 5? (A) nitrogen(II) oxide (B) nitrogen(V) oxide (C) dinitrogen oxide (D) dinitrogen pentoxide PQ-9. What is the formula of sulfur trioxide? (A) so2 (D) soi- PQ-10. What is the name ofHBr04(aq)? (A) bromic acid (B) bromous acid (C) hydrobromic acid (D) perbromic acid PQ-11. What is the fommla ofhydroiodic acid? (A) HI(aq) (B) HIO(aq) (C) HI02(aq) (D) HI03(aq) 6 -~~ · A( S Toolbox (Foundational Concepts) -----------------------··~~.~·Exams------------~----------~~ PQ-12. If a palladium nanoparticle has a density of 12.0 g·cm-3 , what is the mass of a nanoparticle with a volume of 1.84x 1o-21 cm3? (A) 2.2lxi0-20 g (B) 1.63x I0-20 g (C) 6.52x I0-21 g (D) 1.53x]0-22 g ~~.~o\ PQ-13. Which sample has the largest volume? Densities.,.,efl aluminum 2.7 g·cm-3 [ copper 9.0 g·cm-3 (,0'" (A) 1.0 g aluminum (B) 5.0 g aluminum (C) 1.0 g copper (D) 5.0 g copper ~~.~o\ PQ-14. Which diagram represents a heterogeneous mixture? ~,efl «> c,o «> ~ ~ ~ ~..... fiJ : A C 5 Atomic Structure ------------------------1 ·~u.a·Exams--------------~~~~~ Answers to Study Questions l.B 4. A 7. c 2. A 5. c 8. D 3. c 6. c 9. A Answers to Practice Questions l.C 11. c 21. B 2. B 12. A 22. D 3. A 13. B 23. c 4. c 14. D 24.A 5. D 15. c 25. B 6. D 16. c 26.D 7. B 17.A 27. B 8. D 18. c 28. B 9. B 19. B 29. B 10. A 20.C 30.D 23 "··A ~A~t~om~ic~S~t~ru~c~tu~r~e-------------------~ C 5 _ _ _ _ _ _ _ __ Exams 24 Chapter 2- Electronic Structure Chapter Summary: This chapter will focus on electrons and their arrangement in the atom. The relationship between the arrangement of the electrons and energy is also covered. Finally, the relationship between the electronic structure ofthe atom and the atom's location on the periodic table is reviewed. Specific topics covered in this chapter are: Light and the electromagnetic spectrum Quantum numbers and atomic orbitals Orbital diagrams and electron configurations Paramagnetic and diamagnetic species Energies of electron transitions in atoms Periodic properties related to the electronic structure Previous material that is relevant to your understanding of questions in this chapter include: Significant figures (Toolbox) Scientific notation (Toolbox) Metals and nonmetals (Chapter 1) Common representations used in questions related to this material: Name Example Used in questions related to Atomic symbols H Electron configurations Orbital box diagram D ODD DDDDD Electron configurations s p d Bohr model GJ, n~2 The Bohr model and energy levels in the hydrogen atom n~3 Where to find this in your textbook: The material in this chapter typically aligns to "Electronic Structure" or "Quantum" in your textbook. The name of your chapter may vary. Practice exam: There are practice exam questions aligned to the material in this chapter. Because there are a limited number of questions on the practice exam, a review of the breadth of the material in this chapter is advised in preparation for your exam. How this fits into the big picture: The material in this chapter aligns to the Big Idea of Atoms (1) as listed on page 12 of this study guide. 25 Electronic Structure ----------------------~·~w.~·Exams Ai'~,,~ A c5 Study Questions (SQ) SQ-1. What is the wavelength of light (in nm) produced by the electronic transition between levels 4 and 2 of a hydrogen atom? (A) 182 nm (B) 364 nm (C) 486 nm (D) 1450 nm Knowledge Required: (1) Ability to use the Rydberg formula. (2) Understanding of the units of energy. (3) Calculation of wavelength from an energy. (4) Relationship between the AE of the transition and the energy of the photon emitted. Thinking it Through: You know that the energy of the electron energy levels in hydrogen depends on the value of then principal quantum number. Note: It is very likely that you will be provided this equation, and others, and any needed constants. -R E =--H n2 n You also know you can calculate the energy of transition from n = 4 ~ n = 2 using the Rydberg relationship: till =R (...!__...!_]where RH = 2.18xi0- 18 J, ni = 4 and nr= 2 H 2 2 n; n1 AE = RH ( -1--1 ) = 2.18xlo- 18 J ( -1 - - 1 ) = -4.09xlo- 19 J 42 22 16 4 The energy of the transition is negative because the process produces a photon and the atom releases energy. The energy of the photon is the absolute value of the energy of the transition. 