Chemistry, 21.08.2019 01:30 tay3270

Aspherical vessel has internal radius r, and a wall of thickness b and thermal conductivity k pluid at temperature t, is stored in its interior, while its external side is in contact with cold air of temperature t. find the temperature distribution within the wall, as well as the energy that mus be offered to the internal fluid to keep its temperature constant.

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Answer from: Quest

answer: the figure is attached herewith.

step-by-step explanation: we are given to sketch a unit circle and label the terminal points corresponding to

$\theta=\dfrac{\pi}{3},~2\dfrac{\pi}{3},~3\dfrac{\pi}{3},~4\dfrac{\pi}{3},~5\dfrac{\pi}{3}.\\\\\\\rightarrow \theta=60^\circ,12^\circ,180^\circ,240^\circ,300^\circ.$

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.

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.

Answer from: Quest

The answer is tea due to the main ingredient being a herb that grows from the ground and sugar that grows from a stalk, i hope this : ) p.s. can i be brainliest?

Answer from: Quest

the volume of 0.220 m k2s solution required to completely react with 175 ml of 0.115m co(no3)2 calculate the moles of co(no3)2 = molarity x volume/1000175 x0.115/1000= 0.02 moleswrite the reacting equationk2s(aq) + co(no3)2 = 2 kno3(aq) + cos(s)by use of mole ratio between k2s to co(no3)2 which is 1: 1 the moles of k2s is therefore = 175 x0.115/1000= 0.02 molesvolume = moles/molarity=0.02/0.22=0.09 ml of k2s

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