In a compound formed by Xe and F,the mass per | vedclass

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(D) $1$. Calculate the moles of each element in $100 \ g$ of the compound: Moles of $Xe = \frac{63.8}{133} \approx 0.48 \ mol$. Moles of $F = \frac{10

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$+4$

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(D) $1$. Calculate the moles of each element in $100 \ g$ of the compound: Moles of $Xe = \frac{63.8}{133} \approx 0.48 \ mol$. Moles of $F = \frac{100 - 63.8}{19} = \frac{36.2}{19} \approx 1.90 \ mol$. $2$. Determine the molar ratio: Ratio of $Xe : F = 0.48 : 1.90 \approx 1 : 4$. $3$. The empirical formula is $XeF_4$. $4$. In $XeF_4$,the oxidation state of $Xe$ is $x + 4(-1) = 0$,so $x = +4$.

In a compound formed by $Xe$ and $F$,the mass percentage of $Xe$ $(M=133)$ is $63.8\%$. The oxidation state of $Xe$ in this compound is ..........

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