Question

half life of s35 is 87.8 d what percentage of s35 sample remains after 180 days​

Answer 1

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Answer 2

Percentage of S-35 sample remains after 180 days​ is 24.1

Explanation-

Half-life of S-35 = 87.8 days

First we have to calculate the rate constant, we use the formula

$k=\frac{0.693}{87.8\text{days}}$

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

Now we have to calculate percentage of s35 sample remains after 180 days

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

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

t = age of sample  = 180 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = a g

Now put all the given values in above equation, we get

$180=\frac{2.303}{7.89\times 10^{-3}}\times \log\frac{100}{a}$

$\frac{100}{a}=antilog(0.62)$

$a=\frac{23.9}{100}\times 100=24.1\ %$

Learn more about first order kinetics

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