your answer would be C hope this helps
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
I am going to say it is false.
C=0.10 mol/l
pH=-lg[H⁺]
HCl = H⁺ + Cl⁻
pH=-lgc
pH=-lg0.10=1.0
pH=1.0
Answer: 317 joules
Explanation:
The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
In this case,
Q = ?
Mass of aluminium = 50.32g
C = 0.90J/g°C
Φ = (Final temperature - Initial temperature)
= 16°C - 9°C = 7°C
Then, Q = MCΦ
Q = 50.32g x 0.90J/g°C x 7°C
Q = 317 joules
Thus, 317 joules of heat is gained.