Chapter 11 Solutions to Problems

Chapter 11 Solutions to Problems - Chap ter Eleven Practice...

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Chapter Eleven 209 Practice Exercises 11.1 (a) CH 3 CH 2 CH 2 CH 2 CH 3 < CH 3 CH 2 OH < Ca(OH) 2 (b) CH 3 –O–CH 3 < CH 3 CH 2 NH 2 < HOCH 2 CH 2 CH 2 CH 2 OH 11.2 Propylamine would have a substantially higher boiling point because of its ability to form hydrogen bonds (there are N–H bonds in propylamine, but not in trimethylamine.) 11.3 The piston should be pushed in. This will decrease the volume and increase the pressure, and when equilibrium is re–established, there will be fewer molecules in the gas phase. 11.4 The number of molecules in the vapor will decrease, and the number of molecules in the liquid will increase, but the sum of the molecules in the vapor and the liquid remains the same. 11.5 The boiling point is most likely (a) less than 10 °C above 100 °C. 11.6 We use the curve for water, and find that at 330 torr, the boiling point is approximately 75 °C. 11.7 Adding heat will shift the equilibrium to the right, producing more vapor. This increase in the amount of vapor causes a corresponding increase in the pressure, such that the vapor pressure generally increases with increasing temperature. 11.8 Boiling Endothermic Melting Endothermic Condensing Exothermic Subliming Endothermic Freezing Exothermic No, each physical change is always exothermic, or always endothermic as shown. 11.9 For calcium: 8 corners × 1/8 Ca 2+ per corner = 1 Ca 2+ 6 faces × 1/2 Ca 2+ per face = 3 Ca 2+ For fluoride: 8 inside the unit cell × 1 F = 8 F Thus, the ratio is 4 Ca 2+ to 8 F . 11.10 For cesium: 8 corners × 1/8 Cs + per corner = 1 Cs + For chloride: 1 Cl in center, Total: 1 Cl Thus, the ratio is 1 to 1. 11.11 The compound is an organic molecule and the solid is held together by dipole–dipole attractions and London forces. It is also a soft solid with a low melting point, so it is a molecular crystal. 11.12 Because this is a high melting, hard material, it must be a covalent or network solid. Covalent bonds link the various atoms of the crystal. 11.13 Since the melt does not conduct electricity, it is not an ionic substance. The softness and the low melting point suggest that this is a molecular solid, and indeed the formula is most properly written S 8 . 11.14 The line from the triple point to the critical point is the vapor pressure curve, see Figure 11.21. 11.15 Refer to the phase diagram for water, Figure 11.46. We "move" along a horizontal line marked for a pressure of 2.15 torr. At –20 °C, the sample is a solid. If we bring the temperature from –20 °C to 50 °C,
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Chapter Eleven 210 keeping the pressure constant at 2.15 torr, the sample becomes a gas. The process is thus solid gas, i.e. sublimation. 11.16
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Chapter 11 Solutions to Problems - Chap ter Eleven Practice...

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