HW Review-book- Ch 11-key Review for Chapter 11: p.481 #95, 96, 98, 103, 107, 112, 113, 118, 119, 122, 126, 128, 133, 137; Study Guide p. 231 #13 95 Name the kinds of attractive forces that must be overcome in order to: (a) Boil liquid ammonia: requires breaking hydrogen bonds between molecules. Dipole-dipole

and dispersion forces must also be overcome. (b) Melt solid P4: P4 is a nonpolar molecule, so the only intermolecular forces are dispersion. (c) Dissolve CsI in liquid HF: CsI is an ionic solid. Ion-ion interparticle forces must be overcome. (d) Melt potassium metal: Metallic bonds must be broken. 96 Which of the following properties indicates very strong IMFs in a liquid? (a) A low surface tension means the attraction between molecules making up the surface is weak.

Water has a high surface tension; water bugs could not "walk" on the surface of a liquid with a low surface tension.

(b) A low critical temperature means a gas is very difficult to liquefy by cooling. This is the result of weak intermolecular attractions. Helium has the lowest known critical temperature (5.3 K).Don’t worry about understanding this answer.

(c) A low boiling point means weak intermolecular attractions. It takes little energy to separate the particles. All ionic compounds have extremely high boiling points.

(d) A low vapor pressure means it is difficult to remove molecules from the liquid phase because of high intermolecular attractions. Substances with low vapor pressures have high boiling points.

Thus, only choice (d) indicates strong intermolecular forces in a liquid. The other choices indicate weak intermolecular forces in a liquid.

98 Elemental boron has a high melting point (2300°C), poor electrical and thermal conductivity,

is not soluble in water and is very hard. Classify it as one of the crystalline solids. • The conductivity and combination of high MP and insolubility could place boron in either the ionic

or network covalent categories. • However, boron atoms will not alternately form positive and negative ions to achieve an ionic

crystal. • The structure is network covalent because the units are single boron atoms. (Boron can’t be

molecular covalent because then mp would be low.) 103 Explain what happens to water in a vacuum chamber as it

first boils, then freezes, then disappears. • As the vacuum pump is turned on and the pressure is

reduced, the liquid will begin to boil because the vapor pressure of the liquid is greater than the very low external pressure--approximately zero. (Bp decreases when external pressure decreases.)

• Boiling is an endothermic process (+ΔHvap) because bonds are being broken. Thus, the process of boiling absorbs heat from the water, so the water cools.

• Soon the water loses sufficient heat to drop the temperature below water’s freezing point, so the water freezes.

• Finally the ice sublimes (disappears) under reduced pressure.

2 Homework Ch 11 Review Answer key

107 Identify W, X, Y, and Z as gold, lead sulfide, mica (SiO2) or I based on the following tests: (a) W conducts electricity, X, Y, Z don’t; (b) W flattens with a hammer, X shatters, Y is pulverized, and Z is unaffected; (c) When heated, Y melts/sublimes, the others remain solid; (d) X dissolves in 6 M HNO3, the others have no effect.

• W must be a reasonably non-reactive metal. It conducts electricity and is malleable, but doesn’t react with nitric acid. Of the choices, it must be gold(metallic).

• X is non-conducting (and therefore isn’t a metal), is brittle, is high melting, and reacts with nitric acid. Of the choices, it must be lead sulfide (ionic).

• Y doesn’t conduct and is soft (and therefore is a nonmetal). It melts at a low temperature with sublimation. Of the choices, it must be iodine (molecular covalent).

• Z doesn’t conduct, is chemically inert, and is high melting (network solid). Of the choices, it must be mica, SiO2 (network covalent).

112 Select the substance with the higher BP and identify the IMFs: (a) K2S or (CH3)3N: Ionic forces are much stronger in K2S than the dipole-dipole forces in

(CH3)3N. (b) Br2 or CH3CH2CH2CH3: Both molecules are nonpolar; but Br2 has more electrons, so it has

stronger dispersion attractions. (The boiling point of Br2 is 50°C and that of C4H10 is -0.5°C.) 113 A small drop of oil in water assumes a spherical shape. Explain. (Hint: oil is made up of

nonpolar molecules, which tend to avoid contact with water.) To minimize contact with polar water, the nonpolar oil drop assumes a spherical shape. (For a given

volume the sphere has the smallest surface area.) 118 Heat of hydration, that is, the heat change that occurs when ions become hydrated in solution,

is largely due to ion-dipole interactions. The heats of hydration for the alkali metal ions are Li+ = −520 kJ/mol ; Na+ = −405 kJ/mol K+ =−321 kJ/mol

Account for the trend in these heats of hydration values. • Smaller ions have more concentrated charges (charge densities) and so make stronger ion-dipole

interactions. • The greater the ion-dipole interaction, the more favorable (more exothermic) the heat of hydration.

