2008 AP Chemistry Practice Exam - Advanced...

2008 AP Chemistry Practice Exam
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Unformatted text preview: Advanced PlacementProgramAP ChemistryPractice ExamThe questions contained in this AP Chemistry Practice Exam are written to the content specifications ofAP Exams for this subject. Taking this practice exam should provide students with an idea of their generalareas of strengths and weaknesses in preparing for the actual AP Exam. Because this AP ChemistryPractice Exam has never been administered as an operational AP Exam, statistical data are not availablefor calculating potential raw scores or conversions into AP grades.This AP Chemistry Practice Exam is provided by the College Board for AP Exam preparation. Teachersare permitted to download the materials and make copies to use with their students in a classroom settingonly. To maintain the security of this exam, teachers should collect all materials after their administrationand keep them in a secure location. Teachers may not redistribute the files electronically for any reason. 2008 The College Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central,SAT, and the acorn logo are registered trademarks of the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All other products and services maybe trademarks of their respective owners. Visit the College Board on the Web: www.collegeboard.com.ContentsDirections for Administration ............................................................................................ iiSection I: Multiple-Choice Questions ................................................................................ 1Section II: Free-Response Questions .............................................................................. 21Student Answer Sheet for Multiple-Choice Section ...................................................... 32Multiple-Choice Answer Key........................................................................................... 33Free-Response Scoring Guidelines.................................................................................. 34The College Board: Connecting Students to College SuccessThe College Board is a not-for-profit membership association whose mission is to connectstudents to college success and opportunity. Founded in 1900, the association iscomposed of more than 5,000 schools, colleges, universities, and other educationalorganizations. Each year, the College Board serves seven million students and theirparents, 23,000 high schools, and 3,500 colleges through major programs and services incollege admissions, guidance, assessment, financial aid, enrollment, and teaching andlearning. Among its best-known programs are the SAT, the PSAT/NMSQT, and theAdvanced Placement Program (AP). The College Board is committed to the principlesof excellence and equity, and that commitment is embodied in all of its programs,services, activities, and concerns.Visit the College Board on the Web: www.collegeboard.com.AP Central is the official online home for the AP Program: apcentral.collegeboard.com.-i-AP ChemistryDirections for AdministrationThe AP Chemistry Exam is 3 hours and 5 minutes in length and consists of a multiple-choice section and a freeresponse section.The 90-minute multiple-choice section contains 75 questions and accounts for 50 percent of the finalgrade.The 95-minute free-response section contains 6 questions and accounts for 50 percent of the final grade.Part A is timed and is 55 minutes long; Part B is 40 minutes long.A 10-minute break should be provided after Section I is completed. Students should be given a 10-minutewarning prior to the end of each of the Parts A and B in Section II of the exam.The actual AP Exam is administered in one session. Students will have the most realistic experience if a completemorning or afternoon is available to administer this practice exam. If a schedule does not permit one time periodfor the entire practice exam administration, it would be acceptable to administer Section I one day and Section IIon a subsequent day.Many students wonder whether or not to guess the answers to the multiple-choice questions about which they arenot certain. It is improbable that mere guessing will improve a score. However, if a student has some knowledgeof the question and is able to eliminate one or more answer choices as wrong, it may be to the students advantageto answer such a question.The use of a calculator* is permitted ONLY during Section II, Part A of the exam. After time is called forPart A, students must place their calculators under their chairs. The use of any other electronic devices(including a cell phone) is not permitted during any portion of the exam.It is suggested that Section I of the practice exam be completed using a pencil to simulate an actualadministration. Students can use either a pencil or a pen for Section II.Teachers will need to provide paper for the students to write their free-response answers. Teachers shouldprovide directions to the students indicating how they wish the responses to be labeled so the teacher willbe able to associate the students response with the question the student intended to answer.A periodic table of the elements is provided with both Section I and Section II of the exam. For Section II,a table of standard reduction potentials and tables of commonly used equations and constants are alsoprovided.Remember that students are not allowed to remove any materials, including scratch work, from the testingsite.*Calculators cannot have QWERTY keyboards or be designed to communicate with other calculators(such as via infrared ports).-ii-Section IMultiple-Choice Questions-1-MATERIAL IN THE FOLLOWING TABLE MAY BE USEFUL IN ANSWERING THE QUESTIONS IN THISSECTION OF THE EXAMINATION.GO ON TO THE NEXT PAGE.-2-CHEMISTRYSection ITime1 hour and 30 minutesNO CALCULATOR MAY BE USED WITH SECTION I.Note: For all questions, assume that the temperature is 298 K, the pressure is 1.00 atmosphere, and solutions areaqueous unless otherwise specified.Throughout the test the following symbols have the definitions specified unless otherwise noted.T=P=V=S=H=G=R=n=M=m=temperaturepressurevolumeentropyenthalpyGibbs free energymolar gas constantnumber of molesmolarmolalL, mL =g=nm=atm=mm Hg =J, kJ =V=mol=liter(s), milliliter(s)gram(s)nanometer(s)atmosphere(s)millimeters of mercuryjoule(s), kilojoule(s)volt(s)mole(s)Part ADirections: Each set of lettered choices below refers to the numbered statements immediately following it. Selectthe one that is best in each case and then place the letter of your choice in the corresponding box on the studentanswer sheet. A choice may be used once, more than once, or not at all in each set.Questions 5-7 refer to the following molecules.Questions 1-4 refer to the following chemicalcompounds.