chemistry released exam 1994

7/25/2019 Chemistry Released Exam 1994

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1994

CHEMISTRY

Three hours re allotted or th is examination: 1 hour and 30 m inutes or Section , which consists f multiple-choicequestions, nd 1 hour and 30 minutes or Section I, which consists f problems nd essay uestions. ection is printedin this examination oo klet; Section I, in a separate ooklet.

Battery-operated and-held alculators may be used n both sections f the examination. ll calculator memoriesmustbe cleared of both programs nd data; no peripheral evices uch s magn etic ards r tapes will be allowed.Calculatorsmay not be shared.

SECTION I

Time - 1 hour and 30 m inutesNumberof questions 75Percentof total grade 45

This examination contains 75 multiple-choice questions. herefore, please be careful to fill in only theovals that are preceded by numbers 1 through 75 on your answer sheet.

General nstructions

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE INSTRUCTED TO DO SO.

INDICATE ALL YOUR ANSW ERS TO QUESTIONS IN SEC TION I ON THE SEPARATE ANSW ER SHEET.No credit will be given for anything written n this examination oo klet, but you may use he booklet or notes orscratchwork. fter you have decidedwhichof the suggested nswers s best, COM PLETELY fill in the correspondingoval on the answer heet. Give only one answer o each question. f you change n answe r, e sure ha t he previousmark s erased ompletely.

Example: SampleAnswer

Chicago s a

(A) state(B) city(C) country(D) continent(E) village

Many candidates onder whether or not to gu ess he answers o questions bout which hey are not certain. n thissection f the examination, s a correction or haph azard uessing , ne-fourth f the number of q uestions ou answerincorrectlywill be subtracted rom the number of questions ou answer orrectly. t is improbable, herefore, hat mereguessing ill improve your score significantly; t m ay even ower your score, nd t does ake time. f, however, you arenot sure of the correct answer ut have some knowledge f the question nd are able o eliminate one or more of theanswer hoices s wrong, your chance of getting he right answer s improved, nd t m ay be to your advantage o answersuch a question.

Use your tim e effectively, work ing as rapidly as you can without osing accuracy. Do not spend oo much ime onquestions hat are too difficult. Go on to other questions nd come back o the difficult ones ater f you have ime. t isnot expected hat everyone will be able o answer ll the multiple-choice uestions.

Copyright 0 1994 by Educational Testing Service. All rights reserved.


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Note: For all questions nvolving solutions nd/or chemical equations, ssum e hat the system s in pure water and atG temperature unless otherwise stated.

Part A

Directions: Each set of lettered choices below refers to the num bered questions r statements mm ediately followingit. Select the one lettered choice that best answers ach question or best fits each statement nd then fill in thecorresponding val on the a nswer sheet. A choice may be u sed once, more than once, or not at all in each set.

Questions -4

(4(B)K>0)03

1.

2.

3.

4.

Heisenberg uncertainty principlePauli exclusion principleHun ds rule (principle of m axim um multiplicity)Shielding effectWave nature of matter

Can be used to predict that a gaseous arbon atomin its ground state s paramagnetic

Explains the experimental phenomenon of electron

diffraction

Indicates h at an atom ic orbital can hold no morethan two electrons

Predicts hat it is impossible o determine simulta-neously he exact position and the exact velocity ofan electron

Questions -7 refer to the phase diagram below of apure substance.

P(atm)

lb io

do iio i30-

T (Q

(A) Sublimation(B) Condensation(C) Solvation(D) Fusion(E) Freezing

5. If the temperature ncreases rom 10 C to 60 C at aconstant ressure of 0.4 atmosphere, which of theprocesses ccurs?

6. If the temp erature decreases rom 110 C to 40 C ata constant ressure of 1.1 atmospheres, hich ofthe processes ccurs?

7. If the pressure ncreases rom 0.5 to 1.5 atmospheresat a constant emperature of 50 C, which of theprocesses ccurs?

q 11a


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Ques tions -10 refer to the following diatomic species.

(A) Liz(B) I%(C) N0) 02(E) FZ

8. Has the largest bond-dissociation nergy

9. Has a bond order of 2

10. Contains 1 sigm a (CT) nd 2 pi (7~) bonds

Questions 11-13

(4 Pb(W Cacc> n0% As(E) Na

11. Utilized as a coating to protect Fe from corrosion

12. Is added o siliconsemiconductor

to enhance ts properties as a

13. Utilized a s a shield rom sources f radiation


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Part B

Directions: Each of the questions r incomplete statemen ts elow is followed by five suggested ns wers or com pletions.Select the one that is best n each case and then fill in the corresponding val on the answe r sheet.

14. Which of the following is lower for a l.O-molar 16. Comm ercial vinegar was titrated with N aOH

15

aqueouswater?

solution of any olute han it is for pure

(A) PH

(B) Vapor pressure(C) Freezing point(D) Electrical conduc tivity(E) Absorption of visible light

In a m olecule n which the central atom exhibitssp3d2 hybrid orbitals, he electron pairs are directedtoward the corners of

(A) a tetrahedron(B) a square-based yramid(C) a trigonal bipyramid(D) a square(E) an octahedron

17.

solution o determine the content of acetic acid,HC2H 302 For 20.0 m illiliters of the vinegar,26.7 m illiliters of 0.600-molar NaOH solution wasrequired. Wha t w as the concentration f aceticacid in the vinegar f no other acid was present?(A) 1.60 A4(B) 0.800 M(C) 0.600 M(D) 0.450 A4(E) 0.200 M

Relatively slow rates of chemical reaction are asso-ciated with which of the following?

(A) The presence of a catalyst(B) High temperature(C) High concentration f reactants(D) Strong bonds n reactant molecules(E) Low activationenergy


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18.

19.

20.

2 HZ0 + 4 Mn04- + 3 C102- + 4 MnOz +3 ClOh- + 4 OH-

Which species cts as an oxidizing agent n theabove?

(A) Hz0(B) Clod-(C) ClOz-(D) Mn02(E) Mn04-

reaction epresented

In which of the following compo unds s the ma ss atio of chrom iumto oxygen closest o 1.62 to 1.00 ?

(A) Cr-03(B) CrOr(C) CrO(D) Cr,O(E) CrzO,

. . . Ag + . . . ASH,(~) + . . . OH- + . . . Ag(s) + . . . H3As03(aq) + . . . Hz0

When the equation above s balanced with lowest whole-number coefficients, he

coefficient for OH- is(A) 2(B) 4(C) 5(D) 6(E) 7


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2 1. Correct statements bout alpha particles ncludewhic h of th e following?

I. They have a mass number of 4 and a chargeof +2.

II. They are more penetrating han beta particles.III. They are helium nuclei.

(A) I only(B) III only

(C) I and II(D) I and II(E) II a nd III

22. HSO,- + HZ0 =+ H30+ + SO:-

In the equilibrium represented bove, he speciesthat act as bases nclude which of the following?

I. HS04II. HZ0

III. sod2-

(A) II only

(B) III only(C) I an d II(D) I a nd III(E) II a nd III

23. Step 1: Ce4+ + Mn2++ Ce3++ Mn3+

Step 2: Ce4+ + Mn3++ Ce3++ Mn4+

Step 3: Mn4+ + Tl + T13+ Mn2+

The proposed steps or a catalyzed reactionbetween Ce4+and Tl are represented bove. Theproducts of the overall catalyzed reaction are

(A) Ce4+and Tl(B) Ce3+and T13+(C) Ce3+ nd Mn3+(D) Ce3+and Mn4+(E) T13+ nd Mn2

24. A sample of 0.0100 mole of oxygen gas s confinedat 37 C and 0.216 atm osphere. Wha t w ould be thepressure of this sample at 15 C and the samevolume?

(A) 0.0876 atm(B) 0.175 atm

(C) 0.201 atm(D) 0.233 atm(E) 0.533 atm


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25. Hz(g) + $ Wg) + HzO@) AH = -286 kJ

2 Na(s) + i O ?(g) + NalO(s) AH = -414 kJ

Na(s)+$Ol(g)+~H~(g)+NaOH(s) AH=-425kJ

Based on the information above, what is the standard nthalpychange or the following reaction?

NazO(s) + H,O( ?) + 2 NaOH(s)

(A) -1,125 kJ(B) -978 kJ(C) -722 kJ(D) -150 kJ(E) +275 kJ

26. Wh ich of the following actions wou ld be likely tochange he boiling point of a sam ple of a pureliquid in an open container?

I. Placing it in a smaller containerII. Increasing he numb er of moles of the liquid in

the containerIII. Moving the container and liquid to a higher

altitude

(A) I only(B) II only(C) III only(D) II and III on ly(E) I, II, and III

27. Which of the following sets of quantum numbers(n, I, mp, ms) best describes he valence electron ofhighest energy in a groun d-state allium atom(atomic number 3 1) ?

(A) 4, 0, 0, ;

(B) 4, 0, 1, ;

(C) 4, 1, 1, ;

(D) 4, 1, 2,;

(E) 4, 2, 0, ;


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28.

29.

30.

Given that a solution s 5 percent sucrose y mass , whatadditional nformation s necessary o calculate hemolarity of the solution?

I. The density of waterII. The density of the solution

III. The molar mas s of sucrose

(A) I only(B) II only(C) III only(D) I andII1(E) II andIII

When an aqueous olution of NaOH is added oan aqueous olution of potassium dichromate,K&O ,, the dichromate on is converted o

(A) CrO:-

(B) Cfl2-

(C) Cr3+

(D) CrzO&)

(E) Cr(OH)&)

Reaction Coordinate

The energy diagram for the reaction X + Y + Z isshown above. The addition of a catalyst o this reac-tion would cause a change n which of the indi-cated energy differences?

