option of d is the write answer
Answer:
See attachment.
Explanation:
Elements that are in the same group will definitely possess similar characteristics because they tend to have the same valence electron which determines their reactivity.
On a periodic table, elements in the same group can be found arranged on the same column in the periodic table.
Therefore the two elements that have similar characteristics are those two elements you can see on the same column in group 2. See the two elements indicated in the attachment below.
Answer:
B.) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity decreases because the arsenic atom loses the electron.
Explanation:
Silicon is located in the 3rd row and 14th column in the periodic table. Arsenic is located in the 4th row and 15th column in the periodic table. This means that arsenic has one more valence electron than silicon. Since arsenic is located one row down from silicon, its valence electrons occupy higher energy orbitals.
Silicon maintains a crystal-like lattice structure. Each silicon atom is covalently connected to assume this shape. When silicon gains one extra electron from arsenic, it experiences n-type doping. This new electron is not tightly bound in the lattice structure. This allows it to move more freely and conduct more electricity. This can also be explained using band gaps. Silicon, which previously had an empty conduction band, now has one electron in this band. This lowers the band gap between the conduction and valence bands and increases conductivity.
Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= 
= 
= 
The amount of free SO₂ to be added will be:
= 
= 
∵ 1000 mg = 1 g
So,
= 
= 
Thus,
"9.225 g" should be added.
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.
Best regards.