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:
Put up the answer choices
Explanation:
endothermic means absorbing heat
Given:
Q = 9.4 kJ/(kg-h), the heat production rate
c = 4.18 J/(g-K), the heat capacity
t = 2.5 h, amount of time
Note that
c = 4.18 J/(g-K) = 4180 J/(kg-K) = 4.18 kJ/kg-K)
Consider 1 kg of mass.
Then
Qt = cΔT
where ΔT is the increase in temperature (°K)
(1 kg)*(9.4 kJ/(kg-h))*(2.5 h) = (1 kg)*(4.18 kJ/(kg-K))*(ΔT K)
23.5 = 4.18 ΔT
ΔT = 23.5/4.18 = 5.622 K = 5.622 °C
Answer: 5.62 K (or 5.62 °C)
Answer:
Size of the nucleus of an atom is very small as compared to the size of the atom.
According to Rutherford gold foil experiment, nucleus is very small in size as compared to the size of the atom as a whole. Nucleus is very hard, dense and positively charged which consists of protons and neutrons.
Explanation: