Rearrange the following (I to IV) in the orde | vedclass

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(A) To find the order of increasing masses,we calculate the mass of each component:$(I)$ Mass of $0.5$ mole of $O_3 = 0.5 	imes 48 \, g = 24 \, g$.$(

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$II < IV < III < I$

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(A) To find the order of increasing masses,we calculate the mass of each component:$(I)$ Mass of $0.5$ mole of $O_3 = 0.5 	imes 48 \, g = 24 \, g$.$(II)$ $0.5 \, g$ atom of oxygen means $0.5$ mole of $O$ atoms. Mass $= 0.5 	imes 16 \, g = 8 \, g$.$(III)$ Number of moles of $O_2 = \frac{3.011 	imes 10^{23}}{6.022 	imes 10^{23}} = 0.5$ mole. Mass $= 0.5 	imes 32 \, g = 16 \, g$.$(IV)$ Number of moles of $CO_2 = \frac{5.6 \, L}{22.4 \, L/mol} = 0.25$ mole. Mass $= 0.25 	imes 44 \, g = 11 \, g$.Comparing the masses: $8 \, g (II) Thus,the correct order is $II < IV < III < I$.

Rearrange the following ($I$ to $IV$) in the order of increasing masses:
$(I)$ $0.5$ mole of $O_3$
$(II)$ $0.5 \, g$ atom of oxygen
$(III)$ $3.011 \times 10^{23}$ molecules of $O_2$
$(IV)$ $5.6 \, L$ of $CO_2$ at $STP$

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Rearrange the following ($I$ to $IV$) in the order of increasing masses:$(I)$ $0.5$ mole of $O_3$$(II)$ $0.5 \, g$ atom of oxygen$(III)$ $3.011 \times 10^{23}$ molecules of $O_2$$(IV)$ $5.6 \, L$ of $CO_2$ at $STP$