Notice that the effective analytical connection for analysis of hypochlorite anion by the thiosulfate anion in this experiment is the sum of the two reactions: 1) hypochlorite anion with iodide anion to form iodine, and 2) thiosulfate anion with the iodine formed in reaction one. In your laboratory notebook sum these two reactions to find the stoichiometric factor that relates moles of thiosulfate anion needed to react with each mole of hypochlorite anion in the bleach sample. Then select the correct response. a. One mole of thiosulfate anion needed to react with one mole of hypochlorite anion. b. Two moles of thiosulfate anion needed to react with one mole of hypochlorite anion. c. Three moles of thiosulfate anion needed to react with one mole of hypochlorite anion.

b. Two moles of thiosulfate anion needed to react with one mole of hypochlorite anion.

Explanation:

1. Hypochlorite with iodide

                      2I⁻ ⟶ I₂ + 2e⁻

ClO⁻ + 2H⁺ + 2e⁻ ⟶ Cl⁻ + H₂O      

2I⁻ + ClO⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + H₂O

2. Thiosulfate with iodine

     2S₂O₃²⁻ ⟶ S₄O₆²⁻ + 2e⁻

      I₂ + 2e⁻ ⟶ 2I⁻                  

2S₂O₃²⁻ + I₂ ⟶ S₄O₆²⁻ + 2I⁻

3. Sum of the reactions

                2I⁻ + ClO⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + H₂O

                      2S₂O₃²⁻ + I₂ ⟶ S₄O₆²⁻ + 2I⁻                    

2I⁻ + ClO⁻ + 2S₂O₃²⁻ + 2H⁺ ⟶ I₂  + Cl⁻ + S₄O₆²⁻ + H₂O

4. Molar ratio

S₂O₃²⁻:ClO⁻ = 2:1

it's filtration, but this is an assumption so my apologies if i got this

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