1 text{ g} of H_2 contains the same number of | vedclass

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(A) Step $1$: Calculate the number of moles of $H_2$ in $1 	ext{ g}$. Molar mass of $H_2 = 2 	ext{ g/mol}$. Moles of $H_2 = \frac{1 	ext{ g}}{2 	e

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$14 	ext{ g of } N_2$

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(A) Step $1$: Calculate the number of moles of $H_2$ in $1 	ext{ g}$. Molar mass of $H_2 = 2 	ext{ g/mol}$. Moles of $H_2 = \frac{1 	ext{ g}}{2 	ext{ g/mol}} = 0.5 	ext{ mol}$.Step $2$: Calculate the number of moles for each option:$(A)$ $14 	ext{ g of } N_2$: Molar mass of $N_2 = 28 	ext{ g/mol}$. Moles $= \frac{14}{28} = 0.5 	ext{ mol}$.$(B)$ $18 	ext{ g of } H_2O$: Molar mass of $H_2O = 18 	ext{ g/mol}$. Moles $= \frac{18}{18} = 1 	ext{ mol}$.$(C)$ $16 	ext{ g of } CO$: Molar mass of $CO = 28 	ext{ g/mol}$. Moles $= \frac{16}{28} \approx 0.57 	ext{ mol}$.$(D)$ $28 	ext{ g of } N_2$: Molar mass of $N_2 = 28 	ext{ g/mol}$. Moles $= \frac{28}{28} = 1 	ext{ mol}$.Step $3$: Since the number of molecules is directly proportional to the number of moles, $14 	ext{ g of } N_2$ contains the same number of molecules as $1 	ext{ g of } H_2$.

$1 \text{ g}$ of $H_2$ contains the same number of molecules as in

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