台灣大學93年第一學期; 期中考(佔學期總分30%);日期: 2004/11/10 (星期三)
課程名稱:普通化學(課程編號203-101A1/A2 班次04、05、06)

(1)    (a) 87, (b) 0.57, (c) 2.5 x 1024.    (3%)

(2)    (a) SO3, (b) (NH4)2SO4, (c) MnO4-, (d) KH2PO4, (e) Mn(OH)2,
       (f) Cr2(CO3)3, (g) SiCl4, (h) HNO3, (i) O3, (j) Nd3+. (10%)

(3)    (a) hydrogen sulfide, (b) ammonium sulfate, (c) dioxygen difluoride,
       (d) sulfuric acid, (e) potassium dichromate, (f) magnesium oxide,
       (g) nitrogen monoxide, (h) ammonia, (i) sodium hydrogen carbonate,
       (j) carbon dioxide, (k) uranium hexafluoride, (l) sodium hypochlorite,
       (m) barium chromate, (n) hydrogen cyanide or hydrocyanic acid
       (o) calcium hydroxide, (p) magnesium phosphate (q) copper(I) iodide,
       (r) potassium chlorate, (s) gallium oxide (t) unniloctium. (20%)

(4)    補充教材 (網站) (10%)

(5)    66.2%, textbook example 3.11, page 76-77. (6%)

(6)    (a) HClO4(aq) + KOH(aq) a H2O(l) + KClO4(aq),
       (b) HNO3(aq) + CsOH(aq) a H2O(l) + CsNO3(aq),
       (c) 2HI(aq) + Ca(OH)2(aq) a 2H2O(l) + CaI2(aq), (6%)

(7)  (a) Xe, +6; O, -2; F, -1,
     (b) S, +4, F; -1,
     (c) C, +2; O, -2,
     (d) Fe, +3/8;O, -2,
     (e) Li, +1; N, -3,
     (f) N, +2; O, -2,
     (g) N, 0.  (7%) [textbook, page 134]

(8)  Cr(NO3)3(aq) + 3 NaOH(aq) a Cr(OH)3(s) + 3 NaNO3(aq),
     moles of NaOH used = (2.06 / 103) x 3 = 6.00 x 10-2.
     NaOH(aq) + HCl(aq) a NaCl(aq) + H2O(aq),
     moles of NaOH used = 0.400 x 0.1000 = 4.00 x 10-2
     concentration of NaOH = [(6.00 + 4.00) x 10-2] mol / 0.0500 L = 2.00 M
     [textbook, problem 4-77, page 136] (6%)

(9)  (a) n(He) = 1.96 / 4.00 = 0.490, n(O2) = 60.8 / 32.00 = 1.90,
         n(total) = 0.490 + 1.90 = 2.39,
         mole fraction: χ(He) = 0.490 / 2.39 = 0.205,
                        χ(O2) = 1.90 / 2.39 = 0.795,
     (b) P(He) = (0.490 x 0.08206 x 298.15) / 5.00 = 2.40 atm,
         P(O2) = (1.90 x 0.08206 x 298.15) / 5.00 = 9.30 atm,
     (c) total pressure P = 2.40 + 9.30 = 11.70 atm.  (6%)

(10)  補充教材 (網站) (4%)

(11)  補充教材 (網站) (10%)

(12) (a) particles have no volume,
     (b) particles move with constant velocity,
     (c) no interactions among particles,
     (d) average kinetic energy of a collection of gas particles is directly
         proportional to the Kelvin temperature of the gas.
         See textbook, page 155. (4%)

(13) (a) low pressure and high temperature,
     (b) “a” – interactions (or collisions) among gas particles,
         “b” – volumes of gas particles  (4%)

(14) 1.09 x 1026 collisions/s.
     See textbook example 5.9, page 167 (4%)

(15) C: 58.51 / 12 = 4.87, H = 7.37, N = 34.12 / 14 = 2.44.
     Thus, C:H:N = 2:3:1,
     emprical formula = C2H3N. M of this molecule = 4.00 x (3.20)2 = 41.0
     molecular formula = C2H3N. [textbook, problem 5-105, page 186] (4%)

(16) “mean free path” is inversely proportional to “collision frequency”
      (page 170)
      ”collision frequency” is directly proportional to “pressure”
      (page 169)
      Thus, mean free path is inversely proportional to pressure.
      P1λ1 = P2λ2.  (1 x 10-6 Torr)(50 m) = (760 Torr)λ2
     λ2 = 7 x 10-8 m (4%)
      [textbook, page 161, 3rd paragraph, line 4]