HW_1_solutions[1]

HW_1_solutions[1] - “b1 CHEM 150 HOMEWORK #1 Due:...

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Unformatted text preview: “b1 CHEM 150 HOMEWORK #1 Due: Thursday. 6/25/09 . How many grams are contained in 450. mL of 0.250 M methanol (CH30H)? 31003 . A solution was prepared by dissolving 432 mg of K4Fe(CN)6 in water and diluting to 1.5 L. Calculate the K+ concentration in ppm. 12° N“ . Any dilute aqueous solution has a density near 1.00 g/mL. If a given solution is 1.0 ppm solute, determine the concentration of solute in ug/L. 1.0 XIO3M3/L. . 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? 152 c3/ml. . How many milliliters of 3.00 M H2804 are required to react with 4.35 g of solid containing 23.2 wt% Ba(NO3)2 if the net ionic reaction is: Ba2+ + SO42“ —> BaSO4(s). 1.21MQ . Write each answer with the correct number of significant figures. iog(4.128x1012): ? 12. H51 1023?4 = ? 24?. em“ 2 ? 2.132165 I 5.45/antilog(-3.22) = ? 101 go“ . Write the equilibrium constant expression for 2cro42'+ 2H+ = crzofi' + H20 k .- [Cr20?l-] E . Calculate the molar so ubility ofsilver bromide (Ksp = 5.2 x 10'”) in 00250 M KBr. 2.1x my" a 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 (fl: standard deviations) of m— and p—Xylene were: Turnpike: 31.4 i: 30.0 ug/m3 (32 measurements) Tunnel: 52.9 i 29.8 ug/m3 (32 measurements) Determine the 95% COnfldence interval for each experiment? Does the level of xylene differ in the two tunnels at 95% confidence? Justify your answer. 3% 10. Calculate the pH of 0.010 M HCl, 0.035 M KOH, 0.030 M HNO3 and 3.0 M HCl. 0.01014 Hm Flt = 2.00 mm m w m4 0.03m l-NQ fH= 1-5?— 132 am Mu fir: -— 0-48 ISO “Wit 1 Sokm'k‘lén“ j) 0. 450154,“ ons'onbmm K 32-04I9c1tgoH i £3.14 .1 Mal MQM'I' [3&02 HeoHl 0-. 2) 0.4323 macaque Y 1.5 isaln 5LR.?,'+Z> 3K1FBCU°£ l. mflfiMt 1 310?? )6 1 ,QJQL, 7! ID ; 1. le :03 we“ 1 3K Ll) ZINX cgllukcl 33‘,” ‘L W 7( It?“ 5 i 9L l Mal 5&4 jN‘MI 1L.¥ Mk 5| ‘4. Jn‘q m "P 53 Hz 50a} + ’EMINO332 ___§ 73550“;er ZH++ 1MB“ ‘ EMMA 5' “‘9‘ 3w“ 1 ml ML} . S 5 k \ 25.25 3 L g 4 3 f] a In 7K 100 3391.1 ngl. sugggmfll X 1 kamg)‘; )4 3g 103 ML Sula : 3'” M‘ “NH 1 12m q C0») [03 (4.128%:0'23 s. |2_Q|51 2.33‘1 ’0 :- 24?. 4%le- “S ’3’ -'-’ 2.01110 €3.45 F “is ” 5-4? “ games “LA (-31") d 104'” _ Lean!" ' 1» K : ECrZO 23 ,._.._—_—————.. a! - [Cr 042’ “if 8» m : AS+Cfi3 + ’Brpcc‘l‘x * =15 qfi. 0.0m £1 5 X X+D.oz~m K3? miss] aw ; g1 t 4?“— 3\ an; :3 2.040 (—Fomé L1 ahp‘mfis Lew 3mm cm 40 Arm TMnul :2 :- 52-9 “Vi 5-:- Z‘L? “3/”? N: 32. (MS. :— ei P522349 q+ fisz mtw]; 93-7, oz. :..— 52.9 i 5&3 : 52.3 i {2,34%}(293) {3: (157001 : 5251‘: 11.13.? M3/m5 'r‘lm Jm was gm on warm (ML m ior 4% Magic; 3L4+M = 42.2%: Ma 4L low end gr- M ‘32.OI~ “3.4-: L-[2_;_""§/Mi\ \mxl. ohwbb (km-C"? QW‘R? . “HM:ka ’HM. o;— $11;ng {A L 41:00 KAN-0.8. lo\ 0.010 M Hg} a [W]: 0.01M :5) \EH: 2.00 "In! 3 0"” H K” =3 [W]:- 01°3m=3 W]: fl: 2.6mm}l3 1m 0.03’OH Hm3 z?) LH+3=B£3°H =5 3.0H PM :3 LHf]: 3.0H a ‘0-435 ...
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HW_1_solutions[1] - “b1 CHEM 150 HOMEWORK #1 Due:...

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