Answer:
2.83 g
Explanation:
At constant temperature and pressure, Using Avogadro's law
Given ,
V₁ = 2.12 L
V₂ = 3.12 L
n₁ = 0.120 moles
n₂ = ?
Using above equation as:



n₂ = 0.17660 moles
Molar mass of methane gas = 16.05 g/mol
So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g
<u>2.83 g are in the piston.</u>
1. B
2. H
3. G
4. I
5. D
6. C
7. A
8. F
9. E
I am not sure weather it is correct but I wrote what I know
<span>all of the above can be saturated molecules </span>
Answer:
-0.1767°C (Option A)
Explanation:
Let's apply the colligative property of freezing point depression.
ΔT = Kf . m. i
i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1
m = molality (mol of solute / 1kg of solvent)
We have this data → 0.095 m
Kf is the freezing-point-depression constantm 1.86 °C/m, for water
ΔT = T° frezzing pure solvent - T° freezing solution
(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1
T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C