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CHEM 103: Portage Learning Module 1 to 6 Exam and Final Exam
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MODULE 1 EXAM
Question 1
Click this link to access the Periodic Table. This may be helpful throughout the exam.
- Convert 845.3 to exponential form and explain your answer. 2. Convert 3.21 x 10 -^5 to ordinary form and explain your answer.
- Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places = 8. x 102
- Convert 3.21 x 10 -^5 = negative exponent = smaller than 1, move decimal 5 places = 0.
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) 1/100) 1/10) milli (= 1/1000) deci (= centi (=
- 24.6 grams =? kg
- 6.3 ft =? inches
- 24.6 grams x 1 kg / 1000 g = 0.0246 kg
- 6.3 ft x 12 in / 1 ft = 75.6 inches please always use the correct units in your final answer Module 1 to 6 Exam answers Portage learning
3.0600 ÷ 0.0151 =? (give answer to correct number of significant figures)
- 3.0600 contains 5 significant figures.
- 0.0 151 contains 3 significant figures.
- 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.
- Coca cola
- Calcium
- Chili
- Coca cola - is not on periodic table (not element) - no element names (not compound) appears to be one substance = Solution
- Calcium - is on periodic table = Element
- 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
- Charcoal burns
- Mixing cake batter with water
Baking the batter to a cake
- Al 2 (CO 3 ) 3 - nonbinary ionic = aluminum carbonate
2
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.
- Carbon monoxide
- Manganese (IV) acetate
- Phosphorous acid
- Carbon monoxide - ide = binary, mono = 1 O = CO
- Manganese (IV) acetate - Mn+4, C 2 H 3 O -^1 = Mn(C 2 H 3 O 2 ) 4
- Phosphorous acid - nonbinary acid of H + phosphite (PO -^3 ) = H P 3 O (^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.
- Al 2 (CO 3 ) 3
- C 8 H 6 NO 4 Cl 3
- 8C + 6H + N + 4O + Cl = 215.
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.
- 13.0 grams of (NH 4 ) 2 CO 3
- 16.0 grams of C 8 H 6 NO 4 Br
- Moles = grams / molecular weight = 13.0 / 96.09 = 0.135 mole
- 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.
- 1.20 moles of (NH 4 ) 2 CO 3
- 1.04 moles of C 8 H 6 NO 4 Br
- Grams = Moles x molecular weight = 1.20 x 96.09 = 115.3 grams
- 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. Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal.
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.
- C 6 H 6 + O 2 → CO 2 + H 2 O
- As + O 2 → As 2 O 5
- Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4
- 2 C 6 H 6 + 15 O 2 → 12 CO 2 + 6 H 2 O
- 4 As + 5 O 2 → 2 As 2 O 5
- 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
- H 2 SO 4 → SO 3 + H 2 O
- S + 3 F 2 → SF 6
- H 2 + NiO → Ni + H 2 O
- H 2 SO 4 → SO 3 + H 2 O = Decomposition, One reactant → Two Products
- S + 3 F 2 → SF 6 = Combination. Two reactants→ One product
- H 2 + NiO → Ni + H 2 O = Single Replacement, Hydrogen displaces metal ion
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. mole CO 2 1.582 mole CO 2 x (12.01 + 2 x 16.00) = 69.6 g CO 2 MODULE 3 EXAM
Question 1
Click this link to access the Periodic Table. This may be helpful throughout the exam.
- ql↔s = m x ∆Hfusion = 54.3 g x 0.334 kJ/g = 18.14 kJ (since heat is removed) = - 18.14 kJ
Question 3
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the amount of heat involved if 35.6 g of H 2 S is reacted with excess O 2 to yield sulfur trioxide and water by the following reaction equation. Report your answer to 4 significant figures. 2 H 2 S (g) + 3 O 2 (g) → 2 SO 2 (g) + 2 H 2 O (g) ΔH = - 1124 kJ 1 mol H2s = 34.1 g of H2S = 603.2 kj (35.6g/34.1 g) x - 603.2 kJ = (1.0439) x (-603.2 kJ) = - 629. 7 kj 2 H 2 S (g) + 3 O 2 (g) → 2 SO 2 (g) + 2 H 2 O (g) ΔH = - 1124 kJ ΔHrx is for 2 mole of H 2 S reaction uses 35.6 g of H 2 S = 35.6/34.086 = 1.044 mole of H 2 S q = ΔHrx x new moles / original moles q = - 1124 kJ x 1.044 mole of H 2 S / 2 mole H 2 S = 586.7 given off
Question 4
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 3 C (graphite) + 4 H 2 (g) → C 3 H 8 (g) by using the following thermochemical data: C (graphite) + O 2 (g) → CO 2 (g) kJ
ΔH = - 393.
2 H 2 (s) + O 2 (g) → 2 H 2 O(l)
ΔH = - 571.
kJ C 3 H 8 (g) + 5 O 2 (g) Your→^3 CO^2 (g)^ +^4 H^2 O(l)^ ΔH^ =^ -^ 2220.0^ kJ Answer:
