Remember- Show all work for partial credit
1. (10 points) Calculate the number of moles of F ion present in 250 mls of a
0.15M solution of AlF
3
Molarity = moles/liters: .15 = X/.25l;
.15 mole/l × .25 l = .0375 moles of AlF
3
But the ionization reaction for AlF
3
is AlF
3
6 1Al 3+ (aq)+ 3 F (aq)
So .0375 moles of AlF
3
generates
.0375 moles AlF
3
× 3 mole F /1 mole AlF
3
= .1125 moles F -
2. (10 points) Predict the solubility of the following salts. (S=soluble,
I=insoluble, SS= slightly soluble).
NiS _____ ss ______ Co(OH)
3
___ ss or i_
PbCl
2
____ i_ ________ K
2
SO
4
______ s_ _______
Hg
2
Cl
2
_____ _i__ ________

3. (10 points) Write balanced molecular. complete ionic and net ionic equations for the following reactions:
Solid silver hydroxide and hydrobromic acid:
Molecular:
Net ionic:
AgOH(s) + HBr(aq) 6 AgBr(s) + H
2
O(l)
Complete ionic: AgOH(s) + H + (aq) + Br
AgOH(s) + H + (aq) + Br
(aq) 6 AgBr(s) + H
2
O(l)
(aq) 6 AgBr(s) + H
2
O(l)
Aqueous potassium hydroxide and calcium carbonate:
Molecular: 2KOH(aq) + CaCO
3
(aq) 6 Ca(OH)
2
(s) + K
2
CO
3
(aq)
Complete ionic:
2K + (aq) +2 OH (aq) + Ca 2+ (aq) + CO
3
2(aq) 6 Ca(OH)
2
(s) + 2K + (aq) +
CO
3
2(aq)
Net ionic: 2 OH (aq) + Ca 2+ (aq) 6 Ca(OH)
2
(s)
4. (10 points) Assign oxidation state for all atoms in the following compounds:
KMnO
4
K=+1
O=-2
Mn=+7
MnO
2
O=-2
Mn=+4
2
Mg
2
P
2
O
Mg=+2
O=-2
P=+5
7
N
2
N=0
CO
2
O=-2
C=+4

5. A (5 points) Balance the following redox reaction that occurs in acidic conditions:
Cu 6 Cu 2+
Cu(s) + NO
3
(aq) 6 Cu 2+ (aq) + NO(g)
NO
3
6 NO(g)
Cu 6 Cu 2+ + 2e -
4H +
NO
+ NO
3
-
3
-
6
6
NO(g) + 2H
NO(g) + 2H
2
2
O
O
3e + 4H + + NO
3
6 NO(g) + 2H
2
O
×3 ×2
3Cu 6 3Cu 2+ + 6e 6e + 8H + + 2NO
3
6 2NO(g) + 4H
3Cu + 6e + 8H + + 2NO
3
6 3Cu 2+ + 6e + 2NO(g) + 4H
2
O
2
O
3Cu(s) + 8H + (aq)+ 2NO
3
(aq) 6 3Cu 2+ (aq)+ 2NO(g) + 4H
2
O(l)
B (5 points) Balance the following redox reaction that occurs in basic conditions:
4H
2
4H
2
Al(s) + MnO
4
-
Al 6 Al(OH)
4
-
O + Al 6 Al(OH)
4
-
(aq) 6 MnO
2
(s) + Al(OH)
4
(aq)
MnO
MnO
4
-
4
-
6
6
MnO
2
MnO
2
+ 2H
2
O
4H
2
O + Al 6 Al(OH)
O + Al 6 Al(OH)
4
-
4
+ 4H +
+ 4H + + 3e -
4H
3e -
+ + MnO
+ 4H +
4
6 MnO
2
+ 2H
+ MnO
4
6 MnO
2
2
O
+ 2H
2
O
4H
2
O + Al + 3e + 4H + + MnO
4
6 Al(OH)
4
+4H + + 3e + MnO
2
+ 2H
2
O
2H
2
O(l) + Al(s) + MnO
4
(aq) 6 Al(OH)
4
(aq)+ MnO
2
(s)
(No H + or OH so this equation works under all conditions)
3
6. (10 points) Truck tires can be inflated to 45 lbs/in 2 . Convert this to torr and to Pa. ( 1 atm = 14.7 lbs/in 2 , 1 atm= 101325 Pa)
45 lbs in
2
45 lbs in 2
×
×
1 atm lbs in 2
1 atm lbs in
2
×
760 torr
=
1 atm
×
101325 Pa
1 atm
2300 torr
=
Pa

7. (10 points) In class I demonstrated how to make acetylene (C
2
H
2
) by reacting CaC
2
with water using the reaction:
CaC
2
(s) + 2H
2
O(l) 6 Ca(OH)
2
(aq) + C
2
H
2
(g)
If I wanted to generate exactly 2 L of C
2
H
2
gas at STP, how much CaC
2
do I need for my reaction
At STP 1 mole of gas occupies 22.42 l so:
1 mole/22.42L = X mole/2L
X =(2L × 1mol)/22.42L = .08921 mol C
2
H
2
From the above reaction 1 mole of C
2
H
2
is equivalent to 1 mole of CaC
2 so I need .089 mole of CaC
2
4
8. (10 points) What are the basic postulates of the Kinetic Molecular Theory of gases? How do these postulates explain Boyles Law ( PV=k)
(I gave 2 point of each postulate up to 8 points and 2 points for the explanation of Boyles law)
1. A gas is composed of tiny particles (atoms or molecules)
2. Particles size is so small compared to distance between particles it can be ignored
3. Particles bounce off each other elastically (no energy lost or gained)
4. Average kinetic energy of a particle is directly proportional to the temperature in K
5. Pressure is exerted on a surface fro the particles colliding with that surface.
As you increase volume the particles bounce around in a larger volume to less particles hit the surface at a given time so the pressure decreases

9. (10 points) I have a compressed gas cylinder that contains 5 moles of CO
2
,
1 mole of N
2
, and 6 moles of He gas. The cylinder’s volume is 2.5 l, the temperature is 30 o C, and the pressure in the container is 119.4 atm. What is the mole fraction of each gas in the cylinder, and what is the partial pressure of each gas in the cylinder.
5
There are at least 2 ways to do this. I think this way is the easiest
χ =
χ
CO
2 n
1 n
1
+ n
2
+ n
3
=
/
=
.
417
χ
N
2
χ
He
=
=
/
/
=
.
083
=
.
500 also
χ = p
1
+ p
2
+ p
3 for CO
2
χ p
1
=
.
417
; for N
2
χ =
.
083
=
= x for He
χ
.
x / p
1
+ p
2
+ p
3
= x / 119 4
/ 119 4
119 4 x x
= x
=
= atm atm atm atm
10. (10 points) Real gases are gases that do not obey the ideal gas law. Name at least two physical properties of a gas that would make it vary from the ideal gas law, and will this property cause the gas to have a higher or lower P than expected?
A gas has a volume, this volume correction lowers the total volume that a gas molecule can bounce around in and thus increases the pressure of a gas.
Gases molecules or atom may also be attracted to each other this attractive force make the molecules ‘stick’ together slightly during a collision and decreases the pressure of a gas.