Practice_Problems_Answers

Practice_Problems_Answers - Answers to Harvard General...

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Answers to Harvard General Chemistry Practice Problems 1. Math Review: Calculations and Scientific Notation 1. a) 523.4 b) 2.295 c) 19.2 d) 42.2 e) 858 f) 18.15 g) 1.18 × 10 –4 h) 3.601 i) 2.47 × 10 –4 2. Dimensional Analysis I 1. a) 2.38 atm b) 2.05 × 10 3 torr or 2050 torr c) 1.00 × 10 5 Pa 2. 341 mg 3. 3.60 × 10 6 sheets of paper 3. Dimensional Analysis II 1. a) 4.67 × 10 –2 in 3 or 0.0467 in 3 b) 1.92 g 2. 2.52 × 10 10 gallons 4. Atoms, Molecules and Ions 1. a) 18 protons, 22 neutrons, 18 electrons b) 20 protons, 20 neutrons, 18 electrons c) 19 protons, 20 neutrons, 18 electrons d) 19 protons, 20 neutrons, 19 electrons 2. 20 protons, 18 electrons 3. 35 Cl : 17 protons, 18 neutrons, 18 electrons 37 Cl : 17 protons, 20 neutrons, 18 electrons 4. a) 10 protons, 7 neutrons, 10 electrons b) 11 protons, 7 neutrons, 10 electrons 5. Naming Compounds 1. a) Cu 3 P b) Fe 2 (SO 4 ) 3 c) I 2 d) AlCl 3 e) Cr 2 O 3 f) NH 3 g) Li 3 N h) Al 2 (CO 3 ) 3 i) Cs 3 PO 4 j) ZnO k) SnCl 2 l) N 2 O 5 m) AgF n) H 3 PO 4 o) Ba 3 (PO 4 ) 2 p) (NH 4 ) 2 SO 4 2. a) Carbon tetrachloride b) Copper (II) oxide or cupric oxide (either answer would be acceptable) c) Magnesium nitride 6. Chemical Formulas: Percent Composition 1. ii < iii < i 2. a) C: 34.3% H: 7.2% P: 22.1% O: 22.8% F: 13.6% b) If we combust 10.0 grams of Sarin, we'd get 12.6 g CO 2 and 6.5 g H 2 O. This does not agree with the results obtained for the unknown compound; thus it can't be Sarin. 7. Empirical and Molecular Formulas 1. C 11 H 17 N 3 O 8 2. a) C 5 H 4 O b) C 15 H 12 O 3 ; MW = 240.25 g/mol 3. 64450 g/mol
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8. Writing and Balancing Equations 1. a) CO(NH 2 ) 2 + 6 HOCl 2 NCl 3 + CO 2 + 5 H 2 O b) 2 Ca 3 (PO 4 ) 2 + 6 SiO 2 + 10 C P 4 + 6 CaSiO 3 + 10 CO 2. 6 CO 2 + 6 H 2 O C 6 H 12 O 6 + 6 O 2 3. 3 Nb + 4 I 2 Nb 3 I 8 4. C 8 H 18 + 25 / 2 O 2 8 CO 2 + 9 H 2 O 9. Stoichiometry of Reactions 1. 55.9 g NaClO 2. a) NiS + 2 O 2 + 2 HCl NiCl 2 + H 2 SO 4 b) 0.654 g NiCl 2 10. Stoichiometry with Limiting Reagents 1. a) V 2 O 5 + 3 Zn V 2 O 2 + 3 ZnO b) 68.25 g V 2 O 2 2. a) 5 P 4 S 3 + 12 Br 2 3 P 4 S 5 + 8 PBr 3 b) 77.47 g P 4 S 5 11. Stoichiometry of Mixtures 1. 61.1% CO 2 by mass 2. 6.55 g Cu 2 S, 4.25 g CuS 12. Oxidation Numbers 1. Br: +5 H: +1 As: –3 O: –2 As: +3 H: +1 O: –2 Cr: +3 K: +1 S: +2 Cl: –1 Cl: +5 O: –2 O: –2 O: +2 Na: +1 Fe: + 8 / 3 F: –1 O: –1 O: –2 2. N 2 is reduced, H 2 is oxidized FeCl 3 is reduced, KI is oxidized No oxidation or reduction is taking place. Cl 2 is being oxidized (to ClO 3 ) and reduced (to Cl ) 13. Solutions: Molarity 1. 18.0 M 2. 17.47 M 3. 0.216 M 4. 0.052 g 5. x = 4.4 g KNO 3 y = 4.49 g NaCl
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14. Solution Stoichiometry I 1. a) 3 ClO ( aq ) + NH 3 ( aq ) NCl 3 ( l ) + 3 OH ( aq ) b) 0.028 g NCl 3 2. 2.79 g CO 2 15. Solution Stoichiometry II: Acid/Base Neutralizations 1. 1.72 L 2. 71.2 mL 3. 1.26 L 16. Solution Stoichiometry III: Titrations 1. 1.22 g 2. 312 g/mol 3. 119.2 g/mol 17. Solution Stoichiometry IV: Precipitation Reactions 1. 0.0119 M Ba 2+ ( aq ) 2. CaSO 4 = 8.92 g Al 2 (SO 4 ) 3 = 1.08 g 18. Putting It Together: Carminic Acid 1. a) x = 22 y = 20 b) 1 acidic hydrogen (it is a monoprotic acid) 19. The Ideal Gas Law 1. a) 6.0 × 10 5 L b) 0.348 g Cl 2 2. 53.4 g O 2 20. Mixtures of Gases 1. a) 0.0863 mol O 2 b) 0.0211 torr O 2 c) 2.78 × 10 –6 torr O 2 2. 16.3 g butane volume = 9.286 L 21. Reactions Involving Gases I 1. 0.480 atm 2. 1.5 atm 22. Reactions Involving Gases II 1.
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This note was uploaded on 02/08/2012 for the course CHEM 161 taught by Professor Shaklovich during the Spring '10 term at Harvard.

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Practice_Problems_Answers - Answers to Harvard General...

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