CHEM 103 Final Exam.
Final Exam - Requires Respondus LockDown Browser + Webcam
Question 1
Complete the two problems below:
1. Convert 0.0000726 to exponential form and explain your answer.
2. Convert 5.82 x 10 to ordinary form and explain your answer.
3
Answer:
1.Convert 0.0000726 = smaller than 1 = negative exponent, move decimal 5
places = 7.26 x 10 -5
2.Convert 5.82 x 10 = positive exponent = larger than 1, move decimal 3
3
places = 5820
Question 2
Do the conversions shown below, showing all work:
1. 358 K = ? C
o o
2. 53 C = ? F 3.
o o
158 F = ? K
o o
Answer:
1. 358 K - 273 = 85 C
o o
K → C (make smaller)
o o
-273
2. 53 C x 1.8 + 32 = 127.4 F
o o o
C → F (make larger) x
o
1.8 + 32
3. 158 F - 32 ÷ 1.8 = 70 + 273 = 343 K
o o
F→ C→ K
o o o
Question 3
Show the calculation of the number of moles in the given amount of the
following substances. Report your answerto 3 significant figures.
, 1. 12.0 grams of Ca (PO ) 3 4 2
2. 15.0 grams of C H NO Cl 9 8 4
Answer:
1. Moles = grams / molecular weight = 12..18 = 0.0387 mole
2. Moles = grams / molecular weight = 15..61 = 0.0653 mole
Question 4
Show the calculation of the percent of each element present in the following
compounds. Report your answer to 2 places after the decimal.
1. Al (SO )
2 4 3
2. C H NOBr
7 5
Answer:
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
= 8.04%
%Br = 79.90/199.02 x 100 = 40.15%
Question 5
Show the calculation of the heat of reaction (ΔH ) for the reaction: rxn
CH (g)
4 + 2 O (g) →
2 CO (g)
2 + 2 H O (l) by using the
2
following thermochemical data:
ΔH CH (g) = -74.6 kJ/mole, ΔH CO (g) = -393.5 kJ/mole, ΔH H O (l) = -285.8
f
0
4 f
0
2 f
0
2
kJ/mole
Answer:
Final Exam - Requires Respondus LockDown Browser + Webcam
Question 1
Complete the two problems below:
1. Convert 0.0000726 to exponential form and explain your answer.
2. Convert 5.82 x 10 to ordinary form and explain your answer.
3
Answer:
1.Convert 0.0000726 = smaller than 1 = negative exponent, move decimal 5
places = 7.26 x 10 -5
2.Convert 5.82 x 10 = positive exponent = larger than 1, move decimal 3
3
places = 5820
Question 2
Do the conversions shown below, showing all work:
1. 358 K = ? C
o o
2. 53 C = ? F 3.
o o
158 F = ? K
o o
Answer:
1. 358 K - 273 = 85 C
o o
K → C (make smaller)
o o
-273
2. 53 C x 1.8 + 32 = 127.4 F
o o o
C → F (make larger) x
o
1.8 + 32
3. 158 F - 32 ÷ 1.8 = 70 + 273 = 343 K
o o
F→ C→ K
o o o
Question 3
Show the calculation of the number of moles in the given amount of the
following substances. Report your answerto 3 significant figures.
, 1. 12.0 grams of Ca (PO ) 3 4 2
2. 15.0 grams of C H NO Cl 9 8 4
Answer:
1. Moles = grams / molecular weight = 12..18 = 0.0387 mole
2. Moles = grams / molecular weight = 15..61 = 0.0653 mole
Question 4
Show the calculation of the percent of each element present in the following
compounds. Report your answer to 2 places after the decimal.
1. Al (SO )
2 4 3
2. C H NOBr
7 5
Answer:
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
= 8.04%
%Br = 79.90/199.02 x 100 = 40.15%
Question 5
Show the calculation of the heat of reaction (ΔH ) for the reaction: rxn
CH (g)
4 + 2 O (g) →
2 CO (g)
2 + 2 H O (l) by using the
2
following thermochemical data:
ΔH CH (g) = -74.6 kJ/mole, ΔH CO (g) = -393.5 kJ/mole, ΔH H O (l) = -285.8
f
0
4 f
0
2 f
0
2
kJ/mole
Answer:
