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apexamprepkey - AP® Chemistry Multiple-Choice Answer Key...

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Unformatted text preview: AP® Chemistry Multiple-Choice Answer Key Correct Correct No. Answer No. Answer 1 D ' 31 A 2 B 32 C 3 A 33 B 4 E 34 B 5 A 35 B 6 C ' 36 C 7 D 37 E 8 B 38 D 9 C 39 C 10 A 40 E 11 E 41 C 12 D 42 B 13 B 43 A 14 C 44 C 15 C 45 D 16 E 46 C 17 C 47 E 18 E 48 D 19 D 49 A 20 B 50 C 21 E 51 B 22 D 52 E 23 C 53 B 24 C 54 D 25 C 55 D 26 A 56 C 27 D 57 E 28 E 58 D 29 1 B 59 B 30 D 60 C -33- Correct No. Answer ' 61 D 62 B 63 B 64 D 65 E 66 D 67 E 68 C 69 A 70 E 71 C 72 D 73 C 74 B 75 C AP® Chemistry Free-Response Scoring Guidelines In regard to mathematical errors, a 1-point deduction is made; this deduction may be applied only once per question. In regard to errors in reporting significant figures, a 1—point deduction is made; this deduction may be applied only once per question. A leeway of plus or minus one significant figure different 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 answer that derives from the correct use of a previously caiculated incorrect answer should not be counted as wrong. The emphasis of scoring is on process, and an error made early on in a multipart calculation should not jeopardize the earning of full credit for the correct use of that incorrect value later in the same question. AP® Chemistry , Frees-Response. Scoring, Guidelines ,, Question 1 Answer the following questions about the solubility of the salts Li3PO4 and PbClz. Assume that hydrolysis effects are negligible. The equation for the dissolution of Li3PO4(s) is shown below. Li3PO4(s) (t2 3 Li+(aq) + PO43“(aq) KS? = 3.2 x 10‘9 at 25°C (21) Write the equilibrium-constant expression for the dissolution of Li3PO4(s). KS? = {If} 3{PO43‘] One point is earned for the correct expression. (13) Assuming that volume changes are negligible, calculate the maximum number of moles of Li3PO4(s) that can dissolve in (i) 0.50 L of water at 25°C One point is earned for a . _9 correct setup using the saturated solution of L13PO4. Then K51, = 3.2 x 10 = (3x)3(x) = 27x4, KW expression from I -9 thus x = ’ 22%?" = 3.3 x 10'3 M. Therefore the number of moles p art (20‘ of Li3P04(s) that can dissolve in 0.50 L of water is Let 3: represent the molar concentration of Li3P04 present in a One point is earned for a 3,3 x 10“3 mol Li3po __ correct calculation of 05% x “mgr—5‘ = 1-7 x 10 3 "ml moles of 1.131304. (ii) 0.50 L of 0.20 M LiNO3 at 25°C In 0.20 M LiN03, [LY] = 0.20 M. Assume that the amount of One point is earned for Li+(aq) contributed to the solution by the Li3PO4(s) that dissolves is recognizing that {If} is negligibly small compared to 0.20 M. Then let y represent the molar approximately equal to concentration of P043‘(aq) present in the solution due to the Li3P04(s) 0-29 M in the solution 10 that dissolves in the 0.20 M LiNO3. Then Whmh the L13P04(S) was “9 3 3.2 x 104 -7 added. KSP = 3.2 x 10 = (0.20) (y) :3 y = W = 4.0 x10 M. So [PO43‘(aq)] = [Li3PO4(aq)] = 4.0 x 104M in 0.20 M LiNO3(aq). One point is earned for a correct calculation of moles of Li3PO4. Therefore, the number of moles of Li3PO4(s) that can dissolve in 4.0 x 10'7moi Li3po4 = —7 1.0L 2.0x 10 mol 0.50 L of water is 0.50 L x AP® Chemistry ,, ,,,,fireegResponse,,,.$eoting.Guideiines, , , Question 1 (continued) The equation for the dissolution of PbC12 is shown below. PbC12(s) : Pb2+(aq) + 2 Cl‘(aq) KS? = 1.6 x 10’5 at 25°C (c) Calculate the concentration of C1‘(aq) in a saturated solution of PbC12 at 25°C. Let z = [Pub] in a saturated solution of PbC12 . Then Kip = [P1320 [CH2 = (z)( 2z )2 = 423. One point is earned for a correct setup. I 1.6 x 10‘5 l. ‘5 = 4 = 3 ————-—-— _ , Thus 6 X 10 2:3 :> z 4 One pomt is earned for a correct calculation 2 z = [P1324] = 1.6 x 10.2 M of the value of {Cl'}. :2 [Cl‘] = 2{Pb2+] = 2(1.6x10'2M) = 3.2x 10'2M (d) An open container holds 1000 L of 0.00400 M P1302 , which is unsaturated at 25°C. Calculate the minimum volume of water, in mL, that must evaporate from the container before solid PbClz can precipitate. 