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Unformatted text preview: kramer (jrk2286) – Homework 3 – sutcliffe – (90830) 1 This printout should have 31 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This is for Exam 3 and so covers chapters 3, 5 and some parts of 16. 001 10.0 points Calculate the number of ethanol molecules (CH 3 CH 2 OH) that would contain 164 grams of carbon. 1. 1.65 × 10 25 molec 2. 1.89 × 10 26 molec 3. 4.11 × 10 24 molec correct 4. 8.28 × 10 24 molec 5. 3.73 × 10 25 molec 6. 5.93 × 10 26 molec 7. 9.94 × 10 25 molec Explanation: m C = 164 g ? molec = 164 g C × 1 mol C 12 g C × 1 mol C 2 H 6 O 2 mol C × 6 . 022 × 10 23 molec 1 mol C 2 H 6 O = 4 . 11 × 10 24 molec 002 10.0 points Calculate the number of carbon atoms in 0.75 L of acetone (CH 3 COCH 3 ). (density = 1.3 g/mL and boiling point = 48.5 degrees Celsius) 1. 1 . 8 × 10 22 atoms 2. 1 . × 10 22 atoms 3. 3 . × 10 25 atoms correct 4. 1 . 8 × 10 25 atoms 5. 3 . × 10 22 atoms 6. 1 . × 10 25 atoms Explanation: V = 0.75 L density = 1.3 g/mL BP = 48.5 ◦ C ? C atoms = 0 . 75 L C 3 H 6 O × 1000 mL 1 L × 1 . 3 g C 3 H 6 O 1 mL soln × 1 mol C 3 H 6 O 58 g C 3 H 6 O × 3 mol C atoms 1 mol C 3 H 6 O × 6 . 022 × 10 23 C atoms 1 mol C atom = 3 . × 10 25 C atom 003 10.0 points Balance the equation ? CS 2 + ? O 2 → ? CO 2 + ? SO 2 , using the smallest possible integers. The co efficient of O 2 is 1. 1. 2. 2. 3. 3. correct 4. 5. 5. 4. Explanation: A balanced equation has the same num ber of each kind of atom on both sides of the equation. We find the number of each kind of atom using equation coefficients and com position stoichiometry. For example, we find there are 2 S atoms on the product side: ? S atoms = 2 SO 2 × 1 S 1 SO 2 = 2 S The balanced equation is CS 2 + 3 O 2 → CO 2 + 2 SO 2 , kramer (jrk2286) – Homework 3 – sutcliffe – (90830) 2 and the coefficient of O 2 is 3. 004 10.0 points Nitrogen dioxide (NO 2 ) is a serious air pol lutant that is formed when fuels are burned in air. Some of the NO 2 reacts with water in the air as shown below to form nitric acid (HNO 3 ), a major contributor to acid rain. Suppose that 2 . 2 tons of NO 2 are released into the atmosphere, and that 100% of this NO 2 reacts as shown. 3 NO 2 + H 2 O → 2 HNO 3 + NO How much atmospheric HNO 3 would be formed in this way? Correct answer: 2 . 0087 tons. Explanation: m NO 2 = 2 . 2 tons The balanced equation for the reaction in dicates that 2 mol HNO 3 are produced for every 3 mol NO 2 reacted. The molar mass of NO 2 is 46 g/mol and the molar mass of HNO 3 is 63 g/mol. Therefore 138 g (3 mol) of NO 2 would produce 126 g (2 mol) HNO 3 . Simi larly, 138 tons NO 2 would produce 126 tons HNO 3 . (If this doesn’t make sense convert both 138 g and 126 g to tons. Notice the g : g ratio is the same as the ton : ton ratio.) We use this mass ratio to calculate the tons HNO 3 that would be produced from 2 . 2 tons of NO 2 : ? ton HNO 3 = 2 . 2 tons NO 2 × 126 ton HNO 3...
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 Summer '07
 Fakhreddine/Lyon
 Chemistry, Atom, Hydrogen Bond, Intermolecular force, Kramer

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