Answer:
The cube A is magnesium, the cube B is aluminum and the cube C is silver.
Explanation:
Density is defined by the expression [tex]d=\frac{m}{V}[/tex] where m is the mass and V is the volume, therefore:
- Density of the cube A:
[tex]d_{A}=\frac{m_{A}}{V_{A}}[/tex]
- Density of the cube B:
[tex]d_{B}=\frac{m_{B}}{V_{B}}[/tex]
- Density of the cube C:
[tex]d_{C}=\frac{m_{C}}{V_{C}}[/tex]
Solving for mass:
[tex]m_{A}=d_{A}*V_{A}[/tex]
[tex]m_{B}=d_{B}*V_{B}[/tex]
[tex]m_{C}=d_{C}*V_{C}[/tex]
And all the three cubes have the same mass, so:
[tex]m_{A}=m_{B}=m_{C}[/tex]
Therefore:
[tex]d_{A}*V_{A}=d_{B}*V_{B}[/tex] (Eq.1)
[tex]d_{A}*V_{A}=d_{C}*V_{C}[/tex] (Eq.2)
Solving for [tex]d_{1}[/tex] in Eq.1:
[tex]d_{A}=d_{B}\frac{V_{B}}{V_{A}}[/tex]
Replacing values for the volume:
[tex]d_{A}=d_{B}\frac{16.7mL}{25.9mL}[/tex]
[tex]d_{A}=d_{B}*0.64[/tex]
As we know the density of the aluminum is [tex]2.7\frac{g}{cm^{3}}[/tex], so replacing this value for [tex]d_{B}[/tex]:
[tex]d_{A}=2.7\frac{g}{mL}*0.64[/tex]
[tex]d_{A}=1.728\frac{g}{mL}[/tex]
that is the density of the magnesium.
Solving for [tex]d_{C}[/tex] in Eq.2:
[tex]d_{C}=d_{A}\frac{V_{A}}{V_{C}}[/tex]
[tex]d_{C}=d_{A}\frac{25.9mL}{4.29mL}[/tex]
[tex]d_{C}=d_{A}*6.04[/tex]
[tex]d_{C}=1.728\frac{g}{mL}*6.04[/tex]
[tex]d_{C}=10.4\frac{g}{mL}[/tex]
That is the density of the silver.
Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.
