# Question 32b93

Sep 19, 2017

The thickness of the foil is 0.018 mm.

#### Explanation:

Step 1. Calculate the mass

$\text{ Mass" = 12 color(red)(cancel(color(black)("oz"))) × (28.4 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("oz")))) = "341 g}$

Step 2. Calculate the volume

V = 341 color(red)(cancel(color(black)("g")))× "1 cm"^3/(2.70 color(red)(cancel(color(black)("g")))) = "126 cm"^3

Step 3. Calculate the thickness of the foil

The formula for the volume ($V$) of a rectangular solid is

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ }}{V = l w h = A h} |}$

So, the thickness $h$ is given by

$h = \frac{V}{A}$

A = 75 color(red)(cancel(color(black)("ft"^2))) ×((12 color(red)(cancel(color(black)("in"))))/(1 color(red)(cancel(color(black)("ft")))))^2 × ("2.54 cm"/(1 color(red)(cancel(color(black)("in"))))) ^2= "69 700"color(white)(l) "cm"^2

h = (126 stackrelcolor(blue)("cm")(color(red)(cancel(color(black)("cm"^3)))))/("69 700" color(red)(cancel(color(black)("cm"^2)))) = "0.0018 cm"

h = 0.0018 color(red)(cancel(color(black)("cm"))) × "10 mm"/(1 color(red)(cancel(color(black)("cm")))) = "0.018 mm"# (2 significant figures)

Note: The answer can have only two significant figures because that’s all you gave for the area and the mass of the foil.

I carried out all intermediate calculations to three significant figures (two significant figures plus a guard digit) to avoid accumulated round-off error and rounded off at the end.