1.__Thermal Energy Part 2 is another Question! :))
1 answer:
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Answer : The mole fraction and partial pressure of and
gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.
Explanation : Given,
Moles of = 1.79 mole
Moles of = 1.20 mole
Moles of = 3.71 mole
Now we have to calculate the mole fraction of and
gases.
and,
and,
Thus, the mole fraction of and
gases are, 0.267, 0.179 and 0.554 respectively.
Now we have to calculate the partial pressure of and
gases.
According to the Raoult's law,
where,
= partial pressure of gas
= total pressure of gas = 5.78 atm
= mole fraction of gas
and,
and,
Thus, the partial pressure of and
gases are, 1.54, 1.03 and 3.20 atm respectively.
The shape of XeO₂F₂ is Trigonal bi-pyramidal see-saw tetrahedron (see attached pictures)
- As you said the hybridization of Xe here is sp³d so its geometry has to be Trigonal bi-pyramidal in which F atom located on axial positions but for the final shape we exclude lone pair on Xe to give see-saw shape (see second picture)
- Remember that we have 5 pairs (4 bond pairs + 1 lone pair) and we have to place lone pair at equatorial position.
- As you said the hybridization of Xe here is sp³d so its geometry has to be Trigonal bi-pyramidal in which F atom located on axial positions but for the final shape we exclude lone pair on Xe to give see-saw shape (see second picture)
- Remember that we have 5 pairs (4 bond pairs + 1 lone pair) and we have to place lone pair at equatorial position.