1. Triad 2 Test Study Guide Key
West Valley High School
AP Chemistry
Mr. Mata

2. Chapter 4
1. What is the molarity of 88 grams of Ba(NO3)2 in 850 mL of water?
Answer:
88 g Ba(NO3)2 x 1 mole = 0.34 mol g
850 mL x 1L = L
1000 mL
M = 0.34 mol = 0.40 M
.85 L

3. Chapter 4
2. Calculate the molarity of 1250 mL of H2SO4 if mixed with 700 mL of a 6.0 M KOH solution. Answer: Ma = MbVb = (6.0M)(700 mL) = 3.4 M Va 1250 mL

4. Chapter 4
3. Balance the following redox reaction, in acid solution, using the half-reaction method.
Answer:
MnO2  Mn2+
4H+ + MnO2  Mn H2O e- (oxid)
O e-  2O2- (red)
===================================
4H+ + MnO2 + O2  Mn O H2O

5. Chapter 4
4. Write the net ionic equation for the following: silver nitrate and potassium carbonate
Answer:
AgNO3 (aq) + K2CO3 (aq)  Ag2CO3 (s) + KNO3 (aq)
Ag+ + NO3- + K+ + CO32-  Ag2CO3(s) + K+ + NO3-
Net ionic equation: Ag+ + CO32-  Ag2CO3(s)

6. Chapter 5
5. A sample of helium gas occupies 8.0 L at 66 C and 3.5 atm. What volume will it occupy at 177 C and 3.2 atm?
Answer:
V2 = P1V1T2 = (3.5 atm)(8.0L)(450K) = 12 L
P2T (3.2atm)(339 K)

7. Chapter 5
6. A 10.7 gram piece of solid CO2 is allowed to sublime in a balloon. The final volume of the balloon is 3.4 L at 298 K. What is the pressure of the gas? Answer: 10.7 g CO2 x 1 mol = mol g P = nRT = (0.243 mol)(0.0821)(298K) = 1.7 atm V 3.4 L

8. Chapter 5
7. A gas sample is held at constant pressure. The gas occupies 5.25 L of volume when the temperature is 68.5 C. Determine the temperature at which the volume of the gas is 9.0 L.
Answer:
T2 = V2T1 = (9L)(341.5 K) = K
V L

9. Chapter 5
8. Mercury vapor contains Hg atoms. What is the volume of 248 g of mercury vapor at 584 K and 2.85 atm? Answer: 248 g Hg x 1 mol = 1.24 mol g V= nRT = (1.24 mol)(0.0821)(584K) = 20.9 L P 2.85 atm

10. Chapter 6
9. A 25.0 g piece of aluminum is heated to 82.4 C and dropped into a calorimeter containing water initially at 22.3 C. The final temperature of the water is 24.9 C. Calculate the mass of water in the calorimeter.
Answer:
25.0 g Al x 1 mol = 0.93 mol
26.98 g ∆TAl = 24.9C – 82.4C = C
q Al = m x c x ∆T = 0.93 mol x 24.03J/Cmol x C = J
q Al = J = J = q H2O
∆TH2O = 24.9C – 22.3C = 2.6 C
q H2O = m x c x ∆T  m = q = 1285 J = g
c x ∆T (4.18 J/gC)(2.6C)

11. Chapter 6
10. A 40.2 g sample of metal is heated to 99.3 C and then placed in a calorimeter containing g of water. The final temperature of the water is 24.5 C. Which metal was used?
Answer:
∆TH2O = 24.5 C – 21.8 C = 2.7 C
qH2O = m x c x ∆T  (120.0 g)(4.18 J/gC)(2.7C) = 1354 J
qH2O = 1354 J = J = qmetal
∆Tmetal = 24.5 C – 99.3 C = C
qmetal = m x c x ∆T  cmetal = q = J = 0.45 J/gC
(40.2 g)( C)
Answer: b (iron = 0.45 J/gC)

12. Chapter 6
11. At 25 C, the following heats of reaction are known: At the same temperature, calculate ∆ H for the reaction.
Answer: ∆H
ClF + 1/2O2  1/2Cl2O + 1/2F2O kJ/mol
F2 + 1/2O2  F2O kJ/mol
1/2Cl2O + 3/2F2O  ClF3 + O kJ/mol
================================================
ClF + F2 -> ClF3 ∆H = kJ/mol

13. Chapter 6
12. Consider the following data: Use Hess’ law to find the change in enthalpy at 25 C for the following equation:
Answer: ∆H
2C (graphite) + 2O2 (g)  2CO2(g) kJ
CaC2(s)  Ca(s) + 2C (graphite) kJ
2CaO(s) + 2H2O(l)  2 Ca(OH)2(aq) kJ
2CO2 + H2O(l)  C2H2(g) + 5/2O2(g) 1300 kJ
Ca(s) + 1/2O2(g)  CaO(s) kJ
Ca(OH)2(aq)  CaO(s) + H2O(l) kJ
=================================================
CaC2(s) + 2H2O(l)  Ca(OH)2(aq) + C2H2(g) ∆H = kJ