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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Answer: The bold staircase in the periodic table allows us to classify which elements are metalloids.
Explanation: Additionally, it acts like a "divider" that allows us to properly distinguish the metals from the non-metals in the periodic table.
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Density is an intrinsic property, so it is independent of the amount of substance present: one gold coin would have the same density as a solid gold boulder.
So if the density of gold is 19.3 g/cm³, the density of a bar of gold and the pieces into which the bar is cut would all be 19.3 g/cm³.
1. Ca(HCO3)2
2.Ca(HCOO)2
3. Ca(OH)2
4.NaOH
5.KCI
6.MgSO4
7.PbO
8.HCl
9.HNO3
10.H2SO4
11.NH3
12.(NH4)3PO4
13.NaOH
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Answer:
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