<span><span>LiF, LiCl, LiBr, LiI, LiAtNaF, NaCl, NaBr, NaI, NaAtKF, KCl, KBr, KI, KAt</span><span>RbF, RbCl, RbBr, RbI, RbAt CsF, CsCl, CsBr, CsI, CsAt FrF, FrCl, FrBr, FrI, FrAt<span>
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From the calculation as shpwn in the procedure below, the equilibrium constant of the substance is 6.9 * 10^-15.
<h3>What is equilibrium constant?</h3>
The equilibrium constant for the solubility of aa solid in solution is called the solubility product Ksp. The Ksp shows the extent to which a solid is dissolved in solution.
Given that;
Fe(OH)2 ⇄Fe^2+ + 2(OH)^-
Ksp = s(2s)^2
We have s as 1.2 x 10^-5 M
So
Ksp = 4s^3
Ksp = 4( 1.2 x 10^-5 )^3
Ksp = 6.9 * 10^-15
Learn more about Ksp:brainly.com/question/27132799
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I would say b is the correct answer
Answer:
Assign oxidation numbers to all atoms in the equation.
Compare oxidation numbers from the reactant side to the product side of the equation.
The element oxidized is the one whose oxidation number increased.
Explanation: