Answer:
K remains the same;
Q < K;
The reaction must run in the forward direction to reestablish the equilibrium;
The concentration of 
will decrease.
Explanation:
In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.
Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.
The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.
As a result, since is also a reactant, its concentration will decrease.
The answer would be a. C2H4(ethylene)
1C2H4+3O2~>2CO2+2H2O
Fresh water pollutants are substances which pollute fresh water and industrial waste, is most harmful fresh water pollutant to man and aquatic organisms.
<h3>What are pollutants?</h3>
Pollutants are substances which cause harm when they are present in the environment.
Pollutants include chemicals such as petroleum and material such as sewage.
The presence of pollutants in freshwater results in water pollution and make the water unfit for drinking purposes and also harms aquatic life in freshwaters.
Some freshwater pollutants include:
- Farming wastes
- Household pollutants
- Industrial wastes
- Erosion
- Oil and Gasoline
- heat
Of these pollutants, the most dangerous fresh water pollutant is industrial wastes as they kill aquatic organisms most due to the presence of harmful chemicals in them.
Therefore, fresh water pollutants such as industrial waste is most harmful to man and aquatic organisms.
Learn more about pollutants at: brainly.com/question/1235358
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<span>You must balance your equation correctly.
Here is your answer:
294gFeS2 x 1molFeS2/119.99 x 11mols O2/4mols FeS2--> 6.738mol O2
176gO2 x 1mol O2/32gO2 x 4mols FeS2/11mol FeS2--> 2mols FeS2
Now choose the molecule with the lowest amount (Limiting Reagent)
2molsFeS2 x 2molsFe2O3/4molsFeS2 x 159.7g
159.7g Fe2O3 grams produced.</span>
