Advanced Placement Chemistry 1993 Free Response Answers：高級(jí)化學(xué)1993自由回答的答案

1、.Advanced Placement Chemistry: 1993 Free Response Answers· delta and sigma are used to indicate the capital Greek letters. · square root applies to the numbers enclosed in parenthesis immediately following · All simplifying assumptions are justified within 5%. · One point deducti

2、on for a significant figure or math error, applied only once per problem. · No credit earned for numerical answer without justification. 1) average score 4.5-5a) three pointsCH3NH2 + H2O <=> CH3NH3+ + OH¯Kb = (CH3NH3+ OH¯) ÷ CH3NH2 = 5.25 x 10¯4CH3NH2 CH3NH3+ OH¯

3、I 0.225 0 0 C -x +x +x E 0.225 - x x x 5.25 x 10¯4 = (x) (x) / (0.225 - x)neglect the minus x to get 5.25 x 10¯4 = x2 / 0.225x = square root(5.25 x 10¯4) (0.225)OH¯ = 1.09 x 10¯2(Note: quadratic gives 1.06 x 10¯2)b) three pointsCH3NH3+ = 0.0100 mol / 0.120 L = 0.0833 M5

4、.25 x 10¯4 = (0.0833 + x) (x) / (0.225 - x) = 0.0833x / 0.225x = OH¯ = 1.42 x 10¯3 mol/LpOH = 2.85pH = 11.15alternate solution using the Henderson-Hasselbalch EquationpH = pKa = log (base / acid)pKw = pKa + pKbpKa = 10.7pH = 10.7 + log (0.225 / 0.0833)pH = 11.15The solution using the

5、pOH form is left to you, gentle reader.c) two pointsHCl must be added.5.25 x 10¯4 = (0.0833 + x) (0.0010) / (0.225 - x)1.18 x 10¯4 - 5.25 x 10¯4x = 8.33 x 10¯5 + 1.0 x 10¯3xx = 0.0228 mol/L0.0228 mol/L x 0.120 L = 2.74 x 10¯3 mol HClalternate solution based on Henderson

6、-Hasselbalch Equation11.00 = 10.72 + log (base / acid)log (base / acid) = 0.28base / acid = 1.906 = (0.225 - x) / (0.0833 + x)x = 0.0227 mol / L0.0227 mol / L x 0.120 L = 2.73 x 10¯3 mol HCld) one pointThe CH3NH3+ / CH3NH2 ratio does not change in the buffer solution with dilution. Therefore, n

7、o effect on pH.2) average score = 4.5a) two pointscarbon: 49.02 / 12.01 = 4.081 ghydrogen: 2.743 / 1.008 = 2.722 gchlorine: 48.23 / 35.453 = 1.360 gmole ratios: C/Cl = 3; H/Cl = 2; Cl/Cl = 1empirical formula = C3H2Clb) three pointsdeltaTf = Kf m4.38 °C = (5.12) (x / 0.025)x = 0.0214 mol3.150 g

8、/ 0.0214 mol = 147 g / molNote: the scoring standards has this equation rather than the above three lines:The standards then show:MM = (5.12) (3.150) (1000) / (4.38) (25) = 147c) two pointsmole fraction = moles benzene / total molesC6H6 = 78.10825.00 g / 78.108 = 0.32 mol0.32 / (0.32 + 0.0214) = 0.9

9、4d) two pointsPsoln = Ppure x mole fractionPsoln = (150) (0.94) = 141 mm Hg3) average = thre pointsa) one point(1.00 mol O2) (2 mol MnO2 / 1 mol O2) (1 mol I2 / 1 mol MnO2) (2 mol S2O32¯ / 1 mol I2) = 4 mol S2O32¯Note: answer only is sufficient.b) two pointsmol S2O32¯ = (0.00486) (0.0

10、112) = 5.44 x 10¯5 mol S2O32¯mol O2 = 5.44 x 10¯5 / 4 = 1.36 x 10¯5 molc) one pointless I2 therefore less S2O32¯ required therefore lower amount of O2 (both direction and reason required)d) three points (one for M; one for correct use of R; one for correct T)Msoln in (b) = 1

11、.36 x10¯5 mol / 0.050 L = 2.72 x 10¯4 MV = (nRT) / P = (2.72 x 10¯4) (0.0821) (298) / 1= 6.65 x 10¯3 L or 6.65 mLe) two pointsstarch indicatorcolor disappears or blue disappears (violet or purple OK)color change alone is not sufficient for 2nd ptany other color with starch is not

