Exam 2 Fall 2005

Exam 2 Fall 2005 - Chemistry 105a First letter of PLEASE...

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Unformatted text preview: Chemistry 105a First letter of PLEASE PRINT YOUR NAME IN BLOCK LETTERS last name Fall, 2005 Name: Exam #2 Oct. 6, 2005 Dr. Robert Bau Lab: (M aft. / M eve. / Tu morn. / Tu aft. / W aft. / Th morn. / F aft. / none) (please circle lab section above) 5 s 7 B C N 10.811 17.0111 14.0067 13 14 15 Al Si P 26.9815 23.086 30.9737 31 32 Ga Ge 69.72 72.59 50 49 3 4 Li Be 6.941 9.01218 11 12 Na Mg 22.9897 24.305 Te 78 Pt 195.09 8 O 1 5.9994 16 S 32.064 33 34 As Se 749216 78.96 51 52 90 Actinides T 232.038 231.035 (243) (247) 67 68 70 71 Ho Er Y b L u 164.930 167.26 173.04 174.97 99 100 102 103 F m M d No Lw (757) (259) I have observed all the rules of academic integrity while taking this exam signature Question 1 (10 pts) CIRCLE CORRECT ANSWERS (Solubilig table on last page if needed) (a) For the reaction Sr(N03)2(aq) + NiSO4(aq) —) SrSO4(s) + Ni(NO3)2(aq), the spectator ions are: (i) Sr2+(aq) and Neg-(aq) (ii) Ni2+(aq) and SO42'(aq) (iii) Sr2+(aq) and SO42'(aq) a® Ni2+(aq) and N03'(aq) (v) none of the above (b) When Pb(NO3)2(aq) and KBr(aq) are mixed, the net ionic equation is i Pb+(aq) + Br‘(aq) —> PbBr(s) ,a Pb2+(aq) + 2 Br‘(aq) —-) PbBr2(s) (iii) Pb2+(aq) + Br2'(aq) —> PbBr(s) (iv) K+(aq) + N03'(aq) —> KNO3(s) (v) I<+(aq> + Neg-(aq) —>I<Nos<aq) (vi) none of the above (c) When A1(OH)3(s) and H2504(aq) are mixed, the products are: (i) AlSO4(aq) and H20(l) ii A13(504)2(aQ) and H200) a (33 A12(504)3(aq) and H200) (iv) AlSO4(s) and H200) (v) A13(SO4)2<s) and H200) (vi) A12(SO4)3(s> and H200) (d) For the reaction Mn(s) + 2HBr(aq) —) MnBr2(aq)+H2(g), the spectator ions are: (i) H+(aq) and Br’(aq) (ii) Mn2+(aq) and Br'(aq) (iii) Mn2+(aq) and H+<aq> (iv) the only spectator ion is H+(aq) _——>@ the only spectator ion is Br‘(aq) (vi) the only spectator ion is Mn2+ (aq) (vii) there are no spectator ions (e) When LiN03 (aq) and RbZSO4 (aq) are mixed, the products are: (i) LiSO4(S) + RbN03(aq) (ii) Li2502(8) + RbNO3(aq) (iii) LiSO4(aq) + RbN03(s) (iv) LiZSO4(aq) + RbNO3(S) ,[email protected] none of the above Question #2 (8 pts) Cobalt (III) hydroxide neutralizes sulfuric acid according to the equation: 2 Co(OH)3 (s) + 3 H2804 (aq) —) C02(SO4)3 (aq) + 6 H20 (1) How many liters of a 0.75 M solution of sulfuric acid are needed to neutralize (or dissolve) 0.125 moles of cobalt (III) hydroxide? -..: 0./f’75‘ molt! 2 fhl/cl 51.5%» ) Lia-"mm CHM)? (2. x can) ma 4! 0 3 7 ’11. 5'"!!- max 0 WW mau- Inlay , l/olam 5 z o ' 2‘s L’ ( 4.7! mm. (14'*«)Hv"°v :::1 Question #3 (10 pts) (fill in the blanks or circle correct answers) -1“ ti o) +w\ 72 +‘\ 77- In the following reaction: CH4 + 2 02 —) C02 + 2H20 The element C is oxidized from an ox. state of " ‘l' to an ox. state of + 4 . The element 0 is reduced from an ox. state of 0 to an ox. state of "’ Z . The oxidizing agent is (C / H / O / CH4 02 C02 / H20) (2 pts) The reducing agent is (C / H / 0 CH4 Oz / C02 / H20) (2 pts) Question #4 (20 pts) (a) (8 pts) What are the oxidation states of (ii) oxygen in potassium peroxide ? _—’~____~_ K; 0 L (iii) nitrogen in the nitrite ion ? + 3 N 0 If (iV) carbon in the carbonate ion ? i c 9 3 ' (V) phosphorus in the phosphide ion ? 