The aluminum foil on a certain roll has a total area of 18.5 m squared and a mass of 1275 g. using density of 2.7 g per cubic centimeter for aluminum, determine the thickness in millimeters of the aluminum foil.

Answer is: the thickness is 0.0256 millimeters.

m(Al) = 1275 g; mass of aluminium foil.

d(Al) = 2.7 g/cm³; density of aluminium foil.

V(Al) = m(Al) ÷ d(Al).

V(Al) = 1275 g ÷ 2.7 g/cm³.

V(Al) = 472.2 cm³ ÷ 10⁶ cm³/m³.

V(Al) = 0.000472 m³, volume of aluminum foil.

V(Al) = P · l (thickness).

l = 0.000472 m³ ÷ 18.5 m².

l = 0.0000256 m · 1000 mm/m.

l = 0.0256 mm.

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error