(A) 1only(B) II only(C) III only(D) I and II on ly(E) I, II, and II

q 17B


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31. H2C204 + 2 Hz0 ti 2 H30+ + C20$-

Oxalic acid, H2C204, is a diprotic acid withK1 = 5 .36 x lo-* and K2 = 5.3 x 10m5. or thereaction above, what is the equilibrium constant?

(A) 5.36 x lo-*(B) 5.3 x 1O -5(C) 2.8 x lO A(D) 1.9 x 10-l(E) 1.9 x lo-l3

32. CH3C H20H b oils at 78 C an d CH30C H3 boils at-24 C, although both compound s ave the samecomposition. his difference in boiling points maybe attributed o a difference in

(A) molecularmass(B) density(C) specific heat(D) hydrogenbonding(E) heat of combustion

33. A hydrocarbon gas with an em pirical formula CH2has a d ensity of 1.88 grams per liter at 0 C and 1.00atmosphere. A possible ormula for the hydro-carbon s

(B) C2H4

CC> C3He

CD) Cd-b

(E) Cd-&o

34. CH3-CH2-CH2-CH2-CH3 CH3-CH2-CH2-CH2-OH HO-CH2-CH2-CH2-OH

X Y z

Based on concepts of polarity and hydrogen bonding, which of the following sequencescorrectly lists h e compou nds bo ve n the order of their increasing solubility in water?

(A) 2 < Y < X(B) Y < 2


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35. For which of the following processes would OShave a negative value?

I. 2 Fe203(s) -+ 4 Fe(s) + 3 O?(g)

II. Mg2++ 2 OH- + Mg(OH)&)

III. H?(g) + GH&) + GH&)

(A) I only(B) I and II only(C) I and III only(D) II an d III only(E) I, II, and III

36. Zn(s) + Cu2+ Zn*+ + Cu(s)

An electrolytic cell bas ed on the reac tion repre-sented above was constructed rom zinc an d copperhalf-cells. The observed voltage was found to be1.00 volt instead of the stan dard ell poten tial, E,of 1.10 volts. Wh ich of the following could correctlyaccount or this observation?

(A) The copper electrode was larger than the zincelectrode.

(B) The Zn electrolyte wa s Zn(N 03)2, wh ile theCu+ electrolyte wa s CuS04 .(C) The Zn so lution was more concentrated han

the Cu s olution.(D) The s olutions n the half-cells had different

volumes.(E) The salt bridge contained KC1 as the elec-

trolyte.

37. A sample of 3.30 grams of an ideal gas at 15O.O Cand 1.25 atmospheres ressure has a volume of2.00 liters. Wha t is the molar mass of the gas?The gas constant, R, is 0.0821 (L l atm)/(mol l K).

(A) 0.0218 gram/mole(B) 16.2 grams/mole(C) 37.0 grams/mole(D) 45.8 grams/mole(E) 71.6 grams/mole

38. Concentrations f colored substance s re commonlymeasured by means of a spectrophotometer. hichof the following would ensure h at correct valuesare obtained or the me asured absorbance?

I. There mus t be enough samp le n the tubeto cover the entire light path.

II. The instrum ent mu st be p eriodically resetusing a standard.

III. The solutionmust be saturated.

(A) I only(B) II only(C) I and II only(D) II an d III only(E) I, II, and II

39. Samples of F2 gas and Xe gas are mixed in acontainer of fixed volum e. The initial partial pres-sure of the F2 gas s 8.0 atmospheres nd that ofthe Xe gas s 1.7 atmospheres. When a ll of the Xegas reacted, orming a solid compound, he pres-sure of the unreacted F2 gas was 4.6 atmospheres.The temperature emained constant. What is the

formula of the compound?(A) XeF(B) XeF3(C) XeF4(D) XeF6(E) XeFs


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40.

41.

ILacuum

Closed-end Manometer

The system shown above s at equilibrium at 28 C.At this temperature, he vapor pressure of w ater is28 millimeters of mercury. The partial pressure ofO,(g) in the system s

(A) 28 mm Hg(B) 56 mm Hg(C) 133 mm Hg(D) 161 mm Hg(E) 189 mm Hg

A strip of m etallic scandium, SC, s placed n abeaker containing concentrated itric acid. A

brown gas rapidly forms, the scandium disappears,and the resulting iquid is brown-yellow butbecomes colorless when warme d. These observa-tions best support which of the following state-ments?

(A)m(0

09

03

Nitric acid is a strong acid.In solution scandium nitrate is yellow and

scandium hloride is colorless.Nitric acid reacts with metals to form

hydrogen.Scandium eacts with nitric a cid to form a

brown gas.Scandium and nitric acid react in mole propor-

tions of 1 to 3.

42.

43.

44.

Mas s of an empty container 3.0 gramsMas s of the container plus

the solid sample 25.0 gramsVolume of the solid sample 11.0 cubic

centimeters

The data above were gathered n order to determinethe density of an unknow n solid. The density of thesample should be reported as

(A) 0.5 g/cm3

(B) 0.50 g/cm3

(C) 2.0 g/cm3

(D) 2.00 g/cm3

(E) 2.27 g/cm3

Which of the following pairs of c ompounds reisomers?

(A) CH3--Hz---- CHz--CH3 and CH3-CH-CH3

CH3

(B) CH3-FH-CH3 and CH3-y=CH2

CH3 CH3.

::(C) CH3-0-CH3 and CH3-C-CH3

(D) CH3-OH and CH3-CH2-OH

(E) CH4 and CH*=CH;

Which of the following solutions as the lowestfreezing point?

(A) 0.20 m CaH1206,glucose(B) 0.20 m NI-&Br(C) 0.20 m ZnS04(D) 0.20 m KMn04(E) 0.20 m MgCl2


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45. A sample of an ideal gas s cooled from 50.0 C to25.0 C in a sealed container of constant olume.Which of the following values or the gas willdecrease?

I. The average molecular mas s of the gasII. The averag edistance between he molecules

III. The averag espeed of the molecules

(A) I only(B) II only(C) III only(D) I and III(E) II a nd III

46. Which of the following solids dissolves n water toform a colorless solution?

(A) Cc13(B) FeC13(C) coc12(D) CuC12(E) ZnCl?

47. Which of the following has the lowest conductivity?(A) 0.1 A4CuSO4(B) 0.1 it4 KOH(C) 0.1 M BaC&(D) 0.1 MI-IF(E) 0.1 M HNO3

48. PC&(g > + C12(g @ PCls(g + energy

Some PC13 nd Cl2 are mixed in a container at200 C and the system eaches equilibriumaccording o the equation above. Which of the

following causes n increase n the number ofmoles of PCls present at equilibrium?

I. Decreasing he volume of the containerII. Raising he temperature

III. Adding a mole of He g as at constant olume

(A) I only(B) II only(C) I and III only(D) II and II only(E) I, II, an d III

49. The isomerization of cyc lopropane o propylene sa first-order process with a half-life of 19 minutesat 500 C. The time it takes or the partial pressureof cyclopropane o decrease rom 1.0 atmosphereto 0.125 atmosphere t 500 C is closest o

(A) 38 minutes(B) 57 minutes(C) 76 minutes(D) 152 minutes

(E) 190 minutes

50, Which of the following acids can be oxidized toform a stronger acid?

(A) H3PQ

@I HNO3

(c> H2co3

0% H3BO3

09 H2so3

51. 4 HCl(g) + 02(g) * 2 Cl,(g) + 2H&?)

Equal numbers of m oles of HC l and O2 in a closedsystem are allowed to reach equilibrium as repre-sented by the equation above. Which of thefollowing m ust be true at equilibrium?

I. [HCI] mus tbe less han [Cl,].

II. [02] mu stbe greater than [HC l].

III. [Cl,] mus t equa l [H,O].

(A) I only(B) II only(C) I and III on ly(D) II and II only(E) I, II, and II

121 II


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59. When a 1.00~gram ample of limestone wasdissolved n acid, 0.38 g ram of CO* was generated.If the rock containe d no carbonate other thanCaC0 3, what was the percent of CaC03 by ma ss nthe limestone?

(A) 17%(B) 51%(C) 64%(D) 86%(E) 100%

60. 12(g) + 3 Cl2(g) + 2 ICl,(g)

According to the da ta in the table below, what isthe value of AH0 for the reaction represen tedabove?

Bond

I-ICl-Cl

I-Cl(A) -860 kJ(B) -382 kJ(C) +180 kJ(D) +450 kJ(E) +1,248 kJ

Average Bond Energy(kilojoules/mole)

149239

208

61. A l-molar solution of w hich of the following saltshas the highest pH ?

(A) NflO3

@I Na2C03

(C> WC1

(D) NaHSO

(E) Na2SQ

62. The electron-dot structure Lewis structure) orwhich of the following molecules would have w ounshared pairs of electrons on the central atom?

(A) H2S

(B) NH3

Cl cH4

(D) HCN(E) COz

63.

64.

65.

66.

What is the maximum mass of copper that could beplated out by electrolyzing aqueous CuC1 2 or16.0 hours at a constant urrent of 3.00 am peres?(1 faraday = 96,500 coulombs)

(A) 28 grams(B) 57 grams(C) 64 grams(D) 114 grams(E) 128 grams

At 25 C, a sample of NH 3 (m olar mas s 17 gram s)effuses at the rate of 0.050 mole per m inute. Underthe same conditions, which of the following gaseseffuses at approximately one-half that rate?

(A) O2 (molar mass 32 grams)(B) He (molar mas s 4.0 grams)(C) CO2 (molar mass 44 grams)(D) Cl, (molar mas s71 grams)(E) C& (molar mass 16 grams)

Barium sulfate s LEAST soluble n a O.O l-molarsolution of w hich of the following?

(A) A12(S04)3

(W (Nb)2S04

(C> Na2S04

0) NH3

(E) BaC12

Wha t is the pH of a 1.0 x 10-2-molarsolution ofHCN ? (For HCN , Ka = 4.0 x lo-.)

