课程名称︰普通化学二 课程性质︰化学系必修 课程教师︰郑原忠 开课学院：理学院 开课系所︰化学系 考试日期（年月日）︰100/06/20 考试时限（分钟）：120min (延长至130min) 是否需发放奖励金：是 （如未明确表示，则不予发放） 试题 : General Chemistry (II) Final Exam *Refer to the last page for physical constants and structures of amino acids.* 1. (8%) Write balanced wquations that represent the following nuclear reactions. (a) Alpha emission by (224,88)Ra (b) Positron emission by (26,14)Si (c) Electron capture by (82,38)Sr (d) Beta emission by (100,41)Nb 2. (15%) Radioactive gaseous (222,86)Rn and (220,86)Rn can be produced from the decay of uranium and thorium in rocks and soil. (a) (222,86)Rn is a decay product of (238,92)U, while (220,86)Rn comes from (232,90)Th. How many alpha particles are emitted in the formation of these radon isotopes from their uranium or thorium starting points? Can alpha decay alone explain the formation of these radon isotopes? If not, state what other types of decay must occur. (b) Can (222,86)Rn and (220,86)Rn decay by alpha particle emission? Write balanced nuclear equations for these two decay processes, and predict for each reaction whether it is allowed or not. The masses of (222,86)Rn and (220,86)Rn atoms are 222.01757 and 220.01140 u, respectively; those of electron, (4,2)He, (218,84)Po and (216,84)Po are 0.00055, 4.002602, 218.0089 and 216.00192 u, respectively. (c) Calculate the energy released in the alpha decay of one (220,86)Rn nuc- leus, in million electron volts and in joules. (d) The half-life of (222,86)Rn is 3.82 days. Calculate the initial activity of 2.00*10^(-8) g of (222,86)Rn, in disintegrations per second. What will be the activity of the (222.86)Rn after 14 days? (e) The half-life of (220,86)Rn is 54 s. Are the health risks of exposure to a given amount of radon for a given short length of time greater or sma- ller for (220,86)Rn than for (222,86)Rn? Radon, being gaseous, post high health risks. Explain why radon is a gas. 3. (5%) The second-order Bragg diffraction of x-rays with λ= 1.237 anstron from a set of parallel planes ina luminum occurs at an angle 2θ= 35.58 deg. Cal- culate the distance between the sacttering planes. 4. (10%) The diamond unit cell can be considered as an FCC lattice with half of the tetrahedral interstitial sites filled by C atoms. (a) Sketch the diamond unit cell. What is the number of atoms per unit cell? (b) What is the packing fraction of the diamond unit cell? Given that the density of diamond is 3.5 g/cm^3, estimate the C-C bond length in dia- mond. 5. (6%) This question concerns the chemical compositions of solids. (a) Perovskite has a calcium atom at each corner of the unit cell, a titani- um at the center of the unit cell, and an oxygen atom at the center of each face. What is its chemical formula? (b) A chromium oxide has a structure in which chromium ions occupy 2/3 of the octahedral interstitial sites in a hexagonal closed-pached lattice of oxygen ions. What's the d-electron configuration of the chromium ion? 6. (8%) Predict the structure of each of the following minerals (network, sheets, double chains, and so forth.) In some cases, the Al atoms grouped with the Si and O in the formula substitute for Si in tetrahedral sites. Give the oxidation state of each atom. (a) Andradite, Ca3Fe2(SiO4)3 (b) Hardystonite, Ca2ZnSi2O7 (c) Keatite, Li(AlSiO6) (d) Muscovite, KAl2(AlSi3O10)(OH)2 7. The mobilities for Na(+) and Cl(-) in aqueous solution are 5.19*10^(-4) cm^2 V^(-1) s^(-1) and 7.91*10^(-4) cm^2 V^(-1) s^(-1), respectively. (a) (5%) Calculate the conductivity of a 1.0 M aqueous solution of NaCl at 25 degC. (b) (3%) Explain why the ionic mobility of Na(+) is smaller than that for Cl(-). 8. (10%) Sketch a schema of a dye-sensitized solar cell, and describe how it works as much detail as you can. Make sure you clearly depict the electron flow and all the redox reactions in the cell. What are the properties required for a suitable choice of the dye molecule? 9. (12%) Consider the electronic structure of polyacetylene. (a) Sketch the chemical structure of trans-polyacetylene. Use an energy band model for its π-electrons to explain why it is a semiconductor. (b) Suppose we can stretch the C=C double bond and compress the C-C single bond to make the C atoms all equally spaced in the 1-D chain. Derive the band structure and predict whether this structure is a conductor or a semiconductor. (c) (BONUS) Use a simple MO theory to explain why the structure of the ground state of polyacetylene is what you draw in (a). (This is called bond alternation and it is a consequence if Peierls instability.) 10. (5%) The free-radical polymerization of ethylene is spontaneous. What is the sign of entropy change in this process? Describe whether the polymerization reac- tion is exothermic or endothermic. Explain. 11. (8%) At very low pH, alanine is a diprotic acid that can be represented as +H3N-CH(CH3)-COOH. The pKa of the carboxyl group is 2.3, while that of the -NH3+ is 9.7. (a) At pH=7, what fraction of the amino acid molecules dissolved in an aque- ous solution will be in the zwitterionic form? (b) What fraction of the molecules at this pH will have the form H2N-CH(CH3)-COOH? 12. (a) (4%) Depict the three torsional angles (φ, ψ, and ω) in a protein backbone. Explain why ω is almost always 180 deg in a protein. (b) (4%) What is Levinthal's paradox? Explain. Why do proteins fold correct- ly and rapidly. (c) (4%) Two polypeptide chains with very different-looking sequences actu- ally hace quite similar folding structures. Parts of the sequences, af- ter aligned, look like the following: HRKDHFIVCGHSIL-AINTILQ--LNQRGQNVTV----ISNL-PEDDIKQLEQRLGDNAD . :..:::: : .....:. :.. ..:.: . :. :. ... : .: ... SGRKHIVVCGHITLESVSNFLKDFLHKDRDDVNVEIVFLHNISPNLELEALFKRHFTQVE Explain why these two polypeptide chains fold into similar structure. Use the table for the properties of amino acid in the last page to ra- tionalize your answers. [胺基酸结构图请见 http://en.wikipedia.org/wiki/Amino_acid ] ********************* Physical constants: c = 3.0 * 10^8 m/s 1 u = 1.6605 * 10^(-27) kg mass of electron = 9.1 * 10^(-31) kg h = 6.626 * 10^(-34) Js 1 eV = 1.60 * 10^(-19) J R = 0.082 L atm K^(-1) mol^(-1) = 8.3 J mol^(-1) K^(-1) ********************* THIS IS THE END OF THE EXAM --

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