1.000 L of 0.00400 M PbC12(aq) contains 0.00400 mol of PbClztaq), thus it contains 0.00400 mol Ph2+(aq) and 0.00800 moi Cl"(aq). Let V = volume of the solution at saturation, then 0.00400 mol)(0.00800)2 = 2.6 ><1()"7 V V V3 One point is earned for the 2 6 x 10.7 calculation of the saturation :> V = 3" W = 0.25 L = 250mLat saturation volume. . x Thus the volume of water that must evaporate = 1,000. — 250 = 750 mL Ksp=1.6><10’5= ( 0R One point is earned for subtracting the saturation From part (c), the amount of PbClZ dissolved in l L of saturated PbC12(aq) volume from 1 000 mL is 1.6 x 10'“2 M. Let V = volume of the solution at saturation, then 0.016 mol PbClZ __ 0.00400 mol PbC12 1,000mL — V :9 V=250mL Thus the volume of water that must evaporate = 1,000. - 250 = 750 mL AP® Chemistry ,,Free-Responseficoring,,,,Guidelines , , Question 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 150°C. (i) Calculate the total pressure in the cylinder. 3.22 g He x W = 0.805 mol He One point is earned for the 4.00 g He calculations of moles and adding them together. 1.00 mol N2 28.02 g N2 11.56 g N2 x = 0.4126 mol N2 One point is earned for the correct total moles of gas = (0.805 + 0.4126) = 1.218 mol . . . . substltutlon into the gas law equation. 5.25 L One point is eamed for the correct answer. (ii) Calculate the partial pressure of N2(g) in the cylinder. moles N2 gas One point is earned for the calculation —-——-—-————-—— x total ressure in flask . total moles of gas p of the mole fraction of N2 . pressure N2 = One oint is earned for the correct = W x 5.49 atm = 1.86 atm p answer 1.218 mol ' (b) A 1.50 L container holds a 9.62 g sample of an unknown gaseous saturated hydrocarbon at 30°C and 3.62 atm. (i) Calculate the density of the gas. One point is earned for the correct answer. AP® Chemistry ,,,,,,,Fre,esResanse,,,$,cpri,n9,,,Guidelines Question 2 (continued) (ii) Calculate the molar mass of the gas. Let M = molar mass, then PV = nRT = [3-31) RT 2*» 1 One point is earned for the correct answer. M =-——— _ M = 44.1gmol-1 PV (3.62 atm)(1.50 L) (iii) Write the formula of the hydrocarbon. Saturated hydrocarbons have the generic formula CnHZn +2 , One point is earned for therefore let 44.1 g = 120’!) + 1(2)! +2) = 14}! + 2 I) the correct answer. 42.1:1411 :9 3=n :> C3118 (iv) Calculate the root-mean—square speed of the gas molecules in the container at 30°C. (Note: 11 = 1kg st-Z) 2 _2 _1 ‘1 One point is earned for the correct I) = ’3RT = 3(3-31 kg m S “101 K X303 K) setup using the molar mass in ms M 0.0441 kg mol"1 kilograms. One point is earned for the = 414 m 5‘1 correct answer. -33. AP® Chemistry ,, Free-Response,Scoring.Guidelines, Question 3 A student performs a titration in which a 10.00 mL sample of 0.0571 M HCl is titrated with a solution of NaOH of unknown concentration. (a) Describe the steps that the student should take to prepare and fill the buret for the titration given a wet 50.00 mi. buret and the materials listed below. 0.0571 M HCl solution Indicator solution N a0H(aq) (unknown concentration) Distilled water 10.5 M NaOH solution 100 mL beaker Rinse the buret with some distilled water and then pour some of the NaOH solution of unknown concentration into the beaker and use it to rinse the buret two times. One point is earned for rinsing the buret with the titrant solution. Use the beaker to carefully fill the buret with the NaOH solution of unknown concentration. One point is earned for removing air bubbles from the neck and tip of the Put the beaker under the buret and drain enough solution to bur et remove any air bubbles in the neck and tip of the buret. (b) Calculate the pH of the 0.0571 M HCl. HCI is a strong acid :> [11*] in 0.0571 M HCl = 0.0571 M One point is earned for the pH = -log [in] = —log (0.0571) = 1.243 cometanswer- (c) A volume of 7.62 mL of the NaOH solution was needed to reach the end point of the titration. Calculate the molarity of the NaOH solution used in the titration. One point is earned for mol HCI titrated = 10.00 mL x W = 0.000571 mol; calculating the moles of HCl in the sample that was ratio HClzNaOH in neutralization is 1:1, so 0.000571 moi NaOH reacted; titrated. 