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Reconstruction of NMR spectrum by entropy maximization (MaxEnt)
Currently NMR spectra are generated by Fourier transformation of free induction decay curves from time domain to frequency domain. In the late 60 s-80s some researchers proposed reconstruction of NMR spectra using a mathematical technique called "entropy maximization" from the FIDs. Fourier transformation was involved but only during intermediate steps. Just a note that this entropy maximization has nothing to with classical thermodynamics. It is a constraint optimization technique for reconstructing images with least amount of information. Consider entropy as just a mathematical function which shares the same name (thanks to Shannon) as the thermodynamic entropy.
Papers from the 1980s (e.g. Nature volume 311, pages446–447 (1984)) show that the appearance wise signal-to-noise ratio is higher in the MaxEnt reconstructed NMR spectrum as compared to standard FT method. However, in the Journal of Magnetic Resonance, this idea has been contested, as one cannot get something out of nothing. Some authors say, especially Hoch et al., that this is just a cosmetic exercise, because MaxEnt reduces the noise in the baseline but not in the peak. Therefore, the signal-to-noise ratio improves, but the sensitivity is no better. However, others say that MaxEnt is still very useful, and they contested that Hoch’s calculations were not realistic.
Very crudely, graphically the standard approach is:

but MaxEnt reconstructed NMR does the following magic as shown below. Since we have a couple of expert NMR spectroscopists here, my question is why MaxEnt technique never caught up in mainstream NMR? If it were so good at enhancing SNR, conventional FT should have become obsolete. I understand that it is a very involved calculation, but it looks too promising to an outsider.

The topic greatly outclasses my understanding, but this may be related to compressed sensing, which is a method to extract seemingly impossible S/N ratios based on information theory. In the 2000s it was very gainfully exemplified for simulated MRI scans by Emmanuel Candès and Terence Tao. I've wondered if this would somehow make its way into other applications.

@NicolauSakerNeto, I think MaxEnt is not related to compressed sensing (which is something new to me). Apparently, the MaxEnt reconstruction looks very innocent like a constraint optimization but implementation on discrete data must be a nightmare! I cannot find any simple example of MaxEnt which one can try on a simple chemical data.

Maximum entropy techniques are used in mass spectrometry data processing, usually for analysis of intact, large proteins.  The goal is to extract the mass (not the mass-to-charge) spectrum from instrumental data, which is a mass-to-charge ratio spectrum.

@NicolauSakerNeto yes - CS is one of the more popular algorithms for reconstructing NMR data: https://onlinelibrary.wiley.com/doi/10.1002/anie.201100440

The entropy in "maximum entropy" refers to the information entropy concept as introduced by Shannon. The MaxEnt technique referred to by Hoch may be regarded as cosmetic because what it does is rescale amplitudes in a nonlinear way. That is, big signals become bigger, small signals become smaller. Along the way the method seeks to avoid introducing spurious information (for instance peaks that might be interpreted as true signals) into a spectrum by assuming that noise is uniformly distributed across a spectrum and by penalizing excessive scaling of any potential signal.
The MaxEnt algorithm applies a  "regularization" technique, also used in numerical least squares optimization  to avoid issues with non-uniqueness of solutions (the "ill-defined" problem) during inversion of data. Non-uniqueness is not a problem with traditional FT-NMR because the Fourier Transform is lossless, it is a linear technique, all information originally in the time domain data is retained in the spectrum, which is one reason it is so nice, another reason including the existence of fast algorithms to compute the FT, namely the FFT, which introduced the era of modern NMR and (among other reasons) won Richard Ernst a Nobel prize.
There are a number of reasons there has been and continues to be a lot of interest in MaxEnt and related methods such as compressed sensing. These include noise reduction, but perhaps the most important reason - and importantly related to noise reduction by the $S/N \propto \sqrt{\textrm{time}}$ relation in FT-NMR - is potential savings in time. Multidimensional NMR experiments can be very time consuming. The issue of S/N that can hamper NMR can in fact be circumvented by performing long experiments, and since MD-NMR experiments are long, then low concentrations and S/N are not necessarily an issue- time is! (Sample stability of course also plays a role here). Optimization methods that combine FT with other non-linear methods can save enormous amounts of time by allowing data acquisition in the indirect dimension of MD-NMR experiments to be truncated or, even more importantly, sparsely (non-regularly or even non-linearly) sampled.
If an experiment takes on the order of days to complete using traditional NMR, but in practice S/N considerations say that you only should need a few hours, then there is a huge advantage in implementing non-linear methods that allow that. So what is the problem? The main one is probably the "black-box" approach to data processing that is implemented in many cases to reduce the complexity of applying NMR, ideally to save time and to allow researchers to focus on the question they want to answer, not on technical details such as how to implement numerical algorithms. Applying non-linear sampling methods requires additional steps to set up the experiment in advance, and to process the data and interpret the results. Those details matter if you are e.g. organic chemist looking to obtain a spectrum as quickly as possible, ideally via automation.

I should add that NMR time is very expensive. Problems with experiment duration can also be circumvented by purchasing multiple instruments. Not so easy when each costs on the order of £millions.

Right (+1), but let us focus on uniformly sampled data. This is my main concern with non-linearity of MaxEnt. If MaxEnt always change intensities in a non-linear fashion, then the reconstructed NMR intensities cannot be used for integration. What is the use then? This is what Hoch says, who is indeed a world expert in this field...however many others contest that MaxEnt is still useful. I know of a far simpler way of doing this, by a so-called power transform. It is "silently" applied in charged aerosol detectors of HPLC. You just raise the data to an integer power.

Also note the bottom figure, the relative intensities are not preserved that well after MaxEnt, esp. on the right hand side.

@AChem when it was first introduced the signal enhancement aspect of MaxEnt was emphasized, but the ability to apply it to nonlinearly sampled or truncated data is what sets it apart, typically when combined with the signal enhancement. For instance, you can sample nonlinearly where most of the signal in the time domain shows up (in the direct dimension early on, before relaxation dampens it), or simply until you feel sufficient s/n has been achieved (something that can be tricky otherwise in MD NMR).

The nonlinear issue complicates quantitation. One workaround is to inject digital signals where no signal is expected, as a calibration. In general you need to perform calibrations, which is possible with nonlinear data, but full of potential pitfalls (quantitation is generally difficult with noisy and potentially overlapping spectra).

Do you remember what software was introduced to you? I always see Cambridge algorithm everywhere. Skilling et al. were applied mathematicians. Hard to follow their paper. Basically, the problem is the non-linear nature of entropy function, the presence of natural log makes it difficult.

I first encountered MaxEnt as a routine available within a suite of advanced NMR processing tools written by Hoch and Stern (this would be the Cambridge Mass algorithm). I believe it still underlies the newer online versions (at UConn). Bruker has incorporated NUS as a standard tool into their software and there are non-standard add-ons that do not follow with Topspin written by 3rd parties, usually academics.

As someone who is a professional MR(I) physicist, I should state that compressed sensing techniques are widely used clinically in the reconstruction of proton imaging data, in which we have the luxury of two extra bits of information, related to a) obtaining data from multiple independent RF coils, rather than one probe; and b) knowledge of what "real" images look like: sparse in the Wavelet domain, and well-reconstructed by penalised total-variation terms. We acquire noisy, aliased data, and used CS to sort out aliasing. Spectra are different, but much similar work is underway.

The question here seems to be:

why MaxEnt technique never caught up in mainstream NMR?

This is not really true.
MaxEnt and compressed sensing (CS, as mentioned in Nicolau Saker Neto's comment) are just different ways of reconstructing undersampled data - the method is different but the application (in the context of NMR) is very much the same. More broadly speaking, all of these are lumped under the umbrella of non-uniform sampling (NUS).
If we're talking about NUS as  a whole, I would definitely refute the assertion that it hasn't caught on in mainstream NMR. NUS is very, very popular, and it's really a mainstream technique nowadays — virtually every practising spectroscopist will know of it.
Sure, it's not covered in the basic spectroscopy textbooks, but in practice many people use NUS, sometimes even without knowing (the data acquisition and processing is so streamlined nowadays that you wouldn't be able to tell unless you knew where to look). Part of the reason why the processing is so easy is because it's so popular: this creates demand for common software packages to implement it.
However, the main reason why it has caught on is not sensitivity.
One should bear in mind that NUS spectra are typically acquired in much shorter time compared to 'normal', uniformly sampled (US) spectra, as explained in Buck Thorn's answer. Even if the sensitivity (= SNR per unit time) of the NUS spectrum is larger, the fact that the NUS spectrum is acquired for such a short time means that its overall SNR will inevitably be lower than a US spectrum acquired for a longer time.
The real reason why NUS is hugely popular is time savings.
(The SNR per unit time of NUS vs 'normal' uniformly sampled spectra is quite an involved topic. I'm not an expert on it, so I will defer to the literature on this, see e.g Palmer, M. R.; Suiter, C. L.; Henry, G. E.; Rovnyak, J.; Hoch, J. C.; Polenova, T.; Rovnyak, D. Sensitivity of Nonuniform Sampling NMR. J. Phys. Chem. B 2015, 119 (22), 6502–6515. DOI: 10.1021/jp5126415.)

+1 Useful points here but I would like to emphasize for other readers that MaxEnt does not require non-uniformly sampled data. It works very well for uniformly sampled data as well.

One other comment is that uniformly sampled spectra reconstructed via FT have uniform gaussian noise: spectral fitting routines therefore can use techniques such as least squares to fit a model in either the time or frequency domains without worrying about bias, and integration (d delta) works because the noise averages out to 0 (in a well-phased, real mode spectrum). You can also derive Cramer-Rao lower bounds on the variance on peak parameters of interest, such as amplitude, phase or frequency/chemical shift. None of this is true in general for CS acquisitions.

Around 15 years ago I was an intern in the NMR department at a company. Someone higher up had heard about Maximum Entropy and suggested that we try a specific program to test if it is useful for us. I ended up trying this, and in the end simply could not get that software to work. It was buggy and almost entirely undocumented. These methods also tend to have a few parameters that you are supposed to set to reasonable values, which can be rather intimidating especially when they are unclear or even contradictory on what "reasonable values" are and you have no idea what the consequences of setting these values wrong are.
Applying FFT to NMR spectra is very easy, you get a reasonable result out of the box without changing any parameters. And the parameters that you can and should change are pretty straightforward to understand if you read up a bit on them.
You have to keep in mind that many, if not most people measuring NMR spectra are not NMR experts. They're synthetic chemists or other scientists that use NMR as a tool. So to be useful to these users the process of measuring spectra has to work mostly in an automated fashion with reasonably user-friendly software to process and handle the data.
As mentioned in @orthocresol's answer NUS is a somewhat common method today that doesn't use FFT to process NMR. The difference today compared to a decade or two ago is that in newer spectrometers and software this is already integrated and is easy to setup. You could already use this method earlier, but you had to set it up with separate tools and process as well with different software. That is a very large barrier, especially if you consider that any mistake you make is likely to waste hours to days of expensive measurement time.

Interesting. With my limited understanding of the actual MaxEnt protocol the reconstruction does require FFTs during regularization process. I am only focussing on uniformly sampled data. The constraint on the maximum entropy is the chi-square and then people maximize S and implement it as minimization of -S. I have asked at least a score of mathematicians/spectroscopists. Nobody knows the details except the lucky few.


