CHEM2A Practice Midterm KEY

CHEM2A Practice Midterm KEY - Name[k E E Student ID Num r...

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Unformatted text preview: Name [k E E Student ID Num r Lab TA Name and Time Winter 2007 — Enderle CHEMISTRY 2A Exam 11 Instructions: CLOSED BOOK EXAM! No books, notes, or additional 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. (1) Read each question carefully. (2) There is no partial credit for the problems in Part I. (3) The last page contains a periodic table and some useful information. You may remove this for easy access. (4) Graded exams will be returned in discussion sections next week. (5) If you finish early, RECHECK YOUR ANSWERS! Multi Ie Choice U.C. Davis is an Honor Institution __.—___._———————— Exam I Page 2 of 9 Part 1: Multiple Choice, Concepts Circle the correct answer and enter your response on the cover — No partial credit 1. Which of the following laws is/are used to determine the ideal gas law? A. Avogadro’s Law B. Boyle’s Law C. Charles’ Law @ All of the above E. None of the above 2. The measured pressure of a Van Der Waals/Real gas compared to an ideal gas is: A. higher, because molecules occupy volume B. higher, because of intermolecular attraction C. lower, because molecules occupy volume ® lower, because of intermolecular attraction E. the same 3. Which of the following set of quantum numbers is forbidden? A. n= 3, l = 2, m1= 2, ms: +l/2 4. Calculate the density (g/L) of methanol (CH3OH) at 25 °C and 745 Torr? A. 1.43 g/L . 691523;? [trig—.W 2|.9333/L D 140g/L KT masher/3) ‘ Exam I Page 3 of 9 5. Standard conditions of temperature and pressure (STP) are: A. 100 °C, 1 atm @) 273 K, 760 Torr C. 25 °C, 760 mm Hg D. 0 K, 1 atm E. 100 0C, 1 Torr 6. What is the wavelength associated with an electron traveling at one-hundredth the speed of light. ¢ ’L A. 242nm N MW Ligand” B. 412nm . -5i C 798x10'2nm I 67.10? “0 (0.0“< 3H0?) (D) 2.42x10‘lnm _ .. 40 :Jfl'flaxlo m E. 12.5 nm 7. Every electron must have a different set of quantum numbers is explained by: A. the Heisenberg Uncertainty Principle B. Hund’s Rule (C3 the Pauli Exclusion Principle D. the Aufbau Principle E. the Enderle Principle 8. Rank in order of increasing atomic radius: K+, Sez', Cl', Na+, 82' A. (31' < s2' < K+ < Na+ < Sez' B. 82. < Na+ < Cl' < Sez' < K C. Cl' < Sez' < K+ < Na+ < 82' @ Na+ < K+ < Cl' < 32' < Sez' E. K“ < Na+ < C1" < Sez' < 82' 9. Gases tend to behave ideally at: A. low temperature and low pressure B. low temperature and high pressure @ high temperature and low pressure D. high temperature and high pressure E. all gases are ideal Exam 1 Page 4 of 9 10. A gas of twice the molecular weight of C02 will pass through a small hole how much faster than C02? A. 1/2 as fast M 0’1, B. twice as fast h z ——-*A : ‘ ~— tg j Me. i D. 1/‘12 as fast E. 4 times as fast 11. What is the maximum number of electrons that can be accommodated in the shell with principle quantum number n = 5? ® 50 n:5 B' 40 9:4)319’1"0 c. 30 , , ,. D. 20 mfl’q}~f)5; 91' E. 10 14:46: i8+i++l0+é79~ 12. To what precision in nm can the position of an electron traveling at 1.89x105 m/s be measured if the uncertainty of the speed is 2.00%. k A. 0.765nm . ___. A E (is) 15.3nm Ax P 4;: > / C. 83.3 nm AX Mbli ’ 4,“. by D 0 654nm , — AX (Cl 109 ’40 3‘)(0 094418?me 5 4:“ @ volume is proportional to temperature B. volume is inversely proportional to temperature C. pressure is proportional to volume D. pressure is inversely proportional to volume E. pressure is proportional to the number of moles of gas Exam I Page 5 of 9 Part 11: Short Answer 14. (12 points) Fill in the BLANKS with the correct electron configuration for the given atom or ion or