Answer:
The correct answer is: K'= 0.033.
Explanation:
The formation of HI from H₂ and I₂ is given by:
H₂ + I₂ → 2 HI K= 29.9
The decomposition of HI is the reverse reaction of the formation of HI:
2 HI → H₂ + I₂ K'
Thus, K' is the equilibrium constant for the reverse reaction of formation of HI. It is calculated as the reciprocal of the equilibrium constant of the forward reaction (K):
K' = 1/K = 1/(29.9)= 0.033
Therefore, the equilibrium constant for the decomposition of HI is K'= 0.033
1) 2C₆H₆ + 15O₂ = 6H₂O + 12CO₂
2) n(C₆H₆)/2=n(CO₂)/12
n(CO₂)=6n(C₆H₆)
n(CO₂)=6*12.8 mol = 76.8 mol
The compound potassium carbonate(K₂CO₃) is soluble in the water. The formulas for the ions that interact with the water is K⁺ and (CO₃)²⁻.
Dissociation of any compound is defined as the breaking of a compound into a simpler substance which is capable of recombining under different conditions. The compound potassium carbonate(K₂CO₃) is an ionic compound, and ionic compound is formed by a combination of cation and anion. When ionic compounds are dissolved in the water, it separates into the cations and anions.
Potassium carbonate(K₂CO₃) is a strong electrolyte. It means that it is completely dissociate in the water. Its dissociation in the water is shown as
K₂CO₃(aq)→2K⁺(aq)+(CO₃)²⁻(aq)
Therefore, the compound potassium carbonate forms K⁺ and (CO₃)²⁻ ions when interact with the water.
The question looks incomplete but the complete question is
The compound potassium carbonate is a strong electrolyte. Write the reaction when solid potassium carbonate is put into water
To know more about dissociation
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