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When can lone pairs participate in aromaticity?
Basically, I'm confused on when lone pairs can participate in aromaticity, by the Huckel definition. Why can lone pairs on a carboanion be aromatic? I thought carboanions are sp3 hybridized, and the lone pair is not in a p orbital. I understand that by resonance, you can bring the lone pair into the ring, but then you end up with another sp3 carboanion.
I have a similar question for oxygen and nitrogen. Why can their lone pairs be part of aromaticity if they are sp3 hybridized.

Conjugated lone pairs aren't sp3 - it makes them approx. sp2

https://chemistry.stackexchange.com/questions/4390/how-to-rationalise-the-resonance-structures-and-hybridisation-of-the-nitrogen-in

As a general rule, when counting $\pi$ electrons in a conjugated ring to assess aromaticity you need to fill the stable $\sigma$ orbitals first. That means (again, generally, maybe not always) if an atom has no ligand, it has to use a lone pair to fill that ligand orbital and only if there is a second lone pair, as with the sulfur atom in thiophene, can that second lone pair join the $\pi$ conjugation.
For instance, in pyrrole there is not this issue because every ring atom has a ligand, so the lone pair formally on the nitrogen atom is conjugated into the ring. In pyridine, by contrast, the nitrogen atom has no ligand and its lone pair has to occupy the ligand position, so all the conjugated electrons in the pyridine ring have to come from $\pi$ bonds. Unsurprisingly, in both cases the net result is just right to get six conjugated $\pi$ electrons in the ring.

Some carboanions are sp3 hybridized and some aren't.
The simplest carboanion is methide. Methane (CH4) is sp3 hybridized. Remove a proton (H+) to obtain the methide (CH3–) ion, and the central carbon is still sp3. It's just sp3 surrounded by three atoms and one lone pair instead of four atoms. The same thinking would apply to a cyclohexane ring. Whether an H+ is removed to obtain a cyclohexanide ion or or not, the carbons are all sp3 hybridized.
Molecules that obey Hückel's rule for π electrons will have carbon atoms that are sp2 or sp hybridized and aromatic. They are still aromatic after an H+ is removed to obtain a carboanion. Therefore, the benzenide ion (C6H5-) is aromatic.

With the benzenide anion, you cannot bring the lone pair into the ring. I'm almost sure the question isn't about whether an aromatic ring retains its aromaticity, but how an anion can make it become aromatic. But then again the question leaves quite a bit room for interpretation.

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How do I find the minimum amount needed for precipitation?
I want to calculate the minimum amount of solid barium nitrate (g) needed to be added to $\pu{50mL \text{of} 0.05mol/L}$ sodium hydroxide in order to cause precipitation.
My attempt:
$$\ce{2NaOH(aq) + Ba(NO3)2(aq) -> 2NaNO3(aq) + Ba(OH)2(s)} $$
The salt that is going to precipitate is going to be barium hydroxide, which has a $\ce{Ksp = 2.55\times10^{-4}}$. Letting the molar solubility be $s$, I now have $s(2s)^2=2.55\times10^{-4}\Rightarrow s=0.039947\ce{mol/L}$.
$\therefore \ce{ In 50 mL}$, we need $\ce{0.001997 mols}$ for precipitation to occur. However, from the stoichiometry, that means we need $\ce{0.004 mols}$ of $\ce{NaOH}$, whereas there is only $\ce{0.0025 mols}$ to begin with.
So my question is where did my logic go wrong, am I supposed to use the molar solubility at all for this question?
For context, this was a multiple-choice question with options $A=0.339\text{g}, B=0.678\text{g}, C=1.38\text{g} \text{ and } D=2.77\text{g}$.

Do not expect the ratio of concentrations of precipitated ions would be stoichiometric. What matters is the equation for Ksp. [Ba^2+]=Ksp/[OH-]^2

@Poutnik If I have $\ce{[OH-] = 0.025}$, that gives me $\ce{[Ba^{2+}] = 40.8}$ if I plug it into the Ksp which seems way too big so I'm not sure what exactly you mean by what matters is the equation for Ksp, could you explain a little more?

$\frac{\pu{2.55E-4}}{0.05^2}=0.102$

The mistake you made when calculating molar solubility $s$ is that you assumed $\ce{Ba(OH)2}$ was dissociating in pure water, when the truth is it was dissociating in a solution of $\ce{NaOH}$.
Since $\ce{OH-}$ is a common ion between $\ce{Ba(OH)2}$ and $\ce{NaOH}$, it will reduce the amount of $\ce{Ba(OH)2}$ that will dissolve compared to a solution of pure water.
In this case, the molar solubility of $\ce{Ba(OH)2}$ would be calculated using this equation:
$$K_{sp}=\ce{[Ba^{2+}]\;[OH^{-}]^2 = (s)\;(0.05+2s)^2}=\;2.55\;*\;10^{-4}$$
Solving for $s$:
$$s=\pu{0.02525 mol / L}$$
Finally, mass of $\ce{Ba(NO3)2}$ is calculated by:
$$m_{\ce{Ba(NO3)2}}=s\;VM_{\ce{Ba(NO3)2}}=\left(\pu{0.02525 mol/L}\right)\left(\pu{0.05 L}\right)\left(\pu{262 g/mol}\right)=\pu{0.331 g}$$

Thank you, I think the value of s in your answer was typed into the calculator incorrectly but otherwise thanks for spotting the error

You do not dissolve Ba(OH)2. There are no extra OH- from such a dissolution. You dissolve Ba(NO3)2. So Ksp(Ba(OH)2)=[Ba(NO3)2]. (0.05)^2.

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What is the relationship of the following compounds?
What is the relationship of these following compounds? My guess is they are constitutional isomers. However, a couple of people said that they are enantiomers of each other. Which one is correct?

They are enantiomers. It may be difficult to see this in the way you drew it, but if you drew the structures in a proper decagon form, it may be more clear.
Think about it this way. If you were not given the stereoisomerism of the methoxy group, you would think they would be the same molecule by counting the carbons in each direction (you'll notice that methyl and methoxy groups are on carbon 6 and carbon 1 respectively).
Now, if you consider R or S configuration for the methoxy group, you'll see that they are enantiomers. This becomes clear when you apply Cahn–Ingold–Prelog priority rules to the carbon with the methoxy group.


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What other acids can replace hydrochloric acid in the production of galvanized steel?
Galvanized Steel undergoes several processes before being lowered into a molten zinc bath. One is cleaning the steel with "an acidic solution to remove mill scale."
After some more research I found that hydrochloric acid is the most popular choice, because it is so strong and therefore fast.
What I want to know is, can any acid theoretically serve the exact same purpose? And then, how to calculate how much longer it would take to achieve the desired effect.
Believe it or not, the most interesting acid that I'm interested in right now is acetic acid (vinegar). This is because I have seen many videos demonstrating that rust can be removed by soaking the iron in tub of vinegar (and sometimes salt). So I am guessing that it can also remove mill scale.
How would I calculate the time needed from a chemical equation?

Mill scale ( FeO , Fe3O4) is not rust, and is more tenaceaous, so remedies for rust will not necessarily remove it. Sulfuric is cheapest acid for mill scale removal. I assume hydrochloric is used because it has some advantage for galvanizing.

Sulfuric acid has the added advantage that it's not volatile and so can be used in open baths at elevated concentration without quickly corroding everything in the vincinity.


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3-Step Transformation to Humulone

I have some ideas on how this transformation may occur. We're currently studying Ch. 3 (Polar Rxn's under Acidic Conditions) from Grossman's book.
I've found some data online that shows we can perform a standard Friedal-Crafts acylation using -ClCOCH2CH(CH3)2 and AlCl3 to add the substituted ketone alkyl chain.
The next steps I can think of are prenylation using BrCH2CHCH(CH3)2 to add to both of the remaining unsubstituted benzene positions. I've seen data using this bromo-prenyl compound and KOH but our chapter is on reactions under acidic conditions so I don't know if this will fit the theme.
Alternatively, could we begin by nitration with conversion to an alcohol group, making 1,3,4,5-tetrahydroxybenzene, then acylate, and then perform 2 successive alkylations with ferric bromide to add both prenyl groups?
This has been challenging for me to get done in 3 steps. Our last few lectures covered the Prins, Sakurai, Bischler-Napieralski, Vilsmeir-Haack, and just general rearrangement, SN1, SN2, E1, and E2 reactions.

When you include a reference, please make it a proper citation. Could you expand how that is involved with beer, please.

Hey Martin, I have a link to a patent that I've found that involves some of this work: https://patentimages.storage.googleapis.com/f4/ef/3b/36b82a35dd04a8/US9340479.pdf I also found a paper on this using 1,3,4,5-tetrahydroxybenzene though: https://www.journal.csj.jp/doi/epdf/10.1246/bcsj.62.3034

The chemistry on my assignment has to do with compounds found in beer, mainly alpha-hops (humulone) and iso-alpha-hops (iso-humulones). I ultimately need to show an acid catalyzed rearrangement from humulone to the 5-membered iso-humulone molecule but I feel comfortable doing that part so I didn't include it. The alpha-hops convert to the iso-alpha-hops during the "boil process" when beer is produced.

Here is a link to a 2 step transformation relating to the second paper in my original comment: https://tinyurl.com/2fkhfw5m

@PhDeezNutz OP is probably referring to that reaction. However, it involves a base (NaH) and OP wants to get around that and use acid solely I believe.

I have an idea for a 4-step mechanism but I can't get down to 3. I've seen DMP oxidize phenols into adjacent ketones on an aromatic system, and maybe it can do the same here. Then you can do grignard. All before that, you would have done Freidal's Craft selectively for the carbons not connected to an oxygen.

How do we know that three steps are possible?  Is the goal to find a synthetic pathway, or to describe the biosynthesis of humulone?  I'm very unclear what you are actually asking for in this question.


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What is the code in Harmonized System for NGL, LPG and LNG?
I am trying to research trade statistics related to oil and gas using BACI Database. But I am confused about the product codes I should use.
I read an article explaining the difference between Natural Gas Liquid, Liquefied Petroleum Gas, and Liquefied Natural Gas.
Can anyone help with the corresponding HS code?


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How to determine the characters of px atomic orbital in C4v point group?
How to determine the characters of px atomic orbital in $\text{C}_{\text{4v}}$ point group? I understand (px,py) is a set of basis function of E, and px and py are mixed by $\text{C}_4$ and $\sigma_{\text{d}}$. However, based on the figure below, how can I determine the characters of only px orbital? Are the characters the same as those of E?

Operate on the orbital with each of the symmetry operations, $\sigma_v$ etc and compare them with the irreps for each symmetry species in the point group, count 1 for unchanged , -1 if changed,  see https://molecule-viewer.com/ for many examples. A few molecules in each point group also have orbitals included.

@porphyrin Yes, I understand this, but for the px orbital in the figure, $\sigma_\text{d}$ will transform it into py orbital, does this count as zero? Also for the two $\sigma_\text{v}$ operations, one character is 1 and the other is -1, so for the overall $2\sigma_\text{v}$ operation in the character table, does it count as 0?

