Displcement reaction. metal added to solutions containing metal that is less reactive would have visible reaction while less reactive metal will have no visible reaction. Eg, if copper is added to Mgcl2 andZnCl2, there will be no visible reaction. If Mg is added to CuCl2, blue solution will fade to form colourless solution and a reddish brown ppt of Cu will be formed.
This problem is providing us with the molality of a solution of calcium iodide as 0.01 m. So the most likely van't Hoff factor is required and theoretically found to be 3 due to the following:
<h3>Van't Hoff factor:</h3>
In chemistry, the correct characterization of solutions also imply the identification of the ions it will release in aqueous solution. For that reason, the van't Hoff factor gives us an idea of this number, according to the formula the solute has got.
In such a way, for calcium iodide, we write its ionization equation as shown below:

Assuming it is able to ionize due to the low molality, because if it was higher, then it won't ionize. Hence, since we have three moles of ion products, one Ca²⁺ and two I⁻, we can conclude the van't Hoff factor would be 3, although calculations may lead to a different, yet close result.
Learn more about the van't Hoff factor: brainly.com/question/23764376
Answer:
Covalent solids, also called network solids, are solids that are held together by covalent bonds. As such, they need localized electrons (shared between the atoms) and therefore the atoms are arranged in fixed geometries. Distortion far from this geometry can only occur through a breaking of covalent sigma bonds.
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