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exam 1 - Version 071 Exam 1 Mccord(50970 This print-out...

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Version 071 – 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 List in order of increasing atomic size (small- est to largest) the atoms whose atomic num- bers are 6, 32, 7. 1. 7, 6, 32 correct 2. 6, 7, 32 3. 32, 6, 7 4. 32, 7, 6 Explanation: Element Electronic configuration 6 2 · 4 [He] ↑↓ 2 s 2 p Group 4, Period 2 Valence electrons in period 2 7 2 · 5 [He] ↑↓ 2 s 2 p Group 5, Period 2 Valence electrons in period 2 32 2 · 8 · 18 · 4 [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 4 s 3 d 4 p Group 4, Period 4 Valence electrons in period 4 Period 4 is farther from the nucleus than period 2. The half-filled 2 p orbital of ele- ment #7 is energetically more stable and has greater effective nuclear charge than that of element #6. 002 10.0 points Which of the following best describes the range of atomic radii? 1. 1 to 100 ˚ A 2. 0.4 to 3 ˚ A correct 3. 10 to 30 ˚ A 4. 1 to 15 ˚ A 5. 0.05 to 1 ˚ A 6. 5 to 10 ˚ A Explanation: Atomic radii vary from 30 pm (He) to 300 pm (Fr). 1 ˚ A= 10 - 10 m = 100 pm, so the range of atomic radii is 0.3 to 3 ˚ A. 003 10.0 points The n and quantum numbers of the last electron of an element are n = 4 and = 3. The element is 1. a nonmetal. 2. a d -transition metal. 3. a noble gas. 4. a representative element. 5. an f -transition metal. correct Explanation: Where n = 4 and = 3, 4 f electrons are filling. These are the f -transition metals. 004 10.0 points How many s electrons and p electrons does Mn possess? 1. 7 s , 12 p 2. 6 s , 6 p 3. 8 s , 18 p 4. 7 s , 18 p 5. 8 s , 6 p 6. 6 s , 18 p 7. 8 s , 12 p correct
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Version 071 – Exam 1 – Mccord – (50970) 2 8. 6 s , 12 p 9. 7 s , 13 p Explanation: The electron configuration for Mn is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 5 2 + 2 + 2 + 2 = 8 s 6 + 6 = 12 p 005 10.0 points Consider the radial distribution function (RDF) plot shown below. Which of the orbitals given as choices below would corre- spond to the RDF shown. r 4 πr 2 R 2 1. 4 d 2. 5 p 3. 5 d correct 4. 4 f 5. 6 d 6. 4 s Explanation: The plot has 2 radial nodes (spherical nodes). The number of spherical nodes will al- ways be equal to n - - 1. Only the 5 d orbital fits this case. Other orbitals that WOULD fit this would have been the 3 s , 4 p , and 6 f . 006 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. 53.4% correct 2. 69.4% 3. 50.0% 4. 13.2% 5. 76.9% 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.
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