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:
4.36 g of Carbon
Solution:
Step 1: Calculate the %age of Carbon in given Solid as;
Mass of Carbon = 35.8 g
Mass of Hydrogen = 3.72
Total Mass = 35.8 g + 3.72 = 39.52 g
%age of Carbon = (35.8 g ÷ 39.52 g) × 100
%age of carbon = 90.58 %
Step 2: Calculate grams of Carbon in 4.82 g of given solid as;
Mass of Carbon = 4.82 g × (90.58 ÷ 100)
Mass of Carbon = 4.36 g
There are a total of 4 elements
That would be cause part of the sodium is pure and that means it still kind of has it properties when it was an element and that its i think.