0.000571 mol NaOH One point is earned for = 0.0749 M NaOH calculating the molarity of 0.00762 L the NaOH solution. AP® Chemistry ,EreerResppnse,,$,c9rin9, Gu,i,delines,,,, , Question 3 (continued) In a different titration using a different NaOH solution, the concentration of NaOH was determined by the student to be 0.0614 M. ((1) Given that the actual concentration of the NaOH solution was 0.0627 M, calculate the percent error of the student’s result. @0614 —- 0,0627] One point is earned for the correct percent error = X 100 = 2.1 % answer. 0.0627 (e) Calculate the volume of 10.5 M NaOH that is needed to prepare 250.0 mL of 0.0627 M NaOH. 0.0627 moi One point is earned for calculating the 250.0 mL X m = 0.0157 H101 NaOH needed moles 0f NaOI-I needed. 1,000- mL One point is earned for the correct 0.0157 mcl NaOH X = 1.49 mL answer. 10.5 mol NaOH AP® Chemistry Free—Response,ScoringGuidettnes , Question 4 For each of the following three reactions, in part (i) write a balanced equation and in part (ii) answer the question about the reaction. In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may use the empty space at the bottom of the next page for scratch work, but only equations that are written in the answer boxes provided will be graded. W A strip of magnesium metal is added to a solution of silverfl) nitrate. (i) Balanced equation: 149+ 2/433 e—e Mg.“ 1» 2/17. {ii} Which substance is oxidizedin the reaction? (3) Equal volumes of 0.1 M hydrofluoric acid and 0.1 M potassium hydroxide are combined. (i) Balanced equation: HF+OH‘ -—> F‘+H20 (ii) Draw the complete Lewis electron-dot diagram for the reactant that is the Bronstediowry base in the forward reaction. Es a" (b) Solid calcium metal burns in air. (i) Balanced equation: 2Ca+02 —-) 2Ca0 (ii) Predict the algebraic sign of AH” for the reaction. Explain your prediction. The sign of AH° will be negative because AG° is negative {the reaction occurs) and AS" is negative ta solid and a gas react to form a solid). According to the Gibbs-Helmholtz equation, AH° = AG" + T AS". Therefore AH" is the sum of two negative quantities and as such must be negative. ~41- AP® Chemistry , Freer-Resp,¢ns,,e,,,§99rin9, finidelines ,, ,, , Question 4 (continued) (0) Samples of nitrogen monoxide gas and oxygen gas are combined. (i) Balanced equation: 2N0+02 —~> 2N02 (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? rate = kiNOflOZ] General Scoring Notes for Question 4 Five 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 correct product(s), and one point for the correct coefficients in the balanced equation. One point is earned for the correct answer in part (ii). AF” Chemistry iiFrees-Responsesconngsmdalinesei Question 5 Voltmeter 1.00 Mcmnogh 1.00MAgNo3 Answer the following questions relating to the galvanic cell shown in the diagram above. (a) Write the balanced equation for the overall cell reaction. 2 Ag*(aq) + Co(s) ___, 2 Ag(s) + Cozflaq) ' One point is earned for the correct equation. (b) Calculate the value of 12° for the cell. One point is earned for the correct value of $21: = 0.80 - (4128) = 1.08 v , 0 cell ' (c) Is the value of AG° for the overall cell reaction positive, negative, or O ? lust' our answer. , Y The value of AG° for the overall reaction must be negative because the cell reaction occurs (is spontaneous) as the cell operates. One point is earned for the correct OR answer, including a valid justification. Since E5,” is positive and AG° = —nFE°, the value of AG" must be negative. AP® Chemistry Free-ResponseScoring Guidelines, Question 5 (continued) (d) Consider the cell as it is operating. (i) Does Ece“ increase, decrease, or remain the same? Explain. As the cell operates, the concentration of Ag+ decreases and the concentration of (302+ increases :2 {C0241 One point is earned for the correct the ratio Q = increases :2 in Q increases :> [ A g ]2 answer, including a valid justification. Em, = 5:,” - 311:an becomes smaller (decreases). nF (ii) Does AG of the overall cell reaction increase, decrease, or remain the same? Explain. The value of A0 for the system increases (becomes less ' One point is earned