Answer:
Explanation:
it is very important because with out it we will die
The concentration of hydrogen ions in a solution is a measure of its acidity. So the correct option is (b) false.
When an Arrhenius acid is dissolved in water, hydrogen ions are produced:
H+(aq) + A- = HA + H2O (aq)
Here, H+ is the hydrogen cation, A- is the solvated anion, also known as the conjugate base, and HA is the non-dissociated acid. When an Arrhenius base is dissolved in water, hydroxide ions are produced:
BOH + H2O → B+(aq) + OH-(aq)
Is a material with at least one hydrogen atom that has the ability to split apart in an aqueous solution to produce an anion and an H + ion (a proton), creating an acidic solution. Bases are substances that, when dissolved in water, create hydroxide ions (OH) and a cation, resulting in a basic solution.
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There are only two possible structures for an octahedral
molecule with a formula of ax4y2. One is when the two y’s are next to each
other and the other one is when the two y’s are of the opposite side of the molecule.
Different structures can be drawn but only two types of molecule can be formed.
The empirical formula of the substance would be 
<h3>Empirical formula calculation</h3>
The constituents of the compound are as follows:
N = 54/201.6 = 26.78% = 26.78/14 = 1.91 moles
H = 15/201.6 = 7.44% = 7.44/1 = 7.44 moles
Cl = 132.6/201.6 = 65.48% = 65.48/35.5 = 1.84 moles
Divide the number of moles by the smallest"
N = 1.91/1.84 = 1
H = 7.44/1.84 = 4
Cl = 1.84/1.84 = 1
Thus, the empirical formula is 
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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:
