If 50 ml of 1.00 M of H2SO4 and 50 ml of 2.0 M KOH are mixed what is the concentration of the resulting solutes?

0.5 M

Explanation:

First, let us look at the balanced equation of the reaction.

The solute formed is .

Recall that: mole = molarity x volume

Hence,

50 ml, 1.00 M H2SO4 = 0.05 x 1 = 0.05 mole

50 ml, 2.0 M KOH = 0.05 x 2 = 0.1 mole

From the equation

1 mole of H2SO4 reacts with 2 moles of KOH to give 1 mole of K2SO4.

Hence,

0.05 mole H2SO4 reacting with 0.1 mole KOH will give 0.05 mole .

Also recall that: concentration = mole/volume

Total volume of resulting solution = 50 ml + 50 ml = 100 ml or 0.1 liter

Concentration of = mole of /volume of resulting solution

                              = 0.05/0.1 = 0.5 M

The concentration of the resulting solute = 0.5 M

precipitation

explanation:

when it goes from cloud to rain/back to earth in the form of rain

lol for the pointsss