An aqueous solution of iron(II) iodide has a concentration of 0.215 molal. The percent by mass of iron(II) iodide in the solutio
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Answer: Option (d) is the correct answer.
Explanation:
When a hydrogen atom comes in contact with an electronegative atom then it results in the formation of a chemical bond.
More is the electronegativity of combining atom, more stronger will be the bond with hydrogen atom. As a result, the compound formed will not easily give up hydrogen atom upon dissociation.
Whereas less is the electronegativity of atom combining with hydrogen atom, easily it will donate the hydrogen atom upon dissociation.
Since, out of the given option sulfur (S) atom has low electronegativity as compared to oxygen and nitrogen atom.
Hence, will easily donate hydrogen atom.
Thus, we can conclude that molecule would be the best hydrogen bond donor.
Answer : The concentration of at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.
Explanation :
The given chemical reaction is,
First we have to calculate the equilibrium constant for the reaction.
The relation between the equilibrium constant and standard free‑energy is:
where,
= standard free‑energy change = -4.20 kJ/mole
R = universal gas constant = 8.314 J/mole.K
k = equilibrium constant = ?
T = temperature =
Now put all the given values in the above relation, we get:
Now we have to calculate the concentrations of A, B, and C at equilibrium.
The given equilibrium reaction is,
Initially 0.30 0.40 0
At equilibrium (0.30-x) (0.40-x) x
The expression of equilibrium constant will be,
By solving the term x, we get
From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.
The value of x will be, 0.168 M
The concentration of at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M
The concentration of at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M
The concentration of at equilibrium = x = 0.168 M