for the correct negative) as the cell operates and the system approaches answer, including a valid justification. equilibrium (when AG = 0). (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 too small to overcome the electrical resistance of distilled water, which is a poor conductor due to its very low concentration of ions (about to"7 M H+(aq) and 10‘7 M OH“(aq)} that could “carry” the current from one cell to the other. One point is earned for the correct answer, including a valid justification. (e) After a certain amount of time, the mass of the Ag electrode changes by x grams. Given that the molar mass of Ag is 108 g moi“l and the molar mass of Co is 59 g mol‘l, write the expression for the change in the mass of the Co electrode in terms of x. 1 mol Ag 1 A moi Ag = A mass Ag x M = x x E)8_ " One point is earned for using the correct mole ratio of Co to Ag . 1 Incl Co AmolCo = -AmolAgx ~ C One point is earned for A mass Co = A m0} C0 x 3.23.13. __ ' X __ . the COITECI answer 1 “101 C0 (negative sign is not required). AP® Chemistry , ,FreefiewoneeScoring Guidelines Question 6 Answer each of the following using principles of atomic or molecular structure and/or intermolecular or intramolecular forces. (a) Explain why the H—O—H bond angle in H20 is less than the H—N—H bond angles in NH3 , as shown in the table below. H-N~H WBond AngleH 1n H20 Bond Angles 1n NH3 Both molecules have tetrahedral electron—domain geometries and One p.011“ 1s earned for Citing the might be expected to have bond angles of 1095". However, electron difference 1? number Of domains for nonbonding pairs of electrons exert a greater repulsion on nonbondmg p airs 0f electrons. adjacent pairs of electrons than do electron domains for bonding pairs. Thus, in the H20 molecule with its _t____wo nonbonding pairs of electrons, the electron domains of bonding pairs are tfiompressed to a greater repulsion from nonbon din g greater extent than they are 1n the NH3 molecule, w ch has only one pairs as compared with bonding nonhonding pair of electrons. pairs. One point is earned for citing the (b) Explain why the radius of the Br atom is less than the radius of the Br‘ ion, as shown in the table below. Radars of Br Radius of Br: Olllnm Ol96nm One point is earned for recognition that The nuclear charge (+35) is the same for both the Br and Br‘ Br and Br“ have the same nuclear species, but the “extra” electron in Br~ causes the electron charge. cloud to expand due to an increase in mutual repulsions among . . ‘ . . the electrons that make up the cloud. One pornt is earned for Citing mcreased repulsion among electrons. AP® Chemistry Frees—Response,Scoring Guidelines Question 6 (continued) (c) Explain why the dipole moment of HI is less than the dipole moment of HCl , as shown in the table below. Di ole Moment of Ill D1 ole Moment of llCl 0 42 debye l 08 debyes Iodine, having a lower electronegativity than chlorine has, forms a bond with hydrogen that is less polar than the bond between chlorine and hydrogen in HCl . The lower polarity of the H~l bond means that the dipole One point is earned for the comparison of the moment of the bond is smaller than that of the H~Cl polarity of the two bonds. bond. One point is earned for citing the difference in electronegativity between I and C1. (d) Explain why the normal boiling point of Ne is less than the normal boiling point of Kr, as shown in the table below. NOTmal 301119 Point 91’, KT One point is earned for mentioning that intermolecular forces involved are London (dispersion) forces. The intermolecular forces among atoms in liquid Ne are the same type of forces as those among atoms in liquid Kr, namely London (dispersion) forces. However, the magnitudes of these forces are smaller in Ne because the electron clouds of Ne atoms are smaller and less polarizable than the electron clouds of Kr atoms. One point is earned for Note: An explanation that cites only periodic trends or only the mentioning the relative relative masses of Ne and Kr does not earn credit. polarizability 0f the electron clouds of the atoms. ...
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