Quiz 5 (08-09)
Chem 171
20 points
Name______Blue__________
R = 0.08210 L x atm/mol x K
1)
A solution of acetic acid was prepared. It was titrated with 0.340 M NaOH. After addition of 47 mL NaOH, the
solution turned pink. What mass of the acid was present?
The path is vol NaOH to mol NaOH to mol HAc to mass HAc
𝟎.𝟑𝟒𝟎 𝒎𝒐𝒍
𝟎. 𝟎𝟒𝟕 𝑳 𝒙
2)
𝑳
𝟏 𝒎𝒐𝒍 𝑯𝑨𝒄
𝟏 𝒎𝒐𝒍 𝑵𝒂𝑶𝑯
The following reaction is a(n) (
2SO3 (aq)
3)
𝒙
+ H2O(l) →
𝒙
𝟔𝟎.𝟎 𝒈
𝒎𝒐𝒍 𝑯𝑨𝒄
oxidation
2SO4 (aq)
= 𝟗𝟓𝟗 𝒈
reduction
+
+ 2H (aq) + 2e
What is oxidized and what is reduced in the following reaction?
+
2-
2-
8 H (aq) + Cr2O7 (aq) + 3 SO3 (aq) →2 Cr
Oxidized ____S______
4)
) half reaction.
-
3+
Reduced ____Cr______
Balance the following half reaction in acidic media:
-
2-
(aq) + 2 SO4 (aq) + 4 H2O (l)
-
Cl2 → 2Cl
-
2 e + Cl2 → 2Cl
5)
Balance the following half reaction in basic media:
-
-
3 e + 4 H2O + CrO4 → Cr(OH)3 + 5 OH
6)
→ Cr(OH)3
-
A Torricelli barometer is used to measure atmospheric pressure. A modern barometer indicates the pressure is
0.966 atm. How high would the mercury rise in a Torricelli barometer.
0.966 𝑎𝑡𝑚 𝑥
7)
2-
CrO4
760 𝑚𝑚𝐻𝑔
𝑎𝑡𝑚
= 734 𝑚𝑚𝐻𝑔
Consider three 1 L flasks at 1.1 atm and 300 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
d) all contain the same number
8)
As described by Charles’s Law, temperature and volume are ( directly
9)
If temperature and the number of moles remain constant, an increase in the pressure 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
inversely
) proportional.
12)
o
What pressure is associated with 13.7 g of chlorine gas (Cl2) at 45.0 C in a 5.50 L container?
Convert mass to moles and use PV=nRT, rearranged to be P = nRT/V
𝟏𝟑. 𝟕 𝒈 𝒙
𝒎𝒐𝒍
𝟕𝟎.𝟗𝟎 𝒈
= 𝟎. 𝟏𝟗𝟑 𝒎𝒐𝒍 𝑪𝒍𝟐
Rearrange PV=nRT to P = nRT/V and substitute in values
𝑷=
13)
𝒏𝑹𝑻
𝑽
=
(𝟎.𝟏𝟗𝟑 𝒎𝒐𝒍)(𝟎.𝟎𝟖𝟐𝟏
𝑳𝒙𝒂𝒕𝒎
)(𝟑𝟏𝟖𝑲)
𝒎𝒐𝒍𝒙𝑲
𝟓.𝟓𝟎 𝑳
= 𝟎. 𝟗𝟏𝟔 𝒂𝒕𝒎
A sample of gas is contained in a 50.0 mL container at a pressure of 645 torr and a temperature of 25.0 C. The
entire sample of gas is transferred to a new container with a volume of 65.0 mL and heated to 35.0C. What is
the pressure of the gas after these changes?
Can convert mL to L: 50 mL is 0.05L and 65 mL is 0.065 L
Can convert torr to atm: 645 torr x 1 atm/760 torr = 0.85 atm
Must convert C to K: 25 + 273 = 298 K and 35 + 273 = 308 K
Use the relationship: P1V1/n1T1 = P2V2/n2T2
Rearrange, drop n because it doesn’t change, substitute in values:
P2 = P1V1T2/T1V2 =
14)
𝟎.𝟖𝟓 𝒂𝒕𝒎 𝟎.𝟎𝟓𝟎𝑳 (𝟑𝟎𝟖 𝑲)
𝟐𝟗𝟖 𝑲 (𝟎.𝟎𝟔𝟓 𝑳)
= 𝟎. 𝟔𝟕𝟔 𝒂𝒕𝒎
When KClO3 is heated, it decomposes to KCl and O2 according to the reaction below. How much KClO3, in
grams must decompose to form a volume of 200.0 mL O2 at 298 K and 1.00 atm?
2 KClO3(s)  2 KCl(s) + 3O2(g)
path is volume O2 to moles O2 to moles KClO3 to mass KClO3
Volume O2 is obtained using PV=nRT
𝒏=
𝑷𝑽
𝑹𝑻
=
𝟏.𝟎𝟎 𝒂𝒕𝒎 (𝟎.𝟐𝟎𝟎 𝑳)
𝟎.𝟎𝟖𝟐𝟏
𝑳𝒙𝒂𝒕𝒎
𝒎𝒐𝒍𝒙𝑲
(𝟐𝟗𝟖𝑲)
= 0.00817 mol O2
0.00817 mol O2 x 2 mol KClO3/3 mol O2 x 122.5 g KClO3/mol = 0.668 g KClO3