Exam 1-Chem 301 - Version 265 Exam 1 mccord (51600) 1 This...

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Unformatted text preview: Version 265 Exam 1 mccord (51600) 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Which response identifies a possible for the last-filled electron in a chloride ion? 1. 3 , 1 , 1 correct 2. 2 , , 1 3. 2 , 1 , 2 4. 3 , 3 , 1 5. 2 , , 6. 3 , , Explanation: The are written in format n,,m . The electronic configuration for the chloride ion is [Ne] 3 s 2 3 p 6 , so the last-filled electron is in 3 p . For the 3 p orbital, n = 3, = 1, and m can be- 1 , , +1. (The m can be any of those values because the orbitals are degenerate, so may fill in any order.) 002 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. 5 . 90 10 5 J correct 2. 7 . 08 10 6 J 3. 7 . 08 10 3 J 4. 1 . 17 10 17 J 5. 9 . 11 10 21 J 6. 2 . 67 10 19 J 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 . 003 10.0 points How many electrons can possess this set of quantum numbers: principal quantum num- ber n = 4, angular quantum number = 2? 1. 14 2. 6 3. 4. 4 5. 12 6. 16 7. 10 correct 8. 8 9. 18 10. 2 Explanation: Use the rules for the quantum numbers: If n = 4 and = 2 ( i.e. , 4 d ), then m =- 2 ,- 1 , , +1 , +2 are permitted; there are five different orbitals and m s = 1 2 , each holding two electrons. Version 265 Exam 1 mccord (51600) 2 004 10.0 points 2 (2 , 1 , 0) represents which orbital? 1. Different orbitals, depending on the ele- ment in question. 2. The probability of a 2 p orbital becoming a 1 s orbital. 3. 2p orbital. correct 4. 2s orbital. Explanation: 2 (2 , 1 , 0) gives the probability of finding an electron in an orbital with quantum num- bers 2,1,0, which corresponds to n=2, l=1, and ml=0. That is one of the 2p orbitals. 005 10.0 points For the reaction ? C 6 H 6 + ? O 2 ? CO 2 + ? H 2 O 45 . 6 grams of C 6 H 6 are mixed with 159 . 7 grams of O 2 and allowed to react. How much CO 2 could be produced by this reaction? 1. 205 . 1 g 2. 123 . 4 g 3. 323 . 8 g 4. 135 . 3 g 5. 239 g 6. 110 . 1 g 7. 291 . 4 g 8. 192 . 8 g 9. 154 . 2 g correct 10. 175 . 8 g Explanation: m C 6 H 6 = 45 . 6 g m O 2 = 159 . 7 g The balanced equation is 2 C 6 H 6 + 15 O 2 12 CO 2 + 6 H 2 O ....
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This note was uploaded on 05/02/2011 for the course CHEM 301 taught by Professor Wandelt during the Spring '08 term at University of Texas at Austin.

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Exam 1-Chem 301 - Version 265 Exam 1 mccord (51600) 1 This...

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