Derikcastillo810
contestada

If 45.0 mL of a 4.00 M sodium sulfate (Na2SO4) solution is used for the reaction shown above, how many moles of sodium ions were present in solution before the reaction proceeded?​

Respuesta :

Answer:0.360

Explanation:

I had this on ck12 yesterday

abidemiokin

The moles of sodium ions were present in the solution before the reaction proceeded is 0.36moles

Molarity of a solution

The formula for calculating the molarity of a solution is expressed as:

Molarity = moles/volume

Given the following parameters

Molarity = 4.00M

Volume of sodium sulfate = 45mL = 0.045L

Substitute to have;

Molarity = 4 * 0,045

Molarity = 0.18

Since there are 2 atoms of sodium ion in the solution, hence the moles of sodium ions were present in the solution before the reaction proceeded is 0.36moles

Learn more on molarity here: https://brainly.com/question/17138838

Otras preguntas