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larisa [96]
3 years ago
5

On the addition of HI

Chemistry
1 answer:
victus00 [196]3 years ago
8 0

Answer:

CH_3-CH(I)-CH(CH_3)-CH_3

I-CH_2-CH_2-CH(CH_3)-CH_3

Explanation:

Given the compound CH_2=CH-CH(CH_3)-CH_3

The steps of reaction:

HI \longrightarrow H^+ + I^-

CH_2=CH-CH(CH_3)-CH_3 + H^+ \longrightarrow CH_3-CH^+-CH(CH_3)-CH_3

<u>But this intermediate product has a resonance: </u>

CH_3-CH^+-CH(CH_3)-CH_3 \longleftrightarrow C^+H_2-CH2-CH(CH_3)-CH_3

The reaction with I-

CH_3-CH^+-CH(CH_3)-CH_3 + I^- \longrightarrow CH_3-CH(I)-CH(CH_3)-CH_3

C^+H_2-CH2-CH(CH_3)-CH_3 + I^- \longrightarrow I-CH_2-CH_2-CH(CH_3)-CH_3

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Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

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Final temperature of water= 20 ºC

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Now if we remember the <u>heat equation</u>:

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Now we can <u>put the values into the equation</u>:

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Now we can <u>solve for the initial temperature of gold</u>, so:

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T_i~=163.1 ºC

I hope it helps!

5 0
3 years ago
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Answer:

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