Chemistry, 19.04.2021 06:00 AdoreeLayla

S8 +O2 = SO8

Given 80 mols of product, find the amount of each reactant in mols.

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Answer from: iliketurtures

10 moles of S₈

320 moles O₂

Explanation:

S₈ +O₂ → SO₈

The balanced equation is:

S₈ + 32O₂ → 8SO₈

Now we can convert moles of SO₈ (the product) into moles of each reactant, using the stoichiometric coefficients of the balanced reaction:

80 mol SO₈ * $\frac{1molS_8}{8molSO_8}$ = 10 mol S₈80 mol SO₈ * $\frac{32molO_2}{8molSO_8}$ = 320 mol O₂

In order for 80 moles of SO₈ to be produced, 10 moles of S₈ need to react with 320 moles of O₂.

Answer from: Quest

the transfer of heat by electromagnetic radiation is called thermal radiation . it is usually carried out by electromagnetic waves in an energy range just below the visible spectrum, which is the infrared range.

so the answer would be radiation.

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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.

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S8 +O2 = SO8

Given 80 mols of product, find the amount of each reactant in mols....

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