Chem 10 Final Exam - key
Useful information: R = 0.08206 liter atm/ mol K = 8.3145 J/mol K -31 h = 6.626 x 10-34 J s, NA = 6.02 x 1023, Mass of an electron = 9.11 x 10 kg, mass of a proton =
1.673 x 10-27 kg, 1 atm = 101325 Pa = 760 torr, F=96485 coul/mol
Chem 10 Exam 1, version A
Useful information: R = 0.08206 liter atm/ mol K = 8.3145 J/mol K 1 atm = 101325 Pa = 760 torr
1. At 100 C, a 1.20 L flask contains 42 mg of N2, 80 mg of O2, 5 x 10-3 mol Ar and NO at a concentration of 2.50 x1018 molecules/cm3.
Quiz 9 (This is part of problem 16.18) Which has the highest boiling point: HCl, Ar, F2 (only HCl has hydrogen bonds)
Highest freezing point: H2O, NaCl, HF (NaCl is an ionic solid)
Lowest freezing point: N2, H2, CO2 (H2 has only weak dispersion forces bet
Quiz 8 The mechanism for the reaction of nitrogen dioxide with carbon monoxide is: NO2 + NO2 NO3 + NO NO3 + CO NO2 + CO2 slow fast
What is the overall balanced reaction? NO2 + CO NO + CO2 Write the rate law for this mechanism. The first step is rate limit
Quiz 7 What wavelength of light corresponds to the n = 1 to n = 3 transition for an electron in a 0.30 nm box? E = hc/ = (32- 12)h2/(8mL2) hc/ = 5.35 x 10-18 J, solve for . = 3.71 x 10-8 m = 37.1 nm (Its a very small box, so the transition is far into the
Quiz 6. The energy necessary to remove an electron form one type of molecule is 750 kJ/mol. What is the longest wavelength photon needed to do this? What spectral range does this fall in to? (infrared?, visible?, ultraviolet?) Recall: h = 6.626 x 10-34 J
Quiz 5 solution Consider the reaction 3 Mn2+ + 2 Fe (s) 3 Mn (s) + 2 Fe3+ a) Is the reaction favorable as written? b) What is the voltage of a galvanic cell based on the this reaction? You need to know that the standard reduction potentials for manganese
Quiz 4 What is the vapor pressure (in Torr) of water at 40 C? Hvap = 42.5 kJ/mol Svap = 119 J/mol K In general, ln Keq = -H/RT + S/R In this case, ln P = - Hvap/RT + Svap/R So, ln P = -42500 J/ (8.314 J/mol K x 313 K) + 119 J/mol K / 8.314 J/mol K or ln P
Quiz 3 solution 10 grams of water freezes, calculate S. 10 grams / 18 g/mol = .555 mole Hfus = 6.02 kJ/mol qrev = - 3.34 x103J S = q/T = 3.34 x103J/273K = -12.25 J/K (note that this is water ice, S is neg) If this happens at -10 C, calculate G. G = H - TS
Quiz 2 solution (problem 116, page 187) The number of moles that go from the unknown volume to the other container is given by the ideal gas law: n = PV/RT = (1 atm 1.75 cm3)/ RT = (P Vunknown)/RT P = 1.96-1.71 atm = 0.25 atm. Solving for Vunknown we get
Quiz 1 solution P = density x h x g, where the density is 1.3 g/cm3, h is the height (=35 mm) and g = 9.8 m/s2 So, P= 1.3 x 103 kg/m3 x .035 m x 9.8 m/s2 = 445.9 Pa Convert to atmospheres 1 atm = 101325 Pa P = 4.4x 10-3atm OR Use the fact that the density
Chem 10 Exam 3 Useful information: R = 0.08206 liter atm/ mol K = 8.3145 J/mol K h = 6.626 x 10-34 J s, NA = 6.02 x 1023, mass of an electron = 9.11 x 10-31 kg, mass of a proton = 1.673 x 10-27 kg, 1 atm = 101325 Pa = 760 torr, F=95485 coul/mol, E = h = h
Sarah Rutherford
3 December 2012
Dr. Pumure 8AM General Chemistry
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