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8 March, 15:08

Calculate the number of milliliters of 0.587 M NaOH required to precipitate all of the Ni2 + ions in 163 mL of 0.445 M NiBr2 solution as Ni (OH) 2. The equation for the reaction is: NiBr2 (aq) + 2NaOH (aq) Ni (OH) 2 (s) + 2NaBr (aq) mL NaOH Submit AnswerRetry Entire Group

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  1. 8 March, 15:58
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    We need 247 mL of NaOH

    Explanation:

    Step 1: Data given

    Molarity of NaOH = 0.587 M

    Volume of 0.445 M NiBr2 solution = 163 mL = 0.163 L

    Step 2: The balanced equation

    NiBr2 (aq) + 2NaOH (aq) Ni (OH) 2 (s) + 2NaBr (aq)

    Step 3: Calculate moles of NiBr2

    Moles NiBR2 = Molarity NiBR2 * volume

    Moles NiBR2 = 0.445 M * 0.163 L

    Moles NiBR2 = 0.0725 moles

    Step 3: Calculate moles of NaOH

    For 1 mol NiBr2 consumed, we need 2 moles NaOH

    For 0.0725 moles NiBR2, we need 2 * 0.0725 = 0.145 moles NaOH

    Step 4: Calculate volume of NaOH

    Volume = moles NaOH / Molarity NaOH

    Volume = 0.145 moles / 0.587 M

    volume = 0.247L = 247 mL

    We need 247 mL of NaOH
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