Chemistry, 25.06.2021 22:10 hebrew1148

What would be the effect of decreasing the pressure on this system when it is in equilibrium?
2H2 +02 2 2H2O
O A. H, and O2 would react to produce
H20 more quickly.
B. The system would remain in equilibrium.
O C. H20 would react to produce H, and 02 more quickly
.
D. All of the molecules would react more slowly.

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Noble gases is far right

Answer from: Quest

Ithe undersigned only for use your own email and delete any and copies and of the itself and to be there is a good timewhat does a decrease in temperature indicate about the molecules in a sample of liquid water? decreased chemical energy of the moleculesdecreased kinetic energy of the moleculesincreased chemical energy of the moleculesincreased kinetic energy of the moleculeshow does a sample of helium at 15 â°c compare to a sample of helium at 215 k? the helium at 15 â°c has a higher average kinetic energy than the sample at 215 k.the helium at 15 â°c has lower nuclear energy than the sample at 215 k.the helium at 15 â°c has slower-moving atoms than the sample at 215 k.the helium at 15 â°c has smaller atoms than the sample at 215 k.

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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.

Answer from: Quest

Iam pretty sure that the answer is d

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What would be the effect of decreasing the pressure on this system when it is in equilibrium?
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