19 £photon = IMtransitionl = 4.09xl0- J You recall that if you know the energy of the photon, you can find the wavelength, A, using: he.h £photon == -,Wit A h = 6.626x 1o-34 J·S and e = 2.998 X 1o-S m·s- 1. he (6.626x10-34 1)(2.998xl08 m·s· 1) Solving for the wavelength you get: 1c = 4.86x10- 7 m £photon 4.09xl0- 19 J mC 01 ;m) = 486 nm 9 You can convert the wavelength to nm: 4.86x10- 7 The correct choice is (C). Choice (A) is not correct because it is the wavelength calculated when the energy of then= 2level, not the difference in energy of the levels, is used to calculate the wavelength. Choice (B) is not correct because it is the wavelength calculated if when then values are not squared. Choice (D) is not correct because it is the wavelength calculated when the energy of then= 4 level, not the difference in energy of the levels, is used to calculate the wavelength. Practice uestions Related to This: PQ-1, PQ-2, and PQ-3 In the energy diagram, transitions between energy levels are denoted by arrows. Which transition corresponds to the absorption of the shortest wavelength of light? (A) Ez to E1 (B) Ez to E3 (C) £3 to Ez (D) £1 to £3 26 A( S Electronic Structure Exams -------=.=.:::...::=...::::.=-=.c:..:.==--=-.................................................................................-............ Knowledge Required: (I) Difference in absorption and emission. (2) How to interpret an energy level diagram. (3) How the energy of a photon relates to the difference in electron energy levels. (4) Relationship between energy and wavelength. Thinking it Through: You read in the question that the process occurring is absorption. You know that this means energy is being absorbed by the atom. This tells you that you want a transition where the electron is being promoted to a higher energy level. This is represented on the diagram by an arrow pointing upwards. The farther apart the energy levels, the larger the energy of the photon involved in the transition. So, the longer the arrow the more energetic the photon. Remember that photon energy and wavelength are inversely related. he E photon = J....- The question is asking you to find the transition with the shortest wavelength, which means the highest energy (longest arrow). The question also wants the transition to be an absorption (upward pointing arrow). The correct choice is (D). Choice (A) is not correct because the transition indicated would represent the emission of a photon. Choice (B) is not correct because the transition indicated would represent the absorption of a photon, but the energy of this photon would be less than the energy (longer wavelength) of the photon absorbed in transition D. Choice (C) is not correct because the transition indicated would represent the emission of the photon with the longest wavelength. Practice uestions Related to This: PQ-4 and PQ-5 SQ-3. Which set is NOT an allowed set of quantum numbers? n.£ mt ms (A) 3 2 -1 +Yz (B) 4 0 -1 +Yz (C) 3 -1 -Yz (D) 4 0 0 -Yz Knowledge Required: (1) Rules for quantum numbers. Thinking it Through: You are being asked to decide which set of quantum numbers is incorrect. You remember the rules for the allowed values of the quantum numbers: Allowed values of n are any integer from 1 to infinity Allowed values of.£ depend on the value of the n quantum number. Allowed values of.£ are 0 up to the value ofn- 1. Allowed values of mt depend on the value of the.£ quantum number. Allowed values of mt are -.£to 0 to+ e. Allowed values of ms are either+ Yz or- Yz. You next proceed to check each combination. Choice (A) With n = 3, the value of 2 for.£ is allowed. With.£ = 2, the value of mt = -1 is allowed. The given value of ms is allowed. Choice (B) With n = 4, the value of 0 for.£ is allowed. With.£ = 0, the value of mt =-I is NOT allowed. The given value of ms is allowed. Choice (C) With n = 3, the value of 1 for.£ is allowed. With.£ = l, the value of mt = -1 is allowed. The given value of ms is allowed. Choice (D) With n = 4, the value of 0 for.£ is allowed. With t =0, the value of mt = 0 is allowed. The given value of ms is allowed. The choice with the incorrect set of quantum numbers is choice (B). Practice Questions Related to This: PQ-7 27 Electronic Structure.,·Acs -.,Exams SQ-4. Which type of electron is described by the quantum numbers, n = 3, f = 2, mr = 1, and m, = + 'h? (A) 2p (B) 3p (C) 3d (D) 3f Knowledge Required: (!) Relationship between principal quantum number and energy level designation. (2) Letter abbreviations for values of the t quantum number. Thinking it Through: You are given a set of quantum numbers and asked to determine what type of electron would have this set of quantum numbers. You recall that the two parts of an electron description are the energy level (n quantum number) and the orbital type (f quantum number). The first part of the description is the value of n. For example, 2p describes an electron in the second energy leveL The letters