In the given above, we have two densities which are 0.89 g/mL and 0.72 g/mL. We are also given that the liquids are immiscible. After the settlement of the liquids, they will form two layers.
The heavier substance, the one which has a higher density will be at the bottom and the lighter substance, the one which has a lower density will be at the top layer.
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:
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:
How does the energy required to remove an electron from an atom change as you move left to right in Period 4 from potassium through iron? ... A greater nuclear charge pulls the electrons closer to the nucleus, decreasing the atomic radius.
Answer:
The answer to your question is P2 = 170.9 torr
Explanation:
Data
Volume 1 = 12.1 l Volume 2 = 21.1 l
Temperature 1 = 241 °K Temperature 2 = 298°K
Pressure 1 = 546 torr Pressure 2 = ?
Process
To solve this problem use the combined gas law.
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = T1V1T2 / T1V2
-Substitution
P2 = (241 x 12.1 x 298) / (241 x 21.1)
-Simplification
P2 = 868997.8 / 5085.1
-Result
P2 = 170.9 torr