Answer: 23 liter
Explanation: Im pretty sure
1 is B (Just remember to have the same number of atoms on both sides)
2 is B (A precipitate is a solid forming from 2 liquids)
The new pressure : P₂ = 1038.39 mmHg
<h3>Further explanation</h3>
Given
1.5 L container at STP
Heated to 100 °C
Required
The new pressure
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).
So P₁ = 1 atm = 760 mmHg
T₁ = 273 K
T₂ = 100 °C+273 = 373 K
Gay Lussac's Law
When the volume is not changed, the gas pressure is proportional to its absolute temperature

Input the value :
P₂=(P₁.T₂)/T₁
P₂=(760 x 373)/273
P₂ = 1038.39 mmHg
Explanation:
It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.
Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.
So, vapor pressure of mixture = 1% vapor pressure of pure water
Therefore,
=
= 27.785 Pa
Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