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).
Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.
Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.
Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:
Therefore, density of metal is 19.33 g/mL.
Potential energy is energy due to an object's height above the ground.
Potential energy = mass x gravity x height
Kinetic energy is energy due to the motion of the object.
Kinetic energy = 1/2 x mass x velocity²
1.
The ball is not moving and is at a height above the ground so it has only potential energy.
P.E = 2 x 9.81 x 40
P.E = 784.8 J
2.
The ball is moving and has a height above the Earth's surface so it has both kinetic and potential energy.
P.E = same as part 1 = 784.8 J
K.E = 1/2 x 2 x 5²
K.E = 25 J
3.
The ball has no height above the Earth's surface and is moving so it has only kinetic energy.
K.E = 1/2 x 2 x 10²
K.E = 100 J
4.
50000 = 1/2 x 1000 x v²
v = 10 m/s
5.
39200 = 200 x 9.81 x h
h = 20.0 m
6.
12.5 = 1/2 x 1 x v²
v = 5 m/s
98 = 1 x 9.81 x h
h = 10.0 m
I think the answer is B. the sum of the enthalpy changes of the intermidiate reactions
The answer is E and F.
Hope this helps!!
