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CHEM103 / CHEM 103 Module 2: (Latest Update 2026 / 2027) General Chemistry I with Lab | Questions & Answers | Grade A | 100% Correct - Portage Learning

HEAL100 Health Care — exam and revision material · Other

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Health Care exam material, 7 pages. Originally published on StudyMerit.

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7
Academic year
2025/2026
Language
English

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CHEM 103 MODULE 2 EXAM 2022 Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1 . A l 2 ( C O 3 ) 3 2 . C 8 H 6 N O 4 C l 1. 2Al + 3C + 9O = 233.99

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2. 8C + 6H + N + 4O + Cl = 215.59 Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of moles in the given amount of the following substances. Report your answer to 3 significant figures. 1. 13 . 0 g r a m s o f ( N H 4 ) 2 C O 3 2. 16 . 0 g r a m s o f C 8 H 6 N O 4 B r 1. Moles = grams / molecular weight = 13.0 / 96.09 = 0.135 mole 2. Moles = grams / molecular weight = 16.0 / 260.04 = 0.0615 mole Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of grams in the given amount of the following substances. Report your answer to 1 place after the decimal. 1. 1 . 2 0 mo l e s o f ( N H 4 ) 2 C O 3 2. 1 . 0 4 mo l e s o f C 8 H 6 N O 4 B r 1. Grams = Moles x molecular weight = 1.20 x 96.09 = 115.3 grams 2. Grams = Moles x molecular weight = 1.04 x 260.04 = 270.4 grams Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam.

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Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal. 1. ( N H 4 ) 2 C r O 4 2. C 8 H 8 N O I 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% Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the empirical formula for each compound whose elemental composition is shown below. 38.76% Ca, 19.87% P, 41.27% O 38.76% Ca / 40.08 = 0.9671 / 0.6416 = 1.5 x 2 = 3 19.87% P / 30.97 = 0.6416 / 0.6416 = 1 x 2 = 2 4 1 .2 7 % O / 16 . 0 0 = 2 .5 7 9 / 0 .6 4 1 6 = 4 x 2 = 8 → C a 3 P 2 O 8 Question 6

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