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28 April, 11:40

In the laboratory a student combines 44.9 mL of a 0.159 M barium nitrate solution with 16.2 mL of a 0.595 M barium chloride solution. What is the final concentration of barium cation?

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  1. 28 April, 13:56
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    [Ba2+] = 0.2746 M

    Explanation:

    We have a 44.9 mL barium nitrate solution of 0.159M and a 16.2mL barium chloride solution of 0.595M

    To find the concentration of the barium cation we use the following equation:

    Concentration = moles of the solute / volumen of the solution

    [Ba2+] = (44.9 * 10^-3 L * 0.159 M + 16.2 x 10^-3 L * 0.595 M) / ((44.9 mL + 16.2mL) * 10^-3 L)

    After calculating this, we find that the concentration of the barium cation:

    [Ba2+] = 0.2746 M
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