A chemist had a bottle of HCl (aq) of unknown concentration. She performed a titration experiment in order to determine its concentration. She pipetted 12.55 mL of the HCl solution into an Erlenmeyer flask. She then added about 21 mL of deionized water and 3 drops of phenolphthalein indicator dye to the flask. She titrated this mixture with 0.0716 M NaOH (aq), and obtained a permanent, pale pink endpoint after the addition of 23.16 mL of NaOH (aq). Calculate the molarity of the HCl solution and report your final answer with units and the correct significant digits. HCl (aq) NaOH (aq) --> H2O (l) NaCl (aq)

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Ithe undersigned only for use your own email and delete any and copies and of the itself and to be there is a good timewhat does a decrease in temperature indicate about the molecules in a sample of liquid water? decreased chemical energy of the moleculesdecreased kinetic energy of the moleculesincreased chemical energy of the moleculesincreased kinetic energy of the moleculeshow does a sample of helium at 15 â°c compare to a sample of helium at 215 k? the helium at 15 â°c has a higher average kinetic energy than the sample at 215 k.the helium at 15 â°c has lower nuclear energy than the sample at 215 k.the helium at 15 â°c has slower-moving atoms than the sample at 215 k.the helium at 15 â°c has smaller atoms than the sample at 215 k.

Answers are 1,2,4 its correct on edgenuity.

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.