4 1° 3 ’ (b) (12 pts) Balance the following redox reaction, carried out in basic solution: 5b203 + [N03]' —-> [SbO4]3“ + NO / \ I \ +3 )2. +I5 -2. +1, .\_2, +2. -z. 0.1.- “(+2) —-» {505) + 24" x3 fad: N(f:)I-?¢—.——3N(4z) 3’ (M): 2:10;) -> 3:60;) J-é“ xz. 2 (“"02 2N(+5) 4- 60’ ——9 2A: (+2.) “0". 356 (+3) + z~(+s’) —a 2:4 (+:)+ 2~C+Z) Subtl‘i'hd‘: ?/2 5-5103 7‘ 2N0],— —-—9 35605‘3- + zuc —- 3‘ “UH/2,7 6’ 2: 3 $4,103 + H. N03 -——-> 6 573$”F ,4 five 3- 5414..“ 04074: 3 5610? + 4mg; 4 Hon“ ———> 6 $64? 4 lr-ua Ali/‘- 0H1 Lat/an“ h’: __ ' (.71. u‘f‘rl 3 5620; * 4"“; 1- may a 7:35;“ 55““, 3‘ 'f- ‘f-uo 4- 743° fuel. 4-4 4 6‘ 4h /al< q (fie-a1 aura“) Question #5 (10 pts) The valve between a 5-liter tank containing a gas at 9 atm and a 10-liter tank containing a gas at 6 atrn is opnened. Calculate the final pressure in the tanks. (1.161794) (find) it / 1}" 7" ‘4”: pa"; = 4‘9 (7419') (51%) =- 4; (I: m) 95nd fmfl~L/n¢;;uv‘ fan! f4: 1’ ‘ (74M)(J‘L'f) : 3m If”; :zzw 7'4: #2., (104?); Inf (0"?) a, 1; ?"M~L far/5‘64 f’Wu/a- 7 / (édw)(,.x#) 1 ._..———-————-—-‘ "-'— 444(- AVA“! 5:9 mhfu- ffaM/rmur¢ ‘: 3%4—9a‘u = 74/» Question #6 (15 pts) A solution, is prepared by dissolving 15.0 grams of sodium hydroxide in 150.0 mL of 0.250pitric acid (HNO3). Calculate the concentrations of all the ions (H+, OH‘, Na+, N03") present in the solution after the reaction has occurred. Assume no volume change on addition of NaOH. (Note: this is Problem #51 from Chapter 4) N401: I- have) = [Va/V03 + flzo /J' Am"! /.r' man. o7 M401] 2 ———2——-——-—-— 1" ’-—-- 9'37fho/4 (z)+l‘¢l)i/fluo/‘ 4’ g7 [V4011 [MD/44 ‘7 H‘”? = (0.1.70 lif)(d-2.Y0 “‘l") : 0.0?7rrm/4. 4n» 7 ’W ‘3 5° 7m 6514 A“ a“ #14" 1.4 1‘40: fi'm‘: a: mall. 45;" 4; add- (a haf‘ Axuil 47 Am MAG“! flu»! (A 444 «'4 #- A‘mikuf Icaf4uf 4«( will A‘ ‘1’ “4401 a?) Am” «1' H’ a. ’ Mia/1.: 0f" ital—f; 0-37; 0-37.5' 0.0375- 0,537; rho/u he!“ ran“: a -——0JJ7$' —-o,oJ7J' o [Mo/A4 MIR .l-«J 0.774“ 0.3375” 0 6.0375 (nah flu! N4’ 4““ 6" an 7‘«t+ file‘s/5h”) dutch/lulu“: A- »«(z Naf- ___ 4.37! Ada/0.,“ ‘7' 3 1.5.0” 0H = (“3374’ rho/- /o./_w 1:]- : 2.2514 H‘f' g a M [fin-55‘“, rag/#10 N03” 2:. 0~037f malty/0.];a/ff: a-ZS‘M Question #7 (12 pts) Nitric acid is produced commercially by the Ostwald Process. In the first step, ammonia is oxidized into nitric acid according to the formula: 4NH3(g) + 502(g) —> 4N0(g) + 61'120(g) Assume the reaction is carried out in the apparatus below, with the volumes and the starting pressure values of the reactants indicated as shown. The stopcock between the two reactants is opened, and the reaction proceeds. Calculate the partial pressure of H_20 1g) after the reaction is complete. Assume the whole apparatus is kept at a constant high temperature throughout the process, such that the H20 produced remains as a gas (water vapor) instead of a liquid. (Note: this question is practically identical to Problem 102 in Chapter 5) AV, (4.5’014M)(2L7‘) - a 3)] H '. F; r V;- _-_-, 3 - ‘1”. . AV (nta 4A.. [/L‘r) FwoL‘ ,0=——I: ) =d-5'aoabu ‘ V, 3 0+ {0 P! 0 J a -.ro 0 40h“ 144‘. u = a. " '——"—’ = /-50 40/4,) pat/.49) ' 0.3:? ffk‘x'n. Mia’u’r‘, 0;.