1 sublimation condensation freezing 2 evaporation

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1. sublimation, condensation, freezing 2. evaporation, deposition, freezing 3. evaporation, fusion, sublimation 4. condensation, fusion, deposition 5. condensation, freezing, deposition cor- rect Explanation: Phase changes are equilibrium processes and are thus temperature dependent. Also, the ΔH and ΔS always have the same sign. Lowering the temperature makes exothermic phase transitions more likely - freezing, con- densation and deposition. 017 3.8points Water boils at a much higher temperature than all other molecules of similar molecular weight because I) it makes very strong hydrogen bonds; II) it is a very polar molecule; III) it can make more hydrogen bonds per
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casey (rmc2555) – Homework 1 – holcombe – (51395) 6 molecule than any other molecule; IV) the OH bond is the most polar of all common bonds leading to the strongest dipole-dipole interactions. 1. II and IV only 2. IV only 3. III and IV only 4. I and III only correct 5. I only 6. I and II only 7. III only 8. II only Explanation: Not all polar molecules are liquids ( e.g. , ammonia). The most polar bond is H-F. Water’s particular shape (bent) allows it to make several hydrogen bonds to neighboring water molecules. 018 3.8points The vapor pressure of a pure liquid depends on which of the following I. the volume of the liquid II. the volume of the gas III. the surface area of the liquid IV. the temperature 1. all of them 2. only IV correct 3. only II 4. I and II 5. III and IV 6. only I 7. only III Explanation: The vapor pressure of a given liquid de- pends only the temperature 019 3.8points Carbon can exist in two forms, graphite and diamond. The heats of combustion at 25 C and one atm pressure for these two forms were measured and found to be C(graphite) + O 2 (g) CO 2 (g) Δ H = - 393.5 kJ/mol C(diamond) + O 2 (g) CO 2 (g) Δ H = - 395.4 kJ/mol From this information, calculate Δ H at 25 C and 1 atm pressure for the reaction C(graphite) C(diamond) . 1. - 788.9 kJ/mol 2. +788.9 kJ/mol 3. +1.9 kJ/mol correct 4. Totally irrelevant 5. - 1.9 kJ/mol Explanation: The desired reaction C(graphite) C(diamond) is the sum of C(graphite) + O 2 (g) CO 2 (g) Δ H = - 393.5 kJ/mol AND the reverse of the other reaction CO 2 (g) C(diamond) + O 2 (g) Δ H = +395.4 kJ/mol Note that the sign changes because we’ve re- versed the reaction. O 2 (g) and CO 2 (g) cancel out, and then you can sum the Δ H s for each reaction. C(graphite) C(diamond) Δ H = +1.9 kJ/mol 020 3.8points Is the reaction Cl 2 (g) + 2 HBr(g) Br 2 ( ) + 2 HCl(g)
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casey (rmc2555) – Homework 1 – holcombe – (51395) 7 spontaneous at 25 C and one atmosphere pressure? The standard molar Gibbs free en- ergy of formation of HBr(g) at 25 C is - 53 . 43 kJ/mol and that of HCl(g) is - 95 . 30 kJ/mol. 1. No, because Δ G 0 for the reaction is posi- tive. 2. No, because Δ G 0 for the reaction is neg- ative. 3. Yes, because Δ G 0 for the reaction is pos- itive.
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