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Is it "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function"? If so, why?
The recent paper in Nature Independent confirmation of a methane spike on Mars and a source region east of Gale Crater is interesting and the basis is a reanalysis of 2013 data taken by a X spectrometer about the Mars Express spacecraft in orbit around Mars looking down at the Martian atmosphere.
The paper itself is paywalled but the supplemental data describes the reanalysis in great detail.

The Planetary Fourier Spectrometer [38] (PFS) on Mars Express 37 is a double-pendulum infrared Fourier spectrometer optimized for atmospheric studies. It has two distinct spectral channels operating simultaneously and covering the wavenumber range between 200–2000 cm−1 (Long Wavelength Channel, hereafter LWC) and 2000–8300 cm−1 (Short Wavelength Channel, hereafter SWC). Both channels have a sampling step of 1 cm−1 and a spectral resolution of ~1.3 cm−1, when no apodization function is applied, and ~1.8 cm−1 when a Hamming function is applied to the measured interferograms (as in the case of the present work).

Key to the process is an understanding of how the raw data is preprocessed using a Hamming apodization function before spectral fitting.

Step 1: Hamming apodization function is applied to PFS interferograms. It is common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function in order to reduce the amount of ringing present in the resulting instrumental line shape [77]. This reduces the spectral resolution a bit (from 1.3 cm-1 to 1.8 cm-1 in the case of PFS [78]), but also reduces the instrumental noise and the magnitude of the side-lobes in the instrumental line shape, which are a direct result of the finite maximum optical difference in the measured interferograms [79]. The Hamming function H is defined as H(δ/L) = 0.54+0.46cos(πδ/L), where δ is the optical path difference out to a maximum value of L.

Davis, S. P., Abrams, M. C. & Brault, J. W., Fourier Transform Spectrometry, Academic Press (2001).
Giuranna, M., et al., Calibration of the Planetary Fourier Spectrometer short wavelength channel. Planet. Space Sci. 53 (10), 975–991 (2005).
Naylor, D.A. & Tahic, M. K., Apodizing functions for Fourier transform spectroscopy. J. Opt. Soc. Am. A 24, 3644–3648 (2007).

I was surprised to read that raw data is filtered before fitting to spectroscopic models in order to extract concentrations. I'm not an FTIRer, but I would have instead expected that all instrumental errors would be included in the fitted theoretical spectra generation and that the raw data would be fit in its pristine, unaltered form. After all, the only thing you really know for sure when fitting is that the data is the data, it's what you actually measured. Everything else is speculation.
QUESTION: Assuming that the block quote is right and it is indeed "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function in order to reduce the amount of ringing present in the resulting instrumental line shape" why is this considered "safe" to do? Why aren't all instrumental effects incorporated into the fitting function allowing for the raw data to be fit directly instead?

The spectra in question, from Independent confirmation of a methane spike on Mars and a source region east of Gale Crater:

As someone who did his studies in signal processing, this is the first time I've heard of apodizing function. In electrical engineering, we tend to call it windowing function.

@AndrejaKo you can probably guess that it is commonly used in Fourier optics. If you put an aperture at the pupil of an imaging system it has the same effect. In projection lithography for microchips, apodization can get pretty complicated with both amplitude and phase, and sometimes even polarization modulating masks at the Fourier plane.

@uhoh do you know if this is complex data? It seems to me given the noise and potential phase issues that the little blip at 3017 $cm^{-1}$ could be an artifact.

@NightWriter You'll have to read the main paper and especially the procedure outlined in great detail supplemental data link I've provided first, then decide for yourself. I believe they provide links to the raw data, so you can have a look.

You usually post interesting questions, which appear to be deceptively simple yet they are very challenging. Modern day data acquisition and signal processing is so complicated that it is almost like a black-box. It is amusing when I ask electrical engineers some signal processing questions, they don't know the answers and when I ask mathematicians it is too much applied mathematics for them. I am not an FTIR expert either but as an analytical chemist, I can add some remarks. I think apodization is routinely done in order to decrease noise and oscillations when you do an inverse transform. If you do some mathematical operations in the frequency domain, when you do the inverse transform, the level of noise is crazily high. Of course, when you are looking at very small signals do not wish to have noise.
In general, sometimes I need to fit functions to the data to peak shaped functions. I always get a better fit once I digitally filter the data and fit later (moving average, Savitsky Golay, Hamming in the time domain and so on.) Noise is the biggest enemy of any analytical chemist or a spectroscopist.
With any smoothing process, either in the frequency domain or in the time domain, you tend to loose resolution. There is always a sweet spot for digital filtering or if you have heard the story of Goldilocks who entered the house of three bears..."When she had finished the porridge Goldilocks felt tired and went to find somewhere to sit down. The first chair she found was too big, the second chair was still too big, but the third chair felt just right." The same rule of thumb goes for digital filtering. Too little is useless, too much you loose all resolution and just the right filtering gives you the best signal to noise ratio.

As an ex organic chemist, I can confirm that using the Hamming function to apodise the data before fft is common, and I can also confirm we all just nod our heads and smile when the NMR technician explains why this is the case.

If there is a noisy signal that decays away, such as the FID in an NMR experiment, the signal to noise ratio is larger at shorter times than at longer ones where the noise remains but the signal is weaker. Multiplying the FID by a decaying exponential, i.e. an apodising function, thus suppresses the signal where the noise is larger and so increases the signal to noise in the spectrum after the signal is fourier transformed. The downside of this is that some resolution is sacrificed. Both the S/N and resolution depend on the decay time of the apodising function.  (If higher resolution is required an inverse exponential can be used but at the cost of degrading the signal to noise).
In general the shape of an opodising function chosen will depend on the nature of the signal and various shapes are used.
(If a signal is to be fourier transformed it is implicit in this that the signal is one replica of a repetitive series of signals. In practice one does not measure this, just a single decay so it is important that the signal is zero at the end of the data, the fourier transform otherwise folds some of this signal into the result leading to artefacts. The difference between the start and end of the signal appears as a step change and so has frequency components. An apodising function removes this problem.)
The figures below show the raw data and then fourier transformed and below when the same data is rather heavily apodised. Note that the signal to noise is increased but also the spectral resolution is degraded slightly.

"In general the shape of an apodising function chosen will depend on the nature of the signal and various shapes are used." So if for example you "know" that you are looking for a narrow gaussian peak, broadened only by the resolution of your interferometer, then you can choose an apodising function that would be compatible with this scenario. But if you had some completely arbitrary spectrum produced by a phenomenon you didn't understand, you'd be taking some risk choosing any given apodising function?

yes, you would take a risk, but presumably you know something about the data you are analysing and can then try different functions:  'horses for courses' :)

@porphyrin, nice illustration. What software did you use?

I used python/numpy/matplotlib included with Jupyter notebooks via Anaconda. All are free.

Hamming apodization function is also known as a Hamming window.
If you have data that looks like this:
            _________
           |         |
           |         |
           |         |
           |         |
  _________|         |___________

because your sensor only picks up data over a certain window, when you feed that to a FFT you get a pile of artifacts caused by the window.
Then when you reconstruct it (after doing seemingly minor changes), instead of a nice rectangle, you get overshoot/undershoot at the sharp edges.  For example, if you supersample in reconstruction, the intermediate data near those edges will be garbage overshoots.  Those overshoots are artifacts of the method of analysis.
One of the rules of a FFT is that it uniquely reconstructs the original signal if the original signal had no frequency components that where above a certain threshold.
But those really sharp drop offs?  They are, in a sense, frequency components of infinite frequency.  So FFT's assumptions are violated.
A Hamming Window smooths the edges down to zero in a way that doesn't generate too much additional garbage in the frequency domain.
This is "safe" because you can mathematically bound how much "garbage" your windowing function can add to your signal (both in magnitude and in what part of the frequency domain).  Nobody is particularly interested in the ringing artifacts generated by the specific finite window your instrument used to detect the signal.
After applying the window, the impulse will look much like the original signal, but instead of sharp cliffs it will have a smooth slope.  And the signal in the middle will be a bit blurred.
People use specific windowing functions because their effect on the frequency domain is well understood and bounded.  Ad-hoc attempts to fix these artifacts are going to add "worse" and less predictable effects on the resulting frequency domain data.
The main thing cleaned up is, after all,

[...] a direct result of the finite maximum optical difference in the measured interferograms

ie, the signal window.

QUESTION: Assuming that the block quote is right and it is indeed "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function in order to reduce the amount of ringing present in the resulting instrumental line shape" why is this considered "safe" to do? Why aren't all instrumental effects incorporated into the fitting function allowing for the raw data to be fit directly instead?

Apodization amounts to multiplying the data (a function of time or frequency) by an envelope function prior to Fourier transformation. The purpose can be multifold, but the main ones are resolution enhancement, sensitivity (signal-to-noise) enhancement, and suppression of artifacts due to instrumental limitations, particularly signal truncation, which presumably gives rise to the ringing referred to in the article. In fact the term apodization refers to this last task, as the operation can remove the "feet" at the edges of the data window. The various goals are not always mutually compatible, as S/N improvements come at the cost of resolution (peak broadening). In the spectra shown, the resolution looks pretty low but the concern appears to be the s/n, which is also not impressive. The point of the apodization here is mainly to reduce noise and suppress lobes (ringing) due to a limited acquisition bandwidth, at the expense of resolution.
But, does it not matter, as far as quantitation is concerned, whether apodization is applied? Doesn't the apodization function distort the results? Why is it ok to perform such denoising/apodization?
Apodization can significantly improve s/n and therefore improve the precision of parameters derived from the data. The data is assumed to arise from a signal function plus a noise function, and these are typically assumed to be statistically independent, and noise is typically also assumed independent (uncorrelated, but of constant variance) between signals. These are important assumptions but usually safe to make. If the same apodization function is applied to spectra being compared in an analysis (such as a time series), as the effect of the apodization function is linear$^\dagger$ it does not introduce quantiative artifacts. Other denoising algorithms (usually iterative) are nonlinear and can cause problems with quantitation.
Why are methods to account for the noise and truncation (lobes) factors not subsumed into a complex fitting function? Because it's not necessary. Other than performing the apodization operation on the simulation result in one domain prior to Fourier transformation, the simplest method of suppressing truncation/noise during fitting would amount to performing a convolution operation with the Fourier transform of the apodization function. In the same way that FFT provides advantages in terms of acquisition speed and s/n, multiplication in the one domain rather than convolution by a complex function in the alternate domain saves time and headaches, so apodizing in one domain prior to the FFT is preferred.

I was surprised to read that raw data is filtered before fitting to spectroscopic models in order to extract concentrations. I'm not an FTIRer, but I would have instead expected that all instrumental errors would be included in the fitted theoretical spectra generation and that the raw data would be fit in its pristine, unaltered form. After all, the only thing you really know for sure when fitting is that the data is the data, it's what you actually measured. Everything else is speculation.

