text:
Can we calculate final temperature of two gases in thermal contact?
This is a question from IIT-JEE - ADVANCE 2018

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in $\pu{m3}$) of the compartment A after the system attains equilibrium is ___?

I've understood on how to solve the Question, we can equate the number of moles before and after on either side seperately (as gas transfer isn't allowed, so amount of gas remains same), take the volumes by considering one to increase and the other to decrease, and the pressure final to be the same for both to attain equilibrium.
My doubt is how do I find the final temperature on either side, are we allowed to take average ?
And also if the gases on both sides were different are we still allowed to take average, or then there's another way.

The answer is "2.22" == "(20/9)" if anyone is wondering

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. It looks like you are asking for a recipe to solve the problem.

It can be helpful if you calculate how many moles of oxygen are in the compartment A and in comparment B using the ideal gas laws. If I am not mistaken $n_A \neq n_B$.

No, you cannot just take the average. To make it more "intuitive" why not, think about a more extreme situation. Let's say compartment A had 1 mol of gas at 400 K and compartment B was very small and contained only 0,000001 mol of gas at 300 K. Do you expect the final temperature to be 350 K, or something very close to 400 K? You have to write a heat balance: $Q_A = -Q_B \implies m_A c \Delta T_A = -m_B c \Delta T_B$.

@PAEP, I don't think that both the sides have equal moles, because I think both are completely filled (according to the language of the question, and they have different volumes and so if they are fully filled nA ≠ nB. Also, I only asked the question because I wanted to know about the temperature point, now if I asked it directly without context, a lot of people wouldn't understand it, and that would defeat the purpose of asking a question

Images are not searchable. Please write the relevant question as texts. https://chemistry.meta.stackexchange.com/questions/3758/policy-on-picture-questions

It would be nice if I put 1 drop of water 60 °C into 100 L of water 20 °C and would get warm water 40 °C. Unfortunately, it does not happen.

As you have correctly said the final pressure must be the same to balance the forces on the movable partition. Also, note that the partition is a conductor, which means it wont allow any temperature difference across it - i.e. it will keep conducting heat and doing work until the temperature difference is 0.
 Instead of finding the exact number of moles, just find a ratio of the number of moles on either side using n = PV/RT; which gives us a ratio of 5:4. Honestly speaking, you don't need to find the individual temperatures in the final state to attain your answer. Just by equating pressure in final state you get n1RTfinal⁄(1+x) = n2RTfinal⁄(3-x)
, solving which you get  x = 11⁄9 , where x is the increase in volume of the left chamber
However, just pondering back to your original question of the temperatures. Since no energy loss occurs to the surroundings, the change in internal energy of the left gas = change in internal energy of right gas. You might think that some amount of work have been done by the piston, but actually that work is responsible for transferring energy between two chambers. So, n1Cv(400 - Tfinal) = n2Cv(Tfinal-300). The ratio of n1:n2 is sufficient here to guide you towards answer. Had the gases been different in nature - a monoatomic and a diatomic then you had to consider the ratio of n1Cv1:n2Cv2 for finding the temperature difference
 Doing some algebraic manipulation, you can easily find the final temperature. What is interesting to note is - the temperature isn't a mean or an average as you say. Rather it is a weighted mean - something very similar to finding location of center of mass between two unequal masses. The final temperature is at temperature differences from initial temperature of chambers A and B in a ratio of 4:5; closer to A's 400K and farther from B's 500K frankly because of the quantitative dominance of A's molecules. Had the ratio of number of moles been same, the final temperature would have been an average of the two as you said. You can always bring physical analogy of masses, if two equal masses are placed, center of mass is the exact midpoint of the line joining them, which ain't true if the masses are unequal

+1 for discussing the possibility of two gases having different heat capacities, and for saying that you don't need any temperature data to solve the problem, realizing the two gases will be at thermal equilibrium. Giving additional, unnecessary information in a problem makes it much more difficult for many students.

Thanks for taking your time to point out the positive points of my answer. The feedback is really going to be helpful as I am a new user out here..

Looking forward to more!

@Sam Thanks for the answer, your solution does make sense, the only doubt that I have is, how does the temperature stay the same even if heat transfer takes place

@Paras Gupta Temperature doesn't stay the same, read carefully. The heat exchange causes the temperature of one chamber to drop and another to rise, but temperatures of the two chambers must come to a fixed value in the final state - the reason of which I mentioned. Heat flow occurs because of temp difference and in final state heat flow will cease only if the temp difference = 0

@Sam I did understand that, I just misinterpreted the statement, I thought you were talking about the initial condition, rather than the final one. So I thought that you had said, that heat transfer would take place (as it's a conductor) but no change in temperature, so that's why the doubt. Anyways, tysm for the help, this community is great


text:
What happened after fire finished burning?
After fire finished burning, did it merge with the atmosphere and becoming water vapors, oxygen and nitrogen, or did it simply disappear into nothing?

This is typical scenario for parallel accounts, when one of them is forbidden to ask new questions. That would be against the site policy.

One thing is a flame as phenomena, other thing is matter involved in a flame.
A flame as phenomena just disappears. It is a state of gaseous matter at high temperature, when its gaseous and possibly small solid particles are emitting light. When the gas cools down enough, it stops emission and is not perceived as a flame anymore. The flame is  usually caused, but is not limited to, by exothermic combustion of gases or vapors in air or oxygen. ( Small flames over burning coke is combustion of gaseous carbon monoxide.)
The matter, that was forming the flame, mixes with air in the same way how it was mixing during fire when leaving a flame.

If matter mixes with air and if vapour, carbon dioxide, oxygen, nitrogen and minor components are matter then they mix with air.

source: Stack Exchange

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text:
What are some good synthetic samples I can make myself to start getting better at Quantitative X-Ray Diffraction?
I'm a student wanting to become an expert in quantitative X-Ray Diffraction (QXRD) to determine the wt%s of both crystalline and amorphous phases in a solid specimen. At my university I have access to a Bruker D8A X-Ray Diffractometer and use DiffracEVA for PhaseID and TOPAS5 for quantitative analysis.
I've recently asked a question on how to do QXRD on complex fly ash specimens but have been advised by some colleagues that I shouldn't start with quantifying such a complex sample so quickly as it's very difficult and likely not time-efficient. At present I have only done successful QXRD (i.e. with minimal error in Rietveld Refining) with the simple idealised files found in the TOPAS how-to guide.
I'm looking to build my skillset and want to sample prep, test and QXRD analyse some synthetic samples (i.e. where I make the samples myself and know the exact wt%s) in order to get better at QXRD. Does anyone have any advice around this and, more specifically, recommend any good starting points for which material feedstocks to use and sample prep methodologies?

The easiest crystal to synthesize is probably alum, or potassium aluminium sulfate $\ce{KAl(SO4)2·12H2O}$ . It is not expensive and easily soluble in water. Evaporation produces nice colorless octahedral crystals, which may grow up to macroscopic dimensions.
To obtain nice crystals, a hot concentrated solution has to be done in hot water in some beaker. After cooling down, plenty of small crystals are to be found on the bottom of the beaker. Most of them have irregular shape. Select by hand the crystal which looks like a regular octahedron. Paste it at the tip of a cotton wire. Stir and reheat the alum solution so as to dissolve the crystals. Stop the heating when the last crystals is dissolved. Hang the wire from above, so that the small octahedron is in the middle of the solution. Let it cool down. When cold, the crystal dimension has increased. Repeat the operation heating - cooling, as often as required.

This may be a silly question, but could I just mix a 50/50 wt% mixture of lab-grade CaF2 and corundum (Al2O3) together and go from there trying out some QXRD methods?

Try ! But lab grade $\ce{CaF2}$ is obtained by precipitation in aqueous solution. So it may not be really crystallized.

Could you please expand on the above comment? Why/how would CaF2 "not be really crystallised "?

Usually, when preparing a substance in a test tube by a precipitation reaction, the obtained substance is made of micro crystals, which are too small to be analyzed by X-rays.

Wow interesting, what other analytics would help here? NMR?

Forget about X-rays, and try atomic spectroscopy !

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What's a good high precision analysis technique (after GC) to measure trace isopropanol in water samples?
Looking to measure trace IPA content in water samples.
Only GC I have access to doesn't take samples with water. Different columns not an option.

The poor man's path: One could roughly estimate the amount of IPA in pure water by adding a photocatalyst (like Vitamin B12 or TiO2...) with appropriate light source, and collect all formed gases by the displacement over the reaction mix. Use say 20 such mixes with known varying amounts of IPA and the data collected to get a precision estimate by statistical analysis. More information from the precise composition of the gas mix may also be of value in the estimation process. See https://pubs.acs.org/doi/10.1021/jacs.6b06860 as background.

https://www.researchgate.net/post/How_can_I_analyse_alcohols_methanol_ethanol_acetone_2-propanol_in_water_solution_with_GC-FID

You can try headspace analysis.

Good idea unless water vapor is a problem with the GC. If they have access to an FTIR with a multibounce ATR cell such as a circle cell, the C-H stretch bands should be sensitive enough.

Other than finding any way to get access to a compatible GC (and yes, you can change columns), one might:
Extract the IPA into an organic solvent and then try an IPA shoot.  "Salting" the aqueous solution might help, but adding steps for trace analysis adds error, particularly since IPA is volatile.
One might consider speaking to their supervisor about the issue.


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How did we know about the existence of elements before quantum theory?
I'm reading a book about the elements, and it says all the time: discovered/isolated by X in a year.
But how did the people know about the elements before the quantum theory? How did they know they were in front of a new "element"?

Lots of experimental work, and they got it wrong many times as well. See here the "elements" described by Lavoisier in his Elementary Treatise on Chemistry - some of them are not really elements.

Consider for history related chemical questions https://hsm.stackexchange.com

How did we know about the existence of elements before quantum theory?

The short answer is that the old quantum theory had nothing to do with element discovery. Elements were considered elements after a thorough classical chemical analysis which included lots of gravimetry (weighing of precipitates), and observation of their novel chemical or physical properties. Phosphorus, for instance, was discovered by a German alchemist who collected buckets and buckets of urine in search of a recipe to make gold. One fine day, upon strongly heating dried urine residues strong with sand/coal he discovered a waxy substance which glowed in the dark for hours. Quantum theory had no part to play!
Modern textbooks teach science as a success story. In fact, science is also a story of many failures that nobody wants to record. Many elements are pre-historic like carbon, iron, copper.
When Bunsen and Kirchoff developed the spectroscope, it allowed them to see the spectrum of chemical compounds in a flame or in gas discharges. These new colors and patterns in the spectrum revealed brand-new elements. Cesium, rubidium, thallium and many gases were found this way. Helium was discovered in the Sun first! As before, no old quantum theory was involved, since the initial focus was on observation rather than determining the reason for things to happen. Even the electron and the proton were not needed to discover those elements. It was pure chemical analysis of newly discovered rocks and materials.

