Chemistry, 01.04.2020 04:10 crazylife77

Warming a gas at constant volume and a pressure of 1.50 atm from 27 degrees Celsius to 127 degrees Celsius gives a final pressure of

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Answer from: Tips101

Final pressure of gas = 2 atm

Explanation:

Given data:

Initial pressure of gas = 1.50 atm

Initial temperature = 27°C

Final temperature = 127°C

Final pressure of gas = ?

Solution:

Initial temperature = 27°C (27+273 = 300 K)

Final temperature = 127°C ( 127+273 = 400 K)

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

1.50 atm / 300 K = P₂/400 K

P₂ = 1.50 atm × 400 K / 300 K

P₂ = 600 atm. K /300 K

P₂ = 2 atm

Answer from: Quest

False, as a substance changes from a solid to a liquid and from a liquid to a gas the molecules get farther apart from each other

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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Warming a gas at constant volume and a pressure of 1.50 atm from 27 degrees Celsius to 127 degrees C...

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