Where two air masses and do not mix
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Aldehydes and ketones having α-hydrogen atoms, undergoes aldol condensation, in present of base (NaOH).
The initial product formed during this reaction is β-hydroxy alcohol, which then undergoes dehydration to form α,β-unsaturated aldehyde or ketone.
In present case, 3,3-dimethyl-2-butanone has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
5-hydroxy-4,4-dimethly-1-(2-methylcyclopentyl)pent-1-en-3-one.
The complete reaction product formed is shown below.
The initial product formed during this reaction is β-hydroxy alcohol, which then undergoes dehydration to form α,β-unsaturated aldehyde or ketone.
In present case, 3,3-dimethyl-2-butanone has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
5-hydroxy-4,4-dimethly-1-(2-methylcyclopentyl)pent-1-en-3-one.
The complete reaction product formed is shown below.
Answer:
Explanation:
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In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:
Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:
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