Which of the following pieces of deposited if a current of 10A was passed through a solution of copper(ll) salt for 965 seconds? (1F = 96500 C)

ayozy1234

4 years ago

Q=IT, 10*965=9650c
cu+2e=cu^2+
2F contain 64g of cu
2*96500=193,000
193,000 contain 64g of cu
9650 will contain 9650*64/193,000
= 3•2 mass of cu deposited
mole= mass of cu deposited/ m.m
mole=3.2/64
=0.05 mole

Mrnoah

1 year ago

To find the number of moles of copper deposited, we can use the formula:

m = Q / (F×z)
where:

m = mass of substance deposited (in grams)
Q = total electric charge passed (in coulombs)
F = Faraday's constant (96500 C/mol)
M = molar mass of the substance (in grams/mol)
z = number of electrons transferred per ion

Given values:

Q = current (I) × time (t) = 10 A × 965 s = 9650 C
F = 96500 C/mol
M = molar mass of copper (Cu) = 63.55 g/mol
z = 2 (since copper(II) ions have a +2 charge)

To find the number of moles of copper deposited, we can use the formula:

moles = Q / (F × z)

Given values:

Q = 9650 C
F = 96500 C/mol
z = 2

Now, let's plug in the values:

moles = Q / (F × z)
= 9650 C / (96500 C/mol × 2)
= 9650 C / 193000 C/mol
= 0.05 mol

Therefore, the number of moles of copper deposited is 0.05 mol.

Junetwek

5 years ago

vvggg

kikidaniels

5 years ago

there is no formula for the copper salt amd theres mo molar mass