13

It gets very complicated, there are many papers and books on the topic and this little table, from Jensen et al. 1, gives a glimpse of the anisotropic effects: An extremely short answer is this: when linearly polarized light is incident on an optically active medium, it can be thought of as consisting of equal components of right circularly polarized (RCP) ...


11

The figure below shows the situation between configuration for a $p^2$ configuration, terms, levels and states. The word 'state' tends to be used colloquially to mean any of Term, Level or State. The Configuration such as $(1s)^2$ or ...$(2p)^2$ etc. tell us which orbitals are occupied. These are split with electrostatic (Coulomb) coupling to form Terms. ...


10

It depends on a lot of things, software-wise some of these techniques are simply a software option you can implement pulse sequences yourself, if that´s possible with your control software you need to have the right software to transform and evaluate the data ... or write it yourself (not exactly rocket science if you know your way with Matlab/Octave etc.) ...


7

There are multiple aspects to consider here. In terms of the magnet, the ones suitable to record a $300$ or $500\, \pu{MHz}$ $\ce{^1H}$ NMR spectrum may be well used to record 1D or 2D homo- and heteronuclear spectra as well. Other nuclei than $\ce{^1H}$ or $\ce{^{13}C}$ routinely are recorded in 1D with these standard spectrometers as well; for example $\...


7

As a quick fix: To access the IR spectral data of $\ce{HCl}$, I opted for a seach by (Hill) formula on NIST's web page, which yielded three entries here. For $\ce{HCl}$ in particular (here) I opted for the first one here, indeed offering the spectrum as image file or as JCAMP-DX. As in the comment by @EdV, this JCAMP-DX an ASCII file indeed. Note JDXview ...


7

"A signal is produced when a radio frequency causes one state (could be either α or β) to invert" This isn't correct. It really isn't, even though it's often incorrectly described that way. The reality is more complicated, and it involves understanding QM as well as the fact that single spins do not necessarily exist in $\alpha$ or $\beta$ states (any ...


6

Since I was reviewing this in some detail recently, I will try to supplement Prof. Hutchison's already good answer with some more detail as to why one would choose to work with normal coordinates, rather than the already quite simple Cartesian coordinates. The short answer is that, under the assumption of small displacements, normal coordinates decouple the ...


6

There are really two parts to your question - one is to interpret the wikipedia entry, the other is the more general question about the transition of methane from the parent molecule to the ionized $\ce{CH4+}$. For the first question, I believe the author has muddled the explanation a bit (though it's of course just as likely that I'm the one who is ...


6

Ed V's answer above--which I am accepting--is correct and also allowed me to reach the conclusion on my own. I am just posting this because my confusion was not too hard to resolve once I saw the quantities involved and someone else may have the same confusion. The short of it is: student polarimeters measure circular birefringence (CB), usually at a ...


6

Disclaimer: I'm not affiliated with any company However, Ocean Optics cuvette spectrophotometers are a good choice This model is used in some undergraduate labs and is good because the cuvette holder is built in (i.e. it doesn't require a fiber optic cable). As long as the optics don’t get jumbled, the prism detector mechanism work well and reliably ...


6

Good spectrometers in "new, unopened" condition are out of your budget. Their prices start at about 2000 USD. This also includes the spectrometers specifically made for education—like the ones from PASCO (example; they are usually based on Ocean Optics components), and from Ocean Optics themselves (example). One option for your budget is to buy a used ...


6

In a very very broad reasoning, XRF is much more sensitive to the "mid-range" elements. Elements below Na are generally not detectable. The x-ray tubes must have a beryllium window to seal the tubs which absorbs the low energy x-rays. The lower the atomic number the less the fluorescent yield. Low energy x-rays don't penetrate from the sample ...


5

You have assessed the problem correctly. Physicists and chemists use just the opposite directions. Keep in mind that both are arbitrary choices, however the physicist's convention is consistent with electrostatics. All you have to do is to be aware of this direction in the chemistry literature. I have rarely seen any student ever thinking about it which is ...


5

Long time ago, I wrote mass spectroscopy, by mistake, in an undergraduate exam and the instructor told me "You were the only one in my class who wrote mass spectroscopy." The explanation was that MS is mass spectrometry because it does not deal with light or electromagnetic radiation and that's what the textbook said. At that time internet was not that ...


5

If we consider N atoms in a non-linear molecule, then each atom can move independently on the X, Y, and Z axes. That's $3D \times N = 3N$ degrees of freedom. But if I move the whole molecule 3 Å along the x-axis, that doesn't constitute a vibration - it's just a translation. So we have to remove 3 degrees of freedom for translations. Similarly, if I ...


5

The product $xyz$ is an odd function and as such should produce the same selection rules as 1 photon (electric dipole) transitions. Thus CO has a 1 photon transition at 146.7 nm in the vacuum uv that can be more easily reached with a 3-photon transition with a pulsed laser at 440 nm. ( Three photon transitions are also used in bio-imaging where a ir photon ...


5

Yes, of course the Raman scattering can suffer from additional scattering on its way through the sample, just like it will generally be attenuated on its way to the detector by any element in its path (mirrors, lenses, glass surfaces, etc.). If scattering is a problem or not very much depends on your sample. The ideal sample will not scatter your ...


5

This is an interesting question. However, disadvantage is having only $\ce{^1H}$-$\mathrm{NMR}$ to deal with. Yet, it has pretty good resolution (probably using $\pu{400 MHz}$ machine) so we can resolve the splitting pattern very easily and predict the structure. First, OP has correctly assigned two degree of unsaturation. For molecular formula, $\ce{...


5

Your analysis is generally correct. In the case where all the protons are equivalent, you don't need to worry about proton–proton coupling. The main thing you need to worry about is that when you have a term such as $2I_{1z}S_x$ you will get evolution of different proton–carbon couplings, represented by (e.g.) the Hamiltonian $2\pi J_{IS} I_{2z}S_z$. This ...


4

Finding a structure with the correct formula that matches the chemical shifts reasonably well is not so difficult. The multiplet integrals, normalized such that the area corresponding to one $\ce{H}$ is $\approx 33$ suggests the following number of $\ce{H}$ (moving upfield): 1,1,2,2,3. The upfield singlet suggests an uncoupled terminal methyl group, so we ...


4

The problem with computational chemistry software (ORCA, GAMESS etc.), as described in NotEvans' answer and in various comments, is that they only spit out a list of chemical shifts and coupling constants. For example, this is part of a typical ORCA output: -------------- Nucleus 34H : -------------- [...] Total shielding tensor (ppm): 28....


4

If you use R for data analysis, take a look at the R package hyperSpec. Example: install.packages('hyperSpec') library('hyperSpec') spectrum <- read.spc('<your spc file>') ... plot(spectrum) You can also use Sys.glob() function for wildcard search of your files, then read all of them with read.spc(), and call collapse() to build a single hyperSpec ...


4

Firstly, regarding the extra question: Atkins' Molecular Quantum Mechanics (5th ed.) uses "term" for $^2\!S$, and "level" for $^2\!S_{1/2}$. Back to the main question. It's been a long time since I did term symbols, so I am happy to be corrected, but if I am not wrong, your $^4\!P$, $^4\!D$, and $^2\!S$ terms are not allowed because of the Pauli exclusion ...


4

You must first calculate dilution factors. Assuming the volumes are additive, we can calculate dilution using $M_1V_1=M_2V_2$ equation. Also, assuming $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated, we can say: $$\ce{[Fe(NO3)3] = [Fe^3+]} \text{ and } \ce{[KSCN] = [SCN-]}$$ Thus, initial $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ are $0.20$ and $\pu{...


4

"How can it remain secret?". Trivial answer is - it can't (it isn't). WD40 is a Fast Moving Consumer Good (FMCG). It is also a reasonably cheap product - as an industrial chemist this suggests at least two things to me: It doesn't contain anything rare or expensive It isn't difficult to manufacture As an ex-petroleum chemical analyst, with access to a ...


4

Yes, you are right. If you have $n$ components in a mixture each with a spectrum $S_1$, $S_2$,...,$S_n$, the observed spectrum $S_{obsvd}$ would be the sum of $S_1$+ $S_2$+...+$S_n$. In FTIR you either measure the transmittance or the absorbance. The term intensity is more or less used when you deal with molecular or atomic emission of light (fluorescence ...


4

I just did the experiment suggested by @porphyin. I used a 405 nm laser diode built into a toy "Space Gun" and the orange long pass filter shown in the figure below: The results are in the next dark figure: The spots on the paper are not fluorescence of the paper (see the violet fluorescence of the paper around the dark shadow of the filter) and are not ...


4

The NMR signal is generated by magnetization oscillating in the transverse reference plane of the experiment (orthogonal to the principal field). The oscillation induces an alternating current in a pickup coil. From a classical perspective, an oscillating transverse magnetization is generated from an initial longitudinal magnetization to which a torque is ...


3

This answer is based on the educated guess that the green excitation Raman spectrum was acquired with a dispersive Raman instrument*. But why is there general nonzero background in intensity? What caused it? For any real-life instrumentation, you should never assume to measure 0 signal for 0 analyte concentration (see also Does a calibration curve of ...


3

More a set of queries and comments than an answer. I don't understand why you make A and B vary with time irrespective of your reaction scheme. You should really analyse using Fick's diffusion equations with the reaction scheme added and do so for each species. This will have to be a numerical calculation in space and time. (There are well established ...


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