Answer:
Van't Hoff factor is 1.72 ≅ 2
Explanation:
Let's apply the colligative property of boiling point elevation:
ΔT = Kb . m . i
ΔT = Boiling T° of solution - T° boiling of pure solvent
103.45 °C - 100°C = 0.512 °C/m . 3.90 m . i
3.45 °C / 0.512 m/°C . 3.90 m = i
1.72 = i ≅ 2
Answer: At temperature of 269 K the gas would occupy 1.33 L at 217 kPa
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 147 kPa
= final pressure of gas = 217 kPa
= initial volume of gas = 1.8 L
= final volume of gas = 1.33 L
= initial temperature of gas = 
= final temperature of gas = ?
Now put all the given values in the above equation, we get:
Thus at 269 K temperature the gas would occupy 1.33 L at 217 kPa
Llowing certain substances to pass through it but not others, especially allowing the passage of a solvent but not of certain solutes.
Gravity is pulling the skydiver down therefore gravitational force
Answer:
Hydroxide concentration of the sample is 1.3x10⁻⁶M
Explanation:
The equilibrium constant of water, Kw, is:
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
Kw is defined as:
Kw = 1.7x10⁻¹² = [H⁺] [OH⁻]
As the sample is of pure water, both H⁺ and OH⁻ ions have the same concentration because come from the same equilibrium, that is:
[H⁺] = [OH⁻]
We can write the Kw expression:
1.7x10⁻¹² = [OH⁻] [OH⁻]
1.7x10⁻¹² = [OH⁻]²
1.3x10⁻⁶M = [OH⁻]
<h3>Hydroxide concentration of the sample is 1.3x10⁻⁶M</h3>
