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Naily [24]
4 years ago
14

For each of the sites specified in the molecules, select whether the site is nucleophilic, electrophilic, or neither. Compound A

: The indicated site is a carbon in cyclohexane which is bonded to a bromine and a hydrogen. The indicated carbon in compound A is nucleophilic. neither electrophilic nor nucleophilic. electrophilic. Compound B: The indicated site is the double bond in cyclohexene, a 6 carbon ring with an internal alkene. The indicated bond in compound B is nucleophilic. electrophilic. neither electrophilic nor nucleophilic. Compound C: The indicated site is a carbon double bonded to oxygen, and bonded to O C H 3 and ethyl. The indicated carbon in compound C is neither electrophilic nor nucleophilic. nucleophilic. electrophilic. Compound D: THe indicated site is a carbon bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. The indicated carbon in compound D is neither electrophilic nor nucleophilic. electrophilic. nucleophilic. Compound E: The indicated site is an oxygen bonded to a carbon and a hydrogen. The indicated oxygen in compound E is neither electrophilic nor nucleophilic. electrophilic. nucleophilic.
Chemistry
1 answer:
GenaCL600 [577]4 years ago
6 0

Answer:

自分の仕事をするthis is your answe but translate it to english

Explanation:

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A solution's solute and solvent are two different types of substances that can dissolve one another. Different solvents have different levels of solubility for different solutes. For instance, sugar is far more soluble in water than salt. Even sugar, though, has a limit on how much may dissolve.

learn more about solubility refer:

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5 0
1 year ago
If an atom has 11 protons and 13 neutrons what is it’s atomic number
Travka [436]

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the atomic number is 11 and element is Sodium (Na)

Explanation:

Number of Protons = Number of Electrons = 11

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Here, the Number of Protons and Number of Electrons is 11

So,

the atomic number is <em><u>11</u></em> and element is <em><u>Sodium </u></em><em><u>(</u></em><em><u>Na) </u></em>

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