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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions)

Enhance your exam preparation with CHEM103 Portage Learning Module 1 to 6 Exam with Answers, which features solved past exams to help you focus on key areas.

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    Pay it forward. MODULE 1 EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. Convert 845.3 to exponential form and explain your answer. 2. Convert 3.21 x 10 -5 to ordinary form and explain your answer. 1.Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places = 8.453 x 10 2 2.Convert 3.21 x 10 -5 = negative exponent = smaller than 1, move decimal 5 places = 0.0000321 Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. Using the following information, do the conversions shown below, showing all work: 1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts 1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10) 1. 24.6 grams = ? kg 2. 6.3 ft = ? inches 1. 24.6 grams x 1 kg / 1000 g = 0.0246 kg 2. 6.3 ft x 12 in / 1 ft = 75.6 inches please always use the correct units in your final answer Question 3

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    Click this link to access the Periodic Table. This may be helpful throughout the exam. Do the conversions shown below, showing all work: 1. 28 o C = ? o K 2. 158 o F = ? o C 3. 343 o K = ? o F 1. 28 o C + 273 = 301 o K o C → o K (make larger) +273 2. 158 o F - 32 ÷ 1.8 = 70 o C o F → o C (make smaller) -32 ÷1.8 3. 343 o K - 273 = 70 o C x 1.8 + 32 = 158 o F o K → o C → o F Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam. Be sure to show the correct number of significant figures in each calculation. 1. Show the calculation of the mass of a 18.6 ml sample of freon with density of 1.49 g/ml 2. Show the calculation of the density of crude oil if 26.3 g occupies 30.5 ml. 1. M = D x V = 1.49 x 18.6 = 27.7 g 2. D = M / V = 26.3 / 30.5 = 0.862 g/ml Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. 3.0600 contains ? significant figures. 2. 0.0151 contains ? significant figures.

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    3. 3.0600 ÷ 0.0151 = ? (give answer to correct number of significant figures) 1. 3.0600 contains 5 significant figures. 2. 0.0 151 contains 3 significant figures. 3. 3.0600 ÷ 0.0151 = 202.649 = 203 (to 3 significant figures for 0.0151) Question 6 Click this link to access the Periodic Table. This may be helpful throughout the exam. Classify each of the following as an element, compound, solution or heterogeneous mixture and explain your answer. 1. Coca cola 2. Calcium 3. Chili 1. Coca cola - is not on periodic table (not element) - no element names (not compound) appears to be one substance = Solution 2. Calcium - is on periodic table = Element 3. Chili - is not on periodic table (not element) - no element names (not compound) appears as more than one substance (meat, beans, sauce) = Hetero Mix Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Classify each of the following as a chemical change or a physical change 1. Charcoal burns 2. Mixing cake batter with water

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    3. Baking the batter to a cake 1. Charcoal burns - burning always = chemical change 2. Mixing cake batter with water - mixing = physical change 3. Baking the batter to a cake - baking converts batter to new material = chemical change Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the full Nuclear symbol including any + or - charge (n), the atomic number (y), the mass number (x) and the correct element symbol (Z) for each element for which the protons, neutrons and electrons are shown - symbol should appear as follows: x Z y +/- n 31 protons, 39 neutrons, 28 electrons 31 protons = Ga 31 , 39 neutrons = 70 Ga 31 , 28 electrons = (+31 - 28 = +3) = 70 Ga 31 +3 Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Name each of the following chemical compounds. Be sure to name all acids as acids (NOT for instance as binary compounds) 1. PF 5 2. Al 2 (CO 3 ) 3 3. H 2 CrO 4 1. PF 5 - binary molecular = phosphorus pentafluoride 2. Al 2 (CO 3 ) 3 - nonbinary ionic = aluminum carbonate

