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When titrated in neutral solution with 0.05000 N I, a mixture of As203, As2O5, and inert material requires 20.10 mL. The resulting solution is then acidified, and excess KI is added. The liberated Iz requires 29.92 mL of 0.1500 N Na,S,03. Calculate the sum of the weights of As,03 and As205 in the sample.

When titrated in neutral solution with 0.05000 N I, a mixture of As203, As2O5, and inert material requires 20.10 mL. The resulting solution is then acidified, and excess KI is added. The liberated Iz requires 29.92 mL of 0.1500 N Na,S,03. Calculate the sum of the weights of As,03 and As205 in the sample.

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
Chapter18: Representative Metals, Metalloids, And Nonmetals
Section: Chapter Questions
Problem 25E: A chemist dissolves a 1.497-g sample of a type of metal (an alloy of Sn, Pb, Sb, and Cu) in nitric...
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. When titrated in neutral solution with 0.05000 N I2, a mixture of As2O3, AS2O5, and inert material requires 20.10 mL. The resulting solution is then acidified, and excess KI is added. The liberated Iz requires 29.92 mL of 0.1500 N Na,S,O3. Calculate the sum of the weights of As2O3 and As2O5 in the sample.
Transcribed Image Text:. When titrated in neutral solution with 0.05000 N I2, a mixture of As2O3, AS2O5, and inert material requires 20.10 mL. The resulting solution is then acidified, and excess KI is added. The liberated Iz requires 29.92 mL of 0.1500 N Na,S,O3. Calculate the sum of the weights of As2O3 and As2O5 in the sample.
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