For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:
Kd = c </span>× α²
α = √(Kd/c) × 100%
Kd = 6.0×10⁻⁷
c(HA) = 0.1M
α = √(6.0×10⁻⁷/0.1) × 100% = 0.23%
So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>
Overuse of the same chemicals can result in the pest becoming immune to the pesticides.
Answer:
2NaOH + H2So4 》Na2So4+ 2H2O
Explanation:
SODIUM HYDROXIDE, 2NaOH IS (aq)
FULFURIC ACID, H2So4 IS (aq)
SODIUM SULFATE, Na2So4 IS (aq)
WATER,2H20 IS (l)
Answer:
The projection of the Fisher projection of D-Fructose and D-glucose is that The carbonyl carbon in D-glucose is carbon 1 (aldehyde), whereas in D-fructose, the carbonyl group is on carbon 2 (ketone).
Explanation:
An aldehyde is a compound containing a functional group with the structure −CHO, consisting of a carbonyl center and
A ketone is a functional group with the structure RC(=O)R', where R and R' can be a variety of carbon-containing substituents.