Resposta :
Dados:
[tex]Q=0.6\times10^{19}\ \mu C=6\times10^{18}\times10^{-6}=6\times10^{12}\ C\\ n=1.6\times10^{-19}\ C[/tex]
Explicação:
[tex]\boxed{Q=ne}\ \therefore\ 6\times10^{12}=1.6\times10^{-19}e\ \therefore[/tex]
[tex]e=\dfrac{6\times10^{12}}{1.6\times10^{-19}}\ \therefore\ \boxed{e=3.75\times10^{31}\ \mathrm{el\acute{e}trons}}[/tex]