describing the orbital type ares, p, d, and f. The relationship between the letters and the values of the f quantum number are: E = 0 is an s orbital; f = 1 is a p orbital, t = 2 is a d orbital, and e = 3 is a f orbitaL You return to the given set of quantum numbers and decide the type of electron being described: n = 3 and t = 2 describes a 3d electron. The correct choice is (C). Choice (A) is not correct because 2p describes an electron with n = 2, t =1. Choice (B) is not correct because 3p describes an electron with n = 3, £ =1. Choice (D) is not correct because 3f describes an electron with n = 3, t =3 and this is not a possible set of quantum numbers. Practice Questions Related to This: PQ-7 SQ-5. What is the electron configuration of the valence electrons for a ground-state Ge atom? (A) 4p 2 (B) 4s2Jdlo4p2 (C) 3dl04p2 (D) 4s 24p 2 Knowledge Required: (1) How to write electron configurations using the periodic table. (2) Definition of valence electrons. Thinking it Through: You are being asked for the valence electron configuration of Ge. You also make use of the definition of valence electrons: electrons in the outermost energy leveL The atomic number ofGe is 32, which you know means the neutral atom has 32 electrons. Ge is also in the fourth period (row) and group (column) 14 of the periodic table. Because Ge is in the fourth period, the highest energy level containing electrons is n = 4 and these are the valence electrons forGe. Group 14 is the second column in the p-block of the periodic table. You can then write the electron configuration for the valence electrons: 4s 24p 2. You can check this result using the core notation: [Ar] 4s 23d 104p 2. The [Ar] core has 18 electrons, the filled 3d holds 10 more, and the valence shell has 4. You have accounted for the all 32 electrons. The correct choice is (D). Choice (A) is not correct because it does not include the 4s electrons as valence electrons. Choice (B) is not correct because electrons in the filled d block are not considered valence electrons. Choice (C) is not correct because it does not include the 4s electrons and does include the filled d orbitals. Practice Questions Related to This: PQ-8, PQ-9, and PQ-10 SQ-6. What is the ground state electron configuration for the Zr2+ ion? (A) [Kr] 5s2 (B) [Kr] 4d 2 (C) [Kr] 5s 24d 2 (D) [Kr] 5s 24d4 28 >~ (A) Si (Z= 14) (B) P (Z= 15) (C) Ge (Z = 32) (D) As (Z = 33) PQ-21. The species F-, Mg2+, and Na+ all have the same number of electrons. Which is the predicted order when they are arranged in order of decreasing size (largest first). 33 Electronic Structure ~:c.~ A C 5 _______________________ ~Exams PQ-22. Which ion is the smallest? (A) Al 3+ (B) Na+ (D) 0 2- "'"'o\ PQ-23. What happens when a bromine atom becomes a bromide ion? ,e9 co~ (A) A positive ion is formed. (B) The bromine nucleus acquires a negative charge. (C) The atomic number of bromine is decreased by one. (D) The bromide ion is then larger than the bromine atom. PQ-24. Which ion has the largest radius? (A) CI- (B) p- (D) Cu 2 + PQ-25. Which element has the highest electronegativity? (A) cesmm (B) iodine (C) oxygen (D) lithium PQ-26. Which element is more electronegative than arsenic and less electronegative then sulfur? (A) chlorine (B) phosphorous (C) tin (D) oxygen PQ-27. Which equation corresponds to the electron affinity of chlorine? (A) CI-(g)---+ Cl(g) + e- (B) Clz(g) + e----+ Clz-(g) (C) Cl(g)---+ CJ+(g) + e- (D) Cl(g) + e----+ CI-(g) PQ-28. Which pair of elements is listed in order of decreasing first ionization energy? (A) Na, Mg (B) Mg, AI (C) AI, Si (D) Si, P PQ-29. When the species F-, Na+, and Ne are arranged in order of increasing energy for the removal of an electron, what is the correct order? (A) F- < Na+ < Ne (B) Na+ C02 + _HzO (A) 2, 21, 14, 14 (B) 2, 19, 14, 14 (C) 2, 19, 7, 7 (D) 1,10,7,7 PQ-3. What are the smallest whole number coefficients for each substance (respectively) in the chemical reaction when balanced? NaBr+ Cb---> NaCl+_Br2 (A) 1, 1,2,2 (B) 1, 2, 2, 2 (C) 2,1,2,1 (D) 2, 4, 2, 4 PQ-4. When this equation is balanced with the smallest set of whole numbers, what is the coefficient for N 2? _N2H4(g) + _Nz04(g) ---> _Nz(g) + _HzO(g) (A) 1 (B) 2 (C) 3 (D) 4 PQ-5. What is the coefficient for ZnO when the reaction is balanced with smallest whole number coefficients? ZnS + _02 ---> ZnO+_SOz (A) 1 (B) 2 (C) 3 (D) 4 PQ-6. What are the smallest whole number coefficients for each substance (respectively) in the chemical reaction when balanced? _CH4+ (A) 1,2,1,2 (B) 1,3, 1,2 (C) 2, 4, 2, 4 (D) 4, 2, 2, 2 PQ-7. How many moles of iron are necessary to react