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Final Exam - Version 151 Final Exam Lyon(52370 This...

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Version 151 – Final Exam – Lyon – (52370) 1 This print-out should have 60 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. PLEASE BUBBLE YOUR NAME, EID AND FINAL EXAM VERSION NUM- BER CORRECTLY ON YOUR ANSWER SHEET. FAILURE TO DO SO WILL RE- SULT IN DELAYING THE SUBMISSION OF COURSE GRADES. 001 10.0 points Which of the following has the lowest lattice energy? 1. KI correct 2. KBr 3. LiCl 4. KCl 5. NaCl Explanation: K + and I have the smallest charge densi- ties. 002 10.0 points Which of the following is the correct Lewis structure of ozone (O 3 )? 1. O O O 2. O O O ⇐⇒ O O O 3. O O O ⇐⇒ O O O correct 4. O O O 5. O O O Explanation: 003 10.0 points A heteronuclear diatomic has six non-bonding electrons, five anti-bonding electrons, and eight bonding electrons. What is the bond order of the molecule? 1. one 2. one-half 3. one and a half correct 4. zero 5. two 6. three Explanation: e antibond = 5 e bond = 8 Bond order = 8 - 5 2 = 1.5 004 10.0 points What is the density of chlorine gas at 850 Torr and - 12 C? 1. 1 . 85 g · L 1 2. 3 . 39 g · L 1 3. 3 . 70 g · L 1 correct 4. 7 . 40 g · L 1 5. 27 . 8 g · L 1 Explanation: P = (850 Torr) 1 atm 760 Torr = 1 . 11842 atm T = - 12 C + 273.15 = 261.15 K MM Cl = 70.906 g/mol The ideal gas law is P V = n R T n V = P R T with unit of measure mol/L on each side.
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Version 151 – Final Exam – Lyon – (52370) 2 Multiplying each by molar mass (MM) gives n V · MM = P R T · MM , which now has units of g/L (= density). Thus ρ = P R T · MM = 1 . 11842 atm (0 . 08206 L · atm / mol / K) (261 . 15 K) × (70 . 906 g / mol) = 3 . 70055 g / L Cl 005 10.0 points Which of the following molecules is incor- rectly matched with bond angles? 1. SF 6 : 90 and 180 2. BeI 2 : slightly less than 109 correct 3. BF 3 : 120 4. AsF 5 : 90 , 120 , and 180 5. SiCl 4 : 109.5 Explanation: Be I I BeI 2 is a linear molecule with a 180 bond angle; Be is an exception to the octet rule. 006 10.0 points The electronic configurations of most ele- ments are correctly predicted by the standard order of energy of the orbitals. Which of the following statements explains several of the apparent exceptions to this general order (the Aufbau order) of filling? 1. When an electron has n , m , and m s all equal to +1, a special stability results. 2. There is a special stability associated with half-filled or filled sets of equivalent orbitals. correct 3. Sometimes it is possible for two electrons with the same spin to occupy the same or- bitals. 4. Exceptions occur for those elements whose emission spectrum is a continuous spec- trum. 5. When the potential frequency times the Schrodinger quantum equals Planck’s con- stant, paramagnetic multiplicity dominates the orbitals. Explanation: 007 10.0 points Of the following choices, which has a molecule paired incorrectly with either its bond an- gle(s), electronic geometry, or hybridization?
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