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RSB [31]
3 years ago
7

A first-order reaction has a half-life of 29.2 s . how long does it take for the concentration of the reactant in the reaction t

o fall to one-sixteenth of its initial value?
Chemistry
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value

<span> The half-life for the chemical reaction is 29,2 s and is independent of initial concentration.
c</span>₀ - initial concentration the reactant.

c - concentration of the reactant remaining at time.

t = 29,2 s.<span>
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s</span> = 0,0237 1/s.<span>
ln(c/c</span>₀) = -k·t₁.<span>
ln(1/16 </span>÷ 1) = -0,0237 1/s · t₁.

t₁ = 116,8 s.

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What is the H+ concentration when pH is 2, 6, 7, 8, and 12?
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pH = - log (H+)

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The molecules of a substance must have what characteristics in order to be flexible
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Answer:
            <span>The molecules of a substance which must have the <u>a</u></span><u>bility to move past one another</u> are said to be flexible.

Explanation:
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How are the following aspects of a reaction affected by the addition of a catalyst? 1) activation energy of the reverse reaction
snow_tiger [21]

These are four questions and four answers.

Answers:

1) activation energy of the reverse reaction

     b. Decreased

2) Rate of the forward reaction

    a. Increased

3) Rate of the reverse reaction

    a. Increased

4) Activation energy of the forward reaction

    b. decreased

Explanation:

<em>Activarion energy</em> is the energy required by the reactants to form the intermediate transition state and become products.

<em>Catalysts</em> are substances that change the path of the chemical reactions, lowering the activation energy, and thus speeding up the rate of the reactions, since the products can reach the new lower activation energy faster.

The equilibrium reactions are the chemical process in which two reactions, the <em>forward and the reverese reactions</em>, occur simultaneously and at the same rate.  The equlibrium reactions may be represented by:

  • A ⇄ B

Where A → B is the direct or forward reaction, and A ← B is the reverse reaction (note the inversed arrow, from right to left).

For the direct reaction A represents the reactants and B represents the products. On the other hand, B represents the reactants and A represents the reactants of the reverse reaction and A. This, is A is the reactant of the forward reaction and the product of the reverse reaction, while B is the reactant of the reverse reaction and the product of the forward reaction.

Since, <em>the addition of a catalyst</em> lowers the activation energy of the process, the new activation energy is lower for both the forward and the reverse reaction, meaning that:

1. <em>The activation energy of the reverse reaction is decreased</em> (option b. of the first question)

2.<em> The rate of the forward reaction is increased</em> (option a. of the second question)

3. <em>The rate of the reverse reaction is increased</em> (option a. of the third question).

4. <em>Activation energy of the forward reaction is decreased</em> (option b. of the fourth question).

In summary, the addition of a catalyst decreases the activation energy for both forward and reverse reactions, and increases the rate of both forward and reverse reactions.

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3 years ago
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