text:
Why is HCl a stronger acid than H2S on the basis of electronegativity, but HF is a weaker acid than HCl?
We know that $\ce{HCl}$ is a stronger acid than $\ce{H2S}$. This can be explained due to that fact that chlorine is more electronegative than sulfur, which means the hydrogen ion has a much easier time leaving when bonded to chlorine than sulfur. However, when looking at $\ce{HCl}$ and $\ce{HF}$, $\ce{HCl}$ is actually a stronger acid, even though $\ce{HF}$ is more electronegative and polar. Is there is flaw in my reasoning?

Higher electronegativity actually usually means more difficulty to extract a proton from an acid. $\ce{HBr}$ and $\ce{HI}$ are both stronger acids than $\ce{HCl}$, and both halogen atoms are less electronegative than chlorine. Conjugate base stability is a better criterion, as more stable conjugate bases prefer to stay dissociated (like $\ce{Cl-}$,$\ce{Br-}$, and $\ce{I-}$) while unstable conjugate bases (like $\ce{F-}$) prefer to stay associated. Fluorine atoms are too small to effectively carry a negative charge, while chlorine, bromine, and iodine are big enough to do so.

As to why HI is a stronger acid than HF, you might want to look here.

F- is unstable? Is that why fluorine is one of the strongest oxidizing agents known?

@jimchmst It's unstable relative to the bigger halide ions in terms of acid-base equilibria and acid strength. The oxidizing capability of $\ce{F-}$ is irrelevant in this case because an acid dissociating in aqueous solution involves a transfer of protons, not electrons.

Why can the acidic strength between H2S and HCl be explained on the basis of electronegativity but H2S and H2O can't? Note the proper spelling of HCl, too.

There are several factors involved in acidity in water. The most important is probably the X-H bond strength. This effectively explains the increasing acidity from HF to HI The bonds become weaker. It also explains the increasing acidity from H2O to H2S, H2Se and H2Te. A second factor is solvation of the anions. There probably is a sweet spot here for the surrounding water molecules to disperse the charge and minimize crowding. This might also involve the shape of the ion. Another consideration is the ability to hydrogen bond. The hydrogen bond will give a reprotonation reaction more opportunity and should lower the acidity of HF, H2O and possibly a bit for HCl.
All the factors should be considered in making comparisons especially when comparing  between groups. Also the extent of chemical reactions is governed by the Second Law energetics


text:
Which is a better poison to the Haber's process, $\ce{CO}$ or $\ce{H2S}$?
I am studying about catalysts and their promoters and poisons and came across two of the possible poisons for Fe used during Haber's process and wanted to compare their effect on the rates of reaction. Which of them would lower the rate of reaction of Haber's process more, $\ce{CO}$ or $\ce{H2S}$?

Wikipedia states: Sulfur compounds, phosphorus compounds, arsenic compounds, and chlorine compounds are permanent catalyst poisons. Oxygenic compounds like water, carbon monoxide, carbon dioxide and oxygen are temporary catalyst poisons.[3][35] Permanent trumps temporary? I have no access to the publication though, maybe you can get it yourself.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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If I have a mixture of potassium hydroxide and sodium hypochlorite, what will be the result of adding citric acid or ascorbic acid?
I have a cleaning chemical that is $\pu{10\%}$ potassium hydroxide ($\ce{KOH}$}) and $\pu{5\%}$ sodium hypochlorite ($\ce{NaClO}$). What is the result if I add citric acid ($\ce{C6H8O7}$) or ascorbic acid ($\ce{C6H8O6}$)? My goal is to neutralize the high pH of the potassium hydroxide without creating dangerous byproducts. I do not have a background in chemistry. How do I determine the result of the mass balance equation?
$\ce{KOH + NaClO + C6H8O7 -> ?}$
$\ce{KOH + NaClO + C6H8O6 -> ?}$

Chlorine gas maybe?

For non chemists, I rather suggest sufficient washing a/o dilution and neutralization of eventual residues by diluted vinegar. Processing it at high concentration can cause violent effects, mostly being attacked by obsolete yet powerful chemical weapon of chlorine ( Somme 1915 ).

The proposed equations with three substances is not possible, The acids, whatever their structure, react with $\ce{KOH}$ and also with $\ce{NaClO}$ but these two equations are occurring independently from one another. But the reaction with $\ce{NaClO}$ produces chlorine  $\ce{Cl2}$ which is a dangerous gas. Try not to do such a mixture !

The reason hypochlorites are kept alkaline is to prevent evolution of chlorine gas, $\ce{Cl2}$.

Any attempt to make the cleaners less alkaline will release poisonous $\ce{Cl2}$, unless
A reactant combines with the chlorine, preventing its evolution... thus making the cleaner useless as a bleach and disinfectant, or
The worst outcome: the reactants eliminate the chlorine, but produce even more toxic and volatile products. For example, the reaction of hypochlorites and household (aqueous) ammonia produces hazardous chloramine, or even hydrazine. See Science ABC for more detail on that reaction.

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Why do vessels get a better shine when tea-leaves are soaked overnight in it before washing?
The other day I dumped consumed tea leaves into a brass utensil because I needed the tea-pan urgently. The vessel with the tea leaves didn't get around for cleaning until the next day. Ergo, the tea leaves soaked overnight in dirty water. When finally the vessel was cleaned (as in hand-washed with utensil cleaner), the quality of shine was exceptional.
Piqued, I repeated the tea-soaking in another vessel — and was rewarded with the same high gleam. This vessel was made of aluminium.
Why do vessels get a better shine when tea-leaves are soaked overnight in it before washing?

As tea leaves mature, they can oxidize and become more acidic. Leaving it in the water most likely made the solution acidic which cleaned the Aluminum oxide layer off the container, leaving a nice shiny layer of aluminum.

How do you mean mature?

Polyphenols from tea leaves are great chelating agent for aluminium that has high affinity to oxygen. In a way, it is similar like the "be shining again" procedure for old coin being boiled in suspension of EDTA.

What type of tea leaves are being used for this observation? Usually black tea leaves brown stains in metallic vessels. Do you have green tea leaves or oolong leaves etc?

This stuff

@Poutnik: That fits with the Aluminium vessel; would it have been the same with the brass vessel?


text:
What does "acidic" mean in an article on hydrogen boride nanosheets?
From Rojas et al. [1] (emphasis mine):

On the other hand, we found that proton exchange with HB occurs in water with the estimated $\mathrm{p}K_\mathrm{a}$ of $3.5\pm 0.2,$ even after reactive sites for the hydrolysis reaction are terminated. Thus, for the practical application, one should keep in mind that HB sheets have a chemical stability against hydrolysis reaction, but become acidic in the presence of water

Does “acidic” here refer to the solution? Or acidic sheets that are gaining protons?
Reference

Rojas, K. I. M.; Cuong, N. T.; Nishino, H.; Ishibiki, R.; Ito, S.; Miyauchi, M.; Fujimoto, Y.; Tominaka, S.; Okada, S.; Hosono, H.; Arboleda, N. B.; Kondo, T.; Morikawa, Y.; Hamada, I. Chemical Stability of Hydrogen Boride Nanosheets in Water. Commun. Mater. 2021, 2 (1), 1–8. DOI: 10.1038/s43246-021-00184-5. (Open Access)

Exactly what should be expected? Your hydride is an acid.


text:
Why concurrent reading instead of mean in titration?
While doing titration we take concurrent reading like.
Reading..........Concurrent reading

3.6
3.4=====> 3.4
3.4

Why don't we take the mean value as we do in physics?

When you say "physics", what exactly do you mean? Many things follow a standard distribution, readings from a voltmeter e.g. usually don't, operator bias doesn't, ... Without a rigorous error estimation, just taking a mean can be a bad choice.

Could you attempt to add more detail (explain more clearly what you mean, for instance in your example)?

We usually refer to the median rather than the mean. In the case shown, when you have two readings that are identical and one that's different, the median is the reading with two values (here, 3.4) rather than averaging all three (here, that would be 3.5 to two significant figures).
The trade-off goes like this: when we are sure that all the data are normally distributed in one population, the mean represents the best estimate of the true average value. But in many types of experiments, including titration, we have to guard against being influenced by a sample where we went wrong with the procedure or there was some contamination in the sample we couldn't see beforehand. The median is less sensitive to such abnormality in the experiment, and the resulting non-normality in the statistics, than the mean. This is in exchange for being a less optimal estimate if we could be sure everything were perfect/normal.

Nice answer, especially considering that you had to infer that from the question. (I honestly had no clue.)

First! A buret reading should be to the estimated 0.01 mL, the first Insignificant figure not to the gradations of 0.1 mL. One titration is one data point. If you want to check your overall precision the procedure is to run complete replicate samples on the one sample or [even on simultaneous replicate samples to check on the sampling technique]. Since only 2 or 3 replicates are usually run, when readings are to the first insignificant figure a simple average is sufficient and the range is a good approximation to the error. In your case the buret should have been read to 3.61 3.43 3.46 [making up last digit] average is ~3.50 error is ~0.15; ~5% relative error quite unacceptable in titrations. To apply correct statistics more data points are needed. Data such as this simply means that your technique needs serious improvement. Once you have the technique down pat such data could be meaningful and there is something going on with your sample. This is why I strongly advise that standard solutions be standardized by the user. First to ensure the source is competent and second to ensure one's technique and equipment are up to par on a known reaction.
I worked as an applications chemist for a instrument manufacturer and was dismayed on finding out how many of the labs I visited were working with uncalibrated or improperly calibrated instruments.
Conclusion: run 2-3 complete replicates. Read the buret to the first estimated digit. If the data has no variation or if the variation is in the significant digit or relative error is greater than 1%, something is wrong.


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Can the overall order of a reaction ever be strictly less than $1$?
In A-level chemistry we are taught that:

To every reaction: $$\sum_{i=1}^n c_iA_i\to\sum_{j=1}^m d_jB_j$$There is a rate equation: $$r=\kappa[A_1]^{\alpha_1}[A_2]^{\alpha_2}\cdots[A_n]^{\alpha_n}$$With the $(\alpha_i)$ called the partial orders and $\sum_{i=1}^n\alpha_i$ the overall order, $\kappa$ some constant.

Since the case of multiple reactants is confusing (e.g. I have not been able to get a clear answer of how 'rate' is defined in general) I'll stick to a reaction with only one reactant, $A\to bB+cC+dD+\cdots$.
We can then explicitly solve the differential equations, for $[A]_t$ treated as a function in time $t\ge0$, with $[A]_0>0$: $$\frac{\mathrm{d}[A]_t}{\mathrm{d}t}=\kappa\cdot[A]_t^\alpha\implies\kappa t+C=\begin{cases}\frac{1}{1-\alpha}[A]_t^{1-\alpha}&\alpha\neq0\\\ln[A]_t&\alpha=0\end{cases}$$After some shuffling, you get: $$[A]_t=\begin{cases}\{(1-\alpha)(\kappa\cdot t+C)\}^{\frac{1}{1-\alpha}}&\alpha\neq1\\Ce^{\kappa t}&\alpha=1\end{cases}$$Where $C$ is some constant in both cases. If $\alpha=1$, we get $C=[A]_t$: otherwise, $C=\frac{1}{1-\alpha}[A]_0^{1-\alpha}$.
I played with this a little, plotting concentration-time reaction curves on Desmos, and I quickly noticed a significant difference between the cases $\alpha<1$ and $\alpha\ge1$: if the order $\alpha$ is less than $1$, the model predicts the reaction terminates in finite time, at $t=-C/\kappa$. Else, the reaction never terminates!
Intuitively, I don't think a chemical reaction can ever fully terminate, so I'm inclined to believe the following conjecture:

In a reaction with only one reactant, the order of reaction must be greater than or equal to $1$.

