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

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 261.35 g 850 mL x 1L = .85 L 1000 mL M = 0.34 mol = 0.40 M .85 L

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

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

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)

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 P2T1 (3.2atm)(339 K)

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 = 0.243 mol 44.01 g P = nRT = (0.243 mol)(0.0821)(298K) = 1.7 atm V 3.4 L

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) = 585.4 K V1 5.25 L

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 200.59 g V= nRT = (1.24 mol)(0.0821)(584K) = 20.9 L P 2.85 atm

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 = - 57.5 C q Al = m x c x ∆T = 0.93 mol x 24.03J/Cmol x - 57.5 C = -1285 J q Al = -1285 J = + 1285 J = q H2O ∆TH2O = 24.9C – 22.3C = 2.6 C q H2O = m x c x ∆T  m = q = 1285 J = 118.2 g c x ∆T (4.18 J/gC)(2.6C)

Chapter 6 10. A 40.2 g sample of metal is heated to 99.3 C and then placed in a calorimeter containing 120.0 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 = - 1354 J = qmetal ∆Tmetal = 24.5 C – 99.3 C = - 74.8 C qmetal = m x c x ∆T  cmetal = q = - 1354 J = 0.45 J/gC (40.2 g)(- 74.8 C) Answer: b (iron = 0.45 J/gC)

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 83.7 kJ/mol F2 + 1/2O2  F2O -21.7 kJ/mol 1/2Cl2O + 3/2F2O  ClF3 + O2 -170.7 kJ/mol ================================================ ClF + F2 -> ClF3 ∆H = - 108.7 kJ/mol

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) - 787.02 kJ CaC2(s)  Ca(s) + 2C (graphite) 62.8 kJ 2CaO(s) + 2H2O(l)  2 Ca(OH)2(aq) -1306.2 kJ 2CO2 + H2O(l)  C2H2(g) + 5/2O2(g) 1300 kJ Ca(s) + 1/2O2(g)  CaO(s) -635.5 kJ Ca(OH)2(aq)  CaO(s) + H2O(l) 653.1 kJ ================================================= CaC2(s) + 2H2O(l)  Ca(OH)2(aq) + C2H2(g) ∆H = -712.82 kJ