14a1fnlk - CHEM 14A YOUR NAME.A.MAITIM£.K.~.I’...

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Unformatted text preview: CHEM 14A YOUR NAME ...A.MAITIM£.K.~.I’.... Instructor: Dr. Laurence Lavelle STUDENT ID# ............................... FALL 2006 FINAL EXAM (Total number of pages : 13) (Total points : 170) (Total time = 170 mins) YOUR DISCUSSION SECTION ...................................... YOUR TA’S NAME ...................................... Final exams will be available next quarter from 3104 Young Hall. WRITE IN PEN (Show all your work on this paper, check units and significant figures.) Good Luck Q1 A. In medicine it is sometimes necessary to prepare solutions with a specific concentration 01a given ion. A lab technician has made up a 100.0 mL solution containing 0.50 g NaCl and 0.30 g KCl, as well as glucose and other sugars. What is the concentration of chloride ions in the solution? Ans: w (3pt) The total molar concentration of Cl' : ’ .5 g - . .3 1 0 ()gNaCll + 0 OgKCnl X[ 1 120.1%”! 58.44 g-mol 74.55 g~mol 0100 L ~1 i1" no units Q18. A 1.50 g sample 01 metallic tin (Sn) was placed in 212645 g crucible and heated until all the tin had reacted with the oxygen in air to form an oxide. The crucible and product together were tound to weigh 28.35 g. (a) What is the empirical formula of the oxide? Ans: (3pt) (b) Write the name of the oxide. Ans: (3pt) Total mass ()1 tin oxide is: 28.35g — 26.45 g = 1.90 g 1 mol Sn (a)1.50gSn>< — 118.71 g Sn : 1.264 x 10’2 mol Sn \ 1molO Moles of 0: (1.902150% x 16.00 g 0 j = 0.025 mol 0 Mole ratio of Sn:O is 1:2 Empirical l‘ormula: 31103 (b) tin(1V)0xide QlC‘. Consider the reaction NO(g) é Ngtg) + 02(g). If the initial partial pressure of NO(g) is 1.0 bar. and .r is the equilibrium concentration of N3(g). what is the correct equilibrium relation? Ans: m 7 (3})1) (a)K'=.\':/(1.0—_\') (b) 1<=.t-3 (c) 1<=x3/(i.0—2x)2 at) K = 4_\-3/ti.0 — 2.03 (e) K = 2x/(1.0 — .03 (c) K : xI/(1.0—2x)Z because balanced rxn is 2 NO(g) : N3(g) + 03(g) ,, 7 QlD. A compound found in the nucleus of a human cell was found to be composed of carbon, hydrogen, and nitrogen. A combustion analysis of 1.35 g of the compound produced 2.20 g of CO; and 0.901 g of H30. When a separate 0.500 g sample of the compound was analyzed for nitrogen 0.130 g of N; was produced. What is the empirical formula of the compound? Ans: (6P1) Calculate the mass percentage composition. j=0mMgC 2.20gCOZ XL lmol CO2 jx 44.01 g (:02 0.600 g C 1.35 g unknown 11.0 2 1H 1. o 0901:2110 xf I'm] — >< ”‘0 ><{ 00798H] 20101 g H ~ 18.02 g 1130 1 mol H30 1 mol H 0.101gH 1.35 g unknown 7 \ LT 0.130gN1x ”MINI x “mm x[l4'018N]=0.130gN ‘ 28.02 g N3 1 mol N2 1 mol N 0.130 g N 0.500 g unknown 1 molC 2X[12.01gC lmolCO2 lmolC %C : x 100% = 44.5% C \ (/8 H : x 100% = 7.48% H % N = x 100%: 26.0% N Oxygen must be present in the compound because the percentages of C, H, and N only account for 78.0% of its composition. Combustion analysis does not generate data directly for oxygen; we calculate it by difference. %0 : 100 — 78.0 = 22% 0 To find the empirical formula. assume a sample size of 100 g and find the mole ratios. _ " 1 1C ’ 1 , 41>ng “mo 23.71 molx —— =2.69><3=8.()7 \ 12.01 g C 1.38 mol 1 1H 7.48 g H x “mo = 7.42 mol x W1 2 5.38 x 3 =101 1.0079 g H 1.38 mol 1 26.0ng fl =1.86mol>< ; :1.35><3:4.05 \1401 g N 1.38 mol " 1 1o / \ 22.0ng 17m“ J 21.38molx ul— :1.00><3=3.00 \ 10.00 g 0/ .138 mol The empirical formula is CxH16N403. QZA. The electron affinity of thulium (Tm) has been measured by laser photodetachment electron spectroscopy. In this technique a gaseous beam of the anions of an element is bombarded with photons from a laser. The photons eject electrons from some of the anions and the kinetic energy of the ejected electrons are detected. The photons had a wavelength of 1064 nm and the ejected electrons had kinetic energy of 0.137 eV. Although the analysis is somewhat more complicated, we can obtain a rough estimate of the electron affinity as the difference in energy between the photons and the energy of the ejected electrons. What is the electron affinity of thulium in kJ.mol_l ? (lOpt) 2. = 1064 nm : 1.064 x 10" m The energy of a photon of this wavelength is I“, (6.626 x 10‘34J - sec)2.998 x1081” 15 : . 4 6 = 1.867x10”19J A l.064><10_ m The c11c1'g_\ 61‘ the ejected electron is 0.137 eV x (1.602 X 10“”1-ev‘) = 2.195 x 10‘20 1. The difference between these two values is 1.65 X 10‘“) J or 1.65 X 10’22 kJ per atom of thulium. Multiplication by Avogadro’s number gives an electron affinity of 99.2 kJ.mol'1 of thulium. .1 no units —lsf Q28. The average speed of a helium (He) atom at 250C is 1.23 X 103 ms". What is the average wa\ elength of a helium atom at this temperature? Given that the diameter of a helium atom is approximately 10"“) m. Does your answer make sense? (lOpt) The mass of one He atom is given by the molar mass of He divided by Avogadro’s constant: 4.00 g - mor‘ 6.022 ><1033 atoms 1 mol'1 : 6.64 x10“ g or 6.64 x 10 '17 kg mass of He atom = From the de Broglie relationship, [7 = 11/?1 or 12 : HIV/l, we can calculate wavelength. /l : /1(mv)"1 6.626 ()8x10’34 J-s : (6.64x10'27 kg)(1230 m-s") _ 6.626 08x 10"“ kg-m: e" <6.64><10"3" kg)(1230 ms '1 :8.ll><l() H m Since the wavelength is less than the size (diameter), yes it makes sense. —lunits —1sf LII QRACertain gases. called greenhouse gases, contribute to global warming by absorbing infrared radiation. Only molecules with dipole moments (or nonpolar molecules that undergo bending motions which create momentary dipoles) can absorb infrared radiation. Which of the following gases. all of which occur naturally in air, can function as greenhouse gases? In other words which of the following are polar 1’ (lZpt) (a) CO Ans: (Zpt each) (b) O; Ans: (c) O; Ans: (d) SO; Ans: (0) NO Ans: (l) Ar Ans: Look at the Lewis structures for the molecules: :1 + (Ll) 54:30; (e) :N: N :o' (r) IAr: Of these. a: c, d. and e can all function as greenhouse gases **crase lpr e— on the oxygen in (a). QBB. What are the principal and orbital angular momentum quantum numbers for each of the lollowing orbitals: (opt) (a) 25‘ Ans: (2pt each) lb) 4d Ans: (c) 5/7 Ans: (al 1121/20: (b) 11241123; (C) Ii=5Ll=l Q4. For each of the l‘ollowing molecules or ions, draw the Lewis structure, li_st the number of lone pairs on the central atom, identify the shape, and estimate the bond angles. (25pt) (5pt each) 0' '. : BI: HO ..Bi‘\l ii P—B I .. . -‘ / I : F—Xej ..B];. | " ' B. l (a) PBl'j (b) erF2 ”3: no lonc pairs on central atom; trigonal bipyramidyl; 90”, 1200 F- ’F\ :‘F/ E: l S l (c) SF; .F no lone pairs on central atom; trigonal hipyramidyl: 90°, 1200 (c) 810:7 one lone pair on central atom trigonal pyramidyl; 1070 or less than 109.50 two lone pairs on central atom; T—shape. less than 90”. less than 1800 (d) 11:3 two lone pairs on central atom; T—shape, less than 90”, less than 1800 QSA. Write out the valence e- configuration for these three unusual species. (a) Cf; (b) C3; (C) Cf. Based on their valence-shell electron configurations which of the following species would you expect to have the lowest ionization energy? (Hint: You may want to draw a molecular orbital diagram, but we will only look at your answer for the e— configuration.) (lZpt) (3pt each answer) (a) (‘1‘ Ans: C; (0': )1 (0-1“: )1 (”l/K ) (”3” ) (1)) C3 Ans: C: (”A )2 (03‘ :E: )3 (”l/K ) (7th) ) (c) C: Ans: _ ((7; )1 (an )3 (773K ) (712p) j(o*zp)l C: is expected to have the lowest ionization energy because its electron is lost from a higher energy MO ( (73/, ) than either C1" or C: (fill) ). QSB. Which M2+ ions (where M is a metal) are predicted to have the following ground-state