How many electrons are present in 2 times 10^ | vedclass

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(C) The atomic number of oxygen is $8$,so a neutral oxygen atom has $8$ electrons.For the oxide ion ${}_{8}^{18}O^{2-}$,the number of electrons is $8

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$1.2 	imes 10^{22}$

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(C) The atomic number of oxygen is $8$,so a neutral oxygen atom has $8$ electrons.For the oxide ion ${}_{8}^{18}O^{2-}$,the number of electrons is $8 + 2 = 10$ electrons per ion.The number of ions in $2 	imes 10^{-3} \ mol$ is $2 	imes 10^{-3} 	imes N_A$,where $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Total electrons = $(	ext{Number of ions}) 	imes (	ext{Electrons per ion})$Total electrons = $(2 	imes 10^{-3} 	imes 6.022 	imes 10^{23}) 	imes 10$Total electrons = $12.044 	imes 10^{21} \approx 1.2 	imes 10^{22}$.

How many electrons are present in $2 \times 10^{-3} \ mol$ of ${}_{8}^{18}O^{2-}$?

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