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Oksanka [162]
2 years ago
11

The most likely van't Hoff factor for an 0.01 m calcium iodide solution is

Chemistry
1 answer:
monitta2 years ago
5 0

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:

CaI_2\rightarrow Ca^{2+}+2I^-

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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4 0
2 years ago
9. Suppose that 25.0 mL of a gas at 725 mm Hg and 20°C is converted to standard
storchak [24]

Answer:

V₂ = 22.23 mL

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 25 mL

Initial pressure = 725 mmHg (725/760 =0.954 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final pressure = standard = 1 atm

Final temperature = standard = 273.15 K

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁  P₂

V₂ = 0.954 atm × 25 mL × 273.15 K / 293 K × 1 atm

V₂ =  6514.63 mL . atm . K  / 293 K . atm

V₂ = 22.23 mL

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2 years ago
From the list which items of ppe should always be put on first
fenix001 [56]
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3 years ago
If 4.0 mol of NO and 4.0 mol of O2 are combined, how many moles
Masja [62]
4.0


i think it has something to do with molar ratios and finding the limiting reactant

4.0 mol NO * 2 mol NO2/2 mol NO = 4.0 moles of NO2

4.0 mol O2 * 2 mol NO2/1 mol O2 = 8.0 moles of NO2

so the limiting reactant (the reactant that runs out the quickest leaving an excess) is NO

once the limiting reactant is found, we can use that data for that substance to calculate the amount of product

4.0 mol NO * 2 mol NO2/2 mole NO = 4.0 moles of NO2

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3 years ago
Give the electronegativity for the following elements of LI
amid [387]

Answer:

0.98

Explanation:

7 0
2 years ago
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