Anonymous
Anonymous asked in Science & MathematicsChemistry · 3 weeks ago

Help with Chemistry PLEASE!!!?

Given the volumes can someone tell me how to calculate:

-mmol of HCL

-mmol of NaOH

-Molarity  of NaOH

-Average molarity of NaOH

-Deviation

-Average Deviation

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  • 3 weeks ago

    In all cases adjust the number of significant digits as you see fit:

    mmol of HCl used:

    Trial 1: (12.6 mL) x (0.101 M HCl) = 1.2726 mmol

    Trial 2: (13.4 mL) x (0.101 M HCl) = 1.3534 mmol

    Trial 3: (13 mL) x (0.101 M HCl) = 1.313 mmol

    mmol of NaOH:

    HCl + NaOH → NaCl + H2O  [balanced as written]

    Trial 1: 1.2726 mmolTrial 2:  1.3534 mmolTrial 3:  1.313 mmol

    Molarity of NaOH:

    Trial 1: 1.2726 mmol / 13.4 mL = 0.09497 mol/LTrial 2:  1.3534 mmol / 14.4 mL = 0.09399 mol/LTrial 3:  1.313 mmol / 13.7 mL = 0.09584 mol/L

    Average molarity of NaOH:

    all Trials: (0.09497 mol/L + 0.09399 mol/L + 0.09584 mol/L) / 3 =  0.09493

    Deviation:

    Trial 1:  0.09497 mol/L - 0.09493 mol/L = 0.00004 mol/LTrial 2:  0.09493 mol/L - 0.09399 mol/L = 0.00094 mol/LTrial 3:  0.09584 mol/L - 0.09493 mol/L = 0.00091 mol/L

    Average Deviation:

    all Trials: ( 0.00004 mol/L +  0.00094 mol/L +  0.00091 mol/L) / 3 = 0.00063 mol/L

    • kens3 weeks agoReport

      can you help me out plz the concentration is 0.101M HCl

  • Anonymous
    3 weeks ago

    I can only help if you've asked an actual question(which you did not).

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