A cube of platinum of edge length 0.038 mm (density \( \left.=21.4 \frac{\mathrm{~g}}{\mathrm{~cm}^{3}}\right) \). Round your answer to 2 significant digits. mass: \( \square \mathrm{g} \)
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The Deep Dive
To find the mass of the cube, we first calculate its volume. The volume \( V \) of a cube is given by the formula \( V = a^3 \), where \( a \) is the edge length. Given the edge length \( a = 0.038 \, \text{mm} \), we convert this to centimeters: \[ 0.038 \, \text{mm} = 0.0038 \, \text{cm} \] Now we calculate the volume: \[ V = (0.0038 \, \text{cm})^3 = 5.5 \times 10^{-7} \, \text{cm}^3 \] Next, we find the mass \( m \) using the density \( d = 21.4 \, \frac{\text{g}}{\text{cm}^3} \) with the formula \( m = d \times V \): \[ m = 21.4 \, \frac{\text{g}}{\text{cm}^3} \times 5.5 \times 10^{-7} \, \text{cm}^3 \approx 1.18 \times 10^{-6} \, \text{g} \] When rounded to two significant digits, the mass of the cube is: \[ m \approx 1.2 \times 10^{-6} \, \text{g} \]