‘U/(7 rMVJ H 6‘ 51")“ z ['25:] In a“. :4 f Am, a, I met Mr; as I2. 1mm“, rut“. {IM‘M fla" A’A.“ firm-n14 d] fiO/Iaéu‘l = filmy-7) x 9 rho/A. All!) = whim x i 4;. = 5.5—“ am 11199:) Question #8 (16 pts) A bottle with a volume of 4.25 liters is filled with xenon gas at STP (standard temperature and pressure). A small piece of dry ice (solid C02), weighing 2.75 g, is dropped into this bottle. (a) (6 pts) When the dry ice has evaporated, what is the partial pressure of C02 in the bottle? (assume that the temperature remains constant at 0°C). run/4r cf] (’0‘ : :- iZf : 0~0(2J' Ito/.4 :7 a; (Ill-(ltlé)j/mou 4‘“ 7. Ion/:- mAT, no ’3 hR/V 3" - .. (a.o(2:‘m.l¢)(a.af'z I “Mda In”. ’)(27.? K) F150,): ._________—__————l——————— 4.2.J'14‘t : 0.32? ab. (b) (2 pts) What is the total pressure of the gaseous mixture inside the bottle after the dry ice has evaporated? flaw“) a [(1‘) +/(co,.) = la.le o-Jzfi aha: /.327a6~ ____.._——-—-b _____——-——b (c) (4 pts) What is the mole fraction of C02? ‘f-ZJ‘ l-‘l‘ K : = 0' 4. rho/M I7 I. 22.“ A? (£97 mo/ 67 .(e. . awKzJ‘ Mo/ freq-m ‘7 C4 “ 0.0‘21 = —-———’_"' = 0-2‘/5 L 0-151r+9-/I77 0er2 (4v 29"!94) (d) (4 pts) What is the mass percent of C02? [mm 07 Xe : (o 4:77 Wok.) (lg/J'J/nn/c) 1' 24-7'1; . 4' . ‘ Wail 7.07 C0; 2 i—L—t : 2 74 = 7'770 2-734- 2947/ 27-“ _..__.. TABLE 4.1 Simple Rules for the Solublllty of Salts In Water I ,Most salts containing the alkali metal ions rammonium'ion (NI-14*) are soluble. .‘ Most chloride. bromide. and iodide salts are solublo.,Notab1o aim (1110;) salts are some. (Lii Nazi”. Ki, 9 excépiiéhé salts I the ions Ag+,‘sz+, and ngz”. " . 7 f: 5 ‘ sulfate salts are soluble. Notable exceptions . Most hydroxide salts are only slightly soluble. The sclublof ally soluble. ‘ are, NaOH. and KOH. The compounds Ba(0H)2, Mon», mid Comm2 an: . Most sulfide (s21, carbonate (cor), chromatc (0:023, anophosphoto (For) salts am only slightly soluble. Volume of oné mole of any gas at STP = 22.4 L R = 0.082 lit atm K“1 mole-1 PV = nRT P1V1/1"1 = P2V2/T 2 Constants and equations you may need 6 7 C N 12.0111 14.0067 25 26 27 Mn Fe Co 14 15 Si P 28.086 30.9737 28 29 32 33 Ni Cu G As 54.9380 55.847 58.9332 58.69 63.546 72.59 30 Z 11 65.37] 74.9216 138.905 43 44 45 46 47 Tc Ru Rh Pd Ag 98.9062 101.07 102.905 106.4 1 07.868 75 76 77 78 79 Re Os Ir Pt Au 186.2 190.2 192.2 195.09 196.966 Nb . 929064 72 73 178.49 180.948 04 1 58 60 62 63 64 65 66 67 68 69 70 71 Lanthanides S m Eu G d Tb Dy Ho Er Tm Yb Lu 144.24 150.35 151.96 157.25 158.925 162.50 164.930 167.26 168.934 173.04 174.97 92 93 94 95 96 97 98 99 100 101 102 103 Actinides Th U Np Pu Am Cm Bk Cf Es Fm Md No Lw 232.038 231.035 238.029 237.048 244 243 24 24 251 252 25 258 259 ...
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Exam 2 Fall 2005 - Chemistry 105a First letter of PLEASE...

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