Answer:
[tex]m=14,105.71 g Fe[/tex]
Explanation:
Hello,
In this case, the first step is to compute the volume of the block considering the length, height and width:
[tex]V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3[/tex]
Then, we compute the volume in cubic centimetres:
[tex]V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3[/tex]
Finally, as the density is given by:
[tex]\rho =\frac{m}{V}[/tex]
We solve for the mass:
[tex]m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe[/tex]
Best regards.
