Answer:
They become ductile and deform plastically
Explanation:
When rocks are buried by the materials up to a greater depth, then the confining pressure increases significantly. This results in the ductile behavior of the rocks at such depth. These rocks are present in the ductile region where the depth is about more than 20 to 30 km. Here the rocks are subjected to extremely high pressure and temperature conditions, which favors the transformation of rocks into more higher-grade metamorphic rocks. It is also enhanced due to the geothermal gradient.
Under such high pressure and temperature, the rocks show the behavior of plasticity, where the rocks undergo bending, buckling as well as they tend to flow, and there occurs low strain rate, resulting in the permanent deformation of rocks.
Thus, the rocks become ductile and deform plastically at such conditions.
I’m pretty positive the answer is True
The answer is Electron
Hope this helped!!! :)
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:
Mg(OH)₂ ⇄ Mg²⁺ + 2 OH⁻
Ksp = [Mg²⁺] [OH⁻]²
6.0 x 10⁻¹⁰ = 0.10 x [OH⁻]²
[OH⁻] = 7.746 x 10⁻⁵ M
when Mg(OH)₂ 1st precipitates, [OH⁻] = 7.746 * 10⁻⁵ M
Fe(OH)₂ <—> Fe²⁺ + 2OH⁻
Ksp = [Fe²⁺] [OH⁻]²
7.9 x 10⁻¹⁶ = [Fe²⁺] x (7.746 x 10⁻⁵)²
[Fe²⁺] = 1.32 x 10⁻⁷ M
Answer: 1.32 x 10⁻⁷ M
