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Travka [436]
3 years ago
11

The reaction below virtually goes to completion because cyanide ion forms very stable complexes with Ni2+ ion:[Ni(H2O)6]2+(aq) +

4 CN-(aq) → [Ni(CN)4]2-(aq) + 6 H2O(l)At the same time, incorporation of 14C labelled cyanide ion (14CN-) is very rapid:[Ni(CN)4]2-(aq) + 4 14CN-(aq) = [Ni(14CN)4]2-(aq) + 4 CN-(aq)Which statement below is correct with regard to stability and rate of reaction?

Chemistry
1 answer:
MAVERICK [17]3 years ago
8 0

Answer:

Option A and D are correct.

Unstable species react rapidly.

Stable species do not react rapidly.

Explanation:

The complete question is attached to this solution.

The more stable a reactant is, the less reactive it will be. A stable reactant has a very stable structure in which it will avoid any perturbations. And for a reaction to occur, the bonds in the reactant must break down to form the products. A stable reactant has very strong bonds that aren't easy to break down, hence, reactions involving very stable reactants do not proceed rapidly.

And the more unstable a reactant specie is, the more rapidly it reacts. This is why the reaction involving the less stable isotope of carbon; Carbon-14 is very rapid. It is the same reason as explained above that is responsible for this. The bond between unstable species are not strong and are easily breakable, thereby leading to a quick reaction.

Hope this Helps!!!

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What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? G
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Answer:

the wavelength of radiation emitted  is \mathbf{\lambda= 2169.62 \ nm}

Explanation:

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\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}

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\mathtt{= 0.57245\ eV}

Wavelength of the radiation emitted:

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where;

hc  = 1242 eV.nm

\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}

\mathbf{\lambda= 2169.62 \ nm}

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