iF 23.74Ml OF 0.01470m nAoh are required to completely neutrize 25.00mL of gastric juice,calculate the concentration of the hydrochloric acid(in g/L )in the stomach gastric steps and calculations

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  1. 23.73*.01470=25*M

    solve for M, the molarity of the gastric juice.

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  2. You need one mole of NaOH to neutralize one mole of HCl.

    You require 0.02374 L x 0.0147 mole/L = 3.49 *10^-4 moles of NaOH

    That many moles of HCl have a mass of 1.274*10^-2 g

    The concentration is
    (1.274*10^-2 g)/(25*10^-3 L)
    = 0.510 g/L

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