(A)(B)(C)(D)(E)(A) CO(B) CH4CH4CCl3FH2SH2O2K2CrO4(C) HF(D) PH3(E) F25. Contains two -bonds1. Commonly used as a disinfectant for minor skinwounds6. Has the highest dipole moment2. A refrigerant implicated in the thinning of thestratospheric ozone layer7. Has a molecular geometry that is trigonalpyramidal3. A major component of the fuel known asnatural gas4. A yellow solid at room temperature and 1 atmGO ON TO THE NEXT PAGE.-3-Questions 8-11 refer to neutral atoms for which the atomicorbitals are represented below(A)(B)(C)(D)(E)8. Is in an excited state9. Has exactly five valence electrons10. Has the highest first ionization energy11. Forms an aqueous cation that is coloredGO ON TO THE NEXT PAGE.-4-Questions 12-15 refer to the chemical reactions represented below.(A) HC2H3O2(aq) + NH3(aq) C2H3O2(aq) + NH4+(aq)(B) Ba2+(aq) + SO42(aq) BaSO4(s)(C) Zn(OH)2(s) + 2 OH(aq) [Zn(OH)4]2(aq)(D) 2 K(s) + Br2(l) 2 KBr(s)(E) N2O4(g) 2 NO2(g)12. An oxidation-reduction reaction13. A precipitation reaction14. A reaction in which a coordination complex isformed15. A Lewis acid-base reaction that is not a BrnstedLowry acid-base reactionGO ON TO THE NEXT PAGE.-5-Questions 16-17 refer to various points in timeduring an experiment conducted at 1.0 atm. Heatis added at a constant rate to a sample of a puresubstance that is solid at time t0. The graph belowshows the temperature of the sample as a functionof time.(A)(B)(C)(D)(E)t1t2t3t4t516. Time when the average distance between theparticles is greatest17. Time when the temperature of the substance isbetween its melting point and its boiling pointGO ON TO THE NEXT PAGE.-6-Part BDirections: Each of the questions or incomplete statements below is followed by five suggested answers orcompletions. Select the one that is best in each case and place the letter of your choice in the corresponding box onthe student answer sheet.21. Which of the systems in equilibrium representedbelow will exhibit a shift to the left (towardreactants) when the pressure on the system isincreased by reducing the volume of the system?(Assume that temperature is constant.)(A) 2 Mg(s) + O (g) 2 MgO(s)18. Which of the following is the correct name for thecompound with formula Ca3P2 ?(A)(B)(C)(D)(E)Tricalcium diphosphorusCalcium phosphiteCalcium phosphateCalcium diphosphateCalcium phosphide2(B) SF4(g) + F2(g) SF6(g)(C) H (g) + Br (g) 2 HBr(g)19. What mass of KBr (molar mass 119 g mol1) isrequired to make 250. mL of a 0.400 M KBrsolution?(A)(B)(C)(D)(E)22(D) N2(g) + 3 H2(g) 2 NH3(g)(E) SO Cl (g) SO (g) + Cl (g)20.595 g1.19 g2.50 g11.9 g47.6 g22222. The standard enthalpy of formation, DH f , ofHI(g) is +26 kJ mol1. Which of the followingis the approximate mass of HI(g) that mustdecompose into H2(g) and I2(s ) to release20. The value of the acid-dissociation constant, Ka ,for a weak monoprotic acid HA is 2.5 106.500. kJ of energy?The pH of 0.40 M HA is closest to(A)250 g(B)650 g(C) 1,300 g(D) 2,500 g(E) 13,000 g(A)(B)(C)(D)(E)2.03.04.06.08.0GO ON TO THE NEXT PAGE.-7-CaCl2(s) Ca2+(aq) + 2 Cl(aq)23. For the process of solid calcium chloride dissolving in water,represented above, the entropy change might be expected tobe positive. However, S for the process is actually negative.Which of the following best helps to account for the net lossof entropy?(A) Cl ions are much larger in size than Ca2+ ions.(B) The particles in solid calcium chloride are more orderedthan are particles in amorphous solids.(C) Water molecules in the hydration shells of Ca2+ and Clions are more ordered than they are in the pure water.(D) The Ca2+(aq) and Cl(aq) ions are more free to movearound in solution than they are in CaCl2(s) .(E) In the solution, the average distance between Ca2+(aq)and Cl(aq) is greater than the average distancebetween Ca2+ and Cl in CaCl2(s) .. . . CH3OCH3(g) + . . . O2(g) . . . CO2(g) + . . . H2O(g)24. When the equation above is balanced using the lowest wholenumber coefficients, the coefficient for O2(g) is(A)(B)(C)(D)(E)64321GO ON TO THE NEXT PAGE.-8-25. For which of the following processes doesentropy decrease (S < 0) ?28. Which of the following is a formula for an ether?(A)(A) H2O(s) H2O(l )(B) Br2(l ) Br2(g)(C) Crystallization of I2(s) from an ethanolsolution(D) Thermal expansion of a balloon filled withCO2(g)(E) Mixing of equal volumes of H2O(l) andCH3OH(l )(B)(C)26. In a laboratory, a student wants to quantitativelycollect the CO2 gas generated by addingNa2CO3(s) to 2.5 M HCl . The student sets up theapparatus to collect the CO2 gas over water. Thevolume of collected gas is much less than theexpected volume because CO2 gas(D)(A)(B)(C)(D)is very soluble in wateris produced at a low pressureis more dense than water vaporhas a larger molar mass than that of N2 gas,the major component of air(E) has a slower average molecular speed thanwater vapor at the same temperature(E)27. What mass of Cu(s) would be produced if0.40 mol of Cu2O(s) was reduced completelywith excess H2(g) ?GasAmount(A) 13 g(B) 25 g(C) 38 g(D) 51 g(E) 100 gAr0.35 molCH40.90 molN20.25 mol29. Three gases in the amounts shown in the tableabove are added to a previously evacuated rigidtank. If the total pressure in the tank is 3.0 atmat 25C, the partial pressure of N2(g) in the tankis closest to(A)(B)(C)(D)(E)0.75 atm0.50 atm0.33 atm0.25 atm0.17 atmGO ON TO THE NEXT PAGE.-9-30. Which of the following best explains why thenormal boiling point of CCl4(l ) (350 K) is higherthan the normal boiling point of CF4(l ) (145 K) ?33. Which of the following would produce theLEAST mass of CO2 if completely burnedin excess oxygen gas?(A) The CCl bonds in CCl4 are less polar thanthe CF bonds in CF4 .(A)(B)(C)(D)(E)(B) The CCl bonds in CCl4 are weaker thanthe CF bonds in CF4 .(C) The mass of the CCl4 molecule is greaterthan that of the CF4 molecule.(A) CH2F2(B) N2H4(C) CH3OCH3(D) C2H4(E) C2H231. At which of the following temperatures andpressures would a real gas be most likely todeviate from ideal behavior?(A)(B)(C)(D)(E)10020030050050035. In an aqueous solution with a pH of 11.50 at25C, the molar concentration of OH(aq) isapproximatelyPressure(atm)5050.010.011(A)(B)(C)(D)(E)32. After 195 days, a 10.0 g sample of pure 95 Zrhas decayed to the extent that only 1.25 g of theoriginal 95 Zr remains. The half-life of 95 Zr isclosest to(A)(B)(C)(D)(E)CH4CH3OHC2H4C2H6C4H5OH34. Which of the following substances exhibitssignificant hydrogen bonding in the liquid state?(D) The electron cloud of the CCl4 moleculeis more polarizable than that of the CF4molecule.