The pKapKa of cyclopentane is > 60, which is about what is expected for a hydrogen that is bonded to an sp3sp3 carbon. Explai
n why cyclopentadiene is a much stronger acid (pKapKa of 15), even though it too involves the loss of a proton from an sp3sp3 carbon. Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
1 answer:
Answer and explanation:
cyclopentadiene is more acidic than cyclopentane
hydrocarbon compound are weak acid in nature
This relative acidity is explained by the stability of
cyclopentadienyl anion which is aromatic in nature
(check the attached image file 1)
To answer this question we must look at the stability of the anions that are formed when the compound lose proton.
All the electron in the cyclopentyl anion are localized.
In contrast, the aromatic cyclopentadienyl anion is a stable carbanion as a result of its aromaticity therefore making its conjugate acid a very strong acid compare to other compounds with hydrogen attached to sp³ carbons
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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of NaI solution = 0.130 M
Volume of solution = 0.400 L
Putting values in above equation, we get:
The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:
The precipitate (insoluble salt) formed is lead (II) iodide
By Stoichiometry of the reaction:
2 moles of NaI produces 1 mole of lead (II) iodide
So, 0.52 moles of NaI will produce = of lead (II) iodide
To calculate the number of moles, we use the equation:
Moles of lead (II) iodide = 0.26 moles
Molar mass of lead (II) iodide = 461.1 g/mol
Putting values in above equation, we get:
Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams