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slava [35]
3 years ago
10

During which stage of a thunderstorm does rain begin to fall?

Chemistry
2 answers:
damaskus [11]3 years ago
6 0

Answer:

The correct answer is the mature stage.

Explanation:

The majority of the thunderstorms get produced by a cycle that comprises three phases: the cumulus stage, the mature stage, and the dissipating stage. In the mature stage, the cumulus cloud turns very big, and the water in it turns heavy and large. The drops of rain begin to fall via the cloud when the elevating air can no more hold them.  

In the meantime, the cool dry air begins to enter the cloud. As the cool air is denser in comparison to the warm air, it begins to incline in the cloud called downdraft. The downdraft makes the heavy water to fall downward, resulting in rain.  

KiRa [710]3 years ago
3 0
C. The mature stage (this would be the second stage out of three)
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The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

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R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

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Explanation:

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