They also discovered coronium. Only later did the true identity of this element -- iron heated to millions of degrees and thus highly ionized -- come to light, and with it the real discovery of the high temperature in the Sun's corona.

Oscar, there is a very nice paper on forgotten names of the elements: https://link.springer.com/article/10.1007/s10699-013-9326-y

But how did they know it was an "element", I mean we know an element it's an atom with unique number of protons now, but what was an "element" for them before that? how did they know it's an element and not a compound? I think it's answered here: https://chemistry.stackexchange.com/questions/137140/how-did-mendeleev-know-elements-from-compounds-or-mixtures


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How do I calculate q/m?
My chemistry book (Mortimer in German) has the following question:
Calculate the value q/m for the following Ions:

("mit Masse" meaning "with mass")
How do I calculate q/m?
Thanks for any help!

Note that the first part has a charge of $+1$, that is it is a Hydrogen atom, in which an electron (actually it's only electron) is removed. Since the charge of an electron in $\pu{1.60 x 10^-19 C}$, and the mass of the Ion (as given per the question) is $\pu{1.67 x 10^-27 kg}$ [I prefer working in the SI units, you should be able to convert to any desired measurement system].
For this case, $$\frac{q}{m} \approx \pu{9.6 x 10^7 C kg-1}$$ Use the same logic for the other parts of the question.

You can use $\pu{}$ for physical units. Also, I like how you say you prefer working in SI and then use kg.. :o


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What is the energy difference between inner subshells?
For exemple, in silver the energy difference between the 4d 5s subshells is ≈ 4 eV (in the ultraviolet region, that's why it has no color, etc.), but what is the energy difference between some of its inner subshells, like 4s 3d, 3d 4p or 4s 4p?
I assume there is also a energy difference between them, right?

To find the energy difference between any of the subshells you can use the equation for Rydberg's formuals : Delta (E)= -2.17*10^-18J (1/nf^2)-(1/ni^2)


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What Are Angular Nodes , Radial Nodes, Nodal Plane and Nodal Surfaces?
Actually, i am confused between all of them. some says that angular node is same as nodal planes and radial nodes is same as nodal surfaces... and what's the difference between nodal planes and nodal surfaces. other says that angular node is the intersection point of nodal plane. please explain in easy explanation.

Summarize then who says what, and/or focus on single terms. They are rather self-explaining.

could you please share your explanation on the same

Write down your understanding effort first.

what i understood till now is that angular nodes are the intersection point of the nodal plane with the atom..... and radial nodes are the spherical region and the surface which it covers is called as nodal surfance... btw nodes and nodal planes are the areas where e- density  is zero

actually no, i have read that but couldn't understand

You are on the right track.

Nodes are generally 3D surface regions where electron occurrence probability density is zero.

Nodal planes and surfaces are such nodes from geometrical point of view.

Radial and angular nodes are such nodes from point of view of wave functions. As either radial either angular part of a wave function is responsible of the zero density value.

From the geometry of used spherical coordinates, it is quite clear that radial nodes are 3D spherical surfaces with constant radius, like for s orbitals.

Analogically, a constant angle of angular nodes leads to a 3D plane intersecting the coordinate origin, like for p orbitals.

Therefore,

radial nodes = spherical  nodes
angular nodes = planar nodes

For higher values of n and l quantum numbers, nodes are collection of radial and angular nodes, or nodal spherical surfaces and planes.

As wave functions do not have sharp boundary, but converge to zero values in infinity, the intersection of an atom and a nodal plane is – mathematically – the nodal plane. Practically the conventional size of atoms.

could you please differentiate nodal plane and angular node or radial node and nodal surfaces by graphics

I can, but I will not until you read my answer properly. As it is  there in more than 1 instance.

sorry i want to say differentiate them by the use of 3D graphics

Do you need to be provided by 3D graphics to differentiate a plane and a spherical surface? Or to differenciate 2 identical planes or 2 identical spherical surfaces ?

Have you ever tried the picture section of Google search for wave function radial angular nodes ?


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Are the (133,885) molecules in the QM9 database drug-like?
Are all the individual 133,845 molecules of QM9 drug-like molecules?
Can other drug-like molecules be made from the these 133,845 molecules by some combination or permutation which are not part of QM9?

How would you propose the drug-like properties are checked?

@Waylander, by using RDKit and Frechet distance could be a good starting point...

@Waylander, I have made the edit to my question, I wanted to know are all the individual molecules in QM9 are drug candidates, I mean is every molecule is a complete in itself or do we need to combine different molecules of QM9 to find a effective drug ?

I have not worked with QM9 but similar packages I have worked with consisted of fragments which would need to be extended to improve activity

QM9 contains a subset (with up to nine non-hydrogen atoms CNOF) of the GDB-17 database (up to 17 non-hydrogen atoms CNOS and halogen). Here is what the authors of GDB-17 said about drug-like:

While GDB-11 and GDB-13 uncovered impressive numbers of possible molecules, the databases only addressed very small organic molecules (MW < 200 Da), which are of interest as relatively small fragments but rarely correspond to actual drugs. Herein we report the enumeration of organic molecules up to 17 atoms of C, N, O, S, and halogens, forming the chemical universe database GDB-17 containing 166.4 billion organic molecules. GDB-17 reaches into molecular sizes compatible with many drugs (367 approved drugs ≤17 atoms) and typical for lead compounds (100 < MW < 350 Da). Millions of isomers of known drugs are readily identified in GDB-17.
[...] The idea is to enumerate molecules from first principles starting from mathematical graphs irrespective of pre-existing building blocks to avoid a historical bias in structure selection. Geometrical strain and functional group stability criteria are used to ensure that the molecules produced are chemically meaningful.
[...]While molecules up to 17 atoms in the public databases PubChem, ChEMBL, or DrugBrank are mostly achiral, aromatic, and heteroaromatic compounds with rodlike shapes, GDB-17 molecules are mostly nonaromatic heterocycles with many quaternary centers and stereoisomers.
Source: doi.org/10.1021/ci300415d

This seems to answer the OP's questions:

[OP] Are all the individual 133,845 molecules of QM9 drug-like molecules?

No, probably not a single one because they are too small.

[OP] Can other drug-like molecules also be made from the these 133,845 molecules by some combination or permutation which are not part of QM9?

Yes, by combining them in a reasonable way, you would get larger molecules (see Waylander's comment to the question). You would also have to check a parameter related to solubility. The computed data in QM9 would probably be helpful:

We report computed geometric, energetic, electronic, and thermodynamic properties for 134k stable small organic molecules made up of CHONF. These molecules correspond to the subset of all 133,885 species with up to nine heavy atoms (CONF) out of the GDB-17 chemical universe of 166 billion organic molecules. We report geometries minimal in energy, corresponding harmonic frequencies, dipole moments, polarizabilities, along with energies, enthalpies, and free energies of atomization.
[...]
This data set contains small amino acids, such as GLY, ALA, as well as nucleobases cytosine, uracil, and thymine. Also pharmaceutically relevant organic building blocks, such as pyruvic acid, piperazine, or hydroxy urea are included.
Source: https://www.nature.com/articles/sdata201422

No. The molecules in QM9 (i.e., a subset of the GDB-11) are part of an enumeration of possible compounds.

GDB-11 enumerates small organic molecules up to 11 atoms of C, N, O and F following simple chemical stability and synthetic feasibility rules.

The enumeration process is covered in the paper by the Reymond group:
Virtual Exploration of the Chemical Universe up to 11 Atoms of C, N, O, F
My understanding from the von Lilienfeld group, is that the full GDB-11 (26.4 million structures) was too big. So they scaled back to up to 9 "heavy atoms" of C, N, O, F, which was a more reasonable number of DFT optimizations.
Almost none of these are "drug-like" by the usual rules (e.g., Lipinski Rule-of-Five).


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creating hydrogenated castor oil/sebacic acid copolymer
I would like to create a copolymer using Hydrogenated castor oil and Sebacic acid to create a hair conditioning treatment, however I dont know where to begin so any advice will be highly appreciated.
I would like to know how a copolymer is formed, is it as simple as mixing sebacic acid and Hydrogenated Castor Oil under heat?
What formula should I use to work out the proportions of Hydrogenated Castor Oil and Sebacic Acid?
What formula should be used to work out the temperature needed to begin the chemical reaction?
How do I ensure that the chemical reaction has finished? Is there any formulas to work out the time needed for the chemical reaction of let's say 10g of sebacic acid and 50g of hydrogenated castor oil?
Is there any other substances that must be used to start the polymerisation or is sebacic acid and hydrogenated castor oil enough?
Any advise would be appreciated.
Many thanks

You might start with the application of your finished product, and why you would choose hydrogenated oil.  Unsaturated "drying oils" like linseed polymerize into coatings by reacting with O2.  You might be able to make polyester with sebacic and castor, but for what?

The application for the finished product is for a hair conditioning treatment. I would like to test the efficacy.

Well, my advice is to not do it yourself. Sorry, but it appears you don't know first thing about making polyesters :( If you have some chemist friend you could try to do it together.

Thank you for the advice, could you point me in the direction to learn the first thing about making copolymers?

Making esters can be done under nitrogen with a mixer and heat.  Water is the by product of esterification and should begin to distill off when the temperature reaches above 350F.  Xylene can be used to help the water off.  Water will split out from xylene when it condenses.  Since the alcohol group on castor oil is inhibited, the rxn may be slow.  The finished product may have a higher viscosity, which may help for hair products!

Thank you for the information Robert. I will have to learn quite a bit more! It seems more complicated than I initially thought


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Can micelles form in a polar-vacuum interface?
As the formation of micelles depends on, among many factors, the difference of polarity/relative permittivity of the solvents involved with the surfactant, I'm wondering if a polar-vacuum system is possible.
After all, a vacuum has a relative permittivity of 1, the lowest on the dielectric scale. Am I making an incorrect assumption or missing a piece of information in regard to this?
PS: I’m asking from a theoretical framework, not a practical/feasible one.
PS: In regards to the comments asking to define the polar-vacuum system, it is equivalent to the picture micelle; the only difference is that the middle of the normal system is a vacuum, as it has the lowest polarity, even lower than oils.

