Answer:
Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²
Explanation:
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²
Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48
Ca. #Substrate e⁻ = 6F = 6(8) = 48
#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs
#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs
BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²
XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)
1000 kilojoules of electrical energy gets transformed into 1000 kilojoules of kinetic energy.
Answer:
The pressure is 58.75 atm.
Explanation:
From Vanderwaal's equation,
P = nRT/(V-nb) - n^2a/V^2
n is the number of moles of SO2 = mass/MW = 500/64 = 7.81 mol
R is gas constant = 0.0821 L.atm/mol.K
T is temperature of the vessel = 633 K
V is volume of the vessel = 6.3 L
a & b are Vanderwaal's constant = 6.865 L^2.atm/mol^2 and 0.0567 L/mol respectively.
P = (7.81×0.0821×633)/(6.3 - 7.81×0.05679) - (7.81^2 × 6.865)/6.3^2 = 69.30 - 10.55 = 58.75 atm
