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CHEM 103 Final Exam (Latest-2022)/ CHEM103 Final Exam / CHEM 103 General Chemistry Final Exam/ CHEM103 General Chemistry Final Exam: Portage Learning |Verified Q & A|
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Course
CHEM 103: General Chemistry I w/Lab
Institution
Portage Learning
CHEM 103 Final Exam (Latest-2022)/ CHEM103 Final Exam / CHEM 103 General Chemistry Final Exam/ CHEM103 General Chemistry Final Exam: Portage Learning |Verified Q & A|
chem 103 final exam latest 2022 chem103 final exam chem 103 general chemistry final exam chem103 general chemistry final exam portage learning |verified q amp a|
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CHEM 103: General Chemistry I w/Lab
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CHEM 103 Final 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 102
2. Convert 3.21 x 10-5 = negative exponent = smaller than 1, move decimal 5
places = 0.0000321
Question 2
Do the conversions shown below, showing all work:
1. 246oK = ? oC
2. 45oC = ? oF
3. 18oF = ? oK
1. 246oK - 273 = -27 oC o
K → oC (make smaller)
-273
2. 45oC x 1.8 + 32 = 113 oF o
C → oF (make larger) x
1.8 + 32
3. 18oF - 32 ÷ 1.8 = -7.8 + 273 = 265.2 oK o
F → oC → oK
Question 3
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.
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.
1. Al2(SO4)3
2. C7H5NOBr
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
= 8.03%
%Br = 79.90/199.02 x 100 = 40.15%
Question 5
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:
2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l)
by using the following thermochemical data:
ΔH f0 C 2H 6 (g) = -84.0 kJ/mole, ΔHf 0 CO (g) = -110.5 kJ/mole, ΔHf 0 H2 O (l) = -285.8
kJ/mole
2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l)
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