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vesna_86 [32]
2 years ago
12

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

ke for 90% of the chemical to be destroyed? (b) how long will it take for 99% of the chemical to be destroyed? (c) how long will it take for 99.9% of the chemical to be destroyed?
Chemistry
1 answer:
Klio2033 [76]2 years ago
5 0

Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

Here, k is rate constant of the reaction, t is time of the reaction, A_{0} is initial concentration and A_{t} is concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c)  Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

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<u>Explanation:</u>

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\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m

OR

\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

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m_{solute} = Given mass of solute = 10 g

M_{solute} = Molar mass of solute = ? g/mol

w_{solvent} = Mass of solvent = 200 g

Putting values in equation 1, we get:

81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol

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 The  limiting  reagent  is   <u>H₂SO₄</u>

   <u><em>calculation</em></u>

<u><em> </em></u>Step 1 :write the equation for reaction

2 NaOH + H₂SO₄  → Na₂SO₄   + 2 H₂O

Step  2: use the mole ratio to determine the  moles of product  produced from each reactant

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= 10.0 moles x 1/2 =  5.0 moles

H₂SO₄ :Na₂SO₄  is 1:1 therefore the moles  of Na₂SO₄  is also = 3.50 moles


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How a reaction between colorless substances can produce a colored precipitate
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A reaction between colorless substances can produce a colored precipitate by formation of new ionic compound.

<h3>What is an ionic compound?</h3>

An ionic compound is a compound that is made up of two elements, one metal and the other non metal, which are held together by an ionic bonding.

When reaction occurs between two colourless substances, colored precipitate results by formation of new ionic compound.

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Answer:

3.54 mol

Explanation:

Step 1: Given data

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Step 2: Convert "T" to Kelvin

We will use the following expression.

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