text:
Solids to liquids process
Melting point is the critical point at which a solid starts to be transformed into a liquid? Does this mean that the geometric lattice of a solid is broken?

The notion is right, but you have to be careful. Glass does not have a matrix, but it still melts.

I mean for solids that do have a lattice.

I don't understand what [tag:transition-state-theory] tag has to do with melting of crystalline matter. Why have you decided to add this tag? Also, could you please elaborate a bit more on what exactly what exactly you are interested in? Currently the question seems trivial to me, but I suspect you are trying to ask something else.


text:
If magnesium reacts with oxygen to produce magnesium oxide only on the application of heat, then why isn't it categorised as an endothermic reaction?
Why isn't the energy that is used to overcome the activation energy of magnesium metal atoms to react with oxygen, not taken into account?

It is. The gist is that there is a large energy difference between metallic magnesium and magnesium oxide. So if you have a ribbon of magnesium metal that is a yard long you don't have to "activate" the whole yard. You just need to heat up a 1/4 inch and get that burning. Then the burning itself supplies enough heat  to keep the ribbon burning.

Exothermic/Endothermic is based on the overall net energy change.

So since more energy is released than absorbed when the whole yard-long magnesium ribbon is burnt, it is an exothermic reaction. Thank you so much, @MaxW and @Waylander!

Why isn't the energy that is used to overcome the activation energy of magnesium metal atoms to react with oxygen, not taken into account?

Typically, the activation energy is much higher than the difference in energy between reactants and products. The activation energy gets used when you go from reactants to activated complex, and more or less becomes available again when you go from activated complex to product. As Waylander stated in the comments, exo- or endothermic is based on the difference in energy between reactants and products, not the size of the activation energy (otherwise, all reactions would be endothermic).

If magnesium reacts with oxygen to produce magnesium oxide only on the application of heat, then why isn't it categorised as an endothermic reaction?

Just because you are heating something to get it to react does not mean there is a net flow of heat from the surrounding into the reaction. It would be more accurate to say that you are increasing the temperature to allow the reaction to proceed.

[from the comments]: You just need to heat up a 1/4 inch and get that burning. Then the burning itself supplies enough heat to keep the ribbon burning.

Again, this is about the temperature, not about exothermic or endothermic. There are multiple reasons why a reaction would happen at high temperatures but not at low temperatures:

Change in equilibrium constant: For example, water vapor forms dew at lower temperature, and dew evaporates again at higher temperature. The activation energy is sufficiently low that evaporation and condensation take place at either temperature, but the net direction changes.
Reaction rates: Some reactions have a high activation energy and are too slow at low temperature, so you have to raise the temperature to observe them. This does not mean they are endothermic (check out the thermite reaction).
Allowing the reactants to come together: Magnesium is covered in a layer of magnesium oxide, preventing oxygen in air to react with the magnesium. At higher temperature, magnesium metal melts and comes in contact with air.
A slow step in a multi-step chain reaction: In radical reactions, the initial formation of a radical has the highest activation energy, whereas reaction of a radical with other reactants to form another radical (chain propagation) has a lower activation energy. In this case, the first step would require higher temperature (like a spark) or a catalyst (radical starter) and then the chain reaction can proceed at low temperature.

For all but the first, it does not matter whether the reaction is endothermic or exothermic - higher temperature helps or is required to observe a reaction. You can have a endothermic reaction even at lower temperatures. The heat transfer can happen even under low temperature conditions because the molecules have thermal energy nonetheless.
Using day-to-day language, that sounds weird because we associate heat with temperatures higher than body temperature, so we might doubt that there is heat in cold water. However, reactions that cool down the water even more demonstrate that the water can transfer heat to these processes (e.g. cold water can melt ice).

The reaction of magnesium with oxygen gives off a lot of heat, but it is difficult to get the reaction started. The surface of magnesium metal is coated with a thin adherent film of oxide/hydroxide, and this prevents continuous reaction. Also, the bulk of the metal conducts heat away; fine powder or ribbon reacts much more easily - may even be pyrophoric! Essentially, the reaction stifles itself, until the metal gets so hot that the film becomes less protective. Perhaps because of different thermal expansion coefficients or dehydration of hydroxide.
Many metals appear to be similarly inert to oxygen unless heat or a catalyst is present. Iron reacts with oxygen slowly, but water increases the rate of corrosion, and chloride ion increases the rate of reaction even more.

source: Stack Exchange

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

sort: votes


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What methods could be used for determining the concentration of Zn2+ in a solution?
I know that complexometric titration with EDTA is one method. What other methods could be used?

Have you searched Google Scholar? Use these key words atomic absorption spectrophotometer, ICP for zinc.

Mass spectroscopy(MS)
HPLC-MS, if separation is required
Atomic emission spectroscopy(AES)
(Probably)
Polarography/voltammetry

The list is not exhaustive.

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


text:
Melting point of aspirin, contradicting sources
Querying wolfram alpha for the melting point of aspirin (link) returns $\pu{140 °C}$. The Wikipedia page for Aspirin lists $\pu{136 °C}$ instead, citing this book.
How am I supposed to know which figure to cite? And in general, what should I do when I encounter contradicting chemical data?

Like all measurements there's a range. Some ranges are larger than others. If you're unsure, you should report the range.

Could someone expand on what it means for aspirin to melt?  Naively, it sounds like the temp where the little solid tablet becomes a puddle of liquid, but I assume it's more of a medical effectiveness definition (like when the molecule breaks down and "melts" into constituents molecules or something).

@user1717828 Melting is turning into a puddle of liquid; the molecule breaking down is decomposition.

How many significant figures does WolframAlpha use? The number 140 is ambiguous with respect to how many significant figures it has, and if WolframAlpha for some reason decided only to use 2 significant figures, then there is no contradiction.

I might be cautious with the WA data: they also claim the boiling point is $140^\circ \mathrm{C}$.  Further, they don't provide a useful citation to a primary source.

While I'm a fan of WolframAlpha, most of the figures I've gotten from them tend to be fairly approximate.  In general I wouldn't regard them as an authoritative source, but rather just a quick-and-easy source of approximations.

Wolfram company doesn't conduct any experimental determinations of physical constants for chemical compounds and uses literature data sources.
The webpage for ChemicalData Source Information lists numerous sources of chemical information used by the company's products, including Wolfram Alpha.
Wolfram Alpha Knowledge Database is linked with 87th ed. (2006) of CRC Handbook of Chemistry and Physics (m.p. $\pu{135 °C}$), whereas Wikipedia cites newer 92nd ed. (2011) of CRC Handbook of Chemistry and Physics (m.p. $\pu{136 °C}$).

Obviously, Wolfram's sorting algorithm preferred another data source despite it's quoting the CRC Handbook.
Why and how it did that cannot be answered as the server-side engine is proprietary.
Note that even newer 97th ed. (2017) of CRC Handbook of Chemistry and Physics adds measurement error of $\pu{4 °C}$ (m.p. $\pu{136(4) °C}$).
The difficulty of determining the exact value probably arises from the fact that at that temperature range thermal decomposition already begins, which is reflected in m.p. dependence on heating rate and atmosphere and melt solidifying at lower temperatures (down to $\pu{118 °C}$ [1, table 3.4]).
References

Aspirin and Related Drugs; Rainsford, K. D., Ed.; CRC Press; Taylor & Francis: London, 2004. ISBN 978-0-7484-0885-6.

In other words, the two temperatures are within measurement uncertainty of each other.

The CRC Handbook of Chemistry and Physics ($\mathrm{86^{th} Ed.}$, section on Physical Constants of Organic Compounds) provides an MP of $\pu{135^\circ C}$
for aspirin (2-(Acetyloxy)benzoic acid).
According to the Handbook:

The data in the table have been derived from many sources, including both the primary literature and evaluated compilations.

If you need a highly reliable source you'd want to perform a more thorough search rather than rely on a tertiary (encyclopedic) source such as the wikipedia, and in particular ideally find primary sources of data (original literature reporting experimental results) or perhaps secondary (compilation of scientific results).
This is a pretty good guide to different sources of data:

Sources are considered primary, secondary, or tertiary depending on
  the originality of the information presented and their proximity or
  how close they are to the source of information. This distinction can
  differ between subjects and disciplines. In the sciences, research
  findings may be communicated informally between researchers through
  email, presented at conferences (primary source), and then, possibly,
  published as a journal article or technical report (primary source).
  Once published, the information may be commented on by other
  researchers (secondary sources), and/or professionally indexed in a
  database (secondary sources). Later the information may be summarized
  into an encyclopedic or reference book format (tertiary sources).

I would regard the CRC Handbook as a tertiary source of information based on the above definition, but it's generally considered highly reliable. Wolfram Alpha or the wikipedia are also tertiary sources but because of the way they are compiled, unless stated and verified you can't be sure where the data comes from. One of the ongoing efforts at the wikipedia is to include a mechanism to ensure that such information is verified.
As regards the CRC, you should attempt to use the most recently available edition, but the reported melting point of aspirin is not likely to have changed by much in the last hundred years (and yet consider the differences between the values quoted in your question, and this and another answer).


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Does iron nail sink in rubbing alcohol
I have been trying for so long to get answer

Yes, iron sinks in rubbing alcohol, water and in any liquid with density lower than density of iron.
It would float e.g in melted lead or gold.
For a very light iron/steel pieces like needles,, or pieces very flat like a classical Gillette razor blade, they could stay afloat on water due surface tension. But this is seldom case for alcohols, as their surface tension is several times smaller.

I firmly believe that the knowledge of iron's ability to sink in water, when it comes into one's mind, deserves more heft than a mere forum message. One should learn it the hard way, by drowning an axe in the river or something like that. Therefore, though your answer is correct and otherwise good, I have mixed feelings about it.

I prefer dropping a camera from the canoe, but it is indecisive, as cameras are seldom a stainless steel cases these days. Sometimes it is hard to imagine one does not know the answer yet, where one reaches a stack exchange site. BTW, I do like your comments, seen in various topics.

That can be pretty instructive too, though cameras hardly contain much iron, if any at all. My concern, however, is of different nature. Some questions tend to display the lack of certain underlying knowledge that pretty much makes the very question void, together with the answer. Imagine a person asking for a very specific piece of information and its location in a very specific book on the subject, and then adding a casual remark that reveals they are illiterate. That would instantly make you aware that their real problem is not there, but very many levels deep down below. Same thing here.

