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How many additional mL of H2O are required to dilute 135.0ml of 6.50 M HCl solution to 0.750 M?
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is this not the same concept as before? what chapter are you on ?
Chapter 4 and you are probably right.
Chapter 4 being over stoichiometry
You're taking a small amount of something very concentrated, and adding water to dilute it to a larger volume, but a lower concentration. The equation that is useful here is :\[M{_C}*V{_C} = M{_D}*V{_D}\] where C is the concentrated form, and D is the dilute form. You have 3 of these 4 values given in the problem, plug in and solve.
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