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Volume of HCl used 25.0mL 4 lInitial burette reading 0.50mLFinal burette reading 25.60mLConcentration

Posted on October 23, 2021 By Amberwithnell12512 3 Comments on Volume of HCl used 25.0mL 4 lInitial burette reading 0.50mLFinal burette reading 25.60mLConcentration

Volume of HCl used 25.0mL 4 l Initial burette reading 0.50mL
Final burette reading 25.60mL
Concentration of KOH 1.0M
the molarity
HCl solution
Calculate​

Chemistry

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Comments (3) on “Volume of HCl used 25.0mL 4 lInitial burette reading 0.50mLFinal burette reading 25.60mLConcentration”

  1. Expert says:
    October 23, 2021 at 3:34 pm

    You know how the first one if filled up like the objects do the same but with the other categories

    Reply
  2. Expert says:
    October 23, 2021 at 5:01 pm

    i think the answer is terrae but not 100 percent sure. ▓▬▬ §

    Reply
  3. bella2284 says:
    October 23, 2021 at 10:54 pm

    1.0 M

    Explanation:

    Reaction equation;

    KOH(aq) + HCl(aq) > KCl(aq) + H2O(l)

    Concentration of acid CA = ?

    Concentration of base CB = 1.0 M

    Volume of base VB = 25.60 - 0.50 = 25.1 ml

    Volume of acid VB =  25.0 ml

    Number of moles of acid NA = 1

    Number of moles of base NB =2

    CAVA/CBVB =NA/NB

    CAVANB = CBVBNA

    CA = CBVBNA/VANB

    CA = 1 * 25.1 * 1/25.0 *1

    CA = 1.0 M

    Reply

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