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UT CH 302 - Practice Exam 1
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Practice Exam 1 1. In general, increasing the temperature leads to which phase changes? 1. sublimation, vaporization, fusion *correct* 2. sublimation, vaporization, condensation 3. fusion, vaporization, deposition 4. vaporization, freezing, deposition 5. freezing, deposition, condensation Explanation: For each of these the final state is the one favored at the highest temperature 2. Vapor pressure: A. Is an equilibrium process B. increases as IMF increases C. is temperature dependent D. is higher in CH3CH3 than H2 E. decreases with the addition of solutes F. depends on the whole volume of the liquid G. decreases at temp increases 1.A, C, E, G 2. A, B, C, E 3. A, C, D, E 4. B, D, F, G 5. A, C, E *correct* 6. A, C, E, F Explantion: Vapor pressure is an equilibrium process that occurs only at the surface area of the liquid. The VP decreases with the presence of solutes because they will also take positions at the surface of the mixture, thus decreasing the surface area of the liquid. This is VP depression. Additionally, as IMF increases, VP decreases, since those IMF's are intermolecular attractions that are keeping the liquid intact. These require energy to be broken, so evaporation is less spontaneous. Ethane is larger than hydrogen, so it will have a lower vapor pressure (size does matter). 3. Rank the following in decreasing order of solubility in water: Mg(OH)2, NaCl, Al2O3, CsI, BeBr2, KOH, BaO. 1. Al2O3 > BaO> Mg(OH)2> BeBr2> KOH> NaCl> CsI 2. NaCl> CsI> KOH> BeBr2>Mg(OH)2> BaO> Al2O3 3. Al2O3 > BaO> Mg(OH)2> BeBr2> KOH> NaCl> CsI 4. CsI > NaCl> KOH> BeBr2> Mg(OH)2> BaO> Al2O3 *correct* 5. NaCl> CsI> BeBr2> KOH> Mg(OH)2> BaO> Al2O3 6. CsI > NaCl> BeBr2> KOH> Mg(OH)2> BaO> Al2O3 Explanation: Solubility in water will depend on the crystal lattice energy of the salt. Rank the salts based on their charge densities. CsI has a lower charge density than NaCl, and its ionic bond is much weaker. KOH is next because it is still in the +1,-1 charge category, but now there is the ionic bond and additional H-bonds (from the OH) that much be broken for this salt to dissolve. BeBr2 has a greater solubility than Mg(OH)2 for the same reason. Though an oxide, BaO has a much weaker charge density than Al2O3. 4. According to the given phase diagram for carbon, how many triple points are there? What would you expect to see at 0.01GPa, 4500K? 1. 5; metastable liquid, graphite and liquid2. 4; graphite, metastable liquid and vapor. *correct* 3. 5; vapor, metastable liquid, graphite. 4. 4; graphite, liquid, diamond. 5. 3; graphite, metastable liquid and vapor. Explanation: The triple point is the place where three lines (phases) intersect, and at that point, each of those phases is present. There are only 4 here, the point at 10.0 GPa does not count...there is crazy stuff going on there. There are three down where liquid, graphite, vapor and metastable liquid are interacting. The fourth is around 10000K:100GPa, where diamond, liquid and metal interface. 5. According to the given pressure-temperature diagram, a sample of carbon at 0.005GPa and 2000K is graphite. This sample is then heated to 7000K at constant pressure. Then, at constant temperature, the sample is compressed to 1.00GPa. Again, at constant pressure, the temperature is decreased to 1000K. At this temperature, the pressure is increased to 500GPa. How many phase transitions has the carbon sample undergone? 1. 6 *correct* 2. 4 3. 1 4. 5 5. 7 Explanation: Navigating through the pressure-temperature diagram while the sample undergoes the stated manipulations, it starts out as graphite. Then, it is heated to vapor, compressed to liquid, cooled back to graphite and finally to graphite+metastable diamond. Lastly, it is compressed further to diamond+metastable graphite and finally to diamond. Six transitions total. 6. How much heat is generated when 10g steam at 115 oC is cooled to -75 oC? Cice = a J/g oC; Cwater = b J/g oC; Csteam = c J/g oC; ΔHvap = d J/g; ΔHfus = e J/g. 1. q = - [150(5a + c) + 10(100b + d + e)] J*correct* 2. q = [150(5a + c) + 10(100b + d + e)] J 3. q = -[150(a + c) + 10(100b + d + e)] J 4. q = [150 (a + c) + 1000b] J 5. q = - [150 (5a + c) + 1000b] J Explanation: Cooling steam from 115 oC to 100 oC, ∆H = m csteam ∆T = 10g x c x 15 oC = 150c J. Changing steam to water at 100 oC (the boiling point of water), ∆H = m ∆Hvaporization = 10g x d = 10d J. Cooling water at 100 oC to water at 0 oC, ∆H = m cwaterb x 100 oC = 1000b J. Changing water to ice at 0 oC (the freezing point of water), ∆H = m ∆Hfusion = 10g x e = 10e J. Finally, cooling ice from 0 oC to -75 oC, ∆H = m cice ∆T = 10g x a x 75 oC = 150(5a) J. Adding everything together, the ∆Hsys for the entire process is [150(5a + c) + 10(100b + d + e)] J. Since the system is losing heat, the sign is negative. ∆T = 10g x 7. Which solvent would you expect BH3 gas to be most soluble in? 1. C6H6 *Correct* 2. H2O 3. CH3CH2OH 4. CH2Cl2 5. N(CH3)3 Explanation: BH3 is a non-polar gas. According to like dissolves like, it will be most soluble in a non-polar solvent. Benzene (C6H6) is the only non-polar solvent here. If you are unsure about any of them, draw the lewis dot structures. 8. Rank the following in decreasing order of miscibility with water: H2O2, C6H6, HOCH2CH2NH2,C2H6Cl2, and CH3CH2CH2SH. 1. H2O2 > HOCH2CH2NH2 > CH3CH2CH2SH > C2H6Cl2 > C6H6 2. HOCH2CH2NH2 > H2O2 > C2H6Cl2 > CH3CH2CH2SH > C6H6 3. H2O2 > HOCH2CH2NH2 > C2H6Cl2 > CH3CH2CH2SH > C6H6 *correct* 4. C6H6 > CH3CH2CH2SH > C2H6Cl2 > HOCH2CH2NH2 > H2O2 5. C6H6 > CH3CH2CH2SH > C2H6Cl2 > H2O2 > HOCH2CH2NH2 Explanation: following the rule ”like dissolves like” small very polar molecules will be more soluble in water than non-polar ones. Hydrogen peroxide is most polar (and actually looks like water) and so its most soluble. Next is amino-ethanol since it has a small carbon chain and has can participate in 3 hydrogen bonds. Dichloroethane is next, due to the polar chlorine-carbon bonds. Then propanethiol, since there is only a small EN difference between S-H and C-S, followed by benzene which is completely non-polar. 9. At 25 C, the vapor pressure of pure benzene (C6H10) is 0.1252 atm. Suppose 6.4 g of napthalene, C10H8, is dissolved in 78 g of benzene (benzene’s molar mass is 78 g/mol). Assuming ideal behavior, what is


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UT CH 302 - Practice Exam 1

Type: Miscellaneous
Pages: 10
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