A jet takes off from an aircraft carrier starting from rest and travels 93 m in 1.2 s when being pushed by the catapult. What is its final velocity at takeoff?

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.

Following are the  advantages of the parametric equations: 1) they can be used to graph curves that are not functions, like the unit circle.2) the parameter can be used to represent the quantity that is  useful to  provide us with better information about the graph.3)  it is useful to convert between rectangular equations and parametric equations.with specific reference to chemistry,  parametric equations are employ for solving hydrogen and hydrogen-like ions, quantum mechanically. herein, cartesian coordinates are converted in spherical polar coordinates. the solution obtained, when plotted on spherical polar graph paper, provides visualization of atomic orbital such as s,p,d and f.