D Question 7
5 pts
Use the following fake periodic table to answer the question.
For scientific notation numbers, use e in your answer. Example: 6.02x1023 would be entered
as 6.02e23.
A: 1.114
B: 2.017
C: 3.459
D: 4.781
E: 5.991
H: 11.489
I: 13.157
J: 14.998
K: 20.102
L: 21.703
O: 29.487
P: 31.452
Q: 33.333
R: 37.882
S: 40.004
V: 62.804
W: 68.114
X: 72.011
Y: 79.019
Z: 88.476
How many atoms are there in 1.42L of J at STP?
Round your answer to 3 sig-figs.

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.

Answer it is also known as the completion state is untrue of the activation complex;

Answeropposite;

They make up examples of physical properties