Final Review

Final Review - Final Exam W 1 3 PM 11 questions Typical mix...

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Final Exam W 1 – 3 PM 11 questions Typical mix of short answer and Calculations (See Brown 2007 Exam) Includes questions on kinetics and Equilibrium Fundamental questions from all other main topics
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4. Dr. Brown’s new house will need to use propane (C 3 H 8 ) gas to fuel the range, because natural gas service is not available at the location. The propane is stored as a liquid, and the gas to be burned is produced as the liquid evaporates. The propane gas consumed by a typical range burner at high power in 1 hour would occupy roughly 165 L at 25°C and 1 atm, and the range will have 6 burners. If the liquid propane were to be stored in a tank that holds 500-gallons, how many hours would it take for the range to consume an entire tankful of propane? The density of liquid propane is 0.5077 kg/L. 1 gallon = 3.7854 L. n = PV RT = (1 atm)(165 L) (0.08206 L atm mol K )(298.15 K ) = 6.744 mol 500 gal × 3.7854 L gal × 507.7 g 1 L × 1 mol 44.094 g = 2.179 × 10 4 mol 2.179 × 10 4 mol 40.46 mol hr = 539 hr That’s per burner per hour. So with 6 burners, max total consumption would be 40.46 mol/hr. For the tank: Just divide those to get: (Which would be about 22 days of continuous full-power cooking.)
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495,000 J/mol 6.022 × 10 23 atoms/mole = 8.220 × 10 -19 J/ato m E = hc λ so λ = hc E = (6.626 × 10 -34 Js)(2.9979 × 10 8 m/s) (8.220 × 10 -19 J) = 2.42 × 10 -7 m = 242 nm 5. Dissociation of O 2 molecules into oxygen atoms is an important step in the formation and destruction of ozone in the atmosphere. The energy needed for the dissociation is provided by photons of ultraviolet light. The bond energy for the O 2 molecule is 495 kJ/mol. Use this to find the longest wavelength of light for which a single photon is capable of dissociating an O 2 molecule. That is the photon energy, so use it to solve for the wavelength: The bond energy value is for a mole, so we need to divide it by Avogadro’s number to get the dissociation energy per molecule.
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18.9 g NaBH 4 37.832 g mol = 0.500 mol NaBH 4 (0.25 L)(0.400 M) = 0.100 mol H 3 PO 4 0.100 mol H 3 PO 4 × 1 mol B 2 H 6 2 mol H 3 PO 4 = 0.05 mol B 2 H 6 V = nRT P = (0.05 mol)(62.364 L torr mol K )(298.15 K) (760 torr) = 1.22 L Diborane (B 2 H 6 ) can be produced by the following reaction. 2NaBH 4 (s) + 2H 3 PO 4 (aq) Æ B 2 H 6 (g) + 2NaH 2 PO 4 (aq) + 2H 2 (g) Suppose that 18.9 g of solid NaBH 4 is allowed to react with 250 mL of 0.400 M H 3 PO 4 . If all of the B 2 H 6 produced is collected at 25 ° C and 760 torr, what volume would it occupy? Since the mole ratio is 1:1, it’s easy to see that the H
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This note was uploaded on 01/22/2010 for the course ARTS 150 taught by Professor Caffey during the Spring '08 term at Texas A&M.

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Final Review - Final Exam W 1 3 PM 11 questions Typical mix...

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