CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
Answer:
I think im good at it i have an A in the class
Explanation:
lol
Answer:
pH 
Explanation:
For every mole of hydrochloric acid, one mole of hydronium ion is required. Thus, in order to neutralize 0.014 moles of HCL, 0.014 moles of hydronium is required.
![[H_3O^+] = [HCl] = 0.014](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%20%5BHCl%5D%20%3D%200.014)
pH ![= -log [H^+] = -log [H_3O^+]](https://tex.z-dn.net/?f=%3D%20-log%20%5BH%5E%2B%5D%20%3D%20-log%20%5BH_3O%5E%2B%5D)
Substituting the available values in above equation, we can say that the pH of the solution is equal to

pH 
pH of a
M HCL solution 
Answer:
it's food engineering obviously
Answer:
i think the answer is C
Explanation:
not sure plsss dont bash im a beginner