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
It will have traveled 0.78 m. You find this by multiplying .013 by 60
Answer: The rate constant is
Explanation ;
Expression for rate law for first order kinetics is given by:

where,
k = rate constant = ?
t = age of sample = 4.26 min
a = initial amount of the reactant = 2.56 mg
a - x = amount left after decay process = 2.50 mg
Now put all the given values in above equation to calculate the rate constant ,we get



Thus rate constant is [tex]0.334s^{-1}
Answer:
water evaporates from the ocean into atmosphere. water vapour condenses to form clouds. clouds produce rain. rainwater needed for plant growth.
Explanation:
Hydrogen mass is 1, the atom with mass number 7 is Lithium