mid1key - . ‘ -1_‘ 14A1W02-1 Baur ’ CHEMISTRY 14A...

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Unformatted text preview: . ‘ -1_‘ 14A1W02-1 Baur ’ CHEMISTRY 14A LECTURE SECTION 1 WINTER QUARTER, 2002 First Migrm Examinw ‘ Friday, February 1, 2002 Name ' TA: (check one) I (LAST) (FIRST) - Natalie Gassman Section Numbe Jenny Lockard ' ‘ . Kim Miller_____ ' ’ Will Molenkamp Alex Smith . INsTnucTIous . 1. There'are three questions. Answer each in the space prgvidgj; you may write on the backs of sheets if you need more room, but no lodse sheets of paper will be accepted. Your work must be clearly mm and LagitLle; credit will not be given for unintelligible responses. _M_ g g I 7 r r? f nin ‘ must be shown for all computational items; no credit for merely putting down an answer, even it correct. Make sure that all answers have the appropriate number of signifigant figures. Give mils for all answers, unless dimensionless. Open book examination. Any written or printed materials may be used. DO NOT LIFT THIS COVER SHEET AND BEG—IN WORK— UN—TI-L- lNSIRUC-T-ED-TOfi-O-SO! 5 th fll win t ifn : H 1.008 O 16.00 Na 22.99 Cl 35.45 Cr 52.00 Br 79.90 9°.” N991? He 4.003 F 19.00 Mg 24.30 Ar 39.95 ' Fe 55.85 Ag 107.87 C 12:01- . Ne-20.—18- 8-7 32.06» _ K- 39.10» Cu~--63-.55-- Odd-12.44 . N 14.01 ’ t Ca 40.08 Zn 65.39 I 126.90 NA (Avogadro’s Number) = 6.022 x 1023 I Photon energy (Planck'Law): a = hv = hc/A c(velocity of light) = 3.00 x 1013 ms-1 ' . fi(Pla'nck’s Constant) = 6.63% 10134 us Bohr formula: En = -2.178 x 10-180 (22/n2) ‘ ' 1 e.v.'(electron volt) = 1160 x 10-19 J Coulomb. energy. betweeniwo “charges Q1..and 02.3: EC =2.31 x1019 J nm(.0102/r) (Q.in multiples of the elementary charge) The Wand ample are on the back ht this sheet. CHECK To MAKE SURE THAT THIS EXAMINATION PAPER HAS THREE QUESTIONS AND FIVE PAGES -2- 14A1W02—1 ’ Name 1. (35 points) - . i (a) (10 points) A molecular compound 2 has molecular mass 60. When burned in excess oxygen, 1;OO .mole of Z yields 2.00 moles of 002, 4.00 moles of H20, and 1.00 mole osz as the only products. _ Find the m and mm formulas for compound Z. ' fin MOLE/5 607/ a 1:017 Mal/£3 C 9-00 MoL£5 H710 x 2 «new ~; 3400 Maw/5 H . Mun/514;) /'UD MOLE. A/z Mow” __ Mable/.5 N‘ F C} 8’ L r '55 7H5 MOLECULMQ FDMVLA— MUST 3/ CWieM 14/ 1711mol¢£cubm Mk5; éo) EMF/21am, FORMULA” 15 CHq N (b) (11 points) A sample of an ionic compound W is analyzed and found to contain 29.1% Na (sodium) and 40.6% S (sulfur). The remainder is 0, oxygen. Find the simplest (empirical) formula for compound W. M, mung: JF comfowvy’ mus'r' .H’A—VE 7 L4,!flA/g,’ 4101;599Jfifl (fly prafiawcé) ~ ' A4 _ :_ /’ MPL Ma, ‘ .L3-0éMIL" 715 _ 96%.; l = 52,; Mar, 31.16—Mu‘ M0 : 52':th" : /’g? M"!- .- r Vg£EQ:H «(I 41? ; Lie; /,5‘ mosr/M'VE 3 0 f7 . 414/”, /,;7 ) l’. EMF/QIcAL FaEMl/LA' ‘5 A/4—L52—03 (c )‘(14 points) ConSider the redox reaction: , i g , 2Kl03 + 3N2H4 —+ 2Kl + 3N; + 6H20 Give the oxidatiOn number for each atom on each side of the reaction: LEELSLQE. .. aigiflsios Kit. 5 ' K____‘_ |_-l;5_ l—l ' 0; 0-2 N~’2. N 0 HT. H.:l:_L and state which atom is oxidized (reading from left to right) M" 1 To 0) and which atom is reduced (reading from left to right) M1”? To -— l) -3. MM W02-1 Name 2. (35 points) } (a) (5 points)Ca|culate the frequency in 3-1 (Hz) of photons with wavelength 400 nm. 8 MS" ’ ," I4 -| :54; 20700260444 } 7’w W0 s (b) (5 points) Calculate the energy of a single 400 nm photon, in J and in e.v. (electron volts). 