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sergij07 [2.7K]
3 years ago
5

What is the best name for the molecule below? a compound with 10 total carbons with a double bond and the rest single bonds. the

re is a straight chain of seven carbons along the bottom with a double bond between the third and fourth carbons when counting from left to right. there is a propyl group branching off of the fifth carbon from the left. 5-propyl-3-heptene 5-ethyl-3-octene 3-ethyl-5-octene 3-ethyl-4-heptene?
Chemistry
2 answers:
yanalaym [24]3 years ago
6 0
The correct answer is <span>5-propyl-3-heptene. </span>
Alina [70]3 years ago
5 0

Answer:

5-propyl-3-heptene

Explanation:

There is a straight chain of seven carbons along the bottom with a double bond between the third and fourth carbons when counting from left to right.

1  2   3  4  5  6  7

C-C-C=C-C-C-C

There is a propyl group branching off of the fifth carbon from the left.

1  2   3  4  5  6  7

C-C-C=C-C-C-C

                 |

                 C₃H₇

Th final structure is:

1        2      3    4     5      6       7

H₃C-CH₂-CH=CH-CH₂-CH₂-CH  ₃

                              |

                              C₃H₇

The  best name for the molecule is:

5-propyl-3-heptene

The side chain propyl is at the 5th C atom. Also the third bond from the left is a double bond. A seven C main chain is called heptane if there is no double bond. Since there is one double bond in the main chain it is named heptene.

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An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ
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Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

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moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = \frac{112,2 mgC}{200,0mg}×100 = 56,1%C

%H = \frac{11,0 mgH}{200,0mg}×100 = 5,5%H

%Cl = \frac{55,14 mgCl}{200,0mg}×100 = 27,6%Cl

%N = \frac{21,66 mgN}{200,0mg}×100 = 10,8%N

I hope it helps!

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