Either the same apodization function was presumably applied to the raw simulated data as to the raw experimental data, or the peak width was treated as an adjustable parameter. Although I have not read the article, I presume the presence of the small peak at a specific frequency (~$\pu{3017 cm^-1}$) was more important to draw conclusions about the presence of a specific chemical signature, than its exact intensity and width. On the other hand, if the effect of the apodization can be accounted for in the data simulation, then quantitation might even be possible.

$\dagger$ 1. The effect of the apodization function on the noise and the signal can be treated separately; and 2. scaling the raw signal by a constant and performing the apodization returns the original apodized function scaled by that constant.

If you look at the figures above, notice that the FID ends are "square." When this is Fourier transformed, this fast drop off shows up as high frequency components since sharp changes are equivalent to high frequencies. All the apodization functions that are used drop off to zero at the edges and eliminates this artifact. The various apodization shapes used have been found to be the best for various uses by minimizing the distortion given the necessity to approach zero at the edges.


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Are electronic descalers a scam?
It seems there are some electronic kits sold around nowadays by major retailers (e.g., Home Depot and Amazon) claimed to remove built-up scale from water pipes.
But, do they actually work? If so, then how?
Perhaps, inducing high-frequency eddy currents inside copper piping somehow lessens its ability to support limescale? Or, perhaps it gets broken down by some specific radiation (emitted by the kit) in the limescale's absorption spectrum?

I'm sure someone makes money, so in that regard I'm sure they work.

Yes , it is a scam . This has been discussed at National Association of Corrosion Engineers meetings for decades. The vendors have actually sold some to refineries; however they can never demonstrate good results in objective tests. The excuses the vendors find for the failures are a tribute to creative thinking.

A.K. blacksmith37 +1 +1 I was afraid this was a scam, but thought it would be cool if there was indeed some kind of underlying concept or mechanism here beyond my (rather basic!) understanding of physics/chemistry^ What do you think of the theory presented below in Agriculturist's answer?

The "comments" remind me of the NACE meeting discussions.

There is the fact that permanent magnets create conservative fields which cannot do net work (perpetual energy). So if it is doing anything to the water or it’s contents then it’s undone as the water leaves the field or energy comes out of the waters motion.

Magnetic fields are not conservative. They do not have the related potential field and the closed path work on nonzero charge is nonzero.

These devices have been reported to work in some situations, but not in others.
The underlying theory involves altering the growth of calcium carbonate scale with an ordinary magnetic field. The magnetic field changes which form of calcium carbonate crystal (calcite, aragonite, and vaterite) is favored. Certain forms of crystal make tough scale while others can pass through a water system without scaling.
The growth and precipitation of carbonate crystals are affected by many other things such as calcium concentration, carbon dioxide concentration, temperature, pH, and the presence of other metallic ions. Exposure time to the magnetic field, magnetic field strength, and magnet orientation also have an effect.
The different forms of carbonate crystals can be tested to have specific properties under controlled laboratory conditions.  Under uncontrolled conditions, the scientific literature is full of mixed results and mixed explanations.  Because so many things affect the formation of scale, it is impossible to say if a magnet or device generating a magnetic field will work to thwart scaling in any specific situation.  It may work wonderfully in some situations but not at all in others.
The best way to tell if such a device is worthwhile would be to see if several people using your same water supply report it as having worked lately.  If water management practices change (e. g. factors affecting the formation of scale are controlled) then such a device may cease to work.  If no one using your same supply can attest to its operation then this would be a risky purchase.

Interesting, so what kind of magnetic field does the underlying theory then recommend? One that oscillates back and forth, or a constant one? If the latter, does it matter how it's oriented relative to the water flow? (i.e., parallel, antiparallel, or orthogonal)

https://www.sciencedirect.com/science/article/abs/pii/S0304885301004322  - this paper tested the commercial units against a magnet and found the magnet more effective.  I do not think there was anything special about the magnetic field.

https://www.researchgate.net/profile/Vladimir_Zlotopolski/publication/321649716_Magnetic_Treatment_Reduces_Water_Usage_in_Irrigation_Without_Negatively_Impacting_Yield_Photosynthesis_and_Nutrient_Uptake_in_Lettuce/links/5a73b5baaca2720bc0dbd38e/Magnetic-Treatment-Reduces-Water-Usage-in-Irrigation-Without-Negatively-Impacting-Yield-Photosynthesis-and-Nutrient-Uptake-in-Lettuce.pdf  - This paper talks about its use in irrigation.  It mentions contradicting claims surround what type of magnetic field is best.

I don't have access to Elsevier ($$$) for the first publication, but it looks like the researcher who published the second was actually working on behalf on the company behind the magnetic water treatment technology promoted in his paper at the time (so, there may have been a conflict of interest).

Also, that technology uses permanent magnets, while the kits in my original question require electrical power. In either case, it seems only an orthogonally oriented magnetic field (relative to the water flow) could possibly affect the ions' trajectories (via Lorentz force), but the kits in my original question can only produce a parallel (or antiparallel) one given the placement of their solenoids.

I suppose a strong orthogonal magnetic field could theoretically disrupt crystallization by separating the necessary positive and negative ions (again, via Lorentz force on the flowing ions as well), but the effectiveness of this could largely depend on water temperature (which could affect the entropy of a given distribution/placement of ions in a given cross-section of the pipe, as well as the general solubility of the scale altogether) and the Coloumbic interactions opposing this separation of charge in the first place.

The engineer would not want to be liable for this tech; however, the Amazon reviews seem consistent with a hit or miss product.

The linked paper describes using a permanent magnet to cause preferential crystallization of calcium carbonate as aragonite or vaterite instead of the more-scale-prone calcite. Who knows whether it works as a treatment if the field is limited to merely the building inlet.
Electronic descalers theoretically use an electric field to entice the calcium and carbonate ions to interact with other impurities and precipitate instead of each other. Anecdotally - my roommate got one and our house sprang leaks everywhere as years of limescale shed off the pipe internals. Not ideal actually.

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Standard abbreviations for polycyclic aromatic hydrocarbons, similar to those for nucleic and amino acids?
The Nature paper Characterizing particulate polycyclic aromatic hydrocarbon emissions from diesel vehicles using a portable emissions measurement system addresses polycyclic aromatic hydrocarbons (p-PAHs) and both the text and the plot below make extended use of short, mixed-case abbreviations for various specific p-PAHs.
Question: Do these refer to some standardized naming convention for p-PAHs? If so, where can this be found?

Figure 2. Average distance-based emission factors for each p-PAH component according to the engine type.

(electronically-controlled fuel injection (EI) and mechanical pump fuel injection (MI))

For amino acids and nucleic acids for example, standard one- and three-letter abbreviations exist:

Table of standard amino acid abbreviations and properties
Table of Amino Acids and Their Abbreviations
Nucleic acid notation

Here is an example of a plot where the abbreviations are not used. From Figure 20 of polycyclic aromatic hydrocarbons in automotive exhaust emissions and fuels:

Likely for PAH there are indeed common abbreviations, but I don't know how much standardised.  I comment just to say that in a paper you can always use acronyms and abbreviations, even for reporting molecules for the first time.  Just one need to define each. This is useful to avoid clumsy graphs as well fighting characters limits. In chemistry, very often the abbreviations take even the forms of numerals and alphabetical signs. Some abb are very common, like Ph for phenyl- Ac for acetyl, ecc....

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Confusions about inductive effect and acid strength
If one of the hydrogens in the alkyl side chain of ethanoic acid is replaced with an atom of Chlorine, the acid has a larger Ka value. My book says this is since the Cl "borrows" electron density from the Carbon it is connected to. And, in this example, this carbon borrows some electron density from the Carbon in the acid group, and thus the oxygens of the acid have less charge density. The bold part is what I don't get.
The C to which this 1 chlorine is attached gets a bit electron deficient but not as much deficient as the C in the acid group (since this has two oxygen atoms attached). So how can it borrow density away from the C in the acid group? In my mind, to borrow some charge density you have to be more electronegative than what you're attached to, and the more deficient carbon should have a greater pull.
What am I missing in this?

OK, think of it this way: to say that it borrows density away from the acid group is an abuse of language, actually supposed to mean that it gives density to the acid group, but less willingly than -CH3 did. Imagine a guy who says he was robbed because his welfare check is a tad less than it was the other month. Same thing here.

And, though the principle is the same, it is easy to see this effect as stabilization of a real charge.  So it is easy to discuss the conjugated bases situation, and in general that with a real charge.

@Ivan Neretin, I thought about what you said. One question though : the data shows that the acid with the the chlorine is also stronger than methanoic acid. So, all in all, this CH2Cl is a worst donor than both the CH3 and the H?

So it is, then.

@IvanNeretin, one follow up: my teacher said the alkyls groups always donate electron density to whatever they are attached to (regardless of electronegativity between the C of this group and what it is attached to). This seemed odd to me and got me wondering whether my claims in the last paragraph of the question is true. Could you check that and about this always donating claim?

Well, this is pretty much true across a pretty broad realm. I guess you'd have to go for some exotic metal organics to make this untrue.

My arguments or the teacher's?

@Ivan Neretin. Could you please clarify

I mean that alkyl groups are pretty much always donating.

@Ivan, and the cause is the electronegatvity difference between the C and whatever they attach to?

Yes.$\mathstrut$

The result is correct but the explanation is bad.
The main reason is that when you compare CH3COOH and CH3ClCOOH , the main difference is that the alpha carbon in the former has more electron density than the latter one because of which when the conjugate base gets formed, it is more stabilised than ethanoic acid because of low electron density.
You are correct about this "In my mind, to borrow some charge density you have to be more electronegative than what you're attached to, and the more deficient carbon should have a greater pull".
Edit 1 - The orbitals of CH3COOH can provide more electron density than CH2ClCOOH due to the former having more electron rich orbitals which interferes constructively with the orbitals of other C to satisfy its demands. Think in terms of how Hyper conjugation or resonance works in terms of interference of wave functions.

@Sal_99 The reason is overlapping of electron containing orbitals which decreases the deficiency of electrons. Think this like Hyper conjugation or Resonance. Interference of various wave functions take place constructively or destructively.

I will improve the answer over time. First try to think yourself.


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Does increasing the concentration of the anode electrolyte in a galvanic cell increase or decrease the voltage produced
Im in kind of a pickle here. I am using nernst equations to calculate the theoretical voltage produced for different concentrations of $\ce{Al(NO3)3}$.
Reaction: $$\ce{2Al(s) + 3Pb^{2+}(aq)⟶2Al^{3+}(aq) + 3Pb(s)}$$
Electrolytes are $\ce{Al(NO3)3}$ and $\ce{Pb(NO3)2}$.
I am using the Nernst equation $$E_{cell}=E^0-\frac{RT}{nF}\mathrm{ln}(Q)$$  and have the following calculation: $$E_{cell}=1.53-\frac{(8.314×293.7)}{(6×96500)}×\mathrm{ln}\left(\frac{(.01)^2}{(0.1)^3} \right)$$
where $Q=[Al^{3+}]^2/[Pb^{2+}]^3$   is the ratio (products)/(reactants).
I am finding that using the Nernst equation, an increase in the concentration of $\ce{Al(NO3)3}$ leads to a decrease in voltage whereas according to my practical data im finding an increase in voltage.
Also, according to this an increase in concentration should lead to an increase in voltage.
Can anybody tell me what im doing wrong here.
Any help would be appreciated!

