. Write the two resonance hybrids for the carbocation that would be formed by protonation at C-1 of 2-methyl-1,3-pentadiene. Wit
1 answer:
Answer:
Answer is explained in the explanation section below.
Explanation:
This question is requires the diagram, which I have drawn and it is attached in the attachment below. Please for this answer, refer to the diagram attached in the attachment below.
Referring to the diagram, attached. As we know that, Allylic secondary carbocations are more stable than Allylic tertiary carbocations.
Hence,
C2 will have a more positive charge since a tertiary carbocation (C2) is more stable than a secondary carbocation (C4). Therefore, the resonance structure will favor the positive charge at C2.
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Taking into account the Boyle's law, 3.377 L of helium the tank will be able to produce at a pressure of 94.2 kPa.
The gas laws are a set of chemical and physical laws that allow determining the behavior of gases in a closed system. The parameters evaluated in these laws are pressure, volume, temperature, and moles.
Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at constant temperature.
This law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; while if the pressure decreases, the volume increases.
Mathematically, Boyle's law states that the product of pressure and volume is constant:
P×V= k
Studying two different states, an initial state 1 and a final state 2, it is satisfied:
P1× V1= P2×V2
In this case, you know:
- P1= 101 kPa
- V1= 3.15 L
- P2= 94.2 kPa
- V2= ?
Replacing in Boyle's law:
101 kPa× 3.15 L= 94.2 kPa× V2
Solving:
<u><em>V2= 3.377 L</em></u>
In summary, 3.377 L of helium the tank will be able to produce at a pressure of 94.2 kPa.
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