Yes, this was a thoughtful remark.


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Why there is a need of independent definitions for both mass and mole?
Considering that the two quantities are linked by reality, how can be both imposed by independent definitions?
I am aware that the change taking place next May does not effect common weighing in chemistry and engineering, but still the new definition of mole (an exact number of entities, to be called Avogadro number as well) does confuse me, conceptually.
Direct counting is not a practical way of sampling, so already I do not see the good of defining mole as a fixed number of entities (of course I have thought in this way already, but I am not a definition :).
How can I now weigh a mole of substance, in principle? It seems that without a history-of-science background, armed by the mass definition and mole definition, I cannot weigh in moles despite of having both very well defined.
Before, the "only" knowledge required was a table of atomic masses, and one could have done that even ignoring the value in grams for 1 amu, if not interested in spelling out the value of Avogadro number itself. I think is what happened before mass spectroscopy sort of techniques revealed the absolute mass of particles and atoms.
To date, the Mole is conceptually and mathematically linked to a mass measurement, ie. the amu and its multiples. With the new definition the two things are distincts, as 12 g of 12 Carbon is not a Mole anymore.
Please clarify without listing the non-operational pros of the new definitions, as the fact that the latter stresses the difference between matter as mass and matter as substance even more than before, to mention one that, as a chemist,  I like much.
The abstract of this paper by B. P. Leonard,Metrologia 2010, 47 (3), L5–L8 expresses my concerns in a more elegant way: Metrologia 2010, 47 (3), L5–L8

The fundamental concept of the mole requires the number of entities comprising one mole, i.e. Avogadro's number, to be exactly equal to the gram-to-dalton mass ratio. If this compatibility condition is to be satisfied, the mole, the kilogram and the dalton cannot all be defined independently. This note concerns recent Metrologia publications that do, however, propose independent definitions of all three quantities: the mole by fixing the value of Avogadro's number and the kilogram by fixing the value of the Planck constant, while retaining the current carbon-12-based dalton. Adoption of these incompatible definitions would likely cause serious widespread confusion and might even result in a split in scholarly and technical communication between the quantum
physics and chemistry communities. Other entirely compatible alternatives are possible: either retaining the current (inexact) carbon-12-based mole and dalton with an independently redefined kilogram or redefining the mole by fixing the value of Avogadro's number, with a compatible dalton that is exact in terms of the redefined kilogram.

Check the articles in Metrologia (An official society journal dealing with measurements) with key words mole and kilogram. None of the approaches are problem free. See for example, "Comments on recent proposals for redefining the mole and kilogram" by BP Leonard (2010) in Metrologia. One the Avogadro number is redefined, the C-12, will no longer remain 12.00000000000000000 (ad infinitum)

Thanks. I am reading already materials that I found links for here. One point is that we already approximated things all along. For instance, taking an amount in g equal to the molecular mass assumed 1 atomic mass unit to be the same, while is not as every atom has is own mass defect. Ie. We never got an exact number of Avogadro of particles, even in principle. That was for 12 C only. I am always surprised by the depth, practical and experimental, of the reasoning involved in defining units.

True, the atomic masses are not absolute either. I have asked at so many places as to why we don't measure absolute atomic masses, and nobody has an answer including myself despite having decade long experience in the academic/research world. The choice of C-12 is also arbitrary because the true mass of C12 cannot be 12 exactly. A reasonable answer was it that it experimentally difficult to measure absolute masses. The problem is that from schools we start to take all these units for granted and never give a second thought and nobody (teachers) ever bothers to make us give it a second thought.

But I am not in school and I consider myself a chemist, so I am upset with myself. All this started because explaining the absurdity of homeopathy. My friend asked about Avogadro number and mole.  While this was ok, I afterwards realizes starting May something is deeply different (perhaps no but so it appears to me). Anyway I will edit my Q linking to the Leonard abstract that renders my concern in a more elegant way.  @M. Farooq

@ Alchimista, I was sure that this query came from a chemist not a student. My interest also arose because my niece was asking about Avogadro and mole. After explaining the typical story, I had a moment of guilt as I felt I just repeated what was written in her school book.

The downvoter could have left a comment. In order to make clearer why I felt to answer, in this update a paste the abstract of an article focused on the points of concern.

Those keeping down voting this could rather answer instead, or at least down vote and leave a hint in form of a comment.


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Which of the two structures is more stable?

Which of the two structures is more stable?

As far as I know, alkenes with a greater number of hyperconjugating structures are more stable. Hyperconjugation has a greater stabilizing effect than inductive effect. So, (d) should be more stable as it has more α-hydrogens.
But the answer given in the book is (c).
Is the answer wrong? Or am I missing something?

Only sterics may be important here.

The explanation can[1] be given by analysing their respective resonance[2] structures which is as follows:

(Credits: ChemWriter)

Now, inductive effects would work.
In case of first (c's resonance), positive charge would be stabilised by $\ce{+I}$ effect of methyl group; whereas in the second (d's resonance) one, $\ce{+I}$ effect would become dead upto the positive charge.
Hence, less stable would be (d).

Notes:
[1]: Firstly, I'm not sure if hyperconjugation will stabilize carbocation in resonance structures or not, because it will separate the charges even more. Secondly, in the given structures, it is very hard to differentiate the stability, as contribution of inductive and hyperconjugation are resulting in comparable stabilities. Hence, I tried it explaining by other canonical form.
[2]: resonance will happen due to conjugation of $\ce{\pi}$ bonds in both of the molecules


text:
How to recreate bioluminescence in fireflies in a lab?
I have several mg/mL of D-luciferin, luciferase (firefly), ATP solution (25 mmol) and MgSO4 (for the mg2+ ions). The experiment has to be done in cuvettes of 2,5 mL (if I get any results, it will be measured with a spectrophotometer).
I was wondering if you can give me an idea/tips/advice how I can recreate bioluminescence with those chemicals.
I am struggling with the ratio's of the chemicals, pH and whether I should dilute the solution or not.
I don't know how I should mix the ratio's of D-luciferin, luciferase, ATP solution etc.
Next thing is pH, I dont know what the optimal pH will be of this mixture and how I can change the pH accurately. These two points go with dilution as well, because I want to use water as a solvent and for dilution, but I dont know if it is needed or not.
(Also, if you know how the other chemicals might impact the energy that will be absorbed in the mixture and how it will be released when you know that the emission wavelength of D-luciferin is around 530 nm in water, please share you knowledge if you want)
I would greatly appreciate it if someone can give me advice or lead me to the right direction on the problems that I am facing with this bioluminescence experiment. I have tried asking this to other people, but no one was willing to help me, please.

Buffer with physiological pH instead of water. Probably dilute as little as possible for stronger effect imo.

I think with some effort Google could provide a protocol. Did you search for it? Biology SE might also have a community of teachers who did it


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In Ion Chromatography instrument, Does the specific conductivity (peak area) of an analyte depends upon background conductivity?
In Ion Chromatography instrument,
Does the specific conductivity  (peak area) of an analyte depends upon background conductivity ?
Assume Specific Conductivity (peak area) of fluoride ion is 4 unit for 1 ppm fluoride standard with background conductivity of 15 uS/cm (Carbonate/ bicarbonate eluent with suppression.
What will happen if the background conductivity increases to 30 uS/cm or 50 uS/cm.
Pl do not consider the change in retention time.
https://www.metrohm.com/easyDB/Web-61006520_std_carbonate/orig

The heart of an ion chromatograph is the suppressor, whose sole purpose is to minimize the background conductivity and convert analytes into highly conductive ions for example fluoride ion becomes HF, chloride becomes HCl and so on. If your background conductivity is high, it means that the conversion of F(-), Cl(-) into corresponding acids is also not 100% efficient-peak height can potentially decrease as a result. The peak areas will also change. Although the background is higher, one can still make a calibration curve (it will be less sensitive than the one with low background).
Remember that peak area in chromatography is not an absolute measure like mass; it is only true for one particular condition and set-up.

Does the specific conductivity (peak area) of an analyte depends upon background conductivity?

It depends on how the background measurement is made by the instrumentation. (How overlapping peaks are handled would be a different but related problem...)
In this day and age I'd expect the spectra to be digitized and processed by a microprocessor system at least. One way to handle the background would be to sample regions on both sides of the peak and then draw a best fit line through those two regions to serve as the background for the peak. Thus the S/N ratio decreases as background increases and the area would be measured with less precision, but the same accuracy.


text:
Using Roman Numerals to indicate a compound
I don't completely understand it's use. Here's an example I don't understand:
Shortly after discussing $\ce{[CuCl4]^2-}$, an exam paper asked a question about copper(II) chloride. I understand that copper(II) chloride could refer to $\ce{CuCl2}$, calling it copper(II) chloride, but couldn't it also refer to $\ce{[CuCl4]^2-}$? I would understand the question in reference to $\ce{CuCl2}$, but it got confusing when I realised it could equally be referring to $\ce{[CuCl4]^2-}$, especially after it just mentioned it.

No, those are two distinctive species with different names:

$\ce{CuCl2}$: copper(II) chloride;
$\ce{[CuCl4]^2−}$: tetrachloridocuprate(II).

Oxidation number denoted with the Roman numerals in parentheses refers to the element it's placed after and doesn't carry any additional information besides the fact that copper exists as $\ce{Cu^2+}$ in both compounds.
Nomenclature rules were designed that way so that a name would unambiguously define a single compound only (I'd suggest the Red Book's Nomenclature by IUPAC [1] as a reference).
References

IUPAC “Red Book” Nomenclature of Inorganic Chemistry, 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, UK, 2005. ISBN 978-0-85404-438-2. (PDF)


text:
Is there a error at page 566 of Atkins Physical Chemistry?
At page 556 I found this question being analyzed and the rate law is provided.
However, when I apply steady state approximation, I can't find the result without the factor 2.
Is it me being wrong?

Edition number?

I guess I'm a dolt, but I don't understand "I can't find the result without the coefficient 2."

It's edition 11.

