Answer
See the answer below.
Work Step by Step
a) Since doubling H2O2 concentration doubles the initial rate, the reaction is of first-order:
$r=k[H_2O_2]$
b) In the first line:
$5.3\times10^{-5}\ mol/L.min=k\times0.05\ mol/L$
$k=0.00106\ min^{-1}$
c) Previously $r=\Delta[O_2]/\Delta t$, now $r'=-\Delta[H_2O_2]/\Delta t=-2\dot{}\Delta[O_2]/\Delta t=2\dot{}r$
therefore:
$k'=2\dot{}k$
