1. Review of Chapter 10 (condensed states of matter) A.P. Chem Pressure→ Temperature→ SOLID Defined: Structural units held together by attractive forces enough to make a rigid structure Properties to be aware of : melting pts, conductivity, and solubility Types (ranked by bond strength): 1.Macromolecular (atoms—covalent bonds) 2.Ionic (ions—ionic bonds) 3.Metallic (atoms—metallic bonds) 4.Molecular (molecules – intermolecular bonds) Organization? No→ Yes→ Amorphous Crystalline … units arranged in a geometric pattern Found by X-ray Diffraction Bragg’s Law: (n = 2d sin  Same mat’l but diff. pattern: Allotropism Simplest repeating pattern: Unit Cell Types: Cubic, Tetragonal, Orthorhombic, Monoclinic, Triclinic, Rhombohedral, Hexagonal Coord. # Varieties 6 simple (1) 6 simple (1) 8 BCC (2) 8 BCC (2) 12 FCC (4) aka CCP (ABC…ABC stacking in metals) (AB…AB stacking in metals) LIQUID Defined: Clustery with short range order Properties to be aware of : V.P., B.P.,  Hv, Surf. Ten., volatility, viscosity ln (VP 1 /VP 2 ) =  H v /R (1/T 2 -1/T 1 ) Clausius-Clapeyron …all related to attractive forces GAS Defined: Random V.P. solid curve V.P. liquid curve Triple Point… all 3 phases in equilibrium Critical Point… TcTc PcPc KE Time ← q = mc  T  H fus  H vap V.P. solid = V.P. liquid

2. Ch. 10 Practice Problems 1a. The edge length of the unit cell for metallic iron, which has all axes equal in length and right angles between them and exhibits a coordination # of 8 for each atom, can be found using x-ray data. A second order reflection is found (i.e. n = 2) when x-rays of = 1.377 angstroms are incident on the crystal at an angle of 28.72°. If the density of the iron is 7.87 g/cm 3, what is the mass of the individual iron atom? After you complete the calculation see if this gives you the correct molar mass for iron as found on the periodic table. D= 7.87g = M = 2 (x) cm 3 V (2.866x10 -8 cm) 3 2(1.377x10 -10 m)= 2d (sin 28.72°) edge length 2.866x10 -10 m = d=2.866x10 -8 cm x = 9.26 x 10 -23 g (gives a MM of 55.8 g/mol) mass of atom

3. Ch. 10 Practice Problems 1b. Show that the unit cell of iron is 32% intersticial space.

4. Ch. 10 Practice Problems 1b. Show that the unit cell of iron is 32% intersticial space. d 2 +h 2 = (4r) 2 4r d d d h d 2 + d 2 = h 2 d 2 +2d 2 = 16r 2 3d 2 = 16r 2 d = 16/3 r √ V spheres = V cell 0.68 [(4/3)  r 3 ]2 = (2.31r) 3 8.37 = 12.3 0.68

5. Ch. 10 Practice Problems 2. 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  H vap for each substance and then rank the following properties for each substance based on the info given: Volatility Boiling Point Surface Tension  H vap A = 30700 J/mol = 30.7 KJ/mol ln 200. mm =  H vap 1 - 1 600. mm 8.31J 321K 293K mol K Identify the substances as chloroform (CHCl 3 ) or water and say why you chose as you did. For A:  H vap B = 39.6 KJ/mol A is higher B is higher