2H2O2 decomposes into the products 2H2O + O2(g)
Answer:
The value of the equilibrium constant for reaction asked is
.
Explanation:


![K_{goal}=\frac{[C][O_2]}{[CO_2]}](https://tex.z-dn.net/?f=K_%7Bgoal%7D%3D%5Cfrac%7B%5BC%5D%5BO_2%5D%7D%7B%5BCO_2%5D%7D)
..[1]
![K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCH_3COOH%5D%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E2%7D)
..[2]
![K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2%5D%5E2%5BO_2%5D%7D)
..[3]
![K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_3%3D%5Cfrac%7B%5BC%5D%5E2%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BCH_3COOH%5D%7D)
[1] + [2] + [3]

( on adding the equilibrium constant will get multiplied with each other)



![K=\frac{[C]^2[O_2]^2}{[CO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC%5D%5E2%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%7D)
On comparing the K and
:


The value of the equilibrium constant for reaction asked is
.
This question asks to compare the energy emitted by a piece of iron at T = 603K with the energy emitted by the same piece at T = 298K.
Then you need to use the Stefan–Boltzmann Law
That law states that energy emitted (E) is proportional to fourth power of the to the absolute temperature (T), this is E α T^4 (the sign α is used to express proportionallity.
Then E (603) / E (298) = [603K / 298K]^4 = 16,8
Which meand that the Energy emitted at 603 K is 16,8 times the energy emitted at 298K.
Answer:
Option B: It remains unused during the reaction
Explanation:
Just took the test and got it right.
A-covalent because the two elements are non-metals.
B- ionic because one element is metal (K) and the other element is non-metal(O).
I hope that helps :)