Question

Given the density of iron (Fe) is 7.87 g/cm3, determine the mass of iron (in grams) in a rectangle block with the dimensions of 12.5 in long, 3.50 in wide, and 2.50 in high. (1in = 2.54 cm).

Answer 1

Answer:

$m=14,105.71 g Fe$

Explanation:

Hello,

In this case, the first step is to compute the volume of the block considering the length, height and width:

$V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3$

Then, we compute the volume in cubic centimetres:

$V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3$

Finally, as the density is given by:

$\rho =\frac{m}{V}$

We solve for the mass:

$m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe$

Best regards.