Yes zero, also look at the third column in a point group, see entries for x, y, z and $R_x $ etc so you can check your answers.

@porphyrin Thanks. However, for $C_{4v}$ point group $p_x$ is a basis function of the doubly degenerate irreducible representation, is it correct to consider how only one of the basis functions transform? Since the calculators to decompose reducible representatons just return error, if I consider $p_x$ as a reducible representation.

see the last part of this answer for a worked example. https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613 , a fuller example is given by R. Carter, chapter 3 'Molecular Symmetry and group theory' publ Wiley.

Since the question has received 0 answers or votes here, I wonder if you considered asking it at MMSE?


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Nucleophilicity order among the following?
This question is irritating me up. In the given question, the -ve charge on O atom in (1) and (4) is delocalised, in (4) the charge is more delocalised then in (1) due to more resonating structures. Therefore, the (4) is least nucleophilic among the following. Then comes the (1) option. In (2) and (3) the charges are localised and hence they have higher nucleophlicity then 1 and 4. But among (2) and (3), in (2) although the O atom is more electronegative, there is a CH3 group showing +I effect, also the atom is sp3 hybridised, the less the %s character the less stable with the -ve charge on the atom is. On the other hand, in (3) there is no +I group and also the N atom is sp hybridised, %s character more hence more stable the atom is in holding the -ve charge. So the order of nucleophilicity must be 2>3>1>4. But the answer says 3>2>1>4. Help me out and please let me know where I am missing.

The negative charge in $(3)$ is not on the nitrogen atom, it's on the carbon atom and is therefore a carbanion.
Carbanions (carbon atoms with a formal negative charge) are very unstable and capable of easily donating their lone pair of electrons because of carbon's relatively low electronegativity.
In this case, since the carbanion in $(3)$ is competing with the oxyanion in $(2)$, the least electronegative species of the two will donate a lone pair of electrons more easily and therefore work as a stronger nucleophile.
Since carbon is less electronegative than oxygen, $\ce{^-CN}$ is a stronger nucleophile than $\ce{CH3O-}$

Thanks for the answer. Had I knew the charge is on the Carbon atom and not on Nitrogen, there would not have been such a doubt!


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Redox reactions: In order to get as much copper as possible from a reaction, one usually adds an excess of zinc. Why?
In order to get as much copper as possible from a reaction, one usually adds an excess of zinc. Why? We already have a set amount of zinc needed to reduce all the copper(II) ions. Why should we add an excess of zinc? Is it to make sure that there's enough zinc to react with the copper, or will there be a loss of electrons during the reaction?

If stoichiometric ratios are used, the kinetic rate of reactions rapidly drops toward the end . With excess of some of reactants, the drop is much smaller.

An excess of zinc is often added to a solution of copper ions. This is to make sure that there is enough zinc to react with copper, because, quite often, part of the metallic zinc used is already oxidized into zinc oxide, specially when metallic zinc is used in powder. Zinc oxide is unable to reduce copper ions.


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Why do the two layers of a phospholipid bilayer stick together?
Every answer I can find about why phospholipids form a bilayer says the same thing: hydrophilic heads associate with water, hydrophobic tails do not, so they arrange themselves with heads facing outward. Makes sense.
What I have not been able to find an explanation for is why the two layers of a bilayer sheet stick together in a cell. Why aren't the phospholipids of the upper layer pulled away by passing water molecules, like how salt ions are pulled away from a salt crystal? Why doesn't the same thing happen in the bottom layer? And what keeps the two layers bonded when the fatty acid tails have (in my understanding) little to no attraction to each other?

You may want to read about the hydrophobic effect. Also, forces at the molecular level, and the behavior of matter generally, is very different from what we are familiar with in our macroscopic world. The microscopic world of condensed matter is viscous and motion is dominated by diffusion. Of course, once biological systems get involved things can get more complicated.

Also, that picture is a cartoon representation. Things are far more disorderly.


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Is the number of particles with opposite spin roughly equal?
I was reading about the Pauli exclusion principle and this question occured to me about the electron. It can be +1/2 or -1/2 spin. And usually atoms outside stars (where it is cool enough to attach electrons) seem to have their orbitals filled with enough electrons to appear neutral. So I assume there are similar number of electrons with +1/2 and -1/2 spin, and they are quite evenly mixed.
Is this true/generally true?
Is there no assymetry like in matter antimatter?

Electrons have two spin quantum numbers, $S=1/2$ and the second is the azimuthal/projection/magnetic quantum number with $s_m=\pm 1/2$ . The electron with spin $S=1/2, s_m=1/2$ is the same as that with $S=1/2, s_m=-1/2$  unless you put the electron (or other spin particle say a proton) in a magnetic field then the energy levels split and one rises and one falls in energy by an equal amount but now by Boltzmann distribution the numbers of $s_m=1/2$  is no longer equal to that with $-1/2$. This forms the basis of EPR spectroscopy and NMR with protons etc.

Spin of electron as a particle is 1/2, no plus or minus. These signs are only to symbolise that these moments are opposite if two electrons "are in" one orbital.

You may want to check at physics SE for similar questions, in particular whether there are asymmetrical magnetic fields on a universal scale, and why not.

One of the weird things of quantum mechanics is that most properties do not have defined values except when they are measured. Thus most electrons have neither +1/2 or -1/2 spin but rather exist in a superposition of both until we measure them.


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Line Entry system like SMILES for metals?
SMILES is not designed for metallic compounds.
Is there a similar ASCII line entry notation for metallic compounds, including repeating structure in space, &c.?

Representing organometallics by ASCII is notoriously difficult.  Neither InChI nor SMILES does it well.
The most common approach that I've seen is to shoehorn organometallics into the SMILES system by adopting various conventions that don't represent the structure of the represented compounds very well, but do at least work (sort of) in SMILES.  See here for one example.
Folks working on InChI have faced similar problems.  Here's a slide deck from an InChI working group trying to develop principles for representing organometallics in InChI.

Alex Clark is the developer currently working on InChI for organometallics. The latest I can find is a progress update from 2019. IIRC there's been much more recent progress, but there's been no release yet.

@CurtF 1) Wasn't the question about metals rather than organometallics?

Though -- from the side of a potential user -- there is little advance visible, the most recent progress report about InChI on organometallics neither is by 2017 (current reference in the answer, or 2019 (comment by Geoffrey Hutchison), but April 2022 during the remote InChI days by InChI trust (https://www.inchi-trust.org/decks-open-inchi-days-2022/) by Hinnerk Rey and Sonja Herres-Pwalis.  Pointer to the slides deck (a .pdf): https://drive.google.com/file/d/1YayfltZVZDYehU54U8Ccot5WCaBl8Xvy/view?usp=sharing

@Buttonwood I think non-organometallics, or at least simple metallic compounds like iron(III) chloride, are represented just fine in SMILES (and InChI), keeping in mind that neither SMILES nor InChI records arrangement in space, just "connectivity".


text:
Does stoichiometric gas mixture produce the highest explosion pressure (or energy) compared with other explosive limit mixture?
I understand that a stoichiometric mixture contains a balanced mixture of air and fuel and both fuel and air are completed consumed. For example, the ideal stoichiometric mixture for propane is approximately 4.01% of the fuel which is equivalent of 23.91:1 ARF by mass.
I also understand each fuel has its own explosive limit. As an example of Propane, the lower explosive limit is 2.1% (lean mixture) and upper explosive limit is 9.5% (rich mixture).
The explosive pressure (or energy) will be larger with 4.01% of fuel mixture compared with 2.1% of fuel mixture assuming the mixture is in a defined container with no fuel saturation. Following this pattern, if the gas mixture reaches to its rich state, fuel level at 9.5%, would the explosive pressure (or energy) be greater than the stoichiometric mixture? Can this be calculated?

Do you have a way of knowing/measuring %propane before reaction happens and %propane after reaction happens?

@Sam202 I can calculate the %propane based on O2 analyzer before reaction because it's fuel mixing with air. After combustion, it might be difficult because of the product of CO, CO2, H2O and etc.

Do you know the initial mass or amounts (moles) of each gas (oxygen, nitrogen, propane) before reaction?

@Sam202 Yes, I can calculate or estimate nitrogen and propane based on the oxygen level measured.

Yes, the stoichiometric mixture is the most energetic and creates the highest pressure.
If the mix is not stoichiometric, then some of the energy of the explosion goes into heating up and expanding the unreacted component, whether that is fuel or air.
To use your example, the combustion of propane can be represented as
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$
This creates 7 moles of gas (three of $\ce{CO2}$ and four of $\ce{H2O}$) for each 6 moles of gas that react (five of $\ce{O2}$ and one of propane).  Propane has a standard heat of combustion of -2220 kJ/mol.  So if a stoichiometric mix is present, each mole of propane that combusts will release 2220 kJ of energy into the $\ce{CO2}$ and the $\ce{H2O}$ that is formed.
If a non-stoichiometric mix is used, say the propane is double (or the oxygen is half) of the stoichiometric mix, then the 2220 kJ released from burning one mole of propane will be released to (a) the 3 moles of $\ce{CO2}$ and the 4 moles $\ce{H2O}$ that is formed, but also to (b) the unreacted 1 mole of propane.  When the energy of combustion must be "shared" by more moles of gas, the resulting temperature and thus pressure increases from combustion/explosion are less intense.
If the mixture were lean, i.e. with excess oxygen instead of propane, the unreacted oxygen would absorb some of the energy, with a similar decrease in obtained temperature and pressure.

Thanks for your detailed explanation. I understand the chemistry and math side of your explanation. In order word, it is not a linear relation if I plot the graph of explosive pressure against gas mixture from LEL to UEL. I believe it is more of a bell curve where the maximum explosive pressure is measured at the peak of the bell curve which is close to stoichiometric mixture.

@WilliamChao, yes, exactly.  For similar reasons, the temperature and pressure released during combustion with air is much lower than when using a stoichiometric mix of fuel and pure oxygen.  With air, a lot of the energy has to go into heating up and pressurizing the 79% of air that is nitrogen gas.  Here is a somewhat related and definitely entertaining demonstration of a similar concept.

Great video of demonstrating the explosive energy between pure nitrogen and 2/3 nitrogen with 1/3 oxygen adding into the system. Thanks for finding that video. Is there any other literature you recommend that I can read more about?

This is perfectly true in theory, assuming perfect mixing and complete reactions to equilibrium (so that even the very last propane molecule happens to encounter the last few oxygen molecules in time). In practice, the observed highest-energy mixture can be slightly non-stoichiometric if that helps avoid early terminations and incomplete combustion. Basically, it's a balance of two factors: excess unreacted molecules soaking up energy and local lack of reactants causing incomplete combustion.

We were using two different methods for mixing the mixture: natural convection and forced convection with a fan. Since propane is heavier than air, it saturates on the bottom of the container which is where the sparker is located. This means there could be three different levels of mixture: rich mixture on the bottom, balanced mixture in the middle, and lean mixture on the top. It will reach to the point where there is no combustion since the mixture is too rich. Could be interesting to find out the mixture where the sparker is located.