(E) The bonds in the CCl4 molecule arecovalent, whereas the bonds in the CF4molecule are ionic.Temperature(K)10.0 g10.0 g10.0 g10.0 g10.0 g3.2 1012 M3.2 103 M2.5 101 M2.5 M3.2 1011 M36. Which of the following changes to a reactionsystem in equilibrium would affect the value ofthe equilibrium constant, Keq , for the reaction?(Assume in each case that all other conditionsare held constant.)195 days97.5 days65.0 days48.8 days24.4 days(A) Adding more of the reactants to the system(B) Adding a catalyst for the reaction to thesystem(C) Increasing the temperature of the system(D) Increasing the pressure on the system(E) Removing some of the reaction productsfrom the systemGO ON TO THE NEXT PAGE.-10-Questions 37-38 refer to a galvanic cellconstructed using two half-cells and basedon the two half-reactions represented below.Zn2+(aq) + 2 e Zn(s)Fe3+(aq) + e Fe2+(aq)E = 0.76 VE = 0.77 V37. As the cell operates, ionic species that are foundin the half-cell containing the cathode includewhich of the following?I. Zn2+(aq)II. Fe2+(aq)III. Fe3+(aq)(A)(B)(C)(D)(E)I onlyII onlyIII onlyI and IIIII and III38. What is the standard cell potential for the galvaniccell?(A) 0.01 V(B) 0.01 V(C) 0.78 V(D) 1.53 V(E) 2.31 VGO ON TO THE NEXT PAGE.-11-Ionization Energies for Element X1stIonization Energy(kJ mol1)2nd3rd4th5th6th7th7871,5803,2004,40016,00020,00024,00039. The first seven ionization energies of element X are shown in the table above. On the basis of these data,element X is most likely a member of which of the following groups (families) of elements?(A)(B)(C)(D)(E)Alkaline earth metalsBoron groupCarbon groupNitrogen groupHalogen group40. Which of the following particles is emitted by anatom of 39 Ca when it decays to produce an atomof3943. A certain reaction is spontaneous at temperaturesbelow 400. K but is not spontaneous attemperatures above 400. K. If H for thereaction is 20. kJ mol1 and it is assumedthat H and S do not change appreciablywith temperature, then the value of S for thereaction isK?(B)4He21n0(C)1H1(D)(A)(A) 50. J mol1 K1(B) 20. J mol1 K1(C)0.050 J mol1 K1(D)20. J mol1 K1(E) 8,000 J mol1 K1(E) +41. At approximately what temperature will 40. g ofargon gas at 2.0 atm occupy a volume of 22.4 L?44. A sample of a solution of RbCl (molar mass(A) 1,200 K(B) 600 K(C) 550 K(D) 270 K(E) 140 K121 g mol1) contains 11.0 percent RbCl bymass. From the following information, what isneeded to determine the molarity of RbCl inthe solution?42. Which of the following aqueous solutions hasthe highest boiling point at 1.0 atm?(A)(B)(C)(D)(E)I. Mass of the sampleII. Volume of the sampleIII. Temperature of the sample0.20 M CaCl20.25 M Na2SO40.30 M NaCl0.30 M KBr0.40 M C6H12O6(A)(B)(C)(D)(E)I onlyII onlyI and II onlyII and III onlyI, II, and IIIGO ON TO THE NEXT PAGE.-12-. . . Au3+(aq) + . . . I(aq) . . . Au(s) + . . . I2(s)45. When the equation above is balanced using the lowest wholenumber coefficients, the coefficient for I2(s) is(A)(B)(C)(D)(E)8643246. A closed rigid container contains distilled waterand N2(g) at equilibrium. Actions that wouldincrease the concentration of N2(g) in thewater include which of the following?I. Shaking the container vigorouslyII. Raising the temperature of the waterIII. Injecting more N2(g) into the container(A)(B)(C)(D)(E)I onlyII onlyIII onlyI and II onlyI, II, and IIIGO ON TO THE NEXT PAGE.-13-Z X+YCS2(l) + 3 O2(g) CO2(g) + 2 SO2(g)47. A pure substance Z decomposes into twoproducts, X and Y , as shown by the equationabove. Which of the following graphs of theconcentration of Z versus time is consistent withthe rate of the reaction being first order withrespect to Z ?48. When 0.60 mol of CS2(l) reacts as completely aspossible with 1.5 mol of O2(g) according to theequation above, the total number of moles ofreaction products is(A)(B)(C)(D)(E)(A)2.4 mol2.1 mol1.8 mol1.5 mol0.75 molQuestions 49-50 refer to an experiment todetermine the value of the heat of fusion of ice. Astudent used a calorimeter consisting of a polystyrenecup and a thermometer. The cup was weighed, thenfilled halfway with warm water, then weighed again.The temperature of the water was measured, and someice cubes from a 0C ice bath were added to the cup.The mixture was gently stirred as the ice melted, andthe lowest temperature reached by the water in thecup was recorded. The cup and its contents wereweighed again.(B)(C)49. The purpose of weighing the cup and its contentsagain at the end of the experiment was to(A)(B)(C)(D)(E)(D)determine the mass of ice that was addeddetermine the mass of the thermometerdetermine the mass of water that evaporatedverify the mass of water that was cooledverify the mass of the calorimeter cup50. Suppose that during the experiment, a significantamount of water from the ice bath adhered to theice cubes. How does this affect the calculatedvalue for the heat of fusion of ice?(E)(A) The calculated value is too large because lesswarm water had to be cooled.(B) The calculated value is too large becausemore cold water had to be heated.(C) The calculated value is too small because lessice was added than the student assumed.(D) The calculated value is too small because thetotal mass of the calorimeter contents wastoo large.(E) There is no effect on the calculated valuebecause the water adhered to the ice cubeswas at 0C.GO ON TO THE NEXT PAGE.-14-F (aq) + H2O(l) HF(aq) + OH (aq)51. Which of the following molecules containsbonds that have a bond order of 1.5 ?(A)(B)(C)(D)(E)55. Which of the following species, if any, acts as aBrnsted-Lowry base in the reversible reactionrepresented above?N2O3NH3CO2CH2CH2(A) HF(aq)(B) H2O(l)(C) F (aq) only(D) Both F (aq) and OH(aq) act as BrnstedLowry bases.(E) No species acts as a Brnsted-Lowry base.52. Of the following metals, which reacts violentlywith water at 298 K?(A)(B)(C)(D)(E)AuAgCuMgRb56. What is the empirical formula of a hydrocarbonthat is 10.0 percent hydrogen by mass?(A)(B)(C)(D)(E)53. Heat energy is added slowly to a pure solidcovalent compound at its melting point. Abouthalf of the solid melts to become a liquid. Whichof the following must be true about this process?(A) Covalent bonds are broken as the solid melts.(B) The temperature of the solid/liquid mixtureremains the same while heat is being added.(C) The intermolecular forces present amongmolecules become zero as the solid melts.(D) The volume of the compound increases as thesolid melts to become a liquid.(E) The average kinetic energy of the moleculesbecomes greater as the molecules leave thesolid state and enter the liquid state.CH3C2H5C3H4C4H9C9H10Pb(s) Pb(l )57. Which of the following is true for the processrepresented above at 327C and 1 atm? (Thenormal melting point of Pb(s) is 327C.)