Answer:
151.63 g
Explanation:
We first get the number of moles;
Moles = Molarity × volume
= 2.0 × 0.475
= 0.95 moles
1 mole of CuSO4 = 159.609 g/mol
Therefore;
Mass = moles × molar mas
= 0.95 moles × 159.609 g/mol
= 151.63 g
Answer:
molar composition for liquid
xb= 0.24
xt=0.76
molar composition for vapor
yb=0.51
yt=0.49
Explanation:
For an ideal solution we can use the Raoult law.
Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.
For toluene and benzene would be:

Where:
is partial pressure for benzene in the liquid
is benzene molar fraction in the liquid
vapor pressure for pure benzene.
The total pressure in the solution is:
And
Working on the equation for total pressure we have:
Since
We know P and both vapor pressures so we can clear
from the equation.
So
To get the mole fraction for the vapor we know that in the equilibrium:
So
Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.
It’s written by some one that was there as show on google your welcome
Answer:
See explanation.
Explanation:
I highly suggest you watch OChem Tutor's videos on IUPAC nomenclature because the actual naming would take a lot of time to teach in text-based format. But here is how to name them:
1) I think there are two seperate pictures for number 1. The molecule on the left is 1-pentene and the one on the right is 4-methyl-1-pentene. If the whole thing is one molecule but there is just a bond missing where the red marker numbers are, that molecule would be 9-methyl-1,6-decadiene.
2) 4-methyl-2-pentene
3) 2,4-octadiene
4) 1,5-nonadiene
5) 2,5-dimethyl-3-hexene
6) 3,6-dimethyl-2,4-heptadiene
7) 2,5,5-trimethyl-2-hexene
A molecule can<span> possess </span>polar bonds<span> and still be </span>non polar. If the polar bonds<span> are evenly (or symmetrically) distributed, the </span>bond<span> dipoles cancel and </span>do<span> not create a molecular dipole.</span>