The factor of 2 originates from the definition of $k_b$ (I thought it was an error, but it turns out IUPAC guidelines lead to its presence).
If $k_\mathrm{a}' \gg k_\mathrm{b}[\ce{O2}]$ and defining
$$K_\mathrm{a} = \frac{k_\mathrm{a}}{k_\mathrm{a}'}$$
it follows from the equation posted in the OP that
$$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = 2k_\mathrm{b}K_\mathrm{a}[\ce{NO}]^2[\ce{O2}]$$
However, starting from scratch we could write that
$$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = k_\mathrm{b}[\ce{N2O2}][\ce{O2}]$$
(this is not the standard IUPAC convention however).
Then, if
$$K_\mathrm{a} = \frac{k_\mathrm{a}}{k_\mathrm{a}'} = \frac{[\ce{N2O2}]}{[\ce{NO}]^2}$$
it follows that
$$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = k_\mathrm{b}K_\mathrm{a}[\ce{NO}]^2[\ce{O2}]$$
and the factor of $2$ is not there. In the fourth version of Atkins' book (which I parallel in this derivation from "scratch") the factor of 2 is missing, but in that book the analysis does not proceed by dissecting the full rate law before imposing the pre-equilibrium approximation. The OP presumably sources a later version of the book. As noted in a comment, IUPAC guidelines call for incorporation of that factor of 2. One should write
$$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = 2k_\mathrm{b}[\ce{N2O2}][\ce{O2}]$$ because that step of the reaction reads $$\ce{N2O2 + O2->2NO2}$$ The stoichiometric factor of 2 means two moles of
$\ce{NO2}$ are produced for every mole of reaction.

Um. It sort of depends on how you define the rate equation for the third reaction in the series; if you define it instead as $$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = 2k'\mathrm{b}[\ce{N2O2}][\ce{O2}]$$ then you end up with the correct factor of 2. That simply necessitates that my rate constant $k'\mathrm b = k_\mathrm b / 2$. In fact, this is more in line with IUPAC: https://goldbook.iupac.org/html/R/R05156.html There's a question about this on SE, but I don't have the time to find it now.

@orthocresol totally right...

@NightWriter I think it would be helpful if you edited your answer to indicate more clearly that including the factor of 2 is in fact the standard convention and not an error, rather than just acknowledging that at the end.

@orthocresol This one?

@hBy2Py I was thinking of this one https://chemistry.stackexchange.com/q/38167/16683, but yours fits as well and I wouldn't be surprised if there are even more.

@orthocresol Ah, yeah, yours looks rather more canonical and detailed than mine.


text:
How do I make 100 milliliters of 100mM H2O2 using 3% H2O2?
I need to make 100 milliliters of 100mM H2O2 using 3% H2O2.
Do I:

add X milliliters of distilled water to the 3% H2O2? How many ml?

or

add X milliliters of 3% H2O2 to the distilled water? How many ml?

You take a certain volume of 3% $\ce{H2O2}$ and add water until the total volume is 100 mL.  Because 3% $\ce{H2O2}$ is mostly water, the total volume will be very close to the sum of the volume of the two liquids you are mixing. How far did you get in finding the answer, and where did you get stuck?

I am a non-chemist and non-student. I just needed this one question answered for an old-seed germination project I am working on. All of the layman values I have found online have been specified by H2O2 percentage, and all scientific papers have been specified by mM.

It seems that this is some sort of reagent solution so +/- 25% is probably no problem.
Assume density 1 g/ml for 3% solution which gives 0.03 g $\ce{H2O2}$ per ml, or 30 g $\ce{H2O2}$ per liter.
The molecular weight of $\ce{H2O2}$ is about 34 g/mol so:
$\dfrac{30}{34} \approx 1$ molar.
You want 100 mMolar which is 0.100 molar. Thus you need a 10 fold dilution.
So add 10 ml of the 3% to 90 ml water to get 100 ml of solution.
Typically add the chemical solution to pure water.

THANK YOU!! I can't upvote you as a non-member, but I really appreciate your help on this. I am using H2O2 to stimulate germination rates in old seeds. The scientific studies that I am using all use mM.


text:
Is there a name for this algorithm to calculate the concentration of a mixture of two solutions containing the same solute?
There is an algorithm called "Mischungskreuz" (German for "x of mixing") that is sometimes taught as a shortcut to figure out the following problem:
You have two solutions that contain a solute at different concentrations $c_1$ and $c_2$. At what ratio $V_1/V_2$ do you have to mix them so that the mixture has the desired concentration $c_m$?
For example, let's say you want to make a 22% solution from a 35% and a 15% solution. You write the desired concentration in the center and the available concentrations at the left ends of the "x", and get the ratio of volumes on the right side of the "x" as shown below:

So in this case, mix 7 parts of 35% with 13 parts of 15% solution to get the desired 22%.
Source: https://de.wikipedia.org/wiki/Mischungskreuz
My questions are: Is this method taught outside of Germany, and is there a non-German (maybe English?) name for it?

Cute. I never saw that before.

Very interesting!  I have not seen it any English textbook so far. I am an analytical chemist. Most English books teach the dilution formula or mass balance as CiVi=CfVf. The German mixing cross (if this translation is better of Mischungskreuz) is a short cut to solve two problems. If you check Wörterbuch der Chemie / Dictionary of Chemistry: Deutsch/Englisch - English, it also calls it the dilution formula. https://books.google.com/books?id=mfFdDwAAQBAJ&pg=PA104&dq=Mischungskreuz+chemistry&hl=en&sa=X&ved=0ahUKEwjQv8bqpZnhAhUJXa0KHbBAAqYQ6AEIKzAA#v=onepage&q=Mischungskreuz%20chemistry&f=false

I repostt my comment, originally to the answer below, for maximum visibility: I do have an important comment and I am not sure if I should write it in form of answer. From a physical chemical point of view this rules is simply wrong. A fact that likely explain why isn't teached or widespread. Of course by contingencies, though wrong, it can be useful, for instance for preparing cleaning or conditioning solutions, of for purposes with little or no physical chemical and analytical relevance. This likely explains why it was(is?) familiar to those working in dairy or sugar industries.

As was already pointed out, you cannot reliably use the Mischungskreuz to calculate mixtures this way, mostly because volume is not necessarily a conserved quantity - think mixing Ethanol and Water; the sum of the mixture is less than its constituent parts, which really messes up the whole calculation and therein lies the crux of the matter. Which is exactly why the Mischungskreuz is taught (at least where i live) utilizing mass & mass fractions, since those are conserved quantities and the math checks out again. Hope this helped.

@CoffeeAddict This is a comment rather than an answer to the original post.

This is a so-called "Pearson's square" or "Box method" of balancing ratios, originally used extensively in dairy industry (at least since 1920s judging from Google Books search).
Earlier the similar approach has been used in sugar industry by using "Cobenz diagrams" aka spider diagrams.
Widely popularized in Soviet books for analytical chemistry at least since 1940s (probably adapted from the German literature as many other tech novelties of that time were), also used in current Russian literature by the names "Метод креста" ("Cross method"); "Конверт Пирсона" ("Pearson's envelope") or "Диагональная схема правила смешения" ("Diagonal mixing rule scheme").

Very interesting and thanks for sharing this info. I have always been in favor of learning another language besides English for scientific purposes. English is my second language. I am writing one article for the Journal of Chemical Education on the utility of foreign languages in literature search. Your point provides another motivation to finish that article soon.

I remember learning it in the Czech chemistry high school as "křížové pravidlo", What literally means "the cross rule". But the truth is, Czech chemistry was influenced by Germany, as German language was like a Latin in chemistry in late 19th/early 20th century.

I do have an important comment and I am not sure if I should write it in form of answer.  From a physical chemical point of view this rules is simply wrong. A fact that likely explain why isn't teached or widespread. Of course by contingencies is wrong but can be useful, for instance for preparing cleaning or conditioning solutions, of for purposes with little or no physical chemical relevance.  Again, this likely explain why it was (is?) familiar to those working in dairy or sugar industries.


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Why are synthetic pH indicators used over natural indicators?
Synthetic indicators seem to be exclusively used when determining the pH of a substance with an indicator (with the exception of that school experiment where you boil cabbage to demonstrate natural pH indicators) or in acid-base titrations etc. over natural indicators. Why is this the case when naturally occurring indicator can be easily prepared from accessible and sustainable methods, such as boiling cabbage?
Intuition seems to suggest that perhaps synthetic indicators are much more accurate and reliable, being specifically designed for the task, is this the case? Or are there other reasons why natural indicators are regularly used as a cost-effective, environmentally friendly and easily prepared alternative to synthetic pH indicators?
What are the advantages/disadvantages of using synthetic indicators over natural ones (or vice versa)?

Probably it's still cheaper. Also why waste good cabbage ;)

Scientists like to minimize unknown factors. If you can give me a top 3 list of what's in red cabbage, I will be impressed.

Another point is that any kind of analysis is about achieving consistency. It is often better to synthesize a pure compound than to try to extract it from some biological and purify it.

Most of the common synthetic pH indicators were probably discovered somewhere around 1850-1920, when organic chemistry really started to take off with simple aryl compounds. This means the indicators are very easy to make with modern knowledge/facilities/supply lines, so they can be made pure in metric ton scale for pennies. Hard to beat those economics.

As a side note, you shouldn't equate "natural" and "environmentally-friendly." Extracting a bunch of hydrogen cyanide from bitter almonds and dumping it in a lake has the same effect as synthesizing a bunch of hydrogen cyanide using Andrussow oxidation and dumping it a lake.

In acid-base titrations, synthetic indicators are exclusively used to find accurate end-point determinations because they always have a highly defined color change at certain pHs. For example, phenolphthalein ($\mathrm{p}K_\mathrm{a} = 9.7$ at $\pu{25 ^{\circ}C}$) is colorless in acidic solutions (precisely $0 \lt \mathrm{pH} \lt 8.2$), but it is pink in basic conditions when the pH of the solution goes above 8.2 (precisely $8.2 \lt \mathrm{pH} \lt 12.0$), which makes it ideal for strong acid-strong base titrations (Wikipedia).  On the other hand, bromothymol blue ($\mathrm{p}K_\mathrm{a} = 7.1$ at $\pu{25 ^{\circ}C}$) is yellow in acidic solutions (precisely $0 \lt \mathrm{pH} \lt 6.0$), but it is blue in basic conditions when the pH of the solution goes above 7.6 ($7.6 \lt \mathrm{pH}$). However, it shows greenish-blue color in a neutral solution (precisely $6.0 \lt \mathrm{pH} \lt 7.6$; see the picture below).

It also change color from yellow to pink in high acidic conditions such in concentrated hydrochloric acid ($0 \gt \mathrm{pH}$; see most left test tube in the picture below), thus it is ideal to use as a universal indicator (Wikipedia). These two examples would clearly show you, as you correctly put it, synthetic indicators are much more accurate and reliable since they are specifically designed for the task.

