Volume (L)
Pressure
(atm)
0.86
22.9
1.94
10.3
2.96
6.7
4.03
5.0
5.02
3.9
5.59
3.5
In-Class Lab Activity 2.14 b
Homework. Unanswered
In some cases, the relationship between two variables is not linear, but the data can be manipulated to make a linear relationship.
Using the data above, make a graph of the Volume vs. Pressure of a container, with volume being the independent variable. Is the
relationship linear? if it is not make a graph of Volume vs. 1/pressure. What would the pressure of the container be if the volume
were 2.4L? Show your calculation.

The closer the atoms are to becoming ionic the more highly dense the negatively charged side gets.

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.