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Answers to Problem Set 5 Part 1

Answers to Problem Set 5 Part 1 - 5.32 Deuhle mews are not...

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Unformatted text preview: 5.32 Deuhle mews are not used fuI line reaction of: strong acid trim water because the rescfinu is nut in equilibrium. These are Imt reversible readiness, i.e., line resume medium has plastics“)? no tendency to DCCUI. 5.43 [a] {b} {E} {d} 5.44 Hgflsfig 5.53 (a) (b) (1:) {d} hypuehlamus acid sodium hypeehlun'te NsGC] ioduus acid sodium iudite Nslflz bmmix: acid sodium hremste NaBrD; peathlurie acid sodium perchlmate NECK); mm: Cu2+{aq] + swing} + 333mg]. + 2Cl'{ag} —> Ease-4m + Cu2+[ag} + 2C1'I[aqj net Bafimg} + 304mg) —> Rams) Fei’caql + 3N031aql+ 31mg) + 30mm —> {DHJfi} + 3mm + Home net Fefifiaqj + 3GH‘(an —> FefiflHhIEsJ firming) + HOE—{M} + 3Ca2+[aq} + mag) —> CagflPD4HsJ + ISNa+[an + ISCHan net sea-“(cu + Haring) —> Cassatt-412m mammal: sway 24':an + magmas —> mm: + music-{tam + Agzsm net EAE (Hg) + S [aql —> H—gESC-F} 5J3 HCHCIfiaq] + H1011!) :1 H301”) + mug-(aw 5.8? (a) (b) (1:) molecular. NaHSDfiaq) + HBIQJgJ —} 301(k) +NsBI(Hq} + H200!) ionic: Name) + HSGa'iaq} + Hide) + Br'iaq} -> is) + Hams) + BITan + Hnflm net REDS—{:19} +HTGE} -> 302(3) + Hflfl} molecular Wing} + maoucaqj —> 2mm +Na2C03[ag} + mm) ianic: 3111.1an + curling} + 2Na+(aq} + 201mm —> 2mm + 2Na+[ag} + CDszTag} + 2mg net NH4+(aq] + 0mm) —} mg} + Hzflii} molecular momma) + Hamming} —> Bail-33m + ENHst} + EHzoifl ionic: 21411.1an + coflaq) + Bafltag} + EGH'iaq} —> Baa-33m + mg) + 21-1208] net ENE—(m) + cos-{W} + Bfiaq) + 2011qu) —> Bacogm + ENHyEg} + smog) LMPIEI 1'1 II'E ([1] molecular FeSfis] + EHCKHQ} —} FeClgfisg-jl + Hfitfgj ionic: FeSm + 21mm) + 2011an —> Feflmgj + zenaq) + Hflg) net FeS{sJ + 2mm) —> Fez’mq) + Hgfifig] 5.95 (a) H1304—>2H++ 5041“ # 1:11:11 H13014 = 3.00 gstm [fl] = 0.0301r mnl H2804 011.00 g 11230., . 0.030? 1011le304 1000110.H151:14 MHSG 1nan= — — =0.0010MH30 2 45'” [45011011210011 I 1LHISI04 ” ‘ 1:0} Fan-:13); —1 Fe!— + 2011'}; . = 2.0 x103 [:10qu100311 1000 1111103010311] = 1110110103); 51:11:1th [—1213 ELF 21010332 I—ILF 0115103): 0.11 111050103): (1:) Hill —> I-l+ + Cl’ . _ 1.05111111ch _ MHClsoluun-n— [ 2.16LHCI] 0.154MH01 {11] Hill —} F + Cl’ 1 111101 HC] 1+ mul H01 = 10.0 HCl — = 0.404 1HC1 g [35.40 g HCl] m” . 0.494 ml HCl M HCI. SfllutIfl-II. — [m] — 1.32 M HCI. 5.103 Ml‘Jl = M2V2 1U: = [“1“] 111I 1f2= (1.5 M HCIIHD 11.1.): 1110 1111. HE] 1.00 M H01 The 12110 IIJL 0fHCl must be diluted to 130 IIIL. The volume ufwater to be added is: 1812! mLusz— llflmLof‘Ul =00. meater 5.11?| Nai-IDD:+HC1—>N3Cl+ H10+Cflz {1.052 ml HCl X 111.1~.111~13H1:1:13 111101510111003 111110] HCl 111.111103H01013 = 0.11 g1~111111303 ...
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