Make Up Exam 2-solutions - Version 582 Make Up Exam 2 Laude...

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Version 582 – Make Up Exam 2 – Laude – (52390) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. K a = [ H + ] · [ A ] [ HA ] pK a = - log K a pH = - log [H + ] pOH = - log [OH ] [ OH ] = ( K b · C b ) 1 / 2 [ OH ] = K b · p C b C a P 10 14 = K w [H + ] · [OH ] = K w K a · K b = K w pK a + pK b = pK w 14 = pK w ( K a · C a ) 1 / 2 = [ H + ] K a · p C a C b P = [ H + ] [ H + ] = ( K ax · K ay ) 1 / 2 pH = 0 . 5( pK ax + pK ay ) 0 = [ H + ] 2 - C a [ H + ] - K w K sp = [ C ] c · [ A ] a E o cell = E o cathode - E o anode E cell = E o cell - p 0 . 05916 n P · log Q Q = [ C ] c · [ D ] d [ A ] a · [ B ] b Δ G o = - n · F · E o cell = - R · T · ln K E o cell = p R · T · ln K n · F P p I · t n · F P = moles oF product HalF reaction E o Au 3+ + 3 e -→ Au +1 . 50 Cl 2 + 2 e -→ 2 Cl +1 . 36 Ag + + e -→ Ag +0 . 80 O 2 + 2 e + 2 H+ -→ H 2 O 2 +0 . 68 I 2 + 2 e -→ 2 I +0 . 53 Pb 2+ + 2 e -→ Pb - 0 . 13 Ni 2+ + 2 e -→ Ni - 0 . 25 Mn 2+ + 2 e -→ Mn - 1 . 12 Na + + e -→ Na - 2 . 71 Sr 2+ + 2 e -→ Sr - 2 . 89 F = 96 , 485 . 3 C per mole oF e 1 Ampere = 1 C · s 1 R = 8 . 314 J · mol 1 · K 1 N = 6 . 022 × 10 23 001 10.0 points IF the two halF reactions below were used to make a battery, what species would be produced at the anode? HalF reaction E Cu 2+ ( aq ) + 2 e -→ Cu( s ) +0 . 34 ±e 3+ ( aq ) + e -→ 2 ±e 2+ ( aq ) +0 . 77 1. Cu 2+ (aq) correct 2. ±e 2+ (aq) 3. ±e 3+ (aq) 4. Cu (s) Explanation: A battery must have a negative positive cell potential and thereFore the anodic reaction must produce Cu 2+ (aq). 002 10.0 points What would be the pH oF a 6 × 10 9 M solution oF NaOH? 1. 7.007 2. 7.013 correct 3. 7.016 4. 7.021 Explanation: ±or a dilute strong base, [OH ] 2 - Cb[OH ] - K w = 0 [OH ] = x = ± - b ± b 2 - 4ac 2a ² pOH = 6.987 pH = 7.013 003 10.0 points A 40 mL sample oF 0 . 25 M NaCHOO is
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Version 582 – Make Up Exam 2 – Laude – (52390) 2 titrated with 0 . 2 M HCl. What is the pH of the solution after 140 mL of HCl has been added? 1. 3.0 2. -1.0 3. 1.0 correct 4. 2.0 5. 2.1 6. 0 Explanation: After the CHOOH has been neutralized, 0 . 018 mol of HCl remain in a solution with a total volume of 180 mL . [H + ] = 0 . 018 mol 180 mL = 0 . 1 M pH = 1 004 10.0 points Balance the following half reaction in acid: Mn 2+ ( aq ) -→ MnO 4 ( aq ) How many electrons are involved in this pro- cess? 1. 5 correct 2. 9 3. 7 4. 4 5. 8 6. 3 Explanation: 4H 2 O( l ) + Mn 2+ ( aq ) -→ MnO 4 ( aq ) + 8H + ( aq ) + 5e 005 10.0 points What would be the pH of a 0 . 45 M solution of H 2 SO 4 ? Assume HSO 4 has a K a of 2 × 10 2 .
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Make Up Exam 2-solutions - Version 582 Make Up Exam 2 Laude...

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