More generally, I suppose:

In any reaction for which the rate equation holds, the overall order must be greater than or equal to $1$.

Does that make sense? I imagine I'm quite wrong, since I'm doing this on the back of relatively little chemical knowledge. In particular, it's hard to picture what solving the differential equations looks like for multiple reactants.

Aside of photochemical reactions of zeroth order, consider complex, multistep reactions with reaction loopbacks. Things about their order can go rather wild. Even for the simple looking $\ce{Br2 + H2 -> 2 HBr}$.

see https://www.coursehero.com/file/21287225/Chemical-Kinetics-3/

Long story short, if the reaction is truly elementary, then all orders would be positive integers; and if it isn't, then its order is just an approximation that won't hold all the way down to zero concentrations.

:( Even or all this nice formatting, I'm still tempted to downvote. It maybe could be a duplicate target, if well answered, but the topic's being asked so many times before...

@IvanNeretin That essentially answers the question, thank you.

@Mithoron This topic has been asked about before, sure. My question is specific though: from my mathematical manipulation of the differential equations, is my conclusion about the possible orders correct? Ivan suggests an answer, though I might like some more detail

Irreversible reaction with just one substrate can of zeroth order. Like Poutnik said it could be photochemical, but there are also other reactions that simply aren't controlled by concentration. And yes it leads to equation with finite end of recation.

See https://chemistry.stackexchange.com/questions/154182/are-pseudo-zeroth-order-reactions-possible https://chemistry.stackexchange.com/questions/94965/can-a-zeroth-order-reaction-be-reversible https://chemistry.stackexchange.com/questions/263/how-to-identify-zero-order-reactions

@FShrike Since you have got the answer, would you mind self answering it? Self answers are encouraged, it completes the post and it will also be helpful for future readers.

@NilayGhosh I often self-answer over at MSE. However, I don't have the authority to self-answer here, since I am not even close to knowing the details properly.


text:
Show that if H is hermitian, then <Ψ|H|Φ> = <HΨ|Φ>
Since H is hermitian, it is equal to its complex conjugate (shown below), a property which I used to do the proof:

I had to prove that <Ψ|H|Φ> = <HΨ|Φ>, which I did as below:

My question is whether my proof is valid - are the two things equal because the pink bits equal each other?

There are several mistakes. Let's begin with your first line. The second equality is wrong (as per your "definition", you need to take the complex conjugate of the entire integral). On the second line, the second equality is wrong. You've taken the conjugate of the integrand. If you do that, you will get the conjugate of the result. Again, the entire integral needs a conjugate to make that correct.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts. (You have been already informed about this once.)

As the operator $H$ is linear, when taking it out of the bra symbol it must be replaced by its adjoint and placed on the right, i.e. $\langle H\psi|=\langle \psi|H^\dagger$ and so $\langle  H^\dagger\psi|=\langle \psi|(H^\dagger)^\dagger=\langle\psi| H$. In you case the operator is hermitian so  $H=H^\dagger$. Compare with $|H\psi\rangle$ which is another way of designating the ket $H|\psi\rangle$.


text:
Making a reagent preparation simpler that requires pH adjustment
I have a basic question on a reagent (solution) preparation. Here are the steps that I take to prepare the reagent:

Weigh the chemical (solid form, ammonium bicarbonate) and dissolve with DW to make 100 mM solution. The weight (or volume needed to make 100 mM) can vary depending on the # of the samples to be analyzed

Calibrate the pH meter and adjust the pH of the above solution to a certain pH

I would like to make my life easier by not using the pH meter (gets tedious having to wait for the pH calibration and whatnot). I do know that theoretically, if I make the weight of the chemical the same each time I prepare the solution, I can add the same amount of base to set the pH. But is it correct that the volume of base added changes depending on the amount of chemical weighed?
I don't remember my chemistry very well, but from my memory, pH is concentration dependent so I was wondering if adding the same volume of base to a solution of desired concentration (but may have different volumes as I may weigh the chemical differently) is reproducible.
So basically, I want to simply weigh the chemical, dissolve with required volume of DW to make 100 mM concentration, and then add the same volume of base without having to use my pH meter, which will save a lot of time in the morning.

Why not to note the particular volume ratio for the used base or acid neutralization solution and the 100 mM NH4HCO3 solution ? Then you can repeat it without pH measurement. E.g.illustratively,  for 100 mL of 100 mM NH4HCO3 you spend 1.23 mL of 1 M acid/base.  Then you use 100/1.23 mL ratio all the time, occasionally check if all is as expected.

Oh yeah wow I completely forgot about using the ratio!! Thank you!!


text:
Difference between pre-equilibrium and steady-state approximation?
Both are reaction mechanisms in which there are several steps and at least one intermediate present. However, I am confused about the differences of both methods, especially when using them to determine rate laws. Also, in which types of reactions can we apply each of the methods?

See the related page on chem.libretexts.org

See also https://chemistry.stackexchange.com/questions/110794/is-there-a-difference-between-equilibrium-and-steady-state


text:
Is this sulphur?
I was electroplating with a copper sulfate electrolyte, and now a yellow substance has settled at the bottom of the container, is it sulphur from the copper sulfate or something else from the screw I electroplated?
If it is sulphur, how can I safely remove it?
Thanks for any help you can give me.

The yellow precipitate is probably cuprous oxide $\ce{Cu2O}$ which is insoluble, and which is produced in the electrolysis of $\ce{CuSO4}$if the solution is not acidic enough. Check by adding some acidic solution, like concentrated HCl. If the yellow precipitate is sulfur it will not be attacked by the acid. If it is $\ce{Cu2O}$, il will soon be dissolved and produce a colorless solution, due to : $$\ce{Cu2O + 4HCl ⟶ H2CuCl2 + H2O}$$

I do not have hcl, would distilled white vinegar work?

Sulfur can never be formed by electrolytic reduction of copper sulfate.

@UriahPowell  acetic acid (vinegar) is not a strong enough acid to lower the pH sufficiently

@ Uriah Powell. Acetic acid is too weak an acid to redissolve $\ce{Cu2O}$. Why don't you make the same test with ammonia $\ce{NH3}$, or with ammonium chloride $\ce{NH4Cl}$. These substances should be able to dissolve $\ce{Cu2O}$

I would also electrolyte the copper with some added sulfuric acid. This prevents Cu2O from forming.


text:
Why does glass etching cream contain sulfuric acid?
The MSDS for various glass etching creams all show that they're primarily composed of 4 substances (the composition percentages vary):

Ammonium Bifluoride  - 20 - 40%
Sodium Bifluoride    - 5 - 10%
Sulphuric Acid       - 1 - 5%
Barium Sulfate       - 1 - 5%

The purpose of the fluoride ingredients is obvious. They decompose into hydrofluoric acid (presumably when reacting with water in the air) and do the actual etching. But why is there Sulphuric Acid? It doesn't react with glass at all.
My speculation would be that perhaps it's a strong desiccant and it serves to slow the reaction?

$\ce{HF2- + H+ <=> 2 HF}$, resp.   $\ce{F- + H+ <=>  HF}$

Using bifluorides with a mineral acid is technically the second best after using directly HF(aq), as nobody wants to manipulate with stock HF(aq) unless absolutely necessary. (One bigger splash can kill you after few hours or days, unless treated. It is extremely painful and may left permanent damage.)

Both Ammonium bifluoride (NH4HF2) and sodium bifluoride (NaHF2) when reacted with water, produce the primary etchant for glass which turns out to be Hydrofluoric acid (HF). The reaction goes as followed:
NH4HF2 (or) NaHF2 + H2O → HF + NH4OH (or) NaOH
NOw that we got our etching agent Hydrofluoric acid (HF) which is very strong acid capable of etching glass as it reacts with the silicon dioxide (SiO2) present in the glass structure, producing soluble Fluorosilicic acid (H2SiF6) and water:
SiO2 (glass) + 6HF → H2SiF6 + 2H2O
Then coming to your actual question, why is there Sulphuric Acid? there are a couple of reasons like:

It helps in lowering the ph of the solution creating an acidic environment to aid the actual process. This contribution promotes the dissociation of fluorides and thereby supporting the reaction between HF and glass indirectly.

It contributes to overall consistency and stability of the etching cream. Increasing and decreasing the acidity of the cream, sulfuric acid can help to optimize the viscosity, adhesion, and other physical properties of the cream, ensuring uniform application and etching results.

It may be used in the process of production to
purification of raw materials, to increase the span before expiry.

Oddly, Claude and ChatGPT had very similar answers, also giving me three points (about kinetics, viscosity, and moisture control).


text:
Derivation of the Hartree-Fock equation
I have to do some tutoring about basic quantum chemistry and the main purpose is to give a deep insight into Hartree-Fock theory. My plan is to start from expressing the many-electron wavefunction as a Slater determinant and why this is even necessary, etc. From this starting point, one intermediate objective is to give an elaborate and crystal clear derivation of the Hartree-Fock equation and I used my old lecture notes and some textbooks to work out a detailed - and most importantly - comprehensive approach.
But I have to consider some constraints. Firstly, the majority of the students are likely to struggle with more sophisticated aspects like the formalism of second quantization (like used in the textbook by Hegaker, Jorgensen and Olsen). Things like functional variation or Lagrange mulitpliers shouldn't be a problem because they were covered in prequisited math lecture series. Secondly I realised that every textbook I have used for preparation is leaving out some so called "trivial" parts of the derivation that I need for my own understanding.
So far I have worked out the following key steps to arrive at the general form of the Hartree-Fock equation:

Establishing the first approach to the many-electron wave function (the Hartree product)
Considering the indistinguishibility of electrons and thus the antisymmetry of the wavefunction with the regard of the exchange of two particles
Setting up the Slater determinant to come up for the antisymmetry
Deriving the expression for the energy of a Slater determinant (keeping in mind the mean-field approximation)
Minimization of the energy with the constraint that the spinorbitals remain orthonormal

I am struggling to formulate the last part as detailed and clear as possible. The best approach I found so far is in Introduction to Computational Chemistry by Jensen, but even in this textbook, there are some mathematical "leaps" I don't get.
So the essence of my question is: Do you know any resource (textbook or lecture notes, etc.) in which this last step (minimization with the constraint of orthonormality yielding the general form of the Hartree-Fock equation) is explained very detailed and clear ?
Any help is heavily appreciated.


text:
What to do when there is white ppt formation even after adding dil. sulphuric acid in the mixture of aluminium sulphate and water?
I heated the solution containing aluminium sulphate and water with 3 ml dil. sulphuric acid. But I didn't get a clear solution because the solution got milky in color due to the hydrolysis of aluminium sulphate to aluminium hydroxide.
I heated the solution to get a clear solution but couldn't succeed.
I tried to filter the solution but in the filtrate, I got the same milky filtrate.
What to do to get a clear solution?

