Answer:
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Explanation:
Answer:
the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene
Explanation:
The λmax is the wavelength of maximum absorption. We could use it to calculate the HOMO-LUMO energy difference as follows:
For ethylene
E= hc/λ= 6.63×10^-34×3×10^8/170×10^-9= 1.17×10^-18J
For cis,trans−1,3−cyclooctadiene
E= hc/λ=6.63×10^-34×3×10^8/230×10^-9=8.6×10^-19J
Therefore, the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene
Answer:
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Explanation:
Answer:

Explanation:
Let us first take a look at the image below;
In the acid - base reaction; we can see the transfer of electrons that takes place;
We can also see that the reaction goes in the direction which converts the stronger acid and the stronger base to the weaker acid and the weaker base.
The stronger acid is shown with the one with more negative
Value.
∴ The equilibrium constant for the acid-base reaction is expressed as:


From
Value (shown in the image below), it is clear and vivid that hydrobromic acid is a stronger acid than the ethyloxonium ion, therefore the equilibrium lies to the right.
From the chemical equation (shown in the attached image); the equilibrium constant for the acid-base reaction can be expressed as:


