text:
Why isn't the American nickel magnetic?
I never thought that modern American nickels actually contained nickel anymore. However, according to this wiki article, the coins actually do contain 25% nickel, the rest being copper. And yet, no US coin produced today is officially magnetic. Why is this alloy of nickel not attracted to a magnet?
(and yes, I tried time and again to find and answer elsewhere online.)

RE: "... And yet, No US coin is officially magnetic. ..." The 1943 Lincoln penny was steel coated zinc due to a copper shortage during WWII. It is ferromagnetic. There were a few copper blanks that got pressed in 1943 and these specimens are very valuable. A common scam was to plate a steel/zinc penny with copper. Such an altered coin is easily detected with a magnet.

Be aware that even chrome-nickel stainless steel ( typically 16-20% Cr, 8_12% Ni, rest mostly iron ) is not ferromagnetic ( magnetic in common terms ), even if it contains mostly ferromagnetic metals as Fe and Ni. Because Ni keeps iron in the austenitic non ferromagnetic form.

@Poutnik  So why are other Fe-Ni alloys, such as permalloy (which have ~45 - 85% Ni), ferromagnetic?  Is it becauses the Fe is not austenitic in permalloy?  Or because there's enough Ni to overcome the austentic Fe?

@theorist Probably the latter. But I am not expert in metallurgy, relying on googled info and I do as well remember typical CrNi steel is not ferromagnetic.

@Poutnik  I think it is indeed the latter -- through my own googling, I just found that permalloy has an FCC (i.e., austenitic) crystal structure.  Interestingly, I also found that austenitic alloys can be partly converted to ferromagnetic forms (martensitic and ferritic) using cold working (the austentic crystal structure -- which Fe favors in the presence of Ni -- can be restored by melting and recooling).

Back in the day, I worked at Duracell. We spot-welded Nickel tabs to cells and then attached them to our testing computers.
Needless to say, we had a lot of scrap tabs after testing 1000's of cells. When we took the scrap to a dealer, he refused to believe it was (almost pure) nickel because it was magnetic. No amount of arguing could get him to change his mind !

There are many types of magnetic properties, including ferromagnetism, paramagnetism, diamagnetism, antiferromagnetism, ferrimagnetism, superparamagnetism, metamagnetism, spin glasses, and helimagnetism.  Many of these are too weak to cause any noticeable interaction with a magnet. The type of everyday magnetism you're thinking of, which nickel has, is ferromagnetism.
While nickel is ferromagnetic,  copper is not. As you said, the American nickel is currently 25% nickel and 75% copper.  According to this paper (from 1931!), in order for a nickel-copper alloy to be ferrogmagnetic, it must contain at least 56% nickel:

... 56 percent nickel is required before the alloy shows ferromagnetic properties at ordinary temperatures.

https://journals.aps.org/pr/abstract/10.1103/PhysRev.38.828
E. H. Williams, Magnetic Properties of Copper-Nickel Alloys. Phys. Rev. 38, 828 (1931).

So, according to your source, American nickels are only paramagnetic?

@mpprogram6771 Yes, that's what's it's saying (at around room temp).

You have forgotten ferrimagnetism.

@Poutnik  And superparamagnetism, metamagnatism, and spin glasses.

@theorist So it is even worse... :-)

So is it possible to make a ferromagnetic alloy with two or more non-ferromagnetic metals in the right combination?

Yes.  For instance, Heusler compounds, like Cu-Mn-Al, are composed of non-ferromagnetic metals that, together, can be ferromagnetic.  See:  https://en.wikipedia.org/wiki/Heusler_compound

@theorist you could also add helimagnetism in, too. MnSi in the B20 structure is one of the more famous examples. :)

@rcollyer  I have added it to the list :).

None of the US coins are magnetic (ferromagnetic), except for the 1943 Lincoln penny (Steel Cents, made in steel and zinc to save copper for ammunition during wartime), which are considered magnetic. Almost all of those coins other than Steel Cents are made with higher percentage of copper ($\ce{Cu}$) and lower percentages of other metals such as nickel ($\ce{Ni}$), zinc ($\ce{Zn}$), etc. (mostly $\ce{Ni}$).  For example, current 5-Cent US coin (US Nickel) is made of 75% $\ce{Cu}$ and 25% $\ce{Ni}$, as OP noted in the question. You can find some percentages in US coins in this AZO Materials Article.

