# Question #faa56

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you can use a substance's **density** as a **conversion factor** to go from *grams* to *volume* and vice versa.

Simply put, if you know the density of a substance, you can use it to determine the **volume** of a given *mass* of said substance. Likewise, if you are given the *volume* of the substance, you can use its density to determine the **mass** of the sample.

You're dealing with a *rectangular prism*, which means that you can use its dimensions to find its **volume**

#color(blue)(|bar(ul(color(white)(a/a)V = "length" xx "height" xx "width"color(white)(a/a)|)))#

In your case, you will have

#V = "3.00 cm" xx "2.50 cm" xx "1.80 cm" = "13.5 cm"^3#

So, this substance is said to have a density of **every**

Use this as a conversion factor to get the **mass** that will occupy a volume of

#13.5 color(red)(cancel(color(black)("cm"^3))) * overbrace("3.14 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(purple)("the given density")) = color(green)(|bar(ul(color(white)(a/a)"42.4 g"color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.