CHEM 103: MODULE EXAM Q AND A
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
, CHEM 103: MODULE EXAM Q AND A
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Do the conversions shown below, showing all work:
1. 246oK = ? oC
2. 45oC = ? oF
3. 18oF = ? oK
1. 246oK - 273 = -27 oC oK → oC (make smaller)
-273
2. 45oC x 1.8 + 32 = 113 oF oC → oF (make larger)
x 1.8 + 32
3. 18oF - 32 ÷ 1.8 = -7.8 + 273 = 265.2 oK oF → 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.
1. Moles = grams / molecular weight = 12..15 = 0.0908 mole
2. Moles = grams / molecular weight = 15..17 = 0.0837 mole
, CHEM 103: MODULE EXAM Q AND A
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.
, CHEM 103: MODULE EXAM Q AND A
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 f 0 C
2
H
6
(g) = -84.0 kJ/mole, ΔH
f 0 CO (g) = -110.5 kJ/mole,f ΔH2 0 H O (l) = -
285.8
kJ/mole
2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l)
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
, CHEM 103: MODULE EXAM Q AND A
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Do the conversions shown below, showing all work:
1. 246oK = ? oC
2. 45oC = ? oF
3. 18oF = ? oK
1. 246oK - 273 = -27 oC oK → oC (make smaller)
-273
2. 45oC x 1.8 + 32 = 113 oF oC → oF (make larger)
x 1.8 + 32
3. 18oF - 32 ÷ 1.8 = -7.8 + 273 = 265.2 oK oF → 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.
1. Moles = grams / molecular weight = 12..15 = 0.0908 mole
2. Moles = grams / molecular weight = 15..17 = 0.0837 mole
, CHEM 103: MODULE EXAM Q AND A
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.
, CHEM 103: MODULE EXAM Q AND A
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 f 0 C
2
H
6
(g) = -84.0 kJ/mole, ΔH
f 0 CO (g) = -110.5 kJ/mole,f ΔH2 0 H O (l) = -
285.8
kJ/mole
2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l)
