Answer :
We start by finding the number of moles in 9 g of nitrogen. The mass of one mole of nitrogen is approximately 14 g/mol. Thus, the number of moles is
[tex]$$
\text{moles} = \frac{9\ \text{g}}{14\ \text{g/mol}} \approx 0.642857\ \text{mol}.
$$[/tex]
Next, we use Avogadro's number, which is approximately
[tex]$$
6.022 \times 10^{23}\ \text{atoms/mol},
$$[/tex]
to convert moles to atoms. Multiplying the number of moles by Avogadro's number gives
[tex]$$
\text{atoms} = 0.642857\ \text{mol} \times 6.022 \times 10^{23}\ \text{atoms/mol} \approx 3.87 \times 10^{23}\ \text{atoms}.
$$[/tex]
Therefore, there are approximately [tex]$3.87 \times 10^{23}$[/tex] atoms in 9 g of nitrogen.
This corresponds to option d.
Final Answer: [tex]$\boxed{3.87 \times 10^{23}}$[/tex].
[tex]$$
\text{moles} = \frac{9\ \text{g}}{14\ \text{g/mol}} \approx 0.642857\ \text{mol}.
$$[/tex]
Next, we use Avogadro's number, which is approximately
[tex]$$
6.022 \times 10^{23}\ \text{atoms/mol},
$$[/tex]
to convert moles to atoms. Multiplying the number of moles by Avogadro's number gives
[tex]$$
\text{atoms} = 0.642857\ \text{mol} \times 6.022 \times 10^{23}\ \text{atoms/mol} \approx 3.87 \times 10^{23}\ \text{atoms}.
$$[/tex]
Therefore, there are approximately [tex]$3.87 \times 10^{23}$[/tex] atoms in 9 g of nitrogen.
This corresponds to option d.
Final Answer: [tex]$\boxed{3.87 \times 10^{23}}$[/tex].