Sure, natural indicators could be easily prepared from easily attainable row materials, and hence, could be regularly used as a cost-effective and environmentally friendly alternative to synthetic pH indicators, but I regret to say that it can be used only in high school science activities to fascinate science-loving kids. Since those natural sources contain several color changing substances (e.g., flavanones, flavones, flavonols, and anthocyanidins), the color change at the end-points is not sharp, perhaps, due to the difficulty of determining $\mathrm{p}K_\mathrm{a}$ of relevant indicator(s). For example, although boiled red cabbage water can be used as a universal indicator in high school projects and show the end point of an acid-base titration (see the following figure), it cannot be used in a day-to-day analytical lab to do accurate calculations.

Also, another reason for not using natural indicators is that most of these dyes tend to decompose in higher pH values.  The best example is most common natural color indicator, anthocyanidins. The stability of anthocyanidins is dependent on pH. At a low pH (acidic conditions), colored anthocyanidins are present, whereas at a higher pH (basic conditions) the colorless chalcones forms are present (see scheme below) (Wikipedia).

Most importantly, aside from the above, flavonols including anthocyanidins have been studied extensively in drug development due to their plethora of biological activities (Ref.1). Thus they have been targetted in much more important tasks than a usage as pH indicators.
Reference:

M. Rahnasto-Rilla, J. Tyni, M. Huovinen, E. Jarho, T. Kulikowicz, S. Ravichandran, V. A. Bohr, L. Ferrucci, M. Lahtela-Kakkonen, R. Moaddel, “Natural polyphenols as sirtuin 6 modulators,” Scientific Reports 2018, 8, 4163 (11 pages) (DOI:10.1038/s41598-018-22388-5).

The main issue with natural indicators is their purity, long term chemical stability and shelf-life. Note that some natural indicators are pretty good for example the famous litmus paper is obtained from lichens. It is a mixture of several pigments like the red cabbage. Sometimes titrations are carried out at warm temperatures, do you think the pigments in cabbage survive heat in acidic and basic solutions?


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Is NaCl the only precipitate when you add HCl to concentrated NaCl
When aqueous HCl is added to a saturated salt water solution, is NaCl the only precipitate or does some HCl crystalize as well?
Sources online say that NaCl crystallizes but they don't say what happens to the HCl
https://bouman.chem.georgetown.edu/S02/lect8/lect8.htm
http://www.csun.edu/~ml727939/coursework/695/common%20ion%20effect/recrystallization%20of%20NaCl%20solution%20with%20HCl.htm

I would guess that excess HCl will either continue to displace NaCl or be left in the gas phase.

Even if the HCl is aqueous?

HCl is a gas at nominal room temperature and pressure. If you leave a container of HCl acid (HCl in water) open, then the water in the solution will evaporate, and the HCl will go back into the gas phase. There would be no HCl ppt.

HCl and H2O form a high boiling azeotrop at near 20% of HCl. Any HCl solution, when evaporated, converges to final concentration of the azeotrop. So evaporating the concentrated HCl will provide mainly gaseous HCl.

$\ce{HCl}$ is at ambient conditions a gas, very well soluble in water, dissociating as a strong acid.
When there is excess of it in solution, that cannot be kept solved, it does not precipitate, but evaporates or even boils out, depending on scenario.


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Treatment of ambidentate ligands in geometrical and optical isomerism
Are ambidentate ligands considered different entities when they donate through different atoms? For example will a hypothetical square planar complex like [Pt(ONO)2(NO2)2] show geometrical isomerism?

i think you are right

To specify [Pt(NO2)4]2+ would mean the ligand donates electrons through the N-atom, and not the O-atom. If it were to donate through O, it would be written as (ONO) and not (NO2).
Thus, this compound would not show geometrical isomerism. Also note that, if NO2 ligands are replaced by ONO, then that is a form of structural isomerism in complexes, called linkage isomerism.

Edited the complex to [Pt(NO2)2(ONO)2]. This doesn't answer my main question though, as to whether -ONO and -NO2 will be treated differently when considering stereoisomerism

If you write the complex as you have written now, then it does show geometrical isomerism. The point is, if a ligand is written as NO2 then it is nitrito-N and if it is written as ONO then it is nitrito-O. Similarly for other ligands (CN and NC are different, etc). Just like in organic chemistry, when discussing stereo isomers, don't count structural isomers.


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Flame Ionization Detector versus Mass Spectrometer
Why is chromatographic separation more important with a flame ionization detector than with a mass spectrometer?

FID requires separation as it is not specific and cannot distinguish 2 non separated compounds.
MS is specific and can determine the particular compound even qualitatively.
For complex matrices, it is always good to have saparation even for MS, as MS spectrum of mixture would be confusing.


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Conversion of graphite into diamond

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.
The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $\pu{2e-6 m3 mol-1}$.
If C(graphite) is converted to C(diamond) isothermally at $\pu{T = 298 K}$, the pressure at which C(graphite) is in equilibrium with C(diamond), is
(A) $\pu{14501 bar}$
(B) $\pu{58001 bar}$
(C) $\pu{1450 bar}$
(D) $\pu{29001 bar}$

We know that
$$\mathrm{d}G = V\mathrm{d}P - S\mathrm{d}T$$
At constant temperature, $\mathrm{d}T = 0$ and we are left with
$$\mathrm{d}G = V\mathrm{d}P$$
But as the volume is also changing in conversion of graphite to diamond and the pressure has also to be increased, how do I integrate this equation? Please tell me if I'm going wrong.

Why not just use $ΔV$ instead of $V$ since the contraction takes place? You then integrate from 1 bar to the unknown pressure, assuming that $ΔG = 0$ at equilibrium.

But isn't the V in the equation dG=VdP - SdT the absolute volume of the system? Or is it the change in volume?

That equation is derived for the constant volume (partial derivative with constant volume). In your case volume changes, so this has to be taken into account too.

Sorry but I'm not getting it, can you please ellaborate this equation a bit because I don't know that there are partial derivatives in this equation too.

You need to apply the equation dG=VdP to the graphite and the diamond separately.  So, for example, for the graphite, $G_g-G^0_g=V_gP$.  At equilibrium, the free energy of the graphite is equal to the free energy of the diamond.

Thanks.. Now I got it.. I was applying that equation simultaneously for both which wasn't correct..

@ChetMiller how did integral of v*dP become VgP?

@satan29. I don't understand what you are asking.


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When burning Metal Salts such as Lithium Chloride is the emission toxic?
I'm completely new; Just thought this was the place to ask.

Burning is the wrong word. Lithium chloride or sodium chloride cannot burn. If the temperatute is high enough it can melt and later vaporize. For harzard conditions you can read MSDS sheet for the compounds.
https://www.atmos.umd.edu/~russ/MSDS/lithium_chloride.htm


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Basic strength of the compound with an conjugate, and an aromatic amine

According to the source of the question answer is c>a>b but shouldn't the order be c>b>a as (a) is attached to a benzene ring and the lone pair will be delocalized?

B is effectively non-basic as it is part of an amide. The given answer is correct.

I think we can all agree that the aliphatic amine has the greatest basicity since the lone pair is not delocalised into any neighbouring $\pi$ network. You have mentioned about the delocalisation of the lone pair on the $\ce {-NH2}$ substituent into the aromatic ring. However, as Waylander commented, the the lone pair of the $\ce {N}$ in the amide is also delocalised into the adjacent carbonyl group and he even goes to say that it is "effectively non-basic". He is certainly justified in making this comment because the protonation of amides actually occur at the carbonyl $\ce {O}$ and not the $\ce {N}$. To see this better, we need to examine the extent of delocalisation of the lone pairs on $\ce {N}$.
This analysis I believe can be carried out in two ways.
Looking at the resonance structures
Firstly, we could take a look at the resonance structures which can be drawn to illustrate the delocalisation. Shown below the resonance structures that can  be drawn for the delocalisation of the lone pair in aniline, taken from here.

Shown below is the resonance strcutres which can be drawn for the delocalisation in an amide, taken from here.

Yes, more resonance structures can be drawn for the delocalisation of the lone pair into the aromatic ring. But this does not imply that the extent of delocalisation is greater. We have to take a look at the extent of contribution of the resonance structures to the resonance hybrid. Observe that in the three resonance structures, we have $\ce {N}$ bearing a positive formal charge while another $\ce {C}$ is bearing a formal negative charge. Recall that $\ce {N}$ is more electronegative than $\ce {C}$, thus this distribution of charge is less likely to be the case. Hence, the three resonance structures are likely to have small contributions to the overall resonance hybrid.
For the amide, we have a resonance structure where there is $\ce {N}$ incurring a positive formal charge while an $\ce {O}$ takes up the formal negative charge. This is a much more favourable and hence, much more likely charge distribution in the amide. Hence, this resonance structure has quite significant contribution to the resonance hybrid, albeit still less than the neutral structure.
Looking at some data
For there to be maximum delocalisation, the $\ce {N}$ atom must be as close to the $\ce {sp^2}$ hybridised state as possible where it can have an unhybridised $\ce {p}$ orbital being parallel to the adjacent $\pi$ system. How can we assess its hybridisation state? Well... We can look at the deviation from planarity of the $\ce {NH2}$ group. We can look at the value of the pyramidalisation angle between the $\ce {C-N}$ bond and the bisector of the $\ce {H-N-H}$ bond angle. The following values are obtained from Wiki:
\begin{array}{|c|c|c|c|} \hline
\text{Compound} & \text{Pyramidalisation angle} \\ \hline
\ce{Formamide} & 180 ^\circ\\ \hline
\ce{Aniline} & 142.5 ^\circ\ \\ \hline
\ce{Methylamine} & 125 ^\circ\ \\ \hline
\end{array}
Notice that the amide is fully planar! This suggests a hybridisation state of $\ce {N}$ in the amide that is almost fully $\ce {sp^2}$ Although the $\ce {NH2}$ in aniline may not be fully planar, it is still relatively more planar than the pyramidal methylamine, suggesting greater $\ce {sp^2}$ character in the $\ce {N}$ of aniline than the usual amine. More evidence regarding the deviation from planarity can also be obtained from the energetics data for pyramidal inversion given by Ron here.
Conclusion
Both the analysis based on resonance structures and that based on empirical data converge to give the same conclusion:

There is greater delocalisation in the amide than in aniline.

