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Worked-Out Solutions Exam 3
1
...
So the final solution has a more moles in a lower volume,
meaning it’s more concentrated
...

M1 V 1 = M2 V 2

(0
...
968 x where x is the final volume

Solving: x = final volume of solution = 136 mL
...
By definition, ΔHf is the ΔH of a reaction where 1 mole of product is formed from elements
in their standard states at room temperature and pressure
...
If we form one mole of HF, the
equation would be:
½ H2(g) + ½ F2(g) → HF(g) ΔH = -273 kJ
Choice D is the reverse of this equation with a change of sign in ΔH
...
Balanced equation: C3H8 + 5O2 → 3CO2 + 4H2O
2 mol O2 x

1mol C3H 8
kJ
x 2040
= 816 kJ Choice C
mol C3H 8
5mol O 2

4
...
Assume 1
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0 g Ar, 10
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0 g N2 x

1mol
= 0
...
0g

5
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126 mol Ar
39
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141 mol Cl2
70
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0357 + 0
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141 = 0
...
0 g Cl2 x

Mol fraction Cl2 =

mol Cl 2
0
...
466
0
...
466 (6
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8 atm

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3 deg x
= 171
...
3kJ
Heat released per mole of aniline:
= 3
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98g x
93
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Heat released = 1017 g x 4
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ΔH = -3
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Choice B

2 Fe2O3(s) + 3 C(s) → 4Fe(s) + 3 CO2(g)
3
...
0300 mol CO2
159
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0300)(0
...
711 L CO2
770
P
(
)
760

Choice B

Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
ΔH = -1049 kJ
1st equation left as is
3H2(g) + 3Cl2(g) → 6HCl(g)
ΔH = -184
...
8 x 6 2nd equation x 6
Adding:
7
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5(3)) + 323(2) + (-74
...
3 kJ
Choice C

8
...

LiF, LiI, CsI are all ionic lattices with +1 and -1 ions
...
Smaller ions can get closer to each other and
attract more
...
F- is smaller than I-, so LiF is a stronger lattice than LiI
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9
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No change in oxidation number, so not a redox
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10
...


750
(
)(0
...
00825 mol
RT (0
...


13
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801g
= 97
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00825mol

Choice C

PV < nRT for real gases when intermolecular attractive forces are important
...
If the attractive forces are high
enough, the gas will liquefy
...
500 g Cu x

2 mol AgNO3
mL
1mol
1L
x
x
x 1000
= 39
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54g
0
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14
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025 L Mg(NO3)2 x 0
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050 L KNO3 x 0
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00500 mol NO3L
mol Mg(NO3 ) 2

1mol NO−3
mol
x
= 0
...
00500 + 0
...
0100 mol NO3Total volume = 25 mL + 50 mL = 75 mL or 0
...


16
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0100 mol
= 0
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0750 L

Gases at the same temperature have the same average kinetic energy of the molecules
...

Choice D
2Νa(s) → 2Na(g)
½ O2(g) → O(g)
2Νa(g) → 2Na+(g) + 2eΟ(g) + e- → O-(g)
Ο-(g) + e- → O2-(g)
2Νa(g) + O2-(g) → Na2O(s)
Adding the equations:
2Νa(s) + ½ O2(g) → Na2O(s)

2ΔHsubl
½ BE
2 IE1
EA1
EA2
x (lattice energy of Na2O)

ΔHf

Αdding the ΔH expressions:
2ΔHsubl + ½ BE + 2IE1 + EA1 + EA2 + x = ΔHf
17
...
These are strong electrolytes when
dissolved in water
...
This is a weak electrolyte when dissolved in water
...
This is a weak electrolyte when dissolved in water
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3RT
To use this formula properly, R must be in Joules, and the molar mass M
M
must be in kg
...


v=

R = 8
...
0

v=

19
...
028
mol
mol

3(8
...
028
s

rateof gas x
= 2
...
9

g
mol

molar massCl 2
=
molar mass x

Choice B

70
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Choice E

Total Score _________

Open-Ended Questions
Name: ____________________________________

White/Yellow
RUID: _____________________

In numerical problems show all your work, and report your answer to the correct number of
significant digits
...

Include the appropriate state symbols, (s), (ℓ), (g), (aq) in your answer
...


(b) Write a net ionic equation for a reaction releasing a gas, using two of these
solutions
...

2H+(aq) + CO32-(aq) → CO2(g) + H2O(ℓ)

(c) Choose either of these reactions and state whether it is an oxidation-reduction
reaction, and give a reason for your choice using oxidation numbers
...
There is no change of
oxidation number for any element
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Score Q2
______
Question 2 (worth 8 points, or the equivalent of 2 multiple-choice questions)
(a)

N2(g) + 3F2(g) → 2NF3(g)

ΔHrxn = -264 kJ

Assume that 0
...
g block of iron initially at 25
...
What is the final temperature of the iron
block?
The specific heat capacity of iron is 0
...

0
...
0 kJ released = 33000 J
2 mol NF3

33000 = 900 (0
...
6
Final temperature = 25
...
6 = 107
...

N2(g) + 3F2(g) → 2NF3(g)
946 + 3x -6(272) = -264

ΔHrxn = -264 kJ
Solving: x = 141 kJ/mol

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600 mol NO and 0
...
00 L container and the
reaction proceeds with 100% yield
...

Calculate the total pressure in the container after the reaction is over
...

Limiting reactant is NO
...
600 mol NO reacts with 0
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600
mol NO2 with 0
...

After the reaction, there are 0
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100 mol O2 = 0
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700)(0
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46 atm
V
5
...
What is the pressure, in torr, of the gas in the
bulb in the diagram below?

Mercury goes from -3
...
2 cm or h = 6
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