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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. П = MRT П =[(g/MM) / L]RT ( substiture g/molar mass for moles) MM = gRT/ LП = (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. П = 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....
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This note was uploaded on 10/13/2009 for the course CHEM 1442 taught by Professor Rogers during the Fall '08 term at UT Arlington.
 Fall '08
 ROGERS
 Chemistry, Mole

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