The matter is going to have to be in a <em>plasma </em>state! =)
Explanation:
Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.
It is known that at standard condition, vapor pressure is 760 mm Hg.
And, it is given that methanol vapor pressure in air is 88.5 mm Hg.
Hence, calculate the volume percentage as follows.
Volume percentage = 
= 
= 11.65%
Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.
I think the answer is 4 carbon dioxide
Answer:
90.3 L
Explanation:
Given data:
Volume of water produced = 77.4 L
Volume of oxygen required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
It is known that,
1 mole = 22.414 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
There are 6 moles of water = 6×22.414 = 134.5 L
Now we will compare:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.