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:
2 NO (g) → N2 (g) + O2 (g)
2 NOCl (g) → 2 NO (g) + Cl2 (g)
____________________________
2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)
ΔH = [90.3 kJ x 2 x -1] + [-38.6 kJ x -1 x 2] = -103.4 kJ
The ΔH for the reaction is -103.4 kJ
The position of equilibrium lies far to the right, with products being favored.
Answer:
the stabilization of the negative charge in orbitals with higher s character
Explanation:
Acetylide anion is a carbon anion compound or popularly called carbanion. Now Acetylide anion is sp hybridized. However acetylide anion tends to be more acidic as we move from sp³ to sp, hence acidicity increases, which makes sp to have the highest acidity and become the most stable.
So, we can conclude that the acetylide anion is more acidic due to the stabilization of the negative charge in orbitals with higher s character and as the s character increases, acidic nature of acetylide anion also increases.
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