Portage Learning CHEM 103 Final Exam Study Guide Latest Updated 2023

Portage Learning CHEM 103 Final Exam Study Guide Latest Updated 2023 Question 1 Show the calculation of the percent of each element present in the followingcompounds. Report your answer to 2 places after the decimal. 1. (NH4)2CrO4 2. C8H8NOI 1. %N = 2 x 14.01/152.08 x 100 = 18.43% %H = 8 x 1.008/152.08 x 100 = 5.30% %Cr = 1 x 52.00/152.08 x 100 = 34.20% %O = 4 x 16.00/152.08 x 100 = 42.08% 2. %C = 8 x 12.01/261.05 x 100 = 36.80% %H = 8 x 1.008/261.05 x 100 = 3.09% %N = 1 x 14.01/261.05 x 100 = 5.37% %O = 1 x 16.00/261.05 x 100 = 6.13% %I = 1 x 126.9/261.05 x 100 = 48.61% 1. Convert 0.00346 to exponential form and explain your answer. 2. Convert 2.76 x 10 4 to ordinary form and explain your answer. Your Answer: 1. 0.00346 = smaller than 1 = negative exponent, move decimal 3places = 3.46 x 10-3 2. 2.76 x 10 4 = positive exponent = larger than 1, move decimal 4 places = 27600 Question 2 Do the conversions shown below, showing all work: 1. 28 oC = ? oK 2. 158 oF = ? oC f 2 6 f 2 f 2 3. 343 oK = ? oF 1. 28oC + 273 = 301 oK oC → oK (make larger) +273 2. 158oF - 32 ÷ 1.8 = 70 oC oF → oC (make smaller) -32 ÷1.8 3. 343oK - 273 = 70 oC x 1.8 + 32 = 158 oF oK → oC → oF Question 3 Show the calculation of the number of moles in the given amount of thefollowing substances. Report your answerto 3 significant figures. 1. 12.0 grams of Ca3(PO4)2 2. 15.0 grams of C9H8NO4Cl 1. Moles = grams / molecular weight = 12.0 / 310.18 = 0.0387mole 2. Moles = grams / molecular weight = 15.0 / 229.61 = 0.0653mole