No, slightly rich mixtures of propane and air have the highest explosive pressure as reported in DEFLAGRATION PARAMETERS OF PROPANE–AIR MIXTURES IN A CLOSED CYLINDRICAL VESSEL U.P.B. Sci. Bull., Series B, Vol. 73, Issue 3, pages 17-26 (2011).
Stoichiometric is 4.02% by volume propane as you say, while 4.70% is found to give the highest pressure.
This reference considers many different reaction products, not just the main reaction products.  Solid graphite, and gaseous propane, CO2, CO, H2O, O2, N2, CH4, C2H2, C2H4, H2, NO, H, OH and O are all considered.
See also Laminar burning velocity and explosion index of LPG–air
and propane–air mixtures Fuel 87 (2008) 39–57 which explains:

the maximum rate of flame propagation occurs on the rich side since additional fuel is needed to compensate for the effect of dissociation at higher temperature

Assuming you do not want incomplete combustion of propane leading to the formation of poisonous $\ce{CO}$ gas, then propane would be the limiting reagent outside of stoichiometric conditions, according to the reaction:
$$\ce{C3H8 +5O2 -> 3CO2 +4H2O }$$
If you want to calculate how much energy (heat) is released as a function of the initial molar fraction of propane, we will consider the following:
Let:
$A$ represent propane.
$y_{Ao}$ = initial molar fraction of propane.
$n_{Ao}$ = initial moles of propane. [mol]
$n_{o}$ = initial total moles of gas. [mol]
$n$ = final total moles of gas. [mol]
$Q$ = heat released by combustion reaction. [kJ]
$\Delta\overline{H}^{\;°}$ = standard molar enthalpy of propane. [kJ/mol]
$X_A$ = conversion of propane.
The heat released by combustion of propane can be calculated by:
$$Q=n_{Ao}\;X_A\;\Delta\overline{H}^{\;o}$$
Assuming combustion is complete since oxygen is in excess, $X_A=1$, so we have:
$$Q=n_{Ao}\;\Delta\overline{H}^{\;o}$$
Since we're dealing with a reaction in gas phase, and the amount of gas molecules on the products side is higher than on the reactant side, the total final moles $n$ will be higher than the total initial moles $n_o$:
$$n=n_o(1+\epsilon)$$
We can calculate $\epsilon$ with:
$$\epsilon=\frac{\Delta n}{a}\;y_{Ao}=\frac{1}{1}\;y_{Ao}=y_{Ao}$$
Dividing both sides of the second equation by $n$, we get:
$$\frac{Q}{n}=\frac{n_{Ao}}{n_o(1+y_{Ao})}\;\Delta\overline{H}^{\;°}$$
Or equivalently:
$$\frac{Q}{n}=\frac{y_{Ao}}{1+y_{Ao}}\;\Delta\overline{H}^{\;°}$$
Finally, for propane: $\Delta\overline{H}^{\;°}=\;$-2043 kJ/mol
If we recognize that heat will be released to the surroundings , and we're only interested in the amount of heat produced per total moles of final mixture, we can ignore the negative sign, and we get:
$$\frac{Q}{n}=2043\frac{y_{Ao}}{1+y_{Ao}}$$
However, it's important to note that this equation is only valid for the following interval (since propane is only flammable within a certain composition range, and we assumed it was the limiting reagent):
$$y_{AL}\leq y_{Ao}\leq y_{AS}$$
Where $L$ represents the lower flammability limit composition of propane, and $S$ represents the stoichiometric composition of propane.
Substituting the values:
$$0.021\leq y_{Ao}\leq 0.0401$$
Conversely, when oxygen is the limiting reagent, an incomplete combustion of propane takes place, forming $\ce{CO}$ in addition to $\ce{CO2}$:
$$\ce{C3H8 +4O2 -> 2CO + CO2 +4H2O }$$
Since reaction is incomplete, conversion of propane is less than 1:
$$X_A<<1$$
$\epsilon$ for this reaction is different and calculated by:
$$\epsilon=\frac{\Delta n}{a}\;y_{Ao}=\frac{2}{1}\;y_{Ao}=2\;y_{Ao}$$
So $\frac{Q}{n}$ for this reaction is:
$$\frac{Q}{n}=\frac{y_{Ao}}{1+2\;X_A\;y_{Ao}}\;X_A\;\Delta\overline{H}^{\;°}$$
For this reaction, $\Delta\overline{H}^{\;°}=\;$ -1587 kJ/mol
So we have:
$$\frac{Q}{n}=1587\;X_A\;\frac{y_{Ao}}{1+2\;X_A\;y_{Ao}}$$
Which is valid for the interval:
$$y_{AS}< y_{Ao}\leq y_{AU}$$
$$0.0401< y_{Ao}\leq 0.095$$
If we assume only half of propane is converted into products, then $X_A=0.5$, and:
$$\frac{Q}{n}=793.5\;\frac{y_{Ao}}{1+y_{Ao}}$$
If you plot $\frac{Q}{n}$ vs $y_{Ao}$ for both reactions, while respecting the domain of $y_{Ao}$ in each case, you will notice the amount of heat produced per total moles of gas reaches its maximum value when stoichiometric composition is used:
$$y_{Ao}=y_{AS}\implies\frac{Q}{n}=\left(\frac{Q}{n}\right)_{max}$$

Thank you for your derivation on calculating the heat released. I did an spreadsheet calculating the heat released on in each case (before and after stoichiometric mixture). With mole fraction of 0.0218, the heat released is 43.63 kJ. With mole fraction of 0.0401, the heat released is 78.85 kJ. With mole fraction of 0.0947, the heat released is 68.67 kJ. Does this mean that the highest energy released does not necessary need to be at the stoichiometric mixture? I thought the the mole fraction after 0.0401 will decrease as mole fraction increases.

You're welcome. Firstly, it's important to note that those heat values you calculated are in kJ/mol, not kJ. If you want kJ, you need to know how many total moles of gas $n_o$ you had before propane combustion. I did not understand your last sentence, since it seems contradictory.

Thank you for for the correction and apology for the confusion. I have plotted the Heat Released per mole versus Initial mole fraction of propane for both reactions. Please see link here. link. I understand two linear lines on the graph since it's two different reactions (completed and incomplete combustion). I thought the graph will be more of a bell curve shape.

The functions of $\frac{Q}{n}$ vs $y_{Ao}$ are not linear, they're inverse functions. The "true" theoretical curve would be similar to what you're describing (bell with a maximum heat release per mole at stoichiometric composition), but that would be harder to obtain since you'd have to consider that molar enthalpy $\Delta \overline{H}$ is also a function of $y_{Ao}$, as formation of CO gas will alter gas composition and lower the amount of heat released per mol of total gas.

Thank you for the clarification. I thought it's a simple plotting two functions on a same graph. I am not familiar with molar enthalpy as a function of molar fraction and maybe that's a topic for another day. This will give me a general idea of the amount of heat released at stoichiometric composition.


text:
NMR Assignment of cinnamaldehyde
I'm struggling to understand the assignment of the aromatic protons in this compound. Based on the simulated spectra (and lit spectra), the three protons on the left are all overlapping. To the best of my knowledge, you can draw resonance structures with both carbocations or carbanions at the ortho and para positions, so I don't quite get why the meta protons will end up with nearly identical shifts as there's no similar resonance effect (unless I'm missing it).

The spectrum belongs to the actual compound and the checmical shifts depend on local electron density distribution. The ionic resonance structures you can draw are far from reality, unless the aromatic system breaks up, i.e. in a reaction. Why do you expect the distortion of the aromatic ring system by the substituent to be very strong?

Because the electron density distribution can be approximated via resonance structure and chemical shifts can be estimated based on the ortho/meta/para directing nature of substituents as described in every modern organic chemistry textbook.

Drawing resonance structures does not give you any quantitative information how strong the effect on the electron distribution and therefore chemical shift will be. ?! And of course the ortho/meta/para directing nature of substituents correlates somewhat with the chemical shifts increments, but one is at least partially a property of the sigma complex , the other a property of the undisturbed aromatic ring. Pears and apples ... ;)

The system of chemical shift increments is very well worked out (since many decades), and the prediction you get is very likely correct (you mention yourself that literature values agree with it). So obviously the attempt to explain the CS with the resonance structures doesn't carry very far. If modern books (which?) recommend that, phooi.


text:
Is there a general single term for benzylic-like positions in non-benzene aromatic rings?
This is a simple vocabulary question.  Is there a single chemical term in general use that generalizes the concept of "benzyl" to aromatic but not necessarily benzene rings?
For example, in the case of phenyl or benzene rings, people talk about "benzylamine", the "benzylic position", "benzyl bromide", etc.
To be more general about aromatic rings, people talk about "arenes", and this would include not just benzene rings like in toluene or naphthalene, but also aromatic rings like pyrroles, isoxazoles, thiophenes, furans, pyridines, etc.
So what is the "benzylic position" of a general aromatic ring called? Is there a more general term for the "benzylic" position in compounds like 3-aminomethylpyridine, furfuryl alcohol, 3-hydroxymethylpyrrole?  Can I still call it "benzylic"?  Or is there a different word?

(Note: I'm not really interested in an official IUPAC-designated term here, unless that term also happens to be in general use in the chemical community.)

Well, it's not benzylic... couldn't it be just called an alpha position respective the ring?

Since the rings involved are annulenes, "annulylic" maybe?

@OscarLanzi "Annulyl" would be rather an aromatic group connected directly like phenyl not benzyl. There's for eg. furyl, tolyl, etc. but they're aryl not arylmethyl groups like benzyl.

@Mithoron Maybe I just need the word "arylmethyl"?

https://books.google.com/ngrams/graph?content=arylmethyl

Based on the suggestion from @Mithoron it seems like "arylmethyl", and maybe the more specific "arylmethylene" could be contenders.

It appears that aryl methyl is occasionally used. For example, there are references to aryl methyl ketones, which is basically a benzylic ketone if a benzene group is involved [1][2].
However, this 'aryl methyl' reference can also refer to the nomenclature of ethers and sulfides (i.e. aryl methyl ether)
Sources:
[1] https://cen.acs.org/articles/92/i4/Simple-Method-Aryl-Methyl-Ketones.html
[2] https://pubs.acs.org/doi/full/10.1021/jo062483g

Do you prefer "aryl methyl" with a space, or "arylmethyl" with no space?

I would personally prefer aryl methyl for clarity, but it isn't really something I can say as a fact. It isn't a very widely used term, I only found several references to it in published works.


text:
Defining SMARTS reaction template
I want to have a reaction template involves combination of 2 radicals such as this one O=C1CCC(=O)[N]1.[H]>>[H]N1C(=O)C([H])([H])C([H])([H])C1=O.
A generic reaction template would involve [N]  (with 1 radical electron and 2 unpaired electrons) attached to any 2 groups (or in a ring) and a radical [H] and these 2 get combined. So I tried something simple like this *-[N]-*.[H]>>*-[NH]-*  as a SMARTS template but I didn't get what I want.
I would really appreciate it if you could help with this specific example and perhaps also provide some info (links or keywords) regarding this topic! (I have checked out the Daylight website https://www.daylight.com/cheminformatics/ but it's not clear at all)
Code sample
from rdkit import Chem

first_radical = 'O=C1CCC(=O)[N]1'
second_radical = '[H]'

rxn = Chem.AllChem.ReactionFromSmarts('*-[N]-*.[H]>>*-[NH]-*')

ps = rxn.RunReactants((Chem.MolFromSmiles(first_radical),Chem.MolFromSmiles(second_radical)))
print(Chem.MolToSmiles(ps[0][0]))

Desired result: [H]N1C(=O)C([H])([H])C([H])([H])C1=O

I'm not sure exactly what you want to know. What do you mean by "I didn't get what I want"?