6:” =éé3x/o‘3fi‘rs x7.raw0"’$" 7- 1/:q7x/0‘MJ‘ ( "fiTY _____I :- l . 9WD”) Lax/0"":i’ 3" 6M (0 ) (10 points) When monochromatic light of wavelength 400 nm strikes the surface of metallic Ba (barium), photoelectrons with kinetic energy 6.4 x 10-20 J are emitted. What is the maximum wavelength of light which will release photoelectrons from metallic Ba? ' :73. g g 4- goggmu 5/ch 6' 0F A—l-l’dp 4M PM'IZU = ‘f-flhflf r _ A @4546? A“ Om E/NW M '— .20 ' ' ~ » "wk/Hy”) j i]! f ’ '4irq7X/0 I4 33 54501144) K/Il/ETIC MW ,y‘q-mo - ‘M 3/ (Iv/MGM is 63",.(1 loveSTF/mau Mat/r 054le . I ) , ' 3 —i In A V = _‘E affix/0 “3’5 (wax/o an! MA’X émfiu ‘ 4/«33X/0‘ 7 14A1W02-1 ' g ' - I Name 2. (Con’t.) , I (d) (5 points) Below are five configurations of valence (outer shell) electrons among orbitals. For each, you are to state whether the arrangement corresponds to the 9W an em or is I. I. 1 ' . u _ _ 5 yo HEP 31.1; 3p p p ‘ ' ’ . j s LL’LLL‘ L No 24' Ow’W’EX’W uu1L2d2d2d2d2d Impossle EC ‘ .LL 2p 2p 2p . 25 ' _L_L_i__i__L Zis’ ' WM" my“) 3d3d 3d 3d3d- 0K axe/716,3 a b ‘ 4 . ,5 6L 05m} Pie/NO “9 it 515.215.1213 v, Imposglgéa (ix/owes x 23 ' _L .L _L . LL 2p 2p 2p M 25‘ I (e) (10 points) For each of the neutral atom or ionic species below, you are to draw the W (that is, the configuration of electrons in the outer unfilled shell; omit inner closed shell ' electrons), clearly labelling the orbitals and placing arrows to represent the electrons. The first case is done for you as an example. ' m ro n' tin? _'.L_1__ ' ,c it. 292P29 » 2s _ F fl nil?!» L5 w W W S+ 4, i- 3} .2? 3r 30" Na5+w f 43 ___'_ 1%? a»? ’r’ ‘ H? Fe He 953$ 77; ,aavtézaxa Cd2+ . -5- 14A1W02-1 r Name 3. (30 points) 7 g ' (a) (21 points)For each of three molecules containing three fluorine atoms bonded to a central atom, BF3, NF3 and CIF3, draw the Lewis structure for the molecule; state using the VSEPR approach what the electron group and skeletal geometries are. ' ' Maul: _ Me i tr t r Mama ' M ' u my! 1 F- '\ . . BF3 V u l p “ Tammie PWN- WWW“ FM iF”E\F a! ‘ L "F3 ‘ ,‘f: /"‘J n;me Foam/£9 Wall/up \ \‘ ‘ I. ' ' "' Swab . ' ’ " I g I .- ‘ ' g" V AQE EQQA‘IQE/flé) Which of these molecules d_o_e_s_ obey the octet rule? 0A}ij NF: ’ Which of these molecules m have an electric dipole moment overall? 0 flag: 3 (b) (9 points) Consider the following diatomic species built up from second row atoms: I BeB+, BeC, BN, BF, 00+, NO+ ’ Answer the following questions with brief explanation using the orbitral energy'level Scheme developed in lecture and given on p. 148 of the text. ' Which species has the strongest bond? -- 1.. ' 1— / a Numémp a"; 395+ 4) 595 5/. 5M 6:: IO; C0 til/d 517W BF flit/D l0" f/m/g Ng' LIKE/l4, m/p A5041) oeféfjf/ij; So “may Am! Moor EQUAP Mswmflc may do / frawae 307' 5/77/15,: MSVEZ /: 4665773665) . - , -' ‘ . ' . a . flaw/M4 am a, ., Ad W m a e I #6350 P055 MOTH?“ kfl’fl' Which species has the largest paramagnetism (largest number of unpaired electrons)? + I 595 Wm ; e— x; Ll/(é 5],) Wm 2m Far/zap e .‘r (Cd W5 U/l/Pfi'liefip e'i) ...
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This note was uploaded on 02/08/2012 for the course CHEM 14A 142042200 taught by Professor Lavelle during the Spring '10 term at UCLA.

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mid1key - . ‘ -1_‘ 14A1W02-1 Baur ’ CHEMISTRY 14A...

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