(A)(B)(C)(D)(E)54. A steady electric current is passed through moltenMgCl2 for exactly 1.00 hour, producing 243 g ofMg metal. If the same current is passed throughmolten AlCl3 for 1.00 hour, the mass of Almetal produced is closest toH = 0TS = 0S < 0H = TGH = TS(A) 27.0 g(B) 54.0 g(C) 120. g(D) 180. g(E) 270. gGO ON TO THE NEXT PAGE.-15-N2(g) + O2(g) + Cl2(g) 2 NOCl(g)H = + 104 kJ mol158. The equilibrium system represented above is contained in a sealed, rigid vessel. Which of the following willincrease if the temperature of the mixture is raised?(A)(B)(C)(D)(E)[N2 (g)]The rate of the forward reaction onlyThe rate of the reverse reaction onlyThe rates of both the forward and reverse reactionsThe total number of moles of gas in the vessel59. If a metal X forms an ionic chloride with theformula XCl3 , then which of the followingformulas is most likely to be that of a stablesulfide of X?(A) XS2(B) X 2 S3(C) XS6(D) X(SO3 )3(E) X 2 (SO3 )3GO ON TO THE NEXT PAGE.-16-Questions 60-61 refer to the figures below. The figures show portions of a buret used in a titration of an acidsolution of known concentration with a saturated solution of Ba(OH)2. Figures 1 and 2 show the level of theBa(OH)2 solution at the start and at the endpoint of the titration, respectively. Phenolphthalein was used as theindicator for the titration.60. What is the evidence that the endpoint of thetitration has been reached?(A) The color of the solution in the buret changesfrom pink to colorless.(B) The color of the solution in the buret changesfrom blue to red.(C) The color of the contents of the flask belowthe buret changes from colorless to pink.(D) The color of the contents of the flask belowthe buret changes from blue to red.(E) The contents of the flask below the buretchange from clear to cloudy.61. The volume of saturated Ba(OH)2 used toneutralize the acid was closest to(A)(B)(C)(D)(E)6.60 mL22.80 mL23.02 mL23.20 mL29.80 mLGO ON TO THE NEXT PAGE.-17-C(diamond) C(graphite)62. For the reaction represented above, the standard Gibbs freeenergy change, G298 , has a value of 2.90 kJ mol1. Whichof the following best accounts for the observation that thereaction does NOT occur (i.e., diamond is stable) at 298 Kand 1.00 atm?(A) S for the reaction is positive.(B) The activation energy, Ea , for the reaction is very large.(C) The reaction is slightly exothermic (H < 0).(D) Diamond has a density greater than that of graphite.(E) Diamond has a heat capacity lower than that of graphite.8 H2(g) + S8(s) 8 H2S(g)65. By mixing only 0.15 M HCl and 0.25 M HCl, itis possible to create all of the following solutionsEXCEPT63. When 25.6 g of S8(s) (molar mass 256 g mol1)(A)(B)(C)(D)(E)reacts completely with an excess of H2(g)according to the equation above, the volume ofH2S(g) , measured at 0C and 1.00 atm, producedis closest to(A)(B)(C)(D)(E)30 L20 L10 L5L2L66. At 25C a saturated solution of a metal hydroxide,M(OH)2 , has a pH of 9.0. What is the value of thesolubility-product constant, Ksp , of M(OH)2(s)at 25C?(A)(B)(C)(D)(E)2 HClO + 3 O2 2 HClO464. As the reaction represented above proceedsto the right, the oxidation number of chlorinechanges from(A)(B)(C)(D)(E)0.23 M HCl0.21 M HCl0.18 M HCl0.16 M HCl0.14 M HCl5.0 10281.0 10275.0 10195.0 10161.0 10151 to +31 to +5+1 to +5+1 to +7+3 to +7GO ON TO THE NEXT PAGE.-18-71. When a solution is formed by adding somemethanol, CH3OH , to water, processes that areendothermic include which of the following?67. A student weighs out 0.0154 mol of pure, dryNaCl in order to prepare a 0.154 M NaClsolution. Of the following pieces of laboratoryequipment, which would be most essential forpreparing the solution?(A)(B)(C)(D)(E)I. Methanol molecules move water moleculesapart as the methanol goes into solution.II. Water molecules move methanol moleculesapart as the methanol goes into solution.III. Intermolecular attractions form betweenmolecules of water and methanol as themethanol goes into solution.Large crucible with lid50 mL volumetric pipet100 mL Erlenmeyer flask100 mL graduated beaker100 mL volumetric flask(A)(B)(C)(D)(E)68. In which of the following are the chemical speciescorrectly ordered from smallest radius to largestradius?(A) B < C < N(B) Ar < Xe < Kr72. Of the following gases, which has the greatestaverage molecular speed at 298 K?(C) Cl < S < S2(D) Na < Na+ < K(A)(B)(C)(D)(E)(E) K+ < Ca2+ < K69. A large piece of wood can burn slowly, but woodin the form of sawdust can combust explosively.The primary reason for the difference is thatcompared with a large piece of wood, sawdustCl2(g)NO(g)H2S(g)HCN(g)PH3(g)73. Types of hybridization exhibited by carbon atomsin a molecule of propyne, CH3CCH, includewhich of the following?(A) has a greater surface area per kilogram(B) has a greater carbon content per kilogram(C) absorbs more atmospheric moisture perkilogram(D) contains more compounds that act ascatalysts for combustion(E) contains more compounds that have higherheats of combustionI. spII. sp2III. sp3(A)(B)(C)(D)(E)70. Of the following elements, which would beexpected to have chemical properties most similarto those of sulfur, S ?(A)(B)(C)(D)(E)I onlyIII onlyI and II onlyII and III onlyI, II, and IIII onlyIII onlyI and III onlyII and III onlyI, II, and IIIBrClNPSeGO ON TO THE NEXT PAGE.-19-XY2(aq) X2+(aq) + 2 Y(aq)75. In which of the following processes arecovalent bonds broken?74. A soluble compound XY2 dissociates in wateraccording to the equation above. In a 0.050 msolution of the compound, the XY2(aq) speciesis 40.0 percent dissociated. In the solution, thenumber of moles of particles of solute per 1.0 kgof water is closest to(A)(B)(C)(D)(E)(A)(B)(C)(D)(E)Solid silver melts.Solid potassium chloride melts.Solid carbon (graphite) sublimes.Solid iodine sublimes.Glucose dissolves in water.0.150.0900.0700.0400.020END OF SECTION IIF YOU FINISH BEFORE TIME IS CALLED,YOU MAY CHECK YOUR WORK ON THIS SECTION.DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.-20-Section IIFree-Response Questions-21-INFORMATION IN THE TABLE BELOW AND IN THE TABLES ON PAGES 23-25 MAY BE USEFUL INANSWERING THE QUESTIONS IN THIS SECTION OF THE EXAMINATION.GO ON TO THE NEXT PAGE.-22-STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25CE (V)Half-reaction-F2 ( g ) + 2 e-2F2+CoAu(s)2 Cl -O2 (g) + 4H + + 4 e -2 H 2 O(l )Br2 (l ) + 2 e2 Br -2 Hg2+ + 2 e -Hg2+ + 2 e -Hg2 2+Hg(l )Ag(s)2 Hg(l )Fe 2+2 I-Cu(s)Cu(s)Cu+Sn 4+ + 2 e -Sn2+S(s) + 2 H + + 2 e -Co3+Au+e3+-+ 3eCl2 (g) + 2 e+Ag + eHg2Fe2+3+---+ 2e+e--I 2 (s ) + 2 e -Cu+ + e Cu2++ 2eCu2+--+e-H 2S(g)+Pb+ 2e-H2 ( g)2+Pb(s)Sn2++ 2e-Sn(s)-Ni(s)+ 2e-Co(s)+ 2e-Cd(s)Cr 2+Fe 2+ + 2 e -Fe(s)-Cr(s)Zn(s)H 2 ( g) + 2 OH -Mn(s)Al(s)Be(s)Mg(s)2H + 2eNi2++ 2eCo2+Cd2+CrCr3++e3+Zn-+ 3e2++ 2e-2 H 2 O(l ) + 2 eMn2+ + 2 e Al3+Be+ 3e2+Mg+ 2e2+Na + eSr2+2+Na(s)-Ca(s)-Sr(s)-Ba(s)Rb(s)+ 2e+ 2e2++ 2e-Rb + e--+Ba-+ 2e+Ca---K ( s)+-Cs(s)Li + e-Li(s)+K +eCs + e+2.871.821.501.361.231.070.920.850.800.790.770.530.520.340.150.150.140.00 0.13 0.14 0.25 0.28 0.40 0.41 0.44 0.74 0.76 0.83 1.18 1.66 1.70 2.37 2.71 2.87 2.89 2.90 2.92 2.92 2.92 3.05GO ON TO THE NEXT PAGE.