Answer:
y1 = 0.3162
y2 = 0.6838
Explanation:
ok let us begin,
first we would be defining the parameters;
at 25°C;
1-propanol P1° = 20.90 Torr
2-propanol P2° = 45.2 Torr
From Raoults law:
P(1-propanol) = P⁰ × X(1-propanol)
P(1-propanol) = 20.9 torr × 0.45 = 9.405
P(1-propanol) = 9.405 torr
Also P(2-propanol) = P⁰ × X(2-propanol)
P(2-propanol) = 45.2 torr × 0.45
P(2-propanol) = 20.34 torr
but the total pressure = sum of individual pressures
total pressure = 9.405 + 20.34
total pressure = 29.745 torr
given that y1 and y2 represent the mole fraction of each in the vapor phase
y1 = P1 / total pressure
y1 = 9.405/29.745
y1 = 0.3162
Since y1 + y2 = 1
y2 = 1 - y1
∴ y2 = 1 - 0.3162
y2 = 0.6838
cheers, i hope this helps.
Answer: Europe
Explanation: it’s in germany
Answer : The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2
Explanation :
The relation between the equilibrium constant and standard Gibbs free energy is:
![\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Ctimes%20%5Cln%20Q%5C%5C%5C%5C%5CDelta%20G%5Eo%3D-RT%5Ctimes%20%5Cln%20%28%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%29)
where,
= standard Gibbs free energy = -14.1 kJ/mol
R = gas constant = 8.314 J/K.mol
T = temperature = 
Q = reaction quotient
= concentration inside the cell
= concentration outside the cell
Now put all the given values in the above formula, we get:
![-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})](https://tex.z-dn.net/?f=-14.1%5Ctimes%2010%5E3J%2Fmol%20%3D-%288.314J%2FK.mol%29%5Ctimes%20%28298K%29%5Ctimes%20%5Cln%20%28%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%29)
![\frac{[A]_{inside}}{[A]_{outside}}=296.2](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%3D296.2)
Thus, the ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2
Answer:
Noble gases.
Good Luck & Have a good day!
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