Question

a lead cube is 3 cm on each side contains 8.91x10^23 atoms what is the density of this cube in g/cm^3

Answer 1

\[\frac{8.91x10^{23} atoms pb(\frac{1 mole pb}{6.022x10^23 atoms pb})(\frac{207.2g pb}{1 mole pb})}{27cm^3}=density\]

Answer 2

thank you z bay is the answer 0.550 or 0.990 or 11.4 or 26.7 or 34.1

Answer 3

1 sec

Answer 4

11.4

Answer 5

Do you see how I got that?