This study source was downloaded by 100000880877832 from CourseHero.com on 02 -15-2024 03:00:05 GMT -06:00 https://www.coursehero.com/file/189588951/CHEM -103-Final -Examdocx/ CHEM 103 Final Exa m (100 out of 100) Question s and Answers (New Update) Verified Elaborations. Question 1 Complete the two problems below: 1. Convert 0.0000726 to exponential form and explain your answer. 2. Convert 5.82 x 103 to ordinary form and explain your answer. Answer: 1. Convert 0.0000726 = smaller than 1 = negative exponent, move decimal 5 places = 7.26 x 10-5 2. Convert 5.82 x 103 = positive exponent = larger than 1, move decimal 3 places = 5820 Question 2 Do the conversions shown below, showing all work: 1. 358oK = ? oC 2. 53oC = ? oF 3. 158oF = ? oK Answer: 1. 358oK - 273 = 85 oC oK → oC (make smaller) -273 2. 53oC x 1.8 + 32 = 127.4 oF oC → oF (make larger) x 1.8 + 32 3. 158oF - 32 ÷ 1.8 = 70 + 273 = 343 oK oF → oC → oK Question 3 Show the calculation of the number of moles in the given amount of the following substances. Report your answer to 3 significant figures. 1. 12.0 grams of Ca 3(PO 4)2 This study source was downloaded by 100000880877832 from CourseHero.com on 02 -15-2024 03:00:05 GMT -06:00 https://www.coursehero.com/file/189588951/CHEM -103-Final -Examdocx/ 2. 15.0 grams of C9H8NO 4Cl Answer: 1. Moles = grams / molecular weight = 12.0 / 310.18 = 0.0387 mole 2. Moles = grams / molecular weight = 15.0 / 229.61 = 0.0653 mole Question 4 Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal. 1. Al 2(SO 4)3 2. C7H5NOBr Answer: 1. %Al = 2 x 26.98/342.17 x 100 = 15.77% %S = 3 x 32.07/342.17 x 100 = 28.12% %O = 12 x 16/342.17 = 56.11% 2. %C = 7 x 12.01/ 199.02 x 100 = 42.24% %H = 5 x 1.008/ 199.02 x 100 = 2.53% %N = 1 x 14.01/199.02 = 7.04% %O = 1 x 16.00/199.02 x 100 = 8.04% %Br = 79.90/199.02 x 100 = 40.15% Question 5 Show the calculation of the heat of reaction (ΔH rxn) for the reaction: CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2O (l) by using the following thermochemical data: ΔH f0 CH 4 (g) = -74.6 kJ/mole, ΔH f0 CO 2 (g) = -393.5 kJ/mole, ΔH f0 H2O (l) = -285.8 kJ/mole Answer:
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