Answer:
See explanation
Explanation:
The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.
This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).
In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.
Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.
Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.
Answer:Label the parts of this wave.
A:
✔ crest
B:
✔ amplitude
C:
✔ trough
D:
✔ wavelength
Explanation:
For the chemical reactiom to be at equilibrium:
1- The rate of forward reaction must be equal to the rate of the reverse reaction.
2- The mass of EACH element must be equal before and after the reaction (no NET change in mass), otherwise the equilibrium will shift.
Important note: you need to check the mass of each element before and after the reaction (i.e, reactants side and products side) and the not the mass of the system as a whole. This is because the mass of the whole system will be preserved whether the system is at equilibrium or not (this is the fundamental law of mass conservation)
To turn the flow of electricity on or off. Probably wrong
Answer:
Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below
The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.
Explanation:
The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.
In the Left Beaker (Left half cell), their is less concentration
Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side
In the Right Beaker (right half cell), their is more concentration
Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side
Electrons will flow from Left to Right direction.
