Answer:
3.0×10⁻¹³ M
Explanation:
The solubility product Ksp is the product of the concentrations of the ions involved. This relation can be used to find the solubility of interest.
<h3>Equation</h3>
The power of each concentration in the equation for Ksp is the coefficient of the species in the balanced equation.
Ksp = [Al₃⁺³]×[OH⁻]³
<h3>Solving for [Al₃⁺³]</h3>
The initial concentration [OH⁻] is that in water, 10⁻⁷ M. The reaction equation tells us there are 3 OH ions for each Al₃ ion. If x is the concentration [Al₃⁺³], then the reaction increases the concentration [OH⁻] by 3x.
This means the solubility product equation is ...
Ksp = x(10⁻⁷ +3x)³
For the given Ksp = 3×10⁻³⁴, we can estimate the value of x will be less than 10⁻⁸. This means the sum will be dominated by the 10⁻⁷ term, and we can figure x from ...
3.0×10⁻³⁴ = x(10⁻⁷)³
Then x = [Al₃⁺³] will be ...
![[\text{Al}_3^{\,+3}]=\dfrac{3.0\times10^{-34}}{10^{-21}}\approx \boxed{3.0\times10^{-13}\qquad\text{moles per liter}}](https://tex.z-dn.net/?f=%5B%5Ctext%7BAl%7D_3%5E%7B%5C%2C%2B3%7D%5D%3D%5Cdfrac%7B3.0%5Ctimes10%5E%7B-34%7D%7D%7B10%5E%7B-21%7D%7D%5Capprox%20%5Cboxed%7B3.0%5Ctimes10%5E%7B-13%7D%5Cqquad%5Ctext%7Bmoles%20per%20liter%7D%7D)
We note this value is significantly less than 10⁻⁷, so our assumption that it could be neglected in the original Ksp equation is substantiated.
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<em>Additional comment</em>
The attachment shows the solution of the 4th-degree Ksp equation in x. The only positive real root (on the bottom line) rounds to 3.0×10^-13.
If we are to draw the sulfuric acid structure based on
Lewis it would look like:
O
ll
O =
S – O –H
l
O
l
H
So from this, we can see that there are 2 double bonds (O
= S) and 4 single bonds (S – O and O – H).
Therefore the answer is:
4 single, 2 double
Explanation:
a) Cr is <u>acid</u> K is <u>base</u>
b) Ca is <u>base</u> and the other left part is <u>acid</u>
c)Ca is <u>base</u> and F2 is <u>acid</u>
<u>d</u><u>)</u><u> </u>NH4 is <u>acid</u> and SO4 is <u>base</u>
Answer:
BaCl₂ and H₂SO₄
Explanation:
The reaction is the precipitation of BaSO₄ from solution.
This means that our reactants must contain the Ba ion and SO₄²⁻ ion. The reactants that meet this criteria are the; BaCl₂ aand H₂SO₄
The reaction is given as;
BaCl₂ + H₂SO₄ --> BaSO₄ + 2HCl
The two elements that produce background radiation on earth are Radon and Uranium. Airborne radon can decay on its own. Radon undergoes alpha decay to produce Polonium. Uranium naturally undergoes alpha decay to produce Thorium.