The answer is:
E per gram = 0.45 V
The explanation:
when MnO2 is the substance who oxidized here so, the oxidizing agent and the anode here is Li.
and when the molar mass of Li is = 7 g/mol
and in our reaction equation we have 1 mole of Li will give 3.15 V of the electrical energy
that means that :
7 g of Li gives → 3.15 V
So 1 g of Li will give→ ???
∴ The E per gram = 3.15 V / 7 g of Li
= 0.45 V
Sodium/Atomic number
11
Gold/Atomic number
79
Potassium/Atomic number
19
Silicon/Atomic number
14
M = 22.8 g
V = 14.7 mL
ρ - ?
ρ = m/V
ρ = 22.8/14.7 = 1.55 g/mL