completely with 1.75 mol of oxygen gas? 302(g) + 4Fe(s)---> 2Fe03(s) (A) 1.31 mol (B) 1.75 mol (C) 2.33 mol (D) 5.25 mol PQ-8. What mass ofNOz is formed by the complete reaction of2 mol 0 3 with Molar mass I g·moi- 1 excess NO? 03 48.0 N02 46.0 (A) 23.0 g (B) 46.0 g (C) 92.0 g (D) 96.0 g PQ-9. How many moles of Ba3N2 are necessary to react completely with 1.5 mol of water? Ba3N2 + 6H20---> 3Ba(OH)2 + 2NH3 (A) 0.25 mol (B) 1 mol (C) 6 mol (D) 9 mol 52 ----------------------------~u'~~~,~· A C S ----------------------~S~t~o·~·c=h~io~m~e~tr~y_ ,. Exams PQ-10. How many moles of water are necessary to react completely with 8,0 mol PC1 5? PCls + 4Hz0 ----> H3P04 + 5HC1 (A) 2-0 mol (B) 8,0 mol (C) 16 mol (D) 32 mol PQ-11. What mass ofSbF 3 (I 79 g·moJ- 1) is needed to produce 1.00 g of Freon- 12, CCbF 2 (molar mass= 12 I g·moJ- 1)? 3CCJ4 + 2SbF3----> 3CCbFz + 2SbCb (A) 0.667 g (B) 0.986 g (C) l.48g (D) 2.22 g PQ-12. What is the theoretical yield of aluminum chloride that could be Molar mass I g·mol- 1 obtained from 6.00 mol of barium chloride and excess aluminum AlCh 133.3 sulfate? BaCb 208.23 Ab(S04)3 + 38aCb----> 38aS04 + 2A1Cb Ab(S04h 342.15 (A) 1250 g (B) 801 g (C) 534 g (D) 134g PQ-13. A self-contained breathing apparatus uses potassium superoxide, K0 2, f------=--=:::=-=--==--:....=~::..::...._----1 to convert the carbon dioxide and water in exhaled air into oxygen. L___ _.: : :C. : : Oc=- 2 ____.....:....:.=-:._______j 4KOz(s) + 4COz(g) + 4HzO(g)----> 4KHC03(s) + 30z(g) How many molecules of oxygen gas will be produced from 0.0468 g of carbon dioxide exhaled in a typical breath in excess KOz and water? (A) 4.8x 1020 molecules (B) 6.4x 10 20 molecules (C) 8.5x 1020 molecules (D) J.9x 1021 molecules PQ-14. lf365.5 g HzO(l) is combined with excess KOz, what mass of0 2(g) is Molar mass I g·mol- 1 expected to be produced? H20 18.02 2K02(s) + 2H20(I)----> 2KOH(s) + 02(g) + H20z(l) 02 32.00 (A) 11.42 g (B) 162.4 g (C) 324.7 g (D) 659.4 g PQ-15. How many water molecules are produced if0.34 mol of propane, C3 H 8, combusts in excess oxygen? C3Hs(g) + S02(g)----> 3C02(g) + 4H20(g) (A) 5.lxl022 (B) 2.1 X 1023 (C) 8.2x 1023 (D) 2.4x 10 24 PQ-16. What is the maximum mass (in g) ofN0 2 formed when 7.5 g ofN20s Molar mass I g·mol-1 are consumed? N20s 108.0 NOz 46.0 (A) 1.6 g (B) 3.2 g (C) 6.4 g (D) 15 g (A) 104 g (B) 527 g (C) 151 g (D) 301 g PQ-18. What mass ofFeS is necessary to form 1.08 g Fe203 with excess Oz? Molar mass I g·mol- 1 4FeS + 702----> 2Fez03 + 4S02 FeS 87.92 Fe203 159.7 (A) 0.0136 g (B) 0.298 g (C) 1.19 g (D) 2.16 g 53 _S_to_ic_h_io_m_e_t~ry~--------------------~~· ~X~~ PQ-19. A 2.32 g sample ofNazS04·nHzO yields 1.42 g NazS04 upon heating. Molar mass I NazS04·nHzO(s)---+ NazS04(s) + nHzO(g) What is the value ofn? (A) 2 (B) 3 (C) 5 (D) 10 PQ-20. A mixture of9 mol Fz and 4 mol S are allowed to react. 3Fz+ S---+ SF6 How many moles ofFz remain after 3 mol ofS have reacted? (A) 3 (B) 2 (C) 1 (D) 0 PQ-21. What is the maximum amount of ammonia that can be produced when 0.35 mol Nz reacts with 0.90 mol Hz? N2(g) + 3H2(g) ---+ 2NH3(g) (A) 0.18 mol NH3 (B) 0.60 mol NH3 (C) 0.70 mol NH3 (D) 1.4 mol NH3 "'-0 \ PQ-22. Equimolar amounts of nitrogen, hydrogen, and argon are placed in a reaction chamber. The amount of c,o'(\'e~ which substance(s) will determine the amount of ammonia produced in the reaction below if the reaction goes to completion? Nz(g) + 3Hz(g) ---+ 2NH3(g) (A) argon (B) hydrogen (C) nitrogen (D) nitrogen and hydrogen "'-0\ PQ-23. How many total molecules will be present when the reaction of 5,000 molecules of CS2 with 15,000 1(\,e~'li (,0 molecules of 0 2 goes to completion? CSz(l) + 3 02(g) ---+ C02(g) + 2S02(g) (A) 5,000 (B) 10,000 (C) 15,000 (D) 20,000 PQ-24. What amount of the excess reagent remains when 0.30 mol NH 3 reacts with 0.40 mol 0 2 to produce NO and H20? 4NH3 + S02 ---+ 4NO + 6Hz0 (A) 0.10 mol NH 3 (B) 0.10 mol 02 (C) 0.025 mol NH3 (D) 0.025 mol 02 PQ-25. What amount of Ab03 is produced from the reaction of3.0 mol AI with 2.0 mol Fez03? 2Al + Fe203---+ Ah03 + 2Fe (A) 1.5 mol (B) 2.0 mol (C) 3.0 mol (D) 6.0 mol PQ-26. What amount of the excess reagent remains when 0.500 mol Li reacts with 0.350 mol N 2 ? 