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# EXAM1 - Version 643 Exam 1 Mccord(50970 This print-out...

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Version 643 – Exam 1 – Mccord – (50970) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points It takes light with a wavelength of 261 nm, or less, to break the O H bond in wa- ter. What energy does this correspond to (in kJ/photon) and what is the O H bond strength in kJ/mol? 1. 7 . 6 × 10 - 22 kJ/photon, 458 kJ/mol cor- rect 2. 7 . 6 × 10 - 22 kJ/photon, 458,000 kJ/mol 3. 6 . 6 × 10 - 25 kJ/photon, 4 × 10 - 4 kJ/mol 4. 6 . 6 × 10 - 25 kJ/photon, 0.40 kJ/mol Explanation: 002 10.0 points If a particle is confined to a one-dimensional box of length 300 pm, for Ψ 3 the particle is most likely to be found at 1. 50, 150, and 250 pm, respectively. cor- rect 2. 17.3 pm. 3. 100 and 200 pm, respectively. 4. 0 pm. 5. 300 pm. Explanation: 003 10.0 points C 6 H 6 = 78.11 g/mol Cl 2 = 70.91 g/mol C 6 H 5 Cl = 112.55 g/mol HCl = 36.46 g/mol If 39.0 g of C 6 H 6 reacts with excess chlorine and produces 30.0 g of C 6 H 5 Cl in the reaction C 6 H 6 + Cl 2 C 6 H 5 Cl + HCl , what is the percent yield of C 6 H 5 Cl? 1. 13.2% 2. 50.0% 3. 53.4% correct 4. 76.9% 5. 69.4% Explanation: m C 6 H 6 = 39.0 g m C 6 H 5 Cl = 30.0 g Our first step is to determine the theoretical yield of the reaction. The reaction started with 39.0 g of C 6 H 6 . We convert from grams to moles using the molar mass: ? mol C 6 H 6 = 39 . 0 g C 6 H 6 × 1 mol C 6 H 6 78 . 11 g C 6 H 6 = 0 . 4993 mol C 6 H 6 From the balanced equation we see that 1 mole C 6 H 5 Cl is produced for each mole of C 6 H 6 reacted. Therefore, if 0.4993 mol C 6 H 6 were reacted we would expect to pro- duce 0.4993 mol C 6 H 5 Cl. We use the molar mass of C 6 H 5 Cl to covert from moles to grams: ? g C 6 H 5 Cl = 0 . 4993 mol C 6 H 5 Cl × 112 . 55 g C 6 H 5 Cl 1 mol C 6 H 5 Cl = 56 . 196 g C 6 H 5 Cl This is the theoretical yield; the maximum number of grams of C 6 H 5 Cl of that could be poduced from 39.0 g C 6 H 6 . To find the percent yield, we divide the actual yield by the theoretical yield: ? percent yield = 30 . 0 g C 6 H 5 Cl 56 . 196 g C 6 H 5 Cl × 100% = 53 . 4% 004 10.0 points How many electrons could be described by the quantum numbers n = 5 and m s = + 1 2 ? 1. 50

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Version 643 – Exam 1 – Mccord – (50970) 2 2. 1 3. 2 4. 25 correct 5. 5 Explanation: For a given principle energy level, n , a total of 2 n 2 electrons are described. Exactly half of those, n 2 , will have a + 1 2 spin. 005 10.0 points How many electrons can possess this set of quantum numbers: principal quantum num- ber n = 4, angular quantum number = 2? 1. 10 correct 2. 2 3. 8 4. 14 5. 18 6. 0 7. 16 8. 4 9. 12 10. 6 Explanation: Use the rules for the quantum numbers: If n = 4 and = 2 ( i.e. , 4 d ), then m = - 2 , - 1 , 0 , +1 , +2 are permitted; there are five different orbitals and m s = ± 1 2 , each holding two electrons. 006 10.0 points What is the de Broglie wavelength of a bowl- ing ball rolling down a bowling alley lane? Assume the mass of the ball is 4500 g and it is moving at 4.12 m/s. 1. 1 . 229 × 10 - 32 m 2. 3 . 57389 × 10 - 35 m correct 3. 1 . 4725 × 10 - 37 m 4. 3 . 574 × 10 - 38 m Explanation: λ = h p = h m · v = 6 . 626 × 10 - 34 kg · m 2 s (4 . 5 kg × 4 . 12 m / s) = 3 . 57389 × 10 - 35 m 007 10.0 points Calculate the longest-wavelength line in the Balmer series for hydrogen.
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EXAM1 - Version 643 Exam 1 Mccord(50970 This print-out...

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