the correct atom for the given electronic configuration Electron Configuration — mW 15. (10 points) Consider a particle in the box where n = 6. Write the mathematical equation describing the wavefunction and the electron probability density 01162) and draw both. Label the x-axis and the points x = 0 and x = L. ”m = EM?) Weld): % swag) L 0 C) r. 7; Exam I ' Page 6 of 9 Part 111: Long Answer Please show all work — Partial credit 16. (16 total points) A 25.00 mL solution of barium hydroxide is titrated with 0.1694 M hydrochloric acid. A. (8 points) Write the balanced chemical equation for this reaction. (504w); + QHU ——>ng+ 60d; B. (8 points) What is the molarity of the barium hydroxide solution if 42.65 mL of the hydrochloric acid is required to complete the titration? 0, (WM HCl (004015“ W) (W) : 0.00395 was Ba(0H)Z 0.00% mm; mum); 44 M (3 ( H) M : 0, 0 0.09.5 L 60W); 1 5 a 2' MA 17. (16 points) A mixture of 4.0 g H; (g) and 10.0 g He (g) in a 4.3-L flask is ‘ maintained at 0 °C. A. (10 points) Calculate the total pressure in the container. 4-0 Hz [0 He, : longmsl + 4.300%32m01 4433 We a“ W: nKT ’_ _. O; WKT , 4.433 moles 0.0851049 L'aJML'mcI ‘~K )(gz731él \ V _ 4.3L Pressure=‘ 023. am B. (6 points) What is the partial pressure of each gas? 4.0 3H2 . O ' m2»? Q0lb3lm0l : lqg4 meld H1 KPH-L; lTe’l‘Xl—ll: 33( 4.483) ' 2 [0337mm [0 He . . _ 34%, ——3—-——-4 Gala/towel : 51.4518 moles l+c 37M: VMXHe: g3< 4.483) ‘ 1 IBSOlLodm/x Exam 1 Page 8 of 9 Conversions: latm = 14.7 psi = 101,325 Pa = 760 mmHg = 1.01325 bar = 760 Torr Constants: R = 8.3145 J / mo] K = 0.08206 L aim / mo] K Avogadro’s number = 6.022 x 1023 c=2.9979x108m/s h=6.626x10'3‘Js RH=2.179x10'“‘J g = 9.81 m / s2 m (electron) = 9.109 x 10'“ kg am = 1.00 g / cm3 Equations and Various Tablas: - 1 1 —— Z 2R h v=3.2881x10‘5s‘[—2——2 En=—zi 2=—— P=dgh E=hv c=lv 2 n n mu h e sion - rateA M nzh2 1 1 AxApZ— —fii‘_——-—= —B- Ek= 2 AE=RH 7-7— eK=‘/2mu2=3/2RT 47: eflusxon — rateB M 8mL n. n f A 2 P + 11—; (V — nb): nRT ‘I’(x) —_- FSin(—’1E),n =1,2’3,_" um = 52 = 3151 V L L \l M PV= nRT b Vapor Pressure «Water at VariousTemper-tures TABLE 9.1 The Angular and Radial Wave Functions of a Hydrogen-like Atom Vapor Angular Pan 1’10. 0) Radial Pan k, , (1') Temperature Pressure . 1 m z 2.1: (°C) (mulls) Y(.1) = (4—1;) R(ls) = 2(a) (""1“ 15.0 12-79 1 Z Jr: 17.0 14.53 R(2.1-) = (-) (2 — a)?” 19.0 16.48 2? 2" m 21.0 18.65 ROI) = _ (_) 0 _ M + a; «:11 23.0 21.07 9% an 1 ( )6 25.0 23.76 . 3 ""2 . _ | z .1: W. 30.0 31.82 11,5) - (3—0:) s1n0cosd: R(2p) — 2V0" ("l—v) in 500 915' Y( ) 1 3 )m ' Osiné R0 ) I (2)354 "" . = —‘ sm . = — — “ ' ”‘ 14m ” 9 0 a.. ( a)” 3 172 1’ _ s _;__ 050 TABLE 5.1 Common (It) (47) C Strong Acids and ' I 1 1 1 ’ Stmn Bases Hal-.2): (i) [31:050-1) Rad) "—,_.(£) 0’0“" 9 '6‘: 9V” “11 . 15 1/3 ' a Auds Bales )(d,:_..~) = (:4?) 5m Geode) HCl LiOH > ‘ 15 1x: ‘ HBr NaOH rm“ = (-4—) sin-0511124: H1 KOH .l: “I HCIO. RbOH . - . , HNO CsOH . = — 0 " 1111,. i ( 4f.) 51110603 (703(1) H2504. Ca(OH) . ’5 ._,-_7 and SnOH) 11.1,.) = (“2) sinflcosflsimb Ba(0H) ' 22, -H,so. ionizcs in we dislinci slaps. g = —— l1 insuongacidonlyin ilsfim ""0 ionfintionswp (see Section I7-6). Page 9 of9 Exam I $23 22:: 14+ .33 an. .93 :5 8.5 :5 55 ea. .3. :2. 99.3" 93.8" was... 233 - . .2.— OZ :2 En— mm— “U u—fl EU E< an— n-Z e.— FF .. . . . . 8_.I «3 .3 as 8 8 .3. 1mm! 3 II it til] . .842 3.3 3%: as: 89.3. a”: 3.! 3E 33.: 3?: an. 5.: 383. :3: s..— ? Eu. .5 0: .AD ah BU 3m— Em E.— BZ :— 0U . . . . R .3 8 S S 3 8 3 S 3:? 3.5.5:... * Son. .35 .2 m: as .. unda— a8— : 5 R .k 8.2. 5.8 5— 3. no 3. «Roan :52 wh— om w< so 3 mm. . Sana m _ ,. s. . 38.4.. .33.— :3; O 2' U a a c < J ...
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