# The density of gold is 19.3 g/cm^{3}. a. What is the

The density of gold is $$\displaystyle{19.3}\frac{{g}}{{c}}{m}^{{{3}}}$$.
a. What is the volume, in cubic centimeters, of a sample of gold that has a mass of 0.715 kg?
b. If this sample of gold is a cube, what is the length of each edge in centimeters?

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Stuart Rountree

Step 1
In this part, we have been asked to calculate the volume of the given material for which we will use the below relation :
$$\displaystyle{V}{o}{l}{u}{m}{e}={\frac{{{M}{a}{s}{s}}}{{{D}{e}{n}{s}{i}{t}{y}}}}$$ (1)
Given, $$\displaystyle{m}{a}{s}{s}={0.715}{k}{g}$$ and $$\displaystyle{d}{e}{n}{s}{i}{t}{y}={19.3}\frac{{g}}{{c}}{m}^{{{3}}}$$
In order to calculate the density in $$\displaystyle{c}{m}^{{{3}}}$$, we will first convert the mass in gram, for which we will use the below relation :
$$\displaystyle{0.715}{k}{g}\times{\frac{{{1000}{g}}}{{{1}{k}{g}}}}={715}{g}$$ (2)
Step 2
Answer to part (a): Given, $$\displaystyle{m}{a}{s}{s}={715}{g}$$ and $$\displaystyle{d}{e}{n}{s}{i}{t}{y}={19.3}\frac{{g}}{{c}}{m}^{{{3}}}$$
Putting above given values in equation (1) and rounding off the volume to 3 significant digits, we get :
$$\displaystyle{V}{o}{l}{u}{m}{e}={\frac{{{715}{g}}}{{{19.3}\frac{{g}}{{c}}{m}^{{{3}}}}}}={37.0}{c}{m}^{{{3}}}$$ (3)
In this problem, we have been asked to calculate the length of each edge of gold (cube), for which we will use the below relation :
$$\text{Volume of gold=(length of each edge of gold)}^{3}$$ (4)
As calculated in the above subpart that the volume of the $$\displaystyle{g}{o}{l}{d}={37.0}{c}{m}^{{{3}}}$$, putting this to equation (4), we get :
$$length=(37 cm{^3})^{1/3}=3.33 cm$$ (5)

###### Not exactly what you’re looking for?
levurdondishav4

Step 1
a.)
First , convert the given mass from kilograms to grams
$$\displaystyle{0.715}{k}{g{{\left({\frac{{{1}{k}{g}}}{{{1000}{g}}}}\right)}}}={715}{g}$$
To solve for the volume, divide the converted mass by the given density
$$\frac{715g}{19.3g/cm{^3}=37.0m^{3}}$$
b.) The formula for the volume of a cube is $$\displaystyle{v}_{{{c}{u}{b}{e}}}={\left({s}{i}{d}{e}{s}\right)}^{{3}}$$
to solve for the length of each side of a cube, take the cube root of the volume
$$\sqrt[3]{37.0}={3.33cm}$$