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:
7mL of sterile water is the initial amount of the concentrated solution is 3mL
Explanation:
In this problem, the vial must be <em>diluted </em>from 5mg/mL to 1.5mg/mL, that means the solution must be diluted:
5mg/mL / 1.5mg/mL = 3.33 times
If the initial amount of the drug in the vial is 3mL, the final volume must be:
10mL
That means the volume of water that should be added is:
10mL - 3mL:
<h3>7mL of sterile water is the initial amount of the concentrated solution is 3mL</h3>
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(Credits to google)
Answer:
212.304 grams
Explanation:
similar to your other question, use the same formula
q=mCpΔT
23617=m(4.182)(46.6-20)
23617=111.2412m
m=212.304g
Volume is 60 and the area is 94 have a great day