Thus, the given answer is fully correct.


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Gibbs free energy in standard state vs. equilibrium
I have a problem with the definition of the standard Gibbs energy and its connection to the equilibrium constants.
I think, that I've basically understood what the different equation mean but there is one thing, I'm unable to understand:
On the one hand:
One may describe a chemical reaction with $\Delta G=\Delta G^\circ + RT\ln{Q}$. In equilibrium $\Delta G = 0$ and the equation reads $\Delta G^\circ = -RT \ln{K}$.
On the other hand:
The definition of standard state is very clear: pressure = 1 bar and all reactants and products must have activity = 1.
If I consider these two aspects separately, everything seems to be fine. But these two concepts have to be valid at the same time, what leads to $\Delta G^\circ = 0$ (always), since $K=1$ (all activities are per definition = 1).
Therefore, $\Delta G^\circ$ would be always zero. I know that this isn't true, but I don't understand why.
Can anyone explain this to me?
Thanks!

"since $K=1$" not necessarily true; it is $Q = 1$. Nobody said that at standard state the system must be in equilibrium.

ok, but the formulae say: At standard state $\Delta G^\circ = -RT\ln{K}$ and K is the ratio of the activities of reactants and products in equilibrium, due to the standard state definition K = 1, because it says: each $a=1$. Doesn't this definition disagree with your comment? I find this all really confusing.

I find that terribly confusing and wrong if it claims $K = 1$.

I know it is wrong, but I don't get why :( Did you understand my problem, that is the pure formula seems to contradict the general understanding...

@user76122 Orthocresol is right. Your definition of standard state in the context of  $\Delta G^{0}$ is wrong. The standard state of a pure material entails that its activity is 1. But in the context of the equation for $\Delta G$ you deal conceptually with mixtures and thus not pure materials. If you follow my derivation of the formula (see here) you can see which assumptions go into $\Delta G^{0}$. It is true that it is defined for standard pressure/concentration, but activity being equal to 1 is not presumed.

@Philipp I read your linked answer. At the moment I think my problem arises, indeed, from a misunderstanding of $K$ and $Q$. Regarding the activities: In my understanding $a=1$ is automatically true in standard state. Since $a$ is a generalization of terms like $\frac{\varphi_i p_i}{p^0}$. If the system is in its standard state, i.e. $p=p^0$, $\varphi_i=1$ per definition and therefore $a=1$, isn't it? For example, in electrochemistry $a=1$ for every species is explicitly demanded in standard state.

@user76122 I think you're still on the wrong track concerning activities and standard state. The activity being equal to 1 in standard state is only true for pure materials. Take the term $\frac{\varphi_{i} p_{i}}{p}$ for example. The $p_{i}$ is the partial pressure of substance $i$ in the mixture. When you have the mixture at standard conditions with $p=p^{0}$ and $\varphi = 1$ the term $\frac{\varphi_{i} p_{i}}{p^{0}} = \frac{p_{i}}{p^{0}}$ still isn't equal to 1 because $p = \sum_{j} p_{j}$. Only in the limit of enriching the mixture with $i$ until it's nearly pure, you get $p_i = p = p^0$.

@Philipp I relooked at my textbook and you are right. So the "trick" is to know which standard state one should use for a real application? So far I found four standard states: infinit dilute solution, pure substance, ideal gas (standard pressure) and in electrochemistry all activities must equal 1. Are there even more? Is it just me or is this standard state topic rather confusing?!?

@Philipp Hm strange, I just thought I got it and then I found the following video: https://de.coursera.org/lecture/advanced-chemistry/5-10-standard-versus-nonstandard-free-energy-change-vfbI5 at approx. 2:30 it is claimed, that in standard state the partial pressures are at 1 bar (std. pressure). That contradicts your comment, doesn't it?

@user76122 In the answer of mine I linked you can find the equation $\sum_i \nu_{i} \mu_{i}^{0} = \Delta G^{0}$ with $\mu_{i}^{0} = \mu_{i}^{*}(p^{0}, T)$ which is the definition of $\Delta G^{0}$, i.e. you can calculate the value of $\Delta G^{0}$ by adding up the chemical potentials of the pure educts and products at the standard pressure of $1 , \mathrm{bar}$ (and at temperature $T$). So, you kind of start out conceptually with the unmixed educts in their standard states and I think this is what the video is referring to.

@user76122 ...But when you mix the seperate educts - each gas being at standard pressure - and look at the way they mix, i.e. how the partial pressures turn out to be, you can derive the value of $\Delta G^{0}$ from this (via $K$). And this is what the equation $\ln K = -\frac{\Delta G^{0}}{RT}$ tells you. There is only one value of $\Delta G^{0}$ that will lead to this special set of partial pressures (represented by $K$) for the reaction at hand under standard conditions.

@Philipp I really appreciate your patience, so I'm sorry to bother you again (maybe I'm just too stupid^^): Doesn't your explanation mean that there are two $\Delta G^\circ$? One for unmixed reactants and products in the standard state ($\sum_i\nu_i\mu_i^\circ=\Delta G^\circ$ with $\mu_i^\circ=\mu_i^*$) and one for the mixed state, where the latter can be calculated by $\Delta G^\circ = -RT\ln{K}$? I thought these two values must be identical? In this answer of https://chemistry.stackexchange.com/a/41864/76122 it is also mentioned, that all $a_i=1$, but maybe I'm missing something?

@user76122 No, there aren't two values for $\Delta G^{0}$. It's just two ways of expressing the same quantity. One, $\sum_{i} \nu_i \mu^{0}{I} = \Delta G^0$ is the defining equation for $\Delta G^0$, the other $\Delta G^0 = - RT \mathrm{ln} K$ just follows from the equilibrium condition $\Delta G = 0$ when you express $\Delta G$ in terms of $\Delta G^0$. You can look at it like this: The chemical potentials $\mu{i}^{0}$ of educts and products which make up $\Delta G^0$ set up some kind of "force" that will push the equilibrium to a certain position, represented by $K$, once the components...

@user76122 ... are mixed and equilibrate. A bit like an ideal spring with a characteristic spring constant that gets compressed by a certain force and upon release will oscillate at a characteristic frequency.

As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$\Delta G=\Delta G^\circ+ RT\ln{1}=\Delta G^\circ$$ On the other hand at equilibrium $Q=K$ and so $$\Delta G=\Delta G^\circ + RT\ln{K}$$ This of course leads to $\Delta G^\circ = -RT\ln{K}$ since at equilibrium $\Delta G=0$.
So you might want to think of it as multiple statements:

For the conversion of reactants to products in their standard states $Q=1$
At equilibrium  $\Delta G=0$
At equilibrium $Q=K$
$\Delta G^\circ$ is the free energy change for conversion of reactants to products in their standard states.

The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.
$K$ and $\Delta G^\circ$ are very much connected, but $\Delta G^\circ$ does not describe the change in free energy from reactants to products at equilibrium.

So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $\Delta G=\Delta G^\circ + RT\ln{Q}$. My misunderstanding arises from the equation: $\Delta G^\circ = - RT\ln{K}$, since it seems to connect standard ($\Delta G^\circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct?

@user76122 $K$ and $\Delta G^\circ$ are very much connected. $\Delta G^\circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer.

At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(\xi)$, $\Delta G^\circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :)

@user76122 I would not interpret $\Delta G^\circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve.

Yes, of course. Please see the picture: https://i.stack.imgur.com/6MfNR.png What I meant was the following: If the system is in equilibrium, i.e. $\frac{\partial G}{\partial\xi}=\Delta G=0$, it corresponds to the red point on the curve. In standard state ($Q=1$) the system is at a point at which  $\frac{\partial G}{\partial\xi}=\Delta G=\Delta G^\circ$. This corresponds to the green point. I'm not sure if the green point is always somewhere on the blue G curve or whether there are possibilities where the definition of a standard state is rather a hypothetical point?

I guess you answer the question yourself: $\Delta G^\circ$ will be on the curve if Q=1 has an associated corresponding value of the progress coordinate on the curve. But  I would consider that more of a coincidence than truly useful. But it doesn't hurt to continue digging deeper.

Thanks! By digging deeper, I found the following answer: https://chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $\infty$ to zero.

@user76122 Yes, you are right. I'll have to eat my words for not thinking a little harder about this. I take back that last statement.

What you enter into $K$ are not the activities of the pure reactants and pure products at standard state (if you did then, yes, $K$ would be 1).  Rather, it is their activities at equilibrium (raised, of course, to the power of their respective stochiometric coefficients).  And, at equilibrium, these activities are generally not equal to one.


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Carbonation of Grignard reagent with a carbonyl group present in it
Here's the question in my book (see picture)

Option A is a standard reaction which gives a carboxylic acid.
Option B is definitely wrong because an acid base reaction will take place and CH4 will be formed after the first step. CO2 will not react with the methane gas.
So without knowing about option C, I can safely tick option A. But what happens in option C?
Here's the solution provided in the book.

Why cannot a carbonyl group and Grignard reagent be present in same substance? Also, what will be the reaction?

Hmm? Does CH3MgBr react with acetone?

In other words, did you think about why your teacher advised you to not wash your glassware with acetone before doing Grignard reaction? :-)

You should look at two things:
1)  Will the reaction stay on course?  If you add water to the Grignard reagent first, without the carbon dioxide initially, what do you get?  Is it something that will get you your carboxylic acid when you add the carbon dioxide?
2)  Can you make the Grignard reagent to begin with?  Here again you are asking the reaction to stay on course; if the Grignard function can react with another functional group in the molecule, it will do so either within the molecule or as an intermolecular reaction between two of the molecules.
Then you no longer have your Grignard reagent.
Only one of the given possibilities works against both of these questions, so there is your answer.


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Reaction of diacetyl with acid
What happens when diacetyl, that is, butan-2,3-dione is treated with acid?
I guess there would be acid catalysed aldol formation.
But what would the final product be? I can see too many possibilities. Please help me out here.
Thanks!

I think the major product would be 2,5-dimethylcyclohex-2,5-dien-1,4-dione.

what possibilities  do you see please elaborate ? although your thought process is right

@Mathew Mahindaratne: Yes, that's the first thing that came to my mind as well... but how do we know whether the enols from both carbonyls of one molecule would attack both carbonyls of another, or whether the enol from one carbonyl of each molecule would attack one carbonyl of the other molecule? And how do we know that two molecules will react with each other and no more will get involved by further aldol condensation? I'm doubtful on this one.