Define a polar vacuum system. What systems do you have in mind?

You cannot have micelles in vacuum.


text:
What bonds are formed between O2 molecules?
So for context the question preceding this is what bonds form within O2, which I know to be covalent bonds. I am not so sure on the bonds between O2 molecules. My first thought is that the bonds would be Van der Waals interactions but I'm not sure where the interaction would come from because there is nothing to cause a dipole in diatomic oxygen. Am I just overthinking it and there wouldn't be any kind of bonding or interactions formed between O2 molecules?

https://chemistry.stackexchange.com/questions/134694/did-o%e2%82%84-turn-out-to-explain-why-liquid-oxygen-appeared-to-defy-curies-law-if-no

Van der Waals intermolecular forces consist of

London dispersion force between random and induced dipoles, that applies to oxygen molecules. These are the weakest.
Debye force between permanent and induced dipoles
Keesom force between 2 permanent dipoles. These are the strongest.

The induction of dipoles in molecules, bonds or atoms can be caused by an external large scale electrostatic field or by nearby presence of ions, permanent or temporary (random or induced) dipoles.
See also Intermolecular force
For chemistry of low temperatures and high pressures, see tetraoxygen and octaoxygen, with the latter forming a nice brick-like structure:

As the pressure of oxygen at room temperature is increased through 10 gigapascals (1,500,000 psi), it undergoes a dramatic phase transition to a different allotrope. Its volume decreases significantly,and it changes color from sky-blue to deep red.

London Dispersion Forces will dominate for a non-polar molecule like $\ce{O2}$. Because electrons are constantly in motion, the molecule of $\ce{O2}$ can develop a temporary dipole which is the source of interaction between molecules. This type of force is weak, but with larger molecules with more electrons, this interaction becomes more significant.

The ground state for the O2 molecule is a triplet diradical; the 2 electrons are in antibonding orbitals with parallel spins. A higher energy singlet state with paired spins also exists. There is evidence that in the liquid phase there are weak molecular bonds between O2 molecules using these unpaired electrons [O4] because liquid O2 absorbs red light; gaseous O2 is completely colorless [not like water, water really is blue]. These unpaired electrons must also be involved in ozone[O3] formation. Solid O2 has at least 6 different phases involving larger O8 molecules. Wikipedia is a good start here, look it up.


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Is resonance possible in a structure with consecutive double bonds? Like in carbon dioxide (CO2)
With all the knowledge that I have secured reading about resonance, I know that delocalisation of pi electrons is possible only with the p orbitals being in the same plane.
So for two consecutive double bonds, the orbitals wouldn't be in the same plane. So there shouldn't be any hybridization, right? Like in $\ce{CO2}$, the O's are $Sp^2$ hybrids and C is sp, as per the structure, the three unhybridized p orbitals wouldn't be in the same plane, making it not possible to have resonance in it. But that isn't the case with it. It does have resonance structures in it. What am I missing?

In a hybridization model of CO2, a better representation is to have the O atoms sp hybridized rather than sp2, and have one lone pair in sp and one in pure p (which is aligned with the C=O bond to the other O).

Not only can there be resonance with cumulated double bonds, this is much in evidence with carbon dioxide itself. What is formally the pi bond between the carbon atom and each oxygen atom conjugates with a formal $p$-orbital lone pair on the other oxygen atom, which is in the same plane for each respective bond. Thus the pi electrons become delocalized, producing a series of contributing structures:
$\ce{\overset{-}{O}-C#\overset{+}{O}}$
$\ce{O=C=O}$
$\ce{\overset{+}{O}#C-\overset{-}{O}}$
The effect of this delocalization is to make the bonds shorter and stronger in carbon dioxide than they would be in an isolated carbon-oxygen double bond: From Wikipedia, we see the difference in bond lengths:

Bond lengths of C=O bonds are around 123 pm in carbonyl compounds. The C=O bond length in carbon dioxide is 116 pm.

But how can they the orbitals aren't in the same plane ?

They are in the same plane. You have an additional $p$ orbital in the same plane as each apparent double bond.


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Provide a mathematical explanation of why the non-relativistic Schrödinger equation for a Molecular system, cannot be solved analytically?
The non-relativistic Schrödinger Equation is:
$\widehat{H}|\psi\rangle=E|\psi\rangle$
Where $\widehat{H}$ is the Molecular Hamiltonian in Atomic Units and has the following terms:
$$  \widehat{H}
    =
    -\sum_{A=1}^M\frac{1}{2M_A}\nabla^2_A
    -\sum_{i=1}^N\frac{1}{2}\nabla^2_i
    +\sum_{A=1}^M\sum_{B>A}^M\frac{Z_AZ_B}{R_{AB}}
    +\sum_{i=1}^N\sum_{j>i}^N\frac{1}{r_{ij}}
    -\sum_{i=1}^N\sum_{A=1}^M\frac{Z_A}{r_{iA}} $$
Where labels A,B,.. denote nuclei and labels i,j,.. denote electrons. $\nabla^2_A$ and $\nabla^2_i$ are the corresponding Laplacians. $M_A$ is the mass of nuclei. $R_{AB}$ and $r_{ij}$ are the distances between corresponding Nuclear pairs and Electron pairs, respectively. $r_{iA}$ is the inter-nuclear-electron distance. $Z_A$ is the corresponding Nuclear Charge.
After applying the Born-Oppenheimer Approximation and treating the nuclei as stationary, and treating the Electronic Hamiltonian separately, i.e.
$$  \widehat{H}_{ele}
    =
    -\sum_{i=1}^N\frac{1}{2}\nabla^2_i
    -\sum_{i=1}^N\sum_{A=1}^M\frac{Z_A}{r_{iA}}
    +\sum_{i=1}^N\sum_{j>i}^N\frac{1}{r_{ij}} $$
I would like to know a mathematical explaination of why we cannot analytically solve the non-relativistic Schrödinger equation using the above electronic hamiltonian? Is it specifically the 3rd term that causes the problem?
In general many-body problems cannot be solved analytically, that is true. I would like to know the explanation for this specific case I mentioned above- for a Molecular System. I haven't found any resources to explain this point in more mathematical detail.
I have been reading Szabo and Ostlund's book. I would like to read more material regarding this specific point. Whether someone has worked this out for simple cases like say H2+ or H2?

This may be better asked in the Maths group

Even a classical, gravitationally bound system of 3 bodies cannot be generally solved analytically, Neither a helium atom can be. Why do you think molecular systems could be?

No, of course, I don't think it can be solved for a molecular system. I get your point regarding the 3-body gravitational system. I was hoping for whether someone has rigorously worked it out for a molecular system. I will probably take Ian Bush's suggestion and ask it in the Maths group. I will also look at the 3-body gravitational problem in mathematical detail. Thanks.

For the 3-body classical problem, I am not sure if the general solution just is not known, or if it was formally proved it does not exist. That is mathematical magic unavailable to merely mortal chemists.

ASide of this particular case for math, it may better fit the matter modeling SE site  // Matter Modeling Stack Exchange is a question and answer site for Matter Modelers: computational chemists, material scientists, particle physicists, data scientists, and anyone else who uses computational methods to study molecules and materials.

Note that you can get a huge improvement in the mathematical typesetting simply by using $$...$$ instead of $...$ for standalone equations ('display-style'). The latter is meant for equations which are part of text in a paragraph ('inline').

As this is a very complicated  integration, the more general question is why some far simpler integrations have no analytic solution and I guess that is a question for mathematicians.

I wonder if there isn't in fact an analytical solution, but it's an infinite series, of which numerical solutions are only a truncated approximation. The answer probably has to do with the universality of the convergence of the sums in the series (eg in a 2 electron system, because of the three "distances" involved: r1, r2 and r12).

see also: https://physics.stackexchange.com/questions/427663/why-cant-the-schr%C3%B6dinger-equation-be-solved-exactly-for-multi-electron-atoms-d

@BuckThorn In the classical case at least there is an analytic solution which is expressed as an infinite series, due to Sundman. Unfortunately it converges extremely slowly. See https://en.wikipedia.org/wiki/Three-body_problem

@Poutnik When we say there is no solution, we mean there is no solution in closed analytical form (which one can argue is not a very well-defined concept).

@Greg I have implied that since the beginning and it was this way discussed explicitly.

I will show specifically the case for a Helium atom, but it would explain in general the difficulty of solving multi-electronic systems. The Hamiltonian for Helium can be represented as given below:
$$\hat{H} = -\frac{1}{2}\nabla^2_1 - \frac{1}{2}\nabla^2_2 - \frac{Z}{r_1} - \frac{Z}{r_2} + \frac{1}{r_{12}}$$
This can be re-written as
$$\hat{H} = \hat{H_1} + \hat{H_2} + \frac{1}{r_{12}}$$
where $\hat{H_i} = -\frac{1}{2}\nabla^2_i - \frac{Z}{r_i}; i = 1,2$
If it were not for the repulsion term in the hamiltonian, it would be completely separable. Now, you also have to consider the fact that the wavefunction, $\Psi(r_1,r_2)$, essentially is a function of two electrons. If we ignored the electron-electron repulsion term, we could separate the wavefunction into the product of two functions for each electron: $\Psi(r_1,r_2) = \psi(r_1)\cdot\psi(r_2)$ and solve the differential equation operating $\hat{H_i}$ on the corresponding $\psi(r_1)$ or $\psi(r_2)$. However, this would be like solving for two separate hydrogen atoms but with Z = 2.
The electron repulsion term makes this separation impossible as the term depends on $r_1$ and $r_2$. This is why there is a slight burden in solving the non-relativistic Schrödinger equation for a multi-electron system, I believe.

That's true, but your answer only explains why the Schrödinger equation can't be solved through separation of variables, it doesn't explain why the Schrödinger equation can't be solved, full stop. I expect OP is looking for the latter. Of course, proving that something cannot be solved is rather non-trivial, and well beyond my capabilities too...

