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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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yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19 km

tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2

\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees north or west or (180 - 72.65) = 107.35 degrees north of east

3 0
3 years ago
A 5000 kg open train car is rolling on frictionless rails at 22 m/s when it starts pouring rain. A few minutes later, the car’s
Phantasy [73]

Answer:

500 kg

Explanation:

It is given that,

The mass of a open train car, M = 5000 kg

Speed of open train car, V = 22 m/s

A few minutes later, the car’s speed is 20 m/s

We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :

initial momentum = final momentum

Let m is final mass

MV=mv

m=\dfrac{MV}{v}\\\\m=\dfrac{5000\times 22}{20}\\\\=5500\ kg

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3 years ago
How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?
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Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of  0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J

So, the magnitude of work done in lifting the book is 6 joules.

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3 years ago
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

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