Answer:
Reagent A: PBr₃
Reagent B: Mg in Et₂O.
Explanation:
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In this case, your facing a problem in which a carboxylic acid is produced starting by an alcohol. More specifically, cyclopentanol must react with phosphorous tribromide in order to yield bromocyclopentane which is more likely to produce a carboxylic acid, therefore, reagent A is PBr₃.
On the other hand, by means of the production of the specified product, bromocyclopentane must react with carbon dioxide and magnesium in diethyl ether in acidic media to promote the production of the cyclopentanoic acid via the grignard reaction (substitution of the bromine by the carboxyle group), therefore, reagent B is Mg in Et₂O.
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the amount of heat produced from the combustion of 24.3 g benzene (c6h6) is ΔH = -976.5 kJ
There are two moles of benzene involved in the process (C6H6). Since the heat of this reaction is -6278 kJ, the burning of 2 moles of benzene will result in a heat loss of 6278 kJ. This reaction is exothermic.
Enthalpy, or the value of H, is a unit of measurement for heat that relies on the amount of matter present (number of moles).
Thus, 24.3 g of benzene contains:
n = mass/molar mass, where n = 24.3/78.11, and n = 0.311 moles.
2 moles = 6278 kJ
0.311 moles =x
By the straightforward direct three rule:
2x = -1953.08 x = -976.5 kJ
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