-23-ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTSEvlpATOMIC STRUCTUREE = hvc = lvhl=p = mumu-2.178 10 -18jouleEn =n2Boltzmanns constant, k = 1.38 10 -23 J K -1Avogadros number = 6.022 1023 mol -1Electron charge, e = -1.602 10 -19 coulomb1 electron volt per atom = 96.5 kJ mol -1Equilibrium Constants[A - ][HA]K a (weak acid)K b (weak base)K w (water)K p (gas pressure)[HB+ ][B]pK a = - log K a , pK b = - log K bpOH = pK b + logK p = K c ( RT )u = velocityn = principal quantum numberm = massPlancks constant, h = 6.63 10 -34 J spH = - log [H + ], pOH = - log[OH - ]14 = pH + pOHDnenergyfrequencywavelengthmomentumSpeed of light, c = 3.0 108 m s-1EQUILIBRIUM[H + ][A - ]Ka =[HA][OH - ][HB+ ]Kb =[B]K w = [OH ][H + ] = 1.0 10 -14 @ 25 C= K a KbpH = pK a + log====K c (molar concentrations),S = standard entropywhere D n = moles product gas - moles reactant gasH = standard enthalpyTHERMOCHEMISTRY/KINETICSG = standard free energy S products - S reactantsDH = DHf products - DH f reactantsETnmqcCpDS =DG = DGf products - DGf reactantsDG = D H - T D S= - RT ln K = -2.303 RT log K= -n EDG = DG + RT ln Q = DG + 2.303 RT log Qq = mcDTDHCp =DTstandard reduction potentialtemperaturemolesmassheatspecific heat capacitymolar heat capacity at constant pressureEa = activation energyk = rate constantA = frequency factorFaraday's constant,ln [A ] t - ln [ A ]0 = - kt11= kt[A] t [A]0ln k ========= 96,500 coulombs per moleof electronsGas constant, R = 8.31 J mol -1 K -1= 0.0821 L atm mol -1 K -1= 62.4 L torr mol -1 K -1()= 8.31 volt coulomb mol -1 K -1- Ea 1+ ln ARTGO ON TO THE NEXT PAGE.-24-GASES, LIQUIDS, AND SOLUTIONSPVTnDmuPA = Ptotal X A , where X A =Ptotal = PA + PB + PC + ...mn=Mmoles Atotal molesK = C + 273PV1 P2V21=T1T2mD=V3kT3 RTurms ==mM12KE per molecule = mu23KE per mole = RT2r1M2=r2M1molarity, M = moles solute per liter solutionmolality = moles solute per kilogram solventDT f = iK f molalityDTb = iK b molalityp = iMRTA = abcpressurevolumetemperaturenumber of molesdensitymassvelocity=======root-mean-square speedkinetic energyrate of effusionmolar massosmotic pressurevan't Hoff factormolal freezing -point depression constantKbAabcQIqtn a P + 2 (V - nb) = nRTV=======urmsKErMpiKfPV = nRT2=========molal boiling-point elevation constantabsorbancemolar absorptivitypath lengthconcentrationreaction quotientcurrent (amperes)charge (coulombs)time (seconds)E = standard reduction potentialK = equilibrium constantOXIDATION-REDUCTION; ELECTROCHEMISTRYGas constant, R = 8.31 J mol -1 K -1= 0.0821 L atm mol -1 K -1Q=I=[ C] c [D] da[A] [B]= 62.4 L torr mol -1 K -1, where a A + b B c C + d D= 8.31 volt coulomb mol -1 K -1Boltzmann's constant, k = 1.38 10 -23 J K -1qtEcell = Ecell log K =bK f for H 2 O = 1.86 K kg mol -1RT0.0592ln Q = Ecell log Q @ 25 CnnK b for H2 O = 0.512 K kg mol -11 atm = 760 mm Hg= 760 torrnE0.0592STP = 0.00 C and 1.0 atmFaraday's constant, = 96,500 coulombs per moleof electronsGO ON TO THE NEXT PAGE.-25-CHEMISTRYSection II(Total time95 minutes)Part ATime55 minutesYOU MAY USE YOUR CALCULATOR FOR PART A.CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit ifyou do not. Attention should be paid to significant figures.1. Answer the following questions about the solubility of the salts Li3PO4 and PbCl2 . Assume that hydrolysiseffects are negligible.The equation for the dissolution of Li3PO4(s) is shown below.Li3PO4(s) 3 Li+(aq) + PO43(aq)Ksp = 3.2 109 at 25C(a) Write the equilibrium-constant expression for the dissolution of Li3PO4(s).(b) Assuming that volume changes are negligible, calculate the maximum number of moles of Li3PO4(s) thatcan dissolve in(i) 0.50 L of water at 25C(ii) 0.50 L of 0.20 M LiNO3 at 25CThe equation for the dissolution of PbCl2 is shown below.PbCl2(s) Pb2+(aq) + 2 Cl(aq)Ksp = 1.6 105 at 25C(c) Calculate the concentration of Cl(aq) in a saturated solution of PbCl2 at 25C .(d) An open container holds 1.000 L of 0.00400 M PbCl2 , which is unsaturated at 25C. Calculate theminimum volume of water, in mL, that must evaporate from the container before solid PbCl2 canprecipitate.GO ON TO THE NEXT PAGE.-26-2. Answer the following using chemical concepts and principles of the behavior of gases.(a) A metal cylinder with a volume of 5.25 L contains 3.22 g of He(g) and 11.56 g of N2(g) at 15.0C.(i) Calculate the total pressure in the cylinder.(ii) Calculate the partial pressure of N2(g) in the cylinder.(b) A 1.50 L container holds a 9.62 g sample of an unknown gaseous saturated hydrocarbon at 30Cand 3.62 atm.(i) Calculate the density of the gas.(ii) Calculate the molar mass of the gas.(iii) Write the formula of the hydrocarbon.(iv) Calculate the root-mean-square speed of the gas molecules in the container at 30C.(Note: 1 J = 1 kg m2 s2)3. A student performs a titration in which a 10.00 mL sample of 0.0571 M HCl is titrated with a solution of NaOHof unknown concentration.(a) Describe the steps that the student should take to prepare and fill the buret for the titration given a wet50.00 mL buret and the materials listed below.0.0571 M HCl solutionIndicator solutionNaOH(aq) (unknown concentration)Distilled water10.5 M NaOH solution100 mL beaker(b) Calculate the pH of the 0.0571 M HCl .(c) A volume of 7.62 mL of the NaOH solution was needed to reach the end point of the titration. Calculate themolarity of the NaOH solution used in the titration.In a different titration using a different NaOH solution, the concentration of NaOH was determined by thestudent to be 0.0614 M.(d) Given that the actual concentration of the NaOH solution was 0.0627 M, calculate the percent error of thestudents result.(e) Calculate the volume of 10.5 M NaOH that is needed to prepare 250.0 mL of 0.0627 M NaOH .STOPIf you finish before time is called, you may check your work on this part only.Do not turn to the other part of the test until you are told to do so.-27-CHEMISTRYPart BTime40 minutesNO CALCULATORS MAY BE USED FOR PART B.Answer Question 4 below. The Section II score weighting for this question is 10 percent.4. For each of the following three reactions, in part (i) write a balanced equation and in part (ii) answer the questionabout the reaction. In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutionsare aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances areextensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may usethe empty space at the bottom of the next page for scratch work, but only equations that are written in the answerboxes provided will be graded.