A current is passed through a solution for 1.5 hours, and after this time period the mass of metal produced was 6.3 grams. What is the current, in amperes, that is required to produce such an amount of gallium

4.83A

Explanation:

The following were obtained from the question :

Mass of gallium deposited = 6.3g

Time = 1.5hrs = 1.5 x 60 x 60 = 5400secs

Gallium is a trivalent metal. This implies that it requires 3 faraday to deposit 1mole of gallium according to the equation

Ga^3+ + 3e —> Ga

Imole gallium = 70g

3 faraday = 3 x 96500C = 289500C

From the equation,

70g of gallium is deposited 289500C.

Therefore, 6.3g of gallium will be deposited by = (6.3 x 289500)/70

= 26055C.

Now we can find the current as follows

Q = It

I = Q/t

I = 26055/5400

I = 4.83A

the dependant variable is spilled water

the independent is the heat