A copper bar with a mass of 11.925 g is dipped into 241 mL of 0.131 M AgNO3 solution. When the reaction that o

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A copper bar with a mass of 11.925 g is dipped into 241 mL of 0.131 M AgNO3 solution. When the reaction that o

[From: ] [author: ] [Date: 12-07-04] [Hit: ]

The reaction is: Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag | . If 1 mmole Cu uses 2mmole AgNO3, then n2 = n1 /2 = 15.7855mmole Cu will be dissolved, so m = n*M = 15.7855*10^-3 * 64 = 1.......

The amount of AgNO3 present in reaction is n1 = 0.131 * 0.241 mole = 31.57 mmole.
The reaction is: Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag | . If 1 mmole Cu uses 2mmole AgNO3, then n2 = n1 /2 = 15.7855mmole Cu will be dissolved, so m = n*M = 15.7855*10^-3 * 64 = 1.01g of copper.That means delta m = 11.925 - 1.01 = 10.915g of copper will remain

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