AKM: In your response to MM the self-condensation of diacetyl 1 occurs stepwise and not with the double enol of diacetyl 1 (biacetyl; butan-2,3-dione). Although your question relates to the acid-catalyzed condensation of diacetyl, historical condensations of diacetyl were conducted under alkaline conditions. The products formed from the base or acid-catalyzed are numerous and dependent on reaction conditions.
Quinone 2 was prepared under basic conditions by von Pechmann.1 Diels and coworkers2 isolated from an alkaline reaction a dimer, C8H12O4, which was assigned lactonic structure 3 with no spectroscopic evidence.3 The structures are color-coded to display the four carbon moiety of diacetyl. Clearly, lactone 3 shows a branched red carbon framework. An undefined "benzylic acid rearrangement" was suggested without details. The structure was later revised to diketone 4 based upon spectroscopic evidence and a more reasonable normal carbon chain.4
Trimer 5, C12H18O6, has also been isolated under alkaline conditions and the stereochemistry confirmed by NMR and IR.5
An acid-catalyzed condensation of diacetyl in dilute aqueous acetic acid (70oC, 13 days) leads to C10 structure 10.6 The structure was confirmed spectroscopically and by periodate cleavage with loss of acetic acid with concommitant formation of diketone 11. Structure 10 does not form a hemiketal as does structure 5 owing to the resonance stabilizing effect of the α, β-unsaturated ketone. The mechanism proposed is described in the sequence of steps 4 --> 10.
As can be seen from the examples provided --- and that's not all of them --- the self-condensation of diacetyl is not straightforward.

1) Pechmann, H. v., Berichte, 1888, 21, 1417; Otte, R. and Pechmann, H. v., 1889, 22, 2115.
2) Diels, O., Blanchard, W. M., Heyden, H. v. d., Berichte, 1914, 47, 2355.
3) Machell, G., J. Chem. Soc., 1960, 683.
4) Birch, A. J. and Moye, C. J., J. Chem. Soc., 1957, 412.
5) Cresswell, R. M., Smith, W. R. D. and Wood, H. C. S., J. Chem. Soc., 1961, 4882.
6) Shapiro, R., Hachman, J., Wahl, R., J. Org. Chem., 1966, 31, 2710.


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Pressure-dependence of electrolysis voltage
I'm trying to better understand electro-chemistry, so please explain why this second-kind perpetual motion device won't work. I guess it's safe to assume it won't work one way or another.
There is a cycle, which starts with low-pressure water.

The water is pressurized by a compressor. This requires work, but since water is nearly incompressible, the required work is very low: the distance over which the pressurization force is applied is almost zero.
Next, the high-pressure water is split into high-pressure hydrogen and oxygen gas. This requires a certain amount of electrical work.
Next, the high-pressure hydrogen and oxygen gas are expanded through turbines, expanding cylinders or other devices that allow the expansion to apply output work. Heat is added to keep the temperature constant during expansion, to maximize the work output. Note that, unlike the compression stage, expansion deals with gases, which are very compressible; therefore, the amount of work produced is substantial.
To close the cycle, the hydrogen and oxygen gases are recombined in a fuel cell, to end up with the same amount of water, at the same pressure as when we started the cycle. This delivers some electrical work.

Of course there are going to be energy losses; I expect the fuel cell will always yield less energy than is required in the electrolyzer. However, the difference between the compression work and the expansion work can be scaled up arbitrarily by using a higher pressure difference, so at some point this difference in work must yield more energy than is lost in the electrolysis/fuel cell combination. This energy comes from the heat flowing into the expander, but simply turning heat into excess work should not be possible.
So, what is wrong? The only way I see to prevent this to work is if the electrolyzer requires more work per reaction mass (essentially its voltage) than the fuel cell, even if they are both 100% efficient. This would imply that the reaction voltage is pressure-dependent, with a higher voltage at higher pressures. Is this the case? And is this only the case if the output material is more compressible (gas) than the input material (liquid)?
(The same should be true for evaporation/condensation enthalpies, as we could have designed a very similar cycle with evaporation/condensation instead of electrolysis/fuel cell.)

I'm voting to close this question as off-topic because it's rather physics then chemistry.

IF there are losses in the system (as you admit) then a perpetual motion machine is impossible. And, no, scaling up a mechanical process might get you a higherefficiency, but it can't get more energy out than you put in. There are always losses.

@Mithoron my goal is to learn about the (physical) properties of a chemical process (electrolysis). The physical set-up with compression and expansion is just a tool to achieve this.

:/ better get to the point - you can't magically expect to get more energy from decompressing water then compressing. Your gas laws simply don't work for it.

I can expect to get more energy from decompressing water vapor than from compressing water liquid, with the arguments described in the question; these are not each others' exact inverse. If that is not the case, please explain why (but that would indeed be a physics question).

Your only misconception is that getting high pressure liquid isn't easier then getting gas of same pressure - and that's kinda physical.

@Mithoron so your explanation is a misunderstanding in the compression/decompression part of the device. I'd like a detailed discussion about this, since I'm not convinced at all. I'm new to this site; what's the best way to do this? Re-ask on the Physics site, and come back here if it turns out I'm correct about the compression/decompression parts?

Cross-posting is discouraged. If question would be put on hold as off topic, it could be migrated to Physics.SE BTW I think current answer does discuss that.

This is chemical thermodynamics; while it could (since it's thermodynamics) be asked on the physics site, I think this is a better home for it.  2) The OP isn't claiming a perpetual motion machine is possible. He understands it's impossible, and thus wants to understand what the particular flaw is in the above set of steps that reveal it to be impossible.

Electrochemical potential is pressure dependent. It is just the free energy difference of the systems in equilibrium, i.e. if you produce high-pressure gasses from a liquid, you have to put this extra work in the system in form of higher electrochemical potential. The higher the pressure you use to gain that work, the more energy you will need to split the water, while you get back always the same energy when you use the fuel cell after decompressing the gas.

I changed the title to better reflect what I'm actually asking for. Also, since answers seem to be pointing to a "yes", other people are more likely to find these answers with this new title.

I’m voting to close this question because debunking perpetual motion schemes is off topic.

Of course, we're starting by acknowledging a perpetual motion machine is impossible.  The question, then, is how do we understand, through chemical thermodynamics, why your specific set of steps can't constitute a perpetual motion machine.
The answer is that what you're not accounting for is that the electrical energy required for your step 2 is greater than you get back from your step 4, even if we don't have losses.  That's because, everything else being equal, the work needed to create gases against a high pressure is greater than that needed to create gases against a low pressure. I.e., $|w_2| > |w_4|$.
To understand why, let's simplify things by considering the pressure-dependence of the free energy of a pure substance at constant T: dG = VdP.  In step 2, because the pressure is so, high, you have increased the free energies of both the reactants and the products relative to what they'd be at atmospheric pressure.  To get the actual changes, you would need to calculate $\int^{P_f}_{P_i}V(P) dP$ for the reactants and products.  Since the products are gases (oxygen and hydrogen), and the volume of gases is much larger than that of liquids, the free energies of the products would be raised far more than that of the reactant (a liquid).  Thus the separation between the free energies of the reactants and products, (i.e.,  the magnitude of $\Delta G_r$) for step 2 (which is at high pressure), will be substantially greater than it is for step 4 (which is at low pressure).
You could also understand it using Le Chatelier's principle:  The volume of the products is higher than that of the reactants, and thus higher pressure will favor the reactants over the products.

This sounds informative, but I'm not that familiar (anymore) with free energies (especially in relation to chemical reactions). Does this translate into a higher required voltage for the electrolyzer, compared to the fuel cell, as I theorized in my question?

It translates into more energy from the electrolyzer.  Energy = voltage x charge, and charge = current x time, so energy = voltage x current x time.  Thus, for constant current and time, it translates into more voltage.

I think it might be easier to understand in terms of the Nernst equation, which directly relates the voltage to the concentration or pressure of the reactants and products. Higher pressures of product gases means higher voltage required.

As you noted, there will be some definite energy losses, however I think it's important to keep in mind as well, that electrolysis is a slow process, like charging an electric car. You can speed it up by applying more current, but ultimately there is a theoretical limit before your electrodes are toast (I nearly experienced that my first time experimenting with this stuff!).
I'm not sure what you mean by "the oxygen and hydrogen gases are expanded into turbines," as yes, they can be pressurized, but sooner or later they are going to find a way to escape the container. Do you mean simply let loose like air from an air compressor, or something a little more specialized? Either way, I'm excited for you! Keep up the good work and don't forget to keep your friends on Stack Overflow posted on your progress!!!

I arrived to your question and it seems that is an answer to another question I had. Let's examine the processes:
We can increase the pressure of water with very little work, while the volume of water will remain almost constant. This is because ... I will try to explain it naivelly ... there is no much available free space between molecules in order to allow for compression.
Then we divide water into oxygen an hydrogen keeping the volume constant.
We have to understand that at high pressures, both hydrogen and oxygen behave as "water", this means that "there is still no space available".
If you look at the curves of compressibility, you will notice that gases are compressible between 0 and 200 bar even more than an ideal gas, while at 200 bar they arrive to be again as compressible as an ideal gas. Afterwards, they become compressible less than an ideal gas and with rising difficulty as the pressure rises.
So by compressing water you cannot store much energy, and similarly by compressing high pressure gases you cannot store much energy.
Here the "store energy" is wrong but more on that later on.
So suppose you have water at 10.000 bar pretty a high pressure by the way, then you will get H2 and O2 at 10.000 bar or even more, but as water does not give back energy when uncompressed, similarly you will not be able to get energy from ultra high pressure H2 and O2 until you arrive to human pressures of 500bar or less and you start to be able to decompress and get energy.
Thanks a lot for the question as it gives ideas to work on.
The question has to do with physics and chemistry both.
When you increase pressure of e.g. aqueous vapor, you arrive to have water and solid water afterwards. Similarly when you increase the pressure of oxygen you start to have liquid oxygen and solid oxygen but that is valid when you are under the critical temperature ( that for oxygen is −118.6 °C) .
When you are above the critical temperature, there is only a status of gas-liquid oxygen that is in the same phase irrespective of the pressure.