Well, the linked answer is completely wrong. Since this is an equilibrium process, an increase in the concentration of the product will make the reaction less spontaneous. In a galvanic cell, this translates to a smaller potential difference.

I've never heard of eNotes before, but judging by this example, what you did wrong was to trust the content on that site.


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Tricky exam question

The mark scheme says carbon dioxide is produced from carbonate + acid. I know hydrogen can't be produced because both the reactions are endothermic.
But where did the acid come from? Can someone give me an example of Solid B and D which produces a mixture and when added together it gives carbon dioxide

Be careful about trying to analyze a problem like this to deeply. With all the weird chemicals I'll bet there are two salts that have an endothermic dissolution, but create hydrogen gas when mixed.

The obvious choice for the source of carbonate is sodium bicarbonate (UPDATE: sodium bicarbonate is not soluble enough per other answer, but sodium carbonate is a possibility), which has a positive enthalpy of solution even at high concentrations, so let's call that B.
The acid is a little trickier. If you look up the standard enthalpy of solution of most acids, you'll find negative values (meaning the temp would go up when they are added to water). However, these values refer to infinite dilution in neutral water. Once you've added a little bit and the pH has dropped, the reaction $\ce{H+ + HO- -> H2O}$, which has a large negative enthalpy, essentially stops happening and the enthalpy of solution becomes positive. This is easy to tell by looking at the maximum solubility as a function of temperature. If it increases, then the enthalpy change of solution at saturation is positive. So at the high concentrations in the problem (4 g per 30 mL), many acids that are solid at room temp will cause a temperature drop when dissolved. Citric acid is one example.
A solution of citric acid is certainly acidic enough that it will cause rapidly production of carbon dioxide gas when mixed with a solution of sodium carbonate.

Several organic acids are endothermic when dissolved in water: citric and tartaric are two that I have experience with. They would be good candidates for experiment 4.
The difficult one is the carbonate. I think ammonium carbonate is endothermic (the chloride and nitrate are). This could be the solid in experiment 2. It would have an odor of ammonia, which was not mentioned.
We're assuming that all the compounds are colorless. Other gases could be O2 or NO/NO2, but surely other hints would have pointed this way, so CO2 is the most likely.
Sodium bicarbonate is not as soluble as 4 g in 30 mL. I'm not sure about the endothermicity of sodium carbonate (which has several hydrates). No other possibilities come to mind.

Good point on the sodium bicarbonate solubility. Sodium carbonate is a lot more soluble, so that would be a possibility.

There are two equilibria to consider, $\ce{ 2H2O <=> H3O^+ + OH^-}$ and $\ce{ HA +H2O  <=> H3O^+ + A^-}$ where HA must be a weak acid, and for example, NaA its sodium salt. Because of the equilibria there is no real distinction between solutions of (1) pure  HA, (2) mixtures of HA and NaA and (3) pure NaA.
In your question one solid must be a pure salt such as NaA which will dissociate to give A$^-$ and so produce some $\ce{H3O+}$.

I think Andrew's (see above answer) initial choice for the source of carbonate as sodium bicarbonate ($\ce{NaHCO3}$) is correct based on its enthalpy of solution, which is listed as $\pu{+18.7 kJ\:mol^{-1}}$ in literature (Ref.1). I disagree with his final choice of sodium carbonate ($\ce{Na2CO3}$) based on its solubility in water as the carbonate of choice though. My disagreement is again based on its enthalpy of solution, which is listed as $\pu{-26.6 kJ\:mol^{-1}}$ in the same literature (Ref.1), and hence, upon dissolution, the final solution should have shown temperature increment in either experiment 2 or 4, instead of decrements as shown in the diagram. Said that, the only carbonate with positive enthalpy of solution I can find in literature is $\ce{NaHCO3}$. Thus, we don't have any choice other than $\ce{NaHCO3}$ in this case, although its solubility come a little bit short than anticipated (According to PubChem, the solubility of $\ce{NaHCO3}$ in water is $\pu{10.3 g}/\pu{100 g}$ of water at $\pu{25 ^{\circ}C}$). Nonetheless, question didn't elaborate all of solids are dissolved after $\pu{2 min}$. :-)
Now, let's see what is the approximate heat changes for these experiments (assumptions: We count only the mass of water disregarding solid substance in it, and consider its heat capacity is $\pu{4.184 Jg^{-1} ^{\circ}C^{-1}}$, for convenience):
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Experiment} & \Delta T, \pu{^{\circ}C} & \text{mass, }\pu{g} & \text{heat absorbed by }\pu{4 g}\text {, in } \pu{J} & \text{heat absorbed by }\pu{1 g}\text {, in } \pu{J}  \\ \hline
 \text{2 for comp B} & 26-22=4 & 30 & 502.1 & 125.5  \\ \hline
\text{4 for comp D} & 29-23=6 & 30 & 753.1 & 188.3  \\ \hline
\end{array}
$$
Now, tabulate the calculated and literature values of enthalpies of perticular solutions:
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Compound} & \text{molar mass} & \mathrm{\Delta H_{sol}}, \text {if B (calculated)} & \mathrm{\Delta H_{sol}}, \text {if D (calculated)} & \mathrm{\Delta H_{sol}}, \text {(Lit.)} & \text{Reference} \\ \hline
 \ce{NaHCO3} & \pu{84.01 g\:mol^{-1}}  & \pu{10.5 kJ\:mol^{-1}}   & \pu{15.8 kJ\:mol^{-1}}  & \pu{18.7 kJ\:mol^{-1}} & 1,2   \\ \hline
 \ce{C2H2O4} & \pu{90.03 g\:mol^{-1}}  & \pu{11.3 kJ\:mol^{-1}}   & \pu{16.9 kJ\:mol^{-1}}  & \pu{8.86 kJ\:mol^{-1}} & 3   \\ \hline
 \ce{C2H2O4.2H2O} & \pu{126.07 g\:mol^{-1}}  & \pu{15.8 kJ\:mol^{-1}}   & \pu{23.7 kJ\:mol^{-1}}  & \pu{35.2 kJ\:mol^{-1}} & 3   \\ \hline
 \ce{C6H8O7} & \pu{192.12 g\:mol^{-1}}  & \pu{24.1 kJ\:mol^{-1}}   & \pu{36.2 kJ\:mol^{-1}}  & \pu{18.2 kJ\:mol^{-1}} & 3   \\ \hline
 \ce{C6H8O7.H2O} & \pu{210.14 g\:mol^{-1}}  & \pu{26.4 kJ\:mol^{-1}}   & \pu{39.6 kJ\:mol^{-1}}  & \pu{29.1 kJ\:mol^{-1}} & 3   \\ \hline
\end{array}
$$
In the Table, $\ce{C2H2O4}$ represent anhydrous oxalic acid and $\ce{C6H8O7}$ is anhydrous citric acid. All four acids in the Table are higher solubility in water. Since we have questionable assumption in the calculations of $\mathrm{\Delta H_{sol}}$, I'd expect lower calculated value than that in literature. Thus, for proximety of the values for $\ce{NaHCO3}$, we can conclude that the compound D is $\ce{NaHCO3}$ (row 1 in the Table). Similarly, based in values in row 5 of the Table, we can also conclude that the compound B is citric acid monohydrate.

References:

J. P. Rupert, H. P. Hopkins Jr., C. A. Wulff, “The Solution Thermochemistry of Polyvalent Electrolytes. IV. Sodium Carbonate, Sodium Bicarbonate, and Trona,” J. Phys. Chem. 1965, 69(9), 3059–3062 (DOI: 10.1021/j100893a040).
C. E. Vanderzee, D. A. Wigg, “The standard enthalpies of solution and formation of Wegscheider's salt: $\ce{Na2CO3·3NaHCO3(s)}$ and of trona: $\ce{Na2CO3·NaHCO3·2H2O(s)}$ at $\pu{298.15 K}$,” The Journal of Chemical Thermodynamics 1981, 13(6), 573–583 (https://doi.org/10.1016/0021-9614(81)90114-2).
Alexander Apelblat, “Enthalpy of solution of oxalic, succinic, adipic, maleic, malic, tartaric, and citric acids, oxalic acid dihydrate, and citric acid monohydrate in water at $\pu{298.15 K}$,” The Journal of Chemical Thermodynamics 1986, 18(4), 351–357 (https://doi.org/10.1016/0021-9614(86)90080-7).

As I pointed out in my answer, the standard enthalpies of solution refer to infinite dilution in water. Bases like sodium carbonate are similar to acids in that as soon as the pH gets away from neutral and the reaction with water is diminished, the enthalpy of solution often becomes positive. Within the relevant temperature range to this problem, the solubility of Na2CO3 increases with temperature, indicating that its enthalpy of solution at saturation is positive, so it could be one of the compounds here. See http://hydro-land.com/e/ligne-en/doc/Na2CO3.html.

Also, your link to the Apelbat article misdirects. The correct link is:https://doi.org/10.1016/0021-9614(86)90080-7

We all can speculate as we wish since not much data is given to pin point this problem. However, I still disagree with $\ce{Na2CO3}$ being the carbonate of choice here because the temperature change is too big for both cases. And, it has a big negative value for salvation (clearly exothermic). Thanks for the error suggestion and it was fixed.

Did you look at the solubility plot? Definitely endothermic at these temps and concentrations. The negative value is the standard enthalpy at infinite dilution.

I did look at the solubility plot. But I thought it is the % solubility of $\ce{Na2CO3}$ vs temperature. Didn't indicate any heat exchange.

Since maximum solubility is an equilibrium state, the van't Hoff equation tells us that the slope of the curve is a function of the delta H of the reaction. Some people also like to think of this intuitively as a case of Le Chatelier's principle. If adding heat (ie raising temp) increases the product concentration (here the amount of dissolved salt), the reaction is endothermic. You can get a good estimate of the enthalpy change by subtracting the large negative enthalpy change of HO- + H+ -> H2O from the standard value you cited. The result is a positive number.


text:
Determine Empirical Formula from Molar Mass?
I am trying to write a computer program that predicts the empirical  formula of a compound based only on that compound's mass, like what you would get from a high resolution mass spec.
I have the basics of the program figured out. I start with the highest mass element and assume the compound is entirely made of it, then iterate through the lower mass elements until I find combinations that get close to the compound's mass. For example, let's assume the heaviest element I consider is Oxygen and I have a mass of 86.178 g/mol. I begin by assuming the compound is made of 5 Oxygens, but 5 Oxygens doesn't get close enough the goal, so I assume 4 Oxygens and 1 Nitrogen and so on until I land on 6 Carbons and 14 Hydrogens.
The program was successfully able to produce C6H14, but also produced nonsense like C7H2 and C0H85. I decided to check the degree of unsaturation of each potential compound and found that by throwing out everything with negative unsaturations I was able to get rid of most of the junk. But the C7H2 is still considered "valid", with an unsaturation of 7. I know this formula is wrong, but I can't think of a simple rule I can program in to check for these mistakes.
I've considered making sure that there are more non-carbon atoms than carbon atoms, but this clearly doesn't apply to inorganic compounds. I've considered that the oxidation state of the entire molecule must be equal to the charge, but I don't know how to get the oxidation state of the whole molecule based only on elemental composition.
I can't be the first person to attempt this, so I thought I'd ask here if anyone knew a better way to determine the empirical formula from the molar mass.
And I'm aware that there could be situations where 1 mass could correspond to more than 1 formula, but the program should produce both formulas.

