Answer:
the answer is solid change into liquide.
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Answer:
5.6
Explanation:
Because of the gravity of the earth
We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.
Here we have to write a simple equation which describes the action of the enzyme catalase.
The equation is: The concentration of the complex [ES] =
Let us consider an enzyme catalyses reaction E + S ⇄ ES → E + P
Where E, S, ES and P are enzyme, substrate, complex and product respectively.
The concentration of the complex [ES] = , where is the Michaelis constant.
[E]₀ and [S] is the initial concentration of enzyme and concentration of substrate respectively.