Calculating K from Known Initial Amounts and the Known Change in Amount of One of the Species<span>Write the equilibrium expression for the reaction.Determine the molar concentrations or partial pressures of each species involved.<span>Determine all equilibrium concentrations or partial pressures using an ICE chart.</span></span>
Answer:
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<em><u>MARK </u></em><em><u>ME </u></em><em><u>BRAINLIST</u></em>
Answer:

Explanation:
Let's examine this reaction, especially the coefficients.

First of all, remember that reactants are used to make the product. They are found on the left of the arrow usually. Therefore, N₂ and H₂ are the reactants, while NH₃ is the product. We can automatically eliminate NH₃ from our answer choices, because it is simply not a reactant.
Next, look at the coefficients.
- N₂ has no coefficient, so a 1 is implied.
- H₂ has a coefficient of 3.
Therefore, for the reaction to work, there must be 1 mole of N₂ and 3 moles of H₂.
We have 3.2 moles of N₂ and 5.4 moles of H₂.
Divide each amount given by the required amount for completion.
- 3.2 mol N₂/ 1 mol N₂= 3.2 times
- 5.4 mol H₂/ 3 mol H₂=1.8 times
Therefore, there is enough nitrogen to complete the reaction 3.2 times, but only enough hydrogen for 1.8 times. If everything is completely reacted, we will run out of hydrogen and have excess <u>nitrogen</u>.
All of the above
Explanation:
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Learn more:
radioactivity brainly.com/question/10125168
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<u>Answer:</u> The solubility product of mercury (II) sulfide is 
<u>Explanation:</u>
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of mercury (II) sulfide follows:
s s
The expression of
for above equation follows:

We are given:

Putting values in above expression, we get:

Hence, the solubility product of mercury (II) sulfide is 