The concentration of a solution after dilution can be calculated using the formula M1V1 = M2V2, where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume.

Here, M1 = 2.0 M, V1 = 0.6 L, and V2 = V1 + volume of solvent added = 0.6 L + 0.4 L = 1.0 L.

We can substitute these values into the formula and solve for M2:

2.0 M * 0.6 L = M2 * 1.0 L

M2 = (2.0 M * 0.6 L) / 1.0 L = 1.2 M

So, the resulting concentration of the solution after dilution is 1.2 M.

Question

The concentration of a solution after dilution can be calculated using the formula M1V1 = M2V2, where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume.

Here, M1 = 2.0 M, V1 = 0.6 L, and V2 = V1 + volume of solvent added = 0.6 L + 0.4 L = 1.0 L.

We can substitute these values into the formula and solve for M2:

2.0 M * 0.6 L = M2 * 1.0 L

M2 = (2.0 M * 0.6 L) / 1.0 L = 1.2 M

So, the resulting concentration of the solution after dilution is 1.2 M.

Knowee AI · Accepted Answer

The concentration of a solution after dilution can be calculated using the formula M1V1 = M2V2, where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume.

Here, M1 = 2.0 M, V1 = 0.6 L, and V2 = V1 + volume of solvent added = 0.6 L + 0.4 L = 1.0 L.

We can substitute these values into the formula and solve for M2:

2.0 M * 0.6 L = M2 * 1.0 L

M2 = (2.0 M * 0.6 L) / 1.0 L = 1.2 M

So, the resulting concentration of the solution after dilution is 1.2 M.

A solution of Cs2SO4 at 2.0 M is diluted from its original volume of 0.6 L by adding 0.4 L. What is the resulting concentration of this solution?

Question

A solution of Cs2SO4 at 2.0 M is diluted from its original volume of 0.6 L by adding 0.4 L. What is the resulting concentration of this solution?

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