Question

$$\frac{5g/(40g\ mol^{-1})}{(045L)}$$

Answer

l/(360*m*o*L)

Solution


Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\frac{5\times \frac{g}{40gmo\times \frac{1}{l}}}{45L}\]
Simplify  \(40gmo\times \frac{1}{l}\)  to  \(\frac{40gmo}{l}\).
\[\frac{5\times \frac{g}{\frac{40gmo}{l}}}{45L}\]
Invert and multiply.
\[\frac{5g\times \frac{l}{40gmo}}{45L}\]
Cancel \(g\).
\[\frac{5\times \frac{l}{40mo}}{45L}\]
Simplify  \(5\times \frac{l}{40mo}\)  to  \(\frac{l}{8mo}\).
\[\frac{\frac{l}{8mo}}{45L}\]
Simplify.
\[\frac{l}{8\times 45moL}\]
Simplify  \(8\times 45moL\)  to  \(360moL\).
\[\frac{l}{360moL}\]