14a201fl - ...fl.M§'KK.’§/§..§ CHEM 14A—2 YOUR...

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Unformatted text preview: ...fl.M§'KK.’§/§..§ CHEM 14A—2 YOUR NAME Instructor: Dr. Laurence Lavelle |D# ............................. .. WINTER 2001 FINAL EXAM (Total number of pages = 13) (Total points = 100) (Total time = 3 hours) “Carefully remove the first page which is your Periodic Table.“ YOUR TA‘S NAME .................................... .. WRITE IN PEN (Show all your work on this paper.) Constants and Formulas Planck constant, it = 6.63 x 10‘34 J ~ s Avogadro constant, NA = 6.02 x 1023 mol‘1 Rydberg constant, R = 3.29 x 1015 Hz Gas constant, R = 8.314 J.K’1.mol'1 Mass of electron, me = 9.1 x 10‘31 kg Speed of light, 0 : 3.0 x108 ms:1 O°C=273.15K 1L=1dm3 1atm=101.325 kPa n=3.14 h2n2 E=hv E=pc En=m2 p=mv _ h R I} IN INITIAL sag: En — - T12— )» - p pH _ pKA + LOG [AH]iNiTiAL ACID 1 h E=§mv2 c=kv Aprxz—A; _ -B +/- \le - 4AC Solution to Ax2 + BX + C = o is x 2A Qt) For each pair circle the larger of the two. If you think the values are the same then circle both. However points will be deducted for incorrect answers. (a) Atomic radius of Li (lithium) W W,”— (b) 15 orbital of F (Fluorine) twl (c) Ionization energy of Cl ' (g) (d) Energy of light with A = 600 nm (14M) Atomic radius of Li+ (lithium cation) ls orbital of Ne (neon) Ionization energy of Br ' (,3) Energy of light with v = 5.0x10“ Hz .9: géwgmb -(sqflfljq‘; SPtME S 2‘ (pw‘pJ‘mrp 5(0 5 a (e) Energy of an electron with = 107 m/s (net?— E: A "L I if...“ z "‘ V 1 113m. (WOMluDBM no? " 5’. 26 HOP“ I (t) Number of planar nodes in a 2p orbital ‘ I pert 661 I ,_,___..__._ (g) Orbitals with quantum number n = 2 Y) 1 Z ’L r. o a ML 0 'l l +\ (h) Uncertainty int/position of an electron with I %A v = 4.2x10 m/s oe3* ‘ h ._ attend-59 3 3 L130 .60 (‘1.llNU"'l‘Q(‘1.2H(;12> : Zl7<¢flo"lm 911.1%“)1”> k2” E:%?‘f :%(M'V)(Vj V13 ._l11‘(0'u\‘_‘3>:’ fi.ll‘i€to"'}_, Lutaflmk Number of planar nodes in a 45 orbital O d-orbitals in an atom \ LAfibt‘L! W Y'd 'WEM Wavelength emitted by an electron with v = 4.2x107 m/s Ada). A 1 c.91020tu'wl-5 MM 4.129t0q'g)Ca,(Lud“ i=5) ‘ L73 firo’l'm l unflatt— Q2) The chemical compound cyanocobalamin is more commonly known as vitamin B12. It is an essential supplement for humans and has the chemical formula 063H8500N14O14P (note the one cobalt atom in this large molecule). (6pt) (a) What is the mass percent of Cobalt in vitamin B12 ?(2pt) MvJ=C203 ‘17-m‘ 75"».6'5 HES’fHuu‘tCt 2 _ gang-L . 89”“ MASS 7t (ca) = '5 ('0' l)‘ ngsz = $351332 t‘sSS‘ “508M ‘»l‘-t H t '- M 5 ‘0 _ “‘9‘” ~0touw8vi’5 o: U—lxtbm — 22am) P t 1' ’ 1 ’és’SSfl‘t (b) People who have a deficiency of vitamin B12, and cannot get enough in their diet, may be prescribed a routine supplemental injection dose of 10 x Io~ "grams. How. many grams of cobalt are in one dose ? (2pt) MAssao) -— MKSS ‘ACéA x WU mm = (0043‘154'5) memo—Win '« ‘tswmm w (c) A normal laboratory blood value for vitamin B12 is from 350 — 1,000 picograms per milliliter. What molar concentration range does this represent ? (2pt) 35’0 /000mL> ( /’9~><'__‘____M\ —. 