# Question #f69e5

##### 1 Answer

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 **every milliliter** of copper has a mass of

To use this as a conversion factor, rearrange it as

#"9 g"/"1 cm"^3 -># when you needmass

#"1 cm"^3/"9 g" -># when you needvolume

So, to find the **mass** of

#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

#color(green)(bar(ul(|color(white)(a/a)color(black)("mass of 5 cm"^3 = "50 g")color(white)(a/a)|)))#

To find the **volume** of

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

Therefore, you can say that

#color(green)(bar(ul(|color(white)(a/a)color(black)("volume of 63 g " = " 7 cm"^3)color(white)(a/a)|)))#