Pregunta
A cube of platinum of edge length 0.034 mm (density \( =21.4 \frac{g}{\mathrm{~cm}^{3}} \) ), Round your answer to 2 significant digits. mass: \( \square \mathrm{g} \)
Ask by Lane Rose. in the United States
Feb 03,2025
Solución de tutoría real
Respuesta verificada por el tutor
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The mass of the platinum cube is \( 8.4 \times 10^{-7} \) grams.
Solución
To determine the mass of the platinum cube, follow these steps:
1. **Convert Edge Length to Centimeters:**
\[
0.034 \, \text{mm} = 0.0034 \, \text{cm}
\]
2. **Calculate the Volume of the Cube:**
\[
\text{Volume} = (\text{edge length})^3 = (0.0034 \, \text{cm})^3 = 3.9304 \times 10^{-8} \, \text{cm}^3
\]
3. **Calculate the Mass Using Density:**
\[
\text{Mass} = \text{Volume} \times \text{Density} = 3.9304 \times 10^{-8} \, \text{cm}^3 \times 21.4 \, \frac{\text{g}}{\text{cm}^3} = 8.4 \times 10^{-7} \, \text{g}
\]
**Final Answer:**
\[
\text{mass: } 8.4 \times 10^{-7} \, \text{g}
\]
Revisado y aprobado por el equipo de tutoría de UpStudy
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Simplifique esta solución The Deep Dive
To find the mass of the platinum cube, first calculate the volume \( V \) of the cube using the formula \( V = \text{edge length}^3 \). So, \( V = (0.034 \, \text{mm})^3 = 0.000039304 \, \text{mm}^3 \). Convert this to cubic centimeters: \( 0.000039304 \, \text{mm}^3 = 0.000000039304 \, \text{cm}^3 \) (since \( 1 \, \text{cm} = 10 \, \text{mm} \)). Now, use the density to find mass: \[ \text{mass} = \text{density} \times \text{volume} = 21.4 \, \frac{g}{\text{cm}^3} \times 0.000000039304 \, \text{cm}^3 \approx 0.000000000842 \, g \] This rounds to \( 0.00 \, g \) since it’s less than the smallest significant digit.
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