Explanation:
The given data is as follows.
Pressure (P) = 760 torr = 1 atm
Volume (V) =
= 0.720 L
Temperature (T) =
= (25 + 273) K = 298 K
Using ideal gas equation, we will calculate the number of moles as follows.
PV = nRT
Total atoms present (n) =
=
= 0.0294 mol
Let us assume that there are x mol of Ar and y mol of Xe.
Hence, total number of moles will be as follows.
x + y = 0.0294
Also, 40x + 131y = 2.966
x = 0.0097 mol
y = (0.0294 - 0.0097)
= 0.0197 mol
Therefore, mole fraction will be calculated as follows.
Mol fraction of Xe =
= 
= 0.67
Therefore, the mole fraction of Xe is 0.67.
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Answer:
Explanation:
M(s) → M (g ) + 20.1 kJ --- ( 1 )
X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )
M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )
( 3 ) - 2 x ( 2 ) - ( 1 )
M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ
0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ
4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ
heat of formation of M X₄ (g ) is - 773.4 kJ
Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole