Answer
Please see the work below.
Work Step by Step
We know that
$K.E=\frac{1}{2}mv^2$
We plug in the known values to obtain:
$K.E=\frac{1}{2}\times \frac{44.02g}{1mol}\times \frac{1mol}{6.022\times 10^{23}mol}\times \frac{1Kg}{1000g}\times (\frac{379m}{1s})^2$
$K.E=9.249\times 10^{-2}\frac{J}{mol}$
