Answer:
Explanation:
Hello,
In this case, the dissociation reaction is:
For which the equilibrium expression is:
![Ksp=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)
In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
![[Pb^{2+}]=1.39x10^{-3}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D1.39x10%5E%7B-3%7DM)
![[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D1.39x10%5E%7B-3%7DM%2A2%3D2.78x10%5E%7B-3%7DM)
Thereby, the solubility product results:
Regards.
<span>To determine the pH of the solution given, we make
use of the acid equilibrium constant (Ka) given. It is the ratio of the
equilibrium concentrations of the dissociated ions and the acid. The
dissociation reaction of the CH3COOH acid would be as follows:
</span>CH3COOH = CH3COO- + H+<span>
The acid equilibrum constant would be expressed as follows:
Ka = [H+][</span>CH3COO-] / [CH3COOH] = 1.8× 10^–5
<span>
To determine the equilibrium concentrations we use the ICE table,
CH3COOH H+ </span>CH3COO<span>-
I 1.60 0 0
C -x +x +x
----------------------------------------------------------------
E 1.60-x x x
</span>1.8× 10^–5 = [H+][CH3COO-] / [CH3COOH] <span>
1.8 x 10^-5 = [x][x] / [0.160-x] </span>
Solving for x,
x = 1.69x10^-3 = [H+] = [F-]
pH = -log [H+] = -log [1.69x10^-3] = 2.8
Answer:
The answer is C) It is given off as heat
Explanation:
There can be two electrons in one orbital maximum. The s sublevel has just one orbital, so can contain 2 electrons max. The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max.
Hope this helped
Have a nice day :)
