Delta enthalpy = 2x386-3x1x432-3x942=-3350kJ/mol
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
<u>c.</u> 12. preparing data tables and gathering safety equipment
<u>b.</u> 13. reading all instructions before beginning a science lab
<u>a.</u> 14. recognizing what a picture of a hand means
<u>e.</u> 15. wiping your work area with a wet paper towel
<u>d.</u> 16. wearing goggles and an apron
Answer:
Filtration
Explanation:
Filtration would be best because the sand particles would be trapped in the filter paper and the water would go through so the mixture would be separated