6Li(s) + Nz(g)---+ 2LhN(s) (A) 0.0833 mol Nz (B) 0.267 mol Nz (C) 0.150 mol Li (D) 0.417 mol Li PQ-27. The combustion ofC3H 80 with 02 is represented by this equation. Molar mass I g·mol-1 2C3Hs0 + 90z ---+ 6C02 + 8H20 When 3.00 g C 3Hs0 and 7.38 g 02 are combined, what is the excess reagent and how many moles of that reagent remain? (A) 0.006 mol Oz (B) 0.24 mol Oz (C) 0.024 mol C3HsO (D) 0.18 mol C3HsO 54 ,~u -----------------------··~~·Exams------------------~~~~- · A C S Stoichiometry PQ-28. Consider this reaction for the production of lead. Molar mass I g·mol- 1 2PbO(s) + PbS(s)--> 3Pb(s) + SOz(g) Pb 207.2 What is the theoretical yield of lead that can be obtained by the reaction PbO 223.2 of57.33 g PbO and 33.80 g ofPbS? PbS 239.3 (A) 43.48 g (B) 72.75 g (C) 79.83 g (D) 87.80 g PQ-29. What amount of excess reagent remains when 4.0 g zinc react with Molar mass I g·mol- 1 2.0 g phosphorus? p 30.97 Zn 63.39 Zn3Pz 258.1 (A) 0.70 g P (B) 1.3 g P (C) 0.22 g Zn (D) 4.2 g Zn ~0\ PQ-30. An initial reaction mixture of0 2 and F2 is shown to the right..,.,e"9'1l c,o · Which representation shows what would be present after the.. (f) , reaction is complete? 02 + 2F2--> 20F2 & I 8 (A) Q = oxygen = fluorine (B) 'Cb cP.. (C) (D) 8 55 Stoichiometry. ACS _______________________ ------~------------------~··· ~Exams l.B 5. A 2. D 6. A 3. D 7. A 4. c ~--~ Answers to Practice Questions-~~--~----------------~ l.D 11. B 21. B 2. B 12.C 22.B 3. c 13.A 23. c 4. c 14. c 24.D 5. B 15.C 25.A 6. A 16. c 26.B 7. c 17.A 27.A 8. c 18.C 28.C 9. A 19. c 29.A 10. D 20.D 30.B 56 Chapter 5- Solutions and Aqueous Reactions, Part 1 Chapter Summary: This chapter will focus on substances in solutions including qualitative and quantitative properties of solutions and reactions of substances in solution. Specific topics covered in this chapter are: Molar concentration including: o Preparation of solutions o Dilutions o Concentration of ions in solutions Solution stoichiometry including: o Gravimetric analysis o Titrations o Limiting reactants, theoretical yield, experimental yield and percent yield Reactions o Precipitation reactions o Acid/base reactions o Oxidation and reduction reactions Reaction representations o Molecular, total and net ionic equations Previous material that is relevant to your understanding of questions in this chapter include: The mole and formula calculations (Chapter 3) Stoichiometry and balancing equations (Chapter 4) Common representations used in questions related to this material: Name Used in uestions related to Particulate representations Reactions, solution properties (NOTE: Water is often omitted from these representations for clarity.) Where to find this in your textbook: The material in this chapter typically aligns to "Aqueous Reactions" in your textbook. The name of your chapter may vary. Practice exam: There are practice exam questions aligned to the material in this chapter. Because there are a limited number of questions on the practice exam, a review of the breadth of the material in this chapter is advised in preparation for your exam. How this fits into the big picture: The material in this chapter aligns to the Big Idea oflntermolecular Interactions (4) and Reactions (5) as listed on page 12 of this study guide. · 57 Solutions and Aqueous Reactions, Part 1 :~:c: A C5 --------~------~--------~Exams Study Questions (SQ) SQ-1. Which compound is a weak electrolyte when dissolved in water? (B) KOH (D) NaN03.................. ~.........,... _...... Knowledge Required: (1) Definition of an electrolyte. (2) Classification of strong vs. weak electrolyte. Thinking it Through: The definition of an electrolyte is a substance or solution that will conduct electricity. For this question, the solutes are all dissolved in water. You also know the reason the solutions conduct electricity is that the substance (or solute) produces ions when dissolved in water, creating an electrolytic solution. A strong electrolyte will dissociate or ionize completely in water while a weak electrolyte will only partially ionize in water. Considering the responses therefore, you are looking for a substance that would be classified as a weak acid or base as these substances would partially ionize in water. Ammonia or NH 3 is a weak base, and therefore is a weak electrolyte. The correct choice is (C). Nitric acid, HN03, is a strong acid and therefore a strong electrolyte, so choice (A) is incorrect. Potassium hydroxide, KOH, is a strong base and also a strong electrolyte, so choice (B) is incorrect. Potassium hydroxide is also ionic as is sodium nitrate, NaN0 3 or choice (D). Ionic compounds when in solution are present as ions, so are strong electrolytes making choice (D) incorrect as well. Practice Questions Related to This: PQ-1 SQ-2. What mass (in g) ofmagnesium nitrate, Mg(N0 3)2, are required to produce 250.0 mL of a 0.0750 M solution? (A) 0.0188 g (B) 0.0445 g (C) 1.61 g (D) 2.79 g Knowledge Required: (1) Calculation of molar mass. (2) Definition of molar concentration. Thinking it Through: The first thing you recognize when reading this question is that you are given a volume (250.0 mL) and molar concentration (0.0750 M). You know that "M" or molar concentration is:. mol solute 0.0750 mol Mg(N0 3)? molar concentratiOn (M) = ( ). 