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:
Acid(BSA) = CH₃COOH
Base (BSB) = H₂O
Conjugate base (CB) = CH₃COO⁻
Conjugate acid (CA) = H₃O⁺
Explanation:
Equation of reaction;
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
Hello,
From my understanding of the question, we are required to identify the
1) Acid
2) Base
3) conjugate acid
4) conjugate base in the reaction
Acid (BSA) = CH₃COOH
Base (BSB) = H₂O
CA = conjugate acid = H₃O⁺
CB = conjugate base = CH₃COO⁻
Answer:
A. cycas
1. it is thick and scaly
2. it grow relatively slowly and have a large, terminal rosette of leaves.
B. bamboo
1. it is very durable
2. it is both flexible and elastic
C. lemna
1. it grows as simple free-floating thalli on or just beneath the water surface
2. it are small, not exceeding 5 mm in length
D. paddy
1. it is variety purity
2. it degree of purity
E. sugarcane
1. it is bear long sword-shaped leaves
2. it's stalks are composed of many segments, and in each joint there is a bud
What are the answer choices?
