Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have 
as nucleophile. Also, this compound is also in excess. So, we will have as solvent 
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).
What question there’s no question on here
Answer:
20 km north that's the answer hope it helped
Answer:
1, 3, 2
Explanation:
N2 + H2 → NH3
I usually find that the best way to systematically balance an equation by inspection is to start with the most complicated-looking formula and then balance atoms in the order:
- All atoms other than O and H
- O
- H
(a) The most complicated formula is NH3.
(b) Balance N.
We have 1 H in NH3, but 2 N on the left. We need 2 N on the right. Put a 1 in front of N2 and a 2 in front of NH3.
1N2 + H2 → 2NH3
(c) Balance H.
We have fixed 6 H on the right, so we need 6 H on the left. Put a 3 in front of H2.
1N2 + 3H2 → 2NH3
The equation is now balanced, and the coefficients are 1, 3, 2.
There are 5 layers. The Lithosphere, the Asthenosphere, the outer core, the inner core and the mantel.
