This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1442 Answers to final review 1. 11M = 11 moles/L of solution H 2 SO 4 = 11(98.1) = 1079.1 g/ L of solution 1 L of solution has a mass of : (1000ml)(1.59 g/ml) = 1590g per Liter of solution Mass% H 2 SO 4 = Part/whole X 100 = 1079.1 g/1590g X 100 = 67.9 % 2. I. is least soluble nonpolar, II most polar and smaller number of C, III middle because it is polar but more C then I. ( Remember if a molecule contains three different elements it is going to be polar.) 3. P soln = P solv X solv = (88.7 torr)(2.5 mol/ 2.5mol + .52mol) = 73.9 torr 4. First calculate the pressure of the vapor: P t = P o a X a + P o b X b = (745torr) (.425) + (290 torr)*(1.425) = 483.4 torr Mole fraction of toluene above the solution = P toluene / P total = .575 X 290/483.4 = .345 5. Heat given off(exothermic) indicates a strong interaction between the solvent and the solute. this causes less of them to turn into gases and in effect causes the pressure to be less than expected. 6. CaCl 2 breaks into three ions: one Ca 2+ and two Cl . The solution was .22m X 3 that makes it .66 in particles. The more the particles the more the change in freezing or boiling point. 7. T b = imK b = (1) [( 40g/160) / .5 Kg] ( 2.8) = 1.4 BP = 81.0 + 1.4 = 82.4 ( boiling point elevation ) 8. = MRT =[(g/MM) / L]RT ( substiture g/molar mass for moles) MM = gRT/ L = (1.73g)(.0821 L atm/mol K)(298k) / ( 2.72 atm)(.1 L)= 156 9. Gases in liquids increase solubility with high pressure and low temperature. Very different from solids. 10. T b = imK b = (3)*([30/174.3] / .1 Kg)(.51) = 2.63 BP = 100 + 2.63 = 102.6 11. Heat of solution = heat of hydration + lattice energy7.6 kJ = heat of hydration + 686 kJ heat of hydration = 694 kJ 12. See answer 13. = MRT 7.8 = M ( .0821Latm/molK)(298K) M = .32M but it is .2 M so .32/.2 = 1.6. It needed to be 1.6 in order to make the pressure = 7.8 atm. MgSO 4 ionized into 2 particles, so i should be close to 2. 14. For the order of B use exp 1 and 2. Since A is constant it can be left out. [0.015 / 0.030] b = [0.010/0.020] b = 1 For the order of A use exp 1 and 3 [0.033/0.099] a = [0.010/0.090] a = 2 Rate Equation = R = k[A] 2 [B] 15. (0.082 M/s/1)(2 NH 3 / 3 H 2 )* = 0.055 * use coefficients in equation 16. Third order means the concentration is cubed. If you double the concentration: [2] 3 = factor of 8 17. t 1/2 = ln2/k 102s = ln2/k k = .0068s1 ln[A] o /[A] t = kt ln (100/65) = .0068/s t t = 63.4 s 18. 1/[A] t vs t straight line = 2nd order, ln[A] vs t = first order, [A] vs t is zero order 19. 1/[A] t 1/[A] o = kt 1/[.08]  1[.1] = k(40min) k = 0.0625/min t 1/2 = 1/k[A] o t 1/2 = 1/(0.0625min0[.1] t 1/2 = 160 min 20. 2 reactants = bimolecular, intermediate is formed in one step and consumes in another....
View Full
Document
 Fall '08
 ROGERS
 Chemistry, Mole

Click to edit the document details