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    3. H 2 CrO 4 - nonbinary acid = chromic acid incorrect fluoride prefix Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Write the formula for each of the following chemical compounds explaining the answer with appropriate charges and/or prefixes and/or suffixes. 1. Carbon monoxide 2. Manganese (IV) acetate 3. Phosphorous acid 1. Carbon monoxide - ide = binary, mono = 1 O = CO 2. Manganese (IV) acetate - Mn +4 , C 2 H 3 O 2 -1 = Mn(C 2 H 3 O 2 ) 4 3. Phosphorous acid - nonbinary acid of H + phosphite (PO 3 -3 ) = H 3 PO 3 MODULE 2 EXAM 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. Al 2 (CO 3 ) 3 2. C 8 H 6 NO 4 Cl 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 answerto 3 significant figures. 1. 13.0 grams of (NH 4 ) 2 CO 3 2. 16.0 grams of C 8 H 6 NO 4 Br 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. 1.20 moles of (NH 4 ) 2 CO 3 2. 1.04 moles of C 8 H 6 NO 4 Br 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. (NH 4 ) 2 CrO 4 2. C 8 H 8 NOI 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 41.27% O / 16.00 = 2.579 / 0.6416 = 4 x 2 = 8 → Ca 3 P 2 O 8 Question 6

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    Click this link to access the Periodic Table. This may be helpful throughout the exam. Balance each of the following equations by placing coefficients in front of each substance. 1. C 6 H 6 + O 2 → CO 2 + H 2 O 2. As + O 2 → As 2 O 5 3. Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4 1. 2 C 6 H 6 + 15 O 2 → 12 CO 2 + 6 H 2 O 2. 4 As + 5 O 2 → 2 As 2 O 5 3. Al 2 (SO 4 ) 3 + 3 Ca(OH) 2 → 2 Al(OH) 3 + 3 CaSO 4 Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Classify each of the following reactions as either: Combination Decomposition Combustion Double Replacement Single Replacement 1. H 2 SO 4 → SO 3 + H 2 O 2. S + 3 F 2 → SF 6 3. H 2 + NiO → Ni + H 2 O 1. H 2 SO 4 → SO 3 + H 2 O = Decomposition, One reactant → Two Products 2. S + 3 F 2 → SF 6 = Combination. Two reactants → One product

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    3. H 2 + NiO → Ni + H 2 O = Single Replacement, Hydrogen displaces metal ion Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the oxidation number (charge) of ONLY the atoms which are changing in the following redox equations. Na 2 HAsO 3 + KBrO 3 + HCl → NaCl + KBr + H 3 AsO 4 Na 2 HAsO 3 + KBrO 3 + HCl → NaCl + KBr + H 3 AsO 4 Na 2 HAsO 3 : Na is metal in group I = +1 (total is +2), H = +1, each O is -2 (total is -6), so As is +3 H 3 AsO 4 : H is +1 (total is +3), each O is -2 (total is -8), so As is +5 KBrO 3 : K is metal in group I = +1, each O is -2 (total is -6), so Br is +5 KBr: K is metal in group I = +1, so Br is -1 Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the balancing of the following redox equation, including the determination of the oxidation number (charge) of ONLY the atoms which are changing. KMnO 4 + KI + H 2 O → KIO 3 + MnO 2 + KOH Mn compounds x 2 ; I compounds x 1 = 2KMnO4 + 1 KI + 1 H 2 O → 1 KIO 3 + 2MnO 2 + 2KOH KMnO 4 + KI + H 2 O → KIO 3 + MnO 2 + KOH KMnO 4 : K is metal in group I = +1, each O is -2 (total is -8), so Mn is +7

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    MnO 2 : Each O is -2 (total is -4), so Mn is +4 KI: K is metal in group I = +1, so I is -1 KIO 3 : K is metal in group I = +1, each O is -2 (total is -6), so I is +5 Since Mn (on left side) is +7 and Mn (on right side) is +4: Mn changes by 3 Since I (on left side) is -1 and I (on right side) is +5: I changes by 6 Multiply Mn compounds by 2 and I compounds by 1 and after balancing other atoms = 2 KMnO 4 + 1 KI + 1 H 2 O → 1 KIO 3 + 2 MnO 2 + 2 KOH Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the balanced equation and the calculation of the number of moles and grams of CO 2 formed from 20.6 grams of C 6 H 6 . Show your answers to 3 significant figures. C 6 H 6 + O 2 → CO 2 + H 2 O Molar mass of CO2 = 44.01 g/mol mass of CO2 = 1.5822 ml x 44.01 g/mol =69.63 grams mass of CO2 = 69.63 grams moles of CO2 = 1.58 mole 2 C 6 H 6 + 15 O 2 → 12 CO 2 + 6 H 2 O 20.6 g / (6 x 12.01 + 6 x 1.008) = 20.6 / 78.108 = 0.2637 mole x 12/2 = 1.58 mole CO 2 1.582 mole CO 2 x (12.01 + 2 x 16.00) = 69.6 g CO 2

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