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4 December, 03:58

an experiment calls for 50 ml of a 0.50M aqueous solution of a sodium bicarnonate. Work with your group to describe the steps you would use in order to make up such a solution making sure you have none left over

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  1. 4 December, 04:17
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    We have to weight out 2.1 grams of NaHCO3 and dissolve it in 50 mL water.

    Explanation:

    Step 1: Data given

    Volume = 50 mL = 0.050 L

    Molarity sodium bicarbonate (NaHCO3) = 0.50 M

    Molar mass NaHCO3 = 84.0 g/mol

    Step 2: Calculate moles NaHCO3

    Moles NaHCO3 = molarity NaHCO3 * volume NaHCO3

    Moles NaHCO3 = 0.50 M * 0.050 L

    Moles NaHCO3 = 0.025 moles

    Step 3: Calculate mass NaHCO3

    Mass NaHCO3 = moles NaHCO3 * molar mass NaHCO3

    Mass NaHCO3 = 0.025 moles * 84.0 g/mol

    Mass NaHCO3 = 2.1 grams NaHCO3

    We have to weight out 2.1 grams of NaHCO3 and dissolve it in 50 mL water. This gives us a 0.50 M aqueous solution of a sodium bicarnonate
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