Answer:
B) The molecules were closer together when the juice pop was frozen.
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
Answer:
CHO
Explanation:
Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%
Step 1:
Divide through by their respective relative atomic masses
41/ 12, 4.58/1, 54.6/16
3.41 4.58 3.41
Step 2:
Divide by the lowest ratio:
3.41/3.41, 4.58/3.41, 3.41/3.41
1, 1, 1
Hence the empirical formula is CHO
Polarity, cohesion, adhesion, surface tension, high specific heat, and evaporating cooling
The equilibrium for the dissolution of the weak base is ;(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH3^+(aq) + OH^-(aq)
<h3>What is a weak base?</h3>
A weak base is one that does not ionize completely in solution. As such, a weak base will have a very low base dissociation constant Kb reflecting its minimal dissociation in solution.
The question is incomplete hence we are are unable to work out the equilibrium but in solution it will look like this;
(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH3^+(aq) + OH^-(aq)
Learn more about weak base: brainly.com/question/4131966