Chemistry, 30.10.2020 17:00 darriennichole

I have added 12.0 L of air. To a balloon at sea level(760 torr). If I take the ballon with me to Colorado, where the air pressure is 0.650 atm, what will the new volume of the balloon be?

Answers: 3

Show answers

Answers

Answer from: hopecastillo83098

18.46 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 12.0 L

Initial pressure (P1) = 760 torr

Final pressure (P2) = 0.650 atm

Final volume (V2) =..?

Next, we shall express the pressure in the same unit of measurement.

Thus, 760 torr is equivalent to 1 atm.

Finally, we shall the new volume of the balloon as follow:

Initial volume (V1) = 12.0 L

Initial pressure (P1) = 1 atm

Final pressure (P2) = 0.650 atm

Final volume (V2) =..?

P1V1 = P2V2

1 × 12 = 0.650 × V2

12 = 0.650 × V2

Divide both side by 0.65

V2 = 12/0.65

V2 = 18.46 L

Therefore, the new volume of the balloon is 18.46 L.

Answer from: Quest

Read the sentences below in 1962, i tried to enroll as the first black student at the university of mississippi. the governor of mississippi blocked my entrance which best describes the verbs used in the sentences? o they are past tense verbs, which are often used in narrative essays o they are past tense verbs, which are not often used in narrative essays. o they are future tense verbs, which are often used in narrative essays o they are future tense verbs, which are not often used in narrative essays.

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.