(a) Equal volumes of 0.1 M hydrofluoric acid and 0.1 M potassium hydroxide are combined.(i) Balanced equation:(ii) Draw the complete Lewis electron-dot diagram for the reactant that is the Brnsted-Lowry base in theforward reaction.________________________________________________________________________________________________________________________________________________________________________GO ON TO THE NEXT PAGE.-28-(b) Solid calcium metal burns in air.(i) Balanced equation:(ii) Predict the algebraic sign of H for the reaction. Explain your prediction.__________________________________________________________________________________________________________________________________________________________________________(c) Samples of nitrogen monoxide gas and oxygen gas are combined.(i) Balanced equation:(ii) If the reaction is second order with respect to nitrogen monoxide and first order with respect to oxygen,what is the rate law for the reaction?__________________________________________________________________________________________________________________________________________________________________________GO ON TO THE NEXT PAGE.-29-Answer Question 5 and Question 6. The Section II score weighting for these questions is 15 percent each.Your responses to these questions will be graded on the basis of the accuracy and relevance of the information cited.Explanations should be clear and well organized. Examples and equations may be included in your responses whereappropriate. Specific answers are preferable to broad, diffuse responses.5. Answer the following questions relating to the galvanic cell shown in the diagram above.(a) Write the balanced equation for the overall cell reaction.(b) Calculate the value of E for the cell.(c) Is the value of G for the overall cell reaction positive, negative, or 0 ? Justify your answer.(d) Consider the cell as it is operating.(i) Does Ecell increase, decrease, or remain the same? Explain.(ii) Does G of the overall cell reaction increase, decrease, or remain the same? Explain.(iii) What would happen if the NaNO3 solution in the salt bridge was replaced with distilled water?Explain.(e) After a certain amount of time, the mass of the Ag electrode changes by x grams. Given that the molarmass of Ag is 108 g mol1 and the molar mass of Co is 59 g mol1, write the expression for the changein the mass of the Co electrode in terms of x.GO ON TO THE NEXT PAGE.-30-6. Answer each of the following using principles of atomic or molecular structure, and/or intermolecular orintramolecular forces.(a) Explain why the HOH bond angle in H2O is less than the HNH bond angles in NH3 , as shown inthe table below.HOHBond Angle in H2OHNHBond Angles in NH3104.5107(b) Explain why the radius of the Br atom is less than the radius of the Br ion, as shown in the table below.Radius of BrRadius of Br 0.111 nm0.196 nm(c) Explain why the dipole moment of HI is less than the dipole moment of HCl, as shown in the table below.Dipole Moment of HIDipole Moment of HCl0.42 debye1.08 debyes(d) Explain why the normal boiling point of Ne is less than the normal boiling point of Kr , as shown in thetable below.Normal Boiling Point of NeNormal Boiling Point of Kr27 K121 KSTOPEND OF EXAMIF YOU FINISH PART B OF SECTION II BEFORE TIME IS CALLED,YOU MAY RETURN TO PART A OF SECTION II IF YOU WISH,BUT YOU MAY NOT USE A CALCULATOR.-31-Name: ____________________________________AP ChemistryStudent Answer Sheet for Multiple-Choice SectionNo.1AnswerNo.31AnswerNo.612326233363434645356563666737678386893969104070114171124272134373144474154575164617471848194920502151225223532454255526562757285829593060-32-AnswerAP ChemistryMultiple-Choice Answer KeyNo.1CorrectAnswerDNo.31CorrectAnswerANo.61CorrectAnswerD2B32C62B3A33B63B4E34B64D5A35B65E6C36C66D7D37E67E8B38D68C9C39C69A10A40E70E11E41C71C12D42B72D13B43A73C14C44C74B15C45D75C16E46C17C47E18E48D19D49A20B50C21E51B22D52E23C53B24C54D25C55D26A56C27D57E28E58D29B59B30D60C-33-AP ChemistryFree-Response Scoring GuidelinesGeneral Scoring PrinciplesIn regard to mathematical errors, a 1-point deduction is made; this deduction may be applied only onceper question. In regard to errors in reporting significant figures, a 1-point deduction is made; thisdeduction may be applied only once per question. A leeway of plus or minus one significant figuredifferent from the correct number of significant figures is allowed before a 1-point deduction is made.For questions including parts that refer back to previous parts of the same question, a wrong answerthat derives from the correct use of a previously calculated incorrect answer should not be counted aswrong. The emphasis of scoring is on process, and an error made early on in a multipart calculationshould not jeopardize the earning of full credit for the correct use of that incorrect value later in thesame question.-34-AP ChemistryFree-Response Scoring GuidelinesQuestion 1Answer the following questions about the solubility of the salts Li3PO4 and PbCl2 . Assume that hydrolysiseffects are negligible.The equation for the dissolution of Li3PO4(s) is shown below.Li3PO4(s) 3 Li+(aq) + PO43(aq)Ksp = 3.2 109 at 25C(a) Write the equilibrium-constant expression for the dissolution of Li3PO4(s).Ksp = [Li+] 3[PO43]One point is earned for the correct expression.(b) Assuming that volume changes are negligible, calculate the maximum number of moles of Li3PO4(s) thatcan dissolve in(i) 0.50 L of water at 25CLet x represent the molar concentration of Li3PO4 present in asaturated solution of Li3PO4. Then Ksp = 3.2 10 -9 = (3x)3(x) = 27 x 4 ,3.2 10 -9= 3.3 10 -3 M. Therefore the number of moles27of Li3PO4(s) that can dissolve in 0.50 L of water isthus x =4-30.50 L 3.3 10 mol Li3 PO 4= 1.7 10 3 mol1.0 LOne point is earned for acorrect setup using theKsp expression frompart (a).One point is earned for acorrect calculation ofmoles of Li3PO4 .