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3 years ago

In which of the following choices is the oxidation number incorrect?

Chemistry

1 answer:

8_murik_8 [283]3 years ago

4 0

Cl2(s); oxidation number 1 is the incorrect choices in oxidation number.

Explanation:

In the elemental form oxidation state is zero. Here chlorine is present in elemental form so oxidation state is zero.

Oxidation number depends on the number of electrons gained or lost by an atom of the element say in compound formation.

If electron is gained oxidation number becomes negative.

If electron is lost then oxidation number is positive.

If the octet rule is fulfilled that valence shell is filled them atomic number gets zero. Since Cl2 is in neutral state the oxidation number is 0.

Oxidation number in general can be made out by checking the valency of the element as oxidation number is also equal to the valency.

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Answer:

The catalyzed reaction will take 2.85 seconds to occur.

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And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

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Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

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