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Nutka1998 [239]
3 years ago
15

Write the name of this structure ​

Chemistry
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

The name of the compound is:

4–bromo–5,5–dichloro–hex–3–ene–1–yne

Explanation:

To the above compound, we simply do the following:

1. Identify the functional group of the compound.

2. Locate the longest continuous carbon chain. This gives the parent name of the compound.

3. Identify the substituent group attached.

4. Give the substituents the lowest possible count alphabetically.

5. Combine the above to obtain the name of the compound.

Now, we shall determine the name of the compound as follow:

1. The compound has carbon to carbon triple bond (C≡C) yne and carbon to carbon double bond (C=C) ene.

2. The longest continuous carbon chain is 6 i.e hexane.

3. The substituents group attached are:

i. Bromine (bromo), Br

ii. Chlorine (Chloro), Cl

4. In counting, ene comes before yne. But in this case, ene assumes same position from either side, So we shall consider counting from the side that gives yne the lowest count. Thus, ene is at carbon 3 and yne is at carbon 1. Therefore,

i. Bromo, Br is at carbon 4.

ii. Chloro, Cl appears twice and both are at carbon 5.

5. The name of the compound is:

4–bromo–5,5–dichloro–hex–3–ene–1–yne.

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What is the difference between natural and artificial transmutation? Give an<br> example of each.
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Explanation:

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6 0
3 years ago
Assuming complete distribution, what is the molarity of 10 milligrams of lisinopril in 100 liters?
SVEN [57.7K]

Answer:

2.47x10^{-7}   M

Explanation:

Lisoprisil's molecular mass is 405.488g/mol, we'll use this fact to calculate molarity, which units are mol/L, and we proceed to the calculus:

  • First, we'll unify unities, the 10 milligrams of lisinopril we'll transform into grams.

10mg*\frac{1g}{1000mg}=0.01g

  • Now that we have the same unities we'll calculate molarity using the molecular mass, the grams of lisinopril and the liters in which these grams are, let's consider that our final unities have to be mol/L.

\frac{1mol}{405.488g}*\frac{0.01g}{100L}=2.47x10^{-7}   M

I hope you find this information useful and interesting! Good luck!

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