The same quantity of electrical charge deposited `0.583g` of Ag when passed through `AgNO_(3),AuCl_(3)` solution calculate the weight of gold formed. (At weighht of `Au=197gmol)`

The same quantity of electrical charge that deposited 0.583g of silver was passed through solution of gold salt and 0.355g of gold was formed. What is the oxidation state of gold in this salt?

The same quantity of electrical charge that deposited 0.583 g of silver was passed through a solution of gold salt and 0.355 g of gold was formed . What is the oxidation state of gold in this salt ?

Same quantity of electrical charge that deposited 0.583 g of silver was passed through a solution of gold salt. If 0.335 g of gold is deposited, calculate the oxidation state of gold in the given salt. At wt of Au = 197.

_____ quantity of electricity will deposit 10.8 gm of ‘Ag’ at cathode from a AgNO_3 solution.

The same quantity of electricity that liberates 4.316 g of silver from AgNO_(3) solution was passed through a solution of gold salt. If the atomic weight of gold be 197 and its valency in the above-mentioned salt be 3, calculate the weight of gold deposited at the cathode and the quantity of electricity passed.

A current 0.5 ampere when passed through AgNO_(3) solution for 193 sec. Deposited 0.108 g of Ag. Find the equivalent weight of Ag.

A current 0.5 ampere when passed through AgNO_(3) solution for 193 sec. Deposited 0.108 g of Ag. Find the equivalent weight of Ag.

0.35 mole of electron was passed through AgNO_3 solution. Calculate amount of Ag metal deposited at the cathode.