2 5873 “0'” M am At M. w new” —1—«/— , 40 ’—‘ L578 limo M m. Z€5 9M 03) Write the Lewis structures and the VSEPR formula, name the shape, and predict the approximate bond angles in: ’ (12pt) 64’ W Mead/g (C) X904 Q4) (a) Is CH3F polar or nonpolar ? (290 (b) Is trans-dichloroethene polar or nonpolar ’? (2pt) C C / \ H 6L (0) Use hybridization to describe the bonding in benzene. (4pt) Sp2 HYBRIDIZATION BENZENE CeHe BENZENE IS PLANAR EACH C TRIGONAL PLANAR (1209) & MUST BE sp2 HYBRIDIZED EACH C CAN FORM 3 O-BONDS (TWO CspZ—Cspz "19 (ONE Hs—Cspz) 9% EACH 0 CONTRIBUTEs Ie- FROM ITS UNHYBRIDIZED 2p ORBITAL LYING PERPENDICULAR TO THE RING. THESE SIX 8— ARE SHARED EQUALL Y THE 6 p-ORBITALS (FORMING A It MO ABOVE & BELOW THE RING). 05) Draw the correlation diagram for two 1 atoms forming I2. I Label all the atomic and molecular orbitals and show the valence e— in the atomic and molecular orbitals. (6pt) Ans) l2 Molecular Orbitals Atomic Orbitals Atomic Orbitals _TLUMO 0% £0 2 HOMO flsm flsw 1b 10 j 1; 1; 1' 5p 5p 4a a; n 5Px [4 n 5Py GSPZ ENERGY ii 70 4i 1k 55 53 058 -1 pt for incorrect energy level representation —1 pt for incorrect or incomplete labels -1 pt for incorrect or incomplete e- configuration 05) (d) Calculate the bond order for lg. Ans) BOND ORDER = 1/2 (8 - 6) :1 In your diagram above, label the HOMO and LUMO orbitals. Write out the valence e- configuration for lg. Ans) o2 VALENCE e- CONFIGURATION IS (055)2(0'55f(aspz)2(n5p)4(n'5p) 4 (1P0 (1P0 (2P0 QB) (a) (b) Draw and name the cis and trans isomers of CrClBr|(NH3)3. (6pt) 1: m \. / / M9} 4 H A/ / / \MG ; 3 Q ..—- ‘ ‘ _. I Mr] mg List the following acids in order of increasing acidity and explain the trend: HCI HF HI HBr (4pt) HF<H/&€</%/§r< 66 WW HI Q7) 0.100 mol of H28 is placed in a 10.0 L reaction vessel and heated to 1132 °C. At equilibrium 0.0285 mol H2 is present. Calculate the value of K0 for the reaction: H28(9) =-‘ Hzlg) + 32(9) (10m) x: 0.90/g/2, .¢:( [flag]: 0.0/00~ 0.001355 5.00'?2m/€,L./ HQ) 30-00 233’ij [it] 20.00/442M.J’ W) /Tc " V @231 ii) «LL02 357%5-‘70/4‘12 IL : (o, 00‘? 2;," (W WK 10 08) One of the acid-base buffer systems in blood is the bicarbonate system. (10pt) It is dominated by the equilibrium reaction: H2003(aq) (carbonic acid) = H* (aq) + HC03'(aq) (bicarbonate) (a) Given the normal concentration of bicarbonate ion in blood is 24 millimoles per deciliter, the normal pH = 7.4, and Ka(H2C03 (3(1)) is 4.3 x 10”, what is the normal concentration of carbonic acid ? (4pt) ' gotta: CW) (WW/Wt»? WM PH 2 "DKQ + , 24"? “Hr “~ 'lo‘3(“l75“°7>* \°?r CH5”; j‘mx ((033%7': lot) {Vi—ELL C‘fiwsj logo. > cw - cwm ‘ (b) Phosphoric acid, another buffer in the blood, has a Ka of 7.6 x 103. What will be the ratio of phosphoric acid to its conjugate base in blood ? (4pt) Cbau] Pht.’ f [4070(1 (km) 7.