(A) 10(B) Between 7 and 10(C) 7(D) Between 4 an d 7(E) 4

n 23 n


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67.

68.

69.

Substances and Y that were in a solution wereseparated n the laboratory using the technique offractional crystallization. This fractional crystalliza-tion is possible because substances and Y havedifferent

(A) boiling points(B) melting points(C) densities(D) crystal colors(E) solubilities

Which of the following molecules has a dipolemom ent of zero?

(A) C6H6 benzene)(B) NO(C) SO1(D) NH3(E) HS

Correct procedures or a titration include which ofthe following?

I. Draining a pipet by touching he tip to the sideof the container used or the titrationII. Rinsing he buret with distilled waterjust before

filling it w ith the liquid to be titratedIII. Swirling the solution requently during the titra-

tion

(A) I only(B) II only(C) I and III only(D) II and II only(E) I, II, and III

70. To determine the molar mass of a solid mono-protic acid, a student itrated a weighed sample ofthe acid with standardized queous NaOH. Whichof the following could explain why the stud entobtained a m olar mass hat was too large?

I. Failure to rinse all acid from the weighing paperinto the titration vessel

II. Addition of m ore water than was needed odissolve he acid

III. Add ition of somebase beyond he equivalencepoint

(A) I only(B) III only(C) I and I only(D) II an d III only(E) I, II, an d III

71. . . . Fe(OH)z + . . . O2 + . . . Hz0 + . . . Fe(OH)3

If 1 mole of O 2 oxidizes Fe(OH)2 according o thereaction represented bove, how many m oles ofFe(OH)j can be formed?

(A) 2(B) 3(C) 4(D) 5(E) 6

n 24 n


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72. The nuclide 249,Cm is radioactive and decays by theloss of one beta (p-) particle. The product nuclide is

(A) ;;Pu

(B) z:Arn

(C) ZiCrn

(D) ;;& -I

73. 2 sol(g) + 02(g) + 2 SO3(g)

When 0.40 m ole of SO ? and 0.60 m ole of 0 2 areplaced n an evacuated .OO -liter flask, the reactionrepresented bove occurs. After the reactantsand the product reach equilibrium an d the initialtemperature s restored, he flask is found to contain0.30 mole of S03. Based on these results, he equi-librium constant, K,, for the reaction s

(A) 20.(B) 10.(C) 6.7(D) 2.0(E) 1.2

74. A solution of calcium hypochlorite, a comm onadditive o swimming-pool water, is

(A) basic because of the hydrolysis of the OCl- ion(B) basic because Ca(OH)2 is a w eak and nsoluble

base(C) neutral if the concentration s kept below

0.1 molar(D) acidic because of the hydrolysis of the Ca

ions(E) acidic because he acid HOC 1 is formed

75. A direct-current power supply of low v oltage lessthan 10 volts) has lost the ma rkings hat indicatewhich output terminal is positive and which isnegative.A chemist suggests hat the pow er supplyterminals be connected o a pair of platinum elec-trodes that dip into O.l-molar KI solution. Whichof the fo llowing correctly identifies the polarities ofthe power supply erminals?

(A) A gas will be evolved only at the positive elec-trode.

(B) A gas will be evolved only at the negative elec-trode.

(C) A brown color will ap pear n the solution nearthe negative electrode.

(D) A metal will be deposited on the pos itive elec-trode.

(E) None of the methods above will identify thepolarities of the power supply erminals.

c

STOPEND OF SECTION I

IF YOU FINISH BEFORE TIME IS CALLED, YO U MAY CHECK YOU R WORK ON THIS S ECTION.DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.

n 25 n


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m REVISED QUANTITATIVEITEMS FROM THE 1994CHEMISTRY EXAM

24R. A sample of 0.010 mole o f oxygen gas is confinedat 127C and 0.80 atmosph ere. What would be thepressure of this sample at 27C and the samevolume?

There w ere a total of 2 0 quantitative questions n themultiple-choice section of the 1994 AP Chem istryExam . Below are 10 quantitative questions rom thisgroup rewritten to conform to the new format beingintroduced in 1996, in which calculators will not beallowed for the multiple-choice questions. The capi-tal R following the ques tion numb er indicates that itis a Revised question.) The remaining quantitativequestions from the 1994 exam are those for wh ichstudents would not typically need a calculator, andtherefore represent quantitative questions that wouldappear exactly as they are now on the new format ofthe exam.

16R. Commercial vinegar was titrated with N aOH solutionto determine the content of ac etic acid, HC,H ,O,.For 20.0 m illiliters of the vinegar, 32.0 milliliters of0.500-molar NaO H solution was required. What wa sthe concentration of acetic acid in the vinegar if noother acid was present?

(A) 1.60 M(B) 0.800 M(C) 0640M(D) 0.600 M(E) 0.400 M

19R. In which o f the following compound s is the massratio of chromium to oxygen closest to 1.6 to l.O?

(A) CrO,(B) CrO,(C) CrO(D) Cr,O(E) Cr,O,

(A) 0.10 atm(B) 0.20 atm(C) 0.60 atm

(D) 0.80 atm(E) 1.1 atm

25~ H,(g) + l/2 O,(g) - H,O@) AHO=_X

2 Na(s) + l/2 O,(g) - N$O(s) AP=y

Na(s) + l/2 O,(g) + l/2 H,(g) - NaOH (s) AH0 = z

Base d on the information above, what is theenthalpy change for the following reaction?

N%O(s) + H,O@) - 2 Na0I-W

(A) x+y+z(B) x+y-z(C) x+y-22(D) 22-x-y(E) z-x-y

31R. H,C,O, + 2 H,O = 2 H,O+ + C,O,2-

Oxalic acid, H&O ,, is a diprotic acid withK, =5 X lo-*andK,=5 X10-5.Whichofthefollowing is equal to the equilibrium constant for thereaction represe nted above?

(A) 5 X lo-*(B) 5 X10-(C) 2.5 x 10(D) 5 x1O-7(E) 2.5 X10-*


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37R. A sample of 3.0 gram s of an ideal gas at 127C and 63R. Wh ich of the following expressions s correct for1 O atmosphere pressure has a volume of 1.5 liters. the maximum mass of copp er, n grams, that couldWhich of the following expressions s correct for the be plated out by electrolyzing aqueous CuCl, formolar m ass of the g as? The ideal gas constant, R, is 16 hours at a constant current of 3.0 amperes?0.08 (L= atm)/(moleeK). (1 faraday = 96,500 coulombs)

(A) (0.08)(400)(3.0)( 1 O)(1.5)

(B) (1.0)(1.5)(3.0)(0.08)(400)

(C) (0.08)( 1 O)(1.5)(3.0)(400)

(D) (3.0)(0.08)(400)(1.0)(1.5)

(E) (3.0)(0.08)( 1.5)(1.0)(400)

59R. When a 1.25-gram sam ple of limestone was

dissolved in acid, 0.44 gram of C O, was generated.If the rock contained no carbonate other thanCaC O,, what w as the percent of CaCO , by m ass inthe limestone?

(A) 35%(B) 44%(C) 67%(D) 80%(E) 100%

60R. I,(g) + 3 Cl,(g) - 2 ICl,(g)

According to the data in the table below, what is thevalue of AHo for the reaction represented above?

BondAverage Bond Energy

(kilojoules/mole)

I-I 150

Cl - Cl 240

I - Cl 210

(A) - 870kJ

(B) -390kJ(C) + 180 kJ(D) +45OkJ(E) +1,260 kJ

(A) (16)(3,600)(3.0)(63.55)(2)(96,500)

(B) (16)(3,600)(3.0)(63.55)(96,500)(2)

(C) (16)(3,600)(3.0)(63.55)(96,500)

(D) (16)(60)(3.0)(96,500)(2)(63.55)

(E) (16)(60)(3.0)(96,500). (63.55)(2)

73R. 2 SO,(g) + Q,(g) = 2 SO,(g)

When 0.40 mole of SO, and 0.60 mole of 0, areplaced in an eva cuated l.OO-liter flask, the reactionrepresented above occurs. After the reactants and theprod uct reach equilibrium and the initial tempe ratureis restored, he flask is found to contain 0.30 mole ofSO,. Based on these results, the expression or theequilibrium constant, K,, of the reaction is

(A) (o.30)2

(0.45)(0. 1o)2

(B) (o.30)2

(0.60)(0.40)2

(C) (2 x 0.30)(0.45)(2 X 0.10)

(D) (0.30)(0.45)(0.10)

(E) (0.30)(0.60)(0.40)

I Answers toRevised Questions

I

I 16R. B

19R. BI 24R. C

25R. D31R. C37R. D59R. D60R. B63R. B73R. A


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1994CHEMISTRY

SECTION IITime-l hour and 30 minutes

Percent of total grade-55

Parts A, B, and C: Suggested ime-50 minutes

Part D: Suggested ime--40 minutesGeneral nstructions

The suggested imes will not be announced, nd you may proceed reely from one question o the next. Do no tspend oo long on any one problem.

Pages containing a periodic table and the electrochemical eries are printed on the green nsert and in the pinkessay booklet for yo ur use.

You may write your answ ers with either a pen or a pencil. Be sure o write CLEAR LY and LEG IBLY. If youmake an error, you m ay save time by c rossing t out rather than rying to erase t.

When you are told to begin, open your book let, carefully tear out the green nsert, and start w ork. Th e questions

are also printed in your ess ay booklet, but it m ay be easier o w ork from the insert when answering questions.Writeall 0 r answ av booklet. Num ber vour answe rs s the auestions are numbered n theu ers n the ink essexamination book.

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INFORMATION IN THE FOLLOWING TABLES MAY BE USEFUL IN ANSWERING THE QUESTIONS INTHIS SECTION OF THE EXAMINATION.

