Question

Question #f69e5

Answer 1

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 `"9 g cm"^(-3)`. This tells you that every milliliter of copper has a mass of `"9 g"`.

To use this as a conversion factor, rearrange it as

`"9 g"/"1 cm"^3 ->` when you need mass

`"1 cm"^3/"9 g" ->` when you need volume

So, to find the mass of `"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

`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 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

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