Chemistry, 16.10.2020 03:01 kabiletproducciones

Atomic theory question Rutherford’s model

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Answer from: Quest

you didnt put any of the following

Answer from: Quest

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

Answer from: Quest

answer: semi-infinite solid initially at a uniform temperature of 400 ◦c is cooled at its surface by convection. the cooling fluid at a constant temperature of 200 ◦c has a convective coefficient (ha) of 250 w/m2.k. the physical properties of the solid: k = 20 w/m.k and α = 4 * 10-5 m2/s. using δx = 0.04 m and δt = 10 s. write the finite difference equations describing the transient heat transfer of the first five δx and the temperature profile after 30 s.

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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Atomic theory question Rutherford’s model
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