Hi @GeoffHutchison I've updated the question to be more specific!

I'm not an expert at RunReactants but I notice that you're not defining atom mappings. If you look at examples, e.g., the "thiazole synthesis" you'll see that you need to give atom classes in both the reactants and products to indicate which atoms go where. It doesn't just "guess" that your [H] radical goes into your product. You have to tell it.


text:
Find the added volume of HCl so that the ions separate

Suppose we have a 10mL solution that contains 0.02M $\ce{Ag^+}$ and 0.04M $\ce{Ba^{2+}}$. a) Find the necessary volume to be added of $\ce{HCl}$ to separate all the silver. (Assume to be separated if there is less than 0'1% of remaining silver in the solution of what it was initially)

So here I tried to do it using the constant of solubility that is given to be $Ks(AgCl)=1.8*10^{-10}$ noting that that new concentration of silver has to be 0'1% of the initial one, which means to me $[Ag^+]'=\frac{0.1}{100}[Ag^+]_0$ hence using the constant: $$Ks(AgCl)=[Ag^+]'[Cl^-]'=[Ag^+]'(s+c_{\text{added}})$$ because we have added some quantity to the solubility of $\ce{Cl-}$, and then since the constant is of the order of $10^{-10}$ we can approximate to $s\approx 0$: $$Ks(AgCl)=0.001[Ag^+]_0\cdot c_{\text{added}}\implies c_{\text{added}}=\frac{Ks(AgCl)}{0.001[Ag^+]_0}$$ but I don't think that's right since I don't get the answer that my books suggests.

You have found how much Cl- should stay in the solution to achieve the desired result. That's not what the problem asks, though. It asks how much Cl- should be added, and not all of that Cl- would stay in the solution. Some would go down with AgCl.

@Ivan Neretin But the thing is that I don't know how I would do that

is concentration of HCl given?

This is an equilibrium shift problem in which adding $\ce{HCl}$ produces two different effects simultaneously:
(1) Change in total volume of solution.
(2) Change in moles of $\ce{Cl-}$ ions present in solution.
We'll start by defining equations for the equilibrium number of moles of each relevant species in the equilibrium system:
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
Let:
$S$ represent $\ce{Ag+}$
$C$ represent $\ce{Cl-}$
$A$ represent $\ce{HCl}$
At equilibrium, (after HCl is added) the moles of silver and chloride ions are, respectively:
$$N_{\ce{S}}=N_{So}-x$$
$$N_{\ce{C}}=N_{Co}-x$$
We're told that at desired separation:
$$N_S=0.001\;N_{So}$$
Substituting above and solving for x:
$$0.001\; N_{So}=N_{So}-x\;\;\implies\;\; x=0.999\;N_{So}$$
Substituting in $N_C$:
$$N_C=N_{Co}-0.999\;N_{So}$$
We then use the relationship between $K_N$ and $K_{sp}$ to form this expression:
$$K_{N}=K_{sp}\;(V_S+V_A)^2=N_S\;N_C=(0.001\;N_{So})(N_{Co}-0.999\;N_{So})$$
Where $V_S$ is the volume of the solution before $\ce{HCl}$ is added, and $V_A$ is the volume of $\ce{HCl}$ added.
Since $\ce{HCl}$ is a strong acid, we can assume that it's dissociated completely and:
$$N_{Ao}=N_{Co}$$
$$V_A=\frac{N_{Ao}}{C_{Ao}}$$
So, we have:
$$\left(\frac{N_{Ao}}{C_{Ao}}+V_S\right)^2=\frac{(0.001\;N_{So})(N_{Ao}-0.999\;N_{So})}{K_{sp}}$$
Plugging in all the known values we're given:
$N_{So}=2*10^{-4}\;mol$
$K_{sp}=1.8*10^{-10}$
$V_S=0.01\;L$
$$\left(\frac{N_{Ao}}{C_{Ao}}+0.01\right)^2=\frac{(2\times10^{-7})(N_{Ao}-1.998\times10^{-4})}{1.8\times10^{-10}}$$
It's important to note that the volume of $\ce{HCl}$ required depends on how concentrated it is, so $C_{Ao}$ should be given.
For example, if we have $\ce{HCl}$ 0.01$\;$M, then we would get when solving for $N_{Ao}$:
$$N_{Ao}=2.006\times10^{-4}\;mol$$
And finally:
$$V_A=\frac{N_{Ao}}{C_{Ao}}=\frac{2.006\times10^{-4}\;mol}{0.01\;mol/L}=0.02006\;L=20.06\;mL$$
If you would like to observe how $V_A$ varies as a function of $C_{Ao}$, you can plot:
$$V_A=\frac{2000\;C_{Ao}-0.036\;-\sqrt{\left(0.036-2000\;C_{Ao}\right)^2-2.878416}}{3.6}$$
Note that $C_{Ao}$ is in mol/L, and resulting $V_A$ value will be in liters (L).
A plot of $V_A$ (mL) vs $C_{Ao}$ (mol/L) from 0.01M to 0.1M, would look like this:

I've noticed that you've written quite a few MathJax heavy posts recently. Please look into the mhchem package, it'll make typesetting easier and it'll produce proper typesetting. There is more information on our meta site.

@Martin-マーチン Thank you. I will look it up.

BaCl2 is water soluble.  AgCl is not.  Adding HCl to the solution of 0.02M Ag+ and 0.04M Ba+ will produce an AgCl precipitate.
There are 10 ml of original solution.
10 ml × 0.02 mmol/ml Ag+ = 0.20 mmol (millimoles) Ag+
Therefor 0.20 millimoles HCl should be added to get at least 99.9% separation of AgCl.
Interestingly, with AgCl, adding excess chloride will actually increase the solubility of AgCl, so care must be taken to not exceed the proper amount.
1.0 ml of 0.2 M HCl will drop most of the silver as AgCl ppt.

mM is not the proper way to write millimole; most chemists will read it as millimolar, which is mmol/L, and there you also see the proper symbol for millimole: mmol.


text:
What is the origin of the name JOE - the dye used to mark DNA sequences?
I came across an article where JOE - a dye used to mark DNA sequences was mentioned. It was also stated that no one knows the origin of the name for the dye. Does anyone here have any clue?
Thanks!

These are just trade names to be used by common users who do not care about the chemical structure. Just like KitKat is a chocolate...easy to pronounce and easy to remember. Such names have zero scientific or scholarly value. You have FAM, TAMRA, Alexa Fluor dyes etc.

@AChem - sure, but somebody came up with the name somehow. Perhaps there is a back story, perhaps not. By the early 90's they  (FAM, JOE, ...) just appeared in the literature. Somebody active back then may have some insight into just why particular dyes got their particular names.

@JonCuster, The problem with commercial stuff is that they do not publish or are not allowed to publish. All these names have a TM (trademark) sign over them. I am sure there must be an internal story in the company and it might be best to contact the manufacturer. I had checked SciFinder for the OP to look at the synonyms. Nothing useful there. Sometimes companies do respond to such requests provided the email is written seriously with a genuine effort.

As per Wikipedia, at Alexa-Fluor dyes named after their son Alex Haugland, son of Richard and Rosaria Haugland (company owners). So it will not be surprising to find that Joe is also a son of a chemist! Best source would be the company itself.


text:
How do you calculate percentage acidity?
I am looking to calculate the percentage acidity of different drinks for the purposes of devising cocktail recipes. According to Liquid Intelligence by David Arnold, the optimum acidity level is 0.7-0.9, based on analysis of classic cocktails. How might I calculate the acidity percentage level of say lemonade (pH 2.9)?

What is percentage acidity and acidity level? Note that there is no direct relation between pH values and acid content. There are factors as acidity constants of phosphoric or carbonic acid or organic acids like citric or tartaric ones. and related pH buffer system presence.

Good question - I was hoping someone could help clarify. In this cocktail calculator> https://www.cocktailchemistrylab.com/calculator < you can see that different ingredients are given different percentage acidities when you enter them into 'ingredients'. This is the approach used in the above book by D Arnold.

I like to say this in similar context: A asking B, why  C said X, does not usually provide a satisfactory answer. As B does not usually know why C did so either. A should ask C.

pH calculations assume absence of other solvents like ethanol.

I doubt your "percentage acidity" or optimum acidity level, whatever it's supposed to be, has much to do with pH, or even chemistry altogether. Taste is an abstract created by human mind.

It is well known sour taste is partially masked by sweet taste, so diluted orange juice seems more acidic, even if it is objectively less acidic.

After looking at the link you posted, it seems "% acidity" refers to the mass percentage of citric acid in lemon juice, which is approximately 6%.
The percent acidity of your lemonade will depend on the percentage of lemon juice that constitutes it.
Let:
$X_{J/L}$ = the mass fraction of lemon juice in lemonade.
$X_{C/J}$ = the mass fraction of citric acid in lemon juice.
$m_{C/L}$ = mass of citric acid in lemonade
$m_L$ = total mass of lemonade
If you know how much water and lemon juice constitutes your lemonade, you can calculate "percent acidity" with:
$$m_L\;X_{J/L}\;X_{C/J}=m_{C/L}$$
Solving for $X_{C/J}$:
$$X_{C/J}=\frac{m_{C/L}}{m_L\;X_{J/L}}$$
Which is equivalent to:
$$X_{C/J}=\frac{X_{C/L}}{X_{J/L}}$$
Finally, solving for $X_{C/L}$:
$$X_{C/L}=X_{C/J}\;X_{J/L}$$
For example, if you mix 150g of lemon juice with 850g of water, the total mass of the mixture will be 1000g, and ${X_{J/L}}$=0.15, ${X_{C/J}}$=0.06
So "%acidity" for your lemonade would be:
$$X_{C/L}=X_{J/L}\;X_{C/J}=(0.15)(0.06)=0.009$$
Which is 0.9 percent.

Please familiarize yourself with Which symbols are written in roman (upright) font and which are italicized? and note that the unit symbol is separated from the numerical value with a non-breaking space (i.e. use \pu macro provided by mhchem). "Mass fraction" is also the better term than "mass percentage".

Thanks for your observations and link. I will try to keep them in mind for my next post.


text:
Why can we add/substract/cross out chemical equations for Hess law?
Suppose I take the model example:
(1) $\ce{C + 1/2 O2 -> CO}$
(2) $\ce{CO + 1/2 O2 -> CO2}$
Adding both together:
(3) $\ce{C + CO + 1/2 O2 + 1/2 O2 -> CO + CO2}$
(=> $\ce{C + 1/2 O2 + 1/2 O2 -> CO2}$)
Now why is it generally ok to cross out CO on both sites? Clearly it's balanced with or without and it's not part of the reaction either but let's assume for a moment that this reaction (3) containing the reactants and products from (1) and (2) suddenly forms other products:
$\ce{C + CO + 1/2 O2 + 1/2 O2 -> C2O3}$
(Only hypothetically, there are certainly some reactions that then form something other than merely additively the same products), why can Hess law still be true than? The formation must clearly have another enthalpy.
Or do I always have to imagine with such manipulations that they are pure mathematical steps to the actual reaction of interest without the various reaction paths actually reacting in such ways?

