Ferrous oxide has a cubic structure and each | vedclass

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(A) The density formula for a unit cell is given by $d = \frac{Z 	imes M_w}{N_A 	imes a^3}$.Given: $d = 4.0 \ g \ cm^{-3}$,$a = 5.0 \ \mathring{A} =

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Four $Fe^{2+}$ and four $O^{2-}$

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(A) The density formula for a unit cell is given by $d = \frac{Z 	imes M_w}{N_A 	imes a^3}$.Given: $d = 4.0 \ g \ cm^{-3}$,$a = 5.0 \ \mathring{A} = 5.0 	imes 10^{-8} \ cm$,$M_w = 72 \ g/mol$,and $N_A \approx 6.022 	imes 10^{23} \ mol^{-1}$.Substituting the values: $4.0 = \frac{Z 	imes 72}{(6.022 	imes 10^{23}) 	imes (5.0 	imes 10^{-8})^3}$.$4.0 = \frac{Z 	imes 72}{6.022 	imes 10^{23} 	imes 125 	imes 10^{-24}}$.$4.0 = \frac{Z 	imes 72}{75.275 	imes 10^{-1}}$.$4.0 = \frac{Z 	imes 72}{7.5275}$.$Z = \frac{4.0 	imes 7.5275}{72} \approx 0.418$. Wait,re-evaluating the calculation: $Z = \frac{4.0 	imes 75.275}{72} \approx 4.18 \approx 4$.Since $FeO$ has a $1:1$ stoichiometry,there are $4$ formula units of $FeO$ per unit cell.Therefore,there are $4$ $Fe^{2+}$ ions and $4$ $O^{2-}$ ions per unit cell.

Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0 \ \mathring{A}$. Assuming the density of the oxide is $4.0 \ g \ cm^{-3}$,calculate the number of $Fe^{2+}$ and $O^{2-}$ ions present in each unit cell ($M_w$ of $FeO = 72 \ g/mol$).

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