The ratio of the number of oxygen atoms (O) i | vedclass

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(D) For $16.0 \ g$ $O_3$: Molar mass $= 48 \ g/mol$. Moles $= 16/48 = 1/3 \ mol$. Number of $O$ atoms $= 3 	imes (1/3) 	imes N_A = N_A$.For $28.0 \

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$1 : 1 : 1$

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(D) For $16.0 \ g$ $O_3$: Molar mass $= 48 \ g/mol$. Moles $= 16/48 = 1/3 \ mol$. Number of $O$ atoms $= 3 	imes (1/3) 	imes N_A = N_A$.For $28.0 \ g$ $CO$: Molar mass $= 28 \ g/mol$. Moles $= 28/28 = 1 \ mol$. Number of $O$ atoms $= 1 	imes 1 	imes N_A = N_A$.For $16.0 \ g$ $O_2$: Molar mass $= 32 \ g/mol$. Moles $= 16/32 = 0.5 \ mol$. Number of $O$ atoms $= 2 	imes 0.5 	imes N_A = N_A$.Thus,the ratio of oxygen atoms is $N_A : N_A : N_A$,which simplifies to $1 : 1 : 1$.

The ratio of the number of oxygen atoms $(O)$ in $16.0 \ g$ ozone $(O_3)$,$28.0 \ g$ carbon monoxide $(CO)$,and $16.0 \ g$ oxygen $(O_2)$ is (Atomic mass: $C = 12$,$O = 16$ and Avogadro's constant $N_A = 6.0 \times 10^{23} \ mol^{-1}$)

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