What is the context and purpose of this experiment? Hint: Alum is used for cleaning dirty water.

I have to make the Alum


text:
Are solid fuels with calorific values >= 10,000 kcal/kg (42 MJ/kg) physically possible?
I have seen laboratory documentation from a reputable lab in southeast Asia that a company has achieved a solid fuel product made from landfill material with a high heating calorific value ($CV$) of approximately 10,000 kcal/kg (42 MJ/kg). The technology is referred to as Refuse Derived Fuel (RDF). From what I understand, such calorific values are only even possible with fluid fuels such as liquefied natural gas (etc.) which can get up to 12,000 kcal/kg.
Doing some calculations around the fixed carbon composition of landfill as similar to a question I've asked previously, I found that the fixed carbon ($FC$) content of the feedstock composition must equal approximately 130% to have the CV mentioned above. This number is obviously impossible but I'm very surprised that it was reported by this lab. Similar products usually get a CV of around 2500 - 5000 kcal/kg if they're lucky. There are several formulas to calculate CVs, the one I've used is below and notably only a function of the fixed carbon content to calculate the calorific value [kcal/kg], so perhaps my equation is too simplified?
$$CV = 0.196FC + 14.119$$
The company of course masks the idea saying it's their intellectual property (etc.) but I'm just curious of if this is even possible or should the laboratory's reputation be called into question?

It sounds similarly to me like once the cold fusion.

The only chance is for well hydrogenated product and perhaps mistaking gross/net resp. higher/lower CV. See also https://www.engineeringtoolbox.com/fuels-higher-calorific-values-d_169.html

Surely a laboratory who have the instruments and expertise to do these tests wouldn't make a calculation mistake like this...

It may not be a mistake of the lab, but misinterpretation of data at whatever level. Or intention.

Since 2500 kcal/kg is about 10,000 kJ/kg, we could have a mixup of units.

In the industry kcal/kg is common and MJ/kg is less common. I have seen no other units regularly being used so I don't believe it's a mix up of units; especially if it's neither of the ones I've mentioned.

It is clearly possible for fuels to have higher calorific values per kg, even solid ones
Calorific values of fuels are well known and available from many sources. For example, this source list the values of fuel oil as 42.5 MJ/kg (same value if you prefer GJ/tonne) and LPG as 46.3 MJ/kg so there doesn't appear to be a simple obvious limit for fuels. Obviously fuel oil is a liquid and LPG a mildly compressed gas under normal conditions.
But one of the reasons for their higher calorific values is that compounds with larger proportions of hydrogen tend to have higher values than ones with more carbon (which when mostly pure has about 33 MJ/kg).
So solid hydrocarbon compounds should be better. And they are with paraffin wax coming in at about the 42 MJ/kg level.
So there is no good reason to expect even slightly higher values to be impossible. we already have examples fuels with that calorific capacity.

LPG LIQUIFIED ...

So it's not just a function of the amount  of fixed carbon, like with the equation in my answer? Could you elaborate how these CVs are possible?

@jimchmst LPG is a mix of small hydrocarbons (propane, butane etc.) that are gases at STP but are easy to liquefy at mild pressure (butane lighters under normal conditions have pressures of just 2-3 atmospheres to keep butane liquid).

@Hendrix13 In hydrocarbons some of the calorific value comes from the fixed carbon. But quite a lot comes from the hydrogen which burns to water releasing more heat. So you have to take that (and the amount of H) into account to get the total calorific content.

Interesting. Do you have a source with a more reputable and accurate equation than the one in my question?

@Hendrix13 Worth noting that pure hydrogen has a calorific value of 141.2 MJ/kg. But more complete reaction mechanisms for carbon, hydrogen and oxygen compounds and how to calculate the calorific value are given in this engineering toolbox page.

Matt that is what I said. The point is that combustion takes place in the gas phase and LPG has an apparent higher heat because the vaporization energy comes from ambient heat and not from combustion energy. Higher MW fuels require combustion energy to initiate vaporization.

@jimchmst I don't think the differences are mostly determined by the energy needed to vaporise the fuel. Do some calculations if you want to prove otherwise.


text:
Nucleophilic substitution in ethers and Alcohols
In Alcohols(secondary,tertiary,benzylic,allyic) on reaction with HX, X=(Cl,Br,I) SN1 takes place as H2O leaves and forms a carbonation.But in ethers on reaction with HX, X=(Cl,Br,I) (except tertiary,benzylic) SN2 takes place. Is ROH not as good leaving group as H2O.what is the reason for SN1 in Alcohols and SN2 in ethers??

From what I understand, the generalization you have constructed is not based on any gross rule that - "ROH not as good leaving group as H2O". I think what you mean is that the protonated alkoxy group is a poorer leaving group than the OH2+ group, in which case I would still not agree to a generalization unless I know the identity of the R group. There are many ethers that show SN1 substitution reactions as well, especially under acidic conditions. A good example is the highly strained oxirane ring, which opens via an SN1 reaction in low pH.

Solvation of the leaving group is important.  an ether under SN1 conditions will be less efficiently solvated


text:
If liquids absorb heat from their surroundings to become gas, why does gas feel hotter than liquid?
I know what you're thinking, "OP, are you for real!? Gas is obviously hotter because it absorbed heat from the environment!"
But hear me out. Yes it is hotter in absolute terms because it absorbed heat to become that way and if you placed one thermometer in the liquid and one in the gas then you'll get a higher temperature reading from the gas.
However, when you read the question again it says "absorbs heat from its environment" thus heat is removed from the environment, thus the environment is less hot than before, and if that "environment" was the palm of our hand (and it was heavier than air so it didn't fly off) then that gas should feel cooler than the liquid because it absorbed heat from our skin thus the nerve endings in our skin had temperature removed from them and now give us the sensation of being cooled down. (assume that the boiling point of the substance is our body temperature).

You need to revise some of your very basic premises, namely that the things always get hotter when you apply heat to them. That's not always so.

Be aware that being hot, or comparatively hotter and colder, depends on multiple objective parameters and is also a subjective, physiological and psychological parameter. Not only temperature plays role, but also heat capacity, specific and volumetric heat capacity, heat conductivity, at vapors also latent heat of evaporation, state of the human body surface tissues, state of the mind. An aluminium block and a piece of styrofoam, both heated up to 80 deg C, would be perceived very differently. So does 100 deg C water and water vapour.

Ok I think this will go way over my head. I'm only asking this to better understand refrigerants in HVAC, but I guess this is more than I need to know for that job!

If there is gaseous and liquid phase of the same matter, the gas is able to release energy of latent heat of evaporation. still staying at the boiling temperature. That is the reason why burns by water vapor are worse than due boiling water. As 1 g of vapor releases due its condentation to boiling water 5.5 tims more energy than is needed to warm up 1 g of water from 0 to 100 deg C.


text:
How to differentiate Coenzyme and Prosthetic group
Coenzyme is a non protein molecules which bind loosely with enzymes, its function is help catalytic reaction proper functioning.
Prosthetic group covalently and tightly bind with protein.As my understanding, I knew that the Heme group in hemoglobin is a good example explaining prosthetic group.
Here is my confusion:
1.Does Coenzyme and Prosthetic group both exist in Enzyme simultaneously?
2.Is there any example about Enzyme with prosthetic group?
3.Does Enzyme without Prothetic group affect catalytic reaction proceeding?

One example is the enzyme methionine synthase. It has vitamin B12 as a prosthetic group and S-adenosyl methionine as a cofactor. Without vitamin B12 present, you would expect no catalysis (in fact, the protein might not even fold properly). For some reading with interactive 3D figures, see https://proteopedia.org/wiki/index.php/Methionine_synthase.
According to IUPAC, these are the definitions of prosthetic group and cofactor:

Prosthetic group: A tightly bound, specific nonpolypeptide unit in a protein determining and involved in its biological activity. See also cofactor.

Cofactor: An organic molecule or ion (usually a metal ion) that is required by an enzyme for its activity. It may be attached either loosely (coenzyme) or tightly (prosthetic group).

Source: IUPAC Recommendations 1997

Is it the reason that Vitamin B12 has metallic coordination complex so it is a prosthetic group? Same as heme group in hemoglobin?

@GordonChao I added some information to the answer to address this follow-up question. Metal ions can be either tightly bounds or not, so they are not always considered prosthetic groups.


text:
How to calculate the buffer capacity for polyprotic acids? specifically H2PO4 --- HPO42- + H30+
as the title says. Is b= amount of OH-/ H3O divided by volume of buffer x change in pH wrong to use with polyprotic acids? Can I calculate the buffering capacity of a sodium phosphate buffer made up of NaH2PO4 and NaH2PO4 using this formula?

Generally, first thing is to check, if you can afford to treat it like if it were not a multiprotic acid. Typically if other pKa are far enough from the supposed pH.

phosphoric acid has 3 pkas. 2.12, 7.21, and 12.4. I'm considering the second pka and pH 7.4 can I treat it as if it isnt multiprotic?

I take it as your rhetorical question, before you will do the calculations and make the conclusion, with eventual final clarifying questions.


text:
Why does benzaldehyde turn yellow?
We used benzaldehyde for a demonstration of the Cannizzaro reaction in school, but the liquid we used appeared visibly yellow, as opposed to the colourless liquid described by Wikipedia.
A few reasons I could think of include:

Reaction with oxygen to form quinone-type compounds
Polymerisation induced by exposure to light
Impurities

I haven't found much discussion about this, nor do I have any candidate compounds for if my guesses above turn out to be right.

First two reasons are valid for pretty much any organics.

very old paper, but relevant: https://pubs.acs.org/doi/pdf/10.1021/ja01345a018

Tl;DR: All three factors are responsible for coloring of benzaldehyde (and pretty much any organics)

For the first two factors, I would mention the paper1 linked by @Andrew. Benzaldehyde is prone to atmospheric oxidation; the reaction is accelerated in presence of light to form primarily benzoic acid mixed other aromatics and polymerized aromatics causing darkening.
For the impurity aspect, let me quote the following from a paper2 discussing about benzaldehyde purification process:

[...] It is clear that traces of impurities, namely metals, influence markedly
the benzaldehyde activity. The active impurities present in pure benzaldehyde or
in that of anal, grade become visible after distillation leaving the distillation residue brown to yellow coloured. The criterion for the good quality of benzaldehyde is the colorless distillation residue.

