Step 1. Calculate the mass
#" 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
#color(blue)(barul|stackrel(" ")(V = lwh = Ah)|)#
So, the thickness #h# is given by
#h=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.
