The balanced equation will be:
Ca(IO₃)₂(s) ⇄ Ca²⁺(aq) + 2 IO₃⁻(aq)
The Ksp equation will be:
Ksp = [Ca²⁺][IO₃⁻]²
For the answer to the question above, well presumably because the exact concentration of the composition KMnO4 solution doesn't matter. <span>If the concentration of the KMnO4 solution is important (usually in titrations etc.) then it is not allowed to use a wet bottle. The water in the bottle will dilute the KMnO4 solution and change the concentration of the said compound.</span>
Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates only partially.
So, most particles of the solute is going to be in the solution of NaCl,
so<span> the lowest freezing point has the aqueous solution of NaCl.</span>
Test tube of ammonium chloride (NH4Cl) being heated over a bunsen burner flame. Ammonium chloride decomposes readily when heated, but condenses in the cooler area at the top of the test tube. This is a reversible reaction, where the ammonium chloride decomposes into the gases ammonia (NH3) and hydrogen chloride (HCl).