Coal would burn, regardless if there is a stone laying aside at the beginning and the end, or if it is not there. // Get familiar with the concept of bystanders/spectators in chemical reactions, that are optional and unnecessary. BaCl2 + Na2SO4 -> 2 NaCl + BaSO4 is de facto  Ba^2+ + 2 Cl- +  2 Na+ + SO4^2- -> 2 Na+ + Cl- + BaSO4 , what can be written juyst as Ba^2+ + SO4^2- ->  BaSO4

When you add two equations such that the product from 1st equation is a reactant for the 2nd equation, instead of thinking of them "cancelling", think about the product from the first reaction is being used up for the 2nd one, thus being effectively "cancelled".

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
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From your example, you were trying to use Hess's law to find the enthalpy change of the reaction $\ce{C + O2 -> CO2}$ from a two-step process:
$(1) \ \ce{C + \frac{1}{2}O2 -> CO}$
$(2) \ \ce{CO + \frac{1}{2}O2 -> CO2}$
As you said, adding these two processes combine to produce the overall reaction we are interested in as the $\ce{CO}$ on both sides would cancel. Now, what you are saying is that, why can't we just not cancel the two $\ce{CO}$ and then combine the $\ce{CO}$ and $\ce{CO2}$ to make a different hypothetical molecule $\ce{C2O3}$.
Sure, you can do that, and obviously, the overall reaction would have a different enthalpy, but the $\ce{C2O3}$ never appeared in our steps. You can easily fix this by adding an extra step as below:
$(1) \ \ce{C + \frac{1}{2}O2 -> CO}$
$(2) \ \ce{CO + \frac{1}{2}O2 -> CO2}$
$(3) \ \ce{CO + CO2 -> C2O3}$
Now if you combine them, you'd get what you had,
$\ce{C + CO + \frac{1}{2}O2 + \frac{1}{2}O2 -> C2O3}$
or $\ce{C + CO + O2 -> C2O3}$
However, we had to consider an additional step for that overall process which would be the source of the difference in enthalpy that you are pointing to.

Enumerated equations of chemical reactions, regardless of if they reflect really ongoing processes or not,, are just accountant inventories of atoms counts on both sides. They say in your example that numbers of atoms of carbon and oxygen on both sides are equal.
If there are the same terms on both sides, they can be subtracted as in any other mathematical equation.
If atoms counts on both sides are equal for 2 reactions, summing the sides is the equivalent operations and counts for both sides will be equal too.
Not there you cannot imply from the above reaction equations that CO and CO2 form C2O3. Such knowledge is outside of mathematical formalism of these equations.

Fundamentally, this is precisely because of Hess's law, which says that (as long as the reactants and the products are both in consistent standard state) the change in enthalpy depends only on the changes between the initial reactants and final products of the reaction, not on how the reaction actually happens (or even whether it can actually happen).
In particular, since a "reaction" like $\ce{CO -> CO}$ where nothing actually changes has $\Delta H = 0$, it follows from Hess's law that adding such a (non-)reaction to another reaction cannot change its $\Delta H$.  So the reactions $$\ce{C + O2 -> CO2}$$ and $$\ce{C + CO + O2 -> CO + CO2}$$ must have the same $\Delta H$, since they only differ by the addition of one $\ce{CO}$ molecule to both the reactants and the products.  According to Hess's law, it doesn't matter whether the $\ce{CO}$ actually participates in the reaction somehow or not: one molecule of $\ce{CO}$ goes in and one molecule of $\ce{CO}$ comes out, and that's all that matters.  Indeed, for all we know, that $\ce{CO}$ might be sitting in a sealed bottle on a shelf somewhere and might never even come anywhere near the other reactants!
It is, of course, extremely convenient that Hess's law happens to hold, since if it didn't, we might not even be able to define the concept of a standard enthalpy of reaction without having to consider details like the specific pathway the reaction takes and what other chemical species might be present and how they might affect the kinetics of the reaction.  But it does, and indeed that's exactly why reaction enthalpy is such a useful concept in chemical thermodynamics.

Ps. Of course, as others have already noted, a hypothetical reaction like $\ce{C + CO + O2 -> C2O3}$ would have a different $\Delta H$ than $\ce{C + CO + O2 -> CO + CO2}$, since it has different products: $\ce{C2O3}$ instead of $\ce{CO + CO2}$.  There is no contradiction here, since the reaction $\ce{CO + CO2 -> C2O3}$ has a non-zero $\Delta H$ that exactly accounts for the difference.

Now why is it generally ok to cross out CO on both sites?

Every chemical reaction tells you about the amount of reactant used up, and the amount of product made. To make things easier, we will consider a situation where exactly one mole reacts, e.g. a stoichiometric coefficient of "1" in front of a reactant means we use up one mole, and a stoichiometric coefficient of "1" in front of a product means we make one mole.
To understand why we can cancel CO on both sides, imagine we have a thousand moles of all species present (reactants and products). Each reaction will change the amounts in our stockpile. Because there is so much, we will never run out.
The first reaction adds a mole of CO to our stockpile. The second reaction uses up a mole of CO in our stockpile. The net effect is no change in CO. It does not matter whether reaction (1) happens first, second, or at the same time as reaction (2).

Or do I always have to imagine with such manipulations that they are pure mathematical steps to the actual reaction of interest without the various reaction paths actually reacting in such ways?

If the reactions are real, substracting or adding reactions is real, as well. When you add them in a one-to-one fashion, there is usually a physical reason. For example, if you start out with no CO at all, the second reaction can only use up as much CO (or less) as the first reaction produces.

Why can we add/substract/cross out chemical equations for Hess law?

When you substract a reaction, you are saying that it runs in the reverse direction.  In this case, your stockpile of products will diminish, and your stockpile of reactants will increase. Again, this might actual happen, or it might be artificial.
For Hess' law, it does not matter. The quantity you calculate with Hess' law, enthalpy, is a state function. Any path from A to B will result in the same enthalpy change, no matter whether real or fictitious. If you can measure the reaction from B to A, for example, you can also give the enthalpy for the reaction from A to B, even if you can't do the experiment.
UPDATE

[OP in comments] "If the reactions are real, substracting or adding reactions is real, as well." - That is a little unclear to me, although I can have two real reactions, but if I add them together, it can also happen that instead of the products merely being additive, another products are formed because, for example, a reactant from reaction a now suddenly prefers to bond with a reactant from reaction b. A+B->C and D+E->F, instead of A+B+D+E->C+F it is A+B+D+E->I+J, where A+D->I and B+E->J. That shouldn't be simply additive anymore because it doesn't produce the same products as the 1. and 2. reactions alone.

That is true in general. If I don't mix reactant, I don't get a reaction. If I add a third thing (maybe an acid or a base, for example), I might get a different product. Hess's law just says that once you describe a certain set of reactions with some reaction equations, the enthalpies add up in the same way that these component reactions do.

"If the reactions are real, substracting or adding reactions is real, as well." - That is a little unclear to me, although I can have two real reactions, but if I add them together, it can also happen that instead of the products merely being additive, another products are formed because, for example, a reactant from reaction a now suddenly prefers to bond with a reactant from reaction b. A+B->C and D+E->F, instead of A+B+D+E->C+F it is A+B+D+E->I+J, where A+D->I and B+E->J. That shouldn't be simply additive anymore because it doesn't produce the same products as the 1. and 2. reactions alone.

That is true in general. If I don't mix reactant, I don't get a reaction. If I add a third thing (maybe an acid or a base, for example), I might get a different product. Hess's law just says that once you describe a certain set of reactions with some reaction equations, the enthalpies add up in the same way that these component reactions do.

Hess' Law is inferred from the nature of Enthalpy: It is a state function. This is derived through its equation,
ΔH=ΔU(or ΔE depending on how you represent the internal energy)+PΔV.
P, Pressure, and V, volume, are both state functions: their values are only based on the present state (for example you could use the present P/V/n/T values, assuming ideal gas, or using the Van der Waals equation to find P and V respectively). Similarly, change in internal energy is also a state function, since its value is irrespective of the pathway taken to achieve the change in internal energy (q+w is always constant).
There's an example in Atkins' Chemical Principles that illustrate this fact: Consider this: If you're rolling a ball, A, down a hill, it ends up bumping into another ball, B, making B roll up a part of the hill, gaining GPE, and A itself slows to rest at the bottom of the hill.
In this example, A does work w on B by allowing it to gain KE which is transferred into GPE as it climbs the hill, and A loses energy as heat q to the hill surface by frictional heating as it moves down the hill. Therefore, the total loss of Internal Energy of A, ΔU, is equal to q+w. If A had lost transferred more heat instead of work, it means the process' pathway has changed. However, we know that A must have lost an amount of energy equal to its initial GPE, which is constant.
Therefore, the value of Internal energy is irrespective of the relative pathway it takes to achieve it. As we get here, you might now have a clue about why this was brought up, because it's beginning to sound a lot like the statement of Hess' Law: The total enthalpy change of a reaction is irrespective of the path taken.
We know that ΔH=ΔU+PΔV. We also know that U, P and V are all state functions. Therefore, since enthalpy is the sum of these functions, we can conclude that enthalpy is also a state function.
Therefore, we only have to consider the net outcomes of a reaction when calculating enthalpy: the net reactants leading to the net desired products, which justifies the cancelling of chemicals on both sides.


text:
Why are iodine precipitates yellow?
Why do iodine precipitates appear yellow? Learning about precipitates I noticed that compounds like Lead Iodide and Silver Iodide are yellow.

And why wouldn't they be? I guess why you could think otherwise, but that needs to be told in question.