My previous discussions:

Phenoquinone vs benzoquinone
Air- and light-sensitivity of phenol

References

THE ABSORPTION OF OXYGEN BY BENZALDEHYDE, H. J. Almquist and G. E. K. Branch
Journal of the American Chemical Society 1932 54 (6), 2293-2302
DOI: 10.1021/ja01345a018
Benzaldehyde oxidation test, a model reaction with radical
mechanism. II. The purification of benzaldehyde and stabilization of its activity by
J. GAŠPERÍK, Chem. zvesti 29 (6) 808-810 1975 (PDF)


text:
Why is a protonated carboxylic acid less acidic than a protonated ester?
To my understanding, both have analogous resonance structures except that the ester has on -OR and the carboxylic acid has an -OH. Why is a positive charge on C-O-R more stable than on C-O-H?

Nice question!  It could be improved a bit if you added the source you used for the acidity of protonated carboxylic acids and protonated esters.  Did you measure these yourself?  Are you talking about in solution or in the gas phase?   And what do you mean by "stable"? etc. etc.

Thank you. I mean in solution, why the pKa of a protonated carboxylic acid is higher than a protonated ester.

Meh, this example is nothing you could not try to brush off with inductive effect or somesuch. Why not ask about ketone instead of ester. Or about ether or alcohol? It appears this resonance stabilisation is at best useless in such stuff. This is simply way beyond lame high school rationalisations. Evans table puts protonated acid as strongest of all of them. These are more or less debatable values, though. All strong acids without a good way to find reliable data.

related https://chemistry.stackexchange.com/questions/116449/what-atom-would-be-protonated-first-in-a-strongly-acidic-solution https://chemistry.stackexchange.com/questions/30864/protonation-of-ketone-and-ester

@Mithoron Uncle Al is that you?


text:
Why is there a discrepancy between predicted and nominal voltages of alkaline batteries?
According to Wikipedia, the sum of standard reduction potentials for the half reactions for alkaline battery yield $\pu{1.43 V}:$
$$\begin{align}
\ce{Zn(s) + 2 HO–(aq) &-> ZnO(s) + H2O(l) + 2 e–} &\quad E^\circ &= \pu{+1.28 V} \\
\ce{2 MnO2(s) + H2O(l) + 2 e– &-> Mn2O3(s) + 2 HO–(aq)} &\quad E^\circ &= \pu{+0.15 V} \\
\hline
\ce{Zn(s) + 2 MnO2(s) &-> ZnO(s) + Mn2O3(s)} &\quad E^\circ &= \pu{+1.43 V}
\end{align}$$
However, the reported voltage of Duracell or other alkaline batteries is $\pu{1.5 V},$ and the actual zero-load voltage of a new alkaline battery is measured up to $\pu{1.65 V}.$ Why?

Electrodes do not have standard potentials.

The standard potential is calculated or measures at all activities equal to 1. .  The manufacturers have managed to adjust the voltage by changing the activities of some of the components. Since they are all solids it looks like it might be a difficult project, but do not forget that they have a plethora of engineers against solitary you. My guess is that the hydroxide content of the electrolyte is raised converting some of the zinc oxide to zincate ion [HZnO2-] thereby increasing the zinc potential a bit. Try calling a company up, I suggest Ultralife, and ask someone in the lab.

Electrodes do have a standard potential.

Electrode and their corresponding solution have the electrode potentials. Metal alone has no meaning.

the electrode is the substance in contact with its conjugate in aqueous solution not what is stuck in the solution.

@jimchmst It is always important in case of disagreement to verify, if all talk about the same thing. // An electrode(1) consists of an electrode(2) as a conductor and a redox system, while the electrode(2) may or may not be part of this redox system. The former case is typically a metal/metal ions system.  Platinum electrode(2) in hydrogen electrode(1) or old time chinhydron (hydrochinon + p-chinone aduct) pH electrode(1) is the latter case.

@jimchmst Correction: Electrodes(2) (in the alkaline zinc cells) do not have standard potentials of electrodes(1).

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

Standard potentials are valid for $1$ molar solutions, and only for these values. Commercial cells do not have $1$ molar solutions at their electrodes.

@Maurice Standard potentials are valid for unit activities. Activity of ions in 1 M solutions is very different from 1.  E.g. in old practical procedures is said that 2 M H2SO4 has pH very close to 0.

This bickering does not help the questioner obviously conditions are not standard. there are several approaches I suggested changing the zinc potential bit. If someone has another idea lets hear it. perhaps adding a more reactive metal to the zinc, or getting MnO2 to reduce to Mn[OH]2. I always thought there was wasted energy in Mn+3 salts.


text:
Are mixtures of finely ground solid fertilizer compounds stable?
If I had a combination of the following as finely ground powders, would it be relatively stable, or would they react to form other compounds?
Sodium Nitrate
Ammonium Nitrate
Calcium Nitrate
Potassium Nitrate
Magnesium Sulfate
Potassium Chloride
Ammonia Sulfate
Copper Sulfate
Zinc Sulfate
Ferrous Sulfate
Manganese Sulfate
Sodium Molybdate
Monoammonium Phosphate
Boric Acid
Sodium Hydroxide
(Some of course being hydrates)

General rule: Do not mix things, unless you know what you are doing.

I mix these compounds all the time in the presence of water.

I found with sad experience ,some release ammonia gas ( I do not remember which. I do not mix such materials now.)

Please read up on ammonium nitrate.

Quite a lot of those are dangerous just by themselves, never mind when finely ground of mixed. If you want to see why ammonium nitrate is problematic, go visit Beirut which was largely destroyed when a mishandled large store of it exploded.

All depends on amounts, concentrations and ratios. It is a big difference between:

being solid/just being dissolved  together
being together in diluted solution

Even in cooking recipes, conditions of mixing ingredients do matter.
Reactions and precipitations often have concentration thresholds. Higher concentrations lead to more intense reactions or  precipitations, undergoing in higher degree, or even to qualitatively new reactions.

Aside of chemical stability, there is danger it would absorb air humidity and all the fine powder could form a single big block.

Sodium hydroxide releases ammonia gas from all ammonium salts when they are all solid or being dissolved.
When being dissolved:

Sodium hydroxide precipitates hydroxides of transition metals like copper, iron, zinc, manganese.
Sulfates precipitate calcium as calcium sulfate.
Phosphate can precipitate calcium and will precipitate copper, zinc, iron, manganese.
Molybdate may precipitate calcium, copper, zinc, iron, manganese.
Boric acid may precipitate calcium, copper, zinc, iron, manganese.

Generally, it is not good to mix together

alkaline compounds (or ones preferring alkaline    conditions)

sodium hydroxide, sodium molybdate

acidic compounds (or ones preferring acidic conditions)

copper sulfate, zinc sulfate, ferrous sulfate, manganese sulfate, monoammonium phosphate, ammonia sulfate, ammonium nitrate

It may be advantageous to have

major nutritients ($\ce{Ca}$,$\ce{N}$, $\ce{P}$, $\ce{K}$) in solid state, and rather course than fine.
minor ones in a stock solution(s)

If you insist on mixing them, then I would suggest to keep sodium hydroxide alone in its original state and keep the rest in separate compatible mixtures, in the sense of above.

@BuckThorn I do not handle them. :-) // Well, the powder mixture is not supposed to stay in that state forever, but is intended to be dissolved // Mix powdered ammonia salts with alkaline compounds and sniff that. // If the OP insists on mixing them, then I would suggest at least to separate them into several compatible mixtures.

P.S.: I got it now about sodium hydroxide. :-) Sure, aside, and not as powder.

Lookup Galveston explosion or the Halifax explosion on the web. The city of Texas City or Halifax was destroyed, and I am not sure if the chemical was "finely ground". There are oxidizing and reducing agents in the mix, release of ammonia gas or precipitation of metal ions are your least concerns. Knowing the names of chemicals is not sufficient. You must know the properties and you have quite a list.
If you must mix these do so as a dilute solution paying particular attention to the order of mixing to prevent unwanted intermediate reactions and avoid cross contamination. Again learn the properties.


text:
Equilibrium constant magnitude for concentration vs. pressure
When introducing the concept of equilibrium constant, we often teach that the size of $K$ relative to $1$ indicates whether equilibrium lies toward the left (reactants) or right) products) of the balanced equation. However, we also know that there are different equilibrium constants depending on whether you express the numbers as ratios of concentrations ($K_c$) or pressures ($K_P$). So how should I think about the fact that the numerical value of $K$ depends on how I choose to express things?
I can illustrate what I mean with a simple example: Consider the reaction
$$
\ce{CO (g) + 2H2(g) <=> CH3OH(g)}
$$
At $220^\circ$C, this reaction as $K_c = 10.5$ but $K_P = 6\times10^{-3}$. So, $K_c$ indicates an equilibrium lying far to the right and $K_P$ indicates one lying far to the left. How should I interpret this?

This is a classical scaling issue. For example, is a $1$ ton object lighter than a $100$ milligram object?
The rule that $K>1$ or $K<1$ should not be taught because it will lead to exactly the same fallacies as raised in the question. As you know very well that $K_c$ and $K_p$ are related by $$K_p=K_c(RT)^{\Delta n}.$$
Then, at least three scenarios can arise depending on the value of ${\Delta n}$ as being equal to $0,1, -1$ in simple examples. This will lead to  $$K_p=K_c,$$ $$K_p=RTK_c,$$ and $$K_p=\frac{K_c}{RT}.$$
In short the simple rule of the equilibrium is favored if it is greater than unity is fallacious because it breaks down right here!

Or rather, comparison K and 1 should be done only for true and unitless thermodynamic constants, expressed in relative terms of activities (or fugacities), referring themselves to their values at standard state.

Thank you, this is really clear! But it brings up two questions: (1) If I had given a reaction with the same number of moles on each side, would $K_c$ have to equal $K_p$? And (2) if I convert partial pressures and concentrations to activities, do I get to the same value of $K$ in either case? Could you show how I might do that so that I can see the same numerical value results?

I agree with Poutnik. The implication of the conditions K>1 or K<1 should be taught but be properly explained.


text:
Usage of acid in redox titration
I've came to know, diluted H2SO4 is used as medium in KMnO4 titration. Whereas, HNO3 is not used as it is an oxidizing agent. My question is H2SO4 is also an oxidizing agent. Then what's the problem with HNO3?

You need usually hot concentrated H2SO4 to manifest its oxidative properties.

Does diluted HNO3 show its oxidative properties?

Does diluted HNO3 dissolve copper, while HCl does not?

Additionally, HNO3 is often contaminated by traces of NO2 and HNO2.

No, it doesn’t dissolve copper. So It also doesn’t show oxidative properties in diluted condition. Then what's the main problem in using it instead of H2SO4? Contamination?

Try again.......

