Question

Gold can be hammered into decorative gold leaf, which is only 1.00E-07m thick. If 1.35 troy ounce of gold is hammered into gold leaf, what is the area that can be covered when used for decoration? 1 troy ounce = 31.1g, and the density of gold is 19.3 g/cm^3

Answer 1

The area that can be covered when used for decoration is 21.8 m².

To solve this question, we'll by converting 1.35 troy ounce to grams (g). This can be obtained as as illustrated below:

1 troy ounce = 31.1g

Therefore,

1.35 troy ounce  = 1.35 x 31.1

1.35 troy ounce = 41.985 g

Thus, 1.35 troy ounce is equivalent to 41.985 g.

Next, we shall determine the volume of the gold leaf. This can be obtained as follow:

Mass = 41.985 g.

Density = 19.3 g/cm³

Volume =?

$Density = \frac{mass}{volume } \\\\19.3 = \frac{41.985}{Volume }$

Cross multiply

19.3 × Volume = 41.985

Divide both side by 19.3

$Volume = \frac{41.985}{19.3}$

Volume = 2.18 cm³

Next, we shall convert 2.18 cm³ to m³.

1 cm³ = 1×10¯⁶ m³

Therefore,

2.18 cm³ = 2.18 × 1×10¯⁶ m³

2.18 cm³ = 2.18×10¯⁶ m³

Thus, 2.18 cm³ is equivalent to 2.18×10¯⁶ m³.

Finally, we shall determine the area. This can be obtained as follow:

Volume = 2.18×10¯⁶ m³

Thickness = 1×10¯⁷ m

Area =?

Volume = Area × Thickness

2.18×10¯⁶ = Area × 1×10¯⁷

Divide both side by 1×10¯⁷

$Area = \frac{2.18*10^{-6} }{1*10^{-7} }$

Area = 21.8 m²

Therefore, the area that can be covered when used for decoration is 21.8 m²

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