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.
Answer:
You're going to have to convert the grams to moles, and then multiply that with the ratio of heat produced to the ratio of CH4
Explanation:
In first one it should be 2Mg
and in second = 2Mgo
hope it helps....
