Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
I think it's 2 I tried looking it up because I was not sure.
Answer:
See image attached and explanation
Explanation:
The stratospheric ozone layer is very important in absorbing high-energy ultraviolet radiation that is harmful to living systems on earth. The concentration of ozone in the stratosphere is determined by both thermal and photochemical pathways for its decomposition. Nitric oxide, NO, is a trace constituent in the stratosphere that reacts with ozone to form nitrogen dioxide, NO2, and the diatomic oxygen molecule. The nitrogen-oxygen bond in NO2 is relatively weak. When an NO2 molecule encounters an oxygen atom, it transfers an oxygen, forming O2 and NO. The chemical reactions involved are formations of NO2 following by reaction of NO2 with atomic oxygen for form NO and O2. The sum of both reactions show that the overall reaction is simply the reaction of ozone with atomic oxygen to form two molecules of molecular oxygen. Hence, NO only serves as a catalyst, it does not undergo a permanent change itself.
The percent by mass of potassium in K3Fe(CN)6 is 35.62%.
Answer:
The correct answer is D) hydration
