HomeworkExampleSample: NaC1+ NaBr + inert = 1.000 gexcess AgNO; UAg: 107.87,Cl: 35.45AgCl(s) + AgBr(s)= 0.5260 gNa: 22.99Cl2 treatment UAgCl(s) + AgCl(s)Br: 79.90= 0.4260 gNaCl = ? % ; NaBr= ? %Solution if NaCl =x g.NaBr =y gi g of AgBr(s) = y×187.77102.89x×143.32g of AgCl(s) =58.44xx143.32+yx187.77= 0.5260 gxx100% NaCl =58.44102.891.000xx143.32yx143.32= 0.4260 gy×100% NaBr58.44102.891.000

NaCl + AgNO3 → AgCl + NaNO3 NaBr + AgNO3 → AgBr + NaNO3 Let the moles of AgCl formed be x and…

Answered: Br= ? % NaBr =y g i g of AgBr(s) =…

Br= ? % NaBr =y g i g of AgBr(s) = y×187.77 102.89 ×143.32 %3D 58.44 87.77 xx100 = 0.5260 g % NaCl 2.89 1.000 43.32 2.89 = 0.4260 g % NaBr yx100 1.000

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter10: Atomic Emission Spectrometry

Section: Chapter Questions

Problem 10.11QAP

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Question

Homework
Example
Sample: NaC1+ NaBr + inert = 1.000 g
excess AgNO; U
Ag: 107.87,
Cl: 35.45
AgCl(s) + AgBr(s)
= 0.5260 g
Na: 22.99
Cl2 treatment U
AgCl(s) + AgCl(s)
Br: 79.90
= 0.4260 g
NaCl = ? % ; NaBr= ? %
Solution if NaCl =x g.
NaBr =y g
i g of AgBr(s) = y×187.77
102.89
x×143.32
g of AgCl(s) =
58.44
xx143.32
+
yx187.77
= 0.5260 g
xx100
% NaCl =
58.44
102.89
1.000
xx143.32
yx143.32
= 0.4260 g
y×100
% NaBr
58.44
102.89
1.000

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