# How do you get a missing measurement knowing only two measurements and density?

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The dimensions of aluminum foil in a box for sale in supermarkets are 66.66

yards by 12 inches. The mass of the foil is 0.83 kg. If its density is 2.70 g/cm^3, then what is the thickness of the foil in cm?

I know the two given sizes equal 185,806 cm and 30.48 cm and the mass equals 830g. So how do I find the final measurement of the sheet of foil?

The dimensions of aluminum foil in a box for sale in supermarkets are 66.66

yards by 12 inches. The mass of the foil is 0.83 kg. If its density is 2.70 g/cm^3, then what is the thickness of the foil in cm?

I know the two given sizes equal 185,806 cm and 30.48 cm and the mass equals 830g. So how do I find the final measurement of the sheet of foil?

##### 1 Answer

Here's how you can do that.

#### Explanation:

This is more of a *unit conversion problem* than anything else. The problem provides you with the length and height of the aluminium foil, but it does so in **yards** and **inches**, respectively.

Also, the mass of the foil is given in **kilograms**.

On the other hand, the **density** of aluminium is expressed in *grams per cubic centimeters*. This means that you're going to have to convert the length and height of the foil to *centimeters* and its mass to *grams*.

You will have

#66.66 color(red)(cancel(color(black)("yd"))) * (0.9144 color(red)(cancel(color(black)("m"))))/(1color(red)(cancel(color(black)("yd")))) * (10^2"cm")/(1color(red)(cancel(color(black)("m")))) = "6095.4 cm"#

#12 color(red)(cancel(color(black)("in"))) * "2.54 cm"/(1color(red)(cancel(color(black)("in")))) = "30.48 cm"#

#0.83 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = "830 g"#

Now, the trick here is to realize that the foil is actually a very, very thin **rectangular prism**.

As you know, the volume of a rectangular prism is given by the equation

#color(blue)(|bar(ul(color(white)(a/a)V = l xx h xx wcolor(white)(a/a)|)))#

Here

*length* of the prism

*hight*

Your goal here is to find the value of **density** of aluminium, which is said to be equal to

What that means is that **every** *volume*

#830 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(2.70color(red)(cancel(color(black)("g")))) ="307.4 cm"^3#

You now know that the volume of the foil is

#V = "307.4 cm"^3#

Rearrange the above equation to solve for

#V = l xx h xx w implies w = V/(l xx h)#

Plug in your values to find

#w = ("307.4 cm"^color(red)(cancel(color(black)(3))))/(6095.4 color(red)(cancel(color(black)("cm"))) * 30.48color(red)(cancel(color(black)("cm")))) = "0.0016546 cm"#

Rounded to two **sig figs**, the answer will be

#color(green)(|bar(ul(color(white)(a/a)color(black)("the width of the foil " = " 0.0017 cm")color(white)(a/a)|)))#