<span> 1C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ
1mol </span>C3H8(g) ---- –2200 kJ
2mol C3H8(g) ----2(–2200 kJ)=-4400kJ
We have to solve this question using the stoichiometry of the reaction:
The equation of the reaction is;

According to the question;
Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles
From the stoichiometry of the reaction:
Since;
24 moles of CO2 released 15,026 KJ
0.48 moles of CO2 will release 0.48 * 15,026/24
= 301 KJ of heat.
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Answer:

Explanation:
We have the reactions:
A: 
B: 
Our <u>target reaction</u> is:

We have
as a reactive in the target reaction and
is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.
A: 
Then if we add reactions A and B we can obtain the target reaction, so:
A: 
B: 
For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.


The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.
<h3>
What is relative abundance?</h3>
- The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
- When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
- Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.
To learn more about relative abundance with the given link
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Most animals obtain their nutrients by the consumption of other organisms. At the cellular level, the biological molecules necessary for animal function are amino acids, lipid molecules, nucleotides, and simple sugars. However, the food consumed consists of protein, fat, and complex carbohydrates.