As the pressure on the on a gas cofined above a liquid increases, the solubility of the gas will increase
this also happen when we lower the temperature
Answer: C. 25.6 kPa
Explanation:
The Gauge pressure is defined as the amount of pressure in a fluid that exceeds the amount of pressure in the atmosphere.
As such, the formula will be,
PG = PT – PA
Where,
PG is Gauge Pressure
PT is Absolute Pressure
PA is Atmospheric Pressure
Inputted in the formula,
PG = 125.4 - 99.8
PG = 25.6 kPa
The gauge pressure inside the container is 25.6kPa which is option C.
Answer:
33km due south
Explanation:
Morant Bay is 33km south of Port Antonio;
Problem: displacement of Morant bay from Port Antonio;
Displacement is a vector quantity. It is the length of path between two position and the direction inclusive.
So, it has magnitude and must be specified with a direction.
- Therefore, the displacement of Morant Bay from Port Antonio is 33km due south.
Answer is: concentration of ammonium ions are 7,14·10⁻¹⁴ M.
Chemical reaction: 2NH₃(l) → NH₄⁺(am) + NH₂⁻(am).
Kam = 5,1·10⁻²⁷.
[NH₄⁺] · [NH₂⁻] = x; equilibrium concentration of cations and anions.
Kam = [NH₄⁺] · [NH₂⁻].
Kam = x².
x = [NH₄⁺] = √5,1·10⁻²⁷.
[NH₄⁺] = 7,14·10⁻¹⁴ M.
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)

[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:

![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)

=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96