@orthocresol Proving something cannot be solved always looked to me as an advanced math magic, approching proving that something does not exist. :-) (Nobody can prove a unicorn does not exist, by principle absence of evidence is not evidence of absence. )

@orthocresol yes I am looking for the latter. So suppose a person would look at the H atom case and say it's solvable analytically. Then they would look at the H2+ molecule and say well you can't solve it analytically because of the following reasons and therefore we go to approximate methods like Hartree-Fock Method. I am looking specifically, for those mathematical reasons. So for eg what @ M.L answered is right. I want to look at the deeper mathematical reason for it. And this is why I have also posted in Mathematics SE.

Maybe take a look at this paper: https://arxiv.org/pdf/0704.0383.pdf and specifically the conclusion if that helps. I kind of got lost a quarter of the way through.

Part of this may be related to the mathematical tools we have available. Algebra has been known for millenia, but it was impossible to solve cubic equations such as $x^3-3x-1=0$ for three real roots until the concept of imaginary numbers (whose cube roots appear in the solution) was developed. By extension, maybe someday we will find tools to solve at least simpler systems such as the helium atom.

Part of this might be related to the solutions of the three body problem, which are mostly chaotic in nature. A closed-form solution must be "expressed in terms of a finite number of standard mathematical operations". If the solutions are too "crazy", that's not possible. You can get numerical approximations, and they show how complicated the solutions are (so it is the solutions themselves, not the path to them, that preclude a closed-form).

If you think about it, we know the solution to the helium atom problem. We can just measure it, the electrons are perfectly fine in "solving it" within nanoseconds.

@Karsten Well, we have to make approximations for the helium atom solution through perturbation theory or variational method.

Or something else but it is not peer-reviewed. Wait a moment, here it is behind a pay wall but peer-reviewed.

The fact that the Schrödinger equation for a multi-body case cannot be solved, i.e. has no closed form solution, has nothing to do with quantum mechanics itself but is a more general issue with dynamical systems governed by invert square interaction (Coulomb, gravitational force, etc.). This is usually refer to as the three-body problem (or more generally the $n$-body problem).


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How long can dimethyl sulfoxide be safely stored in a HDPE "barrier bottle"?
How long can dimethyl sulfoxide be contained in a HDPE barrier bottle without any reaction between the plastic and the solvent?  I have read that DMSO is not reactive with HDPE, but I would like to know if anyone has alternative data.  I received a shipment of 99.999% pure DMSO, directly from the manufacturer, in specially made HDPE gallon jugs, but I am wondering if I would be well advised to transfer it to glass containers for long-term storage, to maintain purity.  Thank you for any insights!
edit: Someone in the comments has suggested I add this information: the intended use of this DMSO is for personal experimentation and research regarding the pain reducing qualities of DMSO, utilizing myself as the test subject.

I am interested to know if anyone who uses DMSO for research, ie someone who might have the equipment to test whether a solvent has leached something from a plastic container, has ever noticed DMSO being contaminated after storage in HDPE.  People who use DMSO for medical reasons generally have no such equipment and have to rely on trusting the manufacturer.  Just looking for data from people who might know!  I don't think the intended usage here matters, plus it is common for people to divert the attention to whether or not one should use said treatment, rather than answer the question.

There is another aspect, uncontrolled exposure to contamination because of liquid transfer. E.g., but not limited to, presence of sulfur in molecule may support extraction of trace heavy metals from glass. There is general stereotype plastics=bad, glass=good. Sometimes, it is the opposite.

Just wondering, why did this question get a minus one?  What is the problem?

2 years according to the Aldrich data sheet
https://www.sigmaaldrich.com/deepweb/assets/sigmaaldrich/product/documents/303/292/d8779pis.pdf

Almost indefinitely as long as the temperature is below 60 degrees or so. I have seen vats full of DMSO of Poly propylene FRP construction up for years. Although the one we used is 99.1% and more and not 99.999% but yeah one FRP tank is more than 22 years old and has been filled with DMSO all its like used to make TMSO-B all day everyday.

Thank you!  This is helpful, the kind of answer I am looking for.  My storage area can get up to 80 degrees F.  Should I go ahead and transfer the DMSO from the HDPE bottles into amber glass?  The glass bottles are lead-free.

60 degrees I meant C. Keep it in HDPE or Amber Glass don't matter. If you want to keep it even safer, just add some water. Water DMSO is even safer. DMSO doesn't attack things like Epoxy either. Its not like its NMP


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Covalent compounds-What are the reasons for attraction between atoms to be strong...?
In general, most of the covalent compounds are known to have low boiling point and melting point due to the weak Van der Waals forces between molecules in them, requiring lesser heat to break the bond.
The weak VdW forces is due to strong covalent bond within their molecule (or maybe due to the atoms are held tightly to each other...)
Here is the question:
What is the reason for covalent bond or attraction force between atoms in them to be strong?

Have you read covalent bond pages on wikipedia or  libretexts.org ?

https://chemistry.stackexchange.com/questions/165469/why-covalently-bonded-molecules-have-strong-bonds-within-the-molecule https://chemistry.stackexchange.com/questions/11048/are-metallic-ionic-bonds-weaker-than-covalent-bonds

Your definition of "covalent compound" is suspect. Iodine and menthol are both covalent compounds by your definition and are volatile or low melting point solids. But diamond and kevlar are also normally thought of as covalent compounds but are both strong and not remotely volatile. So what was the question again?

@matt_black What is the reason of attraction force between atoms to be strong in covalent compound ?(diamond and silica etc. are exceptions)

Are you asking why a covalent bond is strong? (That has not much to do with boiling points or melting points, so it's likely that your first paragraph is confusing)

@orthocresol Hmm..to be honest,yes.Undeniably,the first paragraph is really messy.(I will try better next time)

It is simple just ignore the QM complications. The electrons in a bond are in lowere energy orbitals when they can be attracted to 2 [or more] nuclei. The amount of shielding  and the orbital shapes determines the bond strengths. Imagine walking with your sig other occasionally bumping each other, weak intermolecular attraction, hold hands a weak molecular bond, the nuclei are still far apart[think

I2 molecules Get in close a good strong Hug,  a 4 armed multiple bond. Now give a good hard squeeze your other goes OUCH and springs back the nuclei got too close together and the bond electrons got squeezed out of their orbitals. You really do not need QM but it does explain things


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What is resonance in actuality? (How does electron sharing, bond formation, and overlapping of orbitals take place in resonance hybrids?)
In this post I got an answer to the question "what is resonance".
What I understand  overall is that "resonance is not something really happening physically, it is just an idea to make people understand a concept."
But a few doubts remain:
Previously I thought that due to the movements of delocalised  electron(s), the pi bond and charge keep shifting rapidly, like various structures physically interchanging into one another very quickly as electrons keep revolving. That's why we use dotted lines to show that bonds can exist in two places at the same time.
But after reading the answer to my previous post I think I was wrong.
Then please explain to me what those dotted lines actually mean. Are they a special type of bond or something else?
Does that mean that when we say, for example, that ozone has one double bond and two single bonds, that this is in fact not true!?
Because in a hybrid structure we have one bond completely different in both structures, somewhat dotted and straight, with bond length inbetween.
Then, how is an electron shared in the case of a hybrid bond?
Normally an octet (generally) is completed by a single bond (2 electron sharing) or a double bond (4 electron sharing). How many electrons are shared in a dotted bond, and how, if at all?

Have you considered browsing the site and looking for similar questions? Also, please check grammar etc to the extent possible before posting.

To address your first question:
You (erroneously) believe that resonance structures are in dynamic equilibrium. Your description “the $\pi$ bond and charge keep shifting rapidly , like various structures are Physically Interchanging into one another very fastly” is exactly what Dynamic Equilibrium is. For eg. in the equilibrium $\ce{N_2 +3H_2\rightleftharpoons2NH_3}$, the nitrogen and hydrogen constantly combine to produce ammonia, while the ammonia also simultaneously decomposes to provide nitrogen and hydrogen gases. The rates of the forward and backward reactions are equal at equilibrium.
Resonance forms are NOT in equilibrium. That is why we use $\leftrightarrow$, and not $\rightleftharpoons$ to denote resonance forms, to indicate that they are, indeed, equivalent forms of the same molecule.

The dotted line refers to the fact that the properties of the bonds in the ozone molecule are intermediate to that of single and double bonds. From Wikipedia, BOTH O-O bonds are equivalent, i.e. have exactly the same properties (bond length, bond enthalpy etc). That wouldn’t be true if there actually were a single bond and a double bond. The dotted line shows that the electron density in each O-O bond in ozone is intermediate to that in single and double bonds.

When we draw the above structure of ozone, we are drawing a Lewis structure. This is the best we can come up with to describe the behaviour of ozone (yes, this is not how ozone is actually). The charges that you see on the oxygen atoms are formal charges. They DO NOT actually appear on the atoms individually. The charges are just written to keep track of the valence electrons of the atoms when drawing the inaccurate description of the ozone molecule by the Lewis structure. The overall molecule is neutral, while the central atom acquires a slightly less electron density (called partial positive charge) and the side atoms acquire greater negative charge density.

Resonance is real Electron orbitals arrange to give maximum attraction to the various nuclei while minimizing electron repulsions. Our simple interpretations while pretty amazing in themselves might not be able to truly represent electron distributions. So we make up canonical contributing structures usually further from reality and worry about their contribution.


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Why do the links in a plastic chain become different colors?
I have a plastic chain, about 5 meters long, sitting between two wooden posts. The link has 1 meter long sections of alternating white/red links. It's been sitting there for 2 years, and the once red links have now become purple/pink. I always just thought it's because of exposure to the sun and "the elements" and didn't think much about it.
But today I noticed something strange - the links between each section of white and red are a very different shade of red than the red links inside the red sections.
How can this be? Is there some chemistry going on between the links which makes the color less faded in the ones that are touching one red and one white link?
I am not 100% sure, but I definitely don't remember any color difference when the chain was new - all the white links were the same white color, and all the red links were the same red color.

The number of red dyes used in painting is huge. They all react differently with light. So we cannot answer this question, without some other information.

At a guess, the links came from different batches of plastics, stored in bins. and became mixed when the chain was made, later. As you surmise, sunlight, ozone, chlorine (if stored near a pool) and other conditions and chemicals attack the dyes, and different dyes at differing rates.

@DrMoishePippik Right, the strange thing is though that it's the same for every link touching one white and one red link. The links touching only other red links have changed into a more purple color.

It might be a temperature (light reflecting) effect where the white link cools the neighboring red link. It also might be that they make red and white chains first, and the use one extra link (processed differently) to put it together.


