KellyCHEM2BWinter2008MT2

KellyCHEM2BWinter2008MT2 - Prof Peter 3 Kelly Spring zoos...

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Unformatted text preview: Prof. Peter 3. Kelly ' Spring zoos CHEMISTRY 2B (Sec. A) Midterm II ' . Instructions: ggdgsim BOOK EXAM! No books. notes, or additionai scrap paper are permitted. All information required is contained on the exam. Place all work in the space provided. If you require additional space, use the back 'of the exam. A scientific calculator may be used (if it is a programmable calculator, its memory must be cleared before the exam). (1) Read each question carefully. (2) ForParts l andflfihereisMflgivenandi ' marked on this cover me will be Md. (3) The last page contains a periodic table and some useful information. You may remove it for easy access. (4) If you finish early, RECHECK YOUR ANSWERS! (5) Circle your choice on the right for the multiple choice questions 1 — 16. Page 7 of 12 ' Part 1]]: Long Answer Please show all work —- Partial credit 17. (16 pts) At 25 °C and under an 02 (g) pressure of1.000 atm, the solubility ofOz (g) in water is 28.311111/(1000 L water‘atm). Under an 142(3) pressure of 1000 aim. the solubility of N; (g) at 25 -°C is 14. 34 mL {(1.000 L water‘atm). The composition of the atmosphere IS 78. 08% N2 and 20. 95% 0;, by volume. a) How much oxygen (mIJL water)’ 13 dissolved 111 water in equilibrium: with 1.000 atm air at 25 °C? ' b) How much nitrogen (mlJL water) 15 dissolved in water in equilibrium with 1.000 aim air at 25 °C? c) What is the composition of air dissolved in water at 25 C, expressed as volume percents of oxygen and nitrogen? .13 9 4,2“? __ (1903:1011 J sot =21; 31LW:M_ ' Va -Qoir-zsm film “L 11919141: (29 3‘a\(o.1oa§kfi3: 9.93:. 1:) Sm Q“ ””1 fiMWCGJsosalm) 07. \ 7501 - ii elk-U = 3'14. 202 Siam“. ”If Initial: PageBole -. Midterm ll 18. (15 points) An acetic acid ! sodium acetate buffer solution is made with a pH of 4.52. The concentration of the acetic acid is 0.100 M. Note Ka = 1.80 x 10-5 for acetic acid. The volume of the solution is 100.0 mL . a) What is the base I acid ratio of the solution? les of base, NaOH, can be consumed before the pH increases to pH = b) Howmanymo 5.02? (assume the additional base does not change system volume) -a: - [M a), lq‘C'L‘: ’[uQ 0”?ng )4- \v%(\m) A -" Ifil’kul Vb ....—-——-"'" ' 2. 1 I451 66 m “0.4 . Gm At OVl ( ”their? ' 0. 3 "‘ L‘sQOV'D—g') 4 ‘0‘, ‘h-—-'—"""' h $.51 ’p ‘3 k '1‘ (‘7: 7- ' _ shew 7' [W l H x. Guile—we \thC‘gg --"_—" k. leg-f“ ‘C "' o.tL 0.1M- 0.95-g ’ (3‘ng 0.14; “31* x \O’OM'L = 11.0.7 MW; 0V thyko—‘i‘fl‘x 0H- \4 \ht. comm-4) Initial: I II '- I I - 'Pasafluflz I 419‘ (1513“) T“ Hm solutions are mad: 5%} mL of M150 M H2304 (lg) m_ 50.0 IIIL of 0.0335 M Nam-l. ' fl) Cakulmflummfimof DH“ '31} Calctflatflnuomenufimofflaw c) GllcfilatefllepI-I . d) Calmlaletfleomnnnfsofi- NumKa-l- lfl'xlfl'zmd K313}! iii] (.5; 317.55%} : .051:ng mt 0H fr .— 34”: KNEE); : ,nmg met" #101 [2 mrfi’ of H 5:) 3'ng 1M6 bf ”2 T EHJSMD mm” M {ffinifl M : .510ng 1H M" (£35? an?) L _ -* Ja’ 4mg” 42 1' { {1%) H319] 3-"! - " ‘-—*fi f 91,353.13 M #33) 19M .05“;qu L____’ fi) PH 7 {CE {$.55IIém) J?“ ”-9" i501 F. a) .MS *— ,1- j if! SM 2 mm: M 301 .fL ...
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