+6
Explanation:
Let the total weight of the compound is 100 g.
So the amount of Xe present is 53.5 g and of F is 46.7 g.
Now we find moles of Xe is
[tex] \frac{53.5}{133} = 0.4[/tex]
and of F is
[tex] \frac{46.7}{19} = 2.45[/tex]
Divide both by 0.4 so that we get a simple whole-number ratio, we get
[tex]Moles \: \: of \: \: Xe = \frac{0.4}{0.4} = 1[/tex]
[tex]Moles \: \: of \: \: F = \frac{2.45}{0.4} = 6[/tex]
[tex]So \: \: the \: \: formula \: \: is \: \: XeF_{6 }.[/tex]
The oxidation state of Xe is +6.
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Answers & Comments
+6
Explanation:
Let the total weight of the compound is 100 g.
So the amount of Xe present is 53.5 g and of F is 46.7 g.
Now we find moles of Xe is
[tex] \frac{53.5}{133} = 0.4[/tex]
and of F is
[tex] \frac{46.7}{19} = 2.45[/tex]
Divide both by 0.4 so that we get a simple whole-number ratio, we get
[tex]Moles \: \: of \: \: Xe = \frac{0.4}{0.4} = 1[/tex]
[tex]Moles \: \: of \: \: F = \frac{2.45}{0.4} = 6[/tex]
[tex]So \: \: the \: \: formula \: \: is \: \: XeF_{6 }.[/tex]
The oxidation state of Xe is +6.