Answer:
in a chemical reaction of NaOH with H2O, after NaOH is completely disassociated, we will find Na+ and OH- ions in the solution. (option C).
Explanation:
In a reaction where NaOH is added to H2O.
NaOH is considered a strong base, this means that in an aqueous solution ( in water) it's able to completely disassociate in ions.
There will not remain any NaOH in the solution. This means option D is not correct.
The ions in which NaOH will disassociate are : NaOH → Na+ + OH-
These ions we will find in the solution.
Not only Na+ because NaOH is a strong base, so there will be a lot of OH- ions as well in solution.
This means in a chemical reaction of NaOH with H2O, after NaOH is completely disassociated, we will find Na+ and OH- ions in the solution.
Answer:
A. ![K=\frac{[N_2O]^2}{[N_2]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BN_2O%5D%5E2%7D%7B%5BN_2%5D%5E2%5BO_2%5D%7D)
Explanation:
Hello there!
In this case, for us to figure out the appropriate equilibrium expression, it will be firstly necessary for us to recall the law of conservation of mass which states that the equilibrium constant of an equilibrium chemical reaction is written by dividing the products and reactants and including the stoichiometric coefficients as exponents. In such a way, for the given reaction, we will have:
![K=\frac{[N_2O]^2}{[N_2]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BN_2O%5D%5E2%7D%7B%5BN_2%5D%5E2%5BO_2%5D%7D)
As N2O is the product whereas N2 and O2 are reactants; thus, the equilibrium expression will be A.
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3700 microL is the answer
Hope it was correct :)
Answer:
D is the answers for the question
Explanation:
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