Question

What is the mass, in grams of a pure iron cube that has a volume of 4.20cm^3

Answer 1

$d_{Fe}=7874\frac{kg}{m^{3}}=7,874\frac{g}{cm^{3}}\\ V=4,2cm^{3}\\\\ m=dV=7,784\frac{g}{cm^{3}}*4,2cm^{3}\approx 32,7g$