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
D) Contain Chemical bonds.
Reaction:
<span>HCl + NaOH ---> NaCl + H2O
</span><span>1 mole of HCl = 36,5 g
</span><span>1 mole of NaOH = 40g
</span><span>so, according to the reaction:
</span><span>1 mol HCl = 1 mol NaOH
</span>so, we need > 36,5 g HCl (<u>hydrochloric acid</u><span>)
</span><u>
answer: 36,5 g HCl (hydrochloric acid)
</u><span> ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
</span><span>next question.
</span><span>
1 mole of NaCl = 58,5 g
</span><span>1 mole of H2O = 18g
</span>
so, according to the reaction:
1 mole of HCl (36,5 g) <span>----------------- - 1 mole of NaCl (58,5 g)
</span><span>(the same for NaOH)
i
</span>1 mole of HCl<span> (36,5 g) ------------------ 1 mole of H2O (18 g)
</span>(the same for NaOH)
<span>so, this reaction is stechiometric
</span><u>
answer: 58,5 g NaCl i 18g H2O</u>
Answer:
During a total lunar eclipse, the Earth lies directly between the sun and the moon, causing the Earth to cast its shadow on the moon.
Explanation:
