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A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution are singly ionized gold ions, and you've calculated that you must deposit 0.560g of gold to reach the necessary thickness.

How much current do you need, in mA, to plate the statue in 4.70hours ?

Respuesta :

Answer:

ampere = 15.95 mA

Explanation:

given data

deposit gold = 0.560 g

time = 4.70 hour  = 16920 s

solution

we get here moles of gold that is

moles of gold = 0.560 g ÷ 196.967 g/mol

moles of gold = 0.0028

moles of gold  =  0.0028 ( 1 Faraday / 1 mol )

moles of gold  = 0.0028 Faradays

moles of gold  = 0.0028 Faradays  × 96500 Coulom/ faraday

moles of gold  = 270.2 coulombs  

and we get here ampere that is

ampere = [tex]\frac{270.2}{16920}[/tex]

ampere = 0.01596 A

ampere = 15.95 mA

The total current needed is 15.95 mA

Current:

Given information:

Amount of gold = 0.560 g

time taken to plate the statue t = 4.70 hour  = 16920 s

One mole of gold contains 196.967 g of gold, so the number of moles in 0.560g is given by:

n = 0.560 g ÷ 196.967 g/mol

n = 0.0028

Now, the gold ions are the charge carriers here, so to calculate total charge we proceed as follow:

Charge of 1 mole of electrons/positive ions = 96500 C

So total charge  Q = 0.0028  × 96500 C

Q = 270.2 C

Current is the rate of flow of charge:

I = Q/t

SO, I = 270.2C/ 16920s

Therefore, current I = 15.95 mA

Learn more about current:

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