21.6 g of silver coin is dissolved in `HNO_(3)`. When NaCl is added to this solution, all silver is precipitated as AgCl. The weight of AgCl is found to be 14.35g then % silver in coin is:
`Ag + HNO_(3) overset(NaCl)to AgCl`

A 5.82 g silver coin is dissolved in nitric acid. When sodium chloride is added to the solution, all the silver gets precipitated as AgCl. The percentage of silver in the coin is :

x g of Ag was dissolved in HNO_(3) and the solution was precipitated. The value of x is

5.82 g of a silver coin were dissolved in strong nitric acid and excess of NaCI solution was added. The silver chloride precipitated was dried and weighed 7.20 g. Calculate the percentage of silver in the coin

A silver coin weighing 2.5 gm was dissolved in HNO_3 and then further treated with excess HCI. The mass of Ag CI formed was 2.99 gm. The percentage of silver in the coin will be:

A silver coin weighing 2.5 gm was dissolved in HNO_3 and then further treated with excess HCI. The mass of Ag CI formed was 2.99 gm. The percentage of silver in the coin will be:

Xg of Ag was dissolved in HNO_3 and the solution was treated with excess of NaCl when 2.87g of AgCl was precipitated. The value of X is

90 g of a silver coin was dissolved in strong nitric acid and excess of sodium chloride solution added. The silver chloride precipitate was dried and weighed 71.75 g. Calculate the precentage of silver in the coil (Atomic mass of Ag = 108) Ag+2HNO_(3)rarrAgNO_(3)+NO_(2)+H_(2)O AgNO_(3)+NaClrarrAgCl+NaNO_(3)

0.501 g of silver was dissolved in nitric acid and HCl was added to this solution, AgCI so formed weighed 0.6655 g. If the equivalent weight of chlorine is 35.5, what will be the equivalent weight of silver?

0.501 g of silver was dissolved in nitric acid and HCl was added to this solution, AgCI so formed weighed 0.6655 g. If the equivalent weight of chlorine is 35.5, what will be the equivalent weight of silver?