At equilibrium the concentrations of:
[HSO₄⁻] = 0.10 M;
[SO₄²⁻] = 0.037 M;
[H⁺] = 0.037 M;
There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.
HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.
R 
⇄ 
I 
C 
E 
× 
for 
. As a result,
![\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B%5D.%20%5BSO_4%5E2%5E-%5D%7D%7BHSO_4%5E-%7D%20%3D%20K_a)
is large. It is no longer valid to approximate that ![[HSO^-_4]](https://tex.z-dn.net/?f=%5BHSO%5E-_4%5D)
at equilibrium is the same as its initial value.
× 
× 
Solving the quadratic equation for 
since represents a concentration;
Then, round the results to 2 significant figure;
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Answer:
Explanation:
Cesium Lewis dot structure would look like this:
·Cs, because it only has one valence electron. But, since it has a plus, that means we lost an electron. So, we have to get rid of the dot and you have:
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
A polar molecule<span> has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from </span>polar<span> bonds arranged asymmetrically. Water (H</span>2<span>O) is an example of a </span>polar molecule<span> since it has a slight positive charge on one side and a slight negative charge on the other.</span>
