88) a) the KE of the gas increases which means that the gas molecules are moving faster.  If the gas molecules are moving faster then they will hit the walls more often thereby increasing the pressure.  b)  decreasing the volume will increase the rate at which the molecules of gas will strike the walls, which will increase the pressure.

95)  a)  An ideal gas obey's the mathematical equation PV=nRT (has no volume or intermolecular interactions).  A real gas does not obey PV=nRT (has volume and intermolecular interactions)  b)  high pressure (gas molecules look larger & greater intermolecular forces) & low temperature (greater intermolecular forces)
 

Chapter 13

10)  CH3OH & CH3NH2

16) a) H-bonding (dipole-dipole), dispersion
      b) dipole-dipole, dispersion
      c) dispersion

27)  a) Intermolecular Forces  b) increasing temperature causes a decrease in the viscosity of a liquid

32)  a) BiBr3  b) CO2  c)  N2  d)  HCOOCH3

74)  a) monoclinic (solid)  b) rhombic (solid)  c) liquid  d)  rhombic (solid)  e)  vapour  f)  liquid
 

Chapter 14

10)  a) high due to a strong ionic-dipole  solute-solvent intermolecular interaction
       b)  high due to similar intermolecular interactions
       c)  low due to a strong solute-solute (ionic) intermolecular interaction
       d)  low due to a very weak solute-solvent intermolecular interaction (non-polar solute and a polar solvent)
       e)  low due to a very weak solute-solvent intermolecular interaction (ionic solute and a non-polar solvent)

22)  ionic compounds have a strong solute-solute bond.  The formation of the solute-solvent bond doesn't always compensate for the energy necessary to break the solute-solute and solvent-solvent bonds necessary for solvation.  That is why the dissolution of some salts are endothermic.  The intermolecular forces of miscible liquids are generally very similar so that the formation of the solute-solvent bond normally releases an adequate amount of energy to make the solvation of miscible liquids exothermic.

72)  the van't Hoff factor for Na2SO4 would ideally be 3 and the van't Hoff factor for CaSO4 would ideally be 2.  Therefore, Na2SO4 would produce a greater freezing point depression.