The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
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Answer: NO2.
You can predict that two atoms with similar electronegativity will form covalent compounds while atoms with pretty different electronegativity will form ionic compounds.
Given that N and O are neighbors on the periodic table, you can predict they have similar electronegativities.
Also you can predict that from the fact that N and O are nonmetals, which also means that they will form covalent compounds.
The other compounds liste are formed by metals and non metals wich is an indication that they will have, at least, some ionic character.
Ergo, N and O will form covalent compounds, in this case NO2.
Answer:
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Answer : The volume in mL of the sodium carbonate stock solution is 364 mL.
Explanation :
According to dilution law:
where,
= molarity of aqueous sodium carbonate
= molarity of aqueous sodium carbonate stock solution
= volume of aqueous sodium carbonate
= volume of aqueous sodium carbonate stock solution
Given:
= 1.00 M
= 1.58 M
= 575 mL
= ?
Now put all the given values in the above formula, we get:
Therefore, the volume in mL of the sodium carbonate stock solution is 364 mL.