Answer:
= 0.134; 
= 0.866
The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr
Explanation:
For each of the solutions:
mole fraction of isopropanol (
) = 1 - mole fraction of propanol (
).
Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.
Furthermole, the partial pressure of isopropanol = *vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr
The partial pressure of propanol = *vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr
Similarly,
In the vapor phase,
The mole fraction of propanol () = 
Where, 
is the partial pressure of propanol and 
is the partial pressure of isopropanol.
Therefore,
= 5.26/(34.04+5.16) = 0.134
= 1 - 0.134 = 0.866
Answer: Volume would be 196.15 mL if the temperature were changed to 
and the pressure to 1.25 atmospheres.
Explanation:
Given: 
, 
= 256 mL,
= 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm
, 
Formula used to calculate volume is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to and the pressure to 1.25 atmospheres.
Answer:
A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one electron. The potassium ion that forms has 18 electrons. What best describes the bromide ion that forms? It is a negative ion that has one more valence electron than a neutral bromine atom.
Explanation:
The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
Answer:
B. False is the rite answer
