It’s most likely missing water
plants need to be watered in order to survive
Explanation:
The given data is as follows.
T =
= (120 + 273.15)K = 393.15 K,
As it is given that it is an equimolar mixture of n-pentane and isopentane.
So,
= 0.5 and
= 0.5
According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.
(393.15 K) = 9.2 bar
(393.15 K) = 10.5 bar
Hence, we will calculate the partial pressure of each component as follows.

= 
= 4.6 bar
and, 
= 
= 5.25 bar
Therefore, the bubble pressure will be as follows.
P =
= 4.6 bar + 5.25 bar
= 9.85 bar
Now, we will calculate the vapor composition as follows.

= 
= 0.467
and, 
= 
= 0.527
Calculate the dew point as follows.
= 0.5,
= 0.5


= 0.101966
P = 9.807
Composition of the liquid phase is
and its formula is as follows.

= 
= 0.5329

= 
= 0.467
Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
Answer:
The advantage of this technique is that purified water as well as deposited metals can be re-used. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode.