Answer:
Im sure the answer is c
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The radioactive decay obeys first order kinetics
the rate law expression for radioactive decay is
![ln\frac{[A_{0}]}{[A_{t}]}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D%3Dkt)
Where
A0 = initial concentration
At = concentration after time "t"
t = time
k = rate constant
For first order reaction the relation between rate constant and half life is:

Let us calculate k
k = 0.693 / 72 = 0.009625 years⁻¹
Given
At = 0.25 A0

time = 144 years
So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**
Simply two half lives
In the electrolytic cell, depending on the polarity of the battery, either a more vigorous reaction (though the same as the voltaic cell) would occur, or the reverse would.
If the crucible wasn't covered with a lid the reactants may have produced a gas that was released into the surroundings, or mass may have been lost in the form of water vapour.