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AlexFokin [52]
3 years ago
6

Draw the structure of the compound identified by the following simulated 1H and 13C NMR spectra. The molecular formula of the co

mpound is C10H12O. (Blue numbers next to the lines in the 1H NMR spectra indicate the integration values.)

Chemistry
1 answer:
Luba_88 [7]3 years ago
6 0

Answer:  4-allylanisole

Explanation: The doublets behind the 7 ppm belongs to the

para-substituted benzene ring. The three single-proton multi-plets around 5−6 ppm predicts that there has to be a single subsituted alkene group

A single plus a doublet around 3-4 ppm belongs to CH3 and CH2 Groups as they could be attached to the subsituted alkene group.

Moreover the interpretation of the NMR  that there is no peak with a higher intensity for >180 ppm represents an absence of Carbonyl group.

The Predicted Number is attached from a chemical database along with their peaks information

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Answer:

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Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

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