Exam 1-solutions

Exam 1-solutions - Version 887 – Exam 1 – Mccord...

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Unformatted text preview: Version 887 – 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. McCord CH301 c = 3 . 00 × 10 8 m/s h = 6 . 626 × 10- 34 J · s m e = 9 . 11 × 10- 31 kg R = 2 . 178 × 10- 18 J N A = 6 . 022 × 10 23 mol- 1 1 atm = 1.01325 × 10 5 Pa 1lbs = 453 . 6g 1 gal = 3.785 L 1 in = 2.54 cm c = λν E = hν λ = h p = h mv 1 2 mv 2 = hν- Φ E n =-R parenleftbigg 1 n 2 parenrightbigg- ¯ h 2 2 m d 2 ψ d x 2 + V ( x ) ψ = Eψ ψ n ( x ) = parenleftbigg 2 L parenrightbigg 1 2 sin parenleftBig nπx L parenrightBig E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , ··· 001 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. 69.4% 2. 13.2% 3. 76.9% 4. 53.4% correct 5. 50.0% 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 Version 887 – Exam 1 – Mccord – (50970) 2 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% 002 10.0 points An electron is confined to a one-dimensional box of length L. It falls from the second en- ergy level to the ground state, and releases a photon with a wavelength of 322 nm. What is the length of the box? 1. 0.151 nm. 2. 2.92 nm. 3. 1.14 nm. 4. 4.32 nm. 5. 0.541 nm. correct 6. 322 nm. 7. 1.61 nm. 8. 0.303 nm. 9. 292 nm. Explanation: 003 10.0 points Write the ground-state electron configuration of a lead atom. 1. [Xe] 4 f 14 5 d 5 6 s 1 6 p 6 7 s 2 2. [Xe] 4 f 14 5 d 10 6 s 2 6 p 2 correct 3. [Xe] 4 f 14 5 d 9 6 s 2 6 p 3 4. [Xe] 4 f 14 5 d 10 6 s 1 6 p 3 5. [Xe] 4 f 14 5 d 10 6 p 4 Explanation: The Aufbau order of electron filling is 1 s , 2 s , 2 p , 3 s , 3 p , 4 s , 3 d , 4 p , 5 s , 4 d , 5 p , 6 s , 4 f , 5 d , 6 p , etc ....
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Exam 1-solutions - Version 887 – Exam 1 – Mccord...

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