119 If water were a linear molecule… (a) …would it still be polar? No, it would not be polar because the two O-H bond dipoles would

cancel each other as in CO2. (b) … would the water molecules still be able to form hydrogen bonds with one another? Yes,

hydrogen bonding would still occur between water molecules even if they were linear.???? 122 A beaker of water is placed in a closed container. Predict the effect on the vapor pressure of

the water when.. . (a) its temperature is lowered. Pvap decreases (because rate of evaporation slows down.)

(b) the volume of container doubles: PVAP does not change (twice the moles in twice the volume) (c) More water is added to the beaker: PVAP does not change

(As long as there is some water, the amount of water does not matter.)

Homework Ch 11 Review Answer key 3

126 Carbon and silicon belong to Group 4A of the periodic table and have the same valence electron configuration. However, why does SiO2 have a much higher melting point than CO2? • SiO2 is a network covalent substance with strong covalent bonds between all atoms. • CO2 exists as discrete nonpolar molecules which are only attracted by weak dispersion bonds. • It will take much more energy to break the strong network covalent bonds of SiO2 than the weak

dispersion attractions between molecules of CO2. 128 A 1.20-g sample of H2O is injected into an evacuated 5.00-L Flask at 65°C. What percentage of

the water will be vapor when the system reaches equilibrium? (Assume ideal behavior of water vapor and that the volume of liquid water is negligible. The vapor

pressure of water at 65°C is 187.5 mm Hg)

nwater =

PVRT

=187.5 mmHg × 1 atm

760 mmHg( )(5.00 L)0.0821 L⋅atm

mol⋅K( )(338 K )= 0.0445 mol

mass of water vapor = 0.0445 mol ´ 18.02 g/mol = 0.802 g

0.802 100%1.20

gg

= ´ =2% of H O vaporized 66.8%

133 A chemistry instructor performed the following mystery demonstration. Just before the students

arrived in class, she heated some water to boiling in an Erlenmeyer flask. She then removed the flask from the flame and closed the flask with a rubber stopper. After the class commenced, she held the flask in front of the students and announced that she could make the water boil simply by rubbing an ice cube on the outside walls of the flask. To the amazement of everyone, it worked. Why did the hot water boil in the flask when ice was rubbed on the outside?

• The ice condenses the water vapor inside. Thus, the pressure inside the flask decreases, causing the

boiling point of the water to decreases. • Since the water is still hot, it will begin to boil even though the temperature is slightly below its

normal bp of 100°C 137 Why do citrus growers spray their trees with water to prevent them from freezing? Main reason: As water freezes, heat is released because bonds are made.

H2O(l) ® H2O(s) DHfreezing = −DHfus = -6.01 kJ/mol

The heat released protects the fruit and keeps the temperature at 0°C instead of going below that. Of course, spraying the trees with warm water is even more helpful.

Another reason: Even once all the water turns to ice, the ice forms an insulating layer to protect the

fruit from the cold outside temperatures.

4 Homework Ch 11 Review Answer key

Study Guide p. 231 #13 Given the following data for N2:

Normal melting point: −210°C Specific heat of liquid: 2.0 J/g °C Normal boiling point: −196°C Specific heat of gas: 1.0 J/g °C Specific heat of solid: 1.6 J/g °C Heat of Fusion: 25 J/g Heat of vaporization: 200 J/g How much energy in kilojoules is required to convert 1000 g of N2 at –206 °C to N2 gas at 20 °C? Step 1: Heat up liquid to its bp--The sample is starting above the melting point, as a liquid. First determine the energy needed to raise the temperature of the liquid to its boiling point:

ql =1000 g×2.0 J/g ⋅°C× −196°C− −206°C( )( ) = 20,000 J = 20. kJ

Step 2: Determine energy it takes to vaporize the liquid: qvap =1000 g×200 J/g ⋅°C = 200,000 J = 200. kJ Step 3: Determine energy needed to raise the temperature of the vapor to 20 °C:

qv =1000 g×1.0 J/g ⋅°C× 20°C− −196°C( )( ) = 216,000 J = 216 kJ

Step 4: Add the energies: qtotal = 20. kJ + 200. kJ + 216 kJ = 436 kJ

Temperature

Heat added