Answer:
Reagent A: PBr₃
Reagent B: Mg in Et₂O.
Explanation:
Hello,
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:
44.01 g/mol
Explanation:
Add each elements atomic mass. For oxygen you will do that twice because their is two oxygens.
- Hope that helps! Please let me know if you need further explanation.
D for sure hope this helps
Answer:
14.434 r.a.m.
Explanation:
- The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.
∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)
atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.
atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.
Answer:
A3+ and B-
Explanation:
Elements in group 13 have outermost electron configuration, ns2np1 hence they form trivalent positive ions.
Elements in group 17 have outermost electron configuration ns2np5 hence they form univalent negative ions.
This implies that, if element A is in Group 13 and element B is in Group 17, the ions formed are A3+ and B-.
