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otez555 [7]
3 years ago
6

A sample of an ideal gas is cooled from 50.0°C to 25.0°C in a sealed container at constant volume.

Chemistry
1 answer:
ozzi3 years ago
3 0

Answer: Option (E) is the correct answer.

Explanation:

Since, it is given that the container is sealed one. So, no gas molecule is able to escape into the surrounding. Therefore, average molecular mass of the gas will remain the same.

Also, it is given that gas is cooled down from 50^{o}C to 25^{o}C. This means that there occurs a decrease in kinetic energy of the molecules of gas as kinetic energy is directly proportional to temperature.

Mathematically,      K.E \propto \frac{3}{2}kT

Hence, sped of molecules will decrease and molecules become closer to each other with decrease in temperature.

Thus, we can conclude that following values for the gas will decrease.

  • The average distance between the molecules.
  • The average speed of the molecules
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Dennis_Churaev [7]
The  keg for the reaction
2  SO2(g) + O2(g)   →  2  SO3(g)  is

 Keg =  [SO3]^2/ {(SO2)^2 ( O2)}

Keg (equilibrium constant)  is the ratio of  of equilibrium  concentration  of the product   raised  to the power  of their  stoichiometric  coefficient  to the  equilibrium  concentration  of the reactant  raised  to the power  of  their stoichiometric  coefficient.

7 0
3 years ago
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Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y
denis23 [38]

Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of CHCl_3 = 1.492 g/mL

The density of CHBr_3 = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

6 0
3 years ago
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Ipatiy [6.2K]

<u>Answer:</u> The correct answer is option A.

<u>Explanation:</u>

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Hope it helps

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