Question

Claim The student hopes the samples are gold, which has a density of 19.3 g/cm3 . A local geologist suggested that samples might be pyrite, which is a mineral with a density of 5.01 g/cm3 . What is the identity of the unknown sample

Answer 1

Answer:

Pyrite.

Explanation:

In this case, for the given samples, we compute the densities considering that the volume of each sample is computed by the difference between the volume of the water + sample and the volume of water:

$\rho_1=\frac{50.25g}{(60.3-50.1)mL}=4.92g/mL\\ \\\rho_2=\frac{63.56g}{(62.5-49.8)mL}=5.01g/mL\\\\\rho_3=\frac{g}{(61.5-50.2)mL}=5.10g/mL\\\\\rho_4=\frac{55.35g}{(56.7-45.6)mL}=4.99g/mL\\\\\rho_5=\frac{74.92g}{(65.3-50.3)mL}=4.99g/mL\\\\\rho_6=\frac{67.78g}{(60.8-47.5)mL}=5.10g/mL$

Therefore, since the density of the samples are closer to the density of pyrite (5.01 g/cm³) we conclude that the samples are more like to be pyrite.

Best regards.