Five valence electrons for one nitrogen and 10 valence electrons for two nitrogens
Answer:
1) HCl contains the Cl^- which is a good nucleophile
2) 2-methyl-2- heptanol > 2-heptanol > 1-heptanol
3) see image attached
Explanation:
If the dehydration of alcohols is carried out using HCl, the chloride ion which is a good nucleophile will attack the substrate to yield an undesirable product.
The dehydration of alcohols is an E1 reaction. Recall that the ease of E1 reaction increases in the order 3°> 2°> 1°. Hence, 2-methyl-2- heptanol forms a tertiary carbocation intermediate during dehydration and has the greatest ease of dehydration.
The three products formed during the dehydration of 3,3-dimethyl-2-butanol are shown in the image attached. Two out of the three are formed by rearrangement reactions.
The answer to this is false.
Answer:
A Cellulose not digested by humans.
b. the storage form of carbohydrates in plants is starch
C amylose contains 1-4 glycosidic bond
D Glycogen and starch are highly branched polysaccharides.
Explanation:
Answer:
\left \{ {{y=206} \atop {x=82}}Pb \right.
Explanation:
isotopes are various forms of same elements with different atomic number but different mass number.
Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are
- Alpha particle emission \left \{ {{y=4} \atop {x=2}}He \right.
- Beta particle emission \left \{ {{y=0} \atop {x=-1}}e \right.
- gamma radiation \left \{ {{y=0} \atop {x=0}}γ \right.
in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.
Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below
\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right + \left \{ {{y=0} \atop {x=0}}γ\right.
Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right
