You need to make an aqueous solution of 0.192 M barium sulfide for an experiment in lab, using a 500 mL volumetric flask. How much solid barium sulfide should you add?

Explanation:

Molecular weight of barium sulphide = 169

500 mL of .192 M barium sulfide = .5 x .192 moles of barium sulphide

= .096 moles of barium sulfide

= .096 x 169 gram of barium sulfide

= 16.22 grams of barium sulfide .

We shall have to add 16.22 gram .

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.