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

abreviatsjw 2022-01-12 Answered
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?

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Expert Answer

Stuart Rountree
Answered 2022-01-13 Author has 4802 answers

Step 1
Answer to part (a):
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)
Answer to part (b):
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)

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levurdondishav4
Answered 2022-01-14 Author has 5828 answers

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
\(\sqrt[3]{37.0}={3.33cm}\)

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