There is certainly some variance between different cases, but a major factor is that bonds to iodine are weaker than bonds to lighter halogens. The large size of the iodine atom decreases electrostatic forces that are ultimately responsible for both ionic and covalent bonding.
The lowered bond energy then leads to lowered electronic transition energies, so that these transitions which are ordinarily in the ultraviolet region end up overlapping into the violet/blue portion of the visible region; absorbing that out of white light then leaves a complementary yellow or greenish yellow color. Silver iodide nanoparticles generated by pulsed electric discharge show this overlap, given that visible light lies between 400 and 800 nm wavelength. From Ref. [1]:

The effect is actually most dramatic with carbon. Steric effects weaken the bonding of the relatively small carbon atom to multiple iodine atoms, leading to decreasing electronic transition energies; we see a deepening in color (absorption moves more into the visible region) as we add more iodine atoms to an iodine-substituted methane:
$\ce{CH3I}$: Colorless liquid
$\ce{CH2I2}$: Colorless liquid
$\ce{CHI3}$: Pale, light yellow, opaque crystals
$\ce{CI4}$: Dark violet crystals, forming a dark red solution in diethyl ether
Reference

Tseng, KH., Yeh, CT., Chung, MY. et al. (2021). "A study of preparing silver iodide nanocolloid by electrical spark discharge method and its properties." Sci Rep 11, 20457. https://doi.org/10.1038/s41598-021-99976-5


text:
How can combusted methane from landfill samples be quantified?
Background:
For my studies I'm wanting and attempting to make a landfill greenhouse gas emissions (GHG) model that predicts the amount of greenhouse gas equivalent emissions ($GHG_\mathrm{eq}$ [tonnes/year]) by analysing a series of landfill core drill samples and extrapolating the data to predict how much GHGs would be produced if the samples can be representative of the landfill as a whole and allowed to naturally degrade. Note that this comes with the assumption of minimal heterogeneity which I know is not the best assumption for most landfills but I'm hoping to get a LOT of data (it's better than nothing).
As the most notorious GHG from landfill is methane, I'm going to simplify this question to just predicting the amount of methane which would be produced from the (combustible) material in landfill assuming it would just sit there forever.
The Data:
I have this data which is the average of 3 landfill drill core samples and shows the type of waste which has been calculated from sorting and weighing different types of material found in landfill (Table 1) and the chemical composition and high heating value (HHV)/calorific value of the material (Table 2) which I've had tested in an independent lab. For Table 1, an assumed "type" of plastic, rubber, wood (etc.) is made with a constant chemical composition to simplify the model (although I do recognise that there are many different types of plastics and woods etc.).

The Method (?):
I've seen the generic methanogenesis combustion reaction as:
$$\ce{CO2 + 4 H2 -> CH4 + 2H2O}$$
I've also seen an elemental formula for hydrocarbon combustions along the lines of:
$$\ce{C_xH_y + {(x + \frac{y}{4})} O2 -> x CO2 + {\frac{y}{2}} H2O + Heat + Light}$$
I know how to calculate the conversion rate going from any GHG such as $\ce{CO2}$ or $\ce{CH4}$ to $GHG_\mathrm{eq}$ but it's this first step - the predicting of the GHGs themselves with what formula - which I'm lost at.
The Questions:

How can I calculate $\ce{CH4}$ from landfill samples with the attached data and which reaction should be used?
(Bonus) What other considerations should be accounted for with any of the other data? (e.g. the Sulfur going into $\ce{SO2}$ emissions (etc.)).

I am afraid it cannot be easily calculated. AFAIK, landfill methane is formed by microbial activity, which depends on many factors. See also https://en.wikipedia.org/wiki/Methanogenesis

I've seen some of the complexity from it from reading around in literature. Perhaps a better approach here is to find and adapt and existing model for microbial methane generation from literature?

I assume such models are in large extent empirical. It may be worthy to migrate the question to https://earthscience.stackexchange.com/

This question lies nearer the topic of bioreactor modeling. Others here may be able to help but I have to agree with Poutnik that this might fit better elsewhere, maybe biology SE?

Similar comments have been made when I've posted it on other Stack Exchange websites about which site is best. There doesn't seem to be a clear consensus. I'll wait out and hope for an answer in the coming week or so, otherwise I'll put a bounty on it and hope for the best because now I'm very curious and it seems to be a great multidisciplinary challenge.

Since soil and waste conditions vary, the types of microorganisms vary, as well. For example, what is the moisture level of the site? How large is it? (A deeper dump may have more anaerobic conditions...) Are there specific chemicals present which would encourage or discourage methanogens? Are there methanotrophic bacteria, consuming CH4 as its generated?

Great points. But rather than focusing on the scope of the whole landfill site as it's certainly too broad, what about the specific landfill drill core samples themselves on a case-by-case basis? Then I can process multiple samples and plot the heterogeneity.

If you're just trying to get within an order of magnitude or so, here's one approach:
Assume that the landfill is completely sealed, so that no oxygen gas gets in and no hydrogen gas escapes. Also assume that the amount of inorganic redox reactions is negligible (for example sulfate or nitrate reduction).
Now, we'll convert your mass percentages into moles per kg and get:
H = 43 mol/kg
C = 44 mol/kg
N = 00.4 mol/kg
O = 6.6 mol/kg
S = 0.03 mol/kg
Converting this into a crude stoichiometry, we have roughly $\ce{C100H100NO15S_{0.1}}$.
Now we assume that all of oxygen has oxidation state -2, N is -3, S is -2 and H is +1, which is a reasonable assumption for organic matter in the absence of elemental gases. (I'm assuming any hydrogen gas is produced from the organic matter). We then find that that carbon averages an oxidation state of -0.7 per atoms. Assuming that at the end of decomposition, all of the carbon is either methane (oxidation state -4) or carbon dioxide (+4), we would have to have about 60% of the carbon as methane. That would be about 310 g of carbon in the original kg of material ending up as methane, which would have a HHV of about 5500 kcal/kg, not very far off your measured value.
Given that some energy is lost during the conversion of the organic matter to methane (for carbohydrates, this value is about 17 kcal per mole of carbon, so about 550 kcal per kg in our case), the numbers match up surprisingly well.
UPDATE: additional detail on calculations as requested in comments
We assume that the organic material is net neutral and treat the oxidation state as the charge of each atom, which must total up to zero, resulting in the following equation:
$$100*Ox(C) + 100*Ox(H) + Ox(N) + 15*Ox(O) + 0.1*Ox(S) =0$$
Substituting in the assumed values for all but C, we have
$$100*Ox(C) + 100*(+1) + (-3) + 15*(-2) + 0.1*(-2) =0$$
$$Ox(C) = - 0.67$$
I rounded that to -0.7.
Then, to find the ration of methane to CO2, we assume that all of the carbon is one or the other and write the equation for the average oxidation state of carbon atoms:
$$\frac{+4*m_\ce{CO2} -4*m_\ce{CH4}}{m_\ce{CO2} + m_\ce{CH4}}=-0.7$$
Since we're only interested in the ratio $m_\ce{CO2}/m_\ce{CH4}$, we can assign $x$ as the fraction of total C that is methane, so we replace $m_\ce{CH4}$ with $x$ and  $m_\ce{CO2}$ with $1-x$:
$$\frac{4*(1-x) -4*x}{1-x+x}=-0.7$$
$$\implies 4-8x=-0.7$$
$$\implies x=0.58$$
which is where I get that about 60% of the carbon is in the form of methane.

Brilliant answer thank you! I'm currently working through the calculations you provided. I've gotten up to the part where we have the same "crude" stoichiometry but, due to a lack of practical experience in redox reactions, I'm not sure how to proceed from this point on (i.e. putting the oxidation states into the model quantitatively to get the final result). Could you please expand on this part with the formulas used so then I can mark this as (likely) the accepted answer?

Thanks for the update! How did you come to the assumption that the reaction products would be either $$CO_2$$ or $$CH_4$$?

@Hendrix13 - That assumption came from the premise of your question - that is, the maximum amount of methane if you had infinite time. The max would be if all the rest of the carbon goes to carbon dioxide.

I see, so you've used both of the chemical equations I've listed in the question? I wasn't sure which one would be the right one or even if those were the right equations to use so I just listed both, are you suggesting it's good enough to use them both simultaneously?

@Hendrix13 - no I actually haven't used either. The hydrocarbon combustion describes burning in air, which is an incinerator not a landfill. The methanogenesis (which isn't a combustion) is just one of many possible final stages of methane production from organic matter, and there are even more ways to convert organic matter initially into substrates for methanogenesis. Luckily for us, the exact pathway doesn't matter. Under anaerobic conditions, they all have to balance out the methane production with CO2 production so that there is no net gain or loss of electrons from the system.

Ok I think I'm starting to see where you're coming from, but what specific chemical equation was used? If none are used then, apologies my redox chemistry is not the best which may be the issue here. Please advise.

The implicit equation that comes from the redox balancing is roughly $\ce{C100H100O15 + 65 H2O -> 60CH4 + 40CO2}$. You can fine-tune the numbers so that it balances properly, but since the calculation is approximate anyway, it doesn't really matter.

Thanks. Why is the $H_2O$ "implicit" here and what is the reasoning for its inclusion? I thought the reaction would be just a decomposition reaction of the CHO landfill matter? I guess we should account for water as a catalyst due to it being the microbial environment which kicks off the reaction right? I'm just trying to follow your complete working and reasoning so I can get to the same answer :)

Microbial decomposition is only going to happen in a wet environment, so I assumed excess water available and that your table was of the dry mass. If there's no water, even the easily degraded material will just sit there.

Last questions to help me get through it: (1) How did you know to include the oxidation numbers in the ratio equation? (3) How are the units the same for mass with oxidation numbers in the numerator of that equation?

I should have clarified that $m$ here is moles, not mass. it's just a basic calculation of the average redox state of the carbon atoms.

Also (4) Balancing the stoichiometric equation gives the ratios of products, why not just do that as oppose to going into the redox calculations?

The redox balance explains why that stoichiometric balance works, and they're both about the same amount of work.

By "why" you mean because it gives a method for inferring charge neutrality for the reaction? Sorry my redox chemistry is not up to scratch. I only ever learned how to get oxidation states, never learned the practicality of it all which is extremely annoying from my POV.

Also, how did you calculate the HHV? I can't seem to find a standardised formula online unfortunately.


text:
Formation of AZ91 Magnesium Alloy

I have been researching on how to form AZ91 Common Magnesium Alloy which consists of 90% Mg, 9% Al, 1% Zn, 0.3% Mn. Referring to the article at here, I have read and highlighted important text and trying to picture it in the Mg_Al Phase Diagram, but it looks very weird because the final composition should consist of 90% Mg which is in hyper-eutecoid region which is not mentioned in the text. Can anyone show a formation of AZ91 Magnesium Alloy using the Mg-Al phase diagram as shown?

In the real world it will likely be solution annealed and age hardened . Rapidly cooled from 415 to give a  relatively uniform material , then aged.

The explaining text is horrible.
The hypo/hyper refers to the amount of Al in the Mg relative to the eutectic point, so mentally flip the diagram if you want. "Hypo" is compositions to the right of the eutectic, "hyper" to the left.
The solidification discussed is starting up near the Mg melt point at nearly pure Mg composition, and coming down the solidus until the eutectic temperature is reached (but way over at the pointy bit on the Mg phase). This leads to the increasing percentage of Al in the solid with time. At that point, any remaining liquid solidifies into (Mg) + ($\gamma$-Mg$_{17}$Al$_{12}$). Until that point, all the solidified material has an Al composition below that of the eutectic. Only the Mg$_{17}$Al$_{12}$ exceeds the eutectic composition.


text:
Confusion regarding what reduction/oxidation means in reactions having covalent compounds
In the formation of NaCl, Na gets oxidised because it loses an electron, while Cl gets reduced because it gains an electron.
But how does this concept work in the case of covalent compounds, especially those compounds that have non-polar bonds?
For example, according to my textbook, in the formation of methane Hydrogen is getting oxidised from state 0 to 1, while Carbon is getting reduced from state 0 to -4.
But how is hydrogen losing an electron here? Doesn't it gain an electron due to sharing? C-H bonds are non polar so we cannot say that carbon pulls hydrogen's electrons towards it either right?

In these cases of nonpolar covalent bonding, oxidation and reduction is formal, similarly as oxidation states/numbers.  If a bond is easily broken, like N-Cl, ON can be formally determined from ON of products

@Poutnik What do we mean by oxidation and reduction is formal? What does formal here?