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Zinc and copper
Is potential energy and ionisation energy are same? Plz explain with respect to zinc
Why zinc is undergoing oxidation when combined with with zinc. copper in electrochemical series

I believe you have some conceptual flaws:
I like to think of potential energy as the energy held in a system but more formally it is the energy possessed by a body by virtue of its position relative to others. It is really a product of work/energy being needed to be expended in order to move an object from one point in a field to another.
Ionisation energy is quite different. It is formally defined as the amount of energy required to remove one mole of outer valence shell electrons from one mole of a gaseous element.
Both of these concepts do not really have much to do with oxidation directly (as far as I know of). You might mean electrical potential or potential for short. This is simply the work done per unit charge. I would avoid using potential energy in the context of redox reactions and instead use terms like electric potential and potential difference.

What a shame the question title is not "iron and copper". Then taking your avatar into account, the Deus Ex pun would be complete:)


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What is the point of POAC? Why can't one just balance?
I have been studying the principle of atomic conservation, but have not understood why it even exists. For example:

What mass of $\ce{O2}$ is produced by 1.23 g of $\ce{KClO3}$?

Couldn't this question be solved very easily by just balancing it, and the using no. of moles to find the mass of $\ce{O2}$ produced?
Am I missing something here?

The principle of atomic conservation just basically says that chemical reactions don't create or destroy atoms. So the gist is that you must have the same number and type of atoms on both sides of the chemical reaction. // Nuclear reactions can create or destroy atoms, but not chemical reactions.

There are generally many ways to solve any problem... you could balance the reactions or you could use POAC. Quite often, POAC turns out to be really useful for quicker calculations, since you can focus on only one element and don't have to bother about the others, especially if all of that element ends up in only one product.

@MaxW I think I understood, thank you for your help :)


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How many moles of methyl iodide will be consumed by one mole of crixivan?

My take on this question : a total of 5 moles of $\ce{CH3I}$ should be consumed.

The oxygen atoms in the $\ce{2 OH}$ groups each have a lone pair of electrons. They should displace $\ce{I-}$ from methyl iodide via an $\ce{S_N2}$ reaction.
The two nitrogen atoms in the tertiary amino moieties also should react with a total of 2 moles of $\ce{CH3I}$ .
The $\ce N$ atom in the pyridine moiety should also consume a mole, as the lone pair of electrons on it is not in conjugation with the benzene nucleus.

However, as I checked the answer, it's provided as 3. What may be the fallacy in my reasoning?

The OH groups will not methylate without being deprotonated first, there is no base present strong enough to do this.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". It makes it even worse seeing this on the picture of a textbook question. I am sad now.

Why so? A mole of CH3I is equal to a gram equivalent of CH3I.


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Do endothermic reactions have a higher absolute change in temperature than exothermic reactions?
or does it depend on the reaction itself. I am conducting an experiment and the endothermic reactions have temperature changes of at least 8 degrees while the exothermic reactions have a change in 1 or 2 degrees. Is there a reason for that or is it just depending on the reaction?

It just depends on the reaction.

Just so we get this right, the endothermic reaction gets colder by 8 K (or Fahrenheit?), and the other(s) warmer by 1-2? What are those reactions?

oh sorry, Im measuring it in Celsius. The endothermic decrease by up to 8 degrees celsius and the exothermic warmer by 1-2 celsius. They are five reactions: Sodium hydroxide solution + dilute hydrochloric acid. Sodium hydrogen carbonate solution + acetic acid. Copper (II) sulphate solution + magnesium ribbon. Dilute sulfuric acid + magnesium ribbon. Sodium hydrogen carbonate solution + citric acid.

Note to answer, just a comment on the way you might think, that seems wrong. There is no absolute change in temperature for a reactiin. Take and treat H of reaction and T of reaction as two different things. You cannot deduced H from a undefined T. Unless the T change you mentioned is that of a calorimeter.

What do you mean?

The absolute change in temperature depends on the absolute (not molar) reaction enthalpy and the heat capacity of the system (including the solvent and beaker!) in its final state and the amount of heat exchange with the environment in the timeframe of the experiment.

It depends on the reaction enthalpy.
Both reaction groups have a wide range of absolute reaction enthalpies to cause negligible or huge temperature difference, supposing the system is isolated.
The problem with some highly endothermic reactions is that they often require quite forced reaction conditions like high temperature, so it may be problematic to determine the temperature change directly, e.g. by calorimetry.
The temperature change can be then estimated from the reaction enthalpy of the reverse exothermic reaction and the molar heat capacities.
The reaction enthalpy can be often calculated from tabelated values for particular compounds.

Can you re-phrase the last part of this answer? It's rather unclear, to the point where I'm not sure it's correct.

@Karl   I have tried to reformulate it.


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Is the conjugate base of every polyprotic acid amphoteric?
Since all compounds with an accessable free lone pair can act as a Brønsted-Lowry base, I was wondering if there was any specific example of a substance that when it loses a proton, it cannot accept a proton but can donate another proton.
Please note that I am not considering something like sulfuric acid which is a strong diprotic acid, as $\ce{HSO^{-}_{4}}$ still accepts a proton even if its to a limited extent.

What you ask for is a contradiction in terms. If your hypothetical acid exists  in the protonated form in the first place, then of course it get protonated after being deprotonated. There is no going around it. True, superacids might be pretty reluctant to protonate, but what I said above still applies.

I thought there might be a compound that upon deprotonation forms a base that is very unstable and breaks down into something else in a very one way manner?

If there were cooperativity you could imagine a case where the acid loses two protons or none, but not one. I don’t know of such a case. https://www.sciencedirect.com/science/article/abs/pii/S0301462200001988?via%3Dihub

@sabhoque That felse like a different thing than what is asked in the question. You may want to look for anionic rearrangements. I think that one could design a molecule from which two hydrogens could theoretically be abstracted, but after the first one, it undergoes rapid rearrangement, possibly preventing the second from being abstracted.

The conjugate base of sulfuric acid, bisulfate ion, shows little amphoteric character in water solution.  Unless you drop the pH to about 2 or below, it is only an acid, and a moderately strong one at that.

what do you mean by > moderately strong one at that
Do you mean that it is only a moderately strong acid at a pH less than 2?


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Reasons for negative iodoform test
Why does 2',6'-dimethylacetophenone not give iodoform test?

A  google search for the specific reaction on 2,6-dimethyl acetophenone failed to find an answer. It did find a number of methylacetophenones that do undergo the reaction e.g. 3,4-dimethylacetophenone, 2,4,5-trimethylacetophenone. This review https://pdfs.semanticscholar.org/4f9f/fe77312fe3a5b27bec937274232f25ccb6a7.pdf  notes some acetophenones that do not undergo the reaction (e,g, 2,4,6 tribromoacetophenone). I can see no reason why it would not.

It came in an exam.. I also thought it should give the iodoform test but the answer key for the exam said it would the triiodide intermediate but will not give the final compound i.e. CHI_3

Possibly there is too much steric hindrance for the OH- to attack the tri-iodo intermediate. Unless it has been specifically covered in your lectures or textbook this is an unfair question.

@Waylander We can debate whether or not it's fair, but I think that if the question states that it does not happen and asks for a rationale, that seems totally fair game. I certainly wouldn't want to predict whether or not the reaction happens in the absence of any data.

Depends entirely on the wording of the question. If states it doesn't happen then fair enough, good question. If it is just pick from a list of possible substrates then a very poor question

The question asked to predict whether it gives iodoform test or not... So it was a pretty ambiguous one..

@Waylander: I did a Chem. Abst. search on any  acetophenone being converted to the triiodo cmpd. It turned up only acetophenone triiodide that was not prepared under haloform conditions. Perhaps, once again, it may be a failure of triiodide formation. Another Tet. Lett. paper for that enterprising young chemist?   https://chemistry.stackexchange.com/questions/135626/do-cyclohexane-1-3-diones-give-positive-haloform-reactions haloform conditions.

As @Waylander pointed out, it appears this reaction has not been performed and/or recorded in any literature, so it is quite dangerous to speculate.
But keeping that aside, A 3D perspective reveals that abstraction of protons from the methyl group in quite unhindered.

Hence, the triiodo intermediate is well anticipated.

However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.

Notice that the Burgi-Dunitz trajectory, which we may assume the incoming nucleophile to take, is hindered by the large iodine atoms and the methyl group.
It is quite safe to assume that the attack at the carbonyl carbon is unfavoured, preventing the release of the $\ce{CI3-}$, and ultimately $\ce{CHI3}$ never appears.
EDIT: Apparently there is some relevant literature available for similar compounds, as mentioned in this answer. Thanks to Mathew for searching and pointing it out.

Sorry, I didn't see your edit until today. You are correctly suggested the steric effect, which encouraged me to go for reference to make your argument legitimate. Thank you as well.

There have been a few answers for this question, which make sense on steric hindrance preventing the idoform formation. However, one thing I wouldn’t agree with is almost everybody’s claim of not having literature evidence to support their theory, thus some may categorize this question as just opinion based and chose to close it (regardless how good it is). To avoid that, I’m going to try to give some acceptable literature evidence to support steric hindrance argument.
Because of original idoform test (introduced by Lieben in 1870) is not reliable for water insoluble compounds, Fuson and Tullock provided improved idoform test in 1934 (Ref.1), which use dioxane as a secondary solvent. They have checked new method with wide variety of compounds that have been analyzed previously with Lieben’s method, but had given mixed results, some of which were misleading. For example, in a majority of these cases, the behavior toward hypoiodite has not been previously reported. Such example is pinacolone, which was previously considered as negative to the test, but given positive result with a longer period of heating (one of the evidence for role of steric hindrance). According to authors, some of the most notable negative results were given by following compounds (Figure 1; Ref. 1 & 2):

Yet, Figure 2 listed the compounds, which have given positive results, but have significant steric hindrance (Ref. 1 & 2):

Most notably, you may compare compounds 5 (negative; Figure 1) vs compound 11 (positive; Figure 2). They both have similar steric hindrance (aromatic $\ce{C8-H}$ vs $\ce{C3-H}$ is the only difference), yet gives contrary results. Only explanation could be acetic acid group on $\ce{C1-O}$ must be cleaved before iodoform formation on compound 11, but similar cleavage on $\ce{C2-O}$ of compound 5 is restricted due to its position, since it was known that 1-naphthyloxy group (conjugate base of 1-naphthol) is more stable than 2-naphthyloxy group (conjugate base of 2-naphthol). Same can be told about compound 12 giving positive test. Nonetheless, these results clearly shows that only restriction to give idiform is di-ortho-substitution on the substrate (cf., 1-4; Figure 1), which gives the additional resistance to the formation of final tetrahedral intermediate as pointed by Jan. Meanwhile, mono o-, p-, or m-substitution OR di-o,p-, o,m-, or m,m-substitution does not effect the idoform formation. This fact is stated in Ref.1 as:

The iodoform reaction is greatly retarded by steric hindrance. The test is negative for all compounds which contain one of the requisite groupings joined to an aryl radical carrying two ortho substituents. As a matter of fact, the reaction is slow, even with pinacolone.

