Answer: 20L of H2O
Explanation:
C3H8 + 5O2 → 3CO2 + 4H2O
Recall 1mole of a gas contains 22.4L at stp
5moles of O2 contains = 5 x 22.4 = 112L
4moles of H2O contains = 4 x 22.4 = 89.6L
From the equation,
112L of O2 produced 89.6L H2O
There for 25L of O2 will produce XL of H2O i.e
XL of H2O = (25 x 89.6)/112 = 20L
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So,
for the given reaction is -99.4 J/K
Answer:
The pH of the solution will be 7.53.
Explanation:
Dissociation constant of KClO=
Concentration of acid in 1 l= 0.30 M
Then in 200 ml = 
The concentration of acid, HClO=[acid]= 0.006 M
Concentration of salt in 1 L = 0.20 M
Then in 300 ml = 
The concentration of acid, KClO=[salt]= 0.006 M
The pH of the solution will be given by formula :
![pH=pK_{a}^o+\log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%5Eo%2B%5Clog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
![pH=-\log[2.8\times 10^{-8}]+\frac{[0.06 M]}{[0.06 M]}](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B2.8%5Ctimes%2010%5E%7B-8%7D%5D%2B%5Cfrac%7B%5B0.06%20M%5D%7D%7B%5B0.06%20M%5D%7D)
The pH of the solution will be 7.53.