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

Do you know Avogadro's number?

Answer 2

Using that, you can get moles.

Answer 3

And then when you have moles, you can use molar mass to retrieve grams.

Answer 4

Then you can cube your cm measurement to get cm^3.

Answer 5

\[8.91*10^{23}\ \cancel{atoms} \cancel{Pb} * \frac{1\ \cancel{mol}}{6.022*10^{23}\ \cancel{atoms}} * \frac{207.21\ g\ Pb}{1\ \cancel{mol}\ \cancel{Pb}} =\ ?\ g\ Pb\]

Answer 6

Do you know what (3 cm)\(^3\) is @onmypagrind ? :-)

Density = \(\large \frac{mass}{volume}\)

Answer 7

@agentx5 How did you cancel the units?