An experiment requires 15.0 ml of a 0.100 m solution of hcl. if you have a 12.0 m hcl stock solution in the laboratory, how would you prepare enough 0.1 m solution to run at least 3 experiments?

Missing question: 

A. add 0.833 mL 12 M HCl stock solution to 99.167 mL water. 

B. add 0.833 mL 12 M HCl stock solution to 49.167 mL water. 

C. add 0.833 mL 12 M HCl stock solution to 199.167 mL water.

c₁ - original concentration of the solution, before it gets diluted.

c₂ - final concentration of the solution, after dilution.
V₁ - volume to be diluted.
V₂ - final volume after dilution.
c₁ · V₁ = c₂ · V₂.
c₁(HCl) = 12.0 M.

c₂(HCl) = 0.100 M.

V₂(HCl) = 3 · 15.0 mL = 45.0 mL; for three experiments.

V₁(HCl) = c₂ · V₂ / c₁.

V₁(HCl) = 0.375 mL.

Make proportion: 0.375 mL : 0.833 mL = 45 mL : V.

V = 100 mL.

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the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.