Chemistry, 20.01.2022 14:00 jtg23

Calculate the relative difference of the real (van-der-Waals) gas pressure to the ideal gas pressure under these conditions in %. Assume the ideal gas pressure to be 100%. By how many % does the predicted pressure increase (positive answer) or decrease (negative answer) upon the use of the van-der-Waals corrections compared to the ideal gas law

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Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.

Answer from: Quest

the formula for the conversion of temperature from degree celsius to kelvin is:

$k = ^{o}c + 273.15$

the sample of water at $38^{o}c$ is converted to kelvin temperature by using the above formula as:

$k = 38^{o}c + 273.15 = 311.15 k$

on comparing the two samples of water at $38^{o}c$ or we can say that at $311.15 k$ and at $295 k$ respectively. the difference between the temperatures of two samples of water is:

$311.15 k - 295 k = 16.15 k$

hence, one can say that the sample of water at $38^{o}c$ is greater than the sample of water at $295 k$ by $16.15 k$ .

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

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Calculate the relative difference of the real (van-der-Waals) gas pressure to the ideal gas pressure...

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