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olga_2 [115]
3 years ago
15

What is the IUPAC name of this compund?​

Chemistry
2 answers:
enot [183]3 years ago
5 0

Answer:

The IUPAC name of the compound is 3-ethyl, 4-methyl, heptane.

Explanation:

IUPAC stands for International Union of Pure and Applied Chemistry.

An IUPAC name is a internationally used name of a compound with same name throughout the world. This compound is named according to the IUPAC rules with the following steps-

1. Identification of longest carbon chain.

2. Estimating the number of carbon in the longest carbon chain that is seven carbon means heptane.

3. The side chains are identified, that are branched off from the parent chain.

4.  The parent chains is labelled from the side of nearest side chain.

5. The nearest side chain that is the third carbon contains ethyl as the side chain at 3 position of parent carbon. Thus, so named 3-ethyl.

6. The fourth carbon of parent chain contain a methyl group, thus named as 4-methyl.

The naming is based on the parent carbon chain and side chain naming are written first. Thus, the compound is 3-ethyl, 4-methyl, heptane.

mr_godi [17]3 years ago
4 0

Answer:

3-ethyl-4-methyl Heptane

Explanation:

longest chain is straight .

we start assigning number from where we are close to the branch to this mean its from left hand side. and we saw two chain 1 is ethyl and other is methyl . In alphabetical order e comes first as compare to m so we write ethyl first after that methyl and than in last we write longest chain name with ane (bcz we only have single vonds).

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Read 2 more answers
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i
alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

Explanation:

The pH of the solution = 2.46

pH=-\log[H^+]

2.46=-\log[H^+]

[H^+]=0.003467 M

HA\rightleftharpoons H^++A^-

Initially

0.0144         0      0

At equilibrium

(0.0144-x)       x       x

The expression if an dissociation constant is given by :

K_a=\frac{[A^-][H^+]}{[HA]}

K_a=\frac{x\times x}{(0.0144-x)}

x=[H^+]=0.003467 M

K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}

K_a=1.099\times 10^{-3}

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

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3 years ago
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