How many grams of Na 2 CO 3 are present in 21.0 mL of a solution that is 36.0 % Na 2 CO 3 by mass? The density of the solution is 1.12 g / mL .

The answer to your question is  8.46 g of Na₂SO₄

Explanation:

Data

mass of Na₂CO₃ = ?

volume = 21 ml

percent by mass = 36%

density = 1.12 g/ml

Process

1.- Calculate the mass of the solution

    density = mass / volume

solve for mass

    mass = density x volume

     mass = 1.12 x 21

     mass = 23.52 g

2.- Calculate the mass of Na₂CO₃

Percent by mass = (mass of Na₂SO₄ / mass of solution) x 100

Solve for mass of Na₂SO₄

               mass of Na₂SO₄ = Percent by mass x mass of solution/100  

               mass of Na₂SO₄ = 36 x 23.52/100

               mass of Na₂SO₄ = 8.46 g

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this

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explanation:

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