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Nana76 [90]
3 years ago
15

Addition of dns at the end of the incubation period stopped the reaction by denaturing sucrase. explain why it is important to d

enature sucrase before measuring product concentration
Chemistry
2 answers:
bixtya [17]3 years ago
7 0
Answer is: i<span>t is important to denature sucrase before measuring product concentration because if sucrase is active that will increase the enzyme activity and lower concentration of sugar.
DNS (</span>3,5-dinitrosalicylic acid)<span> is an </span>aromatic compound<span> that reacts with </span><span>reducing sugars.</span><span>

</span>
STatiana [176]3 years ago
5 0

Answer:

It might react with the product and consequently lower its concentration.

Explanation:

Hello,

When adding dna, we must assure the denaturing of sucrase as it could react with the product and lower its final concentration. Moreover, non-denatured sucrase spurs on the enzymatic activity which also would affect the final product resulting concentration, that is why we must guarantee the complete denaturing of the sucrase.

Best regards.

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The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.

The rate law is: rate = 24.00 × 10²⁵ M/s

The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.

Let's consider the decomposition of ethanol on an alumina surface.

C₂H₅OH(g) ⇒ C₂H₄(g) + H₂O(g)

The plot of [A] vs time (t) resulted in a straight line, which indicates that the reaction follows zero-order kinetics.

The slope, 24.00 × 10²⁵ M/s, represents the rate constant, k.

<h3>What is zero-order kinetics?</h3>

It is a chemical reaction in which the rate of reaction is constant and independent of the concentration of the reacting substances

The rate law for zero-order kinetics is:

rate = 24.00 × 10²⁵ M/s

The integrated rate law for zero-order kinetics is:

[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

<h3>What is the half-life?</h3>

Is the time for the amount of substance to decrease by half.

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, we can calculate the half-life [t(1/2)] using the following formula.

t(1/2) = [C₂H₅OH]₀ / 2 × k

t(1/2) = (1.25 × 10²² M) / 2 × (24.00 × 10²⁵ M/s) = 2.60 × 10⁻⁵ s

We can calculate the time required for all the 1.25 × 10²² M C₂H₅OH to decompose using the integrated rate law.

[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

0 M = 1.25 × 10²² M - 24.00 × 10²⁵ M/s × t

t = 5.21 × 10⁻⁵ s

The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.

The rate law is: rate = 24.00 × 10²⁵ M/s

The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.

Learn more about zero-order kinetics here: brainly.com/question/13314785

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