Respuesta :
Answer: 0.256 g
Explanation:
The deposition of Ga from Ga(III) solution
[tex]Ga^{3+}+3e^-\rightarrow Ga[/tex]
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]
[tex]I=\frac{q}{t}[/tex]
where,
I = current passed = 0.590 A
q = total charge = ?
t = time required = 30.0 min = 1800 s (1min=60s)
Putting values in above equation, we get:
[tex]0.590A=\frac{q}{1800}\\\\q=1062C[/tex]
[tex]Ga^{3+}+3e^-\rightarrow Ga[/tex]
[tex]96500\times 3=289500C[/tex] will deposit = 70 g of Gallium
[tex]1062C[/tex] will deposit = [tex]\frac{70}{289500}\times 1062=0.256 g[/tex] of Gallium
Thus 0.256 g of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.590 A that flows for 30.0 min.
