The chemical compound sodium sulfide has the formula Na2S. It is written as 
. It has a crystalline state. Na is followed by a two or 2. this indicates that the molecule has two of the specified element's atoms.
There is just one of these because there is no number following the S.
In all, this molecule consists of one sulfur atom and two sodium atoms.
A compound's chemical formula, which uses symbols for combining atoms, reveals the composition of the complex. We require the valency of the combining parts in order to write the formula.
The sodium sulfide chemical formula must be written down. So sodium and sulfur make up the two elements in the chemical. Sodium has a valency of +1, while sulfur has a +2.
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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:
C) can give up electrons more readily than others.
This is in the assumption that these are the only 2 isotopes. This means that:
abundance of Li-6 = X
abundance of Li-7 = 1-X
Hence from your equations
6.9409 = (X * 6.015121amu) + ((1-X) * <span>7.016003amu</span>)
Solve for X = 0.075 or 7.50%
Answer:
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