Answer:
See Explanation
Explanation:
Ionization energy refers to the energy required to remove an electron from an atom. Metals have lower ionization energy than non metals since ionization energy increases across a period.
One thing that we must have in mind is that it takes much more energy to remove an electron from an inner filled shell than it takes to remove an electron from an outermost incompletely filled shell.
Now let us consider the case of magnesium which has two outermost electrons. Between IE2 and IE3 we have now moved to an inner filled shell(IE3 refers to removal of electrons from the inner second shell) and a lot of energy is required to remove an electron from this inner filled shell, hence the jump.
For aluminium having three outermost electrons, there is a jump between IE3 and IE4 because IE4 deals with electron removal from a second inner filled shell and a lot of energy is involved in the process hence the jump.
Hence a jump occurs each time electrons are removed from an inner filled shell.
Answer:
When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
Explanation:
Ksp of BaF₂ is:
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)
Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.
As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to ksp just when:
1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
1.7x10⁻⁶ = [0.0144M] [F⁻]²
1.18x10⁻⁴ = [F⁻]²
0.0109M = [F⁻]
That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
D) Transportation is the correct answer
18. the blank should be "
"
19. the question seems to be answered.
20. 
Answer:
New pH = 3.84
Explanation:
First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.
Let's determine the moles of each compound:
0.23 M . 1.3L = 0.299 moles of NaBz
0.38 M . 1.3L = 0.494 moles of HBz
We add 0.058 of HCl, which is the same as 0.058 moles of H⁻
HCl → H⁺ + Cl⁻
As we add the moles of protons, these are going to react to the Bz⁻
In the buffer system we have these dissociations:
NaBz → Na⁺ + Bz⁻
HBz → H⁺ + Bz⁻
So, as we add protons, we have a new equilibrium:
Bz⁻ + H⁺ ⇄ HBz
In 0.299 0.058 0.494
Eq 0.241 - 0.552
Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation
pH = pKa + log (Bz⁻/HBz)
pH = 4.20 + log (0.241 / 0.552) → 3.84
