2 Answers

Machtyn · Answer 1 · 2017-11-06T05:51:01+0000

Answer : The amount left of leutium-176 will be, 2.10 g

Solution :

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

k=(0.693)/(t_(1/2))

$k=\frac{0.693}{3.85* 10^(10)\text{years}}$

$k=0.18* 10^(-10)\text{years}^(-1)$

Now we have to calculate the amount left of the sample.

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

$t=(2.303)/(k)\log(a)/(a-x)$

where,

k = rate constant =
$0.18* 10^(-10)\text{years}^(-1)$

t = decay time =
$1.155* 10^(11)\text{ years}$

a = initial amount of the sample = 16.8 g

a - x = amount left after decay process = ?

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

$1.155* 10^(11)\text{years}=\frac{2.303}{0.18* 10^(-10)\text{years}^(-1)}\log(16.8)/(a-x)$

a-x=2.10g

Therefore, the amount left of leutium-176 will be, 2.10 g

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