Question

use the references to access important values if needed for this question.
a. how many atoms of nitrogen are present in 1.68 moles of dinitrogen tetrafluoride?
atoms of nitrogen
b. how many moles of fluorine are present in 6.60×10²² molecules of dinitrogen tetrafluoride?
moles of fluorine
9 more group attempt remaining

Explanation:

Step1: Recall Avogadro's number

Avogadro's number $N_A = 6.022\times 10^{23}\text{ molecules/mol}$ or $\text{ atoms/mol}$.

Step2: Calculate the number of nitrogen atoms in part a

The formula for dinitrogen tetrafluoride is $N_2F_4$. In 1 mole of $N_2F_4$, there are 2 moles of nitrogen atoms. Given $n = 1.68$ moles of $N_2F_4$. The number of moles of nitrogen atoms $n_N=2\times1.68 = 3.36$ moles. Using the formula $N=n\times N_A$, where $n$ is the number of moles and $N$ is the number of atoms. So $N = 3.36\times6.022\times 10^{23}=2.023\times 10^{24}$ atoms.

Step3: Calculate the moles of fluorine in part b

First, find the number of moles of $N_2F_4$ molecules. Given $N = 0.60\times 10^{23}$ molecules of $N_2F_4$. Using $n=\frac{N}{N_A}$, where $N$ is the number of molecules and $N_A$ is Avogadro's number. So $n=\frac{0.60\times 10^{23}}{6.022\times 10^{23}}\approx0.0996$ moles of $N_2F_4$. In 1 mole of $N_2F_4$, there are 4 moles of fluorine atoms. So the number of moles of fluorine $n_F=4\times0.0996 = 0.3984\approx0.40$ moles.

Answer:

a. $2.023\times 10^{24}$ atoms
b. $0.40$ moles