The final volume V₂=4.962 L
<h3>Further explanation</h3>
Given
T₁=20 + 273 = 293 K
P₁= 1 atm
V₁ = 4 L
T₂=100+273 = 373 K
P₂=780 torr=1,02632 atm
Required
The final volume
Solution
Combined gas law :
P₁V₁/T₁=P₂V₂/T₂
Input the value :
V₂=(P₁V₁T₂)/(P₂T₁)
V₂=(1 x 4 x 373)/(1.02632 x 293)
V₂=4.962 L
Answer is: the partial pressure of the helium gas is 0.158 atm.
p(mixture) = 0.48 atm; total pressure.
m(H₂) = 1.0 g; mass of hydrogen gas.
n(H₂) = m(H₂) ÷ M(H₂).
n(H₂) = 1.0 g ÷ 2 g/mol.
n(H₂) = 0.5 mol; amount of hydrogen.
m(He) = 1.0 g; mass of helium.
n(He) = 1 g ÷ 4 g/mol.
n(He) = 0.25 mol; amount of helium.
χ(H₂) = 0.5 mol ÷ 0.75 mol.
χ(H₂) = 0.67; mole fraction of hydrogen.
χ(He) = 0.25 mol ÷ 0.75 mol.
χ(He) = 0.33; mole fraction of helium.
p(He) = 0.33 · 0.48 atm.
p(He) = 0.158 atm; the partial pressure of the helium gas.
Answer:
B 1,3
Explanation:
Count the atoms on each side of the arrow, they have to be equal
After arrow
2x C
6x O
4x H
Before the arrow
2x C - correct, as it is the only compound with C then there must only be 1 mole of it.
This is also the correct amount of H
We need 6x O total so 6/2 = 3
So there are 3 moles of O