Yes, there are computer programs that analyze mass spectra. However it is a lot more complicated than just finding the peak with the highest mass number. For  mass spectra you also have to consider isotopes for instance.

Doing this from scratch for any type of compounds is going to be very hard. Even with a restricted set (say compounds of C,H,O, and N) it isn't easy. Not least because the combinatorial complexity in organic chemistry will give plenty of overlapping formulae. Even counting isomers of saturated hydrocarbons is non-trivial as the possible number of isomers of each grows extremely rapidly with the number of carbons.

Relevant literature for you to read: https://bmcbioinformatics.biomedcentral.com/articles/10.1186/1471-2105-8-105

https://www.ncbi.nlm.nih.gov/pubmed/17389044

Mass spec has the advantage of breaking the molecule into pieces and also a database of fingerprints. If there is a specific task at hand, this approach may not be the best. If this is a puzzle then perhaps you need multiple subroutines. For example, assess the mass numbers potential as an alkane, alkene, aldehyde, carboxylic acid etc.These could be quickly cleared against an array of known values. In any case, use more granular data. Using mass to 3 decimals places (C 12.011 and H 1.008) makes C6H14 an exact match at 86.178. C7H2 is 86.093 and would not have cleared. If you are limited to integers, multiply these by 1000 for analysis.


text:
Using electrolysis of water as a proton generator
I'm working on a project right now, where the challenge is to try to use electrolysis as a way of controlling the $\ce{pH}$ of water.
I've set up an experiment where I have two separate chambers or volumes of water.
Both chambers are connected to the same electrolytic unit. The flow of water is from a respective chamber, into the unit and then back into the same chamber again, in an attempt to separate the two half reactions. For this reason I call the chamber that flows over the anode for the anodic chamber, and the chamber that flows over the cathode for the cathodic chamber. The bodies of water are separated in the electrolysis cell by a proton-exchange-membrane (PEM) that should, in theory, only allow positively charged ions to cross.

Apologies for the Danish text on my setup, but the figures should help you understand what I'm trying to do.
The anodic chamber (the water that goes to the anode half-reaction of the electrolysis cell) contains tap water while the cathode chamber contains Milli-Q water with added $\ce{NaCl}$ for conductivity purposes.
The $\ce{pH}$ of the water is alkaline in both chambers when the experiment starts, but I've also tried to run the experiment where the anodic chamber is acidic.
Anyway, here is what I hoped would happen in alkaline water:
Anode (oxidation):
$$\ce{4OH- (aq) -> 4e- + O2(g) + 2 H2O(l)}$$
Cathode (reduction):
$$\ce{2H2O (l) +2e- -> H2(g) + 2OH-(aq)}$$
In theory, I was expecting the $\ce{pH}$ to drop in the anodic chamber, since removing $\ce{OH-}$ would necessitate a flow of positively charged ions across the PEM membrane, instead of the $\ce{OH-}$ produced in the cathodic chamber in order to satisfy electronegativity. This would also mean that the $\ce{pH}$ should rise in the cathodic chamber.
However, what I see is invariably that the $\ce{pH}$ rises very rapidly, and continuously, in the cathodic chamber, whereas the $\ce{pH}$ in the anodic chamber remains either stagnant or rises very slowly.
Now, part of the reason for the rapid rise in the cathodic chamber is the complete lack of alkalinity, that much is clear to me. However, I don't understand why only the reduction half cell reaction seems to occur.
The only explanation I have been able to come up with is that something else on the anode half-reaction is donating electrons to the reaction. I tried running the experiment at different voltages in the hopes of finding a "sweet spot" that would prevent this "something" from reacting but as of now I haven't been able to do that. I'm worried it may be that my anodic electrode is donating the electrons.
My electrodes are nickel based.
A cursory glance at Wikipedia shows the standard electrode potential of nickel as
$$\ce{Ni^2+ + 2 e− <=> Ni(s)} \;\;\; \pu{E^{\circ} = −0.25 V}$$
And for my hydroxides
$$\ce{O2(g) + 2H2O + 4 e− <=> 4 OH−(aq)} \;\;\; \pu{E^{\circ} = +0.401 V}$$
When I run the reaction, I also see a yellowish/greenish precipitate forming in the anodic chamber, which according to some lab technicians at my location may be the nickel reacting.
Anyway, the question(s) are then - what do you think is happening  in my system, given this information?
Am I on the right track? should I try to change my electrode?
Have I made some obvious, basic mistakes?
What suggestions do you have, so that I can get this to work?
I hope that you have the time to advice me on this, and I wish you all a pleasant day.

In a chemical sense, "anode" of an electrolytic cell is the most powerful oxidizing reagent known, so much so that it oxidize F(-) to elemental fluorine. Don't forget that the electron balance in an electrolytic cell is always maintained so the gram-equivalents of x reduced at the cathode must equal gram equivalents of y oxidized at the anode. As a result, half cell reactions cannot occur alone. So something is being oxidized at the anode, this is your nickel electrode. Platinum is the best one but it is too expensive. Graphite electrode might be an economical substitute. Secondly, you have to add an electrolyte to the anode chamber as well because tap water resistance is pretty high. You can add a non-corrosive electrolyte such as sodium sulfate. With these changes, hopefully you will see the desired trend.


text:
Isotope labeled experiment analysis
I'm analyzing a dataset generated with LC-MS from cells that were fed $\ce{^13C}$-Glucose. There are 3 groups in total each with a different media conditions. My goal is to identify differences between these groups. When I look at the distribution of labeling (MDV) I don't see any difference across the different metabolites but when I look at the relative abundance I see large differences. How can I interpret this, when the labeling patterns are the same but the relative abundance is significantly different? Thanks.

We need more information to provide better answers.  For example, what types of medium and what types of cells?  How many sources of carbon were there in each aside form glucose?  Were those carbon sources labeled?  Also, what is MDV?

E.g., one potential explanation of the data is that in different media, cells have different levels of metabolites.  And in none of the media did they use any of the labeled glucose you fed them.


text:
Can a Hydrogen Radical form in a chemical reaction?
Original question: During kolbe electrolysis of Formic acid, H2 is released at which electrode?
What I thought: H2 will form at cathode due to reduction of water. But will the Formic acid decompose to give H2 and CO2 by radical mechanism?
My doubt: Does H radical form in chemical reactions?

Yes a H radical form can form in chemical reactions. However it is extremely reactive and will exist for only a fleeting amount of time.

@MaxW  will there be release of H2 at the anode then?

@MaxW https://en.m.wikipedia.org/wiki/Kolbe_electrolysis

$\ce{HCOOH -> CO2 + H2}$

@MaxW That is the problem I am having. This is only possible if H radical forms in the reaction. I'm not sure if it will work as standard kolbe reaction.

If you get anything else how would it be a Kolbe electrolysis?

@MaxW What I mean is that will the reaction even happen? From the mechanism, first step is the formation of HCOO• after which it breaks down to give the products. I wanted to ask will it break down to give H• and those H• will combine to give H2 or will it procede some other way.

I understand the concerns that at the concurrent occuence of $\ce{H., HCOO-}$ at the anode,  the former would be first to oxidize.

@Poutnik I'm sorry but I am unable to understand what you mean.

Hehe, it may be mutual, as I cannot see what may be there not to understand... :-) I mean if formiate is oxidized at an anode to HCOO radical, and if by decarboxylation is created the atomary/radical hydrogen, it seems to me it may take priority in further onodic oxidation. But I may be wrong.

@Poutnik I think I shouldn't waste time on this as I am unable to find anything on the internet about kolbe electrolysis of formic acid. Looks like the question maker did not think through. (This is frequent in the material for the exam I'm preparing for)


text:
Can amide bonds fragment in ESI-MS?
Electrospray Ionization Mass Spectrometry(ESI-MS) is a technique in mass spectrometry to get an idea about molecular ion peak. In the ESI-MS spectra of an amide compound, I observed that the molecular ion peak is much less abundant as compared to two other peaks, which correspond to the carbonyl and amine fragments between which the coupling reaction was carried out. But, TLC confirmed that there was no amine or acid remaining in the mixture after the reaction(EDC coupling).Hence my question is, although amide is a very stable bond, can it fragment in ESI-MS?

Yes, amide bonds can fragment during electrospray ionization.
Intentional fragmentation in a collision cell
Amide bonds are intentionally fragmented in mass spectrometers all the time, during proteomics analysis.  In that case, peptides are ionized, particular ions are isolated (by the quadrupole of the mass spectrometer), and then the intact peptide ions are fragmented intentionally in a collision cell inside the instrument.  After collision with high-energy gas molecules, peptides fragment in mostly predictable ways giving rise to y, b, a, and z ions.
"In-source" fragmentation
The key step of the fragmentation process is collision with high-energy gas molecules.  That can happen accidentally during electrospray ionization in the ion source.  That's because the ion source, being at atmospheric pressure, has many many more gas molecules around than a collision cell, which might be at 10$^{-5}$ torr or lower in pressure.  So after your ions of your peptidic molecule form in the ion source, some of them might bump into these atmospheric gas molecules and be fragmented in the ion source, before they even make it into the mass spectrometer.
Controlling in-source fragmentation
The propensity of this "accidental" fragmentation to happen is somewhat tunable by ion source parameters -- nebulizer and capillary voltages, gas flows and temperatures, etc.  The details depend entirely on your exact model of MS.  If the carboxyl and amine fragments are both more intense than the "molecular ion", it could be a sign that the ion source is tuned too aggressively -- trying lowering a voltage or two and see if you boost the molecular ion signal.  (Alternately, for some applications, in-source fragmentation could be desired.)


text:
How can I prepare a PBS (Phosphate Buffer Solution)? pH=7.4
Phosphate-buffered saline (abbreviated PBS) is a buffer solution commonly used in biological research. It is a water-based salt solution containing disodium hydrogen phosphate, sodium chloride and, in some formulations, potassium chloride and potassium dihydrogen phosphate.

You want $1\times$PBS or $10\times$PBS? Or just a protocol for prep?

