Make up 1-solutions - Version 083 Make up 1 laude (51635)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Version 083 – Make up 1 – laude – (51635) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. Constants R = 8 . 314 JK 1 mol 1 Defnitions Q [C] c [D] d [A] a [B] b G H - TS K w [H + ][OH ] pH ≡ - log [H + ] pOH ≡ - log [OH ] Equations Δ G = Δ H - T Δ S ln p P 2 P 1 P = Δ H vap R p 1 T 1 - 1 T 2 P Π = MRT Δ T b = iK b m Δ T f = iK f m P i = P i χ i P = s i P i q = mc Δ T q = m Δ H Δ G = - RT ln K K = e Δ G RT ln K 2 K 1 = Δ H R p 1 T 1 - 1 T 2 P Q = K [at equil.] [H + ] = ( K a C a ) 1 / 2 [OH ] = ( K b C b ) 1 / 2 [H + ] = K a ( C a /C b ) pH = p K a + log( C b /C a ) [OH ] = K b ( C b /C a ) pOH = p K b + log( C a /C b ) 001 6.0 points Write the equilibrium expression For the Fol- lowing reaction: 2 NO ( aq ) + 4 H 2 O( ) ←→ 3 H 2 ( aq ) + 2 H + ( aq ) + 2 NO 3 ( aq ) 1. K = [[ H + ] 2 · [ NO 3 ] 2 [ NO ] 2 2. K = [ NO ] 2 [ H + ] 2 · [ NO 3 ] 2 3. K = [ NO ] 2 [ H 2 ] 3 · [ H + ] 2 · [ NO 3 ] 2 4. K = [ H 2 ] 3 · [ H + ] 2 · [ NO 3 ] 2 [ NO ] 2 · [ H 2 O ] 4 5. K = [ H 2 ] 3 · [ H + ] 2 · [ NO 3 ] 2 [ NO ] 2 correct Explanation: Set up K , products in the numerator, reac- tants in the denominator, all raised to respec- tive stoichiometric coe±cients. 002 6.0 points 30 . 2 g oF glycerine (C 3 H 8 O 3 , with MW 92 . 1 g / mol) are dissolved in 150 g oF wa- ter. What is the boiling point oF the solution? ( K b oF water = 0 . 515 C / m ) 1. 0.104 C 2. 101.13 C correct 3. 101.52 C 4. 100.10 C 5. 1.13 C Explanation: m glyc = 30 . 2 g MW glyc = 92 . 1 g / mol m water =150 g K b = 0 . 515 C / m Δ T b = K b m = K b mol glycerol kg water = (0 . 515 C /m ) 30 . 2 92 . 1 mol C 3 H 8 O 3 0 . 150 kg water = 1 . 13 C
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Version 083 – Make up 1 – laude – (51635) 2 T b = T 0 b + Δ T b = 101 . 13 C 003 6.0 points What would be the pH of a 0.563 M solution of ammonia (NH 3 ) at room temperature? The K b of ammonia is 1 . 78 × 10 5 . 1. 11.5 correct 2. 9 3. 2.5 4. 4.5 5. 7 Explanation: [OH ] = (K b · C b ) 1 / 2 = (1 . 78 × 10 5 · 0 . 563) 1 / 2 = (10 5 ) 1 / 2 = 10 2 . 5 pOH = - log[OH ] = - log(10 2 . 5 ) = 2 . 5 pH = pK w - pOH = 14 - 2 . 5 = 11 . 5 004 6.0 points What would be molar solubility of a very ex- pensive salt, awesomium, which is composed of 3 cations and 2 anion (X 3 Y 2 ) and whose K sp is 1 . 08 × 10 18 ? 1. 1 M 2. 1 × 10 9 M 3. 0 . 0001 M correct 4. 6 . 1 × 10 7 M Explanation: For a salt of the form X 3 Y 2 molar solubility = x = p K sp 27 · 4 P 0 . 2 x = p 1 . 08 × 10 18 27 · 4 P 0 . 2 = 0 . 0001 M 005 6.0 points Which of the following phase changes is not correctly paired with the sign of its change in
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 9

Make up 1-solutions - Version 083 Make up 1 laude (51635)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online