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Favoured reaction product in acid catalysed addition of methanol to alkene
I would like to add methanol to an alkene RCH=CH2, catalysed by conc. sulphuric acid. I would like to know which of these endproducts will be preferentially produced:

RCH(OMe)-CH3
RCH2-CH2OMe

Many thanks in advance!

Ethers are hardly made like that.

Then kindly suggest an alternative method to produce the endproducts from the starting materials.

As the OP asks in the comments, I shall provide a pathway for producing the end products.
For $\ce{RCH(OMe)CH_3}$ from $\ce{RCH=CH_2}$:

Electrophilic addition of water across double bond following Markovnikov’s Rule: Use either Acid Catalysed Hydration or Oxymercuration-Demercuration. $\ce{RCH=CH_2\longrightarrow RCH(OH)CH_3}$
Add a small amount of sodium or sodium hydride: There is formation of hydrogen gas as the alcohol acts like an acid in presence of $\ce{Na}$ or $\ce{NaH}$: $\ce{RCH(OH)CH_3\longrightarrow RCH(O^-Na^+)CH_3}$
Add iodomethane: The nucleophilic oxide ion attacks the iodomethane  to form $\ce{RCH(OCH_3)CH_3}$ via SN2 reaction mechanism. (Williamson’s Ether Synthesis)

For $\ce{RCH_2CH_2OMe}$ from $\ce{RCH=CH_2}$:

Anti-Markovnikov Water Addition: Use Hydroboration-Oxidation.
Add a small amount of sodium or sodium hydride
Add iodomethane (Williamson’s Ether Synthesis)

Alternative Idea:
You may want to explore adding aq. acidified methanol solution into $\ce{RCH(OH)CH_3}$ and $\ce{RCH_2CH_2OH}$. This will give you $\ce{RCH(OMe)CH_3}$ as the major product.  See here for more details.

Thank you insipidintegrator! My original idea is explained here: https://www.chadsprep.com/chads-organic-chemistry-videos/addition-of-alcohol-to-alkene/  How can I modify this to maximise RCH2CH2OMe ?

Unfortunately, $\ce{RCH_2CH_2OMe}$ always remains the minor product as the mechanism is an Electrophilic addition involving formation of a carbocation intermediate which is much more stable on a secondary carbon than a primary one. Also, you can get a bit of $\ce{RCH(OH)CH_3}$ because of water addition as a side reaction. (Note that you cannot operate in $100$% pure alcohol(as a way of side-stepping the reaction with water) as organic acids do not ionise in non-polar solvents) A pathway to produce the ether you want involves use of diazomethane and $\ce{BF_3}$ with $\ce{RCH_2CH_2OH}$.

Thanks again, Insipidintegrator. What would adding H2O2 do to the reaction mechanism? Would we get a Kharasch type anti-Markovnikov addition to favour RCH2CH2OMe?

If you have studied the mechanism of the Kharasch effect, you’ll notice that the first step involves the homolytic cleavage of the radical initiator and subsequently the reagent, i.e. $\ce{HBr}$ (not $\ce{HCl}$ or $\ce{HI}$). I’m not very sure about the viability of the formation of $\ce{CH_3O•}$ radical from methanol. However, I found this: https://pubs.acs.org/doi/10.1021/ja00328a011 It involves photolysis of methanol.

Pure alcohols will be protonated by organic acids altho the reaction is promoted by traces of water or by bases such as pyridine.


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How are spin magnetic moment and orbital angular moment in opposition in a d3 ion in octahedral environment?

I have two queries firstly what does spin exactly mean and secondly how do i decide whether the spin and orbital moment are in opposition or not or is this an experimental fact.
I would also like to know what does it mean by "when the effect of spin orbit coupling is considered it is found that orbital angular momentum is mixed into the ground state from the first excited state of the system."

Images are not searchable. Please write the relevant portion as text. https://chemistry.meta.stackexchange.com/questions/3758/policy-on-picture-questions

You should add that the image is from Cotton and Wilkinson, 'Advanced Inorganic Chemistry' publ Wiley.

The electron has the intrinsic property of spin and so this can give rise to a magnetic dipole. Also, the electron in its orbit produces a magnetic field as a result of its orbital motion. An interaction between the two is possible and is called Spin-Orbit coupling. The magnitude of the interaction depends on the atomic number (more accurately as effective charge $Z^4$) and inversely to the principal quantum number $n$ and orbital quantum number $\ell$. (For a 3d electron this has a value $\approx 0.014Z^4\mathrm{cm^{-1}}$). The sign of the coupling depends on the product of orbital and spin quantum number vectors $\vec L\cdot\vec S$. Additionally this product has a certain symmetry and so can couple an excited and a ground state if the product of the three symmetries contain the totally symmetric representation. This interaction can therefore enhance or reduce the spin only values when calculating the magnetic moment.
(It should be added that a full calculation is very complicated and beyond undergraduate level which is why only a summary is given in the quote above. More details, but only  a little more can be found in specialist books on spectroscopy and in some phys. chem. textbooks. )


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Is melamine soluble in methanol? if it is then what is the solubility of it ? In grams/mL?
I am trying to dissolve 100 mg of melamine in 35 mL of methanol. But after stirring and heating the solution for more than two hours I couldn't get any clear solution. Where have I gone wrong? I couldn't find the solubility of melamine in methanol. Kindly help

Melamine is described as being "slightly soluble in water and alcohol" data here.
The figure for Ethanol is quoted as 0.6g per litre here.
While I cannot find a figure for methanol, these data lead me to think that you will need more methanol to dissolve 100mg. Ethylene glycol is a better choice


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IR / Raman factor group analysis (site and factor group determination)
I'm currently trying to interpret some IR and Raman spectra of solid state compounds. They show a simple molecular unit of which the expected bands are split due to the coordination in the crystal structure. For a complete overview I'd like to add a factor group analysis, too.
I was reading through the original paper but it's a bit too mathematical for me. So instead I looked at different publications that already did a similar compilation of point group, site group and factor group and tried to recreate their solutions. I think I'm fairly sure it matches their results but since I haven't found an exact statement in literature, that would describe the procedure as I do it, I'd like to get some second thoughts on my resoning.
What isn't entirely clear to me is the description of the site and of the factor group. To my understanding the site group is a non-trivial subgroup of the molecule's point group. Say I have a tetrahedral molecular unit in my crystal structure of point group  $T_{d}$ the possible non-trivial subgroups contain for example  $D_{2d}$,  $S_{4}$,  $C_{2v}$, etc. So I would focus on how the coordination of the molecular unit to the next cations in the crystal structure would reduce the symmetry elements and therefore also reduce the point group to a subgroup.
For the factor group I understand that you have to consider the crystal structure and the optical phonons, which is no problem as services like the Bilbao Crystallographic Server have options to calculate this. The optical phonons determined are also split into the different Wyckoff-sites.
To my reasoning, I would simply take the Wyckoff-sites that are occupied by the central ion of my molecular unit, since it's the point I also define my point group and my site group symmetry on.
I am also pretty certain that I read this definition in a publication but at the moment I'm only able to find examples where the Brillouin zone is discussed. So I'm now a bit sceptical about my approach.
Could someone please give me a hint if I'm doing a mistake somewhere that just happened to work for the given examples?

Since 11 months have gone by with no answers or comments, I think it's fair to let you know about Matter Modeling Stack Exchange which also answers group theory questions, but make sure to delete the question from here if you're cross-posting there (or to at least put a note at the top of your question, saying that it's cross-posted).


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Is polycarbonate resistant to extremely cold temperatures?
I am working on a project which involves the use of polycarbonate. Polycarbonate is quite a new material to me as I haven't worked with it yet. I want to know its characteristics in detail, as it would help me in my project. Here they go:

Is polycarbonate strong enough to withstand pressures up to 20 atm? If the surface has a slight curvature, can it retain the curvature under such high pressures?
Is polycarbonate resistant to inhospitable temperatures as low as -200˚C?

Being "resistant" or "strong enough" is quite not clear. Then again, I doubt PC is right pick for whatever you want to do.

"withstand pressures up to 20 atm?" Certainly -- in a thick enough piece. You need to be more specific, and do some research. Polycarbonate cryogenic supplies are sold (e.g., https://www.grainger.com/category/lab-supplies/lab-diagnostics/cryogenic-supplies?attrs=Material|Polycarbonate&filters=attrs), so consult the literature and manufacturer for specs.


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Non toxic Chemicals that will react in water?
Long story short, I'm planning an in-home escape room and want to create a puzzle where the guests find a vial containing some sort of chemical that they can dump into a water fountain. My intention then is that chemical will react with another chemical that was placed on rocks within the water fountain prior. Then, the reaction causes a code to be spelled out on the rocks within the water fountain.
Obviously, the code on the rocks / in the fountain shouldn't be visible until the other chemical is introduced into the water fountain and the reaction occurs. It also goes without saying that the chemicals should be relatively non toxic or not a major issue if someone accidentally gets some on their hands. My one college chemistry course did not quite equip me for this task! Thanks for any help.

I really don’t think it is viable to spell out letters in water. Eithe the “chemical placed on rocks” will dissolve in water, or if it os insoluble it will simply disperse out into the medium.

Non-chem suggestion: hide the parts to make a UV light, e.g. a small (5 mw) UV-B LED and a fresh Li coin cell (not for use with small children who might put it in mouth!), as part of the game. Write the text in invisible fluorescent crayon, or use naturally fluorescent pebbles to spell it out.
Hmm... actually, there is chemistry involved: the chemistry of fluorescence and phosphorescence. You could make your own fluorescent paint with activated calcium sulfide, zinc sulfide, or other comparatively safe chemicals.
Note: sources for parts and chemicals are only used as examples; you can likely find them elsewhere.

(+1) Good answer! There are plenty of naturally fluorescent minerals available quite inexpensively at Gem and Mineral shows, on ebay, etc. Many are mundane looking in daylight or room lighting, but magnificent when exposed to appropriate UV light. Even the “mineral lights” are fairly inexpensive.

Hold a gender reveal party with milk of magnesia. Stir some phenolphthalein (girl) or bromothymol blue (boy) into the basic (pH 10) suspension and let the guests watch what happens.

Acid-base indicators are often used for this purpose, for example phenolphtaleine is common in "secret message" experiments. I am not too sure, however, on how exactly would you ensure that your message is not disperged in the water.


