Answer:
S = 0.788 g/L
Explanation:
The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:
![Kps = \frac{[product]^x}{[reagent]^y}](https://tex.z-dn.net/?f=Kps%20%3D%20%5Cfrac%7B%5Bproduct%5D%5Ex%7D%7B%5Breagent%5D%5Ey%7D)
Where x and y are the stoichiometric coefficients of the product and the reagent, respectively. Because of the aggregation form, the concentration of solids is always equal to 1 for use in this equation.
Analyzing the equation, we see that for 1 mol of 
is necessary 2 mols of 
, so if we call "x" the molar concentration of 
, for we will have 2x, so:
![Kps = [Fe^{+2}].[F^-]^2\\\\2.36x10^{-6} = x(2x)^2\\\\2.36x10^{-6} = 4x^3\\\\x^3 = 5.9x10^{-7}\\\\x = \sqrt[3]{5.9x10^{-7}} \\\\x = 8.4x10^{-3} mol/L](https://tex.z-dn.net/?f=Kps%20%3D%20%5BFe%5E%7B%2B2%7D%5D.%5BF%5E-%5D%5E2%5C%5C%5C%5C2.36x10%5E%7B-6%7D%20%3D%20x%282x%29%5E2%5C%5C%5C%5C2.36x10%5E%7B-6%7D%20%3D%204x%5E3%5C%5C%5C%5Cx%5E3%20%3D%205.9x10%5E%7B-7%7D%5C%5C%5C%5Cx%20%3D%20%5Csqrt%5B3%5D%7B5.9x10%5E%7B-7%7D%7D%20%5C%5C%5C%5Cx%20%3D%208.4x10%5E%7B-3%7D%20mol%2FL)
So, to calculate the solubility (S) of FeF2, which is in g/L, we multiply this concentration by the molar mass of FeF2, which is:
Fe = 55.8 g/mol
F = 19 g/mol
FeF2 = Fe + 2xF = 55.8 + 2x19 = 93.8 g/mol
So,
[tex]S = 8.4x10^{-3}x93.8
S = 0.788 g/L
Halogens
Explanation:
Halogens are a group of non-metals located in the seventh group on the periodic table. The will only gain one electron during a chemical reaction.
- Halogens have a seven electrons in their outermost shell.
- To complete the number of electrons in this shell, they need to gain an additional electron.
- One more electron makes the halogen similar to the corresponding noble gas which is very stable.
- Halogens are very reactive groups of elements and are highly electronegative.
- They have a high affinity for electrons.
- These elements are fluorine, chlorine, bromine, iodine and Astatine.
learn more:
Halogens brainly.com/question/6324347
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Molality is the number of moles of solutes in 1 kg of solvent.
the molality of solution to be prepared is 2.0 molal.
therefore 2 moles in 1 kg water.
the mass of Li₂S required is - 46 g/mol x 2.0 mol = 92 g
the mass in 1 kg of solvent is - 92 g
Therefore mass of Li₂S required in 1600.0 g is - 92 g/kg x 1.6 kg = 147.2 g
