1-1. The mass of a water droplet: m = V ρ = [(4/3) π r3] ρ = (4/3) π (0.5x10-6 m)3 (1.0 g/cm3) = 5.2x10-16 kg
? 10 marks Average kinetic energy at 27oC: KE = mv2/2 = (5.2x10-16 kg) (0.51x10-2 m/s)2/2 = 6.9x10-21 kg m2/s2 = 6.9 x10-21 J
? 15 marks *. The average kinetic energy of an argon atom is the same as that of a water droplet. KE becomes zero at –273 oC. From the linear relationship in the figure, KE = aT (absolute temperature) where a is the increase in kinetic energy of an argon atom per degree. a = KE/T = 6.9x10-21 J/(27+273K) = 2.3x10-23 J/K ? 25 marks S: specific heat of argon N: number of atoms in 1g of argon S = 0.31 J/g K = a x N N = S/a = (0.31 J/g K) / (2.3x10-23 J/K) = 1.4x1022 ? 30 marks Avogadro’s number (NA) : Number of argon atoms in 40 g of argon NA = (40)(1.4x1022) = 5.6 x1023 ? 20 marks
2-1. ? 30 marks mass of a typical star = (4/3)(3.1)(7x108 m)3(1.4 g/10-6 m3) = 2×1033 g mass of protons of a typical star = (2×1033 g)(3/4 + 1/8) = 1.8×1033 g number of protons of a typical star = (1.8×1033 g)(6×1023/g) = 1×1057 number of stellar protons in the universe = (1×1057)(1023) = 1×1080 Partial credits on principles: Volume = (4/3)(3.14)radius3×density; 4 marks 1 mole = 6×1023; 4 marks Total number of protons in the universe = number of protons in a star ×1023; 2 marks Mass fraction of protons from hydrogen = (3/4)(1/1); 5 marks Mass fraction of protons from helium = (1/4)(2/4); 10 marks 2-2. ? 30 marks ?E(2→3) = C(1/4 - 1/9) = 0.1389 C λ(2→3) = 656.3 nm ?E(1→2) = C(1/1 - 1/4) = 0.75 C λ(1→2) = (656.3)(0.1389/0.75) = 121.5 nm No penalty for using Rydberg constant from memory. 15 marks penalty if answered in a different unit (Hz, etc.) 2-3. T = (2.9×10-3 m K)/1.215×10-7 m = 2.4×104 K ? 10 marks 2-4. ? 20 marks .
λ = 3 × 108 m/1.42 × 109 = 0.21 m T = (2.9 × 10-3 m K)/0.21 m = 0.014 K 2-5. ? 10 marks 14N + 4He → ( 17O ) + 1H 178 O-17, O acceptable
3-1. kdes = A exp(-Edes/RT) = (1x1012 s-1)(5x10-32) = 5x10-20 s-1 at T = 20 K ? 10 marks surface residence time, τresidence = 1 / kdes = 2x1019 s = 6x1011 yr ? 20 marks (full credit for τhalf-life = ln2 / kdes = 1x1019 s = 4x1011 yr) 3-2. The distance to be traveled by a molecule: x = πr = 300 nm. kmig = A exp(-Emig/RT) = (1x1012 s-1)(2x10-16 ) = 2x10-4 s-1 at T = 20 K ? 5 marks average time between migratory jumps, τ = 1 / kmig = 5x103 s the time needed to move 300 nm residence time = 2x1019 s = (300 nm/0.3 nm) jumps x (5x103 s/jump) = 5x106 s = 50 days ? 15 marks (Full credit for the calculation using a random-walk model. In this case: t = τ (x/d) 2 = 5 x 109 s = 160 yr. The answer is still (b).) (a) (b) (c) (d) (e)
10 marks 3-3. k(20 K) / k(300 K) = exp[(E/R) (1/T1 - 1/T2)] = e-112 = ~ 10-49 for the given reaction ).) ? 15 marks The rate of formaldehyde production at 20 K = ~ 10-49 molecule/site/s = ~ 10-42 molecule/site/ yr ? 10 marks (The reaction will not occur at all during the age of the universe (1x1010 yr).) 3-4. circle one (a) (b) (c) (a, b) (a, c) rate = 10-42 molecules/site/yr (b, c) (a, b, c) (15 marks, all or nothing)
4-1. H P Number of atoms ( 11.3 ) 1
? 10 marks Theoretical wt % ( 3.43 )
? 10 marks 4-2. NNNNHguanineHONHHHNNNOcytosineNNNHNHNOHNNOthymine adenine(10 marks on each) 4-3. 7 marks each, 20 marks for three OHNHONOHNHHONNNHNONHNHOOHNNNHONNNOHNNONNNHNNNNNHHcytosinethymineguanineadeninecytosinecytosinethymineOthymineOHNHNNNNNHONOHNNNNNHONNHNHNOHNNOONHHguaninethymineNNOHcytosineHNHadenineNNNthyminethymineHHNNNNNHONHHNNNNHNHNNNguanineguanineadenineadenine 4-4. 2.5 marks for each bracket NH2NNHNNNNHNguanineONHNH2NHUracilONHONHNH2NOadeninecytosineHCN ( 5 ) ( 5 ) ( 4 )
( 4 ) H2O ( 0 ) ( 1 ) ( 2 ) ( 1 )
5-1. (20 marks) 1st ionization is complete: H2SO4 → H+ + HSO4- [H2SO4] = 0 2nd ionization: [H+][SO42-]/[HSO4-] = K2 = 1.2 x 10-2 (1) Mass balance: [H2SO4] + [HSO4-] + [SO42-] = 1.0 x 10-7 (2) Charge balance: [H+] = [HSO4-] + 2[SO42-] + [OH-] (3) Degree of ionization is increased upon dilution. [H2SO4] = 0 Assume [H+]H2SO4 = 2 x 10-7 From (1), [SO42-]/[HSO4-] = 6 x 104 (2nd ionization is **plete) [HSO4-] = 0 From (2), [SO42-] = 1.0 x 10-7 [5 marks] From (3), [H+] = (2 x 10-7) + 10-14/[H+] [H+] = 2.4 x 10-7 (pH = 6.6) [8 marks] [OH-] = 10-14/(2.4 x 10-7) = 4.1 x 10-8 [2 marks] From (1), [HSO4-] = [H+][SO42-]/K2 = (2.4 x 10-7)(1.0 x 10-7)/(1.2 x 10-2) = 2.0 x 10-12 [5 marks] Check charge balance: 2.4 x 10-7 ? (2.0 x 10-12) + 2(1.0 x 10-7) + (4.1 x 10-8) Check mass balance: 0 + 2.0 x 10-12 + 1.0 x 10-7 ? 1.0 x 10-7 Species HSO4- Concentration ** x 10-12 ** x 10-7 SO42- ** x 10-7 H+
国际化学奥林匹克竞赛-国际化学奥林匹克竞赛-第38届ICHO理论试题(中文版)答案
