Question

The density of gold is 19.3 g/cm3. A weight of 1 ounce is equivalent to a mass of about 28.35 g. Which of these is closest to the side length of a cube-shaped block of gold that has a weight of 8 ounces?​

Answer 1

Answer: The side length of the cube is 2.27cm

Step-by-step explanation:

The options are not given, then let's solve this in a general way.

We know that the density of gold is 19.3g/cm^3.

We also know that 1 oz = 28.35g.

And we know the relation:

Density = mass/volume.

Now, let's find the volume of 8 ounces of gold.

First, let's rewrite the density changing the units, from grams to ounces.

We know that:

1 oz = 28.35g

(1 oz/28.35g) = 1.

Then if we multiply the density by this ( that is equal to 1) we can change the units and leave the actual quantity invariant.

d =  19.3g/cm^3 =  19.3g/cm^3*(1 oz/28.35g) = 0.68 oz/cm^3

Now let's use the equation for the density, knowing this time that the mass is 8 ounces.

0.68 oz/cm^3 = mass/volume = 8oz/V

Let's find the value of V

0.68 oz/cm^3 = 8oz/V

V = 8oz/0.68 oz/cm^3 = 11.76 cm^3

Now, we know that this is a cube.

Remember that the volume of a cube of side length L is:

V = L^3

Then we have:

L^3 = 11.76 cm^3

L = ∛11.76 cm^3 = 2.27 cm

The side length of the cube is 2.27cm