Answer:
[tex]2.82x10^{23}molecules[/tex]
Explanation:
Hello.
In this case, given the volume (1cm³=1mL) and density of the bromine we are to firstly compute the mass since it will allow us to compute the representative particles:
[tex]\rho =\frac{m}{V}[/tex]
[tex]m=\rho *V=3.12g/cm^3*24.0cm^3\\\\m=74.88gBr_2[/tex]
Next, since the mass of one mole of diatomic bromine is 159.82 g (one bromine weights 78.91), we can next compute the moles in that sample:
[tex]n=74.88g*\frac{1mol}{159.82g} =0.469molBr_2[/tex]
Finally, via the Avogadro's number we can compute the representative particles of bromine as follows:
[tex]particles=0.469mol*\frac{6.022x10^{23}molecules}{1mol}\\ \\2.82x10^{23}molecules[/tex]
Best regards.
