Question

A sheet of aluminum (Al) foil has a total area of 1.000 ft2 and a mass of 3.636 g. What is the thickness of the foil in millimeters (density of Al =2.699 g/cm^3)?

Answer 1

The thickness of the aluminum foil is approximately 0.0042 mm.

Step-by-step explanation:

To find the thickness of the aluminum foil, we can use the formula density = mass/volume. First, convert the area of the foil from square feet to the square centimeters. Since 1 ft² is equal to the 929.0304 cm², the area of the foil is 929.0304 cm². Next, divide the mass of the foil by the area to find the volume.

The volume is 3.636 g / 929.0304 cm² = 0.003918 cm³. Finally, divide the volume by the area to find the thickness. The thickness is 0.003918 cm³ / 929.0304 cm² ≈ 0.0000042 cm, which is equivalent to 0.0042 mm.

Answer 2

The thickness of the foil is 0.01450 mm

Step-by-step explanation:

1 ft = 30.48 cm

So area of Al foil is
`(30.48)^(2)cm^(2)`

if thickness of foil is d cm then volume of Al foil is
`(30.48)^(2)dcm^(3)`

We know,
`density = (mass)/(volume)`

Here density of Al is 2.699
`g/cm^(3)` and mass of foil is 3.636 g.

So volume of foil is
`(3.636 g)/(2.699g/cm^(3))` or
`(3.636)/(2.699)cm^(3)`

So,
`(30.48)^(2)d=(3.636)/(2.699)`

or, d = 0.001450

So thickness of foil is 0.001450 cm or 0.01450 mm