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Unformatted text preview: Version 232 Exam 1 McCord (53130) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. McCord CH301tt This exam is only for McCords Tues/Thur CH301 class. c = 3 . 00 10 8 m/s h = 6 . 626 10 34 J s m e = 9 . 11 10 31 kg N A = 6 . 022 10 23 mol 1 = R 1 n 2 y 1 n 2 x where R = 3 . 29 10 15 s 1 n ( x ) = 2 L 1 2 sin nx L E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , 001 10.0 points A quantum mechanical particle in a box in its ground state is most likely to be found 1. in the middle of the box. correct 2. is equally likely to be found at all positions except the very center of the box. 3. at both edges of the box. 4. is equally likely to be found at all positions in the box. 5. at the very edge of the box. Explanation: The ground state is 1 which has maximum probability in the middle of the box. 002 10.0 points The graph shows the radial distribution plots for the 1s wavefunctions for H, He, and He + . A B C radius 4 r 2 2 Which plot is the 1s wavefunction for the He + ion? 1. There is no way to know 2. C 3. A correct 4. B Explanation: H has one electron and one proton. He has two electrons and two protons. He + has one electron and two protons. Therefore the elec tron will have the greatest attraction to the the nuclei with two protons. In He there will be a slight reduction in the effective nuclear charge due to the electronelectron repulsion. In He + , there is only one electron so the radius will be the smallest. Plot A is peaked closest to the nucleus so it will be the smallest. 003 10.0 points What are the correct quantum numbers ( n , , m ) for the seventh electron of Cl? 1. 2, 1, 1 correct 2. 2, 2, 1 3. 3, 2, 1 4. 3, 1, 0 5. 2, 1, 2 Version 232 Exam 1 McCord (53130) 2 6. 3, 1, 1 7. 2, 0, 1 8. 3, 0, 0 Explanation: The electron configuration for Cl is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 5 The seventh electron is in a 2 p orbital, so n = 2, = 1, m = 1, 0, or +1. 2, 1, 0 and 2, 1, 1 are not given as choices. 004 10.0 points Carbon emits photons at 745 nm when ex posed to blackbody radiation. How much energy would be obtained if 44 g of carbon were irradiated? 1. 2 . 67 10 19 J 2. 7 . 08 10 6 J 3. 9 . 11 10 21 J 4. 1 . 17 10 17 J 5. 7 . 08 10 3 J 6. 5 . 90 10 5 J correct Explanation: = 745 nm = 7 . 45 10 7 m m C = 44 g Assume each carbon atom emits one pho ton. For each photon E 1 = h = h c = (6 . 626 10 34 J s) (3 10 8 m / s) 7 . 45 10 7 m = 2 . 66819 10 19 J , so the total energy emitted is E = E 1 n = 2 . 66819 10 19 J C atom 6 . 022 10 23 C atoms 1 mol C atoms 1 mol C atoms 12 . 01 g C (44 g C) = 5 . 88663 10 5 J ....
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This note was uploaded on 07/13/2011 for the course CHEM 301 taught by Professor Wandelt during the Spring '08 term at University of Texas at Austin.
 Spring '08
 wandelt
 Chemistry, pH

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