Since $\mathrm{p}K_\mathrm{a}$ of monosodium phosphate ($\ce{NaH2PO4}$) is 7.21, for biological applications, $\ce{NaH2PO4}$ and its conjugate base, disodium phosphate ($\ce{Na2HPO4}$), can be used to generate buffers of $\ce{pH}$ values around 7. Usually, biochemists use the Henderson-Hasselbalch equation, $\left(\ce{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\text{Base}]}{[\text{Acid}]}\right)$, to calculate what ratio of acid to base is required to make a buffer of the desired $\ce{pH}$. Technically, this buffer is called phosphate buffer (PB).
Thus, to prepare buffer with $\ce{pH} = 7.4$, let's calculate the $\frac{[\text{Base}]}{[\text{Acid}]}$ from above equation:
$$\frac{[\text{Base}]}{[\text{Acid}]} = 10^{\left(\ce{pH} - \mathrm{p}K_\mathrm{a} \right)} = 10^{\left(7.4 - 7.21 \right)} = 10^{0.19}=1.549$$
Technically, if you want $\pu{1.0 M}$ buffer, that means $[\text{Acid}] + [\text{Base}] = 1.0$. So, you can calculate $[\text{Acid}]$ and $[\text{Base}]$ using these two relationships to get $\pu{1.0 M}$ buffer of $\ce{pH} = 7.4$:  $[\text{Base}] = 1.549 \times [\text{Acid}]$, and $[\text{Base}] = 1.0 - [\text{Acid}]$. Once calculated, use that amount to prepare your buffer with procedure given by Waylander in his answer. You may use phosphoric acid and $\ce{NaOH}$ solutions to adjust $\ce{pH}$ to desired value (7.4 in this case), if you want to have only phosphate components in your solution. Also, final make up to the &\pu{1.0 L}$, you may use a volumetric flask if desired.
However, phosphate buffer saline (PBS) closely mimics the $\ce{pH}$, osmolarity, and ion concentrations of the human body. Therefore, it is usually, $\pu{0.01 M}$ (or $\pu{10 mM}$) in $\ce{Na2HPO4}$ (base), $\pu{0.0018 M}$ (or $\pu{1.8 mM}$) in $\ce{KH2PO4}$ (acid), $\pu{0.137 M}$ (or $\pu{137 mM}$) in $\ce{NaCl}$, and $\pu{0.0027 M}$ (or $\pu{2.7 mM}$) in $\ce{KCl}$. This is called $1 \times$PBS. If you want $10 \times$PBS, you may multiply each component by 10. Once, all components are weighted into large enough container, use enough deionized (or distilled) water to dissolve them, and ensure to adjust $\ce{pH}$ to $7.4$ using phosphoric acid and $\ce{NaOH}$ solutions before make it up to 1.0 L using deionized (or distilled) water.

Google is your friend!
The preparation of pH 7.4 PBS (with KCl and KH2PO4) is described on this site   here. Preparation of many other buffer systems are also described.
Here are the conditions (lifted from the linked site)
Prepare 800 mL of distilled water in a suitable container.
Add 8 g of NaCl to the solution.
Add 200 mg of KCl to the solution.
Add 1.44 g of Na2HPO4 to the solution.
Add 240 mg of KH2PO4 to the solution.
Adjust solution to desired pH (typically pH ≈ 7.4).
Add distilled water until volume is 1 L.

The internet is replete with recipes of all kinds and many tables; so, I will respond on the practical aspect of the question. In reality, many people just buy a premixed packet, mix it with the correct volume of water, and get on with their research. Additionally some buy it as a liquid ready to go. So, depending on the volume needed, your resources, and the stringency of the experiments, mixing up buffer and manually adjusting the pH may just not be worth the effort.


text:
How to dissolve beeswax on the underside of a porous wood structure
Honeybees, like most insects, communicate through smell. They are attracted to the smell of beeswax. My profession is to rescue and relocate honeybees that have taken up residence inside walls, attics and chimneys. New bees should be discouraged to return by removing all beeswax residue so I recommend scraping down to bare wood and then encapsulating with oil based spray paint. The scraping process is laborious, time consuming and inferior since wood is soft and porous. It would be ideal if a chemical could be applied that would actually dissolve the beeswax without harming the structure. Occasionally bees build over electrical wire and galvanized hurricane ties with hundreds of 8 mm dimples making cleaning very difficult. Heat and a wire brush may work but I sense that heating beeswax only makes the oil spread out. Remember this is on the underside of studs, etc so letting solvent sit to dissolve will not work unless in gel form. Any chemist have suggestions?

The majority of the solvents mentioned in the answers to How to dissolve candle wax (paraffin)? should work for you too. Also, see How to breakdown wax, preferably beeswax

Esters are good for dissolving wax, but use a long carbon chain that is not flammable.
Isopropyl myristate is sold online to people that want to make their own cosmetics. It is relatively cheap. It will leave an oil film that can be removed with liquid detergent in warm water.
For thick wax, warm with a blow dryer and scrape with a plastic spatula. Then try above.

Regarding the above link on candle wax: xylene and toluene are dangerous volatile chemicals - not recommended. Volatile petroleum products may be problematic depending on the volume used and if you are in a confined space. I recommended isopropyl myristate because it is safe on skin and far from volatile - vapor presssure of 9.35e-05 mmHg


text:
How to use Lewis theory to differentiate between an acid and a base?
According to Lewis theory a material is an acid if it accepts an electron pair, and a material is base if it donates an electron pair. But if we compare the structures of $\ce{HNO3}$ and $\ce{NaOH}$, i.e.
$$
\ce{H-\underset{.\!.}{\overset{.\!.}{O}}-\underset{\underset{\huge :O:}{||}}{\overset{+}{N}}-\underset{.\!.}{\overset{.\!.}{O}^}\!:^-}
\quad
\text{and}
\quad
\ce{[Na]+ [:\!\underset{.\!.}{\overset{.\!.}{O}}H]-}
$$
there are so many lone pair of electrons which can be shared, and in both structures there is a positive atom, so they can accept a lone pair. How can I say that one of them is acid and the other one is a base?

First of all, within the Lewis framework, you may have substances that can act as both acid and base (though most of the time as one or the other, not both at the same time).$^{1}$ The adjective for such substances is "amphoteric".
It comes down to a bit of chemical intuition to decide which is which. It is true that almost anything that has a lone pair can act as a base - though it may take a strong acid to form an actual Lewis pair. For acids, one should look for an obvious gap in the electron cloud ($\ce{AlCl3}$) or candidates for complex formation ($\ce{FeCl3}$).$^2$
For nitric acid, the situation is pretty clear: the hindrance provided by three oxygen atoms around the central nitrogen prevents access, so it will hardly act as a Lewis acid. It can act as a base, however, weak Brønsted bases are usually also weak Lewis bases because they are content on their own.
For sodium hydroxide, one must realize that in an aqueous medium, the two ions will fully dissociate. The hydroxide anion will act as a Lewis base, but the sodium cation with its full outer electron shell will hardly act as an acid.$^3$

$^1$ Adducts of e.g. $\ce{CO2}$ in frustrated Lewis pairs, wherein the $\ce{CO2}$ acts as a sort-of bridge between the two Lewis sites, are an example of when "both at the same time" can occur.
$^2$ I am aware that these compounds will form dimers or more complex structures in reality. If you are able to spot this behavior and understand why, you are already able to spot Lewis acids.
$^3$ Note that the sodium cation in water will be solvated, i.e. some water molecules will orient their lone pairs towards the ion. The border between solvation and Lewis adduct formation can be blurry.

@adeshmishra In the question, you wrote "and in both structures there is a positive atom", which I take to mean that you are wondering about the nitrogen as an electron acceptor. $\ce{H^+}$ is, of course, a Lewis acid.

Yes you are right. Your answer has really helped me a lot. One question I wanna ask can hydrogen ion accept electron pair without getting disassociated?

@adeshmishra Both happen simultaneously in proton exchange. In hydrogen bonds hydrogen is coordinated to two or more atoms.

@TAR86 There are orthonitrates though. I don't think sterics is main factor why they're less stable then "normal" ones.


text:
Simplest distillation setup
I would like to start learning practical chemistry. I have not done chemistry before and am seeking help assembling a simple distillation setup.
I do not want to use the cold water tubes etc for the time being, I am trying to keep it as simple as possible.
I have a few questions which I am hoping someone can answer:

What are the plastic clips for?
How do I vacuum seal all the joints as simply as possible (preferably without using grease). Would tape work?
How do I seal the small side arms? (In the liebig condenser tube this isn't a problem because they don't lead to the inner distillation tube.)

You're also going to need 2 ringstands and some clamps. See https://www.southernlabware.com/liebig-condenser-distillation-kit-complete-3000ml.html?utm_source=google_shopping&gclid=EAIaIQobChMI5Ij74Ke54QIVTP_jBx3u0A-IEAQYASABEgLOJvD_BwE

That's a nice bit of kit to get started. I've annotated the picture a little to aid discussion:

I've labelled the "male" and "female" joints that should be connected with the same letter (e.g. B inserts into B'). The thermometer goes in the top of the blue plastic screw joint.
Joint A serves the purpose of connecting your system to a vacuum source. If you are doing vacuum distillation, you should connect the vacuum tubing at this point. However, if you are just distilling at atmospheric pressure you should leave this open. You should NOT seal this joint, as this would mean there is no pressure relief in the system - when you start to heat your sample, gases expand and they need somewhere to go, otherwise you will at best pop a joint, and at worst cause an explosion.
As you say, the joints on the liebig condenser (G) can be left open since they aren't open to the interior of the apparatus. However, you will get better results if you do hook up the condenser to water using the tubing provided, especially if you are distilling at relatively low temperatures.
The clips are there to secure ground glass joints so they don't fall out. You only really need them on junctions that arent held in place by gravity (e.g. you need one on D/D' since the joint could fall apart under gravity, but not on B/B' since gravity holds the male joint in place).
You shouldn't need to vacuum seal your joints - ground glass quick-fit joints should be manufactured to a reasonably high tolerance so the leakage is minimal. It would only become a problem if you needed very strong vacuuum.
Note: if you are going to use vacuum for your distillation you should carefully inspect every piece of glassware for cracks, imperfections etc. - if the glassware is damaged then when you apply the vacuum it can implode. You should always wear eye protection when doing any chemistry at all.
Hope this helps, please ask questions if you aren't sure of anything - people on here will be happy to help a new recruit to develop their chemistry skills!

@Raghib The clips may also be intended to fix the rubber tubing on the nipples. Using just a tiny bit of grease will make your setup tight and also allow you to disassemble the glass joints. Using tape can get messy if you start to heat it up ;-) Going away from grease may be considered if the grease is a big problem in combination with your substances. By the way: what do you imagine to distill with it?

