Chem Exam 1-9/14/16. Frequency. Grade Average = 72, Median = 72, s = 20

Transcription

1 Chem 53 - Exam - 9/4/ Frequecy 0 Grade Average = 7, Media = 7, = 0

2 Exam Chem 53 September 4, 065 Quetio, 7 poit each for quetio -4 poit for awerig quetio 5 correctly DO NOT OPEN THIS EXAM UNTIL YOU ARE INSTRUCTED TO DO SO Pleae prit your ame o the catro o Lat Name, Firt Name o That all that eeded Sit i every other eat a itructed Book & Bag i the frot of the room. No text etry calculator. Ue the exam a cratch paper. Keep the exam whe you are doe. Tur i the catro. x i x i i x i _ x x _ t y ( x ) e z x F t calculated x x pooled pooled d. f

3 x x t calculated / /.. f d

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5 ] Cocetrated perchloric acid ha a molarity of.7 M ad a ma percetage of 70.5%. What i it deity? MW HClO 4 i g/mol. ] What i the cocetratio i ppm of a M olutio of KCl (MW = g/mol)? 3] A olutio ha a deity of.6 g/ml. What i the molarity of 6. molal of that olutio? The olute ha a molar ma of ] A olutio ha [H + ] = M. What i the ph of that olutio? 4 5] Stadard Deviatio i a meaure of 5 a) accuracy b) how cloe the mea i to the true reult c) the mea relative to the true reult d) preciio e) preciio ad accuracy 6] Two et of meauremet were made by differet techicia. The firt ha a mea of 55.6 ppm with a tadard deviatio of 7.3 ppm over 7 meauremet. The ecod had x = 6. ppm with = 8.5 ppm over 6 meauremet. Are the two tadard deviatio igificatly differet from each other? 6 7] A fial quatity, D i calculated by the ratio of D = H/G. If H wa meaured 6 time with a mea of 987. gram ad a tadard deviatio of.9 gram ad G had x = 554. liter with a tadard deviatio of 3.7 liter over 0 meauremet. What i the abolute ucertaity of D? 7 8] Calculate the limit of detectio of Method A give the calibratio curve below. Alo ote that the curve ha 9 data poit each replicated 5 time. The data poit at the lowet cocetratio ha a tadard deviatio of 0. igal uit. 8 9] Replicate ru of a aalyi gave 5 value of.77,.45,.9,.85 ad.8. Ca ay of thee value be dicarded with 95% tatitical cofidece? 9 0] Replicate ru of a aalyi gave 5 value of 9.88, 8.9, 9.6, 9.33 ad 9.7. What i the 95% cofidece iterval of thi et of data? 0 ] Whe i it appropriate to calculate pooled for two et of data? a) Whe the tadard deviatio are tatitically the ame

6 b) Whe t-calculated > t-tet c) Whe the tadard deviatio are tatitically differet d) Whe F-calculated = F-table e) Whe the tadard deviatio are ot equal ] Two differet method of Fe aalyi were compared to a NIST tadard cotaiig 6.50% Fe by ma. The reult follow: Method %Fe 6.33% ± 0.3% Method %Fe 6.55% ± 0.45% Which of the followig tatemet i true? a) Method i le precie ad le accurate b) Method i more precie ad le accurate c) Method i le precie ad le accurate d) Method i more precie ad le accurate e) Method i more precie ad more accurate 3] Trace aalyi were coducted o a ample 5 time. The average cocetratio wa foud to be 0.0 ppb with a tadard deviatio of 5.0 ppb. What i the chace that a igle aalyi will yield a reult that i twice thi average? 3 4] A aalyi for lead i groudwater wa coducted. What i the correct term for the lead ad the water? 4 a) Lead i the ample ad the groudwater i the aalyte b) Both the lead groudwater are the aalyte c) Both the lead groudwater are the ample d) Lead i the matrix ad the groudwater i the aalyte e) Lead i the aalyte ad the groudwater i the matrix

7 (.7 mol/l) (00.46 g/mol) ( L/000 ml) (00g ol/ 70.5 g acid) =.67 g/ml (3.5e-4 mol/l) ( g/mol) (L ol/000 g) 0 6 = 6 ppm 3 Aume kg of olvet I kg of olvet kg olv. (6. mol/kg olv.) (00.0 g/mol) = 6 olute 6 g olute g olv. = 6 g olutio Vol olutio = 6 g (ml/.6g) ( L/000 ml) =.390 L Molarity = 6. mol /.390 L = 4.40 M 4 ph = -log[h + ] = -log ( ) = = d) preciio 6 Ue F-tet = 8.5 ad = 7.3 F = 8.5 /7.3 = df = 6- = 5 df = 7- = 6 F-table 4.39 F < F-table o td. dev. Are ot tatitically differet. 7 (%) (%) (%) (%)... 3 (%) =.9/987. * 00 =.% (%) = 3.7/554. *00 = 5.90% t(%) = ( ) / = 6.0% D = 987. g/ 554. L =.78 g/l 6.0 % of.78 g/l = 0.07 g/l 8 LOD = 3/m = 3(0. igal uit)/ igal/coc = 0.8 coc. uit 9.77,.45,.9,.85 ad.8 mea =.76 = ? Ue Grubb Tet G = / 0.8 =.73 G-Table for = 5 i.67 G > G-table,.45 ca be dicarded , 8.9, 9.6, 9.33 ad 9.7 mea = = d.f. = 4 t =.776 _ t x = x ±.776(0.36)/5 ½ = = 0.45 a) Whe the tadard deviatio are tatitically the ame B) Method i more precie ad le accurate 3 z x = 0-0/5 = look up o z-table. Area = Area above = = 0.07 or.7%

8 4 d) Lead i the aalyte ad the groudwater i the matrix