0.0750 M = - vol L solutiOn 1 L solution ( 1 L solutiOn 3 l (0.2500 L solutwn) = 0.0188 mol Mg(N0 3 - - - - -. - ' - - ---'-"-. Then, given the volume, this allows you to determine the number of moles of solute (Mg(N03)2): 0.0750 mol Mg(NO ) 2 )2 l Use the molar mass of magnesium nitrate to determine the mass: (0.0188molMg(N03 )J ( 148.33 g Mg(NO ) molMg(NOJ: 2 =2.79gMg(N03) 2 This corresponds to choice (D). Choice (A) is not correct because the molar mass was not used (this is the number of moles). Choice (B) is not correct because 250.0 mL was not converted to liters and the number of moles was incorrectly calculated. Choice (C) is not correct because the wrong molar mass was used (for MgN03, not Mg(N03)2). Practice Questions Related to This: PQ-2, PQ-3, PQ-4, and PQ-5 58 -------------ill_:uA_.~,\ ACS..,, Exams Solutions and Aqueous Reactions, Part 1 SQ-3. What mass (in g) of sodium ions, Na+, are there in Molar mass I g·mol-1 25 mL of 0. 7 5 M N azS03? 1--N:-a-:-+-----'2~2=-.9:-9::------1 Na2S03 126.1 (A) O.OI9 g (B) 0.87 g (C) 2.4 g (D) 4.8 g Knowledge Required: (1) Interpreting a chemical formula. (2) Definition of molar concentration. Thinking it Through: Similar to SQ-2 previously, you will use the molar concentration and volume of the solution given in the problem to determine number of moles. Here, you are going to pay special attention to the 0.75 mol Na 2SO 3 ). substance:. ( 0.025 L solutiOn) = 0.019 mol Na 2S0 3 ( I L solutiOn From here, if you used the molar mass ofNa2S0 3, you would determine the mass of sodium sulfite needed to make this solution, which is not what the question is asking. Rather, you are going to use the chemical formula and the ratio of sodium ions (Na+) to the formula unit (Na2S03) where the "2" subscript tells you there are 2 Na+ to every I Na2S03 or 2 mol Na+ to I mol Na2S03. Using this, you determine the number of moles ofNa+ and then the mass ofNa+: (O.OI9 mol Na 2SOJ( 2 mol Na+ )( I mol Na 2 S0 3 22 99 · g Na+ = 0.87 g Na+ which is choice (B). mol Na + ) l Choice (A) corresponds to the number of moles ofNa2S0 3. Choice (C) is the mass of sodium sulfite, Na2S0 3. Choice (D) uses the correct mole ratio but the wrong molar mass. Practice Questions Related to This: PQ-6 and PQ-7 SQ-4. As water is being added to a concentrated NaOH solution ,.o\ : No,- No,- 8 Which species is/are present in excess? €> €> NOf CaCh(aq) + 2AgN03(aq)---+ Ca(N03)2(aq) + 2AgCI(s) (A) CJ- only (B) Ca 2+ and CJ- (C) Ag+ and N03- Knowledge Required: (I) Interpreting particulate representations. (2) Using balanced chemical equations. Thinking it Through: You start this question by evaluating the balanced equation for reactants and products. First looking at the products, you see that AgCI is the solid that is formed and Ca(N0 3) 2 is in solution (or spectator ions): No,- 0 No,- No,- 8 ~ Ca(N03)2(aq) Actually as Ca 2 + andNOF 0 0 No,- AgCI(s) Therefore, on first inspection, looking for excess reactant, you might think that the chloride ion is the only ion in excess. However, the aqueous product of calcium nitrate, Ca(N0 3) 2, yields two nitrate ions for each calcium nitrate or: Ca(N03)2(aq) ---+ Ca 2+(aq) + 2N0 3-(aq). This means that two calcium ions and four nitrate ions are the products, leaving one calcium ion and two chloride ions as excess reactant or choice (B) : 08 8 product --- ,.. 8 €Y 8 @.. s ~ ~--- product Choice (A) omits the excess calcium ions. Choice (C) focuses on the other reactant counting the largest number of ions in solution (the nitrate ions). Choice (D) represents all ions in solution and does not account for the product in solution or the spectator ions. Practice Questions Related to This: PQ-11, PQ-12, and PQ-13 SQ-6. Which reaction will result in a precipitate? (A) FeCh(aq) + NazS(aq)---+ (B) HzS04(aq) + NaOH(aq)---+ (C) KCl(aq) + CaBrz(aq)---+ (D) Zn(N03)2(aq) + Hl(aq)---+ Knowledge Required: (l) Solubility rules. (2) Process for precipitation reactions. Thinking it Through: You recall the criteria for a precipitation reaction to occur: (I) both reactants must be in solution and (2) at least one product must be insoluble (or the precipitate). The choices show that the reactants are all soluble (as noted by the "aq"). To solve this, you will likely go through each set for the insoluble product (for each option bere, these will be written with each reactant separated