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.
Answer: 67.23 L
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
of particles.
To calculate the number of moles, we use the equation:
Putting in the values we get:
1 mole of methane occupies = 22.4 L
Thus 3 moles of methane occupy = 
Thus volume of 48 grams of methane gas is 67.23 L
4)<span>1.24⋅<span>1024</span></span><span> formula units
</span>1 mole Na<span> (sodium)</span>
Moles/MxL
1.709moles/2.10L =<span>0.814M</span>
When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]
when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M
so by substitution we can get [Ca2+]
4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L
∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
= 0.00167 mol