Question #06dc7

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Question #06dc7

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Answer 1 · 2017-10-01T00:46:34

${19.3 g/cm}^{3}$ $"19.3 g/cm"^3$

Explanation:

The idea here is that you need to figure out the volume of the cube first, then use that value to find the density of gold.

As you know, the volume of a cube is given by

$volume cube = l^{3}$ $"volume cube" = l^3$

Here $l$ $l$ represents the side length of the cube.

In your case, the volume of the cube will be equal to

$volume cube = {(1.55 cm)}^{3}$ $"volume cube" = ("1.55 cm")^3$

$volume cube = {3.724 cm}^{3}$ $"volume cube" = "3.724 cm"^3$

Now, in order to find the density of gold, $ρ$ $rho$ , you essentially need to figure out the mass of exactly $1$ $1$ unit of volume of gold.

Since you know that ${3.724 cm}^{3}$ $"3.724 cm"^3$ of gold have a mass of $71.9 g$ $"71.9 g"$ , you can get the density of gold by dividing the mass of the sample by the total volume it occupies.

$ρ = \frac{71.9 g}{{3.724 cm}^{3}}$ $rho = "71.9 g"/"3.724 cm"^3$

This will get you

$rho = 71.9/3.724 * "1 g"/"1 cm"^3 = color(darkgreen)(ul(color(black)("19.3 g/cm"^3)))$

The answer is rounded to three sig figs.

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Question #06dc7

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