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
The given question is incomplete . The complete question is :
In ionic bonding, during the transfer of electrons between two neutrally charged atoms, one electron moves from one atom to another. What are the new relative charges between the two atoms?
a. The giving atom and receiving atom are both negatively charged
b. The giving atom is now positively charged and the receiving atom is now negatively charged.
c. The giving and receiving atom are both positively charged
d. The giving atom is now negatively charged and the receiving atom is now positively charged.
Answer: The giving atom is now positively charged and the receiving atom is now negatively charged.
Explanation:
Ionic compounds are formed by transference of electrons between metals and non metals. The bond formed between a metal and a non-metal is always ionic in nature.
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Thus when one electron moves from one atom to another, the metal acquires a positive charge and the non metal acquires negative charge.
For example: 
is formed by transfer of one electron from sodium to chlorine , thus forming 
and 
It looks all correct to me, great job!
Answer : The equilibrium concentration of 
in the solution is, 
Explanation :
The dissociation of acid reaction is:
Initial conc. c 0 0
At eqm. c-x x x
Given:
c = 
The expression of dissociation constant of acid is:
![K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D)
Now put all the given values in this expression, we get:
![6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}](https://tex.z-dn.net/?f=6.3%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%5B%287.0%5Ctimes%2010%5E%7B-2%7D%29-x%5D%7D)
Thus, the equilibrium concentration of in the solution is,
This is late but for anyone else who needs it...It's D. Far left
