Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
Molarity can be defined as the number of moles of solute in 1 L solution.
Molarity of Na₂SO₄ solution - 0.200 M
this means there are 0.200 moles in 1 L solution
Molar mass of Na₂SO₄ - 142 g/mol
therefore mass of Na₂SO₄ in 1.00 L - 0.200 mol x 142 g/mol = 28.4 g
a mass of 28.4 g of Na₂SO₄ is present in 1.00 L
A. We can calculate the initial concentrations of each by
the formula:
initial concentration ci = initial volume * initial
concentration / total mixture volume
where,
total mixture volume = 10 mL + 20 mL + 10 mL + 10 mL = 50
mL
ci (acetone) = 10 mL * 4.0 M / 50 mL = 0.8 M
ci (H+) = 20 mL * 1.0 M / 50 mL = 0.4 M (note: there is only 1 H+ per
1 HCl)
ci (I2) = 10 mL * 0.0050 M / 50 mL = 0.001 M
B. The rate of reaction is determined to be complete when
all of I2 is consumed. This is signified by complete disappearance of I2 color
in the solution. The rate therefore is:
rate of reaction = 0.001 M / 120 seconds
rate of reaction = 8.33 x 10^-6 M / s
