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Homework 3-solutions

# Homework 3-solutions - kramer(jrk2286 – Homework 3 –...

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Unformatted text preview: kramer (jrk2286) – Homework 3 – sutcliffe – (90830) 1 This print-out should have 31 questions. Multiple-choice 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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Homework 3-solutions - kramer(jrk2286 – Homework 3 –...

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