The Bohr model proposed that electrons could just have characterized vitality levels thus when rotting back to a lower vitality level discharge a specific measure of vitality. Since the measure of vitality could be changed over to a specific recurrence then particular emanation lines were found in the electromagnetic range. Alternate speculations couldn't clarify the discharge lines.
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
In order of reactivity: magnesium, iron, copper, silver
Answer:c which is 70
Explanation:
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What of the gasses is N and the flame changes it's color from Orange to blue