2H2(g)+O2(g)→2H2O(l)
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Answer : The vapor pressure of propane at is 17.73 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of propane at
= ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane =
= normal boiling point of propane =
= heat of vaporization = 24.54 kJ/mole = 24540 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of propane at is 17.73 atm.
Any buffer exists in this equilibrium
HA <=>
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base (
)
When a strong acid is added, it reacts with the large reservoir of the conjugate base () forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.
HA <=>
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base (
When a strong acid is added, it reacts with the large reservoir of the conjugate base (