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Set up the problem in the usual way. You are given

Solution Map :. Solution :. SKILLBUILDER 12.1. Using the Heat of Vaporization in Calculations. Calculate the amount of heat in kilojoules required to vaporize 2.58 kg of water at its boiling point. EXAMPLE 12.1. Using the Heat of Vaporization in Calculations.

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Set up the problem in the usual way. You are given

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  1. Solution Map: Solution: SKILLBUILDER 12.1 Using the Heat of Vaporization in Calculations Calculate the amount of heat in kilojoules required to vaporize 2.58 kg of water at its boiling point. EXAMPLE 12.1 Using the Heat of Vaporization in Calculations Calculate the amount of water in grams that can be vaporized at its boiling point with 155 kJ of heat. Set up the problem in the usual way. You are given the number of kilojoules of heat energy and asked to find the mass of water that can be vaporized with the given amount of energy. The required conversion factors are the heat of vaporization and the molar mass of water. Given: 155 kJ Find: g H2O Conversion Factors: ΔHvap = 40.7 kJ/mol (at 100 °C) 18.02 g H2O = 1 mol H2O Draw the solution map beginning with the energy in kilojoules and converting to moles of water and then to grams of water. Follow the solution map to solve the problem.

  2. A drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25 °C. If the heat released during condensation goes only toward heating the metal, what is the final temperature in Celsius of the metal block? (The specific heat capacity of aluminum is 0.903 J/g °C.) SKILLBUILDER PLUS FOR MORE PRACTICE Example 12.7; Problems 51, 52, 53, 54, 55, 56. EXAMPLE 12.1 Using the Heat of Vaporization in Calculations Continued

  3. Solution Map: Solution: SKILLBUILDER 12.2 Using the Heat of Fusion in Calculations Calculate the amount of heat absorbed when a 15.5-g ice cube melts (at 0 °C). EXAMPLE 12.2 Using the Heat of Fusion in Calculations Calculate the amount of ice in grams that, upon melting (at 0 °C), absorbs 237 kJ of heat. Set up the problem in the usual way. You are given the number of kJ of heat energy and asked to find the mass of ice that absorbs the given amount of energy upon melting. The required conversion factors are the heat of fusion and the molar mass of water. Given: 237 kJ Find: g H2O (ice) Conversion Factors: ΔHfus = 6.02 kJ/mol 1 mol H2O = 18.02 g H2O Draw the solution map beginning with the energy in kilojoules and converting to moles of water and then to grams of water. Follow the solution map to solve the problem.

  4. SKILLBUILDER PLUS A 5.6-g ice cube (at 0 °C) is placed into 195 g of water initially at 25 °C. If the heat absorbed for melting the ice comes only from the 195 g of water, what is the temperature change of the 195 g of water? FOR MORE PRACTICE Example 12.8; Problems 59, 60, 61, 62. EXAMPLE 12.2 Using the Heat of Fusion in Calculations Continued

  5. FOR MORE PRACTICE Problems 67, 68. SKILLBUILDER 12.5 Dispersion Forces Which hydrocarbon, CH4 or C2H6, has the higher boiling point? EXAMPLE 12.3 Dispersion Forces Which halogen, Cl2 or I2 has the higher boiling point? Solution: The molar mass of Cl2 is 70.90 g/mol and the molar mass of I2 is 253.80 g/mol. Since I2 has the higher molar mass, it has stronger dispersion forces and therefore the higher boiling point.

  6. EXAMPLE 12.4 Dipole–Dipole Forces Which of the following molecules have dipole–dipole forces? (a) CO2 (b) CH2Cl2 (c) CH4 Solution: A molecule will have dipole–dipole forces if it is polar. To determine whether a molecule is polar, you must: 1. determine whether the molecule contains polar bonds, and 2. determine whether the polar bonds add together to form a net dipole moment (Section 10.8). (a) Since the electronegativities of carbon and oxygen are 2.5 and 3.5, respectively (Figure 10.2), CO2 has polar bonds. The geometry of CO2 is linear. Consequently, the polar bonds cancel; the molecule is not polar and does not have dipole–dipole forces. (b) The electronegativities of C, H, and Cl are 2.5, 2.1, and 3.5, respectively. Consequently, CH2Cl2 has two polar bonds (C— Cl) and two bonds that are nearly nonpolar (C — H). The geometry of CH2Cl2 is tetrahedral. Since the C — Cl bonds and the C — H bonds are different, they do not cancel, but sum to a net dipole moment. Therefore the molecule is polar and has dipole- dipole forces.

  7. FOR MORE PRACTICE Problems 63, 64, 65, 66. SKILLBUILDER 12.4 Dipole–Dipole Forces Which of the following molecules have dipole–dipole forces? (a) CI4 (b) CH3Cl (c) HCl EXAMPLE 12.4 Dipole–Dipole Forces Continued (c) Since the electronegativities of C and H are 2.5 and 2.1, respectively, the C — H bonds are nearly nonpolar. In addition, since the geometry of the molecule is tetrahedral, any slight polarities that the bonds might have will cancel. CH4 is therefore nonpolar and does not have dipole–dipole forces.

  8. One of the following compounds is a liquid at room temperature. Which one and why? FOR MORE PRACTICE Example 12.10; Problems 69, 70, 71, 72, 73, 74. SKILLBUILDER 12.5 Hydrogen Bonding Which has the higher boiling point, HF or HCl? Why? EXAMPLE 12.5 Hydrogen Bonding Solution: The three compounds have similar molar masses. formaldehyde 30.03 g/mol fluoromethane 34.04 g/mol hydrogen peroxide 34.02 g/mol Therefore, the strengths of their dispersion forces are similar. All three compounds are also polar, so they have dipole–dipole forces. Hydrogen peroxide, however, is the only compound that also contains H bonded directly to F, O, or N. Therefore it also has hydrogen bonding and is most likely to have the highest boiling point of the three. Since the problem stated that only one of the compounds was a liquid, we can safely assume that hydrogen peroxide is the liquid. Note that although fluoromethane contains both H and F, H is not directly bonded to F, so fluoromethane does not have hydrogen bonding as an intermolecular force. Similarly, although formaldehyde contains both H and O, H is not directly bonded to O, so formaldehyde does not have hydrogen bonding either.

  9. FOR MORE PRACTICE Problems 77, 78, 79, 80. SKILLBUILDER 12.6 Identifying Types of Crystalline Solids Identify each of the following solids as molecular, ionic, or atomic. (a) NH3(s) (b) CaO(s) (c) Kr(s) EXAMPLE 12.6 Identifying Types of Crystalline Solids Identify each of the following solids as molecular, ionic, or atomic. (a) CaCl2(s) (b) Co(s) (c) CS2(s) Solution: (a) CaCl2 is an ionic compound (metal and nonmetal) and therefore forms an ionic solid (CaCl2 melts at 772 °C). (b) Co is a metal and therefore forms a metallic atomic solid (Co melts at 1768 °C). (c) CS2 is a molecular compound (nonmetal bonded to a nonmetal) and therefore forms a molecular solid (CS2 melts at –110 °C).

  10. Given: 84.8 g H2O Find: heat (kJ) Conversion Factors: ΔHvap = 40.7 kJ/mol (at 100 °C) 1 mol H2O = 18.02 g H2O Solution Map: Solution: EXAMPLE 12.7 Using Heat of Vaporization in Calculations Calculate the amount of heat required to vaporize 84.8 g of water at its boiling point.

  11. Given: 12.4 g H2O Find: heat (kJ) Conversion Factors: ΔHfus = 6.02 kJ/mol 1 mol H2O = 18.02 g H2O Solution Map: Solution: The heat emitted is 4.14 kJ. EXAMPLE 12.8 Using Heat of Fusion in Calculations Calculate the amount of heat emitted when 12.4 g of water freezes to solid ice.

  12. EXAMPLE 12.9 Determining the Types of Intermolecular Forces in a Compound Determine the types of intermolecular forces present in each of the following substances. (a) N2 (b) CO (c) NH3 Solution: (a) N2is nonpolar and therefore has only dispersion forces. (b) CO is polar and therefore has dipole–dipole forces (in addition to dispersion forces). (c) NH3has hydrogen bonding (in addition to dispersion forces and dipole–dipole forces).

  13. EXAMPLE 12.10 Using Intermolecular Forces to Determine Melting and/or Boiling Points Arrange each of the following in order of increasing boiling point. (a) F2 , Cl2 , Br2 (b) HF, HCl, HBr Solution: (a) Since these all have only dispersion forces, and since they are similar substances (all halogens), the strength of the dispersion force will increase with increasing molar mass. Therefore, the correct order is F2< Cl2 < Br2. (b) Since HF has hydrogen bonding, it has the highest boiling point. Between HCl and HBr, HBr (because of its higher molar mass) has a higher boiling point. Therefore the correct order is HCl < HBr < HF.

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