Chemistry, 09.04.2020 01:19 cupcake3103670

A helium-filled balloon has a volume of 2.75 L at 20.°C. The volume of the
balloon decreases to 2.46 L after it is placed outside on a cold day. What is the
outside temperature in K? in °C?

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Answer from: juliaduenkelsbu

The outside temperature is 262.1K or -10.9°C

Explanation:

Given that,

Original volume V1 = 2.75 L

Original temperature T1 = 20.0°C

Since unit of temperature is Kelvin, convert 20°C to Kelvin by adding 273 to 20°C

(20.0°C + 273K = 293K)

Outside volume V2 = 2.46 L

Outside temperature T2 = ?

Since volume and temperature are given while pressure is constant, apply the formula for Charle's law

V1/T1 = V2/T2

2.75L/293K = 2.46L/T2

To get the value of T2, cross multiply

2.75L x T2 = 2.46L x 293K

2.75LT2 = 720.78LK

Divide both sides by 2.75L

2.75LT2/ 2.75L = 720.78LK/2.75L

T2 = 262.1K

Hence, the outside temperature in Kelvin is 262.1K. Then, obtain outside temperature in Celsius by subtracting 273 from 262.1K

i.e 262.1K - 273 = -10.9°C

Thus, the outside temperature is 262.1K, or -10.9°C

Answer from: Quest

2b2t

explanation:

2b2t

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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A helium-filled balloon has a volume of 2.75 L at 20.°C. The volume of the
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