Question

question 3
the average density of the material in intergalactic space is approximately $2.5 \times 10^{-27}$ kg/m³. what is the volume of a lead sample that has the same mass as $8.0 \times 10^{24}$ m³ of intergalactic space? refer to table 11.1.
\\(\bigcirc\\) $1.0 \times 10^{-6}$ m³
\\(\bigcirc\\) $1.8 \times 10^{-6}$ m³
\\(\bigcirc\\) $5.4 \times 10^{-6}$ m³
\\(\bigcirc\\) $3.6 \times 10^{-5}$ m³
\\(\bigcirc\\) $2.2 \times 10^{-5}$ m³

Explanation:

Step1: Calculate mass of intergalactic space

Mass formula: $m =
ho V$.
$m = (2.5 \times 10^{-27} \, \text{kg/m}^3)(8.0 \times 10^{24} \, \text{m}^3) = 20 \, \text{kg}$.

Step2: Get lead density from Table 11.1

Standard lead density: $
ho_{\text{lead}} = 11300 \, \text{kg/m}^3$.

Step3: Calculate lead volume

Volume formula: $V = \frac{m}{
ho}$.
$V = \frac{20}{11300} \approx 1.77 \times 10^{-6} \, \text{m}^3 \approx 1.8 \times 10^{-6} \, \text{m}^3$.

Answer:

1.8 × 10⁻⁶ m³