Chemistry, 05.09.2019 16:20 gamerhunter425

The four lines observed in the visible emission spectrum of hydrogen tell us that: select one: a. the hydrogen molecules they came from have the formula h4. b. we could observe more lines if we had a stronger prism. c. there are four electrons in an excited hydrogen atom. d. only certain energies are allowed for the electron in a hydrogen atom. e. the spectrum is continuous.

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Answer from: brackendillman

The answer is D, only certain energies are allowed for the electron in a hydrogen atom.

Explanation:

We know that light quanta, also called photons have energy:

E= h f Where h is Planck's constant and f is the frequency of the photon. The frequency of a photon tells us the colour of the light, different colours have different frequencies, and white light contains photons of all frequencies in the visible range. But the light emitted by Hydrogen in a low pressure tube that is excited by an electric current contains only photons from a few frequencies.

Why is that? Quantum mechanics explains us that it is because an electron that is bound to a hydrogen atom can take many configurations, of different energies, the configuration for which the electron is as close as possible to the nucleus (in this case a single proton) is called the ground state, and the other higher energy states are called excited states. if an electron is in an excited state, it will tend to "jump" down into the ground state, emitting light in the process. That light is the light we see, and its spectrum is discrete because only a few energies are allowed for the electron in a hydrogen atom.

That is why D is true.

Why are the other questions wrong?

A: Wrong because hydrogen can only form H_2 molecules, and even so, the great energy levels in the tube destroy the molecule leaving only single free H atoms.

B: The prism's only job is to separate light into different frequencies so that we can see the different colours, it won't block any lines.

C: Wrong because hydrogen atoms only have a single electron.

E: wrong because if the spectrum was continuous we wouldn't see single lines, but a continuous range of colours, like in white light.

Answer from: Quest

Ithink it is quantity of plants

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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