A mixture consists of numerous substances, whereas a substance does not consist of mixtures, but rather of compounds. Mixtures can be separated physically, while substances cannot.
Answer:
76,6 kg
Explanation:
A kg it's equal to 1x10^3 grams
A Gigagrams it's equal to 1x10^9 grams
Knowing this, a kg it's equal to 1x10^6 gigagrams
![7,66*10^{-5}[gigagram]*\frac{1*10^6 [kg]}{1 [gigagram]}= 76.6 [kg]](https://tex.z-dn.net/?f=7%2C66%2A10%5E%7B-5%7D%5Bgigagram%5D%2A%5Cfrac%7B1%2A10%5E6%20%5Bkg%5D%7D%7B1%20%5Bgigagram%5D%7D%3D%2076.6%20%5Bkg%5D)
Answer:
so that the can fiuger out what it is for the answer
Explanation:
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:

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