which resonance form would be the most stable contributor to the intermediate arenium ion produced by electrophilic bromination
of methoxybenzene
1 answer:
The question is incomplete as the options are missing, however, the correct complete question is attached.
Answer:
The correct answer is option A. ( check image)
Explanation:
The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.
Thus, the correct answer is option A. ( check image)
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The smallest observable object for an electron microscope is 6.6nm.
<h3>Calculation of smallest observable object for an electron microscope :</h3>Given,
Speed of electron (u) = 5.5 x 10⁴ m/s
De Broglie's formula,
wavelength, λ =h/mu
where,
h = Planck's constant = 6.626 x Js
m = mass of electron, 9.11 x kg
λ =h/mu
= 6.626 x Js /9.11 x
kg x 5.5 x 10⁴ m/s
= 0.132 x m
= 13.2 x m
It is well known that the smallest thing that can be observed under any microscope is identical to the radiation's wavelength for the smallest object that can be seen under an electron microscope.
Therefore, 13.2 x m/2
= 6.6 x m
= 6.6 nm
Hence, the smallest observable object for an electron microscope is 6.6nm.
Learn more about electron microscope here:
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