Pure $\ce{Ni}$ is magnetic at room temperature while $\ce{Cu}$ is not. Yet, why US Nickel is not magnetic?

The magnetic properties of $\ce{Cu/Ni}$ alloys have been studied by several researchers (e.g., Ref.1 and 2), perhaps because their use in alloys goes back at least two thousand years, at the time, the knowledge of the composition of alloy was  unknown (Note: the elemental nickel was only discovered relatively late on). According to this research, pure $\ce{Ni}$ is magnetic with its Curie temperature being around $\pu{527 K}$ (Curie temperature or Curie point is where a metal loses its ferromagnetism with elevating temperature). When $\ce{Ni}$ mixes with $\ce{Cu}$ to make an alloy, this Curie temperature decreases with increasing amount of $\ce{Cu}$ (Ref.3):

Apparently, the Curie point has become $\pu{0 ^\circ C}$ or lower for $\ce{Cu/Ni}$ alloys with compositions 67:33 or lower $\ce{Ni}$ amount. Thus, it is safe to say alloys with less than 65% $\ce{Ni}$ in $\ce{Cu/Ni}$ alloys do not show magnetic properties at room temperature. Therefore, the US Nickel with 25% $\ce{Ni}$ would not act on permanent magnets.
However, Kaufmann and Starr (Ref.2) have shown that these alloys with less amount of $\ce{Ni}$ would gain their their magnetism at way lower temperatures (about $\pu{14-77 K}$).
References:

E. H. Williams, "Magnetic Properties of Copper-Nickel Alloys," Phys. Rev. 1931, 38(4), 828 (DOI: https://doi.org/10.1103/PhysRev.38.828).
A. R. Kaufmann, C. Starr, "Magnetic Properties of Solid Solutions. III. The Paramagnetic Alloys of Copper and Nickel," Phys. Rev. 1943, 63(11-12), 445 (DOI: https://doi.org/10.1103/PhysRev.63.445).
E. A. Brandes, G. B. Brook, In Smithells Metals Reference Book, Seventh Edition; Butterworth-Heinemann: Woburn, MA, United States, 1992, Chapter 11: Equilibrium Diagrams, pp. 11-236.

Nice graph ; Shows why Monel ( 67 Ni , 33 Cu,  + little Mn, Si ,etc) is sometimes magnetic and sometime not at room temp.

Why is there no eutectic point?

so you can freeze a nickel and it will become ferromagnetic?

@Jim: That't the idea (and theory). Even any gas can be magnetic near absolute zero, I read. See ref.2.

@Vladimir F: This alloy might not have one.

No eutectic or anything else, just mutual solubility ( as shown in the diagram).

To the best of my knowledge they are not magnetic to make it very hard to profitably counterfeit our coins. If they are magnetic the vending machines will remove them with a magnet. This is the explanation I was given in the mid 1960's. Yes machines in our town had the magnets, I tried one.

Why would be important for the government, how the vending machines work?

Same reason the bills are very hard to duplicate and pass inspection. The easier it is to spot counterfeit money the less chance it will be counterfeited.  Same for vendors, if they get a counterfeit coin or bill it is there loss, the government does not buy them back.

It probably isn't a very compelling reason. In the UK small "copper" coins used to be  made of a mostly copper alloy and were not magnetic. For cost reasons these were replaced after 1992 with coins that look the same but are made from copper-coated steel. These are are now magnetic and stick to magnets.

and most other UK coins are made from non-magnetic cupro-nickel alloys but were widely counterfeited. The design of the pound coin was changed in 2016 to a bimetallic pattern with a more complex overall design to defeat counterfeiting.

authors: Vladimir F Героям слава


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Molecular similarity using Tanimoto coefficient
I'm trying to use a method for evaluating similarity of molecules. I have seen many publications in the machine learning community have used Tanimoto coefficient. However, when I read about it, it states that this is a measure of similarity for molecular fingerprint representation.

