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forsale [732]
4 years ago
5

Which statement is true according to the kinetic theory?

Chemistry
2 answers:
uranmaximum [27]4 years ago
8 0
The correct answer is option D. i.e.<span>All states of matter show random motion of particles.ding to the kinetic theory

According to the kinectic theory of matter, all the particles show random motion at the room temperature. Decreasing the temperature to absolute zero temperature i.e. 0 K will cease the motion of particles.</span>
love history [14]4 years ago
8 0

Answer:

The correct answer is the option d:All states of matter show random motion of particles

Explanation:

Hello!

Let's solve this!

According to kinetic theory, all states of matter have particle motion.

In the gas state the molecules are free and have a lot of movement.

In the liquid state the molecules are a little closer together but still have movement.

In the solid state, the molecules are more compact and although they do not have so much movement, they all vibrate in place. So they also have movement.

The correct answer is the option d:All states of matter show random motion of particles

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According to the collision theory, what two factors must be true for a given collision to successfully create products?
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For collisions to be effective, there must be proper orientation of the particles and right concentration of the reactants.

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reaction is found to have a rate constant of 12.5 s-1 at 25.0 Celsius. When you heat the reaction up by ten degrees Celsius, the
lisabon 2012 [21]

Answer : The activation energy for the reaction is, 52.9 kJ/mol

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

T_1 = initial temperature = 25.0^oC=273+25.0=298.0K

T_2 = final temperature = 25.0^oC+10=35.0^oC=273+35.0=308.0K

K_1 = rate constant at 25.0^oC = 12.5s^{-1}

K_2 = rate constant at 35.0^oC = 2\times K_1=2\times 12.5s^{-1}=25.0s^{-1}

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

Now put all the given values in this formula, we get:

\log (\frac{25.0s^{-1}}{12.5s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{298.0K}-\frac{1}{308.0K}]

Ea=52903.05J/mole=52.9kJ/mol

Therefore, the activation energy for the reaction is, 52.9 kJ/mol

4 0
3 years ago
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