Answer:
Kc = 0.075
Explanation:
The dissociation (α) is the initial quantity that ionized divided by the total dissolved. So, let's calling x the ionized quantity, and M the initial one:
α = x/M
x = M*α
x = 0.354M
For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:
2HI(g) ⇄ H₂(g) + I₂(g)
M 0 0 <em> Initial</em>
-0.354M +0.177M +0.177M <em>Reacts</em>
0.646M 0.177M 0.177M <em>Equilibrium</em>
The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):
![Kc = \frac{[H2]*[I2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BH2%5D%2A%5BI2%5D%7D%7B%5BHI%5D%5E2%7D)
Kc = 0.075
<h3>
Answer:</h3>
Al- [Ne] 3s²3p¹
As- [Ar] 4s²3d¹⁰ 4p³
Explanation:
- Electron configuration of an element shows the arrangement of electrons in the energy levels or orbitals in the atom.
- Noble-gas configuration involves use of noble gases to write the configuration of other elements.
- This is done by identifying the atomic number of the element and then identifying the noble gas that comes before that particular element on the periodic table.
- Aluminium: The atomic number of Al is 13. The noble gas before Aluminium is Neon which has 10 electrons. Therefore the remaining 3 electrons fills up the 3s and 3p sub orbitals.
- Thus, the noble-gas configuration of Al is [Ne] 3s²3p¹
2. Arsenic, Atomic number is 33
- Noble gas before Arsenic is Ar,. Argon has 17 electrons, then the remaining electrons fills up the 4s, 3d and 4p sub-orbitals.
- Thus, the noble-gas configuration of As is [Ar] 4s²3d¹⁰ 4p³
Sodium Sulfate
= Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M
stands for Molarity, defined as Molarity = (moles of solute)/(Liters of
solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65
moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+
ions in solution for every one Na2SO4.
Therefore if
3.65 moles of Na2SO4 was to dissolve, it would produce 7.3 moles of Na+, and
since this is still a theoretical solution, we can assume 1 L of solution.
Finally we find
[Na+] = 2*3.65 = 7.3M
Use the same
logic for parts b and c
