Use the References to access important values if needed for this question. Please note that "geometry" refers to the molecular or ionic geometry.
In the VSEPR model, the geometry of the regions defined by the electron pairs in the valence shell of an atom is called the electron-pair geometry.
A. The Lewis diagram for SCl2 is: C-5-c:
The electron-pair geometry around the Satom in SCl2 is
There are
unshared pair(s) around the central atom, so the geometry of the SCl2 molecule is
H
B. The Lewis diagram for SiH, is:
H-Si-H
H
The electron-pair geometry around the Si atom in SiH, is
There are unshared pair(s) around the central atom, so the geometry of the SiH, molecule is
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The independent variable is the type of music.the dependent variable is the pulse rate.the intervening variable is the age of the subject.if the mother does not like rock, her pulse is likely to not increase.the same might be true for the sister.the friend may like rap and gets excited.to establish cause/effect relationship, all of the subjects must be consistent.hope i .

Heat is involved in chemical change in two ways; chemical changes either result in heat being added to or removed from an object, and chemical changes require a certain amount of heat before they can take place.

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.