Answer:
[tex]\large \boxed{\text{b. 1.5 h}}[/tex]
Explanation:
1. Calculate the rate constant
The integrated rate law for first order decay is
[tex]\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt[/tex]
where
A₀ and A_t are the amounts at t = 0 and t
k is the rate constant
[tex]\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}[/tex]
2. Calculate the half-life
[tex]t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}[/tex]