Universal gas constant: R = 8.31 joules (mole - K) = 0 .0821 liter-atm/(mole - K)= 62.4 liter-mm Hg/(mole - K) = 1.99 calories/(mole - K)= 8.3 1 (volt)(coulombs)/(mole - K)

1 faraday ( 3 ) = 96,50 0 coulombs = 23.06 0 calories/volt = 96,5 00 joules/volt1 caloric = 4.184 joules1 electron volt/atom = 2 3.1 kilocalories/mole = 9 6.5 kilojoules/mole

Speed of light in vacuu m = 2.998 x lo8 m/setIn, = 2.303 log,,Plancks cons tant It = 6.63 x 10 -34 oule * setBoltzmanns constant k = 1.38 x 10 joule/KAvogadros num ber = 6.022 x 10 13 molecules/mole

At 25 C, R T In Q =0.059 1

n3 ~ log Qn

STANDARD REDUCTION POTENTIALS, E, IN WATER SOLUTION AT 25 C(in V)

Li. + e

Cs + e

K++eRb feBa2+ + 2eSr + 2e

Ca + 2 eNa + em

Mg + 2eBe + 2 e

Al + 3eMn + 2e

Zn + 2 eCr + ?e

Fe + 2 eCr- + e

Cd + 2 e

Tl + eCo + 2 eNi + 2 eSn + 2ePb + 2 e2H +2e

S (s ) + 2 H + 2 eSn4 + 2eCu + eCu+ + 2e

Cu. +e

I:(s) + 2 eFe+ + e

Hg, + 2eAg +eHg + 2em

2Hg + 2e

Br,(Q) + 2eO ,(g ) + 4H + 4 e

Cl,(g) + 2 eAu* + 3 eCo+ + e

FJg) + 2 e

Li(s) - 3.05

Cs(s) - 2.92

K(s) - 2.92Rb(s) - 2.92

Ba(s) - 2.90Ws) - 2.89

Q(s) - 2.87

Na(s) -2.71

Mg(s) - 2.37Be(s) - 1 I0Al(s) - 1.66Mn(s) - 1.18

Zn(s) -076

Cr(s) - 0.74Fe(s) - 0.44Cr -0.41Cd(s) - 0.40

Tl(s) - 0.34Co(s) - 0.28Ni(s) - 0.25

Sri(s) -0.14

Pb(s) -0.13

Hz(g) 0.00

H$ 0.14S n + 0 . 15Cu 0.15

Cu(s) 0.34Cu(s) 0.5221 0.53Fe 0.772 Hg(Q) 0.79

Ag(s) 0.80

Hg(Q) 0.85

Hgz + 0.922 Br- 1.072 H,O I 232 Cl 1.36Au(s) 1.50co:+ I 822F 2.87

n 31 n


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CHEMISTRY

SECTION II

Time-l hour and 30 minutes

The percentages iven for the parts represent he score weightings or this section of the examination. Spend about50 m inutes on Parts A, B, and C combined and about 40 minutes on Pa rt D.

THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS MUST BE SHOWCLEA RLY. It is to your advantage o do this, since you may obtain partial credit if you do and you will receivelittle or no credit if yo u do not. Attention should be paid to significant igures.

Be sure to write your answ ers n the s pace provided following each question.

Data necessary or the solution of the problems may be found in the tables on the preceding pages.

Part A

(20 percent)

Solve the following problem.

1. MgW) + Mg*+(aq) + 2 F-(aq)

In a sa turated olution of Mg F7 at 18 C, the concentration f M g*+ is 1.21 x 10m 3 olar. The equilibrium srepresented y the equation above.

(4

(b)

(4

w

Write the expression or the solubility-product constant, Ksp, and calculate ts value at 18 C.

Calculate he equilibrium concentration f Mg in 1 OO O iter of sa turated MgF 2 solution at 18 C towhich 0.100 mole of solid KF has been added. The K F dissolves ompletely. Assume he volume changeis negligible.

Predict whether a precipitate of MgF 2 will form whe n 100.0 milliliters of a 3.00 x 10-3-molarMg(N03)2solution s mixed with 2 00.0 m illiliters of a 2 .00 x 10-3-molarNaF solution at 18 C. Calculations o sup portyour prediction must be shown.

At 27 C the concentration f Mg in a saturated olution of MgF 2 is 1.17 x 10T3molar. Is the dissolving ofMgF 2 in water an endothermic or an exothermic process? Give an explanation o support your conclusion.

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Part B

(20 percent)

Solve EITHER problem 2 OR problem 3 in this part. (A second problem will not be scored.)

2 2 NO(g) + 2 Hz(g) + N(g) + 2 I-W(g)

Experiments were conducted o study he rate of the reaction represented y the equation above. nitial concen tra-tions and rates of reaction are given in the table below.

(a> (i) Determine the order for e ach of the reactants, NO and HZ, from the data given and show your reasoning.(ii) Write the overall rate law for the reaction.

(b)

w

Calculate the value of the rate con stant, k, for the reaction. nclude units.

For experiment 2, calculate he conce ntration f NO remaining when exactly one-half of the original amo untof H2 had been consum ed.

Cd)The following sequence f elementary steps s a proposed mec hanism or the reaction.

Initial Concentration Initial Rate of Formation(mom) of N2

Experiment [NOI B321 (mol/L*min)

1 0.0060 0.0010 1.8 x 1O-42 0.0060 0.0020 3.6 x 1O-43 0.0010 0.0060 0.30 x 1o-44 0.0020 0.0060 1.2 x 1o-4

I. NO + NO+=N 202II. N202 + H2 + H20 + N20

III. N20 + H2 + N2 + H20

Based on the da ta presented, which of the above s the rate-determining step? Show that the m echanism sconsistent with

(i) the observed ate law for the reaction, and(ii) the overall stoichiome try of the reaction.

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rr as SampleWater

3. A student ollected a sample of hydrogen gas by the displacement f water as shown by the diagram above. Therelevant data are given in the following table.

GAS SAMPLE DATAI

Volume of sample 90.0 mL

Temperature 25 c

AtmosphericPressure 745 mm Hg I

EquilibriumVapor Pressureof Hz0 (25 C) 23.8 mm Hg

(a) Calculate he number of moles of hydrogen gas collected.

(b) Calculate he number of molecules of water vapor in the sample of gas.

(c) Calcu late he ratio of the average speed of the hydrogen molecules o the average speed of the water vapormolecules n the sam ple.

(d) Which of the two gases, H2 or H20, deviates more from ideal behavior? Explain your answer.

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Part C

( 15 percent)

4. Answ er FIVE of the eight options n this part. (Answers o more than five options will not be scored.)

Give the formulas to show he reactants nd the products or FIVE of the following chemical reactions. Each ofthe reactions occurs n aqueous olution unless otherwise ndicated. Represent ubstances n solution as ions f thesubstance s extensively onized. Om it formu las for any ions or molecules hat are unchanged y the reaction. nall cases a reaction occurs. You need not balance.

A strip of m agnesium s added o a solution of silver nitrate.xample:

(4

0-9

(4

Cd)

Cd(0

(g)

(h)

Mg + Ag -+ Mg*+ + Ag

Excess sodium cyanide solution s added o a solution of silver nitrate.

Solutions of mangan ese(I1) ulfate and ammo nium sulfide are m ixed.

Phosphorus(V) oxide powder s sprinkled over distilled water.

Solid am monium carbonate s heated.

Carbon dioxide gas s bubbled hrou gh a concentrated olution of potassium hydroxide.A concentrated olution of hydrochloric acid is added o so lid potassium permanganate.

A small piece of sodium metal is added o distilled water.

A solutionof potassium dichromate s added o an acidified solution of iron(I1) chloride.


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Part D

(45 percent)

Spend about 40 minutes on this part of the examination. Answ ering hese questions rovides an opportunity todemonstrate our ability to present your material in log ical, coheren t, and convincing English. Your responses ill bejudged on the basis of accuracy and importance of the detail cited and on the appropriateness f the descriptivematerial used. Specific answ ers re preferable o broad, diffuse responses. llustrative examples and equations may behelpful.

ANSW ER THE FOLLOWING ESSAY QUESTION.

5 Discuss h e following phenom ena n terms of the chem ical and physical properties of the substances nvolved andgeneral p rinciples of chemical and physical change.

(a) As the system shown on the rightapproaches quilibrium, what changeoccurs o the volume of water inbeaker A ? What happens o theconcentration f the sugar solutionin beaker B ? Explain why thesechanges occur.

(b) A bell jar connected o a vacuum pumpis shown on the right. As the airpressure nder the bell jar decreases,what behavior of water in the beakerwill be observed? xplain why this occurs.

ToVacuum

Pump

PureH2O

(c) What w ill be observed on the surfacesof zinc an d silver strips shortly afterthey are placed n separate olutions fCuS04, as shown on the right? Account forthese ob servations.

(d) A water solution of I2 is shaken with anequal volume of a nonpolar solvent suchas TTE (trichlorotrifluoroethane). Describethe appearance f this system after shaking.(A diagram may be helpful.) Account for thisobservation.

I, in Water

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SELECT TWO OF THE FOUR ESSAY QUESTIONS, NUM BERED 6 THROUGH 9.(Additional essays will not be scored.)

6. 2 HIS(g) + SO*(g) + 3 S(s) + 2 HzO(g)

At 298 K, the standard nthalpy change, AH for the reaction represented bove s -145 kilojoules.

64

(b)

(4

(4

Predict the sign of the standard ntropy change, ASo, for the reaction. Explain the basis or your prediction.

At 298 K , the forward reaction i.e., toward the right) is spon taneous. hat chan ge, f any, would occur inthe value of A G for this reaction as the temperature s increased? Explain your reasoning using thermody-namic p rinciples.

Wh at change, f any, would occur in the value of the equilibrium constant, Keq, for the situation describedin (b)? Explain your reasoning.

The a bsolute emperature at which the forward reaction becomes nonspontaneous an be p redicted. Write theequation hat is used o make the prediction. Why does this equation predict only an approximate value forthe temperature?


