Answer:
1.427x10^-3mol per L
Explanation:
I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L
Answer :
The rate law expression for zero order reaction will be:
The rate law expression for first order reaction will be:
The rate law expression for second order reaction will be:
Zero order reaction : There is no affect on the rate law.
First order reaction : The rate law becomes doubled.
Second order reaction : The rate law becomes quadrupled.
Explanation :
Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
The given reaction is:
The rate law expression for zero order reaction will be:
The rate law expression for first order reaction will be:
The rate law expression for second order reaction will be:
Now we have to determine that if doubling of the concentration of A then the rate of reaction will be:
As we know that the zero order reaction does not depend on the concentration of reactant. So, there is no affect on the rate law.
As we know that the first order reaction depend on the concentration of reactant. So, the rate law becomes doubled.
As we know that the second order reaction depend on the concentration of reactant. So, the rate law becomes quadrupled.