Even tri-substituted compounds such as compound 8 give positive idoform test as long as two of substitutions were not in di-ortho positions. In their conclusion, Fuson and Tullock provided a generalization to the reaction (Ref.1):

The test is positive for compounds which contain the grouping $\ce{CH3CO}$-, $\ce{CH2ICO}$-, or $\ce{CHI2CO}-$ when, joined to a hydrogen atom or to a carboy atom which does not carry highly activated hydrogen atoms or groups which provide an excessive amount of steric hindrance. The test will, of course, be positive also for any compound which reacts with the reagent to give a derivative containing one of the requisite groupings. Conversely, compounds which contain one of the requisite groupings will give a negative test in case this grouping is destroyed by the hydrolytic action of the reagent before iodination is complete.

The reference 3 would give some insight to the formation of sterically hindered idocompounds at $\alpha\ce{C}$ without giving idoform (formation of  the triiodide intermediate as stated in OP's comment). Also read Ref.4 for review on idoform test.

References:

R. C. Fuson, C. W. Tullock, “The Haloform Reaction. XIV. An Improved Iodoform Test,” J. Am. Chem. Soc. 1934, 56(7), 1638–1640 (DOI: 10.1021/ja01322a061).
M. Večeřa, J. Gasparič, In Detection and Identification of Organic Compounds; Plenum Press: New York, NY, 1971, “Chapter XII: Carbonyl Compounds,” pp. 208-246.
C. H. Fisher, H. R. Snyder, R. C. Fuson, “The Haloform Reaction. VI. Alpha-Halogen Derivatives of Hindered Ketones,” J. Am. Chem. Soc. 1932, 54(9), 3665–3674 (DOI: 10.1021/ja01348a025).
R. C. Fuson, B. A. Bull, “The Haloform Reaction,” Chem. Rev. 1934, 15(3), 275–309 (DOI: 10.1021/cr60052a001).

Methyl hexyl ketone doesn't give this test because of steric crowding or is there some other factor at play?

@ Avnish Kabaj: According to Ref.1, 2-octanone (methyl hexyl ketone) is positive for their improved method (even methyl cyclohexyl ketone is positive). If you didn't dissolve your water insoluble substrate in dioxane (authors' choice as the best for this test) before testing, it may give misleading results.

During my mechanisms and named reactions lecture at bachelor’s level, a number of different mechanisms were introduced for transesterification of carboxylic acids and we were required to select the correct mechanism for a given set of reaction partners.
One of the typical special cases was a 2,6-disubstituted benzoic ester—basically your starting material but with an ester group rather than an acetyl group. It was noted that nucleophilic attack on the carbonyl carbon was not possible as the methyl groups sterically blocked the Bürgi-Dunitz trajectory. Instead, these esterifications would proceed via an acylium cation formed by displacement of the $\ce{OR-}$ residue; once the linear acylium cation ($\ce{Ph-C#\overset{+}{O}}$) was formed, this could be attacked from any angle without the ortho substituents interfering.
It is likely, as William outlined, that the same problem exists for 2,6-dimethylacetophenone. While there should be no problem replacing the methyl hydrogens with iodides, there is no angle of attack for the hydroxide ion to generate the tetrahedral intermediate.
In the case of transesterification, the $\ce{OR-}$ group is a sufficiently good leaving group so that the acylium cation may form. $\ce{CI3-}$ is a much stronger base, so it seems highly unlikely to generate a corresponding acylium in this reaction. In the usual haloform reaction, the regeneration of the $\ce{C=O}$ double bond provides a driving force to promote the liberation of $\ce{CI3-}$ but that pathway is not possible if an acylium must be generated.

I'd agree with Waylander's 2nd comment - hydrolysis to form the carboxylic acid requires formation of the tetrahedral intermediate (from OH- attack on the keton), which the presence of two ortho substituents on the benzene ring doesn't allow, sterically.
A related effect is that methyl 2,6-dimethylbenzoate doesn't undergo basic hydrolysis by the normal BAc2, since it would involve that same tetrahedral intermediate. Instead, it goes via the BAl2 mechanism (i.e. SN2 attack at the methyl group and rate-determining cleavage of the alkyl, rather than the acyl bond).


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Stability and Resonance
The linked answers do attempt to clarify on 'stability' but I wanted an answer with specific connection to resonance.
For example, the ethanoate ion can exist in many forms but, as we know from data, it exists in a form that is a hybrid of all the possible forms. Why is this?
I have heard teachers say that this is since that form has 'less energy' or is 'more stable.' I understand - from the linked answer- that these terms are open to interpretation. The criteria my teacher used was that in each of the contributing mesomers, the electrons are localised. But in the hybrid, they are delocalised. Now I understand this results in more 'stability' in some sense of the word.
But is that only criteria ? Should we not also consider the electronegativities of the atoms involved, for example? Basically, why is the hybrid form the form in which the structure exists?

I point you to my A in https://chemistry.stackexchange.com/questions/111005/define-actually-define-the-stability-and-energy-of-a-compound though all the answers are correct I think that is the simpler. Lower the energy, the stable it is. Resonance is a recurrent question. Of course you are right. Resonance isn't stability per se.

Stability is a meaningless word if used alone in chemistry, organic chemistry textbooks make the case even worse. The key question is "stability with respect to what?" We need a reference. For example we say this iron block is heavy. This statement is scientifically of no use. Is it heavy with respect to a sand grain or a 70 kg human being? Please see my response in the  post suggested above by Harsh Wasnik.

With regard to resonance, see this question: What is resonance, and are resonance structures real? It is important to understand, that the structures with localised electron pairs are hypothetical cases, they do not exist; the expectation value of the (electronic) energy of such hypothetical structures would be higher in energy than the 'observed' structure; or in lab-jargon: less stable/unstable. By reversing the argument, the observed structure is more stable. This of course is not really true.

@Sal_99 — The thing between "real-world" chemistry vs quantum chemistry is that, in the real-world, we need some tricks to determine electron-densities in a molecule, resonance... without having to spend 3-week calculations on a super computer. And we have difference ways to explain how we ended with a given conclusion. In the real-world chemistry, it is easier to consider a molecule as a weighted "mix" of all of its mesomers. Though, in quantum chemistry, it would just be: "the calculations show that one species is lower in energy".

"But is that only criteria ? Should we not also consider the electronegativities of the atoms involved, for example? Basically, why is the hybrid form the form in which the structure exists?". Note that comparing mesomers nothing change in atom composition and configuration.  Whatever changes, e.g. electronegativity of a moiety, it will be reflected in the stability of that specific mesomer, thus contributing to determine the importance ("weight") of it in the final hybrid description. Having resonance or not isn't a criterion to compare totally unrelated molecule. It is that sometimes a singl

..single Lewis structure doesn't represent reality. After this, we try to rationalize and predict using as simple as possible rules and sketches.

And as @Martin - ... said,  your question is, or became, more a duplicate of the Q he has linked in a comment.


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Are pKa values of acid and conjugate base the same?
When you look at $\mathrm{p}K_\mathrm{a}$ table, you can read $\mathrm{p}K_\mathrm{a}$ values of acids. Do conjugate bases also have $\mathrm{p}K_\mathrm{a}$ values?
If both can have $\mathrm{p}K_\mathrm{a}$ values, why $\mathrm{p}K_\mathrm{a}$ value of conjugated bases are same as acids?
Can acid have only $\mathrm{p}K_\mathrm{a}$ values, not $\mathrm{p}K_\mathrm{b}$ values?

Let consider the case of the phosphoric acid.
The phosphoric acid $\ce{H3PO4}$ has $pK_\mathrm{a1}$.
Its conjugate base $\ce{H2PO4-}$ has $pK_\mathrm{b1} = 14- pK_\mathrm{a1}$.
But as $\ce{H2PO4-}$  is at the same time a ( weaker ) acid as well,
it has its own acidity constant $pK_\mathrm{a2}$ independent  on $pK_\mathrm{a1}$ and $pK_\mathrm{b1}$.
Generally, acids like $\ce{HA-}$ have $pK_\mathrm{a2}$ and $pK_\mathrm{b1}$ values.
The former associated with $\ce{HA- <=> H+ + A^2-}$,
the latter associated with $\ce{HA- + H2O <=> H2A + OH-}$

thank u for answering my question, but i have a few more questions.  1. pKa2 values exist only weak acid and base, am I right? ex) HCl(strong acid) has no second ionization like H2PO4. 2. I got that a lot of substance have pKa, pKb values both. but when you look at pKa table of Your book, like organic chemistry.  then Acids and conjugate bases are same values and honestly, i don't know why they take same values. why hydroxide ion's value is 15.74? is there any relation with mole concentration?

About 1., it is not true. $\ce{HSO4-}$ or $\ce{(COOH)_2^-}$ are still strong acids.
About 2., If an acid dissociates only once, both the acid and conjugate base may refer in the table to the same reaction, equilibrium and constant.
It would help, if you provide a particular quatation of such a source.

Here the link is. https://www.masterorganicchemistry.com/2010/09/29/how-to-use-a-pka-table/     Is there any particular classfication of determining strong or weak acid? Once I learned from book, strength is relative which substance  dissociate in water

And what is the problem there ? The mentioned $\mathrm{p}K_\mathrm{a}$  always relates to the acid side of the pair, not matter if we do not consider it normally as an acide, like methane.

ah.. that means hydroxide also has pKa value of 15.74, and there's no problem with it, Table shows acid side of conjugate bases. is it right? i'm not good at chemistry so it's hard to understand your saying.... so i felt diffcult to say my words maybe...

No, $\ce{H2O}$ as an acid has $pKa$ of $15.74$. $\ce{OH-}$ is the conjugated base to $\ce{H2O}$ as acid.   Similarly, $\ce{H3O}$+ is conjugated acid to $\ce{H2O}$ as base.

P.S.:  The table shows the conjugated pairs, acids on the left, bases on the right.