Cold $\ce{H2SO4}$ and $\ce{HCl}$ do not react with copper, whatever their concentration. Cold $\ce{HNO3}$ does react with copper, whatever the concentration. Of course, the reaction rate depends on the $\ce{HNO3}$ concentration. It decreases at low values of concentration $\ce{[HNO3]}$. So $\ce{HNO3}$ is an oxidizing agent, and not cold $\ce{H2SO4}$

It depends on the use and exact reaction. Different oxidizers have different potentials [strengths] and react by different mechanisms and cause different reactions. If the oxidizing properties of permanganate and the reducing properties of the other reactants are paramount it is imperative that no other oxidants or reductants are present. In the analytical use of permanganate, a strong oxidizer like nitric acid would interfere, atmospheric O2 might interfere, Cl-, Br-, I- HS-would be oxidized. Complete reduction of permanganate to manganous ion requires an excess of Hydronium ion. The only common acids that supply H3O+ and have low oxidation potentials and cannot be oxidized are sulfuric and possibly phosphoric and some sulfonic acids; so that is what is used.


text:
Can we interpret the number of micro states calculated from molar entropy?
Forgive me if this is not the correct place to ask this. When I encounter a molar entropy, is it possible to rearrange the Boltzmann entropy formula $$S=k\log W$$ and meaningfully interpret the number of micro-states thus calculated?
In particular, I found the molar entropy of quartz $(\ce{SiO2})$ in a table: $$S=41.36\  \text{J K}^{-1} \text{mol}^{-1}.$$
Using the Boltzmann formula for a single formula unit of $\ce{SiO2}$ (rather than a mole), the number of micro-states $W$ is: $$W=\exp\left(\frac{S}{kN_A}\right)\approx145.$$
Is there a sense in which a single formula unit of $\ce{SiO2}$ within a quartz crystal can be arranged into one of $145$ different micro-states? Is there any way of describing what those states are? What else am I missing?

Welcome! Great question - one thing that would be helpful (via a comment here) would be to know a bit about your background. Have you had thermodynamics or statistical mechanics? Trying to figure out the best level to frame the answer...

I'm actually a geology student; I found the table in my textbook but couldn't understand the meaning of entropy other than a tool to construct phase diagrams for interpreting rock formation conditions. I haven't taken either of the courses you mentioned. I found the Boltzmann equation on Wikipedia as I was trying to better understand entropy.

As you note, entropy can be a measure of the number of possible accessible microstates. In other words, it's a measure of disorder.
One common example I give in class is if I toss 2-3 cards on the floor there are only a few ways they could be arranged and it's not a big mess. If I throw a whole deck of cards randomly on the floor, there are a lot of possible microstates (arrangements).
The third law of thermodynamics indicates that at absolute zero, for a perfect crystal, there is only one microstate:
$S=k \ln W=k \ln 1=0$
So how do we get the standard molar entropies? Well, there must be some disorder. In a molecule, you get some disorder because the molecule can rotate, translate, vibrate, and in many cases flop around (i.e., have different conformations).
For a real solid, you can have some entropy due to defects (i.e., the positions of the defects could be different = microstates).
For an ideal solid at finite temperature, without defects, the entropy derives from the vibrations / movements of the atoms. That is, the atoms will move around based on the vibrations of the solid. We often consider these as little harmonic oscillator springs. As a rule of thumb, weaker bonds tend to have weaker spring constants, so vibrations (and the corresponding entropy) tend to be higher.
In solids, we often compare different ionic compounds to see that the stronger ionic bonds have lower molar entropy (e.g., $\ce{LiF}$ vs. $\ce{NaCl}$ vs. $\ce{KCl}$):
From Chem LibreTexts:

$\ce{LiF}$ - 35.7 J/(mol•K) vs. 577 kJ/mol $\Delta H_f$
$\ce{NaCl}$ - 72.1 J/(mol•K) vs. 410 kJ/mol $\Delta H_f$

Note that the entropy is smaller for the more ionic $\ce{LiF}$. It's not a perfect correlation because it's about the shape of the vibrational energies rather than the heat of formation, but it's a useful heuristic.
In other words, for this type of entropy, it's not a function of equivalent microstates like throwing cards on the floor. It's a function of the vibrational entropy at finite temperature.
This implies not the Boltzmann formula (which requires equal weightings of the different microstates) but the Gibbs entropy formula:
$S=-k_{\mathrm{B}} \sum_i p_i \ln \left(p_i\right)$
in which $p_i$ is the probability of a particular micro state $i$ due to different energies $E_i$ and $k_{\mathrm{B}}$ is the Boltzmann constant.
Note this is essentially just a modification of the Boltzmann form, accounting that different microstates have different probabilities due to different relative energies.
So it's not that there really are 145 (or whatever) equivalent microstates at finite temperature. Some vibrations in a solid are lower energy and will contribute more to the entropy through what's called the zero-point energy. Some vibrations will be higher energy and will contribute less:
$p_i \propto e^{-\varepsilon_i /(k_B T)}$

I am still curious about the dimensionless number $W$ that would emerge from this calculation even for a defect-free crystal at non-zero temperature (if I understand correctly). If not a measurement of equivalent micro-states, is there another interpretation for $W$ as a measurement of the vibrational states you describe? Is the Boltzmann equation simply not appropriate for this situation?

Boltzmann would be fine if all the vibrations were equally likely. They're not, because some will be higher energy than others. So you need to use a somewhat different formula to calculate entropies.

@Henry, to add on to Geoff's answer, the Boltzmann entropy formula is applicable for a system with, among other things, a fixed energy E. In most (and your) circumstances, we are interested in the entropy at fixed temperature T instead, so the Boltzmann formula does not apply.


text:
Why doesn't the volume of NH3 = 325 ml in this problem?
I'm struggling with the problem below, and don't understand why

A buffer solution with pH = 9.0 is to be prepared from a 0.2 M ammonia
solution and a 0.15 M $\ce{HCl}$ solution. The sum of the molar
concentrations of $\ce{NH3}$ and $\ce{NH4+}$ in the buffer should be exactly 0.1
mol/ liter. Calculate the volumes of the two solutions needed to
prepare 650 ml buffer solution

So I set up my equations and solved, but get a strange result
650 ml = 0.65 liter, $pK_a$($\ce{NH4+}$) = 9.25
c($\ce{NH3}$) + c($\ce{NH4+}$) = 0.1 mol/liter
$\mathrm{pH}$ = $pK_a$ + $\log$(c($\ce{NH3}$) / (0.1 -c($\ce{NH3}$)))
9 = 9.25 + $\log$(c($\ce{NH3}$) / (0.1 - c($\ce{NH3}$)))
Solving this, we get c($\ce{NH3}$) = 0.036 mol/ liter
c($\ce{NH4+}$) = 0.1 - c($\ce{NH3}$) = 0.064 mol/ liter
n($\ce{NH4+}$) = 0.65 liter * 0.064 mol/ liter = 0.0416 mol
The $\ce{HCl}$ will be converted to $\ce{NH4+}$, so n($\ce{NH4+}$) = n($\ce{HCl}$) = 0.0416 mol
v($\ce{HCl}$) = n($\ce{HCl}$) / c($\ce{HCl}$) = 0.0416 mol / 0.15 mol/liter = 0.28 liter
n($\ce{NH3}$) = 0.65 liter * 0.036 mol/liter = 0.0234 mol
Part of the $\ce{NH3}$ will react with $\ce{HCl}$ to form $\ce{NH4+}$, and the rest will remain $\ce{NH3}$
 v(NH3) = v(HCl) + n(NH3) / c_initial(NH3) = 0.28 liter + 0.0234 mol / 0.2 mol/liter = 0.397 liter = 397 ml
v($\ce{NH3}$) = (n($\ce{HCl}$) + n($\ce{NH3}$)) / c_initial($\ce{NH3}$) = (0.0416 mol + 0.0234 mol) / 0.2 mol/liter = 0.325 liter = 325 ml

Now here is what I don't understand. We start from a concentration of 0.2 M, the sum of the concentration of $\ce{NH3}$ and $\ce{NH4+}$ should be 0.1 mol/liter, and $\ce{NH3}$ is required to create NH4+. With that in mind, shouldn't the volume of $\ce{NH3}$ be equal to half of the buffer solution, in this case 325 ml ?
 Am I missing something ?

"shouldn't the volume of NH3 be equal to half of the buffer solution, in this case 325 ml?" - Yes it should, and it seems you were given a terrible question. By the way, the working you show is largely correct until the last four lines, which are meaningless. You aren't adding NH4+ in any way, so the "volume of NH4+" doesn't mean anything.

@orthocresol So it should be v(HCl) instead of v(NH4+) in the last four lines ? What causes the result to be different ? I just noticed that if I multiply n(HCl) / c(HCl) in the fourth last line by 0.15 M / 0.20 M, the rest adds up to 325 ml for NH3, is this a coincidence ?

If it was just a half of it that would be like a lamest question (yes, I presume there are lamer ones but I don't want to ponder them). For starters concentrations aren't right for that. Another issue is density change in reaction. BTW don't write any "Thanks for your help !" in question - that's not a chat or discussion forum.

@Mithoron What do you mean ? Why does the density change need to be considered ? I removed the "Thanks for your help !" in question

@orthocresol What makes you think the question they gave us is terrible (just curious) ?

I mean, you found the contradiction yourself. A question which asks for a precise number, like this one, should have exactly one answer. It shouldn't have two possible answers depending on which part of it you read first. Nor should it expect you to come up with creative ways around the contradiction such as "adding water", which is not explicitly specified in the question. None of this is good.

@orthocresol the question is "calculate the volumes of the two solutions needed to prepare 650 ml buffer solution".  Very simple.

@RobertDiGiovanni Regardless of how simple it is, the student's job isn't to play Sherlock Holmes and figure out what the question is missing. Their job is to answer the question.

@Mithoron Why was this question closed ?

Why ask me, not the others? Well, I can see issues already in the title - apparently it really is 325, and you claim it's not - false premise, and it's altogether vague and doesn't really describe the question.

Actually, that should have been one of your first equations, leaving you to calculate volume of 0.15 M HCl needed to create the proper amount of NH4Cl in the buffer solution: 280 ml.
What was not asked, but implied, was the balance of 650 ml, 45 ml, would have to be water.


text:
Is the number of atoms per unit cell for Hexagonal Close-packed (HCP) really 6?
I am questioning whether the conventional N=6 for HCP is accurate. This question is spawned from an image in ZYbooks Material Science and Engineering textbook by Callister and Rethwisch, where they show an animation of a HCP. When oriented the unit cell along a plane of a vertical face, the interior atoms clearly bulge out beyond the walls of the unit cell.

This seems to imply the interior atoms are less than 3, thus N would be less than 6.
Can you demonstrate that the 3 interior atoms are completely within the Unit Cell?

Are we talking a conventional cell, or unit cell? Either way, an atom is a point, without extent, when assigned to be inside or outside a given cell volume.

A unit cell. In Materials Science and Engineering, by Callister and Rethwisch, they state without proving that HCP has 6 atoms. I am comfortably able to see that FCC has 4 atoms and BCC has 2 atoms. For HCP it seems like they are assuming the interior atoms are within the bounds of the unit cell. As far as I can tell the interior atoms extend beyond the walls of the Unit Cell.

It just occurred to me that the bits of the interior atoms that extend beyond the walls, are canceled by adjacent unit cells that extend beyond their own walls in the same way. I'm not 100% sure this is the solution, but it seems to make sense intuitively.

