Em tham khảo nha:
\(\begin{array}{l}
a)\\
{N_{Ca}} = n \times 6 \times {10^{23}} = 0,12 \times 6 \times {10^{23}} = 7,2 \times {10^{22}} \text{ nguyên tử}\\
b)\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{20}}{{40}} = 0,5\,mol\\
{N_{NaOH}} = n \times 6 \times {10^{23}} = 0,5 \times 6 \times {10^{23}} = 3 \times {10^{23}}\text{ phân tử}\\
c)\\
{n_{S{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
{N_{S{O_2}}} = n \times 6 \times {10^{23}} = 0,25 \times 6 \times {10^{23}} = 1,5 \times {10^{23}} \text{ phân tử}
\end{array}\)
