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N76 [4]
2 years ago
15

In uncompetitive inhibition, the inhibitor can bind to the enzyme only after the substrate binds first. (T/F)

Chemistry
1 answer:
Veronika [31]2 years ago
4 0

Answer:

True

Explanation:

In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].

The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.

This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.

Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.

       

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The chemical equation below shows the decomposition of nitrogen triiodide (NI3) into nitrogen (N2) and iodine (I2). 2NI3 Right a
yuradex [85]

The moles of I₂ will form from the decomposition of 3.58g of NI₃ is 0.0136 moles.

<h3>How we calculate moles?</h3>

Moles of any substance will be calculated as:

n = W/M, where

W = required mass

M = molar mass

Given chemical reaction is:

2NI₃ → N₂ + 3I₂

Moles of 3.58g of NI₃ will be calculated as:

n = 3.58g / 394. 71 g/mol = 0.009 moles

From the stoichiometry of the solution, it is clear that:

2 moles of NI₃ = produce 3 moles of I₂

0.009 moles of NI₃ = produce 3/2×0.009=0.0136 moles of I₂

Hence, option (3) is correct i.e. 0.0136 moles.

To know more about moles, visit the below link:

brainly.com/question/15303663

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Explanation:

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