The reaction A + 2B ® products has been found to have the rate law, rate = k[A] [B]2. While holding the concentration of A const
1 answer:
Answer:
The rate of the reaction will increase by a factor of 9.
Explanation:
Hello,
In this case, considering the given second-order reaction, whose rate law results:
We easily infer that at constant concentration of A but tripling the concentration of B, we are going to obtain the following increasing factor while holding the remaining variables constant:
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The molecule with higher dipole moment is COFH because the geometry of the molecule in the COF2 nearly cancel the dipolar moment of each other. To be more clear:
The dipolar moment is the vectorial sum of all bond moments in the molecule or dipolar moment of each bond. The dipolar moment of a molecule with three or more atoms is determined by bond polarity as their geometry.
COF2 has a trigonal planar structure which are symmetric. The electronegativity of oxygen is slightly different regarding fluor. So as you can see in the image, the electronic density is specially displaced to the fluor atoms, but either to the oxygen atom.
COFH has a trigonal structure but differs from COF2 because there is an hydrogen who is donating it's electronic density, so in this zone the electronic density is less than over oxygen or fluor. That makes bond angles be different between them.
Answer:
Explanation:
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In this case, we need to remember that for the required time for a radioactive nuclide as radium-226 to decrease to one half its initial amount we are talking about its half-life. Furthermore, the amount of remaining radioactive material as a function of the half-lives is computed as follows:
Therefore, for an initial amount of 100 mg with a half-life of 1590 years, after 1000 years, we have:
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