Gold has a density of 19.3 g/cm^3. Find the thickness of a piece of gold leaf weighing 1.93 mg and covering an area of 14.5 cm^2.
Question:
Gold has a density of {eq}rm 19.3 g/cm^3 {/eq}. Find the thickness of a piece of gold leaf weighing {eq}1.93 mg {/eq} and covering an area of {eq}14.5 cm^2 {/eq}.
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Alma April 15, 2023 в 14:35Gold has a density of 19.3 g/cm?, which means that 1 cm? of gold weighs 19.3 g. To find the thickness of a piece of gold leaf, we need to know its volume. We can find this by dividing its mass by its density: Volume of gold leaf = (mass of gold leaf) / (density of gold) Volume of gold leaf = 1.93 mg / 19.3 g/cm? Volume of gold leaf = 0.0001 cm? Since we know the area that the gold leaf covers (14.5 cm?), we can use this to find its thickness. If we assume that the gold leaf is a rectangular prism, then: Volume of gold leaf = (area of gold leaf) x (thickness of gold leaf) 0.0001 cm? = 14.5 cm? x (thickness of gold leaf) Solving for thickness of gold leaf, we get: Thickness of gold leaf = 0.0001 cm? / 14.5 cm? Thickness of gold leaf = 0.000006897 cm Therefore, the thickness of the gold leaf is \approx imately 0.000006897 cm or 0.06897 micrometers.
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