At high temperatures, sulfur combines with iron to form brown-black iron(II) sulfide: Fe(s) + SD-Fes(s)
In an experiment, 5.00 g of iron reacts with excess sulfur to produce 6.29 g of iron(II) sulfide. Calculate the percent yield if the maximum
amount of iron(II) sulfide that can be produced is 7.87 g.
79.796
79.9%
100%
63.5%
125%

The percent yield : 79.9%

Further eplanation

Percent yield is the comparison of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

Fe(s)+S(s)⇒FeS(s)

The reaction produces 6.29 g of iron(II) sulfide⇒an actual yield

The maximum  amount that can be produced is 7.87 g ⇒ A theoretical yield

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.