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The % w/w in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach end p AgNO3 + Agl (s) + NO3 Ag+SCN-AgSCN What is the % w/w in the sample? O 0.9400 119.3 17.76 0.6860

The % w/w in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach end p AgNO3 + Agl (s) + NO3 Ag+SCN-AgSCN What is the % w/w in the sample? O 0.9400 119.3 17.76 0.6860

Chemistry by OpenStax (2015-05-04)
Chemistry by OpenStax (2015-05-04)
1st Edition
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter15: Equilibria Of Other Reaction Classes
Section: Chapter Questions
Problem 79E: In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The...
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Question 8 The % w/w I in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach end p AgNO3 + Agl (s) + NO3 Ag+SCN-AgSCN What is the % w/w in the sample? O 0.9400 119.3 O 17.76 O 0.6860
Transcribed Image Text:Question 8 The % w/w I in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach end p AgNO3 + Agl (s) + NO3 Ag+SCN-AgSCN What is the % w/w in the sample? O 0.9400 119.3 O 17.76 O 0.6860
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