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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: \( \square \mathrm{g} \)

Ask by Gibson Montgomery. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The mass of the platinum cube is approximately \(1.2 \times 10^{-6}\) grams.

Solution

To find the mass of the cube of platinum, we can use the formula for the volume of a cube and then multiply it by the density of platinum. Given: - Edge length of the cube = 0.038 mm - Density of platinum = 21.4 g/cm³ First, we need to convert the edge length from mm to cm to match the units of the density. Edge length in cm = 0.038 mm * (1 cm / 10 mm) = 0.0038 cm Now, we can calculate the volume of the cube using the formula: \[ \text{Volume} = \text{edge length}^3 \] \[ \text{Volume} = (0.0038 \, \text{cm})^3 \] Next, we can calculate the mass of the cube by multiplying the volume by the density of platinum: \[ \text{Mass} = \text{Volume} \times \text{Density} \] \[ \text{Mass} = (0.0038 \, \text{cm})^3 \times 21.4 \, \text{g/cm}^3 \] Let's calculate the mass of the cube of platinum. Calculate the value by following steps: - step0: Calculate: \(0.0038^{3}\times 21.4\) - step1: Convert the expressions: \(\left(\frac{19}{5000}\right)^{3}\times 21.4\) - step2: Convert the expressions: \(\left(\frac{19}{5000}\right)^{3}\times \frac{107}{5}\) - step3: Multiply the numbers: \(\frac{733913}{5\times 5000^{3}}\) The mass of the cube of platinum is approximately \(1.174 \times 10^{-6}\) g. Rounding to 2 significant digits, the mass of the cube of platinum is approximately \(1.2 \times 10^{-6}\) 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: \[ m \approx 1.2 \times 10^{-6} \, \text{g} \]

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