Question

When t1/2 is given to be 3 hrs , how many grams of a substance will remain after 18 hrs from 300 gm of a substance ?

Answer 1

Given that  

t ½  , half time  = 3 hr

No, initial amount =300 gm,  t= 18 hr  

t1/2 = 0.693/k

so, k = 0.693 /3 hr

= 0.231 / hr

Therefore  

N=No e^ -kt

N=300 8e^ - 0.231*18

= 300*e^- 4.158

= 300* 0.01563

=4.69  grams

Answer 2

4.6875 grams of substance will remain after 18 hrs from 300 gm of a substance.

Explanation:

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

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) to find rate constant:

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

$k=\frac{0.693}{3hrs}=0.231hr^{-1}$

b) to find out amount remaining after 18 hrs

$18=\frac{2.303}{0.231}\log\frac{300}{a-x}$

$1.8055=\log\frac{300}{a-x}$

$63.990=\frac{300}{a-x}$

${a-x}=4.6875g$

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