HW_1_solutions

HW_1_solutions - .3 an a"; HOMEWORK #1 Due:...

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Unformatted text preview: .3 an a"; HOMEWORK #1 Due: Monday, 10/9/06 CHEM 150 . How many grams are contained in 450. mL of 0.250 M methanol (CH3OH)? If]: 1t } 0” . A solution was prepared by dissolving 432 mg of K4Fe(CN)6 in water and diluting to 1.5 L. Calculate the KJr concentration in ppm. arm 11"” 1.2.5 ‘ ’ . Any dilute aqueous solution has a density near 1.00 g/mL. Ifa given solution is 1.0 ppm solute, determine the concentration of solute in ug/L. l. 0 K10} . What is the density of 53.4 wt% aqueous NaOH if 16.7 mL of the solution diluted to 2.00 L gives 0.169 M NaOH? 1. :1 3/341- . How many milliliters of3.00 M H2804 are required to react with 4.35 g of solid containing 23.2 wt% Ba(NOg)2 if the net ionic reaction is: Ba2+ + $0..” —> BaSO4(s). I _ 1m 2.. “UL {1. Write each answer with the correct number of significant figures. log(4.128x1012) = ? iii- iflff'? (La) 101:3? = ? .2 a; i :3) e' - =? Mm (a) 5.45/antilog(-3.22) = '2 .15. m; (A ; Write the equilibrium constant expression for __--. I ..-.-~. 2CrO42'+ 2H+ = Cr2072' + H20 ‘3 ' . Calculate the molar solubility of silver bromide (KSp = 5.2 x 10'”) in 0.0250 M KBr. /}. i .M n L ‘ “Iv-ii / I. \ >< w 1.1;. U' 9. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln Tunnel connecting New Jersey and Manhattan. The total concentrations (at standard deviations) of m— and p-xylene were: Turnpike: 31.4 at 30.0 ug/m3 (32 measurements) Tunnel: 52.9 at 29.8 ug/m3 (32 measurements) Do these results differ at the 95% confidence level? Do they differ at the 99% confidence level? Justify your answers. HO. 'l'tltef‘f CI 3 ova-l at]? 10. Calculate the pH of 0.010 M HC1,0.035 M KOH, 0.030 M HNo3 and 3.0 M HC1. (inform: EH :2 2.00 (“3.035HKDH PH :1 12.5% 0.336 ETHNQ 43H = L512 09M. #2 Lb 32 x (o’gfwréew : ‘l 6 ‘ + [WE-551m \ X Lf k > X / €750!” XlOé a 36339260 atmagegw‘ 6 122. 275034 10 ‘2 mfmkf‘ m v.9”) 'S-‘nnann: Smurf MAJ/2006 __ (R06 #5 ‘ CHEW“ KO Hka # 85kt) + 50%;) ——> mot/5) CASSEJJ 23.2%; lmL—W X [snark X (00% 5 Salvx m. #5 M0!“ Salu‘zlz‘ m 0. 0250 :1 K6, 2.35” 5) lo = 242, ma 6) e"W : 2.76m; no“ of) 5 #5 _ 5. £6 Ad—i’gfizzj g (6-122 amp—£22 Am) + erg) Kr = {mm : 52m“ 7 ‘ [A37 L "X" st = X(‘X+o.o.25o) = 571K104} x" +— 0.025% ~§.2X(o“’ : o /__5___ __.___ K 10. 3361064411192, WY mow—56$: : ml. HJSOY :0)“ s: a) lo}fq.nflto'z:( =12.61573‘{7 7: 2.61 x16y ) ‘Conflaf ‘Hfi ( L60!“ 2 mam L? SE4 VF“ = mm 390?; z SE3 = 10m;qu ; {omog .2 SE3 {laugh-n] Hrrmn‘ { y x = 2. 07%? m‘" Mal/L @XODJSo) X9 2.03m” 3 HMO—"H (2,3 EB} Tumrl’kQ I i 30.0 /|P\3 ( 52 weaaqwm‘h) Tam‘ 52.1: zaz/j/m’ Jr in “I”, (/2) [ ' # _,___...// .0 _.L.m7\1i;' Clam = T = 3” i lag/«JAM; Q 45% Co (I. , ; (2.75o)(3ao) ‘\ “(1/9 T: 31.4 .t (WM €14 A a» ew //___ __ 1mm: 7033222152 2 52.61 i 10. W WM ..._~-"" 31 ../" J; (I 524 r (4.5 jg/fi ‘7‘! X @ 66% (Mag/(%_ gawk/2*43141753): 42.2%) M ,/--- [TIM/Mel 7 >/((52.q—l0.$) : “WV/*3) / @ 77% (“MW [MW] 48w +14.e)= m 1%) ZTIAMM-e‘ >/((52.4(I—N.§): q’“%3) / _s_ \\ of _../,// W/ q—S-Z; 03" W (onffid’c‘tce Mm! (ML/mm I CNS 0?- {50 H wk. #1 - emu! prbé Calculate ‘He PH #3 flu €910“).an recau 1+ lg—IOgZHfl [Hflfou‘] 3 )0"q 4) 0010 H HCI f0” PIoétzofl 19”)“ F0" : ’4 0.010 m Ha : 0.0mm I+* 00H : 4009 [0.0m] sfizm 2.51:} b) 0.035 N KOH OH: "I . : .q55 : — .‘f : .9”: = .' P 0&f00353 I 7 W M I $54 1;? f c) 0.030 N HNQ garb #0040130) = I522” = A) 3.0 H Han pH: 400130) :‘0.‘(77l2 : ~045 25F? ‘ ar‘foafi_'a€' w/g was"? flab; N igw fiw. Ff.3,'f.7o ’ ' ...
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This note was uploaded on 08/06/2008 for the course CHEM 150 taught by Professor Buratto during the Summer '08 term at UCSB.

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HW_1_solutions - .3 an a&amp;quot;; HOMEWORK #1 Due:...

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