Question

A compound of xe and f is found to have 53.3% xe (at. Mass = 133). Oxidation number of xe in this compound is

Answer 1

Compound of Xe and F:

Percentage of Xe = 53.3%

Percentage of F = 100 - 53.3 = 46.7%

To find relative moles:

Xe = Percentage of compound/Atomic mass=53.3133 = 0.4

F = Percentage of compound/Atomic mass=46.719 = 2.45

NOW Mole ratio:

For Xe = 0.40.4= 1

For F = 2.450.4= 6

So the formula is XeF6.

Since F is present in -1 state therefore Xe exists in +6 state.

Hope you get answer of your question

Answer 2

Answer:

Percentage composition of Xe(given)=53.3%

Percentage composition

of F = (100-53.3)%

To find the relative moles

of Xe= 53.3/133=0.4(%compo./at.mass)

To find the relative moles

of F=46.7/19=2.45

Mole ratio for Xe=0.4/0.4=1

Mole ratio for F=2.45/0.4=6

So the formula is XeF6

It works dude

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