Chemistry, 04.05.2021 20:20, chy2313

Calculate the density of CF3Cl at 18 °C and a pressure of 114.0 kPa

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answered: Guest

some of the qualities most frequently associated with the baroque are grandeur, sensuous richness, drama, vitality, movement, tension, emotional exuberance, and a tendency to blur distinctions between the various arts.

answered: Guest

answer:

technetium is the most reactive. this is followed by silver. the least reactive out of these three elements is strontium.

answered: Guest

Answerd. may include all of these are many definitions of bases: the brønsted-lowry definition of acids and bases says that bases are proton acceptors.the arrhenius definition of acids and bases says that bases increase [oh⁻] in solution (thereby producing oh⁻ ions)the lewis definition of acids and bases says that bases donate electron pairs.therefore, the sensible answer is that a base may include all of the described characteristics.

answered: Guest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ 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: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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Calculate the density of CF3Cl at 18 °C and a pressure of 114.0 kPa...