2018年AP化学考试真题+解析（下）
2018/12/14 17:36:43来源：新航道作者：新航道
摘要：2018年AP化学考试真题+解析（下）
上一篇小编给大家分享了2018年AP物理2真题解析(FR),今天上海新航道AP培训中心接着分享.2018年AP化学考试真题+解析（下）。接下来 我们会更新 2018年AP化学考试真题+解析（上）请有需要的同学定期关注的AP真题栏目。
第一题
解析：
(a) For nonpolar CS2 molecule, there is London dispersion force; for polar COS molecule, there are London dispersion force and dipole-dipole force. Since CS2 has a larger molecular mass and more electrons than COS, CS2 has stronger London dispersion force than the combination of London dispersion force and dipole-dipole force in COS. Therefore, CS2 has a higher boiling point than COS.
(b) PV = nRT
P = nRT / V = 10.0g x (1mol/76.13g) x 0.08206 L atm mol-1 K-1 x 325K / 5.0L = 0.70 atm
第二题
解析：
(a) Since percent ionization of HF is 13.0%, most particles containing F should be in the form of unionized form HF at equilibrium, which means that HF is a weak acid and only partially ionized in water. Figure 1 shows partial ionization of HF. In figure 2, all HF molecules are completely ionized into H3O+ and F-, which corresponds to a strong acid.
(b)
Ka = [F-][H3O+]/[HF] = 0.004552/0.03045 = 6.8 x 10-4
(c) The percent ionization of HF in the solution will increase. When the solution is diluted, the concentration of the “particles” in the solution is reduced. According to Le Chatelier’s principle, this reduction in particle concentration is counteracted by shifting the reaction to the side with more particles.The equilibrium shifts from the nonionized acid side to the side containing the H+ ion and the conjugate base. The percent ionization of the acid increases.
第三题
解析：
(a) The missing component of the cell is salt bridge. It allows electrical contact between the two solutions and the cell can form a closed circuit.
(b) (i) = = 0.80V - E0(Cr3+/Cr) = 1.54V
E0(Cr3+/Cr) = 0.80V – 1.54V = -0.74V
(ii) 3Ag+(aq) + Cr(s) -----> 3Ag(s) + Cr3+(aq)
(iii) ?G0 = -nFE0 = - 3mol x 96485 C/mol x 1.54V = -4.46 x 105 J
第四题
解析：
(a) N (7 electrons)
(b) t1/2 = ln2 / k
k = ln2 / t1/2 = ln2 / 10.min = 0.069 min-1
(c) 1/64 = (1/2)6, 6 x t1/2 = 6 x 10.min = 60.min
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