as ions): Reactants Products I Choice (A) I Fe 2+(aq) 2Cl-(aq) sz-caq) I I FeS insoluble NaCl so luble I correct J 60 ----------m~·--~· A C 5 Solutions and Aqueous Reactions, Part 1 ···Exams Reactants Products H20 NazS04 Choice (B) W(aq) HS04-(aq) Na+(aq) OH-(aq) incorrect liquid soluble KBr CaCb Choice (C) K+(aq) CJ-(aq) Ca2+(aq) 2Bc(aq) incorrect soluble soluble HN03 Znh Choice (D) Zn 2+(aq) No3-(aq) W(aq) t-(aq) incorrect soluble soluble Practice uestions Related to This: PQ-14, PQ-15, and PQ-16 SQ-7. What is the net ionic equation for the reaction between Na2S(aq) and Pb(N03)z(aq)? (A) S2-(aq) + Pb 2+(aq)---* PbS(s) (B) NazS(aq) + Pb(N03)z(aq)--*2NaN03(aq) + PbS(s) (C) S2-(aq) + Pb 2+(aq) + 2N0 3-(aq)---* 2N03-(aq) + PbS(s) (D) 2Na+(aq) + S2-(aq) + Pb 2+(aq) + 2N0 3-(aq)--*2Na+(aq) + 2N0 3-(aq) + PbS(s) Knowledge Required: (I) Precipitation reactions. (2) Definition of net ionic equations. Thinking it Through: You start this problem by first determining the molecular equation. To do this, you recognize that the reactants could participate in a precipitation or double-displacement reaction and then determine the products based on solubility rules: NazS(aq) + Pb(N03)2(aq) ---* 2NaN03(aq) + PbS(s) soluble soluble soluble insoluble.... The next step will be to determine the total ionic equation. You do this by separating the soluble compounds into the dissociated ions (but keep the insoluble compound or the precipitate together): 2Na+(aq) + S2-(aq) + Pb 2+(aq) + 2N03-(aq)--*2Na+(aq) + 2N0 3-(aq) + PbS(s) Finally, you cancel all ions that appear the same as both reactants and products: ~(aq) + S2-(aq) + Pb 2 +(aq) + ~-(aq)--*~(aq) + 2~-(aq) + PbS(s) You have canceled the spectator ions (Na+ and N0 3-), leaving the net ionic equation of: S2-(aq) + Pb 2+(aq)--*PbS(s) which is choice (A). Choice (B) is the molecular equation and choice (D) is the total ionic equation. Choice (C) cancels only one spectator ion (Na+) but not the nitrate ion. Practice Questions Related to This: PQ-17, PQ-18, and PQ-19 SQ-8. Which best illustrates how the weak acid, HF, exists in aqueous solution (water molecules not ~J>o\ shown)? ~,e~'IJ c.o @t (!J fiil ~ cff9 ~ ® ~ @/ (A) (B) ~. (C) (ijfiJ (D) GT Cr20l-(aq) + H20(!) Cr201 2-(aq) -1490.3 W(aq) 0 H20(l) -285.8 (A) 272.1 kJ-mol- 1 (B) 13.7 kJ·mol- 1 (C) -13.7 kJ·mol- 1 (D) -272.1 kJ-mol- 1 PQ-25. Calculate the enthalpy of combustion of ethylene, C2H4, at 25 oc and one AH"r I kJ·mol-1 atmosphere of pressure. C2H4(g) 52.3 C2H4(g) + 302(g) --> 2C02(g) + 2H20(l) C02(g) -393.5 H20(!) -285.8 (A) -1411 kJ·mol- 1 (B) -1254 kJ-mol- 1 (C) -732 kJ·mol- 1 (D) -627 kJ·mol- 1 PQ-26. Use the given heats of formation to calculate the enthalpy change for this ~or I kJ·mol- 1 reaction. B20 3(g) -1272.8 B203(g) + 3C0Ch(g)--> 2BCh(g) + 3C02(g) COCh(g) -218.8 C02(g) -393.5 BCh(g) --403.8 (A) 649.3 kJ·mol- 1 (B) 354.9 kJ-mol- 1 (C) -58.9 kJ ·mol- 1 (D) -3917.3 kJ·mol- 1 PQ-27. Using the given thermochemical data, what is the !:iH0 for this reaction? AH"r I kJ·moi-1 2CH30H(l) + 02(g)--> HC2H302(1) + 2H20(l) CH30H(l) -238 HC2H302(1) --487 H20(!) -286 (A) 583 kJ ·mol- 1 (B) 535 kJ·mol- 1 (C) -583 kJ-mol- 1 (D) -535 kJ ·mol- 1 PQ-28. What is the !:iH0 for this reaction? AH"r I kJ·moi- 1 Hg(l) + 2Ag+(aq)--> Hg 2+(aq) + 2Ag(s) Ag\aq) 105.6 Hg 2+(aq) 171.1 (A) 65.5 kJ·mol- 1 (B) 40.1 kJ·mol- 1 (C) --40.1 kJ-mol- 1 (D) -65.5 kJ·mol- 1 PQ-29. What is the enthalpy of formation ofhydrazine, N 2H4(1)? 3N2H4(1)--> 4NH3(g) + N2(g) !:iH0 rm = -336 kJ-mol- 1 NH3(g) --46.3 (A) -521 kJ·mol- 1 (B) -112 kJ-mol- 1 (C) 50.3 kJ-mol- 1 (D) 290 kJ-mol- 1 78 i&fti ACS ----------------------------~·~u.ww· Exams--------------------~~~~~~- Heat and Enthalpy PQ-30. What is the standard enthalpy change for this reaction? Ml"r I kJ·moi- 1 Mg(s) + 2HCl(aq)---+ MgCb(aq) + H2(g) HCl(aq) -167.2 MgCb(aq) -641.6 (A) -307.2 kJ·mol- 1 (B) 307.2 Hmol- 1 (C) -474.4 kJ·mol- 1 (D) 474.4 kJ·mol- 1 79 Heat and Enthalpy --------~~--------------~~Exams :Yi; ~ A C 5 _______________________ Answers to Study Questions l.C 4. D 7. B 2. c 5. A 8. B 3. B 6. B 9. A Answers to Practice Questions l.B ll.A 21. c 2. c 12. c 22. c 3. A 13. c 23. B 4. B 14.A 24. c 5. c 15. B 25.A 6. B 16. B 26.C 7. A 17. c 27.C 8. D 18. c 28. c 9. B 19. D 29. c 10. D 20.C 30.A 80 Chapter 7- Structure and Bonding Chapter Summary: This chapter will focus on concepts related to chemical bonding. This includes both ionic and covalent bonding extending into bonding theories and applications. Although this chapter is traditionally included in first- term chemistry, the concepts in this chapter are critical to your understanding