Those are Keck clips, they’re definitely not for securing tubing. Jubilee clips are best suited to that purpose.


text:
Unit analysis for neutralization reaction

What volume of $\pu{0.500 mol L-1}$ $\ce{HCl}$ (in $\pu{mL}$) is required to neutralize $\pu{2.02 g}$ of solid $\ce{NaOH}$ pellets?

So far I have tried regular unit analysis:
2.02 g NaOH * 1mol NaOH/40g NaOH * 1mol HCl/1mol NaOH * 500g HCl/1mol HCl
= 25.25 g HCl --> 25.5 mL HCl
This answer doesn't make sense in the context of this neutralization reaction. Is there a type of unit analysis that is used specifically for determining the volume required for a reaction?

Could you explain your calculations?

There is no special unit analysis.
It's a recommended practice to solve the problem algebraically first using proper notations for physical quantities, and plug the numeral values at the end minding the units — this way you reduce the chance of making the erroneous calculations and keep track of all units; as a bonus, you simplify handling significant figures.
The majority of problem solving in chemistry begins with writing down the balanced chemical reaction:
$$\ce{HCl + NaOH → NaCl + H2O}$$
So, to find the volume of hydrochloric acid $V(\ce{HCl})$, you may use its molar concentration $c(\ce{HCl})$ and the amount $n(\ce{HCl})$:
$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{c(\ce{HCl})}$$
Unknown amount $n(\ce{HCl})$ can be found from the reaction's stoichiometry:
$$n(\ce{HCl}) = n(\ce{NaOH}) = \frac{m(\ce{NaOH})}{M(\ce{NaOH})}$$
where $m(\ce{NaOH})$ and $M(\ce{NaOH})$ are mass and molecular mass of sodium hydroxide, respectively.
Putting everything together and plugging in the appropriate numerical values, you arrive at the volume of approx. $\pu{100 mL}$:
$$
\begin{align}
V(\ce{HCl}) &= \frac{m(\ce{NaOH})}{M(\ce{NaOH})\cdot c(\ce{HCl})} \\
  &= \frac{\pu{2.02 g}}{\pu{40.00 g mol-1}\cdot \pu{0.500 mol L-1}} \\
  &= \pu{0.101 L}
\end{align}
$$

The Unit/dimension analysis is in this case rather "using a cannon against sparrows".
The first spotted error is $\pu{500 g}$ of $\ce{HCl}$. Where did it come from ?
Note that the molar mass of $\ce{HCl}$ is about $\pu{36.5 g / mol}$.

Get the molar mass of $\ce{NaOH}$.
Calculate the amount of mols of $\ce{NaOH}$.
Calculate the equivalent amount of mols of $\ce{HCl}$.
Calculate the equivalent volume of $\pu{0.500 mol / l}$ $\ce{HCl}$.

For a beginning chemistry student using dimensional analysis is a good thing.

I agree. If he did not introduce those 500 g, he could land at the correct result with the analysis as an added value..


text:
Why does molecular oxygen absorb in the IR range?
I am studying spectroscopy and know that $\ce{CH4}$ and $\ce{CO2}$ are known as greenhouse gases because they are IR active. This is because they have asymmetric stretch or bend vibrational modes which result in electric field changes and involves emission or absorption of IR light.
But, $\ce{O2}$ is a symmetric molecule and should be IR inactive. I am wondering why there are still strong $\ce{O2}$ absorption bands in the IR range if it is expected to be transparent to IR light?

Oxygen has strong electronic transitions in the uv but very, very weak IR transitions as these are dipole forbidden as you mention. Perhaps you could edit and add some data to go with your question

There are two low lying and 'forbidden'  electronic transitions at approx 7800 ($^3\Sigma \to ^1\Delta$) and 13000 wavenumbers $^3\Sigma \to ^1\Sigma$. Additionally there are weak collision induced transitions $\ce{O2:O2}$ and $\ce{O2:N2}$ in a similar spectral region.


text:
Naming alkenes with double bond branches
After reading the IUPAC rules, I found the parent chain to be from the $\ce{CH2}$ numbered 1 to the one numbered 9, and I considered both methyl groups as branches. However, my confusion lies in naming the branch in the middle (depicted in red color) since it contains a double bond:

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the traditional name ‘vinyl’ for $\ce{CH2=CH-{}}$ is no longer used for preferred IUPAC names and is only retained for general nomenclature. The preferred name is the systematic name ‘ethenyl’.
Therefore, the name for the compound that is given in the question ‘3,7-dimethyl-5-vinylnona-1,8-diene’ could be found in the literature; however, the preferred IUPAC name in accordance with current recommendations is 5-ethenyl-3,7-dimethylnona-1,8-diene.


text:
Interchangeable isomers
I am a clinical research professional. I have a question regarding clinical chemistry. I would like to know that whether it is possible to get testosterone undecanoate from ursolic acid and vice-versa by the mean of specific chemical reaction or mechanism? Testosterone undecanoate and ursolic acid both have the same chemical formula C30H48O3 and thus molar mass of 456.7g/mol. But the chemical structures, position of functional groups and arrangement of atoms differ, Therefore,can be considered as isomers of each other.

They're obviously isomers, but there's not much "interchangeable"  here.

There is no simple reaction that would interchange the two.
Ursolic acid (from Wikipedia)

Testosterone undecanoate (from Wikipedia)


text:
Why are some reactions exothermic and some are endothermic?
Some reactions turn out to be exothermic or endothermic. But why does this occur? Why do some reactions need energy to bond but some release energy to bond?

Some reactions turn out to be exothermic or endothermic. -- Some(!?!) -- It is a rare reaction that would be exactly isothermal.

When you walk around, sometimes you have to go uphill. Sometimes you have to go downhill.

@Zhe - and if you walk uphill to school, you walk downhill going back home (in spite of what our parents told us).

@JonCuster That's true. The reverse reaction brings you to the same place.

A chemical reaction occurs from reactants to products, which are both characterized by their energies. If you go from reactants to more stable products, the reaction may be exothermic by releasing the energy difference between the two species. This is a simple example but some systems could be a lot more complex.

A way to predict such a behaviour would be using computational chemistry to compute the energies of the reactants and the products.
It's also possible to play on some effects such as the solvent to stabilize (higher energy) or destabilize (lower energy) the system. Thus, it could make the reaction more exothermic in a specific solvent than in another one for instance.

Chemical bonding is just an electrostatic. But what happens is that we get so many charges I.e. electrons and nuclei that it everything becomes very complicated.
If the reactant is more stable than product ( I must clarify stability is just about electric potential energy, but due to so many other charges with complicated arrangements it’s not possible to say who is more stable just by looking) than we have to apply some energy to break the bonds of reactants and when new bonds will form it’s not gonna release much energy so as to nullify my given energy and hence the net energy will be transferred to the system and we will an endothermic reaction.
On the other hand if product is more stable so sometimes (spontaneous reaction) reactants will react by themselves and hence potential energy gonna drop so we will get energy out of this reaction. Even if I have to apply some energy to make the reaction go then also product would release energy due to drop in potential energy and this release of energy would be more than what I applied for the reaction to occur and hence we will get an endothermic reaction.
Chemistry is a very experimental subject, we can find about endothermic or exothermic reactions only by experiments. But if we know few data we certainly can make predictions whenever those reactions gonna occur. For example an acid base reaction is always exothermic so whenever there is a reaction between any whatsoever base and an acid we can say theoretically that it is exothermic.
Hope it helps


text:
What are the conditions for using skeletal replacement ('a') nomenclature?
The Blue Book 2013 in P-63.2.4.1 gives “1-methoxy-2-(2-methoxyethoxy)ethane” as the IUPAC preferred name for $\ce{CH3OCH2CH2OCH2CH2OCH3}$ and comments,

Skeletal replacement (‘a’) nomenclature [method (4)] generates preferred IUPAC names, when the conditions for using this type of nomenclature are met (see P-15.4); otherwise substitutive nomenclature must be used.

It then names $\ce{CH3OCH2CH2OCH2CH2OCH2CH2OCH3}$ as “2,5,8,11-tetraoxadodecane.”
My question is why skeletal replacement nomenclature was used in the latter case but not the former—why the conditions for using skeletal replacement nomenclature are not met by the former molecule.
P-15.4 states that

“Skeletal replacement (‘a’) nomenclature is used in two ways:
“ (a) …
“ (b) to generate simpler names for heteroacyclic structures than those formed by following the principles of substitutive nomenclature, for example, in naming polyamines, polyethers, etc., by replacing carbon atoms in corresponding acyclic hydrocarbons” (89).

The former molecule seems like an instance of a polyether, and the name “2,5,8-trioxononane” seems simpler than the given preferred name, so it seems that this molecule meets the this criterion.
What factor led to the use of substitutive nomenclature where replacement nomenclature seemed to apply?

Rule P-51.4, not P-15.4, specifies the conditions under which skeletal replacement nomenclature generates preferred IUPAC names.

P-51.4.1.1 Skeletal replacement (‘a’) nomenclature rather than substitutive or multiplicative names must be used to generate preferred IUPAC names for acyclic structures when four or more heterounits are present in an unbranched chain containing at least one carbon atom and when none of the heteroatoms constitute all or part of the principal characteristic group of the compound.

The former compound has only three oxygen heteroatoms, not enough to require the use of replacement nomenclature.


text:
How are these two molecules identical?

I can see that these compounds are identical as we have basically just done a 180 degree rotation of the molecules.
But if we go by the concepts of chirality, the carbons 2 and 4 are chiral centers and since they have inverted configurations at both chiral centers, they should be enantiomers. Or is it just that I cannot assume the hydrogens not shown to be out of plane?

Have you heard of meso compounds?

No, you just said that they're identical. Why would you then say that they are enantiomers? The concepts of chirality don't say that. They have to be nonsuperimoposable mirror images, and they are clearly superimposable.

@Mithoron I have heard of them but I don't exactly know what they are.

@Zhe but they have opposite configurations at both chiral centres. Doesn't that qualify as being enantiomers?

An enantiomer is a non-superimposable mirror image. The definition does not say anything about configurations. True, having opposite configurations at all chiral centres often makes two things enantiomers, but not always.

The basic question: Yes the molecules are identical. Optically activity None, this is a meso compound. The two chiral centers cancel each other out so the molecule as a whole is not optically active.

On a mildly related note, the general relativity books I'm reading keep saying don't use coordinates, just use coordinate-free descriptions because that will be true in every coordinate system. This is exactly like that. The definition of chirality is fully independent of identifying the configurations of centers. Does this molecule have an improper axis of rotation? If yes, it's not chiral, and you don't need to assign stereochemistry to any center. This is the equivalent coordinate-free language.

Sameer. The key to understanding this is that the two asymmetric carbons are indistinguishable because of the symmetry in the molecule. When you do your inversions on them, the molecule looks different, and should be an enantiomer, but after you do the 180 deg rotation, you have essentially switched them over (now the molecule looks the same as the first molecule, but only because you can’t differentiate the two asymmetric carbons).
Imagine that you switched one of the terminal methyl groups with an ethyl, then try the same trick. Because you have broken the symmetry of the molecule, the 180 deg rotation doesn’t get you back to the start. Hence, this meso isomerism relies on the mirror plane of symmetry (or a point of inversion) existing in addition to multiple asymmetric centers.

