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3 years ago

If a solution containing 24.0 g of a substance reacts by first-order kinetics, how many grams remain after three half-lives?

Chemistry

1 answer:

7nadin3 [17]3 years ago

8 0

Answer:

3.0g remain

Explanation:

The half-life is defined as the time required for a reactant to decrease its concentration in exactly the half of the initial amount of reactant. Having this in mind:

In one half-life, the mass will be:

24.0g / 2 = 12.0g

In two half-lifes:

12.0g / 2 = 6.0g

And in three half-lifes, the mass that remain is:

6.0g / 2 =

<h3>3.0g remain</h3>

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3 years ago

an aqeous solution of oxalic acid h2c2o4 was prepared by dissolving a 0.5842g of solute in enough water to make a 100 ml solutio

vampirchik [111]

The question is incomplete, here is the complete question:

An aqeous solution of oxalic acid was prepared by dissolving a 0.5842 g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml. How many grams of oxalic acid are in 100. mL of the final solution?

<u>Answer:</u> The mass of oxalic acid in final solution is 0.0234 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of oxalic acid = 0.5842 g

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of oxalic acid solution}=\frac{0.5842\times 1000}{90\times 100}\\\\\text{Molarity of oxalic acid solution}=0.0649M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated oxalic acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted oxalic acid solution

We are given:

$M_1=0.0649M\\V_1=10mL\\M_2=?M\\V_2=250.0mL$

Putting values in above equation, we get:

$0.0649\times 10=M_2\times 250.0\\\\M_2=\frac{0.0649\times 10}{250}=0.0026M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of oxalic acid solution = 0.0026 M

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0026=\frac{\text{Mass of oxalic acid solution}\times 1000}{90\times 100}\\\\\text{Mass of oxalic acid solution}=\frac{0.0026\times 90\times 100}{1000}=0.0234g$

Hence, the mass of oxalic acid in final solution is 0.0234 grams

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