1.calculate the final concentration of a solution made by mixing 250.mL of 5.00

Question

1.calculate the final concentration of a solution made by mixing 250.mL of 5.00 M HCl with enough water to make a final volume of 600.mL of solution 2.calculate the final concentration of a solution made by mixing 150.mL of 10.0%(m/v) KCl with 475mL of water 3.how many millimoles of magnesium ions are present in 0.300L of a solution that contains 85.0mEq/L the of magnesium ions 1.calculate the final concentration of a solution made by mixing 250.mL of 5.00 M HCl with enough water to make a final volume of 600.mL of solution 2.calculate the final concentration of a solution made by mixing 150.mL of 10.0%(m/v) KCl with 475mL of water 3.how many millimoles of magnesium ions are present in 0.300L of a solution that contains 85.0mEq/L the of magnesium ions 2.calculate the final concentration of a solution made by mixing 150.mL of 10.0%(m/v) KCl with 475mL of water 3.how many millimoles of magnesium ions are present in 0.300L of a solution that contains 85.0mEq/L the of magnesium ions

Explanation / Answer

1.calculate the final concentration of a solution made by mixing 250.mL of 5.00 M HCl with enough water to make a final volume of 600.mL of solution

V1 * S1 = V2 * S2

250 * 5.00 = 600 * S2

S2 = 2.08 M

so final concentration is 2.08 M.

2.calculate the final concentration of a solution made by mixing 150.mL of 10.0%(m/v) KCl with 475mL of water

V1 * S1 = V2 * S2

150.0 * 10.0 = (150 + 475) * S2

S2 = 2.40 %

so final concentration is 2.40 %

3.how many millimoles of magnesium ions are present in 0.300L of a solution that contains 85.0mEq/L the of magnesium ions:

meq of Mg ion = 85.0 * 0.300 = 25.5 meq

milimole of Mg ion = 25.5 / 2 = 12.75 mmole Mg ion is present.

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