Chemistry, 06.03.2020 03:52 oakleylynn

A 3 L gas filled vessel contains HCl gas with a pressure of 9 atm at standard temperature. The contents are bubbled into water until the pressure of the gas container drops to 4 atm. The final volume of the HCL solution is .5 L. How many moles of HCL were used?

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Answer from: Quest

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.

Answer from: Quest

Aquantity of that isotope to decay to 1/2 of its original value

Answer from: Quest

answer: - first one is the correct answer. $k_2s(aq)+cocl_2(aq)\rightarrow 2kcl(aq)+cos(s)$ .

explanations: - potassium sulfide is $k_2s$ and cobalt(ii) chloride is $cocl_2$ .

charge on k is +1, charge on s is -2, charge on co is +2 and charge on cl is -1.

in a double replacement reaction the ion exchange takes place. in general the reaction looks like:

$ab+cd\rightarrow ad+cb$

if one of the product is insoluble as per solubility rules then a precipitate of it is formed.

when we mix the aqueous solutions of potassium sulfide and cobalt(ii) chloride then chloride ion goes with potassium ion and sulfide ion goes with cobalt ion and the balanced equation is written as:

$k_2s(aq)+cocl_2(aq)\rightarrow 2kcl(aq)+cos(s)$

k and cl has equal charges, similarly co and s has equal charges and hence no subscripts are used for the product side.

as per the solubility rules, cos is insoluble and so it's precipitate is formed.

looking at the given choices, first choice is correct.

Answer from: Quest

The answer would be b

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A 3 L gas filled vessel contains HCl gas with a pressure of 9 atm at standard temperature. The conte...