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:
bonding driven by ionic interactions.
Explanation:
C. 50%
Unless the question is saying he only gets heads once, in which case it would be 0%. Or the coin could have 2 heads. Then it would be 100%.
But I'm pretty sure it's 50%.
Answer:
17.1 mol
Explanation:
(8.68g/mL * 125 mL) = 1085 g
1085 g/ (63.55 g/mol) = 17.1 mol
Answer:
To the best of my knowledge, it is because of the amount of gamma rays is given off.
Explanation:
While both are isotopes, Potassium 40 gives off fewer gamma rays compared to Cobalt 60. Potassium 40 isn't really harmful to humans, but Cobalt 60 (I believe) is used in chemotherapy.