explain why the crystal structure of barium fluoride, BaF2, differs from that of magnesium fluoride, MgF2
1 answer:
You might be interested in
Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka=
Ka=
Ka=
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= =
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
Answer:
Kp = 0.022
Explanation:
<em>Full question: ...With 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. </em>
<em />
The equilibrium of ammonia occurs as follows:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where Kp is defined as:
<em>Where P represents partial pressure of each gas.</em>
<em />
As initial pressure of ammonia is 2.3atm, its equilibrium concentration will be:
P(NH₃) = 2.3atm - 2X
<em>Where X represents reaction coordinate</em>
<em />
Thus, pressure of hydrogen and nitrogen is:
P(N₂) = X
P(H₂) = 3X.
As partial pressure of hydrogen is 0.69atm:
3X = 0.69
X = 0.23atm:
P(NH₃) = 2.3atm - 2(0.23atm) = 1.84atm
P(N₂) = 0.23atm
P(H₂) = 0.69atm