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d. (5 Points) A 150 mL portion of 0.250 M HCl solution was titrated with NaOH; It took 100. mL of a 0.375 M base to neutraliz
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* Now, total volume of solution =(150+70) me - 220 мм, о дох), то о At first, moles of Hel present = Іго — . О3 мод ro, moke)

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