electron conl'igurations. (6pt) (3pt each answer) (a) [Ar13d7 Ans: («01+ (h) [KI'HdeSl Ans: Sn‘l‘ QSC. Name each of the following complexes and determine the oxidation number of the metal. (lZpt) (4pt each answer) (a) [FNOHllHJOlsl—+ pentaquah ydroxoiron( 111) ion (Fe)+ l ><(—l)+5><(0)=+2 (Fe) = +3 (h) lCoBrgtl\lH:)z('H:O)]+ triaminincaquadibromocobalt(lll) ion (Co)+2><(—l)+3><(0)+ l ><(0)=+l (Co) = + 3 m [Ni(en):)l3T trislethylenediamine)nicltel(lll) ion (Ni)+3><(()):+3 (Ni) : + 3 on. The equilibrium constant K.. = 1.1 X 10’2 for the reaction PC15(g) 3 PC13(g) + Clgtg) at 400. K. (a) (ii\cn that 1.0 g 01. PC15 is placed in 21250. mL reaction vessel. determine the molar concentrations in the mixture at equilibrium. (l4pt) 77170:: 13le» ,vi . , , _ _ 208.22g-mol’I PClS ,1 Concentration ol PC]. initially 2 ———————~’—— 2 0.019 mol - L ‘ 0250 L Concentration (mol-14“) PCl5(g) a PCl.‘(g) + Cl: initial 0.019 0 0 change —_\' + .r + ,t' final 0.019 — .t' +.\‘ +,\' tPCI; 11031 _ (no) x3 ‘ wmq (omgao (om9—o .\' (om9—o t9:uixm¢nom9~o 19+uixm9n—2ixm4=0 :rixm4 aroma»: uixm4f—hmnezbuoo .\‘: 71 i—UJXHV):00M _____7T—"*— 2 +0010 or —0.021 X The negative root is not meaningful, so we choose x : 0.010 mol - L’l. [PC1{|:[Cllj: 0.010 mol-L 1;[PC15] : 0.009 mol - L”. rlttnits —1sl‘ (b) What percentage 01‘ the PC15 has decomposed at equilibrium at 400. K‘? (3pt) The percentage decomposition is given by 0,010 0.019 X 100: 53% 10 Q7A. A student added solid NaZO to a 200.0 mL volumetric flask, which was then filled with water, resulting in 200.0 mL of NaOH solution. 5.00 mL of the solution was then transferred to another volumetric flask and diluted to 5000 mL. At 25 0C the pH of the diluted solution is 13.25. (a) What is the concentration of hydroxide ion in the diluted solution? (4pt) 7. szl325 unon=10””=56xio”mm L“ [Hgotiiou’pio‘4 ioni=0i8mm-L‘ (h) What is the concentration of hydroxide ion in the original solution? (4pt) J: 18 mol L’1 s . . l0Hl=(H8nmlle£iygflL 5.00mL (c) What mass of Nago was added to the first flask? (4130 The reaction is: NL130(S) + H20“) 9 2 NaOH(aq) 'l‘hc mass of NagO: 18 mol 'L'1 x (0.200 L) x [ lnm/NaZO j x [61.9851 NaZO = 110 grams Nile 2})10/ NaOH 1171(lech —lpt for no units or incorrect units Q7B. Write the Lewis structures of each reactant, identify the Lewis acid and the Lewis base, and then write the Lewis structure of the product for the following Lewis acid—base reactions. (Spt each answer) (lOpt) MWR+FH I'F'I j _ :'F 'F': :'F' F: ..\P/.. ..\1L./.. IE2 IE2 IE2 lxm» acitl Lewis base product (h) Cl + SO; 7% o. “I _ 2C]: ” a n / _Q=S 0:3: \ .. 7 " H \ =9: =9“ :9: Lents acid Lewis haw product ll 08A. The main bul‘l'er in the blood consists primarily of hydrogen carbonate (HCOg') and HgO+ ions in equilibrium with water and C03: H30+(aq) + HC03'(aq) : 2 H300) + COZ(aq) K : 7.9 X 10’7 Suppose that 1.0 L of blood is removed from the body and brought to pH : 6.10. ll the concentration of HCOg' is 5.5 umolL’l. calculate the amount (in moles) of C03 present in the solution at this pH. (10 pt) The amount ol’ CO: present at equilibrium may be found using the equilibrium expression for the reaction ol‘ interest: H 10‘ (aq) + HCO,» :2 HIGH) + COl(aq) [C03 J [Hgotnncog] K = 7.9 x10 7 2 Solving for [C03]: ['CO: I : (7.9x1()"7)[H3O+][HCO;] Given: iH.o“i:10*‘*l = 7.9><10’7 M, and 1HCO;J : 5.5 amoi-L'l = 5.5 x10” M [C03] = (7.9x10’7)(7.9><10’7)(5.5 x10”) = 3.4x10"‘”inoi-L*‘ In 1.0 L of solution there will be 3.4 ><10'18 mol of C02(aq) 4| it no units ()88. The i‘ollirming salts were Obtained L11 the stoichiometric point in titrations. Slate if the pH is 7.21b0ve 7, 01‘ below 7 211 the stoichiometric point. (lOpl) (Ll) NH4BI‘ pH is bclow 7 (h) N113CO: pH is Lime 7 (c) KF p11 is above 7 (L1) KBi‘ pH is7 (C) AlCl; pH is below 7 l3 ...
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