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bija089 [108]
3 years ago
15

At 80 degrees celsius the vapor pressure of benzene is 753 torr and that of toluene is 290 torr. What is the total pressure and

what is the composition of the vapor of a solution that is 2/3 benzene and 1/3 toluene?
Physics
1 answer:
nexus9112 [7]3 years ago
6 0

Answer:

The total pressure of the solution is 598.66 torr

while the vapor pressure of the components is vapor pressure of benzene is =502 torr and that of toluene = 96.6 tor

Explanation:

To solve this question, we need to look at the known variables, the unknown variables, and then we select the appropriate relations between the known variables and the unknown

Here we have the temperature of the gases t = 80°C

the pressures of benzene = 753 torr

the pressures of toluene = 290 torr

mole fraction of benzene in solution = 2/3

mole fraction of toluene in solution = 1/3

the unknown variables = composition of the solution  and the total pressure of the solution

From here given the known variables and the required variables, the appropriate relation bstween the known and unknown variables is Route's Law

Raoult's relates the vapor pressure of a mixture of gases to the mole fraction of solute gases introduced in the  gas solution. Raoult's Law is expressed by the formula: Psolution = Χsolvent1×Psolvent1 + Χsolvent2×Psolvent2+ ... .

where. Psolution = vapor pressure of the solution and

Χsolvent1 = mole fraction of solvent 1

Psolvent1 = vapor pressure of solvent1

Thus we have by Raoult's Law

P_{solution} = X_{benzene} × P_{benzene}+ X_{toluene}×P_{toluene}

which is =2/3×753 + 1/3×290 = 502 + 96.6 = 598.66 torr

and composition vapor pressure of benzene is 502 torr

while the composition vapor pressure of toluene is 96.6 torr

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Alex17521 [72]

Answer:

C. Electric energy

Explanation:

A generator is designed to convert chemical energy to mechanical energy  and them electrical energy.

  • It is an alternate source of electricity for most homes and industrial establishments.
  • The source of energy for the generator is a fuel usually fossil fuels.
  • This energy source is ignited and it produces heat energy which drives the motion of the pistons in the set.
  • As the piston moves, an magnetic field is cut through by a conducting wire.
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3 years ago
An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=584 kg, m2=838 kg, and m3=322 kg have blocke
Elena-2011 [213]

Answer:

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

Explanation:

Total force required = Mass x Acceleration,

F = ma

Here we need to consider the system as combine, total mass need to be considered.

Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg

We need to accelerate the group of rocks from the road at 0.250 m/s²

That is acceleration, a = 0.250 m/s²

Force required, F = ma = 1744 x 0.25 = 436 N

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

8 0
3 years ago
A circular loop of wire 78 mm in radius carries a current of 114 A. Find the (a) magnetic field strength and (b) energy density
Finger [1]

Explanation:

It is given that,

Radius of loop, r = 78 mm = 0.078 m

Current, I = 114 A

(a) Magnetic field strength of the circular loop is given by :

B=\dfrac{\mu_oI}{2R}

B=\dfrac{4\pi \times 10^{-7}\times 114}{2\times 0.078}

B = 0.000918 T

or

B=9.18\times 10^{-4}\ T

(b) Energy density at the center of the loop is given by :

U=\dfrac{B^2}{2\mu_o}

U=\dfrac{(0.000918)^2}{2\times 4\pi \times 10^{-7}}

U=0.335\ J/m^3

Hence, this is the required solution.

7 0
3 years ago
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a
ValentinkaMS [17]

Answer:

a. 11.5kv/m

b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v

to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m

b.the expression for the charge density is expressed as

σ=ξE

where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2

If we insert the values we have

8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2}  \\102nC/m^{2}

c.

from the expression for the capacitance

C=eA/d

if we substitute values we arrive at

C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF

d. To calculate the charge on each plate, we use the formula below

Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC

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