12、 sufficient for 2nd pt4) average = 4.8a) Cu + H+ + NO3¯ -> Cu2+ + NO + H2O (1 pt for either Cu2+ or NO; NO2 also accepted; 2 pts for all three.)b) MnO4¯ + H2O2 -> Mn2+ + O2 + H2O (1 pt for either Mn2+ or O2; 2 pts for all three)c) H+ + MnS -> H2S + Mn2+d) Fe + Cl2 -> FeCl3e) Mg

13、3N2 + H2O -> Mg(OH)2 + NH3 (Mg2+ + OH¯ also accepted)f) SO2 + OH¯ -> HSO3¯g) AgCl + NH3 -> Ag(NH3)2+ + Cl¯ (other coordination numbers also accepted)h) Zn2+ + PO43¯ -> Zn3(PO4)21 pt for reactants2 pts for products; 1 pt per product where two occur; 2 pts for sing

14、le product5) average = 2.9a) two pointsZn is oxidized to Zn2+ by H+ which in turn is reduced by Zn to H2Identify H2(g) or Zn dissolving as Zn2+Explicit: Reox or e¯ transfer or correctly identify ox. agent or red. agent inconsistency among these VOIDS the pointb) two pointsH2SO4 dissociates, for

15、ms ions or hydration "event." Bonds form, therefore energy given off (connection)c) two pointsBaSO4 (ppt) forms or H+ + OH¯ form water. Newly formed water and ppt remove ions lowering conductivity.d) two pointsFirst 10 mL produces solution of SO42¯ and OH¯ or excess OH¯

16、 partial neutralization (pH: 13.0 -> 12.6)Presence of HSO4¯ in solution voids this pointSecond 10mL produces equivalence where pH decreases (changes) rapidly (pH: 12.6 -> 7.0)pH "rises" or wrong graph, fused, voids this point6) average = 2.1a) three pointsElectron configuration

17、of Na and Mg (1 pt)Any one earns a point:Octet / Noble gas stability comparison of Na and MgEnergy difference explanation between Na and MgSize difference explanation between Na and Mg Note: If only Na or Mg is used 1 point can be earned by showing the respective electron configuration and using one

18、 of the other explanationsShielding/effective nuclear charge discussion earns the third point.b) one pointCorrect direction and explanation of any one of the following:shielding differencesenergy differences# of proton/ # of electron differences c) two pointsAny one set earns one point:(i) Ni unpair

19、ed electrons. paramagnetic(ii) Zn paired electrons/ diamagnetic(iii) Ni unpaired electrons/ Zn paired electrons(iiii) Ni paramagnetic/ Zn diamagnetic Orbital discussion/ Hund's Rule/ Diagrams earns the second point.d) two points Expanded octet or sp3d hybrid of phosphorous (1 pt)Lack of d orbita

20、ls in nitrogen (1 pt)ORnitrogen is too small to accomodate (or bond) 5 Fluorines or 5 bonding sites (2 pts)7) average = 5.1a) three points2 Cr + 3 Cu2+ -> 2 Cr3+ + 3 CuCr = reducing agent; Cu2+ = oxidizing agentb) three pointsi) Cu is cathodeii) salt bridgeiii) tranfer of ions or charge but not e

21、lectrons c) two pointsNernst equation useE decreasesGuidelines: (c) Le Chatlier type argument okay less spontaneous, less formed rxn, more reverse rxn.If wrong rxn. written, consistency with incorrect rxn. is required. If wrong rxn. is not a redox reaction, points in (bi) and (c) can only be earned

22、if a detailed explanation accompanies. If rxn does not have both an oxidation and a reduction, then no credit can be earned for agents or cathode.If in part a, reduction and oxidation are correctly labeled, but agents are not addressed, 1 pt can be earned from the "agent" points.8) average

23、 = 4.5a) one pointdeltaS < 0The number of moles of gaseous products is less than the number of moles of gaseous reactantORa liquid is formed from gaseous reactants.b) one pointdeltaG < 0deltaG becomes less negative as the temperature is increased since deltaS < 0 and deltaG = deltaH - Tdelt

24、aS. The term - TdeltaS adds a positive number to deltaH.c) one pointdeltaH < 0The bond energy of the reactants is less than the bond energy of the products.d) one pointThe reaction has a high activation energyORis kinetically slow,ORa specific neuton of the needs for a catalyst or spark.9) averag

25、e = 3.1a) one pointReducing the temperature of a gas reduces the average kinetic energy (or velocity) of the gas molecules. This would reduce the number (or frequency) of collisions of gas molecules with the surface of the balloon (or decrease the momentum change that occurs when the gas molecules strike the balloon surface.) In