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Colt1911 [192]
3 years ago
10

Which law is based on the graph that is shown below? A graph is shown with pressure on the horizontal axis and volume on the ver

tical axis. A curve starts high on the horizontal axis, curves toward the origin, and then starts to level out as it approaches the horizontal axis.
Chemistry
2 answers:
DIA [1.3K]3 years ago
5 0

<u>Answer:</u>

"Boyle's Law" is based on the graph that is shown below.

<u>Explanation:</u>

Boyle's law or Boyle – Mariotte law or Mariotte's law, is an experimental gas law that discusses how a gas's pressure tends to rise as the container volume start declining. This shows the relationship between pressure and volume for a fixed mass at a constant temperature, i.e., number of a gas molecules.This rule visualizes the actions of gas molecules in a confined space. This law can be understood from following equation:

p₁V₁ = p₂V₂

Above the product of the initial volume and pressure is equal to the product of the volume and pressure after a change.

Paladinen [302]3 years ago
5 0

Answer:

A) Boyle's law on edge

Explanation:

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Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

            k = Ae^-(Ea/RT)

where,   k = rate constant

              A = pre-exponential factor

          Ea  = activation energy

             R = gas constant

              T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

         

      T1= (244 + 273.15) K = 517.15 K

      T2= (324+ 273.15) K =597.15 K

    K1  = 6.7 ,     K2 = ?

         R = 8.314 J/mol K

     Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

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