Chem 161-2010 exam 1 + solutions

Chem 161-2010 exam 1 + solutions - Chem 161-2010 Exam I...

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Chem 161-2010 Exam I th edition Chapter 3A Atoms, ions, moles and molecular weights 1. How many molecules of propanol are present in 55.8 mL of propanol (density = 0.8034 g/mL)? A . 4.49 x 10 23 B. 7.56 x 10 23 C. 2.18 x 10 23 D. 9.07 x 10 23 E. 4.54 x 10 23 Plan: mL → g → mol → molecules 55.8mL P x (0.8034g P /mL P ) x (1mol P /60g P ) x (6.022x10 23 molecules P /mol P ) = 4.50x10 23 molecules P Chem 161-2010 Exam I th edition Chapter 2 Periodic Table 2. Which element is a transition metal? A. Ba B. Sm C. Pb D. Sb E . Ag All of the elements between groups IIa and IIIa are transition metals. This includes Ag. Chem 161-2010 Exam I th edition Chapter 3A Empirical and molecular formulas and related stoichiometry 3. A compound contains only C, H, N and O. Combustion of 12.0 g of the compound produces 18.4 g CO 2 , 6.81 g H 2 O and 3.52g N 2 . What is the empirical formula of the compound? A. C 7 H 9 NO 2 B . C 5 H 9 N 3 O 2 C. C 7 H 6 N 2 O 2 D. C 6 H 8 N 3 O 2 E. C 5 H 7 N 2 O 3 C m H n N p O q → CO 2 + H 2 O + N 2
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18.4g 6.81g 3.52g Plan: g A → mol A → mol B → g B 18.4gCO 2 x (1 mol CO 2 /44.01gCO 2 ) x (1molC/1molCO 2 ) = 0.418molC 0.418 mol C x (12.01gC/molC) = 5.02gC 6.81gH2O x (1 mol H2O/18.02gH2O) x (2molH/1molH2O) = 0.7558 mol H 0.7558 mol H x (1.01gH/molH) = 0.763gH 3.52gN 2 x (1molN2/28.02gN 2 ) x (2molN/1molN 2 ) = 0.2512mol N 0.2512 mol N x (14.01gN/molN) = 3.517gN 12.0g – (5.02 + 0.763 + 3.517) = 2.695 g O 2.695gO x (1 mol O/16.00gO) = 0.1684 mol O C 0.418 H 0.7558 N 0.2512 O 0.1684 Divide by smallest number: C 2.48 H 4.48 N 1.49 O 1 Multiply by 2 to get rid of fractions: C 5 H 9 N 3 O 2 Chem 161-2010 Exam I th edition Chapter 3A Atoms, ions, moles and molecular weights 4. How many moles of Cu(NO 3 ) 2 would contain 7.86 x 10 21 atoms of O? A. 6.54 x 10 −3 B. 8.73 x 10 −3 C . 2.18 x 10 −3 D. 7.54 x 10 −3 E. 4.36 x 10 −3 Cu(NO 3 ) 2 → Cu + 2N + 6O atoms o → mol o → mol cu(no3)2 (7.86 x 10 21 atomsO/6.022 x 10 23 atoms/mol) = 0.01305 mol O. 0.01305molO x (1 mol Cu(NO 3 ) 2 /6mol O) = 0.00218 mol Cu(NO 3 ) 2 . Chem 161-2010 Exam I th edition Chapter 1 Units/unit conversions/dimensional analysis 5. In the SI metric system, which power of ten is associated with the prefix µ? A. 10 9
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B . 10 -6 C. 10 -9 D. 10 6 E. 10 -12 µ = micro = 1x10 -6 1 µL = 1 x 10 -6 L Chem 161-2010 Exam I th edition Chapter 1 Classification of matter 6. Indicate whether each of the following changes are physical changes or chemical changes. W. The sublimation of iodine solid in an enclosed chamber X. The evolution of carbon dioxide gas when sugar is added to activated yeast Y. The dissolving of NaCl in water Z. The crystallization of sugar crystals from a sugar solution. W
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Chem 161-2010 exam 1 + solutions - Chem 161-2010 Exam I...

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