1.(10 points) Define and identify the difference between: Strong Electrolytes, Weak
electrolytes, and nonelectrolytes
Strong Electrolytes, like soluble salts, and strong acids and bases ionize completely when
dissolved in water, and the ions conduct electricity. Weak Electrolytes, like weka acids
and bases on partially ionize in solution. Because of the limited number of ions, solutions
of weak electrolytes do not conduct electricity as well as strong electrolytes.
Nonelectrolytes do not ionize at all in water, but still dissolve quite well because they are
highly polar. Solutions of nonelectrolytes do not conduct electricity.
2. (10 points) I need .35 moles of NaOH for a reaction. How many mLs of a .254 M NaOH
solution should I use to obtain this amount of NaOH ?
Molarity = moles/volume(L)
.254M = .35/X
X = .35 moles/.254 (mole/l) = 1.38L
3. (10 points) Identify the following salts as soluble (s), insoluble (i) or slightly soluble (ss)
KNO3
______S______
(NH4)2SO4
______S______
CaS
______SS____
Na2SO4
__________
PbBr
______I______
LiOH
_____S______
S__________
2
4. (10 points) Write the I am going to mix a potassium hydroxide solution with a
magnesium nitrate solution.
Will a chemical reaction occur?
Yes, a Precipitation reaction
If so write:
The balanced molecular equation
2KOH(aq) + Mg(NO3)2(aq) 6 Mg(OH)2(s) + 2KNO3(aq)
The complete ionic equation
2K+(aq) + 2OH-(aq) + Mg2+(aq) +2NO3-(aq) 6Mg(OH)2(s) + 2K+(aq) + 2NO3-(aq)
The net ionic equation
2OH-(aq) + Mg2+(aq)6Mg(OH)2(s)
5. (10 points) I have 4 liters of 2.5M H2SO4 that I wish to neutralize before I pour it down the
drain. Suggest a base that I can neutralize the acid with, and calculate how much of the
base I will need for the neutralization reaction.
Molarity = moles/volume(l); Molarity X volume = moles
moles H2SO4 = 4(2.5) = 10 moles H2SO4
Note that 1 mole H2SO2 has 2 moles of H+ so this is 10 X 2 or 20 moles of H+
Therefore we need 20 moles of a base, say NaOH, or KOH.
(From there you could calculate grams of base or volume and molarity of a base)
6. (10 points) Balance the following redox reaction under acidic conditions
NO2-(aq) + Al(s) 6 NH3(aq) + AlO2NO2- 6 NH3
Al 6 AlO2-
NO2- 6 NH3 + 2 H2O
2H2O + Al 6 AlO2-
7H+ + NO2- 6 NH3 + 2 H2O
2H2O + Al 6 AlO2- + 4H+
6e- + 7H+ + NO2- 6 NH3 + 2 H2O
2H2O + Al 6 AlO2- + 4H+ +3eX1
X2
6e- + 7H+ + NO2- + 4H2O + 2Al 6 NH3 + 2 H2O + 2AlO2- + 8H+ +6eNO2- + 2H2O + 2Al 6 NH3
+ 2AlO2- + 1H+
3
(Problem 6 - continued)
What is the oxidation ½ reaction?
`
2H2O + Al 6 AlO2- + 4H+ +3e-
What is the oxidizing agent in the reaction?
NO2-
Balance the equation under basic conditions
NO2- + 2H2O + 2Al 6 NH3 + 2AlO2- + 1H+
+OH+OHNO2- + 2H2O + 2Al + OH- 6 NH3 + 2AlO2- + H2O
NO2- + H2O + 2Al + OH- 6 NH3 + 2AlO2-
7. (10 points) I have a pressure of 5,000,000 Pa. What is that in atm?
5,000,000 X 1 atm/101.324kPa X 1kPa/1,000Pa = 49 atm
8. (10 points) If I fill a balloon with 1.5 L of He gas at 1 atm, and 25oC, and then take the
balloon to the top of Crow peak when the 650 mm Hg and the temperature is 20o C, what
is the volume of the balloon.
P,V, and T are variables, n&R constants. Rearrange PV=nRT to PV/T =nR
P1V1/T1 = P 2V2/T2
P1 = 1atm, V 1 = 1.5L, T1 = 25+273 = 298, P 2 = 650/760 = .855 atm, T2 = 20+273 = 293
1atm(1.5L)/298K = .855 atm(X)/294K; 1atm(1.5L)293K/.855atm(298K) = X
X = 1.72 L
4
9. (10 points) Airbags in a car are inflated by the explosive reaction 2NaN3(s) 6 2Na(s) +
3N2(g). What is the volume of gas at STP that will be made from 3.5 g of NaN3?
Molar mass of NaN3 = 23 + 16(3) = 65
Moles of product
3.5 g X 1 mole/65g X 3 mole N2/2 mole NaN3 = .081 mole
Volume of product
22.42 L/mole X .081 mole = 1.81 L
10. (10 points) I have a mixture of two gases, one is CH4(g) with a partial pressure of .185
atm, and the other is O2(g) with a partial pressure of .350 atm:
A What is the mole fraction of each gas in the mixture
PO2 = P O2 /P total = .350/(.350+.185) = .654
PCH4 = P CH4 /P total = .185/(.350+.185) = .346
B. If the mixture occupies 9.5L at 55oC calculate the total number of moles of gas in the
mixture
PV=nRT
n = PV/RT
.535 atm × 9.5 L
.08206l ⋅ atm / L ⋅ mol × 328 K
= .188 mol
n=