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7. A chemical reaction occurs when 100. milliliters of 0.200-molar HCl is added dropwise o 100. milliliters ofO.lOO-molarNa3P0 4 solution.

Write the two net ionic equations or the formation of the major products.

Identify the species hat acts as both a Bronsted acid and as a Bronsted base n the equations n (a). Draw theLewis electron-dot diagram for this species.

Sketch a graph using the axes provided, show ing he shape of the titration curve that results when 100. millilitersof the WC1solution s add ed slowly from a bu ret to the Na3P0 4 solution. Account for the shape of th e curve.

0 mL HCl

Write the equation or the reaction hat occurs f a few additional milliliters of the HCl solution are added othe solution esu lting rom the titration in (c).

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8. For each of the following, use appropriate chemical principles o explain the observation.

(a) Sodium chloride may be spread on an icy sidewalk n order to m elt the ice; equimolar amo unts of calciumchloride are even more effective.

(b) At room temperature, NH3 is a g as and HZ0 is a liquid, even though NH3 has a m olar mass of 17 grams andHz0 has a molar mass of 18 grams .

(c) C (graphite) s us ed as a lubricant, whereas C (diamond) is used as an abrasive.

Cd)Pouring vinegarphenomenon.

onto the white residue nside a kettle used for b oiling w ater results n a fizzing/bubbling

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9. Use principles of atomic structure and/or chemical bonding o answer each of the following.

(a) The rad ius of the Ca atom is 0.197 n anometer; he radius of the Ca2+ on is 0.099 nanom eter. Account orthis difference.

(b) The lattice energy of CaO (s) is -3,460 kilojoules per mole; the lattice energy for M20(s) is -2,240 kilojoulesper m ole. Account for this difference.

Ionization Energy(kJ/mol)

First Second

K 419 3,050

Ca 590 1,140

(c) Explain the difference between Ca a nd K in regard o

(i) their first ionization energies,(ii) their second onization energies.

(d) The first ionization energy of M g isAccount o r this difference.

738 kilojoules per mole and that of Al is 578 kilojoules per mole.

END OF EXAMINATION

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Chapter III

Answers to the

994 AP Chemistry Examination

n SECTION I: MULTIPLE-CHO ICEanswered an individual q uestion in this section also

Listed below are the correct answers o the mu ltiple- achieved a h igher m ean score on the test as a wh olechoice questions and the percentage of AP candidates than candidates who did not answ er that questionwho attempted each question and answered it cor- correctly. An answer sheet gridded with the correctrectly. A s a general rule, candidates who correctly responses ppears on the next page.

Section I Answer Key and Percent Answering Correctly

c tem Correct Percent Item Correct Percent Item Correct PercentNo. Answer Correct No. Answer Correct No. Answer Correct1234567

89

10111213141516171819202122232425

C

EBAA

BBCDC

CDACEBDEBDDEBCD

32%

38%

44%

82%

73%

74%

66%

21%47%

57%

52%

21%

78%

46%50%

69%82%

62%

83%

58%58%

: 62%71%

76%

63%

26272829303132

333435363738394041424344454647484950

C

C

EADCD

CEDCDCC

C

DDAEC

EDABE

61%48%

58%

36%

55%

39%77%

52%39%54%

46%

81%

41%

38%

67%

58%

39%

55%31%

44%

46%

38%35%

49%

58%

51525354555657

585960616263646566676869707172737475

DBEDCBA

BDBBABDA

DEACACEAAB

29%

29%

55%

43%35%

35%

46%

29%29%

47%

21%

64%

24%

23%21%

64%25%

41%

36%

15%33%

53%

24%34%

16%


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H SECTION II: FREE-RESPONSE

Report of the Chief Faculty Consu ltantRobert W. GleasonMiddlebury College

Grading the ExaminationThe free-response section of the AP ChemistryExamination is read an d scored by faculty consultants- AP Chemistry teachers and college chem istry pro-fessors who are under the direction of a chemistryteacher designated as the chief faculty co nsultan t. Thefaculty consultants are from secondary schools andcolleges throughout the U nited States, and also fromCanada. The faculty con sultants o not have access o,and therefore are not influenced by, the multiple-choice section of the examination, which is scoredseparately by machine. Student scores on both parts ofthe examination are com bined and used by the chieffaculty consultant to determine the levels of studentperformance on the AP 1 to 5 grading scale.

The ReadingIn June the faculty con sultants meet for six days on acollege cam pus o score the free-response sections ofthe AP Chemistry Examination. The chief facultyconsultant divides the faculty cons ultants nto groups,each under the direction of a designated aculty con-sultant called a table leader. One or m ore table leaderswith their group of faculty con sultants s assigned oscore each free-response question, depending on thenumb er of studen ts who chose to answer t. The tableleaders train their groups to score their designatedquestion, and scoring of student papers comm encesaccording to standards eveloped as described below.Each a nswer booklet is circulated among the variousgroups until all the student responses n that booklethave been scored. The finished booklets are thenremoved from the Read ing site and the scores areentered nto computers and m atched with the students

multiple-choice section scores. Comp osite scores arecalculated and these and other data are provided to thechief faculty consultant or the grade-setting session,which occurs shortly after the Reading is over.

Developing Free-ResponseScoring StandardsScoring standards or the free-response questions area consideration throughout the development of the

examination. Mem bers of the AP Chemistry Develop-ment Committee sub mit suggested scoring standardswith each question that they write. Scoring standardsare discussed urther when questions are revised andchosen by the comm ittee to be included in an exam i-nation. At this stage, consideration s given to potentialdifficulties that might interfere with the reliable scoring

of a question, and the scoring standard may b e revisedaccordingly. Prior to the Reading, the chief facultyconsultant generates a draft of the scoring standard oreach of the nine questions n the free-response section,taking into con sideration ssues aised during previousreviews. The general scoring guide or the free-responsequestions s as follows:

Problems 9 points eachChem ical Reactions 3 points each (15 p oints otal)Essays 8 points each

Two days before the R eading begins, the chief fac-ulty consultant meets with the table leaders to reviewthe draft standards, nd the group reaches a consensuson a possible standard or each question. The tableleaders break into groups o test the standards gainsta n umber of student responses. During this phase ofthe process, he standards may be modified som ewhat.Meeting again as a whole group, the chief facultyconsultant and the table leaders reach an other consen-sus on the stand ards, after which each table leader isassigned o a particular question and is also assignedlist of faculty consultants with whom he or she willscore the qu estion during the R eading.

On the first day of the Reading, the table leaderstrain the faculty consultants n app lying the scoringstandards o a set of sam ple student esponses electedfor that purpose. During this process, he standards rerefined and may be modified slightly once again.After the group is proficient in applying the scoringstandards, the actual scoring of the student papersbegins. The final results of the rigorous standard-setting procedure described above are standards h atcan be applied eliably not only to the common methods

of solution seen in the studen t responses, but a lso tocommon errors and o alternative or unusu al pproach es.

Because various student esponses o a given ques-tion are scored over a six-day period by more than onefaculty con sultant, t is important to monitor the appli-cation of the standards during the Reading. This isdone by h aving a certain numb er of student papersindependently graded more tha n once, either by dif-ferent faculty consu ltants, or by the same consultant at


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a later time. The original sco res are concealed fromsubsequ ent aculty consultants, and the two sets ofscores are then com pared. The checking of one con-sultant against another quickly identifies any rem ain-ing ambiguities that may exist in the standards andallows their further refinement, to help assure hat astudents score is independen t of the person scoring

the pap er. Checking a consultant against his or herown work also helps assure hat a students score isindependent of what day or time the paper is scored.Rarely is there a discrepancy of more than 1 point onthe scale. Other procedures help maintain con sistentscoring standards. After the Read ing is underwa y, thetable leaders select and score another series of ques-tions for yet another kind of consistency check. Thisone does not compare the faculty consultants withthemselves, but rather it com pares hem w ith the otherfaculty consultants in their group and individuallywith the table leaders.

The philosophy of the faculty consultants n sco ringthe free-response questions s to aw ard credit for cor-rect work. Wh en questions nvolve calculations, mostof the points awarded are given for setting up thesolution correctly rathe r than actually carrying ou t thecomputation. Partial credit is awarded within eachpart of a question, so students hould be encouraged oshow their work. Faculty consultants ry to determinewhether an incorrect answe r to a p revious part hasbeen correctly used n a sub sequent art of a question.Full credit for the latter part may be a warded if the

consultant can successfully trace the students workto make that determination. Students should also beencouraged o c ontinue on to later parts of a questionif they get stuck at some point. Parts of a qu estion areoften independ ent of each other; even when they aredependent, credit can be earned on later parts whenearlier answe rs are missing. Also, a stud ents explan a-tion of w hat he or she would do, if possible, could earnsome credit. Finally, when final answ ers are numeri-cal, students hould pay attention to significant figuressince 1 point is deducted (once per problem) if thenumbe r of significant figures in a students answ er

differs by more than one from the appropriate number.

n FREE-RESPONSE QUESTIONS,SCORING GUIDES, AND SAMPLESTUDENT ANSWERS

On the pages that follow are a selection of studentresponses o each of the questions that made up thefree-response section of the 1994 Adv anced Place-

ment Examination in Chemistry. Also included arethe stand ards hat were applied in the scoring process,and an explanation of why each response eceived thescore t did. For each question, two student esponseshave been selected to illustrate a superior answ er andone of some what ower quality.

From our experience in read ing Advanced Place-ment Examinations we know that no set of scoringstandards an possibly anticipate the creativity of highschool students n developing solutions to problems.Therefore, you should understand hat readers makeevery pos sible effort to give credit for every respo nse

that reflects an understanding of basic principles ofchemistry regardless of how far the approach useddeviates from wha t might be a more conv entionalroute developed in the standa rds.