So, conventional unit cells. Which really aren’t that useful for understanding symmetries. Such as seeing that hcp and fcc are closely related.

Thanks Jon Custer, that is good to know that the conventional unit cells are not very useful for understanding symmetries. I ended up answering my own question... I'm gonna fix my question back to the original state and provide my answer as a solution. Right now they are kind of jumbled together...

The solution I came up with is yes, the interior atoms do extend beyond the walls of the Unit Cell. However, adjacent Unit Cells also have interior atoms that extend beyond their walls. These bits of volume of the interior atoms, the parts that extend beyond their unit cell, sum to 3 whole interior atoms per unit cell.
Initially I only found it obvious that this would happen for the corner and face atoms (extending into their neighbor cells), but upon closer inspection it is clear the same happens for the interior atoms.
Here is a Desmos file that illustrates this happening with two adjacent Unit cells.
https://www.desmos.com/calculator/zbpiaqxghi


text:
What is the best chemical for cleaning fossils and rocks
I am new to fossil collecting and would like to know what chemical is best for cleaning fossils and rocks of clay deposits. Specific acids/bases

There can't really be a general answer. The variety of rocks and fossils ot too large for a single, optimal, solution to exist.

You may want to specify the type of fossil. If it contains pyrite then it will probably require more care. See eg https://www.zoicpalaeotech.de/pages/pyritefossils

Great comment, thank you

Water is a great chemical with universal usage. Use other chemicals only if justified and you know, what you are doing. Aggressive chemicals like acids may destroy or damage surface layers.
Specifically fossils as various shells may be based on calcium carbonate, what could dissolve in acids.
Consult your more experienced offline or online colleagues or amateurs about their procedures or specific advises.
Consider also practical advises at

Earthscience SE site or
geo.libretexts.org - Bookshelves

Thank you for you comments


text:
Do aromatic canonical structures make the overall compound aromatic?

Here, the compound itself appears to be non-aromatic, but if the double bond between the ring and the oxygen is broken into a single bond, with oxygen pulling the electrons towards itself and acquiring a negative charge, and the ring acquiring a positive charge, it appears to be aromatic. Does that make the overall compound aromatic?

One thing to consider is there are multiple definitions of aromaticity. So to some degree it depends on your definition of aromaticity. Number of $\pi$ electrons in the ring? NMR peaks? Heat of formation / thermal stability? etc.

Related: https://chemistry.stackexchange.com/questions/163990/are-2-pyridone-and-cycloheptatrienide-antiaromatic/163991#163991. Note the discussion of carbonyl group polarization in an answer.

https://chemistry.stackexchange.com/questions/62588/is-cycloprop-2-en-1-ylidenepropanedinitrile-aromatic-or-not

Not always.
Consider, for example, 4-pyrone. It is possible to draw canonical structures containing the pyrylium heterocycle, which is aromatic. However, the contribution of this aromatic resonance form is rather small, as it requires creating separate opposing charges which could otherwise cancel out, and in particular places a formal positive charge on an electronegative oxygen atom. But more than just argumentation or application of simplistic rules of thumb (e.g. Hückel's rule), certain physical measurements can give an indication for or against aromaticity. The response of compounds to magnetic fields, in particular, is often used. In the case of 4-pyrone, there is indication it may not be aromatic.

In the case of cyclopropenone, however, it seems that aromaticity is important. Even though its reactivity is completely different to most standard aromatic compounds (due to its ring strain), the heavily deshielded $\mathrm{^1H}$ NMR signal at a chemical shift of 9.11 ppm in $\ce{CDCl3}$ (see note 7) is suggestive of an aromatic ring current (another magnetic field effect), with the particularly high value being attributable to the positive charge in an aromatic cyclopropenium ring.

Hückel's simplistic rule arrives at the same result in a fraction of the time.

@KarlsMaranjs ..... but as pointed out in the answer you just replied to, Hückel's rule is hardly a foolproof method. It has arrived at the right conclusion purely by chance. If I came up with a "rule" that said "all molecules are aromatic", it would also arrive at the same (correct) conclusion in an even shorter time, but you surely wouldn't think it sensible to use such a rule.

@orthocresol by chance? Hückel’s rule is derived from the MOT. A huge portion of theoretical chemistry was developed based on this theory, that you just called chance. Still, it’s not the only criteria to meet to determine if a molecule is aromatic. Of course every rule comes with exceptions, but in most cases if all 4 criteria are met then the molecule is almost certainly aromatic. Even for this one which is a very particular case, with the right analysis you have the right answer. And yes, exp methods are always the way to go, but without extra information Hückel is your best ally here.

@KarlsMaranjs Huckel's 4n+2 rule is a single, highly simplified, rule-of-thumb obtained from the generalisation of MO theory to monocyclic compounds. It does not have the same fundamental importance that MO theory itself has, so the wider application of MO theory within chemistry that you cite is not relevant. For the rest of your comment, I have nothing to say except what I already said, so I won't repeat myself - e.g. "with the right analysis you get the right answer" - I can claim the same about my obviously-false rule. I am not saying that Huckel is false, btw, just too simplified.

It is cyclic, planar and conjugated, so the last criteria to meet is the Huckel rule.
In this case you have 2 $\pi$ electrons, because you can consider the O takes one out of the ring so the O - C bond will be polarized as $\ce{O-}-{C}+$:
$$\ 4n + 2 = 2 $$
$$\ 4n = 0 $$
$$\ n = 0 $$
So I would say the molecule is aromatic


text:
Can I use potassium ferricyanide instead of potassium ferrocyanide for the identification of calcium?
I have to perform the identification test of calcium ion in calcium gluconate. I do not have potassium ferrocyanide for the test but I do have potassium ferricyanide. Can I take the ferrocyanide form ?

It looks like you follow the idea similar name = similar behavior. Such an idea may accidentally work, but I would not rely on such luck.

No. Ferricyanide is Fe3+, ferrocyanide is Fe2+. Different compounds with different chemical properties

Why do you use calcium ferrocyanide ? This substance is soluble in water. Calcium ferricyanide is also soluble in water.

@Maurice After reading the question, I have found some youtube video that shows ferrocyanide forms white precipitate with calcium ions. But I assume it would not be very selective. The old good school method of bubbling CO2 through alkalized solution could be better.

According to the Handbook of Chemistry and Physics, Calcium ferricyanide is deliquescent and "very soluble" in water. Calcium ferrocyanide has a solubility of $86.8$ g in $100$ mL water at $25$°C.

Although potasium ferrocyanide is recommended for the calcium test, you can technically use potassium ferricyanide, but there is a chance you will get a mixed salt containing ferriferrocyanide, giving you false positive


text:
Can amino acids exist in the completely non charged state?
Typically amino acids can exist in three states, as far as I’m aware, which depend on the pH:

However, most of the time I see amino acids in a completely uncharged state when shown in textbooks or on the internet:

The issue is, I cannot seem to think of any possible pH where such a state can exist. Please correct me if I’m wrong on this and tell me under what conditions such a state can occur. However, if I’m right then why do they show it as such… for simplicity?

That may occur uncharged in pure form or dissolved in polar aprotic solvents, not in water, where such form exist in trace equilibrium concentrations.

If I had a button to turn comments into answers…

@Karsten Kinda did that for you ;) Duping is close enough for me.


text:
Why is water present in "pure" nitric acid and RFNA and WFNA?
Various types of nitric acid have been used in rocket fuels as oxidisers (RFNA is red fuming nitric acid, WFNA is white fuming nitric acid) as they are often hypergolic with a wide variety of fuels.
RFNA is "pure" nitric acid with a lot of dissolved $\ce{N2O4}$ and WFNA is supposed to be "anhydrous" nitric acid, $\ce{HNO3}$ .
But early research found a variety of problems with suitable nitric acid oxidiser mixes that seemed to depend on their water content. Ignition delays and storage problems both seemed to be influenced by water content.
The chemists developing the fuel mixtures spent a great deal of effort trying to produce nitric acid that did not contain any water but struggled despite taking considerable efforts to prevent the mixtures absorbing water from the environment. But they kept observing water in carefully produced products which, superficially, seems odd for "anhydrous" WFNA acid or RFNA.
So why is water still present in those mixtures? Or, more generally, what sorts of species are present in highly concentrated nitric acid mixtures?

Nature abhors purity as much as it abhors a vacuum.

@OscarLanzi I perhaps should have added more about why that is not,apparently, the major reason especially for RFNA.

While these solutions are highly concentrated, nobody claims they're 100%. Not to mention even then there would be traces of H2O.

@Mithoron If you read some of the history of rocket fuel development, they certainly thought they should be able to get essentially water free products and they had a desperate need to do so for multiple reasons. It looks far more interesting than a simple view would expect.

Hmm, I don't think I got your point in there. If you ask why they don't actually make 100% even if it would be better, than it would be good if you made it more straightforward.

slightly related Why would red fuming nitric acid become white fuming nitric acid if left out at low temperature?

@Mithoron I added some clarification that the chemists did everything they could to make water free products but still failed. There is a genuine and interesting chemical issue with these mixtures as Putnik's answer suggests.

Aside of hygroscopicity and residual initial water, additional water is formed at $\ce{HNO3}$ decomposition to $\ce{NO2}$ ( it would not be red with just $\ce{N2O4}$ ). It is also created by the acid autoionization and due forming nitryl cation. Note that due excess of the acid, practically all water molecules are protonated.
\begin{align}
\ce{HNO3(l) + H2O(solv)&-> NO3-(solv) + H3O+(solv)}\\
\ce{6 HNO3(l) &-> 4 NO2(solv) + O2 + 2 H3O+(solv) + 2 NO3-(solv)}\\
\ce{2 HNO3(l) &<=> H2NO3+(solv) + NO3-(solv)}\\
\ce{H2NO3+(solv) + HNO3(l) &<=> NO2+(solv) + H3O+(solv) + NO3-(solv)}
\end{align}

Those reactions are a much better explanation than hygroscopicity. Is there any recent work on other "internal" equilibria that generate other species even in closed containers of fuming acid? I know some was done in the 1950s.


text:
Why does Le Chatelier's Principle hold when the volume of a system is decreases?
I understand that when an equilibrium system is subjected to a change that it will adjust to counteract the imposed change and that thus a decrease in volume (an increase in pressure) will cause the system to shift in the direction with fewer moles of gas. But why does this work in terms of collision theory and/or kinetics?
The answer I most often see to this is that the side of the reaction with more moles of gas is 'affected more', but wouldn't the collision frequency for both reactants and products increase by the same factor and thus no shift in equilibrium?
I have tried to understand this in terms of the Arrhenius equation; k=Zρ*exp(-Ea/RT) where Z is the frequency of collisions per cubic meter, but the same problem persists. Doesn't this value of Z just increase by the same factor for both forwards and reverse reactions?

Make mental analysis of $\ce{2 A <=> B}$  with $k_\mathrm{f} \propto p^2_\ce{A}$ and $k_\mathrm{b} \propto p_\ce{B}$.