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Does the ace- or acet- prefix denote simplest forms of functional groups?
For example Acetone has the shortest carbon chain in which a ketone functional group can be added. Other examples include Acetaldehyde, Acetylene. Was this some sort of older naming system in which the prefix ace was used for simplest examples of functional groups or is this some sort of large coincidence. There do seem to be some contradictions to my theory such as acetaminophen or paracetamol but these usually contain more than one functional group so they could just be coincidentally named with ace- or acet- at the start.

Your hypothesis falls apart at acetaldehyde; it's not the simplest aldehyde.

Formaldehyde

No. It has a historical origin, going back to the 19th century. Acetone was first discovered by pyrolysis of sodium acetate $\ce{CH3COONa}$. When heated in the Bunsen burner flame, sodium acetate produces a new volatile substance, according to $$\ce{2 CH3COONa -> CH3COCH3 + Na2CO3}$$ As the first chemist who discovered this volatile substance did not know that it was the first member of a new series of compounds, the ketones, he decided to name it as if it was a derivate of an acetate, without any sodium. So he created the name acetone, by changing the last letters of "acetate".


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How to draw delayed picture on the concrete wall?
I have a concrete (stone) vertical wall. I want to put a picture on it.
I have several limitations:

I want to make it in a security, so I want to place something on the wall, that would be transparent or very similar to the original color.
I am not a scientist and have no access to industrial materials, so want to make it out with everyday materials.

The wall itself has properties:

It looks on the south, so sun lights on it almost full day (as a consequence, it is certain temperature and low humidity)
Overall direction is vertical; while it is not fully straight, it has relief.
There is little overhand above the wall, so it would not get heavy water stream in case of straight rain

So I did think about colored bacteria or mold (or other unicellular), that i can put on the wall with nutrients, but in case of mold it requires warm but wet environment. In the case of bacteria, I have no enough expertise to figure out which are needed and how to make them grow.
So maybe there is some chemical reactives that I can put on the wall, and under sunlight it would become colored in some time. Or something that becomes colored after becoming wet after rain?

Do you know these mugs that reveal a picture when you fill in a hot liquid?

With such a security need, others must have the wall. ;-)


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NiCr wire and water
Suppose I'm a very impatient person and I want to heat up water by getting rid of the layer of isolation between a NiCr wire and the water (e.g. skinny-dipping). The idea is that, since the wires can be heated up to >1000°, it can even be running water in high(ish) volumes. Assuming I know how to stay within electrical safety, what disadvantages are there for both the people wanting to potentially ingest water (e.g. make tea), and for the NiCr wire, in the short or long run? I'm interested in both DC and AC conditions.

@Poutnik That's true but, just because it can be heated until 1000+ deg, it doesn't mean it will be (on the electrical side, assume solid protection & co). The purpose is to get the water heated up to usable levels. Plus, the water debit will have an influence in the heat transfer. Assume the heating is "doable", I'm interested in the outcomes of such a thought at the end of the pipe (safe to drink, doesn't cause rash/diseases/etc), and on the wire (corrosion/pollution/etc if the wires will be in water all the time, ...).

I'd be happy to improve the question if there was an explanation attached to the downvote. If I had known what to correct, I would have done it since the beginning. If I haven't done it, it's probably because I am missing something. An explanation would be more than welcome.

@Poutnik Think of a pipe in which the wire is introduced and heated, while the water is running. That pipe will be left with water inside after the usage, the wires will not be conducting (only during usage). Quite literally a running-water heater but, instead of using an isolated pipe, give up the isolation, have the wire straight in the water.

@Poutnik While I appreciate the concern (and agree that accidents can happen, anytime), I'm fairly active on ee.se, so the electrical part can be considered safe: no abrupt heating, no overcurrent/overvoltage issues, all within IEC specs, controlled regulation, etc. As I said: from the electrical side, consider it solid. I'm interested in the chemical aspect of the thought.

I have deleted my prior comments.

Impatient people have no business fooling around with chemistry and electricity.

@jimchmst Oh, definitely, and presumptious people should also avoid pretending to sound smart, wouldn't you agree? I don't know what your comment has to do with my question, though.

The main objections I can see, purely from POV of chemistry:

Intense phase transitions or temperature gradients, together with localized mechanical shocks may lead to surface abrassion, intense corrosion or metal micro particle releases, affecting both the wire and water quality.
Big potential gradient along the wire may lead to electrolytic activities, affecting water quality.
Limescale will not probably have even chance to settle down on the wire, but if it does, it would strongly affect local wire temperature.
Water may get unpleasant taste, not speaking about heavy metal traces in context of requirement for water to be drunk.

They fit quite well with what I had in mind. My reason for asking was to get a more informed version than my suspicions. I'll leave this opened for a day or two, maybe there are others who'd like to take a stab at it.

Better to get stabs here than from angry tea drinkers. :-)

Oh, I'm usually calm when I drink my tea...


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Can the charges at the ends of a dipole in a polar molecule exceed unit electronic charge?
According to my current understanding, charges at the end of the dipole of a polar molecule are formed due to the shifting of the electron cloud towards the more electronegative atom.
In my book it is written "Ends of dipole possess partial charges. Partial charges are always less than the unit electronic charge $(\pu{1.6x10^-19 C})$".
Can the charges at the end of the dipole exceed $\pu{1.6x10^-19 C}$ under any condition? (single bond, double bond, triple bond). If not, then why?

it is 1.6x10^-19 C

The said charges are not quite a thing that exists, hence they can't be measured accurately.

The "partial charges" are an notional explanation, not the thing that is observed. What is observed is the strength of the dipole in a molecule. Whether the observed dipole appears to need a particular partial charge is the issue.

You would deal then rather with ionic than covalent bonds.

"You would deal then rather with ionic than covalent bonds."suppose in a double bond(two electron is shared by each atom) or triple bond(three electrons are shared by each atom), can the electronegative atom pull the shared electron cloud to the extent that the charges at the end of the dipole exceed 1.6×10−19 C.why is that not possible? Is it because the anions are not electronegative to that extent, hence they cannot pull the   electron cloud to the extent that the charges at the end of the dipole exceed 1.6×10−19 C?


text:
Can glycerine be degraded by UV light?
I would like to know, if UV degradation of glycerine is possible and what is the reaction mechanism? It is explicitly about glycerine and I could not find anything useful in the literature.

Like any organic compound can be, but I don't think it's particularly relevant to this one.

Simple alcohols and hydrocarbons have no electronic transitions in the common instrument ultraviolet above ~190-210 nm. Unless you have access to the vacuum uv  any degradation or reaction would be from photoinitiation from impurities present that absorbed the uv light available. Run a uv spectrum and let us know. It seems that Sigma Aldrich has discontinued their spectra grade glycerin so you will have to search round or purify some. [do not ask me how; it seems to be one of those compounds with a good melting point but does not refreeze; I was going to suggest zone refining!]

Yes, glycerine can be degraded by UV light. When glycerine is exposed to UV light, it breaks down into smaller molecules that are then more easily broken down by other means.

Related propylene glycol: https://chemistry.stackexchange.com/questions/75988/propylene-glycol-degradation-due-to-sunlight?noredirect=1&lq=1

While some may have a difference in opinion, but per a source, "Darker colors absorb more UV than lighter colors like whites and pastels. This means the UV rays are less likely to reach your skin. But bright colors such as red can also absorb UV rays. The more vivid the color, the greater the protection—a bright yellow shirt is more protective than a pale one."  http://depts.washington.edu/cmditr/modules/lum/color.html . So, adding say Vitamin B12, could promote a photolysis, as I have actually performed.


text:
Stoichiometric coefficients in reaction between vanadium(II) oxide and iron(III) oxide

Calculate the weight of $\ce{V2O5}$ produced from $\pu{2g}$ of $\ce{VO}$ and $\pu{5.75 g}$ of $\ce{Fe2O3}$

Now we write the reaction as following - $$\ce{2 VO + 3Fe2O3-> V2O5 + 6FeO}$$
Now, by the Principle of Atom Conservation on vanadium, we get,
$$n_\ce{VO} = 2\times n_\ce{V2O5}$$
$$\implies \pu{2 g} / \pu{67 g mol-1} = 2\times n_\ce{V2O5}$$
$$\implies n_\ce{V2O5} = \frac{\pu{0.029850 mol}}{2} = \pu{0.014925 mol}$$
$$m_\ce{V2O5} = \pu{0.0149250 mol} * \pu{182 g mol-1} \approx \pu{2.71 g}$$
I know that this can be done by limiting reagent, but what is wrong in this? The correct answer is $\boxed{\pu{2.18 g}}$

Your calculation is correct if all $\ce{VO}$ is transformed into $\ce{V2O5}$. But here, the amount of $\ce{Fe2O3}$ is not sufficient to transform all $\ce{VO}$ into $\ce{V2O5}$. So please start your calculation again starting with the amount of $\ce{Fe2O3}$

@Karsten, the edit that balanced the reaction given by the OP deviates from the original intent. They did not actually plan to balance and solve.

@Aqua Note that your task analysis has been very rushed. There is a saying "Measure twice, cut once". Applied to practical tasks - "Think twice, calculate once."

Let's start with what POAC, (also known as law of conservation of mass) really talks about,

Within a chemical reaction, no new atoms are created or destroyed. More formally,
$$\sum{m_\mathrm{before\,rxn} = \sum{m_\mathrm{after\, rxn}}}$$

Moving onto what you have done

Now, by the Principle of Atom Conservation on vanadium, we get,
$$n_\ce{VO} = 2 \times n_\ce{V2O5}$$
$$\implies 2/67 = 2\times n_\ce{V2O5}$$

This is correct (well kind of correct). However, this makes one assumption that doesn't really exist in this scenario. It assumes that all the $\ce{VO}$ reacts within the reaction. Which unfortunately doesn't happen here.
Taken from Rafael L's comment, the right method to do this (and the most accurate method) is as follows

[...] convert the masses of $\ce{VO}$ and $\ce{Fe2O3}$ to moles, figure out which is the limiting reagent (balanced eqn: $\ce{2VO + 3Fe2O3 → 6FeO + V2O5}$), then calculate the moles of $\ce{V2O5}$ produced and convert moles of $\ce{V2O5}$ to mass.

