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zlopas [31]
3 years ago
14

Find the original concentration of a solution diluted from 15.00 mL to 125.0 mL with a final concentration of 0.00383 M

Chemistry
1 answer:
BabaBlast [244]3 years ago
8 0
<h3>Answer:</h3>

0.03192 M

<h3>Explanation:</h3>

We are given;

  • Initial volume before dilution, V1 = 15.0 mL or 0.015 L
  • Volume after dilution, V2 = 125.0 mL or 0.125 L
  • Concentration after dilution, M2 = 0.00383 M

We are required to calculate the initial concentration, M1 before dilution.

  • To answer the question, we will use the dilution formula;
  • Dilution formula is given by; M1V1 = M2V2

Rearranging the formula;

M1 = M2V2 ÷ V1

    = ( 0.00383 M × 0.125 L) ÷ 0.015 L

   = 0.03192 M

Therefore, the original concentration of the solution before dilution is 0.03192 M

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Please review the attachment
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Answer: The correct answer is -297 kJ.

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To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

2SO3 —> O2 + 2SO2 (196 kJ)

Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:

SO3 —>1/2O2 + SO2 (98 kJ)

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Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

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Read 2 more answers
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