3.44x10^2
you move the decimal over to get a single digit number with change. The number of times you move the decimal is the number for the 10 power
I think you divide something
Answer:
4.96E-8 moles of Cu(OH)2
Explanation:
Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.
Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.
![pH= -log[H]\\pH= -log (\frac{kw}{[OH]})](https://tex.z-dn.net/?f=pH%3D%20-log%5BH%5D%5C%5CpH%3D%20-log%20%28%5Cfrac%7Bkw%7D%7B%5BOH%5D%7D%29)
![8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}](https://tex.z-dn.net/?f=8.23%20%3D%20-%20log%28%5Cfrac%7BKw%7D%7B%5BOH%5D%7D%20%5C%5C10%5E%7B-8.23%7D%20%3D%20Kw%2F%5BOH%5D%5C%5C%20%5BOH%5D%20%3D%20Kw%2F10%5E%7B-8.23%7D)
![[OH]=1.69E-6](https://tex.z-dn.net/?f=%5BOH%5D%3D1.69E-6)
This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":
The expression for Kps is:
![Kps= [Cu^{2+}] [OH]^2](https://tex.z-dn.net/?f=Kps%3D%20%5BCu%5E%7B2%2B%7D%5D%20%5BOH%5D%5E2)
The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

"s" is the soluble quantity of Cu(OH)2.
The solution for this third grade equation is 
Now, let us calculate the moles in 1 L:

Answer:
1.5 moles of H₂SO₄ needs 3.0 moles pf KOH to be neutralized.
Explanation:
- KOH is dissociate according to the equation:
KOH → K⁺ + OH⁻.
- H₂SO₄ is dissociated according to the equation:
H₂SO₄ → 2H⁺ + SO₄²⁻.
<em>So, every 1.0 mole of KOH produces 1.0 mol of OH⁻.</em>
<em>While, every 1.0 mole of H₂SO₄ produces 1.0 mol of H⁺.</em>
<em />
Thus, every mol of H₂SO₄ needs 2.0 moles of KOH to be neutralized.
<em>So, 1.5 moles of H₂SO₄ needs (2 x 1.5 mol) = 3.0 moles pf KOH to be neutralized.</em>