Now, following this, we see the number of equivalents on $\ce{VO}$ and $\ce{Fe2O3}$ are as follows,
$$n_\ce{Fe2O3} = \pu{0.03593 mol} \implies \mathrm{equiv}_\ce{Fe2O3} = \pu{0.011979 eq}$$
$$n_\ce{VO} = \pu{0.029850 mol}  \implies \mathrm{equiv}_\ce{VO} = \pu{0.0149250 eq}$$
From here, you can notice that the amount of $\ce{Fe2O3}$ that reacts is more than the $\ce{VO}$ but the number of equivalents is less. What this implies is that there is some $\ce{VO}$ left over. The rest is just multiplication.

[...] doesn't then POAC fail entirely? I used it many times which involved such reactions.

The answer to this question is no. POAC still holds. It's just a bit more complicated. Let's check this with the formal definition,
$$\sum{m_\mathrm{before \, rxn} = \sum{m_\mathrm{after \, rxn}}}$$
\begin{align}
m_\ce{VO} &= \pu{0.395 g}\\
m_\ce{Fe2O3} &= \pu{0 g}\\
m_\ce{V2O5} &= \pu{2.180 g}\\
m_\ce{FeO} &= \pu{5.174 g}\\
\end{align}
\begin{align}
m_\mathrm{before\,rxn} &= \pu{7.75 g} \\
m_\mathrm{after\,rxn} &= \pu{7.749 g}
\end{align}
Therefore, both are approximately equal, and POAC holds.

[OP] Now, by the Principle of Atom Conservation on vanadium, we get,
$$n_\ce{VO} = 2\times n_\ce{V2O5}$$

This is not universally true. In the extreme case where you don't provide any iron, it should be easy to see: There is no reaction, so none of the product is made even though there is reactant. Also, the notation is a bit sloppy. You have to know that the first term is supposed to be about "before the reaction", and the second term about "after the reaction". Some textbooks use ICE tables to force you to make this explicit, by writing the amounts in the appropriate box.
The clearer way to write this is as follow:
$$- \Delta n_\ce{VO} = 2\times \Delta n_\ce{V2O5}$$
In words, for each mole of $\ce{V2O5}$ produced, 2 moles of $\ce{VO}$ are consumed in this reaction. Or, and this is easier for me personally because it visually corresponds to the statement just made, you can set up the ratio:
$$ \frac{\Delta n_\ce{VO}}{\Delta n_\ce{V2O5}} = \frac{\pu{-2 mol}}{\pu{1 mol}}$$
It is possible that the reaction does not happen at all or does not use up all the  $\ce{VO}$, then this relationship still holds. However, if there is a different reaction (or a side reaction), you can no longer rely on this modified relationship either.

[OP] I know that this can be done by limiting reagent, but what is wrong in this?

If your reactant is limiting, you can set $\Delta n_\ce{VO}$ to $-n_{0,\ce{VO}}$, i.e. all gets used up when the reaction goes to completion. If your reactant is in excess, you have to figure out how much reacts, i.e. what $\Delta n_\ce{VO}$ is. In either case, no atoms are created or destroyed, though.


text:
Reason for exceptions in I.E. trend in group 13
In general, we would expect IE to decrease down a group. This is because atomic radius increases $\implies$ valence electrons are further from nucleus $\implies$ less effective nuclear attraction on valence electrons, so they are easier to remove. Also, shielding effect due to more subshells also contributes to this. In group 13, for example, going down from B to Al, we do observe this trend:

Boron IE = 801 kJ/mol

Aluminium IE = 577 kJ/mol (numbers from my textbook)

Now when we go from Al to Ga, the IE increases:

Gallium IE = 579 kJ/mol

Question 1) Shouldn't there be a decrease in IE from Al to Ga?
When we go from Ga to In, IE decreases, as it should:

Indium IE = 558 kJ/mol

From In to Tl it increases again:

Thallium IE = 589 kJ/mol

Why?
My teacher said that:

IE increases from Al to Ga because in Ga there are now electrons in the $3d$ subshell, which are poor shielders of nuclear charge.

I don't understand how the poor shielders can INCREASE ionization energy. I get that they do not shield as much as $s$ or $p$ subshells, but after all when we go from Al to Ga, Ga not only has the $4s$ electrons for shielding effect, but now there are even MORE electrons that will shield nuclear charge (from the $3d$ subshell). Thus effective nuclear charge on valence electrons should decrease, and the IE trend should still follow. (According to me, even though $3d$ electrons don't shield as well as others, they should not "unshield" the valence electrons just because of being poor shielders.) There are 12 non-valence electrons in Al and 30 in Ga, so how can shielding effect decrease overall just from 10 of those 30 being poor shielders? Can someone explain this to me?
(I am in 11th grade so please try to explain in simple terms, thanks.)

I think irregularities in this group are discussed often here. Try targetted Google searching for keywords, limiting search to this site by  the search modifier site:chemistry.stackexchange.com

Duplicates: https://chemistry.stackexchange.com/questions/117076/ionization-enthalpy-for-group-13-elements ... https://chemistry.stackexchange.com/questions/133177/why-is-ionization-energy-of-indium-less-than-gallium

Here is one textbook explanation of the trends:

[Averill] In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Zeff) that results from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Zeff increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium.

This explanation uses the same argument as the one the OP cited. However, the OP was not convinced:

[OP] There are 12 non-valence electrons in Al and 30 in Ga, so how can shielding effect decrease overall just from 10 of those 30 being poor shielders?

You have to consider that the nuclear charge goes up from 13 (Al) to 31 (Ga). Most of that charge is shielded, but the effective nuclear charge is a small difference between two large numbers (the nuclear charge and the "effective number of electrons shielding").
In the end, the ionization energies are very similar for the non-boron elements, and the explanation for the exact order will be more subtle than a neat set of rules that works well for other parts of the periodic table.
You can also take a look at similar questions (some with answers) posted in the comment by Nilay Gosh:

Ionization enthalpy for group 13 elements
... Why is ionization energy of indium less than gallium?

Ah, I completely missed the fact that the absolute charge from nucleus is increasing anyway, and the $Z_{eff}$ only decreases if there is no reduced shielding effect.

the d-subshell is not effective at shielding, increasing the effective nuclear charge and ionization energy. According to Wikipedia, The 3d10 electrons do not shield the outer electrons very well from the nucleus and hence the first ionisation energy of gallium is greater than that of aluminium.


text:
Why is the entire 3p sublevel peak shifted to the right on a PES of Sulfur vs Phosphorus?
Edit: Please see answer. I am going to leave the question here in case someone else has a similar misunderstanding.
I am currently a student in AP Chemistry. I was working on an assignment with these two PES spectrums.

One of the questions is:

Explain why peak E in the sulfur spectrum is shifted slightly right comparted to peak E in the phosphorus spectrum.

The answer key says:

Sulfur has four electrons in the outermost shell, compared to three electrons in that shell for phosphorus. Electrons repel each other so there is a slight reduction in ionization energy because of the repulsion effects of the surrounding electrons. The additional sulfur electron occupies a half-filled 3p orbital, thus causing electron-electron repulsion.

I would think that one of the electrons in the full 3p orbital would be first to go, so after that, there would be no full 3p orbital. Then, since sulfur has another proton in its nucleus than phosphorus, the attraction would be stronger so therefore the ionization energy would be higher to remove the remaining 3 electrons in the 3p sublevel.
In summary, I think there should be a peak to the right of the Phosphorus E peak with a height of 1 electron, and then a peak to the left of Phosphorus E peak with a height of 3 electrons. The same could be applied to the C peak: why is it one peak instead of two distinct peaks(one when it has full orbitals, one when it has half-filled orbitals).

Comments are not for extended discussion; this conversation has been moved to chat.

My question was rooted in a misunderstanding with the process of Photoelectron spectroscopy.
Originally, I thought that photons were beamed until all electrons were lost from an atom, which would report the first, second, third, etc. ionization energy, where the second ionization energy starts with the +1 cation, and the third with the +2 cation, etc.
However, Photoelectron spectroscopy actually finds the binding energy for each electron in the ground state. The reason it can do this is that:

it uses a photon energy that can knock off any electron in the atoms, and
since there is a large amounts of atoms and finite photons, very few atoms get struck by a photon and lose even a single electron, so it would be even more rare for an atom that already lost an electron to be struck by a photon and lose another electron.

For others struggling with this concept, the answer here does a good job of explaining the difference between first ionization energy and binding energy.
Here is a Khan Academy link that explains the process of Photoelectron spectroscopy in simple terms.
Thank you to @Karsten for helping with this question!

Greetings fellow AP Chem student:
The general trend for Sulfur is that, yes, as you identified earlier, all of its peaks are shifted right because effective nuclear charge is greater than phosphorus (and thus corresponding ionization energies are higher). However, E is the exception because firstly, it is in the outermost energy shell/level, and it prone to additional nuclear shielding that causes those electrons to feel less electrostatic attractions than electrons that penetrate the core shell (due to the co-electron shielding that one extra 3p electron adds to the mix), making the respective effective nuclear charges for sulfur and phosphorus about the same FOR THAT SPECIFIC SUBSHELL, despite the fact that sulfur has one more proton. The answer key is correct, since for peak E 3p subshell sulfur has an additional electron, and thus additional electron-electron repulsions in THAT SUBSHELL. The COMBINATION of a localized nuclear charge in the outer shell and additional electron-electron repulsions causes the ionization energy of electrons in the 3p subshell in sulfur to have slightly lower ionization energy than that of phosphorus, which doesn't have to deal with as much electron-electron repulsions in the same subshell but with about the same effective nuclear charge. Hence the deviation of peak E from the normal trend.
I am from New York, by the way.
School starts about a month later! ;)


text:
How to make clean water from urine in a wastewater treatment plant?
Referring to a video here, describing a filter that treats wastewater, wouldn't urine in the wastewater not get filtered out because a sand filter alone does not separate urine from water? How is clean water produced from urine in water treatment facilities where large volumes of waste water are being treated everyday?

Filtration removes suspended solids. Urine is in solution not suspension.

It would need some big biotopic reservoir and then final fine filtration and desinfection. Other methods to get drink water would be much cheaper and easier, if available.

In the years 1970, urine was electrolyzed on board space capsules (Apollo) by solar cells. The $\ce{H2}$ and $\ce{O2}$ gases produced were then recombined in galvanic cells where they were producing first pure water, and then electric current necessary to charge the batteries on board. This is a way to obtain drinkable water.

