CHEM
Ex_2A_KEY.pdf

# 0 c when the temperature was increased the pressure

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A sealed bulb containing helium gas had a pressure of 345 torr at 10.0 ° C. When the temperature was increased, the pressure rose to 582 torr. What was the final temperature? (A) 17 ° C (B) 86 ° C (C) 149 ° C (D) 176 ° C (E) 205 ° C P 1 T 1 = P 2 T 2 , so T 2 = P 2 P 1 ×T 1 = 582 torr 345 torr ×283.15 K=478 K=205℃ (6 pts) 23. What is the frequency of violet light with a wavelength of 425 nm? (A) 7.05 ´ 10 5 Hz (B) 1.27 ´ 10 11 Hz (C) 1.27 ´ 10 2 Hz (D) 7.05 ´ 10 14 Hz (E) 4.61 ´ 10 15 Hz c=λν, so ν= c λ = 2.9979×10 8 m/s 425×10 –9 m =7.05 × 10 14 s –1

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NAME:_________________________________ © 2018 L.S. Brown A7 (10 pts) 24. The ionization energy of a particular element is 899 kJ/mol. Suppose an atom of this element has already absorbed a photon with a wavelength of 345 nm so that it has been promoted to an excited state. What would be the longest wavelength of light that would be able to provide enough energy to eject an electron from the excited atom? (This is very similar to a homework problem from Set #5. It may help you to try to draw an energy level diagram summarizing the process.) First convert the IE from kJ/mol to J/atom: Get energy of the 345 nm photon, then subtract that from the IE. 1.49 ´ 10 –18 J – 5.76 ´ 10 –19 J = 9.17 ´ 10 –19 J (actually 9.2 ´ 10 –19 J if we do the sig figs right…) ion Ex. state Gr. state
A8 © 2018 L.S. Brown (10 pts) 25. When a photoelectric effect experiment was carried out using lithium metal and light at frequency n , electrons were emitted with a kinetic energy of 6.40 ´ 10 –19 J. The frequency was reduced to = > n and the experiment was repeated, still using the same lithium target. This time electrons were emitted with a kinetic energy of 8.78 ´ 10 –20 J. Find the electron binding energy for lithium. (The actual frequencies used were not recorded, but it is still possible to find the binding energy.) One way to start is to realize that the difference in the 2 KE’s is the smaller photon E. But I think it is likely that more people will start with the energy equation for the PE effect, and then write 2 versions: E photon = BE + KE h n = BE + 6.40 ´ 10 –19 J (for the 1 st experiment) ࠵? ࠵? ࠵? n = BE + 8.78 ´ 10 –20 J (for the 2 nd experiment) The 2 BE’s are the same. Subtract the 2 nd eq from the 1 st to get: ࠵? ࠵? ࠵? n = 5.52 ´ 10 –19 J Put that back into either equation above to find BE. I will do the 2 nd one. ࠵? ࠵? ࠵? n = 5.52 ´ 10 –19 J = BE + 8.78 ´ 10 –20 J And BE = 4.64 ´ 10 –19 J

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NAME:_________________________________ © 2018 L.S. Brown A9 2.00 L NH 3 0.500 atm 1.00 L O 2 1.50 atm (12 pts) 26. The first step in the production of nitric acid involves the reaction of NH 3 and O 2 to produce NO and H 2 O.
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