What is the solubility in pure water of ba(io3)2 in moles per liter at 25 ˚c? [ksp (25 ˚c) = 6.0  10–10]?

Ba(IO₃)₂(s) partially dissociates in water  as Ba²⁺(aq) and IO₃⁻(aq).
                Ba(IO₃)₂(s) ⇄ Ba²⁺(aq) + 2IO₃⁻(aq)
Initial          
Change        -X                +X             2X
Equilibrium                        X              2X

           Ksp = [Ba²⁺(aq)] x [IO₃⁻(aq)]²
6.0 x 10⁻¹⁰  = X * (2X)²
6.0 x 10⁻¹⁰ = 4X³
              X = 5.313 x 10⁻⁴ mol/L

Hence, the solubility of the Ba(IO₃)₂(s) is 5.313 x 10⁻⁴ mol/L

1. and 2. question: in which equation? or i don't understand you, 3. a), 4. c), 5. c), 6. b), 7. d), 8. d) if small number is underlined(behind symbol) , b) if big number is underlined(in front of symbol)9. what is underlined there? 10. c)

Ithinks its false. tell me if im wrong