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A sample of impure tin of mass 0.530 gg is dissolved in strong acid to give a solution of Sn2+Sn2+. The solution is then titrated with a 0.0448 molL−1molL−1 solution of NO3−(aq)NO3−(��), which is reduced to NO(g)NO(�). The equivalence point is reached upon the addition of 3.91×10−2 LL of the NO3−(aq)NO3−(��) solution.   Part A Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents

A sample of impure tin of mass 0.530 gg is dissolved in strong acid to give a solution of Sn2+Sn2+. The solution is then titrated with a 0.0448 molL−1molL−1 solution of NO3−(aq)NO3−(��), which is reduced to NO(g)NO(�). The equivalence point is reached upon the addition of 3.91×10−2 LL of the NO3−(aq)NO3−(��) solution.   Part A Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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A sample of impure tin of mass 0.530 gg is dissolved in strong acid to give a solution of Sn2+Sn2+. The solution is then titrated with a 0.0448 molL−1molL−1 solution of NO3−(aq)NO3−(��), which is reduced to NO(g)NO(�). The equivalence point is reached upon the addition of 3.91×10−2 LL of the NO3−(aq)NO3−(��) solution.
 

Part A

Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents.
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