# Question f69e5

Oct 23, 2016

Here's what I got.

#### Explanation:

The thing to remember about density is that it gives you a way to convert between the mass of a substance and the volume it occupies and vice versa.

In this case, copper is said to have a density of ${\text{9 g cm}}^{- 3}$. This tells you that every milliliter of copper has a mass of $\text{9 g}$.

To use this as a conversion factor, rearrange it as

${\text{9 g"/"1 cm}}^{3} \to$ when you need mass

$\text{1 cm"^3/"9 g} \to$ when you need volume

So, to find the mass of ${\text{5 cm}}^{3}$ of copper, write

5 color(red)(cancel(color(black)("cm"^3))) * "9 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "45 g"

Rounded to one significant figure, the number of sig figs you have for the volume of the sample, the answer will be

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{mass of 5 cm"^3 = "50 g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

To find the volume of $\text{63 g}$ of copper, write

63 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(9color(red)(cancel(color(black)("g")))) = "7 cm"^3#

Therefore, you can say that

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{volume of 63 g " = " 7 cm}}^{3}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$