Which shows the electron dot formula for hydrogen chloride? An electron dot formula shows a hydrogen symbol with one dot to its right followed by a plus sign and a Chlorine atom with one dot to its left and two dots to its right, top, and bottom.

An electron dot formula shows a chlorine symbol surrounded by six dots, two above, two below, and two to the left. A hydrogen symbol to the left of the chlorine symbol with two dots in between. There are two dots to the left of the hydrogen symbol.

An electron dot formula shows a chlorine symbol surrounded by eight dots, two on each side. A hydrogen symbol is to the left of the chlorine symbol. There are two dots in between the hydrogen symbol and the chlorine symbol.

An electron dot formula shows a hydrogen symbol and a Chlorine symbol with two dots in between.

An electron dot formula shows two dots followed by a hydrogen symbol, another set of dots, a Chlorine symbol, and another set of two dots.

Iwould you figure it out but you don't have a picture or in your question as to what the physical property you're looking for is

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.

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