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2 years ago

What best describes the kinetic energy of two gas particles before and after a collision?

1 answer:

7 0

Kinetic energy is the energy possessed by the body when in motion. It is calculated by 1/2mv² where m is the mass of the body and v is the velocity.
Baring in mind that energy can not be created or destroyed but can only be transformed from one form to another (law of conservation of energy) then when two bodies collide the total kinetic energy of the two particles stays the same (kinetic energy is preserved). This is in the case of perfectly elastic collision, however, during inelastic collisions the kinetic energy is transformed to other forms of energy such as heat, sound and energy used in bond breakage.

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True or false?

Virty [35]

The answer would be true

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3 years ago

Which form of energy is directly related to the measure of the average kinetic energy of the particles in a substance?

marshall27 [118]

<span>The form of energy related to measurement of kinetic energy is heat energy. Basically, thermal energy is related to the measurement of molecula kinetics of a substance. It is due to the thermoagitation of the partices of the substances, which keep on moving. </span>

4 0

2 years ago

Read 2 more answers

The temperature in mark’s town is 30 degrees Fahrenheit. The temperature in Vikki’s town is 82 degrees Fahrenheit. In both towns

allochka39001 [22]

Answer:

sorry don't know the answer but i really need the points sorry

Explanation:

7 0

3 years ago

Which of the following are the bark, roots, seeds, buds or berries of an aromatic plant?

djverab [1.8K]

Answer:

Spices

Explanation:

Herbs are the leaves.

Vegetables are the seeds.

Fruits are the seeds.

So spices are the only option left.

8 0

2 years ago

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>

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

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!

4 0

3 years ago