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3 years ago

Which two substances bind using a lock-and-key mechanism?

Chemistry

2 answers:

Firlakuza [10]3 years ago

8 0

It would be A: enzyme and substrate

Katen [24]3 years ago

4 0

<span>The answer is the first option. Enzyme and substrate bind using a lock-and-key mechanism. Enzymes act on a specific substrate and a substrate needs a specific enzyme, this is what is called a lock-and-key mechanism. Enzymes and substtates are like a key and a lock, one is for each other.</span>

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3 years ago

Read 2 more answers

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

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3 years ago