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ValentinkaMS [17]
3 years ago
14

I need this answer quick please show work

Chemistry
1 answer:
Ainat [17]3 years ago
4 0

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

                        x        ----------------------- 0.4195 moles

                        x = (0.4195 x 60.05) / 1

                        x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

               

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Answer #1. A 2.5% (by mass) solution concentration signifies that there is 2.5 grams of solute in every 100 g of solution. 
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Answer #2. therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution. 
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     mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
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Answer#3. a solution mass of 1 kg is 10 times greater than 100 g, thus 1kg of a 2.5% ki solution would contain 25 grams of ki.
Since 1000 grams is 1 kg, we multiply 10 to each mass so that 100 grams becomes 
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                                                          = 25g solute/(25g solute + 975g solvent)
                                                          = 25g solute/1000g solution
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<h3>Answer:</h3>

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<h3>Explanation:</h3>

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We are required to calculate the change in temperature;

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