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HW_1_solutions

# HW_1_solutions - DW HOMEWORK#1 Due Monday CHEM 150 K How...

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Unformatted text preview: DW HOMEWORK #1 Due: Monday, 10/9/06 CHEM 150 K? . How many grams are contained in 450. mL of 0.250 M methanol (CH3OH)? [email protected]‘% “c5” . 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. Eff» Kt llo . 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. 1.0 200’ AM— . 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? 1.52 ML . 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). l. 201 ML “450% . Write each answer with the correct number of signiﬁcant ﬁgures. log(4.128x1012) = ? "3-5157 (‘4) 102~384=? 2cm (3) e'10'45 = ? 2‘1 “0*? (1) 5.45/antilog(-3.22) = ? {o {‘03 (2 ) . Write the equilibrium constant expression for EC,- 0 2-] a. 7 2CrO42'+ 2H+ = Cr2072' + H20 k? 3 [My ”+1 2 . Calculate the molar solubility of silver bromide (KSp = 5.2 x 10'”) in 0.0250 M KBr. :2-I _, ﬁx I0 I “'L 203.3r 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. l’IO. Hun CI': OVCTllf 10. Calculate the pH of 0.010 M HCl, 0.035 M KOH, 0.030 M HN03 and 3.0 M HCl. Oﬂlbﬂl‘l‘Cl f)” :2 2.00 0.033an PH: )2.“ 0.03:. MM 5H = Liz 0.3.0 n ”q. (all 31 "0.513 r -.-.---33 I 5.-:- 5: ins Ere EMF 5. sot-m: , !-'/..'u=s.? !-'.-:- SHEETS EYE-EASE 5 Etna I wﬁ‘ahﬂﬂaf Branﬁ -I:-.m so -h5—-EET'.-y' L're-E—m—i: 351'J‘.L'-51E [0195112006 I CHEW (50 HWLK #L @E ’/ t « I M20" AM 32, 0 me} new 3' 604% 0.299 ”91% o. 94505—94» We” W , H— ‘ x lw 'X )W W“ W (2;?g3-6 co» 09M. #2 Lb 32 x (O’Wréew : ‘l 6 ‘ + [WE-551m \ X Lf jq'wqﬁso k > I??? X / €750!” XlOé a 36339260 fkvaFeéN)‘, 6 122. 275034 10 ‘2 WKMKY‘ m (2 v.9”) ill-F“. 1:91:90- ”At-Nations}! 'Brana' 1';- [5..qu [0/01/1006 __ ‘ CHEM Is’o HM (PM #5 3176) + my ——> we) LUSH/Y 232%fLF1m4—se— (L50!n K 161 3362’OWX ,W X 30034—565: M. #5 MWJ-iﬂ—M Afar) + emf) MOI“ Salu‘zh‘? : [Ag—7 ‘1 “X” in 0. 0150 11 K6, #) ‘Conflaf =-—"1.257z;zz f :8“ I. 24 ML H360Y so!“ s: a) IO}I‘J.m/Xlon:( = 12.61573‘17 = )2.6(S7 L? SE4 VP“ 5) 20235” = 242. ,qu = 242 3 SE; c) e"W : 2.7?983x10-S- 7: 2.61 no“ 1 51E; 03.3%,) 5%; _ 5% M‘ “(ff-522] lam — w“; 7% :jﬂﬂLhé—L. -13 M» = Warm ; 522% M = X(‘X +0.02%) = 572K158 )(2‘ +— 0-0250): “SZMO‘U : O f 3“,. E» I ; 1 5L _ 42-38? mil 5h wﬂarmnal Brand mm m :5.- x = 2. 07%? Mo‘" Mal/L = 5.2 M603 @Xaono) x9 2.03m“ 3 2.rx(o’”n (mm) Tumrfh: I 31-“ i 30.0 ”J /m3 ( 32 Manrwa{() TENT : 52.1: zaz/j/m’ W 4—: WT , , 3‘” / (lomﬂé) ' ' :Tz 3/.‘1i/05/J/W3Q‘Uzco : Egg-egg: 31H : @IV.6/u//m3 €74%Cm "W _s_ If 1““! : 33W; 70‘4‘2’2Z2w} : 5.2.61 1“ [0, 3y ,y/M; (aw—”F M J; ‘7‘! X @m (MU/W [what] @910.X):492’/’> .9er / //€/JZTW¢FI 7>/§(524 20\$):‘4'9W/3) @ qu CowCioXeuceZTmrm€i\‘1]é/(5[ q +146);- Lt/(O 14%,) m [Tuml ”Z >/((52«J- M): 35 q :43) V VPMM 5w w/ WK or 777% (GAfFoXETcFTTMT He ' M‘s—W €329; ZJFWMH-Miw Mm! r .i2-33‘ 5:] 3P'CET; Ef: FJSE '-ﬁD‘.."U-‘.E , «2 ha) “3': :5: Isms: m E 54-55 a saw-me "I Haunnaf Brand 4:: H1 .1:-.1 SHEE IS- 5'.'E.|:- r-SE =_- :::;I_I.wrﬁ (ML/me I CM 0?- {50 H wk. #1 - emu! prbé #{0 Calculate ‘He PH #3 ﬂu €910“).an recau 1+ €‘IOgZHq [Hﬂfou‘] 3 )0"q 4) 0010 H HCI f0” 5-[06120H‘1 19”)“ F0" : ’4 0.010 m Ha : 0.0mm I+* 00H ; "06g [0.0m] sﬁZOO ] 2.51:} b) 0.035 N KOH OH: "I . z .1455 : — .‘f : 2.5””: =(/ .' f P 0&f00353 I 7 67H M I 354 1 :23: c) 0.030 N Mg, 00% #0040130) = I522” :- J) 3.0 H Han pH: 400130) :‘0.‘(77l2 : ~045 25F? ...
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