A chemist weighed out 5.14 g of a mixture containing unknown amounts of bao (s) and cao (s) and placed the sample in a 1.50-l fl
1 answer:
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Answer:
Equilibrium constant for is 0.5
Equilibrium constant for decomposition of is
Explanation:
dissociates as follows:
initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
Substitute the values in the above formula to calculate equilibrium constant as follows:
Therefore, equilibrium constant for is 0.5
Now calculate the equilibrium constant for decomposition of
It is given that is decomposed.
decomposes as follows:
initial 1.0 M 0 0
at eq. concentration of is:
Expression for equilibrium constant is as follows:
Substitute the values in the above expression
Equilibrium constant for decomposition of is
Answer:
Here's what I get.
Explanation:
The frequency of a vibration depends on the strength of the bond (the force constant).
The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.
Acetophenone
Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.
p-Aminoacetophenone
The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.
The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.
p-Nitroacetophenone
The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.
As electron density moves toward C-1, the double bond character of the C=O group increases.
The bond length decreases, so the bond becomes stronger, and wavenumber increases to 1693 cm¹.
