Question

Half life of 35^S is 87.8d. What percentage of 35^S sample remains after 180d?

Answer 1

Answer:

1/4 Of intial amount

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Answer 2

The percentage of the sample remain is 24.17 %

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

We are given:

$t_{1/2}$ = 87.8 days

Putting values in above equation, we get:

$k=\frac{0.693}{87.8days}\\\\k=7.89\times 10^{-3}days^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,  

k = rate constant = $7.89\times 10^{-3}days^{-1}$

t = time taken for decay process = 180 days

$[A_o]$ = initial amount of the reactant

[A] = amount left after decay process

Putting values in above equation, we get:

$7.89\times 10^{-3}=\frac{2.303}{180}\log\frac{[A]_o}{[A]}\\\\\frac{[A]}{[A_o]}=0.2417$

To calculate the percentage of the sample, we use the equation:

$\text{Percentage of sample remain}=\frac{[A]}{[A]_o}\times 100\\\\\text{Percentage of sample remain}=0.2417\times 100=24.17\%$

Learn more about first order kinetics:

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