Forgive me but how do we know this? I am not so familiar with the Arrhenius equation and haven't seen anything resembling this relationship before.

The Arrhenius equation affects the temperature dependence of proportionality constants of the above proportionalities. For cases at the same temperature, we can ignore the equation. I made a mistake above, should put there rather rate: $\frac{\mathrm{d}p_\ce{\mathrm{b}}}{\mathrm{d}t} = k_\mathrm{f}(T)p_\mathrm{A}^2 - k_\mathrm{b}(T)p_\mathrm{B}$, where both kinetics constants are results of the Arrhenius equation, that are constant or T dependent.

Where can I do further reading on this subject, I still don't completely grasp why the forward k is multiplied by the square of pressure.

If the pressure of A is doubled, how would change frequency of A-A collisions?

I would think doubled, but from the equation you gave above it would increase by a factor of 4?

Think about skating people on Winter stadion, randomly colliding with each other. Now double their number. Each would collide twice more often, as there is twice as much people. But twice more often collisions with twice as many people means 4 times more collisions.

I think I kind of understand what you’re saying although it may take some thinking over. Is there somewhere I can do some further reading?

try google searching with domain narrowing term site:chem.libretexts.org  The site is dedicated on aducation texts.

In the Arrhenius Equation, the value, Z, doesn't change with a change in volume. This is because, in collision theory, it represents the frequency of collisions at a specific or standard concentration of the species (i.e. $\pu{1.0 M}$).
One way to understand this is by using just simple forward and reverse rates. Say you have:
$$\ce{2A <=> B}$$
If the volume is decreased, and thus the pressure is increased, the equilibrium would shift right according to Le Chatelier's Principle. However, we can observe this too by breaking it into separate forward and reverse rates:
$$\ce{r_{for} = k_f[A]^2}$$
$$\ce{r_{rev} = k_r[B]}$$
Let's say the volume is halved. This means both $\ce{[A]}$ and $\ce{[B]}$ would double. However, when we consider $\ce{r_{for}}$ and $\ce{r_{rev}}$, $\ce{r_{for}}$ increases by more as the $\ce{[A]}$ is squared in the $\ce{r_{for}}$ expression. So, instantaneously after the volume decrease, $\ce{r_{for} > r_{rev}}$, which accordingly suggests that the system would shift right as previously predicted.
As this happens, $\ce{r_{for}}$ would decrease while $\ce{r_{rev}}$ would increase as more of $\ce{B}$ is formed until a new equilibrium is re-established.


text:
How does vinegar free deposits from a surface?
I use vinegar to clean/descale an electric kettle.
How does this work? i.e. how does the vinegar free the deposits (calcium, manganese?) from the surface?

$\ce{CaCO3 + 2 CH3COOH -> (CH3COO)2Ca + H2O + CO2}$

A test for limestone is to put a drop of vinegar on it.  If it fizzles, it's limestone.
"Scale" can build up in tea kettles, boilers, pipes, etc. as metal salts, commonly carbonates.  Vinegar acid reacts with the carbonate, forming CO2 (fizz) and dissolving the metal ions.
Other acids, such as phosphoric, sulfuric, and muriatic (HCl) can be used for descaling, but care must be taken that they do not harm the substrate.

Bear with me, not a chemist but trying to understand the chemistry. So Calcium carbonate (CaCO3) is bonded to the metal of the kettle. Vinegar comes along and separates a CO2 from the CaCO3. So the CO2 fizzes out, leaving a Ca and an O freely drifting in the vinegar/water solution. Is this correct? Can you describe in more detail the reaction of vinegar and the carbonate?

@AlLelopath  Poutnik has it in comments.  Actually, vinegar acid bonds to O and separates it from CO2 as H2O (O will not separate under basic conditions).  CO2 escapes as a gas, and the metal, now more soluble, is dissolved in the water.  Scale originally forms because heating and combining with CO2 makes metals much less soluble, hence being common in heating vessels such as boilers, water heaters and ... yes, tea kettles.  Having "soft" water means scale is less likely to form.

metal ... is dissolved in water ... Does that mean, theoretically at least, if you cleaned it enough times, you could wear through the metal?

@AlLelopath depends what the pipes are made of, but a rinse after treatment is not a bad idea. Metals referred to here are those already in solution (as "hard" water precipitating onto the surface), but you make a very good point, the acid, if it is too strong, may also attack the tea kettles as well.  This is why vinegar is a good choice.

A little more internet search and TIL calcium is a metal


text:
magnesium citrate to carbonate conversion - for oral consumption?
I've been instructed to drink a 10-oz bottle of magnesium citrate saline solution, which label says has about 17.5g mag citrate in it. (1.74g/fl oz)
I prefer to make my own solution using a powdered product called Natural Calm, which has "350mg mangnesum (as magnesium carbonate)" per serving (2 tsp). (It also has citric acid to make mag citrate in water - https://www.naturalvitality.com/products-naturalcalm-unflavored)
What's the roughly equivalent amount of the mag carbonate powder to get 10oz of mag citrate solution?
thank you!

1 tsp is not an exact unit.

Requests for medical advises are explicitly off-topic on Chemistry SE site. The best if you ask your physician if you can replace the drink by your formula and how.

If it's Magnesium you are after, the molecular weight of magnesium citrate (C6H6MgO7) is 214, and magnesium carbonate (MgCO3) is 84.  Magnesium has a weight of 24.
84/214 × 17.5 g = 6.87 g magnesium carbonate
6.87 g x 24g Mg/84g MgCO3 x 2 tsp/.350 grams Mg = 11.2 tsp !
This will give you the same amount of magnesium as the 17.5 g magnesium citrate in the entire 10 oz bottle and possibly a burp or two.
However, if a doctor or any other medical professional instructed magnesium citrate, please follow their instructions and discuss the matter with them before doing anything else.

Thanks so much Robert. Despite what I said, this is not for a medical procedure. Self-administered cleanse.


text:
Why is guanine sometimes referred to as "Acyclovir Impurity B"?
"Acyclovir Impurity B" is the name for guanine used by a handful of relatively low-traffic websites. Googling the term shows this. It is nearly always with a capital letter at the beginning of each word, which is another slightly strange thing, since it is a generic name, and therefore should be lower case except for the "B".
What seems especially remarkable is that This page was the relevant hit that was highest up on Page One of Google when I searched for "2-amino-1,9-dihydro-6H-purin-6-one" without quotes. The page linked to states the name as "Acyclovir Impurity B (sic)" and only in the fine print below does it mention any other names, where it says, "Synonyms-
2-Amino-1,9-dihydro-6H-purin-6-one
Guanine".
This is another website that confirms that "Acyclovir Impurity B" is a name for guanine.
There's nothing to indicate that guanine is ever referred by this name in Wikipedia. The closest I could find in Wikipedia by Googling "acyclovir impurity B" was in the entry for acyclovir where it is written

The aciclovir metabolite 9-Carboxymethoxymethylguanine (9-CMMG) has been shown to play a role in neurological adverse events, particularly in older people and those with reduced renal function.[34][35][36]

which contains the string "guanine".
It looks like many to all sites using the term "Acyclovir Impurity B" are selling it.
This paper  says that guanine is the major impurity of acyclovir, but doesn't explain why some others selling it avoid using the term "guanine". It also doesn't explain what the "B" means (doesn't that suggest that it is the second most major impurity?).
PubChem has this page that lists 122 "depositor-supplied synonyms" for "guanine". Included in the list are strange names like "mearlmaid" and "pearl essence" and various variations on "Acyclovir Impurity B" such as "Aciclovir EP Impurity B" though not the exact same name, as far as I could tell. I mean, to me, adding "EP" makes it a different name, though I could be wrong because I don't know what "EP" means.
Googling "mearlmaid" brought up the Wikipedia article on guanine, but searching the article for the word "mearlmaid" didn't detect it, so I guess it's in the meta text or whatever it's called.

I assume it may be just operational sequentional naming of unknown impurities  on HPLC chromatograms, until the identity was determined later. Then others may just refer to an original paper

If you want to see all the synonyms used for a compound, check PubChem. You'll note there are many, many terms. I suspect some of the low-traffic websites take every synonym in Wikipedia and list them in case someone is searching for "Mearlmaid" in which case I see a sales link in the top list.

@GeoffHutchison Awesome link, have added it to the question. I also added some text clarifying that I searched for  "2-amino-1,9-dihydro-6H-purin-6-one" without quotes using Google and the top relevant hit was "Acyclovir Impurity B".

Guanine is guanine. It was discovered centuries ago and named as such. No decent scientific paper will accept Acyclovir Impurity B as its synonym (without its context). In the pharmaceutical industry, however, this label is acceptable. EP stands for European Pharmacopia (set of rules for the pharma industry). The nomenclature A, B, C, D, there makes sense. Any medicine/drug that gets commercially prepared undergoes a very stringent chemical analysis for quality control.
In fact, an analytical expert who is able to analyze known and unknown peaks in a pharmaceutical product is highly valued (very high salary). In this case, acyclovir is the active pharmaceutical (AP) and then if someone analyzes it, e.g., on a liquid chromatography-mass spectrometry system, he/she might see signals (peaks) from other impurities. In order to assign those impurities in the prepared product analysis result, one can run an impurity standard of that particular AP. Instead of having a mouthful of impurity names in the standard, it is easier to designate those peaks by English letters "AP Impurity A", "AP Impurity B" etc. This is just pharma jargon, so there is nothing unusual about it. Guanine will remain guanine.

That is plausible. It makes perfect sense. On the other hand, I think the answer would be better still if it had a citation.

@MatthewChristopherBartsh, The hint is right there! EP = "The European Pharmacopoeia (Ph. Eur.) is a single reference work for the quality control of medicines." Chapter 5.10. CONTROL OF IMPURITIES IN SUBSTANCES FOR
PHARMACEUTICAL USE.


text:
Gibbs Free Energy for solvation using Gaussian09
I am trying to compute the $\Delta G$ for the solvation of a molecule using the continuum dielectric model in Gaussian09. Following the recomendation in the manual of the software, I am using the SMD (solvation model based on density), but I am haveing difficulty to interpret the output. This is the comand line I am using
#P MP2/6-311++G** SCRF(SMD,Solvent=Acetonitrile)

In SMD, the Gibbs free energy is divided in two terms
$$
G_{sol} = G_{EP} + G_{CDS},
$$
where $G_{EP}$ is the bulk-electrostatic contribution arising from a SCRF treatment and $G_{CDS}$ is the contribution arising from short-range interactions between the solute and solvent molecules in the first solvation shell. In the Gaussian output, there is the following
Error on total polarization charges =  0.04321
 SCF Done:  E(UHF) =  -263.928410150     A.U. after   22 cycles
            NFock= 22  Conv=0.84D-08     -V/T= 2.0012
 <Sx>= 0.0000 <Sy>= 0.0000 <Sz>= 0.5000 <S**2>= 0.8777 S= 0.5619
 <L.S>= 0.000000000000E+00
 KE= 2.636170281119D+02 PE=-1.036154210633D+03 EE= 3.034189165458D+02
 SMD-CDS (non-electrostatic) energy       (kcal/mol) =       1.57
 (included in total energy above)

and
     E(PUHF)=      -0.26394237020D+03        E(PMP2)=      -0.26487244445D+03

I can identify what is $G_{SMD}$ (1.57 kcal/mol in this case), but I don't know what is the $G_{EP}$ contribution. would it be the E(PMP2)? Or Gaussian is not computing this?

