Answer
See the answer below.
Work Step by Step
Plotting the data, of $ln[N_2O]$ vs t, we get the linear regression:
$ln[N_2O]=-0.0128\times t-2.3038$ with a $r^2=0.999$, so it's definitely a first-order process.
This regression leads to $k=0.0128\ min^{-1}$
The rate is given by:
$r=k[N_2O]=0.0128\ min^{-1}\times 0.035\ mol/L=4.48\times10^{-4}\ mol/L.min$
