The most likely van't Hoff factor for an 0.01 m calcium iodide solution is
1 answer:
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
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Hello!
In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.
We have the following data:
m(Fe) - mass of iron = 28 g
m(S) - mass of sufur = 16 g
MM(Fe) - molar mass of iron ≈ 56 g/mol
MM(S) - molar mass of sulfur ≈ 32 g/mol
n(Fe) - number of mol of iron = ?
n(S) - number of mol of sulfur = ?
Solving:
* to n(Fe)
* to n(S)
The stoichiometric reaction will be in the same proportion (1 : 1), let us see:
1 mol of Fe -------------- 1 mol of FeS
0.5 mol of Fe ------------ 0.5 mol of FeS
Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:
n(FeS) - number of mol of iron sulfide = 0.5 mol
m(FeS) - mass of iron sulfide = ? (in grams)
MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol
Solving:
Answer:
44 grams of iron sulfide
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