Elements with atomic numbers of 104 and greater are known as super-heavy elements. None of these elements have been found in nature but instead have been made in a laboratory. They are very difficult and expensive to create, and they break down into other elements quickly. There currently are no practical applications for any of these elements. However, some scientists believe that, with further work, they may discover some isotopes of super-heavy elements that are more stable and that could possibly have practical implications. Do you think that scientists should continue to try to create super-heavy elements and expand the periodic table? Explain why or why not.

The density would be 1.622

It could possibly be beneficial if the experiment is successful.

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.