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

Pyrite.

Step-by-step 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=(50.25g)/((60.3-50.1)mL)=4.92g/mL\\ \\\rho_2=(63.56g)/((62.5-49.8)mL)=5.01g/mL\\\\\rho_3=(g)/((61.5-50.2)mL)=5.10g/mL\\\\\rho_4=(55.35g)/((56.7-45.6)mL)=4.99g/mL\\\\\rho_5=(74.92g)/((65.3-50.3)mL)=4.99g/mL\\\\\rho_6=(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.

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