This molecule has a plane of symmetry. Imagine a vertical plane go through $\ce{C}$-3 atom of pentane chain. One side of that plane is the mirror image of other side. Hence, this is not optically active molecule by definition. It is called meso-isomer. If you look at (R,S)-configuration and number the chains as they are drown, first compound is (2R,4S)-2,4-dichloropentane and second compound is (2S,4R)-2,4-dichloropentane. But, since each have plane of symmetry, you may number second compound from opposite direction. Then it is (2R,4S)-2,4-dichloropentane. Therefore, they are identical.
The two enantiomers relevant to this meso-isomer are (2R,4R)-2,4-dichloropentane and (2S,4S)-2,4-dichloropentane.
To understand more on enantiomers and relevant meso-isomer, study halogenation of cis-/trans-2-butene or 1,2-diphenylethene (stilbene).


text:
How does the room gets heated by leaving a fridge open?
Question:

If the door of the refrigerator is kept open, the room in which the refrigerator is kept:

gets cooled
gets heated
neither gets cooled nor gets heated
gets cooled or heated depending on the initial temperature of the room.

The answer mentioned is option b) gets heated
It doesn't seem convincing that on opening refrigerator's door the room gets heated. Also the radiator emits out the heat in the surrounding. Thus the heat absorbed from the room is released in it and the must neither get cooled nor heated.
Just before posting this question I have tried to experimentally verify this and observed no change in temperature of the room.

I think at first pass this looks like homework but is not. I have edited and believe the question should remain open.

Not sure if you'd call it duplicate(from a different site), but I mention it anyway: Opening the fridge door to cool a room

Also the radiator emits out the heat in the surrounding. Thus the heat absorbed from the room is released in it and the must neither get cooled nor heated.

You have forgotten a few parts to your system in how a refrigerator pumps heat. You have assumed a spontaneous steady state cycle of:
$$\text{radiator}\underset{\text{conduction}}{\overset{\text{heat pump}}\leftrightharpoons} \text{fridge}$$
But that is not true to move heat from a low-temperature body body to a high-temperature body requires an input of work which is added to the heat dumped into the high-temperature body.

In this case work is done by the compressor which takes in electrical energy and transforms it into work and heat. Since electrical energy is 100% efficient at heating any electrical energy consumed by the compressor is eventually converted into heat either by ohmic heat, efficiency losses or work being converted into heat at the high-temperature body. Since your high-temperature body and low-temperature body are allowed to conduct/convect/radiate/transfer heat between the two sinks, heat overall is added to the room.

Just before posting this question I have tried to experimentally verify this and observed no change in temperature of the room.

There are a number of problems with this, and why you may not have observed anything.

Rooms are not closed nor isolated systems thus are subject to many thermal variations from light (radiation), drafts (convection), and conduction of heat through walls to/from the outside which can make your results unreliable.
A the heat capacity of a room is very large and it will take a while for your fridge to impart a change that a thermostat/thermometer can detect.
The energy consumed by a refrigerator is relatively small* and it will again take a while for your fridge to impart a change that a thermostat/thermometer can detect.

All and all, such an experiment would require better control than your average room/kitchen. If you did do such a controlled experiment you would observe that the temperature of the room would rise with time due to the heat losses of the compressor.
 *I understand some parents/utility customers (possible yours) may find this statement controversial. Kids do not quote me at home.

Think of a refrigerator as a device that transfers heat from inside a box to its surroundings. The room around a refrigerator is warmed as it receives the heat removed from inside the box.
If you leave the door open, heat is merely recycled from the room into the refrigerator, then back into the room. A net room temperature increase would result from the heat of the motor that would be constantly running to move energy around in a circle. It would be like hooking up a water pump to remove water from your basement and routing the discharge tube back to the basement.

Energy balance
No matter if you leave the door open or closed, if you run a refrigerator in a room (system) that is thermally insulated with its surrounding, the energy of the room will increase.
The reason is that your are performing electrical work on the system (the refrigerator is plugged in and is drawing power). According to the first law of thermodynamics, energy is conserved. If you are looking at a closed system (no transfer of matter), the change in energy of the system is equal to the work done on the system and the heat transferred into the system. If you are looking at an isolated system (no transfer of matter or heat), the only change in energy is through work done on or by the system.
Temperature
Whether the temperature goes up or not depends on what else is in the room. If there is nothing else but the refrigerator in the room, and the refrigerator was turned off before, as soon as you turn on the refrigerator the room temperature will increase. If the refrigerator door is closed, the inside of the refrigerator will decrease in temperature, and the increase in temperature of the rest of the room will be even higher (because of the heat transfer from refrigerator to rest of the room).
However, if your room is filled with water with some ice cubes in it (don't try at home, danger of electrocution), the temperature of the room will remain at about 273K. Instead, the increase in energy will be used to melt some of the ice.

The "cool" in the refrigerator is made by putting the "heat" into the room. However the process isn't 100% energy efficient. The electric motor that drives the pump generates heat. So overall you're putting heat into the room.


text:
What is the percent of potassium as K₂O in the fertilizer sample?

The solubility of sodium tetraphenylborate in water is $\pu{470 g/L}$ while the solubility of potassium tetraphenylborate in water is $\pu{1.8e–4 g/L}$.  Due to the orders of magnitude difference in aqueous solubility, $\ce{NaB(C6H5)4}$ is used to gravimetrically analyze the potassium content of fertilizer:
$$\ce{K+(aq) + NaB(C6H5)4(aq) -> KB(C6H5)4(s) + Na+(aq)}$$
The water soluble components of a $\pu{10.7875 g}$ sample of 15-25-15 fertilizer were separated from the water insoluble components using vacuum filtration.  The filtrate was treated with an excess of $\ce{NaB(C6H5)4}$ and the resulting solid was isolated, dried, and the mass of the solid was determined.  Potassium tetraphenylborate $(\pu{12.4393 g})$ was recovered from the gravimetric analysis of $\pu{10.7875 g}$ fertilizer.
What is the percent of potassium as $\ce{K2O}$ in the fertilizer sample?

I know how to get to the answer; I am just unsure how to use the solubility that was given to get there. I also am not sure why the resulting dried mass is higher than the initial mass of fertilizer:
$$\ce{K+(aq) + \underset{\pu{10.7875 g}}{NaB(C6H5)4(aq)} -> \underset{\pu{12.4393 g}}{KB(C6H5)4(s)} + Na+(aq)}$$
$$M(\ce{NaB(C6H5)4}) = \pu{342.22 g mol-1}$$
$$M(\ce{KB(C6H5)4}) = \pu{358.35 g mol-1}$$
$$\pu{12.4393 g~\ce{KB(C6H5)4}}\cdot\frac{\pu{1 mol}}{\pu{342.22 g}}\cdot\frac{\pu{1 mol}~\ce{K}}{\pu{1 mol}~\ce{KB(C6H5)4}}\cdot\frac{\pu{1 mol}~\ce{K2O}}{\pu{2 mol}~\ce{K}}\cdot\frac{\pu{94.2 g}~\ce{K2O}}{\pu{1 mol}~\ce{K2O}} = \pu{1.712 g}~\ce{K2O}$$
$$1~\text{mass} = \frac{\pu{1.71 g}~\ce{K2O}}{\pu{10.7875 g}}\cdot 100\% = 15.9\%$$

You made a minor calculation mistake in the MW of the salt. It should be 358.35 g/mole not 342.22 g/mol. You want the MW of the potassium salt, not the sodium salt.

The solubility has been given solely to justify the application of the given method of gravimetric analysis.
Since the majority of potassium salts are well or moderately soluble in water, for this analysis it's assumed that entire potassium content is brought to the solution.
Since the solubility value of potassium tetraphenylborate is over a million times smaller than the corresponding value for the analyte (sodium tetraphenylborate), it's safe to speak of the completeness of the analysis as all dissolved potassium species were isolated as tetraphenylborate salt and analyzed.
As for the mass difference, that shouldn't be of your concern as the content of the fertilizer isn't given in the first place and you cannot make any assumptions regarding molecular masses of its constituents.
Plus, tetraphenylborate $\ce{B(C6H5)4-}$ is a bulky anion of substantial molecular mass $(M(\ce{B(C6H5)4-}) = \pu{319.23 g mol-1})$, which is much higher in comparison to "usual suspects" found in fertilizers, such as phosphates $(M(\ce{PO4^3-}) = \pu{94.97 g mol-1})$ and nitrates $(M(\ce{NO3-}) = \pu{62.00 g mol-1})$.
Long story short, your answer seems correct to me, but I propose to solve the problem algebraically first, starting with what has to be determined and omit intermediate calculations, unless you were explicitly asked to perform any.
To illustrate my approach and use of the notations for physical quantities and units, I leave my solution here.
Content of potassium expressed via potassium oxide is a mass fraction $ω(\ce{K2O})$, which is the ratio between the mass of potassium oxide $m(\ce{K2O})$ and the total mass of fertilizer taken for analysis $m_\mathrm{tot}$:
$$ω(\ce{K2O}) = \frac{m(\ce{K2O})}{m_\mathrm{tot}}$$
Unknown mass $m(\ce{K2O})$ can be found from the corresponding amount $n(\ce{K2O})$ and molecular mass $M(\ce{K2O})$:
$$m(\ce{K2O}) = n(\ce{K2O})\cdot M(\ce{K2O})$$
In turn, amount $n(\ce{K2O})$ can be determined from the reaction's stoichiometry:
$$n(\ce{K2O}) = \frac{1}{2}n(\ce{K+}) = \frac{1}{2}n(\ce{KB(C6H5)}) = \frac{m(\ce{KB(C6H5)})}{2\cdot M(\ce{KB(C6H5)})}$$
Now, since we are at the point when no unknowns are left, lets put everything together so that the first equation can be rewritten as such:
$$
\begin{align}
ω(\ce{K2O}) &= \frac{m(\ce{KB(C6H5)})\cdot M(\ce{K2O})}{2\cdot m_\mathrm{tot}\cdot M(\ce{KB(C6H5)})} \\
  &= \frac{\pu{12.4393 g}\cdot\pu{94.20 g mol-1}}{2\cdot\pu{10.7875 g}\cdot\pu{358.32 g mol-1}} \\
  &= \pu{0.152}~\text{or}~15.2\%
\end{align}
$$

Think you should clarify this a bit in that the assumption is that all the potassium species are soluble.

@MaxW Good point; I left a link to the Wikipedia solubility chart for potassium salts. Thanks!

We tend to be sloppy about assumptions on the site. I had a high school teacher who absolutely considered the assumptions as being part of the answer. She was really a fantastic teacher. I took one year of high chemistry under her and exempted two quarters of freshman college chemistry.

