8 g of O_2 has the same number of atoms as t | vedclass

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(B) Step $1$: Calculate the number of atoms in $8 \ g$ of $O_2$.Molar mass of $O_2 = 32 \ g/mol$.Moles of $O_2 = \frac{8 \ g}{32 \ g/mol} = 0.25 \ mol

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$7 \ g$ of $CO$

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(B) Step $1$: Calculate the number of atoms in $8 \ g$ of $O_2$.Molar mass of $O_2 = 32 \ g/mol$.Moles of $O_2 = \frac{8 \ g}{32 \ g/mol} = 0.25 \ mol$.Number of $O_2$ molecules $= 0.25 	imes N_A$.Number of $O$ atoms $= 2 	imes (0.25 	imes N_A) = 0.5 \ N_A$.Step $2$: Calculate the number of atoms in the options.$(A)$ $14 \ g$ of $CO$: Molar mass $= 28 \ g/mol$. Moles $= \frac{14}{28} = 0.5 \ mol$. Total atoms $= 0.5 	imes 2 	imes N_A = 1.0 \ N_A$.$(B)$ $7 \ g$ of $CO$: Moles $= \frac{7}{28} = 0.25 \ mol$. Total atoms $= 0.25 	imes 2 	imes N_A = 0.5 \ N_A$.$(C)$ $11 \ g$ of $CO_2$: Molar mass $= 44 \ g/mol$. Moles $= \frac{11}{44} = 0.25 \ mol$. Total atoms $= 0.25 	imes 3 	imes N_A = 0.75 \ N_A$.$(D)$ $22 \ g$ of $CO_2$: Moles $= \frac{22}{44} = 0.5 \ mol$. Total atoms $= 0.5 	imes 3 	imes N_A = 1.5 \ N_A$.Conclusion: $7 \ g$ of $CO$ contains $0.5 \ N_A$ atoms,which is the same as $8 \ g$ of $O_2$.

$8 \ g$ of $O_2$ has the same number of atoms as that in:

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