Astudent mixes 0.50 l of 0.0020 m ba(no3)2 solution with 0.50 l of 0.0080 m na2so4 solution. which equation shows the proper set-up to calculate molar concentration of ba2+ ions in the new solution?
(0.0020 m)(1 l) = (? m)
(0.0020 m)(0.5 l) = (? m)(1 l)
(0.0020 m)(0.5 l) = (0.0080 m)(1 l)
(0.0020 m)(0.5 l) = (0.0080 m)(0. l)

(0.0020 M)×(0.5 L) = (? M)×(1 L)

Explanation:

To find the concentration of the final solution when you concentrate or dilute a particular solution is done by using the following formula:

initial concentration × initial volume = final concentration × final volume

The question is asking about the concentration of barium ions in the final solution.

initial concentration of Ba²⁺ = 0.0020 M

initial volume = 0.50 L

final concentration = unknown (?)

final volume = volume of the initial solution + volume of the second solution

final volume =  0.50 + 0.50 = 1 L

You insert the values in the equation and you find the answer asked by the question.

B

it's B

Explanation:B

B

h let me think

pure substance: iron,copper

impure substance: water,charcoal