Chemistry, 12.10.2020 14:01 ryanbransky

PLEASE HELP ASAP ONLY HAVE 10 MINUTES. A certain photon of light has a wavelength of 422 x 10 -9 m. What is the frequency of the light? Show work

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Answer from: Kode9595

7.11 x $10^{14}$ Hz

Explanation:

The equation for converting wavelength to frequency is ν=c/λ, or the speed of light over wavelength. The speed of light is 3.00 x $10^{8}$ , and our wavelength here is 422 x $10^{-9}$ . All we have to do now is substitute our given values:

$\frac{3.00x10^{8} }{422x10^{-9}}$

After reduction, our answer comes out to be about 7.11 x $10^{14}$ Hz.

Answer from: Quest

the answer is b because, investigating whether another species of moth also drinks tears might be interesting, but it would not directly explain why this species of moth does.

to explain why this species of moth drinks tears, the scientist could ask: "what nutrients that moths need are in elephant tears? "

the scientist could experiment with moth repellent or oter ways to keep the moths away, but that would not answer his original question.

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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