Question

PLEASE HELP ME!!!!

Answer 1

Answer: B) about 7.4 hrs

Explanation:  

Rate law expression for first order kinetics is given by:

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

where,

k = rate constant

t = time taken for decay process =6 hours

a = initial amount of the reactant = 5.6 g

a - x = amount left after decay process = 3.2 g

Putting values in above equation, we get:

$k=\frac{2.303}{6}\log\frac{5.6g}{3.2g}$

$k=0.09hours^{-1}$

To calculate the half life, we use the formula:

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

$t_{1/2}=\frac{0.693}{0.09}=7.4hours$

Answer 2

Over a period of 6 hours, the substance went from 100% to about 60%, which is slightly less than one half life. One half life is therefore slightly more than 6 hours, with 7.4 hours being the most logical answer.