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Exam 2-solutions - Version 159 Exam 2 mccord(51625 This...

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Version 159 – Exam 2 – mccord – (51625) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points CH 3 COO is a weaker base than NH 3 . Which acid would be stronger: CH 3 COOH or NH + 4 ? 1. NH + 4 2. Neither CH 3 COOH nor NH + 4 would be acids. 3. CH 3 COOH correct 4. Both acids would be equal in strength. 5. Cannot tell from the information given. Explanation: 002 10.0points The plain K given in this problem is the same as K p . At 25 C, K = 6 . 9 × 10 5 for the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g) . Calculate K c at 25 C for this reaction. 1. 1 . 1 × 10 3 2. 1 . 7 × 10 7 3. 4 . 1 × 10 8 correct 4. 2 . 8 × 10 4 5. 6 . 8 × 10 5 Explanation: K is the same as K p in this problem. K p = K c ( RT ) Δ n You must use R = 0 . 08206 L atm/mol K for this problem. The value for Δ n is -2. So solving for K c gives the following equation: K c = K p ( RT ) Δ n K c = 6 . 9 × 10 5 (24 . 47) 2 K c = 4 . 13 × 10 8 003 10.0points Calculate the [OH ] of a solution of 0 . 25 M HClO 4 . 1. 4 . 7 × 10 13 2. 4 . 0 × 10 14 correct 3. 1 . 3 × 10 10 4. 2 . 5 × 10 13 5. 1 . 1 × 10 12 Explanation: [OH ] = 0 . 25 M HClO 4 -→ H + + ClO 4 [H + ] = [HClO 4 ] = 0 . 25 M [OH ] = K w [H + ] = 1 × 10 14 0 . 25 M = 4 × 10 14 M 004 10.0points A 20 mL sample of 0.20 M nitric acid solution is required to neutralize 40 mL of barium hydroxide solution. What is the molarity of the barium hydroxide solution? 1. 0.025 M 2. 0.0025 M 3. 0.050 M correct 4. 0.100 M 5. 0.200 M Explanation: V HNO 3 = 20 mL [HNO 3 ] = 0.20 M V Ba(OH) 2 = 40 mL The balanced equation for this neutraliza- tion reaction is
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Version 159 – Exam 2 – mccord – (51625) 2 2 HNO 3 + Ba(OH) 2 Ba(NO 3 ) 2 + 2 H 2 O We determine the moles of HNO 3 used: ? mol HNO 3 = 0 . 020 L soln × 0 . 20 mol HNO 3 1 L soln = 0 . 0040 mol HNO 3 Using the mole ratio from the chemical equa- tion we calculate the moles Ba(OH) 2 needed to react with 0.0040 mol of HNO 3 : ? mol Ba(OH) 2 = 0 . 0040 mol HNO 3 × 1 mol Ba(OH) 2 2 mol HNO 3 = 0 . 0020 mol Ba(OH) 2 There are 0.0020 moles Ba(OH) 2 in the 40 mL sample. Molarity is moles solute per liter of solution: ? M Ba(OH) 2 = 0 . 0020 moles Ba(OH) 2 0 . 040 L solution = 0 . 050 M Ba(OH) 2 005 10.0points Suppose we put 1.0 mol of HI(g), 1.0 mol of H 2 (g), and 1 mol of I 2 (g) in a 2.0 liter reaction vessel and the following equilibrium is established: 2 HI(g) H 2 (g) + I 2 (g) If K c = 10 for this reaction at the tempera- ture of the equilibrium mixture, compute the equilibrium concentration of HI. 1. 0.429 M 2. 0.240 M 3. 0.295 M 4. 0.260 M 5. 0.205 M correct 6. 0.102 M 7. 0.071 M 8. 0.145 M Explanation: [HI] ini = 1 mol 2 L = 0 . 5 M [H 2 ] ini = 1 mol 2 L = 0 . 5 M [I 2 ] ini = 1 mol 2 L = 0 . 5 M Q = [H 2 ] [I 2 ] [HI] 2 = (0 . 5) (0 . 5) (0 . 5) 2 = 1 < K c = 10 Therefore equilibrium moves to the right. 2 HI H 2 (g) + I 2 (g) ini, M 0 . 5 0 . 5 0 . 5 Δ, M - 2 x x x eq, M 0 . 5 - 2 x 0 . 5 + x 0 . 5 + x (0 . 50 + x ) (0 . 50 + x ) (0 . 50 - 2 x ) 2 = 10 0 . 5 + x 0 . 50 - 2 x = 10 0 . 5 + x = 0 . 50 10 - 2 10 x x = 0 . 148 [HI] = 0 . 5 - 2 x = 0 . 205 M 006 10.0points A solution with a higher pH is more acidic.
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