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Alex Ar [27]
3 years ago
8

In the hydrogenation of ethylene using a nickel catalyst, the initial concentration of ethylene is 2.00mol?L?1 and its rate cons

tant (k) is0.0020mol?L?1?s?1 . Determine the rate of reaction if it follows a zero-order reaction mechanism.
Express your answer to two significant figures and include the appropriate units.
Chemistry
1 answer:
gavmur [86]3 years ago
8 0

Answer:

The rate of reaction of a zero-order reaction  is 0.0020 mol/L.

Explanation:

The rate expression of the zero order kinetic are :

R=k[A]^o

[A]= initial concentration of reactant

k = rate constant

R = rate of reaction

We have :

Rate constant of the reaction , k = 0.0020 mol/L s

R=0.0020 mol/L s\times 1

R = 0.0020 mol/L s

The rate of reaction of a zero-order reaction  is 0.0020 mol/L.

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3,200 joules

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What mass of mgo is produced from 2.00 moles of mg ? Mg +02 mgo
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What factor, according to the data in the table above, has the greatest impact on species loss?
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A radiator contains 10 quarts of fluid, 30% of which is antifreeze. how much fluid should be drained and replaced with pure anti
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Place the following transitions of the hydrogen atom in order from longest to shortest wavelength of the photon emitted. Rank fr
gizmo_the_mogwai [7]

Answer:

7 to 4     (higher  \lambda)

5 to 3

3 to 2

4 to 2   (lower  \lambda)

Explanation:

We can use the Rydberg formula which relates the wavelength of the photon emissions to the principle quantum numbers involved in the transition:

\frac{1}{\lambda}=R((\frac{1}{n_1^2})-(\frac{1}{n_2^2}))

with n_1 final n, and n_2 initial n

evaluating for each transition:

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4 to 2 \frac{1}{\lambda}= R(1/4-1/16)=R(0.1875)

3 to 2 \frac{1}{\lambda}= R(1/4-1/9)=R(0.139)

Note that the above formula is written for \frac{1}{\lambda}, so lower \frac{1}{\lambda} value obtained involves higher \lambda.

So we should order from lower to higher \frac{1}{\lambda}

7 to 4     (higher  \lambda)

5 to 3

3 to 2

4 to 2   (lower  \lambda)

Note: Take into account that longer wavelength involves lower energy (E=\frac{hc}{\lambda}).

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3 years ago
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