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Maslowich
2 years ago
10

In the enzyme reaction above, catechol is the substrate, oxygen is a reactant, catechol oxidase is the enzyme, benzoquinone is t

he product, and water (H2O) is a by-product. Which of the following will decrease the rate of this reaction? a. Reducing the amount of oxygenb. Reducing the amount of catechol c. Increasing the amount of catechol oxidase d. Increasing the amount of benzoquinone
Chemistry
1 answer:
hammer [34]2 years ago
3 0

Answer:  d. Increasing the amount of benzoquinone

Explanation:  The given situation can be written in terms of the equation as follows-

Catechol + Oxygen ⇆ Benzoquinone + water (catechol oxidase is the catalyst)

Le chatleir Principle for the system of equilibrium explains that whenever a reaction is being applied to the stress , the reaction will move in that direction in which the effect of stress is being lessened.

For example if there is a decrease in the concentration of the reactants, the equilibrium will shift to that direction in which the reactants concentration is being increased in order to maintain the equilibrium.

Thus the correct option is (D) as increasing the concentration of the benzoquinone, the equilibrium will shift in the direction in which its concentration is getting decreased, that is in backward reaction, thus decreasing the rate of reaction.

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Answer:

negative, positive, increase

Explanation:

From the given question,  

During the formation of bond, between two atoms with difference between their electronegativity-

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2 years ago
An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
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Vedmedyk [2.9K]

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Explanation:

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