abreviatsjw

2022-01-12

The density of gold is $19.3\frac{g}{c}{m}^{3}$ .

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?

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?

Stuart Rountree

Beginner2022-01-13Added 29 answers

Step 1

Answer to part (a):

We must determine the volume of the material given in this part, and to do so, we will apply the relationship shown below:

$Volume=\frac{Mass}{Density}$ (1)

Given, $mass=0.715kg$ and $density=19.3\frac{g}{c}{m}^{3}$

In order to calculate the density in $c{m}^{3}$, we will first convert the mass in gram, for which we will use the below relation :

$0.715kg\times \frac{1000g}{1kg}=715g$ (2)

Step 2

Answer to part (a): Given, $mass=715g$ and $density=19.3\frac{g}{c}{m}^{3}$

Putting above given values in equation (1) and rounding off the volume to 3 significant digits, we get :

$Volume=\frac{715g}{19.3\frac{g}{c}{m}^{3}}=37.0c{m}^{3}$ (3)

Answer to part (b):

In order to determine the length of each edge of the gold (cube) in this problem, we will use the relationship shown below:

${\text{Volume of gold=(length of each edge of gold)}}^{3}$ (4)

As calculated in the above subpart that the volume of the $gold=37.0c{m}^{3}$, putting this to equation (4), we get :

$length=(37cm{}^{3}{)}^{1/3}=3.33cm$ (5)

levurdondishav4

Beginner2022-01-14Added 38 answers

a.) The given mass should first be converted from kilograms to grams.

$0.715kg\left(\frac{1kg}{1000g}\right)=715g$

To solve for the volume, divide the converted mass by the given density

$\frac{715g}{19.3g/cm{}^{3}=37.0{m}^{3}}$

b.) The formula for the volume of a cube is $v}_{cube}={\left(sides\right)}^{3$

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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