There is no gaining nor loosing electrons in $\ce{C(s) + 2 H2(g) -> CH4(g}$). There is speculative gaining and loosing electrons in thought heterolytic $\ce{H3C-H -> H3C- +  p}$ bond breaking, on which the formal oxidation number is based on.

@Poutnik Oh, Ok. Thank you so much.. So in throry,  if we take an isolated ch4 atom then on breaking one of its bonds we will end up with a methyl anion and a hydrogen cation? has this been experimentally checked too? And pardon me for asking a slightly silly question  but umm, is it possible to just break one bond while leaving rest of the three untouched?

@Poutnik another doubt that I have is, that according to this speculation, if we keep on breaking the bonds, then we keep on creating more H cations or in other words protons. But I have been taught that CH4 disassociates into H2O and CO2 but here how will protons form H20?

Have you invented transmutation available in chem labs? How can CH4 dissociate to CO2 and H2O ?

@Poutnik I am very very sorry. The reaction I gave u was that of ch4 burning in air. Ignore that comment. Just read the one above it. And no, I dont even know whay transmutation means sorry

Redox is just changing oxidation numbers https://chemistry.stackexchange.com/questions/69798/an-introduction-to-oxidation-state-oxidation-number

@Mithoron Then why do we define oxidation and reduction this way.. Also the answer did not mention how oxidation and reduction makes things easier for chemists.

Transmutation is supposed possibility of chemical way of turning one element to another, mostly common or less precious metals to gold. It was widely believed as possible in medieval era in Europe.

@Poutnik So just to confirm we are speculating that  CH4 will break into CH3- and H+ and that is why the oxidation state of carbon is negative here right?

@Poutnik Also why can't we  perform this reaction in the lab? Like maybe in vacuum?

It is thought formal operation, based solely on electronegativity difference. In reality, bond C-H usually breaks homolytically, forming radicals.

@Poutnik So basically, CH4 breaks homolyticaly, but since we think that if it were to break heterolyticaly it would form CH3- and H+, we write the oxidation states of H as 1 and C as -4. Have I understood this correctly?

Yes, that is correct.

@Poutnik Thanks! This helped me  a lot!! :)

Carbon is slightly more "electronegative" than hydrogen, so by convention C in CH4 is -4 and each hydrogen is +1, even though the bonds are considered covalent.
Differences in electronegativity are important in understanding polar molecules.  More reading here.
With CO2, the carbon oxidation state is +4, because oxygen has a stronger affinity for electrons than carbon.

So the definition that oxidation is the loss of electrons is wrong in the case of CH4 right? The only reason H gets oxidised is because its oxidation state increases from 0 to1. And the reason  we write its oxidation state as 1 is that carbon is more electronegative than hydrogen. Nothing about electrons right?

@Obinna  we can observe that the net charge of elemental H, C, and molecular CH4 remain 0, and the bonding is "covalent", or (more or less) sharing the electrons to fill vacant orbitals.


text:
Compare acidic strength of two different hydrogen in a Compound (2-hydroxycyclohex-1-encarboxylicacid)

I want to compare acidic strength between hydrogen-(a) and hydrogen-(b), I am getting answer as (a) but my classmate and my teacher said that (b) will have more acidic strength. I think (a) should have more acidic strength because it will have more resonance and negative charge will present on oxygen two times

I think you're either confusing (a) with (b), or the relationship between acid strength  and conjugate base stability with resonance. If a conjugate base has resonance, it is more stable (weaker) than a base that does not, and it's parent acid is therefore stronger.

Yes I know that, I think conjugate base of (a) will be more stable because it will have more resonance and therefore (a) will be more acidic.

Going by stabilization of conjugate base through resonance, the resonance structures of ($a$) would be:

And the resonance structures of ($b$) would be:

So both ($a$) and ($b$) have two resonance structures, or equivalently, two different ways of placing the negative charge on an oxygen atom.
However, in terms of inductive effects, the carbonyl group ($\ce{C=O}$) acts as an electron-withdrawing group, which means it pulls electron density towards it.
If you look closely, in the resonance structures of ($a$), the carbonyl group is farther away from the negative charge than it is on $\;$($b$).
This means that the two oxygen atoms in ($b$) where resonance takes place in ($b$) can share the negative charge more efficiently than the two oxygen atoms in ($a$) where resonance takes place in $\;\;$($a$).
Therefore, the conjugate base of ($b$) is more stable, and it follows that ($b$) is a stronger acid.


text:
Why does Co(CO)4 exist as dimeric Co2(CO)8?
I was wondering why is this equilibrium below established at all... Why can't Co(CO)4 exist on its own, as a monomer?

Reference 5 in the corresponding Wikipedia article gives a lot of information if not the complete answer.

Also note that the monomer you speak of is not a neutral molecule. It is actually charged, $\ce{[Co(CO)4]-}$. See: https://chemistry.stackexchange.com/questions/97571/why-is-the-oxidation-state-of-co-1-in-coco4

@NilayGhosh The monomer in this case is the (17 electron) neutral Co(CO)4 molecule; [Co(CO)4]− is a different entity. c.f. NO2, N2O4 and NO2-


text:
How to Choose the Appropriate Spindle and RPM in Rotary Viscometer for Viscosity Measurement of a Suspension Sample?
I am currently conducting a method validation for the analysis of a suspension. This would be the first time we will be conducting viscosity analysis via rotary viscometer and the below statements are what we know about the analysis:

P (Percentage) should be between 10 to 90% for an accurate reading of viscosity.
Suspension and other Non-Newtonian liquids have different readings as per the speed and temperature of the analyte.
The variables we can change in the analysis are the Spindle type, Spindle RPM, Sample temperature

What I am confused with is I get different viscosity reading for different P values even if we keep it within the 10 - 90% allowable. How can I choose the proper spindle and speed for my analysis and what is the basis for it.
Attached image is the data we have gathered so far. The data set was acquired for a single sample. In red are the data set within the allowable P values.

What is $P$?  Is it some kind of setting on the instrument?  Or is it a range of allowed values that measured viscosities must lie in?  Is it a sample property?

Is it the percentage of solids in your suspension?  If so, is it a volume % or a mass %?

@CurtF. P is displayed on the rotary viscometer and it stands for percentage. If I search for percentage on the internet, the ff is shown: When the viscosity of a fluid is not known.
If the reading is below 10% or above 100%, the user should choose a different speed to obtain a reading in the recommended range. If changing speeds doesn't provide readings between 10% and 100%, the user should try another spindle.

You alluded to non-Newtonian fluids in your question.  Some examples of these are dilatant fluids and pseudoplastic fluids.  For these types of fluids, apparent viscosity is a function of the shear stress.  This is because the slope of the shear rate vs. the shear stress is not constant.  But at least for these non-Newtonian fluids, viscosity remains essentially a "state function"; all you need to know is (a) the fluid in question and (b) the sheer stress in order to obtain an apparent viscosity.
However, things can be worse.  Thixotropic and rheopectic fluids have complex, time-dependent shear behavior.  This means that the vicosity not only depends on (a) the fluid in question and (b) the shear stress at a particular time, but also (c) the entire history of shear stress applied over the lifetime of the fluid sample in order to be able to characterize its viscosity.  For such fluids, viscosity is no longer really a state function.
The problem you are having may be that your suspension may be thixotropic or rheopectic.  For example, if the particles of solid suspended in your liquid phase may have complex shapes whose net average orientation changes over time, or those particles may aggregate together slowly over time, and these aggregates may be (slowly) disrupted by shear stress.  This means that repeated measurements may not give you the same viscosity.
Of course, there are many other possible causes of this issue, but without a detailed description of your sample types, it will be hard to examine most of these.

So getting the density of my sample would only be possible if I keep some of my variables as a constant and set only one variable to be read in the analysis?


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Standard state clarification
I have seen several questions on standard states, but none quite get at the issue I'm having in understanding them. In Elements of Physical Chemistry by Atkins and de Paula, they describe standard states for pure substances as the state of the substance at 1 bar pressure at a given temperature and in a given phase of matter. As I read their explanation, standard states exist for all phases of matter at a given temperature, regardless of whether they are thermodynamically stable or not.
For example, Argon would have three standard states at 100 K: one for the solid at 1 bar and 100 K, another for the liquid at the same $p$ and $T$, and another for the gas at the same $p$ and $T$. This makes sense to me, because even though two of these states are unstable, they still can be defined in principle; these standard states, while having the same $p$ and $T$ have different energies, densities, enthalpies, etc. In my mind, this also agrees with the fact that we have non-zero enthalpies of phase transitions.
However, Wikipedia states

The standard enthalpy change of formation for an element in its standard state is zero

which seems to contradict the idea that standard states exist for all of the phases of matter for a given temperature. As I read this, it seems to imply that only the thermodynamically stable phase at the given temperature is the standard state, and other phases of matter for the same element at 1 bar and the same temperature do not have a standard state. I have seen a similar definition in Principles of Chemistry by Munowitz.
Can someone clarify what "standard state" actually means? Every time I read a new definition I feel less certain I really know. Is this just a case where different authors use the same term to refer to different things?

It should say element in the reference state, see https://chempedia.info/info/reference_state_of_an_element/

@Karsten That is the terminology that Atkins and de Paula use. If that is the correct terminology (which I feel is likely the case), then it seems that many (perhaps even most) other books and online resources are very sloppy in their use of terminology and freely interchange standard state and reference state.

What they mean is the natural thermodynamic equilibrium  state of the substance as it exists at 1 bar and 298 K.

Wikipedia now uses the term "reference state" for choosing the allotrope in defining enthalpy of formation, and "standard state" to describe the state of that allotrope and all other species in the defining chemical equation.

[OP] Can someone clarify what "standard state" actually means?

Standard state has a common definition across the field of thermodynamics. Each species, in combination with the physical state, has a standard state; often, these standard states are hypothetical (e.g. a solute at 1 mol/L concentration behaving ideally like an infinitely diluted solute).

[OP] As I read this, it seems to imply that only the thermodynamically stable phase at the given temperature is the standard state, and other phases of matter for the same element at 1 bar and the same temperature do not have a standard state.

This is imprecise language, skipping one step. For the definition of enthalpy of formation, you need the elements in a defined state (without giving the chemical equation). This should be called "reference state" (IUPAC Gold Book entry) of the element, and it includes which allotrope of the element is used, and in which physical state. For ions, it also includes using the aqueous hydronium ion as reference state. Once you know which reference state you agree on (i.e. white or red phosphorus, iodine in the solid or gaseous state, hydronium ion rather than another ion), you also stipulate that this reference species is at standard state, and is at the temperature for which the thermodynamic data is given.
The term reference state is mentioned explicitly in a IUPAC recommendation from 1981:

The use of subscript f was discussed in paragraph 3.3. When ° is additionally used, as in $\Delta_fH°$, the implication is that both the compound in question and its constituent elements are in standard states and that the elements, moreover, are in their reference states; for any given temperature the reference states of the elements will normally be those that are stable at the chosen standard-state pressure and at that temperature.

Stable refers to the state (allotrope and physical state) with the lowest Gibbs energy, in this context.