% : 403/7tx10'3) F /°5 (“"43 (bait) 9:78, 1 /0} -9221 .tflfl_}___a/ -1. ' (kmd) / tr ' lac)ng " (c) Carbonic acid is also involved in another, heterogeneous, equilibrium: H2003 (aq) == H200) + 002(9) Given the interrelationship of the two equilibria, what will happen to the pH of a person’s blood if they start to hyperventilate ? Explain why, using the concepts of equilibrium. (2pt) (Hint, hyperventilation = “fast breathing” will cause you to exhale more 002 (9)) (l) cLLCrcasiN) (07, 007“ 51.744“ 24”” FF“ "7‘0 mée (002, W [Vb/0g} “NU w i? (was? t , wt mt er at“ 51ml ‘ln 1M, iv “GHQ 09) When a 100 mg sample of an organic base of molar mass 31.06 g.mol‘1 dissolved in 60.0 ml of water, the pH was found to be 9.85. What is the % protonation of the base ? Calculate the pr of the base and the pKa of its conjugate acid. (10pt) EH} ‘1’ Kflfl ’5 /0~70flfl:/é“¢i/f: SQ/X/aii/(di/ \ a») We]: x it”: = J,&/004 ffloéyfl (O.&fj/._%/Xm'{': fl'affqu/(g/ (2%) @‘r W to ~(f/X’75) K 5i, M? Q? 0.0(35’7 ‘ 44209 .3 ‘ZZ/ (7,{p)</:y) : 03 ffl/M ’: /$4—00« 9-0,? : A417. 2: -?'/X/”.; X/HZ: 0.3% @0536 12 010) While swimming you cut your hand and the pool water causes the out to burn. If the pool water contains a weak acid (0.020 M with a KA = 1.2 x 10‘s) and you do a titration, with the pool water as the sample, what is the pH at the stoichiometric point ? In your titration you use 20.00 mi of pool water and a 0.150 M NaOH(aq) solution as titrant. Also suggest a suitable indicator for this titration. (1 Opt) Ans) AT STOICHIOMETRIC POINT: # MOLES OH' ADDED = # MOLES ACID IN SAMPLE MOLES OF WEAK ACID = (2.000 x 10‘2 L) (0.020 moI.L“) = 4.0 x 10" mol 1pt # MOLES NaOH ADDED = 4.0 x 10“1 moi 1pt VOLUME OF NaOH ADDED =% = 2.7 x 10‘3 L 1pt TOTAL VOLUME AT THE STOICHIOMETRIC POINT IS: 20.00 + 2.7 ml: 22.7 ml SAMPLE IS HA AND THE BASE IS NaOH, AND AT THE STOICHIOMETRIC POINT THERE IS SALT AND WATER IN EQUILIBRIUM: Na*(aq) + A“ (aq) + HgO(l) 2 HA(aq) + OH~ (aq) + Na+(aq) NEED TO CALCULATE [OH'] AND CONVERT TO [H+] TO OBTAIN pH. LET [OH'] = X AND SET UP EQUILIBRIUM TABLE. - 4.0 x 10'4 mol _2 .1 INITIAL MOLARITY OF A = —"——-s~22-7 X10. L = 1.8 x10 moI.L 1pt A' HA OH‘ INITIAL MOLARITY 0.018 0 0 CHANGE IN MOLAFIITY -X +X +X EQUILIBRIUM MOLARITY 0.018 - X X X [HA] [OH‘] KB : _ [A I GIVEN, KA=1.2X10‘8 KA X KB=KW=1O"4 KB=8.3x10'7 1pt _7 _ [HA] [OH'] _ X2 8.3x10 _ [M -———-0'018_X 1pt AT THE STOICHIOMETRIC POINT X MUST BE SMALL 2 8.3 x 10" = 0):)18 X = 1.2 x 10‘4 (approx is valid) 1pt pOH = -LOG1.2 x 10‘4 = 3.97. 1pt pH =14.00-3.9Z=10.o: 1pt PHENOLPHTHALEIN (pKin = 9.4) or other indicator with pKin = 10 +/-1 1pt 13 ...
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This note was uploaded on 06/15/2009 for the course CHEM 14A taught by Professor Lavelle during the Spring '09 term at Mt. SAC.

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14a201fl - ...fl.M§'KK.’§/§..§ CHEM 14A—2 YOUR...

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