Answer:Butane > ethane > methane, because between bigger molecules there are stronger van der Waals forces and also higher molar mass means they need to be given more energy to have enough kinetic energy to move quickly, freely in gas.
There are multiple butene isomers (Butene) and some (2-Butenes - cis and trans) actually have higher boiling point than n-Butane (there is also Isobutane, of course, with quite much lower boiling point than all of them) and some (1-Butene, Isobutylene) have lower, so this isn't really a fair or simple question. But on simplest level, it can again be said that 1-butene has lower boiling point because it has very similar shape but slightly lower molar mass (2H less) than n-butane.
Explanation:
a strong acid fully dissociates in water to form H+ ions in water while a weak acid only partially dissociates to form H+ ions in water
The answer that h are looking for is c
Answer : The equilibrium concentration of 
will be, (C) 
Explanation : Given,
Equilibrium constant = 14.5
Concentration of 
at equilibrium = 0.15 M
Concentration of 
at equilibrium = 0.36 M
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the values in this expression, we get:
![14.5=\frac{[CH_3OH]}{(0.15)\times (0.36)^2}](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%280.15%29%5Ctimes%20%280.36%29%5E2%7D)
![[CH_3OH]=2.82\times 10^{-1}M](https://tex.z-dn.net/?f=%5BCH_3OH%5D%3D2.82%5Ctimes%2010%5E%7B-1%7DM)
Therefore, the equilibrium concentration of will be, (C)
