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
No, physics does not suggest an exact pace in which a chemical compound will travel. It will matter in external forces as well as the median it is travelling through.
Answer:
It lead to diseases like lung cancer
Explanation:
Answer:
The answer to your question is:
1.- CO
2.- 0.414 moles of CO2
Explanation:
Data
2CO + O2 ⇒ 2CO2
CO = 0.414 moles
O2 = 0.418
Process
theoretical ratio CO/O2 = 2/1 = 1
experimental ratio CO/O2 = 0.414/0.418 = 0.99
Then the limiting reactant is CO
2.-
2 moles of CO --------------- 2 moles of CO2
0.414 moles of CO --------- x
x = (0.414 x 2) / 2
x = 0.414 moles of CO2
Answer: 35.4 grams
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
where,
Molality = 2.65
n= moles of solute =?
= volume of solution in ml = 445 ml
Putting in the values we get:
Mass of solute in g=
Thus 35.4 grams of 
is needed to prepare 445 ml of a 2.65 m solution of .
