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).
Answer: 670K
Explanation:
Given that,
Original volume of gas V1 = 1.22 L
Original temperature T1 = 286 K
New volume V2 = 2.86 L
New temperature T2 = ?
Since volume and temperature are involved while pressure is constant, apply the formula for Charles law
V1/T1 = V2/T2
1.22 L/286 K = 2.86 L/ T2
Cross multiply
1.22 L x T2 = 286 K x 2.86 L
1.22T2 = 817.96
Divide both sides by 1.22
1.22T2/1.22 = 817.96/1.22
T2 = 670.459 K (Round to the nearest whole number as 670 K)
Thus, the temperature of the gas is 670 Kelvin
Answer:
if an atom gains an electron, the ion has negetive charge
