Answer:
As solute concentration increases, vapor pressure decreases.
Step-by-step explanation:
As solute concentration increases, the number of solute particles at the surface of the solution increases, so the number of <em>solvent </em>particles at the surface <em>decreases</em>.
Since there are fewer solvent particles available to evaporate from the surface, the vapour pressure decreases.
C. and D. are <em>wrong</em>. The vapour pressure depends <em>only</em> on the number of particles. It does not depend on the nature of the particles.
53.3% + 6.7% = 60%, 100% - 60% = 40%. 40% of glucose is made of carbon. Since there are only three types of atoms in glucose, and the amount of hydrogen and oxygen is already given, this means that whatever percentage is left (40%) has to be carbon.
Answer:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
Explanation:
For the reaction:
H₂C₂O₄(g) → CO₂(g) + HCOOH(g)
At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:
H₂C₂O₄(g) = P₀ - x
CO₂(g) = x
HCOOH(g) = x
P at t=20000 is:
P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x
For 1st point:
x = 92,8-65,8 = 27
Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8
2nd point:
x = 130-92,1 = 37,9
H₂C₂O₄(g): 92,1 - 37,9 = 54,2
3rd point:
x = 157-111 = 46
H₂C₂O₄(g): 111-46 = 65
Now, as the rate law is :
v = k P[H₂C₂O₄]
Based on integrated rate law, k is:
(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k
1st point:
k = 2,64x10⁻⁵
2nd point:
k = 2,65x10⁻⁵
3rd point:
k = 2,68x10⁻⁵
The averrage of this values is:
k = 2,66x10⁻⁵
That means law is:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
I hope it helps!
Answer:
The difference in temperature is significant means that the lower-boiling liquid finishes distilling at a temperature that is too low for the higher-boiling liquid to be in vapor form yet.
Explanation:
The temperature will rise as the vapor of lower-boiling liquid rushes into the distillation head. However once the lower-boiling liquid is done distilling, there is a temperature drop because while the lower temperature liquid is done distilling, the temperature is still too low for the higher-boiling liquid to be rushing in as a vapour, so the temperature drops.
