# The density of gold is 19.5 g/cm. What is this value in kilograms per cubic meter?

Mar 6, 2017

$\rho = 19.5 \times {10}^{3} \cdot k g \cdot {m}^{-} 3$

#### Explanation:

$\rho = 19.5 \cdot g \cdot c {m}^{3} = \frac{19.5 \cdot {10}^{-} 3 \cdot k g}{{10}^{-} 2 m} ^ 3$

$\frac{19.5 \times {10}^{-} 3 \cdot k g}{{10}^{-} 6 {m}^{3}}$

$\left(19.5 \times {10}^{-} 3 \cdot k g \times {10}^{6}\right) {m}^{-} 3$ because $\left(\frac{1}{x} ^ - 1 = \frac{1}{\frac{1}{x}} = x\right)$

$= 19.5 \times {10}^{3} \cdot k g \cdot {m}^{-} 3$

The price of gold is currently $1232-10*USD*"troy ounce"^-1. Given that a $\text{troy ounce}$$\equiv$$31.1 \cdot g\$, how much is your nugget worth?