As noted in video, processes differ from place to place. Here’s one approach:https://news.climate.columbia.edu/2011/04/04/from-wastewater-to-drinking-water/

This filtration will principally work (provided you aerate the inlet water sufficiently), because a biofilm grows on the sand that removes nutrients. Generally speaking. The output is certainly better than what many people drink day to day on this planet. Of course it is nowhere near real drinking water, and magnitudes away from what I have as (non-chlorinated) tapwater at home. How long it will work is hard to tell, and depends strongly on how clean the input water is. The problem are human pathogens, and how much (or little) the input is diluted with nutrient-free (rain) water.

https://en.wikipedia.org/wiki/Sewage_treatment

Urine is mostly water plus small organic compounds, the most common being urea. "The basic methods of urea removal of wastewater streams are electrochemical oxidation, biological treatment, adsorption and hydrolysis using thermal treatment or enzymes and decomposition with the aid of catalyst, strong oxidant." https://iopscience.iop.org/article/10.1088/1757-899X/1046/1/012021/pdf

Aerobic and Anaerobic processes are used to remove the organic content of the urine containing mainly urea and uric acid. The NH groups are oxidized to NO3 groups by the nitrobacter bacteria of the ETP culture. This nitrate can then be used by the Sludge of the ETP itself or be in the water as TDS. This water can then go through RO, MEE or TVR to remove the TDS down to a satisfactory level and then use urine as potable water. A simple sand filter won't do much. Activated carbon would help but would be Over the Limit quite soon. Adding stuff like NaOCl, peroxide, UV etc can help too but ultimately, its the aerobic degradation that does the majority of work

There are many ways to make clean water from urine in a wastewater treatment plant. One way is to use reverse osmosis. This process uses a semipermeable membrane to remove impurities from water. Another way is to use activated carbon filters. These filters remove impurities by adsorbing them onto the surface of the carbon.

Here is a source Applications of ozone for industrial wastewater treatment — A review, to quote:

Ozone coupled with ultraviolet radiation and/or hydrogen peroxide (advanced oxidation) is being utilized to destroy organic contaminants in groundwaters at munitions manufacturing plants and at Superfund sites (hazardous wastes).

Another source, Recent advances in ozone-based advanced oxidation processes for treatment of wastewater- A review, to quote:

Advanced oxidation processes (AOPs) are considered as a good option for removing recalcitrant organic materials in wastewater by oxidation reactions with powerful, non-selective hydroxyl radical (OH•). Ozone alone does not cause complete oxidation of some refractory organic compounds and has a low reaction rate. The ozone is combined with H2O2, UV light, catalyst, ultrasound to enhance the generation of hydroxyl radicals to increase the efficiency of the treatment process. The ozone-based AOPs have been proved to be effective in detoxifying an ample range of industrial effluents containing recalcitrant organics, pharmaceutical products, pesticides, phenols, dyes, etc.

So, basically, the ozone facilitate the introduction of powerful oxidizing species on contact with the wastewater that can chemically attack complex molecules, breaking them down into more simple compounds, like water, carbon dioxide, nitrogen based oxides, etc., thereby permitting there removal.
I hope this helps.


text:
Is there any transparent see material that can endure temperatures of 2000 °C?
I was wondering if there is a transparent material that can endure temperatures up to 2000 °C without deforming or degrading. This material doesn't have to be glass, it can be plastic or a weirdly transparent alloy.

The 1978 Pioneer Venus probe had sapphire windows (and a diamond window) with melting points above that temperature.

Although it was a plot point in a fictional movie, ... there is now such a thing as 'transparent aluminium' - Aluminium oxynitride. More exactly, it is an aluminium ceramic composed of Al, O, and N in some proportions.
Looks like it has a melting point of ~2150 °C - so fits the temperature range.
edit: As various commenters have pointed out, sapphire (aluminium oxide) would also do, as might the suggestions from this answer:
https://engineering.stackexchange.com/questions/16177/transparent-material-which-could-withstand-high-temperatures

At 95% of melt I would not bet it doesn’t deform easily…

The patent for it was issued before the movie came out, so I would guess they based it on science rather than imagination.

More commonly available would be sapphire which is  pure aluminium oxide and melts just above 2000 C. It is also a very tough material. Its used in the lab for its excellent optical quality and ability to withstand v. high pressures.

Magnesium oxide, MgO, melts at 2800C Possibly could be built up by vapor deposition. IS there an actual application


text:
Is there a chemical compound which does not have a liquid state but a solid and a gaseous state?
Is there a chemical compound which does not have a liquid state but a solid and a gaseous state? Meaning no matter the temperature or pressure it will never be in its liquid form. According to my rather basic knowledge
of chemistry this means that it has a critical temperature  lower than the melting point.
I have seen this question already, but it only talks about elements and not compounds.

Lots of them. They decompose on heating before they would melt, like hydrate salts, most hydroxides and carbonates, etc.

@OscarLanzi I updated the question. I meant things that have solid and gaseous forms but no liquid forms

Most organic compounds decompose before melting. So is there many unstable inorganic compounds, that decompose if the frost is not deep enough, or before reaching melting point.

https://chemistry.stackexchange.com/questions/31050/can-you-melt-wood https://chemistry.stackexchange.com/questions/62539/why-don-t-all-objects-melt-and-liquefy-when-heated-sufficiently

https://chemistry.stackexchange.com/questions/94003/what-are-some-examples-of-pure-substances-with-no-liquid-state

Neither of these actually answer the question if any of these compounds have a solid & and a gaseous state but no liquid state, as they decompose before they would be either gaseous or liquid. So no these do not actually answer my question. @Mithoron

That still has nothing to do with any critical points, only chemical reaction.

Then my rather basic chemical knowledge was just plain wrong :D

It could be summarized that compounds with sufficient thermal and chemical stability do have their liquid phase, if there is high enough pressure available. A special case is liquid carbon, where 100-1000 atm at 4500-5000 K is needed.

.....uhh, naphthalene

@NilayGhosh Only if sealed piece of naphtalene wouldn't melt because of decomposition.


text:
What’s the physical meaning of (∂v/∂T )P?
I'm solving this exersice:
The equation of state of a Van der Waals gas is
\begin{equation}
  \left(P + \frac{a}{v^2}\right)(v-b) = RT
  \tag{1}
\end{equation}
Where $a$,$b$ and $R$ are constants.

Calculate $(\partial P/\partial v)_T$ and $(\partial P/\partial
    T)_v$
Use the results of 1 to calculate $(\partial v/\partial
    T)_P$. What's the physical meaning of this partial derivative?

From equation 1 we have:
\begin{equation}
  P  = \frac{RT}{(v-b)} - \frac{a}{v^2}
\end{equation}
Solutionf for 1.1.
\begin{equation}
  \begin{aligned}
    \left(\frac{\partial P}{\partial v}\right)_T & = \frac{\partial}{\partial v} \left[\frac{RT}{(v-b)} - \frac{a}{v^2}\right]                                         \\
                                                 & = \frac{\partial}{\partial v} \left[\frac{RT}{(v-b)}\right] - \frac{\partial}{\partial v}\left[\frac{a}{v^2}\right]
                                                 & = \frac{\partial}{\partial v} \left[\frac{RT}{(v-b)}\right] - \frac{\partial}{\partial v}\left[\frac{a}{v^2}\right]
  \end{aligned}
\end{equation}
\begin{equation}
  \boxed{
    \left(\frac{\partial P}{\partial v}\right)_T= \frac{RT}{(v-b)^2} - \frac{2a}{v^3}
  }
  \tag{2}
\end{equation}
Solution 1.2.
\begin{equation}
  \begin{aligned}
    \left(\frac{\partial P}{\partial T}\right)_v & = \frac{\partial}{\partial T} \left[\frac{RT}{(v-b)} - \frac{a}{v^2}\right]                                         \\
                                                 & = \frac{\partial}{\partial T} \left[\frac{RT}{(v-b)}\right] - \frac{\partial}{\partial T}\left[\frac{a}{v^2}\right]
                                                 & = \frac{\partial}{\partial T} \left[\frac{RT}{(v-b)}\right] - \frac{\partial}{\partial T}\left[\frac{a}{v^2}\right]
  \end{aligned}
\end{equation}
\begin{equation}
  \boxed{
    \left(\frac{\partial P}{\partial T}\right)_v= \frac{R}{(v-b)}
  }
  \tag{3}
\end{equation}
Solution 2.
\begin{equation}
  \left(\frac{\partial P}{\partial v}\right)_T
  \left(\frac{\partial v}{\partial T}\right)_P = -
  \left(\frac{\partial P}{\partial T}\right)_v
  \tag{4}
\end{equation}
substituting 2 and 3 in 4
\begin{equation}
  \left(\frac{RT}{(v-b)^2} - \frac{2a}{v^3}\right)
  \left(\frac{\partial v}{\partial T}\right)_P = -
  \frac{R}{(v-b)}
\end{equation}
\begin{equation}
  \begin{aligned}
    \left(\frac{RTv^3 - 2a(v-b)^2}{v^3 (v-b)^2}\right)\left(\frac{\partial v}{\partial T}\right)_P & = - \frac{R}{(v-b)}                                                   \\
    \left(\frac{\partial v}{\partial T}\right)_P                                                   & = - \frac{R}{(v-b)}\left(\frac{v^3 (v-b)^2}{RTv^3 - 2a(v-b)^2}\right)
  \end{aligned}
\end{equation}
\begin{equation}
  \boxed{
    \left(\frac{\partial v}{\partial T}\right)_P  = - \frac{R v^3 (v-b)}{RTv^3 - 2a(v-b)^2}
  }
\end{equation}
Which I think is all OK. But I don't find any good answer to the question "What's the physical meaning of $(\partial v/\partial T)_P$?"

Isobaric thermal expansion of the real gas. For ideal gas, as v=RT/p, (dv/dT)p = R/p. If you take a=b=0 for ideal gas, you get from your last equation v/T, which is equal R/p.

It's not that mysterious. The mathematical expression is just shorthand for "the change in volume with change in temperature, when holding pressure constant". For ideal gases it is a property independent of the identity of the gas. For non-ideal gases, it depends on the degree of attraction between the molecules and on the excluded volume.

as an aside its hard to calculate $dv/dT$ but easy to calculate $dT/dv=1/(dv/dT)$

Isn't that the same as minus the partial derivative of molar entropy with respect to pressure at constant temperature?