Chapter 10 Homework answers (Zumdahl and Zumdahl)
34.
a) London Dispersion
b) dipole-dipole
c) Hydrogen bonding
d) ionic
e) London Dispersion
f) dipole-dipole
g) ionic
36.
a) OCS since it is polar and has dipole-dipole vs. CO2 which is nonpolar and has weaker
London Disp forces.
b) SeO2 since it is larger and more polarizable than SO2. Both are nonpolar and have
London Disp forces.
c) H2NCH2CH2NH2 because it has more Hydrogen bonds per molecule
d) H2CO because it is polar and has stronger dipole-dipole forces.
e) CH3OH because it has Hydrogen bonds while the other molecule does not.
40. a) CBr4 has a higher boiling point. All choices are nonpolar but since CBr4 is most massive,
it is most polarizable.
b) F2 has the lowest freezing point. F2 is the only nonpolar option. It has weak London
Disp forces.
c) CH3CH2OH has the lowest vapor pressure. It is the only molecule with Hydrogen
bonds.
d) H2O2 has the greatest viscosity. It has more Hydrogen bonds per molecule than HF
does, so it has stronger imfs.
e) H2CO has the highest Hvap. It is the only polar option so has stronger dipole-dipole
forces.
f) I2 has the lowest Hfus because it is the only nonpolar option. The other two have
strong ionic bonds.
44. CSe2 has the strongest London Disp forces. Due to its increased molar mass, it is more
polarizable than CS2 which is more polarizable than CO2.
80.
a) network covalent
b) molecular
c) molecular
d) metallic
e) ionic
f) network covalent
g) ionic
h) molecular
i) noble gas/atomic
j) metallic
k) molecular
89. (answer in back of book)
90.
a) if bp = 115 oC, P = 1.66 atm (aka 1260 torr)
b) If P = 3.50 atm, then T= 412K = 139 oC
104. a) least dense (gas)  liquid most dense (solid)
b) liquid at room temp
c) >320 oC and 100 atm
d) melting and vaporization
126. N2 can never be a liquid at room temperature. Its critical temperature is far below room
temperature whereas NH3 has a critical temperature above 100oC, although P must be
over 100 atm to liquefy it at.