of other concepts throughout general chemistry. Specific topics covered in this chapter are: Lattice energy Electronegativity and ionic vs. covalent bonding Structures Formal charge Resonance Bond enthalpy, bond length and bond strength Molecular shape and geometry Polarity Valence bond theory Molecular orbital theory Previous material that is relevant to your understanding of questions in this chapter include: Atomic orbitals and number of valence electrons (Chapter 2) Balancing equations (Chapter 4) Enthalpy and sign conventions (Chapter 6) Common representations used in questzons re ated to this material: Name Example Used in questions related to ··a·· Structures, formal charge, Lewis dot structures ~N-N=C=o"·.. /...· resonance, bond enthalpy, shape,._o.. polarity, valence bond theory Valence bond theory, molecular Atomic orbitals orbital theory Molecular orbital diagrams Molecular orbital theory ,---,. ,/ r choice (C). Choice (A) was incorrectly determined by using half of the carbon/carbon double bond. Choice ( ~)was incorrectly determined by not considering the carbon-carbon bond (neither breaking the double bond nor n aking the single bond). Choice (D) was incorrectly determined by not including the number of bonds broken (so 154 kJ + 614 kJ) or number of bonds made (so - 347 kJ + -485 kJ). Practice Ouestions Related to This: PQ-13 and PQ-14 An 0=0 bond is than an 0-0 bond. (A) longer and stronger (B) longer and weaker (C) shorter and stronger (D) shorter and weaker Knowledge Required: ( 1) Understanding of covalent bonds. Thinking it Through: You could approach this by knowing facts about bonds, but you also know that you can approach this by considering a single versus a double bond in the context of valence bond theory. A singlf bond is often the overlap of two hybrid atomic orbitals, such as shown below for the 0-0 bond with sp hybrid at01 1ic /~, orbitals to form a o bond : 0 0 A double bond is often the overlap of two unhybridized p orbitals to form an: bond in addition to 11e o bond: Therefore, now you can explain how a double bond is stronger than a single bond because there is )P=Ro additional overlap ofunhybridized p orbitals (thus eliminating (B) and (D)). What is not immediately obvi us from this is that double bonds are shorter than single bonds (and stronger bonds are typically shorter) or ch tee (C). Practice uestions Related to This: PQ-15 and PQ-16 SQ-8. What is the molecular geometry of C0 32-? (A) bent (B) T -shaped (C) trigonal bipyramidal (D) trigonal planar.............................................. ·····-······--···-·--·----···-·····-·-···-···-···· ···········-···-···-···-···-···-···-·""''-----··-···· ······-···-···-···-···-···-······ ················-··· ··············-···-········--·-···-···-···-· Knowledge Required: ( I) Draw Lewis dot structures. (2) Molecular geometry. Thinking it Through: The first thing you notice when you read this is that you are going to need to determ 1e the shape, and the next thing you notice is that this is for an ion. Earlier, you walked through the process to de :.rmine the Lewis dot structure for a molecule, now you will walk through this with an ion (including using formal harge to "place" the charges). Again, you know that to begin the process to determine a Lewis dot structure, you viii need to determine the number of valence electrons for the ion: In one ion of CO/ -: I atom of C x 4 valence electrons for each C atom = 4 electrons 3 atoms of 0 x 6 valence electrons for each 0 atom = 18 electrons - 2 negative charge = 2 electrons Total = 24 electrons Now you will need to place the atoms. For this simple ion (only one central 0 atom), the central atom is listed first/the least electronegative atom and all other atoms are terminal. Following this, bond all of the terminal atoms t o the central o- I c - -o atom with sin le bonds. 86 Structure and Bonding.. ·····- ·····························-···-······ ························································································-··········································- ··········-······. Next, complete the octets of each atom by adding lone pairs of electrons. :o: For en h oxygen atom, you will add three lone pairs and for carbon, you will add one. :o-c-o: I Now, you count the number of electrons shown in your structure so far. Forth ; one, you now have 26 which is more than 24. To decrease to 24, you will need to make a multiple bond. You can add the :o: doubl1 bond between carbon and any oxygen. Then you will again complete the octets for all atoms. :o-c

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