@MaxW True that, I do misjudge quite often based on assumption that since it's something I'm aware of then it must be a trivial fact not worth mentioning. Indeed, I need to be reminded that every life's experience is different.


text:
Determining the formula of tin(II) chloride hydrate

$\pu{1.20 g}$ of hydrated tin chloride decompose to form $\pu{1.01 g}$ of anhydrous tin chloride on heating. Calculate the value of $x$ given
$$\ce{SnCl2.xH2O -> SnCl2 + x H2O}$$

I done it and came out with $1.666477009$ using:

$M(\ce{H}) = 1.008$
$M(\ce{O}) = 15.999$
$M(\ce{Cl}) = 35.45$
$M(\ce{Sn}) = 118.710$

This is a "challenge" question set by my chemistry teacher (GCSE Chemistry), but I won't see any of my teachers for another 2 weeks so I would really like someone to help check if I done this right.

How did you get that value? (Also it has way to many significant figures.)

In addition to the issue with sigfigs, it would be nice to see units. Plain numbers are rarely meaningful in natural sciences.

1.20g of hydrated tin chloride decompose to form 1.01g of anhydrous tin chloride on heating. Calculate the value of x

1.20 Stannous chloride hydrate = 1.01 anhydrous stannous chloride
So 1.20 - 1.01 = 0.19 grams of water
Moles of anhydrous stannous chloride
$$\dfrac{1.01}{189.60} = 0.0053270$$
Moles of water
$$\dfrac{0.19}{18} = 0.01065$$
ratio of stannous chloride to water
$$\dfrac{0.01056}{0.0053270} = 1.982 \ce{->[rounding]} = 2.0$$
So the formula is $\ce{SnCl2\cdot2H2O}$

Notice that I carried some extra digits in the intermediate calculations but rounded the final result to two significant figures since the grams of water only has two significant figures.

With respect to using "moles", "number of moles" etc. instead of amount of substance you might want to check out a recent Meta post:)

Yes I know about IUPAC's moving to amount of substance but to me it seems stupid, after all I am calculating moles.

Ehhm, technically speaking, it's amount (physical quantity) you are calculating. Mole is a unit, and units are not to be calculated.

@andselisk - If you do a dimensional analysis I setup the problem to calculate moles. It seems stupid to introduce a level of indirection in the text. Why we let some fool English major create this brouhaha  is beyond me.

@andselisk - Know the difference between a terminologist and a terrorist? -- You can negotiate with a terrorist.

Terrorists only understand negotiation done the Korben Dallas way. Terminologists are more like inventors and preservers of the uniform language so that the people could still understand each other and keep up building the tower of Babel regardless of the "god's will" (I bet conspiracy theory adepts will also read between the lines:) ).

@andselisk - Made a note to myself. Do not fool with the master blaster.


text:
What is the name of "cis/trans" isomerism when all four groups are different?
How would you call a pair of isomers on the picture below?

They look a lot like regular cis/trans isomers, except all four groups are different. I can tell they are spacial isomers (stereoisomers) but is there a more specific name for this kind of isomerism?

Check : Cahn-Ingold-Prelog priority rules

Cis-trans is generalised as E/Z.

When all four substituents are different cis-trans type of isomerism cannot be applied.  Moreover the Cis trans system is often ambiguous because configurational descriptions have not been defined according to any general and clear set of rules. For these types,  there is system called E-Z system of nomenclature and is based on the sequence rules of Cahn, Ingold  and Prelog (CIP) rule, originally developed for naming optical isomers on R-S system


text:
How to determine the boiling point of substances?

Which of the substances has the lowest boiling point?
a) $\ce{H2O}$ b) $\ce{H2S}$ c) $\ce{H2Se}$ d) $\ce{H2Te}$

I've been searching for any formula or rule to determine this, but I read that it didn't exist.
Do I have to remember some boiling points by heart?

Lookup Hard and Soft (Lewis) acids and bases.

Molecules don't have boiling points: BP is a bulk property when lots of molecules get together.

The key here is to figure out what factors affect boiling point. Specifically, anything that affects intermolecular forces. You should consider the molecular weight (which relates to dispersion forces) and the ability to form hydrogen bonds.

There are basically 3 ways how to know particular data 1/remembering them or their patterns 2/calculating them 3/Looking them up.  In this case, 1/ is not needed - how often ( but H2O ) you need them ? 2/Difficult and generally not possible 3/Easy

While it may be difficult to determine the exact boiling point of a substance, many factors play to compare if a substance has a higher boiling point than the others.

Intermolecular force. Notice that it is "intermolecular force" instead of
"intramolecular force". The reason being when you boil something you want to transform the substance from liquid state into gas state, and that can only be achieved by weakening the connection of molecules, not by breaking the intramolecular bond of a molecule. There are generally three types of intermolecular force : London dispersion force [weakest], dipole-dipole, and hydrogen bonds [strongest]. The stronger the force, the larger the amount of energy needed to break off the connection between the molecules, thus the boiling point is higher.

London dispersion force (LDF/Van der Waals) usually occurs when there exists no significant dipole in the molecule (propane, hexane)

Dipole-dipole occurs when there exist an observable amount of charge difference between atoms in a molecule. This can be seen in the case of hydrogen chloride. The chlorine atom is more electronegative, thus it attracts the more electropositive hydrogens of other molecules too (but doesn't bond with them).

The hydrogen bond is the strongest one. It occurs if a molecule has a hydrogen atom bonded to either $\ce{F}$ (fluorine), $\ce{O}$ (oxygen), or $\ce{N}$ (nitrogen). These atoms are highly electronegative. This happens in the case of water $\ce{H2O}$. I recommend you to think why the hydrogen bond is the strongest and Van der Waals (London dispersion force) is the weakest intermolecular force.

Molecular mass. Suppose you have two substances, both of which happen to interact through LDF. Another thing to consider is the mass of the molecule. If the molecule is bigger, then the surface area is larger, which results in a bigger LDF. This leads us to the conclusion that the mass of the molecule is proportional to the boiling point.

Branches. In alkanes (consisting of $\ce{C}$ and $\ce{H}$ only), generally a straight chain alkane has a higher boiling point than similar branched alkanes due to the surface area between two adjacent molecules. This would be harder if the alkane has branches. Compare: 2,3-dimethylbutane and hexane. Both have 6 carbons, but the boiling point of 2,3-dimethylbutane is $\pu{331.15 K}$ whereas hexane is $\pu{341.15 K}$.

Take a look at your case. Water, $\ce{H2O}$, includes oxygen and hydrogen, and forms hydrogen bonds. Although the difference of electronegativities of hydrogen, selenium, sulfur, and tellurium equates to $0.1$ to $\pu{0.48 units}$, it doesn't really contribute much to the overall dipole of the molecule. So these three must interact with LDF. We can see that the molecular mass of sulfur < selenium < tellurium ($\ce{S} < \ce{Se} < \ce{Te}$). From what we know, the bigger the mass, the larger the boiling point.
From this point we can tell that the trend of the boiling points is
$$ \ce{H2S} < \ce{H2Se} < \ce{H2Te} < \ce{H2O} $$
A brief consultation of Wikipedia's property boxes of compounds for example lists these values as

compound
boiling point (assuming atmospheric pressure)
see

$\ce{H2S}$
$\pu{−60 °C}$
ref

$\ce{H2Se}$
$\pu{−41.25 °C}$
ref

$\ce{H2Te}$
$\pu{−2.2 °C}$
ref

$\ce{H2O}$
$\pu{99.98 °C}$
ref

Boiling point or melting points of compounds depend on the intermolecular forces acting between them. Now here we can see hydrogen bonding is present in $\ce{H2O}$ resulting in highest boiling point among them. In the other compounds dispersion forces act between the molecules which goes on increasing as the molecular mass increases. So $\ce{H2Te}$ has strongest dispersion force so among the rest three it has high boiling point followed by $\ce{H2Se}$ and $\ce{H2S}$. Hence the order of increasing boiling point is
$$ \ce{H2S} < \ce{H2Se} < \ce{H2Te} < \ce{H2O} $$
And to answer your question $\ce{H2S}$ has the lowest boiling point.


text:
Why does copper jewelry & wire smell bad - even when I'm not wearing it?
I make jewelry by weaving a frame from copper wire for a gemstone, then making copper coils or spirals, etc for further decoration. Even 'on the roll', the copper wire smells bad - and the smell increases as I work with the wire.  The smell also clings to my hands, and I have to wash it off.  I read about the chemistry that is created when copper touches skin. But the problem is, copper jewelry that has been hanging, without being worn, has begun to smell even worse - to the point that I can't even consider hanging it around my neck. Can you give me a reason for this odor (not a chemical reaction after touching my skin), and hopefully a remedy for it? I would hate to imagine what other sensitive customers would think after having one of my jewelry pieces for a few months.  Thanks for helping.

Do you observe any change to the copper other than the smell? Does it become dark, or flaky, or does the surface roughen?

see: https://chemistry.stackexchange.com/a/7917/22102

I've never heard of jewelry made with pure copper, only as a part of alloys — and that is probably the reason as it gets corroded quickly when exposed to sweat, not to mention it's cheap, and looks and feels accordingly.


text:
How does strength of boric acid solution increase in presence of salicylic acid?
A while ago, I read that salicylic acid can make boric acid solution strongly acidic when it's added to it.
To my knowledge, I know that boric acid becomes a strong acid in presence of cis-diols (except ethylene glycol). I also know that it happens due to increased stability of conjugate base by chelation using polyhydroxy compounds.
But I can't really make out the case with salicylic acid. I will be really thankful if you could shed some light on this topic.

see: https://www.nrcresearchpress.com/doi/pdf/10.1139/v77-421

Why not ethylene glycol?

Equilibrium will be far to the right, as a stable six membered ring is formed. The proton speaks for itself.

References: Queen, A. The kinetics of the reaction of boric acid with salicylic acid. Can. J. Chem. 1977, 55 (16), 3035–3039 DOI: 10.1139/v77-421.

By googling "boric acid salicylic acid"
I have found the salicylic acid acts like it was a diol toward the boric acid, but just 1:1.
https://www.nrcresearchpress.com/doi/abs/10.1139/v77-421


text:
Influence of protic and aprotic solvents on SN reactions
How does protic and aprotic solvents influence reaction mechanism considering SN1 vs SN2?

You will find your answer here and here

As you know, $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions have different mechanisms. $\mathrm{S_N1}$ occurs in 2 steps: formation of a carbocation and addition of the nucleophile. However, $\mathrm{S_N2}$ has a one-step mechanism where the addition of the nucleophile and the elimination of the leaving group happen simultaneously.
A protic solvent would participate in hydrogen bonding with the nucleophile since it can easily share one proton. A solvent shell is thus formed around the nucleophile, making him less able to undergo a $\mathrm{S_N2}$ reaction.

I believe that a protic solvent also helps stabilise the carbocation in a Sn1 mechanism due to the dipole interactions.