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# answers_to_practice_exam_I - M = 1.41 g/mL X 242 mL = 341 g...

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Multiple choice: 1. A 2. D 3. C 4. A 5. A 6. E 7. D 8. B 9. B 10. C 11. B 12. E 13. D 14. A 15. B 16. E 17. D 18. B 19. A 20. B 21. A 22. B 23. C 24. D 25. E 26. B 27. B 28. C 29. D 30. A Short answer: 1. (a) The balanced equation is given in the problem No conversion is needed because the amount of the starting material, Li, is given in moles. Since 2 moles of Li produce 1 mole of H 2 , we calculate moles of H 2 produced as follows: Moles of H 2 produced = 6.23 mol Li X 1 mol H 2 /2 mol Li = 3.12 mol H 2 (b) moles of Li = 80.57 g Li x 1 mol Li/6.941 g Li = 11.61 mol Li moles of H 2 produced = 11.61 mol Li x 1 mol H 2 /2 mol Li = 5.805 mol H 2

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mass of H 2 produced = 5.805 mol H 2 x 2.016 g H 2 /1 mol H 2 = 11.70 g H 2 2. D = M/V M = D x V D= 1.41 g/mL V = 242 mL
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Unformatted text preview: M = 1.41 g/mL X 242 mL = 341 g 3. The molar masses of Cu and CuFeS 2 are 63.55 g and 183.5 g, respectively, so the percent composition by mass of Cu is: %Cu = 63.55 g/183.5 g x 100% = 34.63% To calculate the mass of Cu in a 3.71 x 10 3 kg sample of CuFeS 2 , we need to convert the percentage to a fraction (that is, convert 34.63% to 34.63/100, or 0.3463) and write: Mass of Cu in CuFeS 2 = 0.3463 x 3.71 x 10 3 kg = 1.28 x 10 3 kg This calculation can be simplified by combining the above two steps as follows: Mass of Cu in CuFeS 2 = 3.71 x 10 3 kg CuFeS 2 x 63.55 g Cu/183.5 g CuFeS 2 = 1.28 x 10 3 kg Cu...
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answers_to_practice_exam_I - M = 1.41 g/mL X 242 mL = 341 g...

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