the percent silver of the dime in the oxidation reaction
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For my AP chemistry lab, I was supposed to find the percent of silver in a dime. I started out with the formula: Ag + NaCl → AgCl +Na. The dime was put in a covered beaker with NaCl. However, during the experiment oxygen was added as one of the reactants, accidentally. The precipitated silver chloride was obviously contaminated with oxygen, suggesting an oxidation reaction took place. The dime weighed 2.522g, but the weight of the precipitated silver chloride was 6.00g. What is the new formula for the reaction, once the solution is oxidized? How can I find out the percent silver of the dime in this oxidation reaction?
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Solution Summary
The solution finds the percent silver of the dime by the oxidation reaction. The oxidation reactions for a percent of silver is given.
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plz see attachment
Solution:
The dime weighed 2.522g and the weight of the precipitated silver chloride was 6.00g.
Ag + NaCl → AgCl +Na
With the help of above equation we can write that
143.4 gm of AgCl will formed by = 107.8 gm of Ag
Therefore ...
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