Let's solve this step-by-step:
1. Given Information:
- The side of the cube is 2.0 cm.
- The mass of the cube is 14.56 g.
- We need to determine the element based on the given densities and classification.
2. Calculate the Volume of the Cube:
The volume (V) of a cube can be calculated using the formula:
[tex]\[
V = \text{side length}^3
\][/tex]
Given the side length is 2.0 cm, we calculate:
[tex]\[
V = 2.0 \, \text{cm}^3 = 2.0 \times 2.0 \times 2.0 = 8.0 \, \text{cm}^3
\][/tex]
3. Calculate the Density of the Cube:
The density (ρ) of a substance is defined as its mass (m) divided by its volume (V):
[tex]\[
\rho = \frac{\text{mass}}{\text{volume}}
\][/tex]
Given the mass is 14.56 g and the volume is 8.0 cm³, we calculate:
[tex]\[
\rho = \frac{14.56 \, \text{g}}{8.0 \, \text{cm}^3} = 1.82 \, \text{g/cm}^3
\][/tex]
4. Compare the Density to Known Densities:
We are given the following densities for different elements:
[tex]\[
\begin{aligned}
&\text{Barium (Ba)}: 3.6 \, \text{g/cm}^3 \\
&\text{Beryllium (Be)}: 1.8 \, \text{g/cm}^3 \\
&\text{Chromium (Cr)}: 7.2 \, \text{g/cm}^3 \\
&\text{Phosphorus (P)}: 1.8 \, \text{g/cm}^3 \\
\end{aligned}
\][/tex]
The calculated density of the cube (1.82 g/cm³) doesn't exactly match any of the given densities. However, it is closest to the density of beryllium (1.8 g/cm³) and phosphorus (1.8 g/cm³).
5. Account for Classification and Appearance:
- The unknown element cube is described as having a shiny, silvery color.
- Phosphorus is a nonmetal and typically not shiny and silvery. In contrast, beryllium is a metal and matches the description of being shiny and silvery.
6. Conclusion:
Given the density of the cube is closest to 1.8 g/cm³ and the appearance described fits that of a metal, the cube is most likely made of beryllium (Be).
Thus, the element that makes up the cube is:
B. Beryllium (Be)
