Question

You are given three cubes, A, B, and C; one is magnesium, one is aluminum, and the third is silver. All three cubes have the same mass, but cube A has a volume of 25.9 mL, cube B has a volume of 16.7 mL, and cube C has a volume of 4.29 mL. Identify cubes A, B, and C.

Answer 1

Answer:

The cube A is magnesium, the cube B is aluminum and the cube C is silver.

Explanation:

Density is defined by the expression $d=\frac{m}{V}$ where m is the mass and V is the volume, therefore:

- Density of the cube A:

$d_{A}=\frac{m_{A}}{V_{A}}$

- Density of the cube B:

$d_{B}=\frac{m_{B}}{V_{B}}$

- Density of the cube C:

$d_{C}=\frac{m_{C}}{V_{C}}$

Solving for mass:

$m_{A}=d_{A}*V_{A}$

$m_{B}=d_{B}*V_{B}$

$m_{C}=d_{C}*V_{C}$

And all the three cubes have the same mass, so:

$m_{A}=m_{B}=m_{C}$

Therefore:

$d_{A}*V_{A}=d_{B}*V_{B}$ (Eq.1)

$d_{A}*V_{A}=d_{C}*V_{C}$ (Eq.2)

Solving for $d_{1}$ in Eq.1:

$d_{A}=d_{B}\frac{V_{B}}{V_{A}}$

Replacing values for the volume:

$d_{A}=d_{B}\frac{16.7mL}{25.9mL}$

$d_{A}=d_{B}*0.64$

As we know the density of the aluminum is $2.7\frac{g}{cm^{3}}$, so replacing this value for $d_{B}$:

$d_{A}=2.7\frac{g}{mL}*0.64$

$d_{A}=1.728\frac{g}{mL}$

that is the density of the magnesium.

Solving for $d_{C}$ in Eq.2:

$d_{C}=d_{A}\frac{V_{A}}{V_{C}}$

$d_{C}=d_{A}\frac{25.9mL}{4.29mL}$

$d_{C}=d_{A}*6.04$

$d_{C}=1.728\frac{g}{mL}*6.04$

$d_{C}=10.4\frac{g}{mL}$

That is the density of the silver.

Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.