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.If 1.36 mL of a solution of Na2S2O3 is mixed with 10.0mL of solution A and 12.30 mL of solution B, the concentration of Na2S2O3 is found to be 4.74*10^-4M. What is the concentration of the original Na2S2O3 solution?

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  • Anonymous
    2 months ago

    You'd get better help if you posted in "Chemistry".

  • Ash
    Lv 7
    2 months ago

    C₁V₁=C₂V₂

    C₁ = C₂V₂/V₁

    C₁ = (4.74 x 10⁻⁴ M)(1.36 mL + 10.0 mL + 12.30 mL)/(1.36 mL)

    C₁ =  8.25 x 10⁻³ M

  • Bryce
    Lv 7
    2 months ago

    1.36 + 10.0 + 12.30= 23.66 mL= 0.02366 L

    x/0.02366= 4.74*10^-4 M; x= 1.12*10^-5 mol Na2S2O3

    1.12*10^-5/0.00136= 0.00824 M

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