Abottle is half-filled with a 50: 50 (mole-to-mole) mixture of heptane (c7h16) and octane (c8h18) at 25 degrees c. what is the mole ratio of heptane vapor to octane vapor in the air space above the liquid in the bottle? the vapor pressures of octane and heptane at 25 degrees c are 11 torr and 31 torr, respectively.

2.8

Explanation:

Dalton´s law of partial pressures tell us that for a binary mixture:

Ptot = Pa + Pb, and

Pa = Xa Pºa where Pa is the partial  vapor pressure of companent A above the solution

Now the gas phase which is formed will have a composition given by :

Ptot = Pa + Pb where Ptot is the total vapor pressure for the components A and B

So lets calculate the partial vapor pressures  in the mixture:

P(C7H16) = 0.50 x  31 torr = 15.5 torr

P(C8H18) = 0.50 x  11 torr =  5.5 torr

Ptot = 15.5 torr + 5.5 torr = 21 torr

Now the composition of the vapor pressure will obey the relation:

γ (C7H16) = P(C7H16) / Ptot where γ is the mol fraction of C7H16 in the vapor. (Note use γ to refer to mol fraction in the vapor to differentiate it from Xa which is the mol fraction in the liquid)

Plugging our values:

γ (C7H16) = 15.5 torr/ 21 torr = 0.74

and  since γ (C7H16) + γ (C8H18) = 1, we have

γ (C8H18) = 1 - 0.74 = 0.26

which is correct: the mol fraction of the more volatile C7H16 should be greater than C8H18.

So the mole ratio of heptane to octane will be given by the ratio:

0.74/ 0.26 = 2.8

which is correct since  the mole ratio  of heptane to octane should be greater than one since the vapor is richer in  the more volatile C7H16 (almost 3 to 1 which is in the same ratio as the pure vapor pressures)

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.