The correct definition of density is the degree of compactness of a substance.
I think is 6.588579795x10^13 Kg because the equation is E=mc^2 and E is your Joules and c^2= 9x10^16 so m=(c^2)/E
Answer:
A reaction is non-spontaneous at any temperature when the Gibbs free energy > 0.
Explanation:
There is a state function, that determines if a reaction is sponaneous or non spontaneous:
ΔG = Gibbs free energy
A reaction is non spontaneous when it does require energy to produce that reaction. It will be spontaneous, when the reaction does not require energy to be occured.
The formula is: ΔG = ΔH - T.ΔS
ΔH → Enthalpy → Energy gained or realeased as heat.
ΔH < 0 → <em>Exothermic reaction. Spontaneity is favored
</em>
T → Temperature
ΔS → Entropy → Degree of disorder of a system.
When the system has a considered disorder ΔS > 0, disorder increases.
When the system is more ordered, ΔS < 0, disorder decreases.
The reaction will be non spontaneous if, the enthalpy is positive (endothermic reaction) and the ΔS < 0 (disorder decreases). It will not occur if we do not give energy.
ΔG < 0 → Spontaneous reaction
ΔG > 0 → Non spontaneous reaction
ΔG = 0 → System in equilibrium
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:
