The solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What is the Ksp of barium carbonate

Ksp = 2.58x10⁻⁹

Explanation:

The equilibrium that takes place is:

BaCO₃(s) ↔ Ba⁺² + CO₃⁻³Ksp = [Ba⁺²] [CO₃⁻³]

The molar concentration of Ba⁺² at equilibrium is the solubility. The same can be said about CO₃⁻³:

Ksp = s * sKsp = s²

So now we can calculate the Ksp from the solubility, after converting the solubility from g/L to mol/L:

0.0100 g BaCO₃/L ÷ 197.34 g/mol = 5.07x10⁻⁵ mol/L

Ksp = (5.07x10⁻⁵ mol/L)²

Ksp = 2.58x10⁻⁹

This is what i found on google “when one phosphate group is removed by breaking a phosphoanhydride bond in a process called hydrolysis, energy is released, and atp is converted to adenosine diphosphate (adp). likewise, energy is also released when a phosphate is removed from adp to form adenosine monophosphate (amp).” i hope this you .