In developing the standards or the 1994 AP Chem -istry Exam , the chief faculty consultant had the assis-tance of 11 faculty consu ltants who served as tableleaders, and two test development sp ecialists fromETS. The group met for two days prior to the Reading.After a draft set of standard s was e stablished, theapplication of the stand ard for each question wastested with abo ut 100 papers. The standards were thenreviewed in the light of that experience and eitherrevised accordingly or adopted.

If a student made an error in part (a) of a four-partproblem and the answ er to part (a) wa s essential toworking the rest of the problem, the reader of thepaper was obliged to work through the solutions o thesubsequent parts of the question with the erroneousanswer to part a. Thus, every effort was made toreward the students with points for the appropriateapplication of chemical principles. Students were, how-ever, penalized for mathem atical errors and errors in

significant figures (exceeding one too m any or one toofew) to the maximum extent of 1 point for an error ofeach kind on any one problem.

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Question 1

This required problem provided students with an opportunity to demonstrate heir understanding of an ionicequilibrium and the common ion effect, to make a prediction based on their calculations, and to relate solubilitydata to thermodynamics.

Scoring Standards

(4 Ksp - [Mg*+J[F-]* (1 pt.>- (1.21 x 10 3)(2 x 1.21 x 10 3)2- 7.09 x 10-9 U pt.1

Note: if number of significant figures in final answer differs bymore than one from the appropriate number, 1 point isdeducted ONCE PER PROBLEM.

(W KsP - [M$+](2x + O.lOO)* 2x IQ*+] . 100.0 x 3.00 x 103 - 300 . 0 x [Mg2+]I%*+1 - 1.00 x 1O'3 M

[F] : 200.0 x 2.00 x 1O-3 - 300.0 x [F-l

WI - 1.33 x loo3 M

Q - Ion Product - [Mg*+][F']*- (1.00 x 103)(1.33 x lo-S)*- 1.77 x 10'9

Since Q < Ksp , no precipitate will form I**Note:**

conclusion must be consistent with Q value.

Correct substitution and calculation of the wrongconcen tration values earns the second point, but notthe first.

(1 pt.) if bothconcentrations arcorrect

(1 pt.>

(1 ptJ


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(d) Solubility of MgF, decreases with increasingtemperature, thus dissolution process is exotherm ic

MgFz(s) + Mg2+ + 2 F- + Q (cr H)

(1 pt.>

Reason:

EITHER

i) Increased tempe rature puts a stress o n the system(Le Chatelier). The system will reduce the stressby shifting the equilibrium in the endoth ermic (left)direction

0%(1 pt*)

ii) a data supported argument such as comparing ion

concentrations, calculating second Ksp and givingproper interpretations. i

n 46m


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Comment: his p aper earned all of the 9 possible points. Although the student did not explicitly indicate that thesimplifying approximation was being made in part b, it clearly was; and the substitution nd calculation were madecorrectly. The calculations n part c were clear and correct and the conclusion based on the relationship betweenQ, and K,, was unambiguous. The analysis and conclusion n part d were straightforward and to the point.


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Sample Student Response 2

Comm ent: n part a of this question, the student ost a point for failing to squ are he concentration of the fluorideion in the calculation. This w as a common flaw in part a. Although the student had an incorrect value for K,, inpart a, he or she earned ull credit in part b for using t correctly in the calculations. The student ost a point in partc for the mathem atics error made in calculating Q,. Although the analysis n part d is somewhat unusu al, t is clearthe student understan ds he implications of raising the temperature on a system n equilibrium in wh ich the forwardreaction is exothermic, thus full credit was awarde d for this part. The total score for this paper was 7.


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Question 2

Students requently did parts a and b correctly (although many omitted units n part b). Credit w as given in (ii) ofpart a if the rate law given was consistent with the kinetic orders of (i). The fact that the initial concentration s ofthe reactants n p art c were not equal led many students o miss this part. Many students who lost points in part csimply failed to take advantage of the simp le stoichiometry but tried to p lug concentrations nto the rate law.Part d w as difficult for many students who tried to show that the reaction stoichiometry was consistent with therate-determining step. Many students who correctly chose Step II of the mechan ism as the rate-determining stepsimply stated hat because 2N0 = N202, then [N202] = [NO ]2 a nd lost a point as a result.

Scoring Standards

a) (i) From exps. 1 and 2: Doub ling [HZ] while keeping [NO]constant doubles the rate, therefore the reactionis first order in [HZ].

From exps. 3 and 4: Doub ling [NO] while keeping [Hz ]constant quadruples the rate, therefore the reactionis second order in [NO].

(ii) Rate - k[Hz I WI2

Note: full credit earned for (ii) as long as rate expres sionis consistent with orders in (i).

b) k - [Hz;:O]LFrom exp. 1: k -

1.8 x 10s4 M/min(1.0 x 1O-3 M)(6.0 x 1O-3 M)2

- 5.0 X lo3 I'* min'l

Note: same result from initial rate d ata from all 4 experiments

c) Stoichiometry: NO:H 2 is 1:l

Wh en 0.0010 mole of Hz had reacted, it mus t havereacted with 0.0010 mole of NO; thus

[NO] remaining - 0.0060 - 0.0010 - 0.0050 M.

(1 pt.) forjus

(1 pt.) forjus

(1 pt.>

order andtification

order andtification

(1 pt.) for value(2 pt.) for units

(1 pt.>


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IN202 1(d) (i) For I: Keq - INW

For II: Rate = w, I [NZOZIN2021 - KeqlNOl'

Rate - WH21[W2 (1 pt.1Note: there must be some clear algebraic manipu lation

showing that [N202] is proportional (NOT equal) to [NO]*.

Step II is the rate-determining step. (1 pt.1

(ii) I: NO + NO N2O2

II: N2O2 + H2 - H,O + N20

III: N,O + H, - N2 + H,O (1 pt.>

I + II + III: 2 NO + 2 H2 - N2+ 2H20


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Sample Student Response


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Comment: n part a, the kinetic orders derived are well justified and the rate law is consistent with them . In partb, the studen t derives an expression or the rate constant rom the rate law an d even specifies the numb er of theexperiment from which he or she takes the experimental data in evaluating k. In part c, the student systematicallytabulates he concentrations of the reactants n a fashion similar to that used in equilibrium problems and clearlyrecognizes the stoichiom etric ratio of hydrogen to nitric oxide. The s tudent earns all 3 of the possible points n partd by choosing the second step of the mecha nism as the rate-determining step, showing that its rate is consistentwith the rate law, an d demonstrating h at the sum of the three steps n the m echanism s equal to the overall reactionstoichiome try. This wa s an excellent answ er that received a perfect score of 9.


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Comment: lthough the student expresses he rate law in part (ii) of a as a proportionality, the equation is w rittencorrectly in part b and full credit is given in part a. The solution o part b is clear and straightforward. The studentruns into trouble in part c by substituting concentrations nto the rate expression a comm on error) and failing tonote the 1: 1 stoichiometry in th e reaction. In part d, a point was deducted because he student did not show that themechan ism was co nsistent with the reaction stoichiometry. The total score for this answer was 7.


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Question 3

Although the difficulty of this question was udged by the readers o be similar to that of qu estion 2, significantlyless han half of the examinees chose o wo rk on th is problem. Points were frequently lost in part a when studentsfailed to con sider the water vapo r in the collected gas. Failure to read part b carefully led many students ocalculate the num ber of molecules of hydrogen rather than water vapor. In part c credit was awarded for answ ersderived from equating the average kinetic energies of the g as molecules or by calculating the root-mean-squarespeeds of the m olecules of the two gases. n part d many students cited the difference in the masses of the two

molecules and lost credit for the explanation part of the question.

Scoring Standards

(a) z2 - -$ - (721)(0.090)(62.4)(298) - 3.49 x low3 mol Hz

25C - 298 K745 - 24 - 721 mm Hg

calculation of moles of Hz

tb) (23*8)(o*ogo) _ 1 15 x 10-4 mol H 0(62.4) (298) 2

(1.15 x 10B4)(6.03 x 1023) - 6.92 x 1019 molecules H20

(c) The average kinetic energies are equal, so

(+ )H,O - (+&H2 .

Note: credit also given for correct use of vrms 3RT- -M

(d) H20 deviates mo re from ideal behavior.

Explanation:EITHER

i) The volume of the Hz0 molecule is largerthan that of the Hz molecule

W

ii) The intermolecular forces among Hz0 molecu les arestronger than those among Hz molecules

(1 pt.>(1 pt.>(1 pt.>

(1 pt.1

(1 pt.>

(1 pt.) for formula(1 pt.) for calculatiorz

(1 pt.>

I 1 pt.>


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Comment: n part a, the conversion of Celsius emperature o Kelvins, the application of D altons Law, and the useof the ideal gas equation were all do ne correctly. In part b, the studen t clearly recognized the relationsh ip betweenthe pressure of a g as and the num ber of moles present and mad e the correct calculation using NA. The solution topart c is clear and straightforward. The analysis n part d wanders somewhat, but credit was awarded because hestudent was aware that the hydrogen bonding in w ater is an intermolecular force and that ideal gases experience nointermo lecular forces. This resp onse eceived a score of 9 points .

n 55m


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Comment:The correct answers were obtained in bo th parts a and b in routine fashion. Although the a nswer inpart b was expressed n too many significant figures, no points were deducted because he number of significantfigures given exceeded the appropriate numbe r by only one. The student earned no credit in part c since theresponse eflected no recognition of the relationship between molecular spe ed and the squa re oot of the molecularma ss nor the inverse relationship between m olecular mass and spee d. n part d, the student correctly identifieswater as the gas hat deviates more from ideal behavior, but, like many students, ttributed hat greater deviation towaters larger molar m ass. The total score was 6 po ints.


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Question 4

The ability to write the formulas of chemical species rom their names and knowing the reactions hat commonlyencountered chemical systems undergo apparently continues o confound AP Chemistry tudents. istorically, heperformance of students on the required equation question has reflected heir lack of exposure o enoughdescriptive hemistry. The performances f students n this years required equation question, however, weresignificantly mproved over 1993 mean scores f 5.3 versus .4). Only 0.7% of the students aking he examinationhad perfect scores f 15 on this question.

Scoring Standards

Guiding principles :

1) Each reaction is worth a total of 3 points2) Reactants +l point; products +2 points3) Ignore balancing and states4) Inappropriate ionization - maximum 1 point penalty per equation

(a) CN + Ag+ - Ag(CN)z

Note: any complex ion of Ag with cyanide with consistent charge earns 3 points;AgCN given as product earns 1 product point

w Mn*+ + 3 - MnsNote: If Mg used instead of Mn, maximum possible score is 2 points

(4 PbOjO (or PZOS) + Hz0 - H3 PO4

Note: Acidic species (H+ or oxyacid of phosphorus) earns 1 product point;P in +5 oxidation state in oxyanion earns 1 product point;anions of oxyacids of phosphorus require H+ for full credit

for products

W lNH4 )ZcC3 - NH3 + Hz0 + CO2

Note: Any one product earns 1 point; all three earns 2 product pointsNH4OH + co* earns 1 product point

NH3 + HtCO3 earns 1 product point

(e) CO, + OH- - HC%

Note: C+2 + H20 as products earns 2 product pointsCaj2 alone as product earns 1 product point

HCOj - + Hz0 earns 1 product point


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(f) Ii+ + Cl' + KMrlO& - K+ + Mn*+ + Cl2 + Hz0

Note: HCl and MnO4' acceptable as reactantsAny valid redox product earns 1 pointAll four products earns 2 pointsK+ and/or Hz0 only as products earns no credit

If both H+ and Hz0 omitted, then maximum of 2 points possible

W Na + Hz0 - Hz0 + Na+ + OH'

Note: All three products earns 2 product pointsAny vaid redox product earns 1 product point

W Crz+2' + Fe2+ + H+' - Cr3+ + Fe3+ + Hz0

Note: All three products earns 2 product pointsAny valid redox product earns 1 product pointHz0 only earns no creditIf Cl' - Cl, instead of Fe2+ - Fe3+, then maximum

of 2 points possible

n s8m


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Comment: his respon se ailed to earn a perfect score for omitting the w ater in equation h. This was a common

error in the redox reactions. The total score for this answer was 14 points.


Comment:A point was lost in equation g for failure to represent sodium hydroxide as an ionized sp ecies. nequation a, a point w as deducted for the failure to recognize that a complex ion was formed (see the ScoringStandards). n equation d, a product point was deducted or the hydroxide on (rather han water). n equation e, thec?udent ost a p oint for the incorrect charge on the carbonate on product. The score or this response was 11 points.

n 59 n


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Question 5

For this required essay question nvolving simulated aboratory exercises, students were asked to explain the resultof a change in conditions on an illustrated experimental situation. The committee hoped it would encourage asignificant increase n the laboratory component n AP Chemistry courses.

Scoring Standards

(a) Volume decreases in beaker A; the concentrationdecreases in beaker B (either observation earns 1 pointprovided other one is not wrong)

(1 pt.>

The vapor pressure of pure Hz0 is greater than the vaporpressure of Hz0 in solution, 1OR, 7 (1 pt.>the rate of evaporation of Hz0 molecules from pure Hz0 isgreater than that from the sugar solution, while thecondensation rates are the same. J

(b) The water will begin to boil (or evaporate).

The external pressure on the water will become equalto the vapor pressu re of the water, causing it to boil,

OR, 7 (1 pt.>the drop in external pressure causes the boiling pointto drop to the temperature of the water. /

(c) Solid copper is deposited on the zinc strip; the zincstrip goes into so lution. No reaction occurs with silver.

Zinc is a better reducing agent or a more active metalthan copper and will be oxidized. Silver is a lessreactive metal than coppe r is.

(d) Two layers will form, one of which is colored.

Iodine is nonpolar and will dissolve in TTE.

Wa ter-is polar and will not dissolve in TTE.

(1 pt.>

(1 pt.>

(1 pt.>

(1 pt.>

Note: placement of 12 must be correctly indicated for 2nd point.

n 60 H


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n 61 n


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Comment: he response n part a presents an unusual combination of a reference to an increase n entropy an d thedifferential rates of evaporation of pure water an d an aqueou s solution of sugar. n part b, the response does notmatch the standards but clearly indicates that the student understan ds what is required for this system to reachequilibrium and why this one never can. The students response o part c essentially embod ies the ideas in thestandards. he response o part d quite simply covers all of the bases. One would have to look hard for a place todeduct points from this response, which received the maximu m score of 8 points.


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Comment: This paper earned a score of 6. The respon se n part a was seen n m any papers. The awarding of theobservation point w as routine, but simply stating that there is a tendency for the con centrations of the twosystems o become equal w as insufficient to earn the explanation point. Both the observation point and theexplanation point were clearly a rticulated n the response o part b. The students response n part c also clearlyearns credit for bo th the observation an d explanation points. In part d a point w as lost since there was a clearstatement hat the solutions would separate and two layers would form. Failure to specify that the iodine will be inthe TTE cost this student a point.


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Question 6

The importance of thermodynamics n even an introductory chemistry course is reflected by the fact that it isalmost a standard essay opic on the AP Chemistry Examination. n part a of this question, tudents ere given heopportunity o relate he entropy change or a reaction o the stoichiometry nd o the phase hange hat akes placeas reactants are converted o products. Parts b and c gave the student an opportunity to demonstrate anunderstanding of the implications of a change n the temperature on the spontaneity of a reaction and the K,, of a

system at equilibrium. In each of the parts a, b, and c, the writing of an equation lone was not suffkient to earnthe explanation oint; he student was equired o connect he equation o the ssue o be explained. n part d, aknowledge of the Gibbs-Helmholtz equation and the dependence f the enthalpy and entropy changes ontemperature were required. Points were not deducted or sloppiness r confusion elative to the differencesbetween he standard nd nonstandard alues of changes n free energy, entropy, or enthalpy.

Scoring Standards

(a) Statement that ASo is negative (1 pt.>

3 moles of gas - 2 moles of gas plus solid,(3 moles - 2 moles earns no points)

0%2 gases - 1 gas + solid, (1 pt.>

0%

use of A@ -A@ - TAS" with Ati - 0

Note: If statement is that AS0 is positive, then explanation of3 moles gas - 5 moles of gas earns I point

(3 moles - 5 moles earns no points)

If correct explanation for ASo being negative is givenbut wrong sign is stated, 1 point is earned.

(b) A@ is less negative, goes to 0, goes +, gets larger

Explanation using A8 - A@ - TAS

Note: if answe r to (a) is that AS o is positive, then fullcredit can be earned here for correct reasoning basedon that assum ption. An explanation that usesLe Ch&teli.er's principle based on sign of @ is NOTvalid here since system not at equilibrium.

(1 pt.>

(1 PU


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(4 &q decreases (exponent - more negative) as 2' increases

CR.

Keq g oes from > 1, to 1, to < 1, as T increases

Correct explanation using the equationA - -RT In Keq (or In&/K; ) - AH% l/T2 - WI)

OR.

(1 pt.>

(1 pt.>higher 2' favors the reverse reaction (Le Chatelier)because the forward reaction is exothermic

Note: if answer for (a) is that AS0 is positive then statementthat Ke

3will decrease or increase depending on the relative

magnitu e of T and A8 change earns 2 points. Recognition thatBOTHA@ and T are changing in A8 - -RT In Keq is necessary.Or, ignoring part (a), use of AH O < 0 explanation to co rrectlypredict that Qq will decrea se earns 2 points.

(d) Since A@W

- 0 at this point, the equation is T - f@/AS.- & - TAS is NOT sufficient without Afl - 0.) (1 pt.>

Prediction is not exact because & and/or AS0 vary with T (1 pt.)

General Note: For parts (a), (b), and (c), just writing an equa tion is notsufficient for the "explanation" point. To earn credit, thestudent must connect the equation to issue to be explained.


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Comment: ull credit is given for part a since he student correctly evaluates he chang e n disorder as he reactantsare converted to products. In part b, the student appropriately discusses he G ibbs-Helmholtz equation andcorrectly notes how the two state functions will change as temperature s increased. The answer in part c couldhave been clearer; nonetheless, he answer does reflect an und erstanding of the relationship between temp erature,K,,, and free energy change, so both points were awarde d. n part d, the student correctly notes hat at the change-over point the free energy change becomes zero and that both cha nges in entropy and enthalpy are slightlytemperature dependen t. The total score for this answer was 8


Comment: his paper earned 4 of the possible 8 po ints. No points were given in part a, because he student ailedto recognize that changes n phase are more important in determ ining changes n entropy than changes n thenumb er of m oles of products versus he number of mo les of reactants. Full credit was given in part b, since theanalysis using the Gibbs-Helmho ltz equation s correct using the incorrect sign of entropy change rom part a. Nopoints were awarded n part c, although there is some merit in the analysis of the exothermic nature of the reactionand the effect of increasing he temperature on the position of the equilibrium. In fact, the student did not answerthe question, What is the change n the value of the equilibrium constant Two points were awarded n part d forthe correct use of the Gibbs-Helmholtz equation and the recognition that (at least) one of the state functions isdependent on temperature.


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Question 7

The students performance on this question s reflected in the liberal nature of the scoring standards. t seem s hatmost students are not exposed appropriately to titration curve s or taught the properties of polyprotic acids.

