The general equation for radioactive decay is;
N = N₀e^(-λt)
x - decay constant (λ) - rate of decay
t- time
N - amount remaining after t days , since we are calculating the half life, amount of time it takes for the substance to to be half its original value, its N₀/2
N₀ - amount initially present
substituting the values
N₀/2 = N₀e^(-0.081t)
0.5 = e^(-0.081t)
ln (0.5) = -0.081t
-0.693 = -0.081t
t = 0.693 / 0.081
= 8.55
half life of substance is 8.55 days
Answer and Explanation:
The balanced chemical equations are as follows:
The chemical formula of oxalic is 
In the case when oxalic acts reacted with the water so here the oxalic acid eliminates one proton that leads to the development of mono acids
After that, the second step derives that when oxalic acid is in aqueous solution eliminates other proton so it represent the polyprotic acid
Now the chemical equations are as follows:
Elimination of one proton

Now the elimination of other proton

Answer:
i think the answer is C
Explanation:
not sure plsss dont bash im a beginner