Quiz 5 (08-09)
Chem 171
20 points
Name____Salmon________
R = 0.08210 L x atm/mol x K
1)
A solution of NaOH was prepared. After 10.25 mL of this was added to a solution containing 0.500 g of acetic
acid containing PTH, the solution turned pink. What is the concentration of the NaOH solution?
mol
1 mol NaOH
x
= 0.00833 mole HAc
60.0 g
1 mol HAc
0.00833 mol
in 0.01025 L 
= 0.813 M
0.01025 L
0.500 g x
2)
The following reaction is a(n) ( oxidation
reduction ) half reaction.
+
25e + 8H (aq) + MnO4 (aq) + S (aq) → MnS (s) + 4 H2O (l)
3)
Balance the following half reaction in acidic media:
+
As2O3 → H3AsO4
-
5 H2O + As2O3 → 2 H3AsO4 + 4 H + 4e
4)
Balance the following half reaction in basic media:
-
-
-
-
CN → CNO
-
-
2 OH + CN → CNO + H2O + 2e
5)
What is oxidized and what is reduced in the following reaction?
+
-
3 Cu (s) + 8 H (aq) + 2 NO3 (aq) → 3 Cu
Oxidized __Cu_____
6)
2+
(aq) + 2 NO (g) + 4 H2O (l)
Reduced ____N______
A Torricelli barometer is used to measure atmospheric pressure. The mercury rises to 738 mm. What is the
pressure of the atmosphere in standard atmospheres (atm)?
738 mmHg x
1 atm
= 0.971 atm
760 mmHg
7)
As described by Boyle’s Law, pressure and volume are ( directly
8)
Consider three 1 L flasks at 1 atm and 273 K.. Flask A contains NH3 gas, flask B contains Ne gas and flask C
contains N2 gas. Which flask contains the largest number of molecules?
a) flask A
b) flask B
c) flask C
inversely
) proportional.
d) all contain the same number
9)
If pressure and the number of moles remain constant, an increase in the temperature will result in an
( increase no change decrease ) in the volume.
10)
For an ideal gas, a plot of V vs T shows vol = 0 at
0
a) 37 C
11)
0
b) 0 C
c) 0 K
d) 273 K
For an ideal gas, which of the following is the correct PV vs P plot?
B, where PV vs P is a line of slope 0, constant PV at any P
12)
What would the temperature be of 13.7 g of chlorine gas (Cl2) placed in a 5.50 L container at 0.884 atm?
mole
= 0.193 mol Cl2
70.90 g
PV
Rearrange PV=nRT to T =
and substitute in values
nR
(0.884 atm)(5.5 L)
T=
= 306 K
(0.193 mol)(0.0821 L x atm/mol x K)
13.7 g x
13)
A sample of gas is contained in a 60.0 mL container at a pressure of 555 torr and a temperature of 25.0 C. The
entire sample of gas is transferred to a new container with a volume of 45.0 mL and heated to 45.0C. What is
the pressure of the gas after these changes?
Can convert V to liters → 0.060 L and 0.045 L
Can convert P to torr → 555 torr x 1 atm/760 torr = 0.73 atm
MUST convert T to K → 25 + 273 = 298 and 45 + 273 = 318 K
In this problem, n is constant and will drop out.
P1V1
PV
= 2 2
n1T1
n 2 T2
Drop n and rearrange:
PVT
(0.73 atm)(0.060L)(318 K)
P2  1 1 2 =
= 1.04 atm
T1V2
(298 K)(0.045 L)
14)
When KClO3 is heated, it decomposes to KCl and O2 according to the reaction below. If 17.6 g KClO3,
decomposes at 298 K and 1.00 atm, what volume of O2 will it form ?
2 KClO3(s)  2 KCl(s) + 3O2(g)
Path is mass KClO3 to mol KClO3 to mol O2 to volume O2