Scoring Standards

(a) POb3' + H+ * HPObt'

HP04*- + H+ _ H2P0,-only

Note: any proton transfer to any PxOY species earns 1 point.

wL

:;:

H : 'b' 0.;:b':**53':

l *.#

3-

explicit 32 eexplicit 2 - charge (somewhere)not more than 1 double P-O bond

Note: HPOb*- (formula only) or other PxOr species withcorrect diagram earns 1 point.

(c) Graph goes from uppe r left to lower right (pH decreases)

Two protons

c

Two "buffers"

7

in eithertransferred Two "equivalence" direction

Explain/correctly label at least one "buffer" or

"equivalence" region

w H2P04- + H+ * H3f04Note: other p roton transfer earns 1 point if

consistent with produ ct in part (a)

(2 p-4

(2 pts*>

(1 pm

(1 pt.>

(1 PC*>

(1 pt.1


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(c) Sketch a graph using the axes provided, showing he shape of the titration kurve that results when 100. millilitersof the HCl solution s add ed slowly from a buret to the Na3P 04 soiution. Account for the sh ape of the curve.

t

0 nL HCI

(d) Write the equation or the reaction that occurs f a few additional milliliters of the H Cl solution are added othe solution e sulting rom the titration in (c).

-Jr3Y I- t I@$ H-,?OY * Ofl

Comment: his was one of a very few perfect responses a score of 8 points) to this question. Both points wereclearly ea rned in part a an d in pa rt b. The nature of this studen ts titration curve (part c) reflects a realunderstanding of the nature of this system, he some what diffuse equivalence points, and the buffering regions. Theresponse o part d is exactly right.


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.


0 rnL HCI1s

(d) Write the equation or the reaction that occurs f a few additional milliliters of the HCl solution are added othe solution resulting from the titration in (c).

Comment: The student learly earns he two points n part a, but he or she ails o identify he amphoteric pecies,and he Lewis structure as ittle validity n reference o the octet ule. This student arns point n part c for theshape of the titration curve and another point for identifying he equivalence oints as drops occurring with theformation of each successive ole of acid. The student was also awarded point n part d for correctly writing theappropriate quation or a total score of 5.


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Question 8

This question ave the students n opportunity to relate chemical rinciplestoeveryday xperiences.The ttempttomakechemistryrelevant"hasbecomemo reandmoreapparentinrecentyearsandthattrendisreflectedineachofthe four parts to this question.

Scoring Standards

(a) The addition of a solute lowers the freezing pointof water.

A mole of NaCl contains (dissociates into) 2 moles ofions/particles, whereas a mole of CaC12 contains (dissociatesinto) 3 moles of ions. Therefore CaC12 is more effective.

w Hydrogen bonding is the most important intermolecularattractive force between molecules of Hz0 and betweenmolecules of NHs.

Water is a liquid because the hydrogen-bond ing forces arestronger between adjacent Hz0 molecules than between adjacentNH3 mo lecules.

(1 pt.1

(1 pt.)

(1 pt*)

(1 l-4

Further explanations for the stronger hyd rogen bonding in Hz0include the larger dipole mom ent (or more polar character)of H z0 compared to NH, and the fact that 0 is more electronegativethan N is.

(c) Graphite's structure consists of 2-dimensional sheetsof covalently bonded carbon atoms. The attractive forcesbetween sheets (layers) are weak London (dispersion) forces, (1 pt.1

which allow the sheets to slide easily over one another.

Note: must indicate layers and sliding to earn point.

Diamond consists of an extended 3-dimensional covalent networkof carbon atoms. This makes diamond a very hard substance. (1 pt.)

(d) Vinegar, a dilute solution of acteic acid, reacts with thewhite solid, which contains metal carbonates,

in a neutralization reaction to form gaseous CO2.

(1 pt.1

(1 PW

n 7i n


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Comment: Since students were not required to relate freezing-point dep ression o vapor-pressure owering, thestatement hat the added salt owers the freezing point of w ater earned the first point in part a. The second point inthis part was awa rded for the clear recognition of the difference in the i factor for the two solutes. Mentioninghydrogen bonding and relating the strength of the hydrogen bonds in the two molecules to electronegativity inoxygen and nitrogen earned both points in part b. In part c, the response learly earned he first point in explainingthe lubricating properties of graphite as a function of sliding layers and, even though the three-dimensional natureof the covalent network was not m entioned, the second point was also awarded. The response o part d is right onthe mark. The total score was 8 points.

n 72 n


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Comment: Although the response s rather cryptic, since the stud ent notes in part a that the freezing point isreduced by dissolved solutes and that calcium chloride produces more ions in solution than sodium chloride, theresponse earned both points. n part b, the student oses a point for mistakenly identifying hydrogen bonding as anintramolecular force and in part c, a point w as deducted because he studen t ailed to note the covalent networkstructure of diamon d. In part d, the s tudent gets back on track by concisely explaining the effervescence whenvinegar is added to the white residue in a tea kettle.


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Question 9

Atomic structure nd chemical bonding, requently he topics of essay questions n the AP Chemistry Examina-tion, were the foci in this question. In addressing he four parts, students were required to relate differences natomic size, crystal-lattice energy, and ionization energy to chemical principles. The fact that very few studentsscored well on part b is indicative, perhaps, hat his opic s often glossed ver n AP courses and should receivemore attention.

Scoring Standards

(a) Cat+ has fewer electrons, thus it is smaller than Ca (1 pt.)

The outermost electron in Ca is in a 4s orbital, whereasthe outermost electron in Ca*+ in in a 3p orbital (1 pt.)

Note: The first point is earned for indicating the loss ofelectrons, the second point for indicating the outermost

electrons are in different shells -- mus t accoun t forthe magnitude of the size difference between Ca and Ca2+.

(b) ZI for CaO is more negative than U for K2 0, so it is more (1 pt.)difficult to break up the CaO lattice (stronger bonds in CaO).

Ca*+ is smaller than @, so internuclear separations (betweencations and 02-)are less,

OR,

Ca*+ is more highly charged than K+, thus cation--02 bondsare stronger

(1 pt.>

Note: understan ding what "lattice energy" is earns 1 point; sizeor charge explanation needed for the second point. Responsesthat use Lewis structures or otherwise indicate molecules ratherthan ionic lattice earn no points.

n 74 n


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(c) (i) Ca has more protons e is smaller. The outermostelectrons are more strongly held by the nuclear chargeof Ca.

(ii) The outermost electrons in Ca are in the 4s, which isa higher energy orbital (more shielded) than the 3pelectron in K.

Note: for (i), the idea of attraction between nucleusand electrons must be present; for (ii), a"noble-gas configuration" argument must be tiedto an energy argument in order to earn credit.

(d) The highest energy (outermost) electron in Al is ina 3p orbital, whereas that electron in Mg is in a 3s orbital. (1 pt*)

The 3p electron in Al is of higher energy (is more shielded)

than is the 3s electron in Mg.(1 pt*)

Note: noting that different orbitals are involved earns the firstpoint; a correct energy argument earns the secon d point.

Responses that attribute the greater stability of Ca over K(or K+ over Ca+,or M g over Al) to the stability of a completelyfilled (vs. half or partially filled) orbital earn NO credit.


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Comm ent: his response earned 4 points. Both po ints in part a were given for noting that electrons are lost in theionization w hile implying that the nuclear charge emains he same, ncreasing he force of attraction of the nucleusper electron. In part b, the student ost both points since t is clear he or she does not und erstand he nature of latticeenergy. In part c, the student earned only 1 of the 2 points, for invoking a comm on error in respon ses o thisquestion (that filled subshells re especially stable) in the first part. The point for part c (ii) was awarded since thestudent clearly kn ew th at the second electron lost from potassium comes from a 3p orbital. In part d, a point wasearned by noting the electron configurations of the species nvolved, but the explanation point wa s ost by using thesame fallacious argument as in part c.


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Chapter IV

Statistical Information

n SECTION II SCORES at each score point is listed, along with the total num -

Table 4.1 show s the score distribution for the free-response section of the 1994 AP Chemistry Examina-tion. Students were required to complete Questions 1,4, and 5, an d then had to choose a fourth question romQuestions 2 and 3, and two more from Questions 6-9.Questions l-3 were scored on a 9-point scale, Ques-tion 4 o n a 15-point scale, and Questions 5-9 on an 8-point scale. For each question, the num ber of students

ber of candidates attempting that essay.Question 2 had the highest mean as percent of

maximum possible score, indicating that the scoreearned by this question s typical student was closer tothe questions highest score than for any other ques-tion. Question 1 had the greatest standard deviation,indicating that the scores ended to be spread out morethan for the other questions,

Table 4.1 - Section II Sco res

Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9(9) (15) (8) (8) (8) (8) (8)

800735963

1,0961,4011,5431,5781,3561,018

624173

215527710892

1,2261,4731,7402,1652,3442,5932,8803,1833,3253,4062,1071,7501,094

486 337 35 165 41938 1,787 57 412 175

1,722 2,300 101 712 2932,601 2,320 152 1,242 8463,840 2,573 268 1,673 1,8395,197 2,470 825 2,165 3,4466,131 2,992 2,162 2,396 5,1045,642 2,072 2,99 1 2,78 1 4,7603,076 1,598 887 3,191 1,4221,997 2,305 1,840 283 242

Tuestion 1(9)* Question 2(9)core161514131211109876543210

NR**

Number ofCandidates

1,5322,1162,6182,6742,5252,5182,8633,87 14,94 13,9831,989

I

3571,1552,2103,3984,3903,1021,4591,137

7981,1111,226

I

3 1,630 19,117 1 11,114 31,630 31,630t (

I *x(,

2.61

18,449 7,478 14,737 1 17,926

:(219/3.45 1 4.71 1 4.32 3.60 1 1.73 2.41 1 2.24

* Numbers in brackets ndicate the maximum possible score.__ -*y No response. Students gave either no response or a response not on the topic. Responses hat fall into this category were not included in the calculation

of number of candidates, mean, standard deviation, or means as a percentage of maximum.


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chemistry released exam 1994

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