Choose an expression for the acid ionization constant (Ka)(Ka) for HFHF. Choose an expression for the acid ionization constant for . Ka=[H3O+][F−][HF]Ka=[H3O+][F−][HF] Ka=[F−][H3O+][HF]Ka=[F−][H3O+][HF] Ka=[HF][H3O+][F−]Ka=[HF][H3O+][F−] Ka=[H3O+][HF][F−]Ka=[H3O+][HF][F−]

Explanation:

Let's consider the acid dissociation of hydrofluoric acid according to Brönsted-Lowry theory.

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)

The equilibrium constant for this reaction is known as acid ionization constant (Ka). It only includes aqueous species and gases if there were. The acid ionization constant is:

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

it's c (engineer's do their work with mechanical stuff not on any living being)