A sample of gold weighs 1.2 oz. The sample is pounded into a thin rectangular sheet with an area of 400. sq. ft. The density of gold is 19.3 grams per cm3. What is the thickness of the foil in centimeters?Given: 28.35 g = 1 oz, 1 ft = 12 inches, 1 inch = 2.54 cm

thickness of the gold sheet = 4.74 * 10⁻⁶ cm

Explanation:

mass of gold sample = 1.2 oz,; area of rectangular gold sheet = 400 sq. ft

Converting mass of gold from oz. to g

1 oz. = 28.35 g

mass of gold sample in grams = 1.2 * 28.35 g = 34.02 g

Converting from feet to cm;

1 feet  =  12 * 2.54 cm = 30.48 cm

1 sq. ft = (30.48)² = 929.0304 cm²

area of gold in cm² = 400 * 929.0304 cm² = 371612.16 cm²

Since the density of a solid is constant

Density = mass/volume

Volume = mass/density

where volume =area * thickness

therefore, area * thickness =mass/density

thickness = mass/(density * area)

substituting the value; thickness = 34.02 g/(19.3 gcm⁻³ *371612.16 cm²)

thickness of the gold sheet = 4.74 * 10⁻⁶ cm

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