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dmitriy555 [2]
3 years ago
8

We have a sample of water vapor which cools from 150 °C to 125 °C. What happens to the motion of the molecules during this time

of cooling? A. The motion of the molecules stops. B. The motion of the molecules increases. C. The motion of the molecules decreases. D. The motion of the molecules remains the same.
Chemistry
1 answer:
pentagon [3]3 years ago
5 0

The motion of the molecules decreases.

<u>Explanation</u>:

  • Gases are formed when the energy in a system overcomes the attractive forces between the molecules. The gases expand to fill the space they occupy. In this way, the gas molecules interact little. In the gaseous state, the molecules move very quickly. As the temperature decreases, the amount of movement of the individual molecules also decreases.
  • The fast-moving particle slows down. When a particle speeds up, it has more kinetic energy. When a particle slows down, it has less kinetic energy. The particles in solid form are commonly connected through electrostatic powers. They don't get enough space to move around, therefore, their speed diminishes, they can't keep their standard speed like in the vaporous or fluid state.
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1 year ago
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6 0
3 years ago
In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
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Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

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