Combustion is a type of exothermic chemical reaction that generates light and heat in most cases, and occurs fairly quickly.
<h3>What is an exothermic reaction?</h3>
An exothermic reaction is one that when it occurs releases energy in the form of heat or light to the environment.
<h3>Characteristics of an exothermic reaction</h3>
- When this type of reaction occurs, the products obtained have lower energy than the initial reactants.
- Most of the exothermic reactions are oxidation, and when they are very violent they can generate fire, as in combustion.
- Combustion is a very fast oxidation reaction that occurs between materials called fuels and oxygen.
Therefore, we can conclude that combustion is an example of an exothermic reaction that releases large amounts of heat, which can lead to fire.
Learn more about an exothermic reaction here: brainly.com/question/14969584
The first statement is False... as
For exothermic reaction :
A+B》 C+D + HEAT..(heat is considered as a product)... as for endo.. heat is a reactant.
So tjey can't be of the same energy...
2nd one...based on the
A+B》 C+D+HEAT...For exo reaction... the product have more Heat energy than potential...so its false
Recall...energy can nither be created nor destroyed but converted from one form to another....
The 4th one however is true for heat...the reactants have nore energy than the products..
A+B+HEAT》C+D
Answer:

Explanation:
Hello,
In this case, for the given reaction at equilibrium:

We can write the law of mass action as:
![Keq=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
That in terms of the change
due to the reaction extent we can write:
![Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7Bx%7D%7B%28%5BCO%5D_0-x%29%28%5BH_2%5D_0-2x%29%5E2%7D)
Nevertheless, for the carbon monoxide, we can directly compute
as shown below:
![[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\](https://tex.z-dn.net/?f=%5BCO%5D_0%3D%5Cfrac%7B0.45mol%7D%7B1.00L%7D%3D0.45M%5C%5C)
![[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\](https://tex.z-dn.net/?f=%5BH_2%5D_0%3D%5Cfrac%7B0.57mol%7D%7B1.00L%7D%3D0.57M%5C%5C)
![[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\](https://tex.z-dn.net/?f=%5BCO%5D_%7Beq%7D%3D%5Cfrac%7B0.28mol%7D%7B1.00L%7D%3D0.28M%5C%5C)
![x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M](https://tex.z-dn.net/?f=x%3D%5BCO%5D_0-%5BCO%5D_%7Beq%7D%3D0.45M-0.28M%3D0.17M)
Finally, we can compute the equilibrium constant:

Best regards.
Answer:
All elements with 84 or more protons are unstable; they eventually undergo decay. Other isotopes with fewer protons in their nucleus are also radioactive.