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Enzymes catalyze the chemical reactions, they act upon the reaction substrates and speed up the reaction. Enzymes have active sites, the places where the reaction substrates interact with the enzyme bringing about the conversion of substrates to products. So, as the enzyme concentration increases the rate of reaction increases till a point where the rate is leveled off. The rate does not further increase, as the substrate might have become limiting at that point. All the available amount of substrate would have been associated with the active sites of the enzymes. So, at that point although there is enough catalyst, lack of substrate would limit the rate of reaction.
Answer:
Q = 0.061 = Kc
Explanation:
Step 1: Data given
Temperature = 500 °C
Kc=0.061
1.14 mol/L N2
5.52 mol/L H2
3.42 mol/L NH3
Step 2: Calculate Q
Q=[products]/[reactants]=[NH3]²/ [N2][H2]³
If Qc=Kc then the reaction is at equilibrium.
If Qc<Kc then the reaction will shift right to reach equilibrium.
If Qc>Kc then the reaction will shift left to reach equilibrium.
Q = (3.42)² / (1.14 * 5.52³)
Q = 11.6964/191.744
Q = 0.061
Q = Kc the reaction is at equilibrium.
Answer:
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Answer:
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