Review of Chapter 10 (condensed states of matter) A.P. Chem Review of Chapter 10 (condensed states of matter) ln (VP1/VP2) = DHv/R (1/T2-1/T1) Clausius-Clapeyron Properties to be aware of : V.P., B.P., DHv, Surf. Ten., volatility, viscosity, crit. temp. Properties to be aware of : melting pts, conductivity, and solubility (▲Semi-conductors) …all related to attractive forces Pc Critical Point… V.P. solid= V.P. liquid SOLID LIQUID V.P. liquid curve Defined: Structural units held together by attractive forces enough to make a rigid structure Defined: Clustery with short range order Pressure→ GAS Types (ranked by bond strength): Macromolecular (atoms—covalent bonds) Ionic (ions—ionic bonds) Metallic (atoms—metallic bonds) Molecular (molecules – intermolecular bonds) Triple Point… all 3 phases in equilibrium Defined: Random (◄Alloys) V.P. solid curve Organization? No→ Yes→ DHvap Amorphous Same mat’l but diff. pattern: Allotropism KE Crystalline…units arranged in a geometric pattern DHfus ←q = mcDT Simplest repeating pattern: Unit Cell Time Coord. # Varieties TwoTypes: Temperature→ Tc 6 simple (1) 8 BCC (2) 12 FCC (4) aka CCP (ABC…ABC stacking in metals) Cubic, Hexagonal (AB…AB stacking in metals)

carbon, hydrogen, lithium Solid Substances: 3 1 2 carbon, hydrogen, lithium Arrange in terms of M.P. (lowest to highest ,1-3) Which, if any of these, will dissolve in H2O? (*) 3 * Underline any of these that dissolve in water and produce a conductive solution? ethanol, methane, dichlorofloromethane, magnesium sulfate 1 2 * 4 * ______________________________

A is higher B is higher B is higher For A: ln 200. mm = DHvap 1 - 1 The vapor pressure of substance A is 200. mm of Hg at 20.0°C and 600. mm of Hg at 48.0°C. Substance B’s vapor pressures at those temperatures are 23.0 and 95.0 mm of Hg respectively. Calculate an estimated DHvap for each substance and then rank the following properties for each substance based on the info given: Volatility Boiling Point Surface Tension A is higher B is higher B is higher For A: ln 200. mm = DHvap 1 - 1 600. mm 8.31J 321K 293K mol K ▼chloroform – weaker dipole-dipole attractions DHvap A= 30700 J/mol = 30.7 KJ/mol ▼water – hydrogen bonding DHvap B= 39.6 KJ/mol Identify the substances as chloroform (CHCl3) or water and say why you chose as you did.