Answer: 116 g of copper
Explanation:

where Q= quantity of electricity in coloumbs
I = current in amperes = 24.5A
t= time in seconds = 4.00 hr =
(1hr=3600s)

of electricity deposits 63.5 g of copper.
352800 C of electricity deposits =
of copper.
Thus 116 g of Cu(s) is electroplated by running 24.5A of current
Thus remaining in solution = (0.1-0.003)=0.097moles
It is water soluble so is also soluble in aqueous solutions of NaOH or NaHCO3.
All phosphorus atoms have the same atomic number. Hope i helped
For this problem, the solution is exhibiting some colligative properties since the solute in the solution interferes with some of the properties of the solvent. We use equation for the boiling point elevation for this problem. We do as follows:
<span>
ΔT(boiling point) = (Kb)mi
</span>ΔT(boiling point) = (0.512)(1.3/2.0)(2)
ΔT(boiling point) = 0.67 degrees Celsius
<span>
T(boiling point) = 100 + 0.67 = 100.67 degrees Celsius</span>