(ii) 0.50 L of 0.20 M LiNO3 at 25CIn 0.20 M LiNO3 , [Li+] = 0.20 M. Assume that the amount ofLi+(aq) contributed to the solution by the Li3PO4(s) that dissolves isnegligibly small compared to 0.20 M. Then let y represent the molarconcentration of PO43(aq) present in the solution due to the Li3PO4(s)that dissolves in the 0.20 M LiNO3 . ThenKsp = 3.2 10 -9 = (0.20)3(y) y =3.2 10 -9= 4.0 10 -7 M.3(0.20)One point is earned forrecognizing that [Li+] isapproximately equal to0.20 M in the solution towhich the Li3PO4(s) wasadded.So [PO43(aq)] = [Li3PO4(aq)] = 4.0 10 -7 M in 0.20 M LiNO3 (aq).Therefore, the number of moles of Li3PO4(s) that can dissolve in0.50 L of water is 0.50 L 4.0 10 -7 mol Li3 PO 4= 2.0 10 7 mol1.0 L-35-One point is earned for acorrect calculation of molesof Li3PO4 .AP ChemistryFree-Response Scoring GuidelinesQuestion 1 (continued)The equation for the dissolution of PbCl2 is shown below.PbCl2(s) Pb2+(aq) + 2 Cl(aq)Ksp = 1.6 105 at 25C(c) Calculate the concentration of Cl(aq) in a saturated solution of PbCl2 at 25C.Let z = [Pb2+] in a saturated solution of PbCl2 .Then Ksp = [Pb2+] [Cl] 2 = ( z )( 2 z )2 = 4z3.Thus 1.6 105 = 4z3 z=3One point is earned for a correct setup.1.6 10 -54One point is earned for a correct calculationof the value of [Cl]. z = [Pb2+] = 1.6 102 M [Cl] = 2[Pb2+] = 2(1.6 102 M) = 3.2 102 M(d) An open container holds 1.000 L of 0.00400 M PbCl2 , which is unsaturated at 25C. Calculate theminimum volume of water, in mL, that must evaporate from the container before solid PbCl2 canprecipitate.1.000 L of 0.00400 M PbCl2(aq) contains 0.00400 mol of PbCl2(aq),thus it contains 0.00400 mol Pb2+(aq) and 0.00800 mol Cl(aq).Let V = volume of the solution at saturation, then(0.00400 mol 0.00800Ksp = 1.6 105 = VVV=)2=2.6 10 -7V3-732.6 10= 0.25 L = 250 mL at saturation1.6 10 -5One point is earned for thecalculation of the saturationvolume.Thus the volume of water that must evaporate = 1,000. 250 = 750 mLORFrom part (c), the amount of PbCl2 dissolved in 1 L of saturated PbCl2(aq)is 1.6 102 M . Let V = volume of the solution at saturation, then0.016 mol PbCl 20.00400 mol PbCl2=1,000 mLVV = 250 mLThus the volume of water that must evaporate = 1,000. 250 = 750 mL-36-One point is earned forsubtracting the saturationvolume from 1,000 mL.AP ChemistryFree-Response Scoring GuidelinesQuestion 2Answer the following using chemical concepts and principles of the behavior of gases.(a) A metal cylinder with a volume of 5.25 L contains 3.22 g of He(g) and 11.56 g of N2(g) at 15.0C.(i) Calculate the total pressure in the cylinder.3.22 g He 1.00 mol He= 0.805 mol He4.00 g He11.56 g N2 One point is earned for thecalculations of moles and addingthem together.1.00 mol N 2= 0.4126 mol N228.02 g N 2One point is earned for the correctsubstitution into the gas law equation.total moles of gas = (0.805 + 0.4126) = 1.218 molP=nRT(1.218 mol)(0.0821 L atm mol1 K 1 )(15 + 273) K=V5.25 LP = 5.49 atmOne point is earned for thecorrect answer.(ii) Calculate the partial pressure of N2(g) in the cylinder.pressure N2 ==moles N 2 gas total pressure in flasktotal moles of gas0.4126 mol 5.49 atm = 1.86 atm1.218 molOne point is earned for the calculationof the mole fraction of N2 .One point is earned for the correctanswer.(b) A 1.50 L container holds a 9.62 g sample of an unknown gaseous saturated hydrocarbon at 30C and3.62 atm.(i) Calculate the density of the gas.D=m9.62 g== 6.41 g L11.50 LVOne point is earned for the correct answer.-37-AP ChemistryFree-Response Scoring GuidelinesQuestion 2 (continued)(ii) Calculate the molar mass of the gas.mLet M = molar mass, then PV = nRT = RT M1mRT(9.62 g)(0.0821 L atm K mol1 )(303 K)M=== 44.1 g mol1PV(3.62 atm)(1.50 L)One point is earned forthe correct answer.(iii) Write the formula of the hydrocarbon.Saturated hydrocarbons have the generic formula CnH2n + 2 ,therefore let 44.1 g = 12(n) + 1(2n +2) = 14n + 242.1 = 14 n3=nOne point is earned forthe correct answer.C3H8(iv) Calculate the root-mean-square speed of the gas molecules in the container at 30C.( Note: 1 J = 1 kg m2 s2 )rms =3RT=M3(8.31 kg m 2s 2 mol1 K 1 )(303 K)0.0441 kg mol1One point is earned for the correctsetup using the molar mass inkilograms.One point is earned for thecorrect answer.= 414 m s1-38-AP ChemistryFree-Response Scoring GuidelinesQuestion 3A student performs a titration in which a 10.00 mL sample of 0.0571 M HCl is titrated with a solution ofNaOH of unknown concentration.(a) Describe the steps that the student should take to prepare and fill the buret for the titration given a wet50.00 mL buret and the materials listed below.0.0571 M HCl solutionIndicator solutionNaOH(aq) (unknown concentration)Distilled water10.5 M NaOH solution100 mL beakerRinse the buret with some distilled water and then pour someof the NaOH solution of unknown concentration into thebeaker and use it to rinse the buret two times.Use the beaker to carefully fill the buret with the NaOHsolution of unknown concentration.Put the beaker under the buret and drain enough solution toremove any air bubbles in the neck and tip of the buret.One point is earned for rinsing theburet with the titrant solution.One point is earned for removing airbubbles from the neck and tip of theburet.(b) Calculate the pH of the 0.0571 M HCl .HCl is a strong acid [H+] in 0.0571 M HCl = 0.0571 MpH = log [H+] = log (0.0571) = 1.243One point is earned for thecorrect answer.(c) A volume of 7.62 mL of the NaOH solution was needed to reach the end point of the titration. Calculatethe molarity of the NaOH solution used in the titration.mol HCl titrated = 10.00 mL 0.0571 mol HCl= 0.000571 mol;1,000 mLratio HCl:NaOH in neutralization is 1:1, so 0.000571 mol NaOH reacted;One point is earned forcalculating the moles ofHCl in the sample that wastitrated.One point is earned forcalculating the molarity ofthe NaOH solution.0.000571 mol NaOH= 0.0749 M NaOH0.00762 L-39-AP ChemistryFree-Response Scoring GuidelinesQuestion 3 (continued)In a different titration using a different NaOH solution, the concentration of NaOH was determined by thestudent to be 0.0614 M.(d) Given that the actual concentration of the NaOH solution was 0.0627 M, calculate the percent error of thestudents result.percent error =0.0614 - 0.0627 100 = 2.1 %0.0627One point is earned for the correctanswer.(e) Calculate the volume of 10.5 M NaOH that is needed to prepare 250.0 mL of 0.0627 M NaOH.250.0 mL 0.0627 mol1,000. mL0.0157 mol NaOH = 0.0157 mol NaOH needed1,000. mL= 1.49 mL10.5 mol NaOH-40-One point is earned for calculating themoles of NaOH needed.One point is earned for the correctanswer.AP ChemistryFree-Response Scoring GuidelinesQuestion 4For each of the following three reactions, in part (i) write a balanced equation and in part (ii) answer thequestion about the reaction. In part (i), coefficients should be in terms of lowest whole numbers. Assume thatsolutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances areextensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may usethe empty space at the bottom of the next page for scratch work, but only equations that are written in theanswer boxes provided will be graded.(a) Equal volumes of 0.1 M hydrofluoric acid and 0.1 M potassium hydroxide are combined.(i) Balanced equation:HF + OH F + H2O(ii) Draw the complete Lewis electron-dot diagram for the reactant that is the Brnsted-Lowry base in theforward reaction.(b) Solid calcium metal burns in air.