Usually Gaussian gives you the total energies. So E(PMP2) is what you need. However, you're missing the freq stuff so that value won't help you at all. Do a gp opt then freq and do the same with the solvent. Subtract. There's a white paper on thermochemistry you might want to enjoy.


text:
What is the precipitation reaction?
I asked my teacher if the reaction
$$\ce{(NH4)2CO3 (aq) + MgBr2 (aq) <=> MgCO3 (s) + 2(NH4)Br (aq)}$$
can be considered a displacement reaction. She answered that it was a precipitation reaction instead.
So what happened in this kind of reaction?

Precipitation reaction is just reactions that forms precipitates and is a subset of displacement reaction,

In fact, it is just
$$\ce{\ce{Mg^2+(aq) + CO3^2-(aq) -> MgCO3(s)}},$$
so the double displacement is very formal, as $\ce{NH4+}$ and $\ce{Br-}$ ions are just "spectator ions".
There are no real $\ce{(NH4)2CO3(aq)}$ nor $\ce{MgBr2(aq)}$, it is just a way of inventory of particles.
Precipitation reactions are, intuitively and not surprisingly, reactions forming precipitates.

Thanks!  However can you explain what do you mean by MgBr2 is not real?

I mean nothing by that as I have not said it. MgBr2(aq) is just a shortcut for Mg^2+(aq) + 2 Br-(aq).

It's a double displacement reaction, but it forms a solid precipitate which is $\ce{MgCO3}$

When you have a solution where the mathmatical product of the activity of two ions which form a 1:1 solid are greater than the solubility product then a solid will form.
A first approximation can be made that the activity is equal to the concentration.
The combination of ammonium carbonate and calcium bromide can be regarded as being equal to a mixture of ammonium bromide and calcium carbonate. The ionic solids dissociate when they are immersed in water thus forming mobile ions. The ions are then in an "ionic atmosphere" which alters their ability.
We normally write for a ZX salt
log fz = -A sqrt(I)/(1+(1.5*sqrt(I)) + B[X] + C[X^2]
This is a Pitzer equation. We can then use it for a poorly soluble salt to predict the concentration of the salt, when we have a more soluble salt EF present
fz = 10^(Bze[E] + Cze[E^2] - {A sqrt(I)/(1+(1.5*sqrt(I))}
So we can then write
Ksp = [X][Z]fx fz
Which becomes
Ksp = [X][Z] (10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})
We can simply things a little when [X] = [Z]
Ksp = [X]^2 (10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})
Rearrange to give us
1/[X]^2 = {(10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})} / Ksp
Rearrange further to give us
[X]^2 = Ksp / {(10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})}
From which we get
[X] = sqrt (Ksp / {(10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})})
This might seem like a lot of maths, but it is not so bad. It explains why we should expect the solubility of CaCO3 to increase as the concentration of the NH4Br increases.
This is a rather interesting problem in environmental chemistry, geochemistry and the chemistry of radioactive waste stores. One concern in the future is the release of radium from spent fuels store thousands of years into the future. The solubility of radium sulfate will be higher in salt water than it will be in distilled water.
If we ignore the activity issues then we can write a solubility product which is based on concentrations alone, this is an apparent solubility product. But the apparent and real solubility products will deviate more and more as the ionic strength (I) of the system increases.

Why is an answer using algebra so unpopular ?

I guess the issue is formatting, or rather a lack thereof. As a friendly reminder, we use MathJax as markup for literate typesetting of math expressions. Currently I find the math pretty much unreadable.

I have got no idea of how to use MathJax, I normally write algebra as if I was writing an expression in excel. How do I use MathJax ?

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, see Which symbols are written in roman (upright) font and which are italicized?


text:
Aqueous solution of FeCl3 is yellow whereas other metals having the same d5 configuration are pale violet coloured. Explain the origin of the colour
The aqueous solution of the other metals with the same d5 configuration give pale violet colour.
Pale colour could be caused by the d-d transitions.
i got to know that Ferric chloride forms a self ionizing complex in aqueous medium as [Fe(H2O)4Cl2]^+[FeCl4]2-
Now i dont know what type of transitions are taking place in this compound which is giving out the yellow colour.

If, in fact, ferric chloride firms $\ce{FeCl4^-}$, then this ion could indeed be responsible for the different color of the chloride in solution.
Textbooks often describe high-spin $d^5$ complexes as only pale colored because the $d-d$ transition is spin-forbidden (in a nonrelatvistic theory), and what little transition (and thus coloration in white light) cones from (relativistic) spin-orbit coupling. Less attention is paid to the fact that with the most common metal centers that give this configuration, we typically have octahedral complexes, whose center of symmetry or near-symmetry (with nonidrntucal ligands) also renders the $d-d$ transitions forbidden due to lack of an inherent dipole component in the transition (Laporte forbidden). Imperfect symmetry with different ligands, or molecular vibrations, are necessary to get around this barrier.
$\ce{FeCl4^-}$, however, has tetrahedral geometry instead, so no center of symmetry. The spin limitation is still there, but without the Laporte forbidden character we see more color from the spin-orbit coupling. The more intense color of the tetrahedral complex then overwhelms anything the proposed octahedral solution complex may offer.


text:
Equality of partial derivatives (Schwartz’s theorem) and total differential
I'm reviewing some Physical Chemistry concepts and the two books I'm using show the total differential and right away they present the Schwartz's theorem. I don't get the relationship. In the screenshot below you can see that the author mentions that the mixed partial second derivatives can be obtained from the total differential. What is the relationship between equations like 1.7 and 1.8, in general?
I got the screenshot from this excellent book: https://peverati.github.io/pchem1/SystemVariables.html#thermodynamic-variables
Atkins does the same think on Mathematical Background 2, pg 91, ninth edition.

There is no real "direct" connection between 1.7 and 1.8. Equations 1.8 are manifestation of the Schwartz's theorem, which states that you can (if the function is "well-behaved") change the order of differentiation. That means that given the function $f(x,y)$, the second derivatives are the same: $\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$. In the equation 1.8, this equality is then used for the function $P(T,V)$.

I thought there was a connection because of this statement "the mixed partial second derivatives that can be obtained from eq. (1.7) are the same"
On Atkins is similar " The quantity df is also called the differential of f. Successive partial derivatives may be taken in any order:"

I read the surrounding context and I don't see the connection myself - I think if you understand 1.7 and 1.8 separately you're doing fine.


text:
Resonance structures of benzene
Does benzene have more than two resonance structures? I draw a third resonance structure keeping 1 bond fixed and rotating other two bonds I get opposite charges on two para position. Is this structure wrong?

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

There are many resonance structures which contribute to the overall reality of benzene.
Some of these contain charges, the higher the energy of a resonance form the less it contributes to the overall reality. Consider the "normal" resonance forms of benzene which most people think of.

Next we have some with two charges in them, I have not drawn all the possible resonance forms with two charges. Sadly if I draw all of the resonance forms possible I have to stay up very late and I will miss out on my beauty sleep (maybe at midnight I will turn into a pumpkin).

Here are the two highest energy resonance forms which I can think of, these will make a minimal contribution to the overall reality. But it is good to be aware that they exist.


text:
Standard suffixes for compounds?
Long ago, I learned that suffixes like -ide, -ate, -ose, etc. had specific meanings.
Now, I'm seeing all these drug ads on TV with generic names that all end in -ab or -ib.  Do these have a standard meaning, or is it a matter of conforming to the crowd?

You can add a few examples. This is not a universal ending.

Related: https://chemistry.stackexchange.com/questions/161667/why-does-almost-every-drug-that-causes-dependence-have-this-ine-suffix

Note that pharmacological and chemical nomenclature conventions are independent.

Anything ending in -ib is a kinase inhibitor e.g. Dasatinib
Anything ending in -mab is a monoclonal antibody e.g Rituximab


text:
Can ICE tables give two answers?
I was recently given the question:

The equilibrium constant, K, for the following reaction:
N₂(g) + O₂(g) ↔ 2NO(g)
is 4.00 x 10⁻² at a very high temperature. The reaction is at equilibrium at this temperature with [N2] = [O2] = 0.100 M and [NO] = 0.0200 M in a 2.00 liter flask. If 0.120 mol of NO is suddenly added to the reaction mixture what will be the concentrations of all species when equilibrium is re-established?

I think filling out the ICE table would look like this:

N₂
O₂
2NO

I
.100
.100
.080

C
+x
+x
-2x

E
.100+x
.100+x
.080-2x

Which would make the the equilibrium formula:
$$
4.00\cdot 10^{-2}=\frac{(.080-2x)^{2}}{(.100+x)(.100+x)}
$$
However, when solving for $x$ I get $x=\frac{8.2\pm\sqrt{7.84}}{198}$ or $.0\overline{5}$ and $.0\overline{27}$.
nvm I get it now but as this is my first time using stackexchange and this syntax took me too long I feel  like posting it because the TeX looks nice to look at :)
noo why does the table only work in preview mode :(

Something wrong, I guess.

The equation as written can be simplified and so not a quadratic

@porphyrin It can be done in this case, as Sam202 explicitly shows in his answer. But it does not address a general case when it cannot be done. Then more or less obvious constraints of valid values must be considered.

Scenarios leading to a quadratic equation have two mathematical solutions, real or complex. But, one of them often does not have interpretation in scientific context, as it's value is out of the acceptable value range.

Typically, it provides, directly or indirectly, a negative value of a quantity that cannot be negative. A trivial example:
What is the length of the shorter side of a rectangular with area $\pu{30 cm2}$, if the other side is longer by $\pu{1 cm}$?
$$x(x+1)=30$$
$$x^2 + x - 30 = 0$$
$$x=\frac{-1 \pm \sqrt{1+120}}{2}=\frac{-1 \pm 11}{2}$$
A side of the length $\pu{-6 cm}$ does not obviously make physical sense.
But $\pu{5 cm}$ does.

In your particular case, the greater result value would lead to negative $\ce{NO}$ concentration, so it is not a valid result. There is the constraint $x \le \pu{0.04 mol L-1}$ for $[\ce{NO}] \ge \pu{0 mol L-1}$.

$$4.00\cdot 10^{-2}=\frac{(0.080-2x)^{2}}{(0.100+x)(0.100+x)}=\frac{(0.080-2x)^2}{(0.100+x)^2}=\left(\frac{0.080-2x}{0.100+x}\right)^2=0.040$$
Taking the square root on both sides:
$$\frac{0.080-2x}{0.100+x}=0.2$$
Solving for x:
$$x=0.0273$$
Notice we did not consider the negative root above, since an equilibrium constant (or its square root) having a negative value is nonsense.