# Elemental mercury has rho=13.6*g*mL^-1. What is the density expressed in kg*m^-3?

Feb 21, 2016

This is dimensional analysis, and we use the $\text{centi}$ $=$ ${10}^{-} 2$, and the $\text{kilo}$ $=$ ${10}^{3}$ prefixes.

#### Explanation:

$13.6$ $g \cdot c {m}^{-} 3$ $=$ $13.6$ $\cancel{g} \times {10}^{-} 3 \cdot k g \cdot \cancel{{g}^{-} 1} \times {\left({10}^{-} 2 \cdot m\right)}^{-} 3$

$=$ $13.6$ $\times$ ${10}^{-} 3 \times {10}^{6} \times k g \times {\left(m\right)}^{-} 3$

$=$ $13.6$ $\times {10}^{3} \cdot k g \cdot {m}^{-} 3$.

This is a lot of mass; of course, mercury metal is pretty dense stuff. I think I am right in this analysis. If you (or someone else) finds an error, let me know.