The ksp of copper(ii) carbonate, cuco3, is 1.4 × 10-10. calculate the molar solubility of this compound.

As,
                              CuCO₃    ⇆  Cu²⁺  +  CO₃²⁻
So,
                             Kc  =  [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
                             Kc (CuCO₃)  =  [Cu²⁺] [CO₃²⁻]
Or,
                             Ksp  =  [Cu²⁺] [CO₃²⁻]
As,
Ksp  = 1.4 × 10⁻¹⁰
So,
                            1.4 × 10⁻¹⁰  =  [x] [x]
Or,
                             x²  =  1.4 × 10⁻¹⁰ 
Or,
                             x  =  1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
                             x  =  1.18 × 10⁻⁵ mol/L × 123.526 g/mol

                             x  =  1.45 × 10⁻³ g/L

The nucleus inside an atom consists of positively-charged  protons  and neutrally-charged  neutrons. a  proton  is a subatomic particle with a positive electric charge. allprotons  are identical.

Idid the calculation and i get 875,000,000 which can be written as 8.75 x 10^6?