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blsea [12.9K]
3 years ago
15

How much energy is released when 0.40 mol C6H6(g) completely reacts with oxygen?

Chemistry
2 answers:
podryga [215]3 years ago
6 0

Answer:

The right answer is -1268 kJ

Explanation:

Vsevolod [243]3 years ago
4 0
 <span>This question asksyou to apply Hess's law. 
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C. 
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3 
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12 
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation. 
Then, 
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ 
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ 
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ 
______________________________________... 
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ 
I hope this helps and my answer is right.</span>
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Explanation:

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Answer:

Butan-2-one

Explanation:

1. 1700 cm⁻¹

A strong peak near 1700 cm⁻¹ is almost certainly a carbonyl (C=O) group.

2. Triplet-quartet

A triplet-quartet pattern indicates an ethyl group.

The 2H quartet is a CH₂ adjacent to a CH₃. The peak normally occurs at δ 1.3, but it is shifted 1.2 ppm downfield to δ 2.47 by an adjacent C=O group.

The 3H triplet at δ 1.05 is the methyl group. It, too, is shifted downfield from its normal position at δ 0.9. The effect is smaller, because the methyl group is further from the carbonyl.

3. 3H(s) at δ 2.13

This indicates a CH₃ group with no adjacent hydrogen atoms.

It is shifted 0.8 ppm downfield to δ 2.13 by the adjacent C=O group.

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The identified pieces are CH₃CH₂-, -(CO)-, and -CH₃. There is only one way to put them together: CH₃CH₂-(C=O)-CH₃.

The compound is butan-2-one.

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