Answer: i think your answer is<u> The giant green anemones, the ochre sea stars, and the red octopuses</u> because an ecosystem means all the organisms and the physical environment with which they interact. If not then your other option would be <u>A a school of fluffy sculpins.</u>
Hope this helped you!
Answer:
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The appropriate answer is they decrease the potential energy difference between reactant and product. They do this by bringing products and reactants together at the active site on the enzyme molecule. Enzymes are biological catalysts. They increase the rate of a chemical reaction without being used up in the reaction itself. So one molecule of an enzyme is used many times to catalyze the reaction of other molecules.
Answer:
Explanation:
GIven that:
The activation energy = 250 kJ
k₁ = 0.380 /M
k₂ = ???
Initial temperature
1001 K
Final temperature
298 K
Applying the equation of Arrhenius theory.

where ;
R gas constant = 8.314 J/K/mol





/M .sec
Half life:
At 1001 K.


1.82368 secc
At 298 K:


Answer:
14 mol O₂
Explanation:
The reaction between CO and O₂ is the following:
CO + O₂ → CO₂
We balance the equation with a coefficient 2 in CO and CO₂ to obtain the same number of O atoms:
2CO + O₂ → 2CO₂
As we can see from the balanced equation, 1 mol of O₂ is required to react with 2 moles of CO. Thus, the conversion factor is 1 mol of O₂/2 mol CO. We multiply the moles of CO by the conversion factor to calculate the moles of O₂ that are required:
28 mol CO x 1 mol of O₂/2 mol CO = 14 mol O₂