Using the following general equation for a nuclear reaction, complete the statements below. the atomic number for uranium is? the mass of the reactant neutron is ? the atomic mass for krypton is? there are ? barium neutrons. the number of neutrons this reaction emits is ?

Atomic number is written in subscript:
Uranium atomic number =92

Mass of neutron is the superscript value times the quantity (moles):
3*1= 3

Atomic mass is approximately equivalent to the number of protons and neutrons in the atom (the mass number). It's written in superscript:

Kr=92

No. of neutrons = Mass number (in superscript) - Atomic number ( in subscript)

= 141-56= 85

No. of neutrons in reactants are: 3

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.