(i) Balanced equation:2 Ca + O2 2 CaO(ii) Predict the algebraic sign of H for the reaction. Explain your prediction.The sign of H will be negative because G is negative (the reaction occurs) and S is negative_ __(a solid and a gas react to form a solid). According to the Gibbs-Helmholtz equation, H = G + TS.Therefore H is the sum of two negative quantities and as such must be negative.__-41-__ ___AP ChemistryFree-Response Scoring GuidelinesQuestion 4 (continued)(c) Samples of nitrogen monoxide gas and oxygen gas are combined.(i) Balanced equation:2 NO + O2 2 NO2(ii) If the reaction is second order with respect to nitrogen monoxide and first order with respect tooxygen, what is the rate law for the reaction?2__ __rate = k[NO] [O2]________________________________________________________________________________________________________________________ _ ____General Scoring Notes for Question 4Five points are earned for each of parts (a), (b), and (c), distributed as follows.Four points are earned for part (i): one point for the correct reactants, two points for the correctproduct(s), and one point for the correct coefficients in the balanced equation.One point is earned for the correct answer in part (ii).-42-AP ChemistryFree-Response Scoring GuidelinesQuestion 5Answer the following questions relating to the galvanic cell shown in the diagram above.(a) Write the balanced equation for the overall cell reaction.One point is earned for the correct equation.2 Ag+(aq) + Co(s) 2 Ag(s) + Co2+(aq)(b) Calculate the value of E for the cell.One point is earned for the correct value ofEcell = 0.80 (0.28) = 1.08 VEcell .(c) Is the value of G for the overall cell reaction positive, negative, or 0 ? Justify your answer.The value of G for the overall reaction must be negativebecause the cell reaction occurs (is spontaneous) as the celloperates.ORSince Ecell is positive and G = nFE , the value of Gmust be negative.-43-One point is earned for the correctanswer, including a valid justification.AP ChemistryFree-Response Scoring GuidelinesQuestion 5 (continued)(d) Consider the cell as it is operating.(i) Does Ecell increase, decrease, or remain the same? Explain.As the cell operates, the concentration of Ag+ decreasesand the concentration of Co2+ increases [Co2+ ]increases ln Q increases the ratio Q =[Ag+ ]2Ecell = Ecell -One point is earned for the correctanswer, including a valid justification.RTln Q becomes smaller (decreases).nF(ii) Does G of the overall cell reaction increase, decrease, or remain the same? Explain.The value of G for the system increases (becomes lessnegative) as the cell operates and the system approachesequilibrium (when G = 0).One point is earned for the correctanswer, including a valid justification.(iii) What would happen if the NaNO3 solution in the salt bridge was replaced with distilled water?Explain.The cell would not operate. The voltage of the cell is toosmall to overcome the electrical resistance of distilled water,which is a poor conductor due to its very low concentrationof ions (about 107 M H+(aq) and 107 M OH(aq)) thatcould carry the current from one cell to the other.One point is earned for the correctanswer, including a valid justification.(e) After a certain amount of time, the mass of the Ag electrode changes by x grams. Given that the molarmass of Ag is 108 g mol1 and the molar mass of Co is 59 g mol1, write the expression for the change inthe mass of the Co electrode in terms of x. mol Ag = mass Ag 1 mol Ag1x=x=108 g Ag108108 mol Co = mol Ag x1x1 mol Co==1082162 mol Ag259 g C o59x= 59 = x mass Co = mol Co 2161 mol Co216-44-One point is earned for using thecorrect mole ratio of Co to Ag .One point is earned forthe correct answer(negative sign is not required).AP ChemistryFree-Response Scoring GuidelinesQuestion 6Answer each of the following using principles of atomic or molecular structure and/or intermolecular orintramolecular forces.(a) Explain why the HOH bond angle in H2O is less than the HNH bond angles in NH3 , as shown in thetable below.HOHBond Angle in H2OHNHBond Angles in NH3104.5107Both molecules have tetrahedral electron-domain geometries andmight be expected to have bond angles of 109.5. However, electrondomains for nonbonding pairs of electrons exert a greater repulsion onadjacent pairs of electrons than do electron domains for bonding pairs.Thus, in the H2O molecule with its two nonbonding pairs ofelectrons, the electron domains of bonding pairs are compressed to agreater extent than they are in the NH3 molecule, which has only onenonbonding pair of electrons.One point is earned for citing thedifference in number ofnonbonding pairs of electrons.One point is earned for citing thegreater repulsion from nonbondingpairs as compared with bondingpairs.(b) Explain why the radius of the Br atom is less than the radius of the Br ion, as shown in the table below.Radius of BrRadius of Br 0.111 nm0.196 nmThe nuclear charge (+35) is the same for both the Br and Brspecies, but the extra electron in Br causes the electroncloud to expand due to an increase in mutual repulsions amongthe electrons that make up the cloud.-45-One point is earned for recognition thatBr and Br have the same nuclearcharge.One point is earned for citing increasedrepulsion among electrons.AP ChemistryFree-Response Scoring GuidelinesQuestion 6 (continued)(c) Explain why the dipole moment of HI is less than the dipole moment of HCl , as shown in the table below.Dipole Moment of HIDipole Moment of HCl0.42 debye1.08 debyesIodine, having a lower electronegativity than chlorinehas, forms a bond with hydrogen that is less polar thanthe bond between chlorine and hydrogen in HCl. Thelower polarity of the H I bond means that the dipolemoment of the bond is smaller than that of the H Clbond.One point is earned for citing the differencein electronegativity between I and Cl .One point is earned for the comparison of thepolarity of the two bonds.(d) Explain why the normal boiling point of Ne is less than the normal boiling point of Kr, as shown in thetable below.Normal Boiling Point of NeNormal Boiling Point of Kr27 K121 KThe intermolecular forces among atoms in liquid Ne are the sametype of forces as those among atoms in liquid Kr, namely London(dispersion) forces. However, the magnitudes of these forces aresmaller in Ne because the electron clouds of Ne atoms aresmaller and less polarizable than the electron clouds of Kr atoms.Note: An explanation that cites only periodic trends or only therelative masses of Ne and Kr does not earn credit.-46-One point is earned formentioning that intermolecularforces involved are London(dispersion) forces.One point is earned formentioning the relativepolarizability of the electronclouds of the atoms....
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