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

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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every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.