Chemistry, 02.10.2019 02:00 christensenlandon15

Hydrogen gas is allowed to permeate through a thin plastic membrane that is 83 microns thick. concentrations of 0.025 moles/m3 and 0.0025 moles/m3 are maintained on the high and low pressure sides of the membrane, respectively. if the diffusivity of hydrogen gas molecules in the plastic is 4.2x10-8 m2/s, how many moles of hydrogen gas pass through a 6.0 square meter area of the plastic membrane every hour ? hint: we are really computing a flow rate here, rather than a flux. a flow rate is equal to a flux multiplied by an area, and its units are generally count / time.

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answer: the primary difference between a hypothesis and a theory is that b hypothesis is falsifiable, but a theory has been proven to be correct. theory is an explanation of something that is happening but hypothesis is a guess of something that can happen

explanation:

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Answer : option a) 84explanation :

considering the question as per the attachment

$^{222}_{86}> ^{218}_{84}po+^{4}_{2}he$

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.

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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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