CHEM 103: MODULE EXAM WITH ANSWERS

CHEM 103: MODULE EXAM WITH ANSWERS

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 WITH ANSWERS
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

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 WITH ANSWERS
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 2C H
6
(g) = -84.0 kJ/mole, fΔH 0 CO (g) = -110.5 kJ/mole,
f
ΔH 0 H O (l) =
2


-285.8
kJ/mole
2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l)


ΔH f 0 2C H
6
(g) = -84.0 kJ/mole, fΔH 0 CO (g) = -110.5 kJ/mole,
f
ΔH 0 H O (l) =
2


-285.8
kJ/mole
ΔHrxn = 2(+84.0) + 5(0) + 4(-110.5) + 6(-285.8) = - 1988.8 kJ/mole

Question 6

, CHEM 103: MODULE EXAM WITH ANSWERS
Click this link to access theperiodic tableThis may be helpful throughout
the exam.


Show the calculation of the number of moles of a 1.25 liter gas sample
collected at 740 mm and 28oC.


P x V = n x R x T
740 mm/760 = 0.974 atm = P R = 0.0821

1.25 liters = V 28oC + 273
= 301oK = T


(0.974) x (1.25) = n x (0.0821) x (301) n = 0.0493
mole

Question 7
Click this link to access the periodic table.This may be helpful throughout
the exam.


Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the
Fe26 atom and then identify the last electron to fill and write the 4
quantum numbers (n, l, ml and ms) for this electron.

Fe26 = 1s2 2s2 2p6 3s2 3p6 4s2 3d6 : n=3, l=2, ml = -2, ms
= -1/2

Question 8
Click this link to access the Periodic Table.This may be helpful
throughout the exam.


1. List and explain which of the following is the smaller
atom. C or N

2. List and explain which of the following atoms holds its valence
electrons more tightly.
Br or I