I was wondering if this means that the measure of distance is not for the molecule itself but for its representation?
If so, is there a measure that compares two molecules based off of their inherent structure rather than the particular representation of choice?

Also, any other suggestions on the choice of measure would be appreciated.

Related: https://jcheminf.biomedcentral.com/articles/10.1186/s13321-015-0069-3

I'm not sure I understand what you mean by "inherent structure rather than the particular representation." If you're storing on the computer, you necessarily are making choices about the representation.

Of course the similarity measure depends on the representation. If I picked some 3D distance-based representation, I'd want some other similarity calculation (e.g. Euclidian)

@GeoffHutchison By inherent I mean the structural similarity. Taking two images as an example, I can find the Euclidian distance between the two images by subtracting corresponding pixel values. But does it give any measure of how similar these two images are? Do these images belong to the same category? Are both apples or one is apple the other one orange?

@GeoffHutchison In the molecular space, let's take alanine dipeptide as an example. Each configuration can be categorized by the characteristic conformation. So I find dihedral angles and I can find a measure of similarity in the space of alanine dipeptide molecules with different configurations.

@GeoffHutchison But this measure can't be extended to every dataset of molecules. So, what I meant is that while many deep learning publications use such measures as Tanimoto to discuss the validity of their method (continuity of latent space, etc.), is it really ok to use it regardless of what my algorithm uses? (eg: I have a VAE based on smiles representations. Then they use Tanimoto to justify their method. Of coarse I can take my smiles, turn it into mol object, turn it into fingerprints, then find the measure, but does this actually make sense given that my VAE's trained based on SMILES?)

I was wondering if this means that the measure of distance is not for
  the molecule itself but for its representation?

Inherently, you must decide on a representation of a molecule in cheminformatics. In most cases, the representation is a "2D" valence-bond connection table (e.g., SMILES, SD file, ChemDraw, etc.) familiar to anyone who has taken an organic chemistry class.
So yes, a similarity measure depends on the representation. Tanimoto coefficients are usually based on comparing binary fingerprints.

If so, is there a measure that compares two molecules based off of
  their inherent structure rather than the particular representation of
  choice?

I'm not sure how to answer this question since a representation forces choices about an "inherent structure" of a molecule. If I pick SMILES, I have no coordinates, implied hydrogen atoms, and choices about implied bond orders aromaticity, etc.
I think perhaps your question is about fingerprints - that "Is is there a similarity measure between a molecular graph - without generating fingerprints as an intermediate step?"
In that case, I think graph similarity measures are what you want, e.g. from Peter Willett, "RASCAL: Calculation of Graph Similarity using Maximum Common Edge Subgraphs" The Computer Journal (2002) pp 631–644
The reason people use fingerprint similarity measures is that they're easy to implement, and have worked fairly well for a wide variety of tasks.

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Are aqueous (aq) elements involved in equilibrium?
The reaction was something like this:
A (s) + B (l) yields to C (aq) + D (aq)
The question asked: If you added another D, which way would the equilibrium shift to?
I answered that it would have no effect because solids and liquids don't have any effect on the shift of equilibrium, but apparently the correct answer was, it shifts left. I know that would work for gases, but I thought that aqueous means liquid, so that is why I said no effect.
Anyways, I don't understand why it counts. I searched up that aqueous solutions happen when a substance is dissolved in a liquid. So from the term "dissolved", doesn't that mean it is a liquid?

Aqueous means dissolved in water, not just any liquid.

@Ello In your interpretation, the term pH would have no sense, as it is just "liquid" H+ with activity equal 1 and all water solutions would have pH=0. :-)

The three simple states of matter are solid noted by (s), liquid noted by (l) and gas noted by (g). The notation (aq) means the ion or molecule is dissolved in water. So A is a solid, B is a liquid not miscible with water but reacting with A, and the products C/D then dissolve into the water phase, not into liquid B.

What is "B (I)", namely capital letter "i" in brackets? Did you mean "B(l)" for a liquid component B? Anyway, try to write down an expression for $K_c$ and see how $[\ce{D(aq)}]$ affects its value.

@andselisk I changed it to (l).

The amount of pure solid or liquid does not affect equilibrium, having chemical activity independent on amount, formally assigned to 1.
If a solvent is in abundance wrt solutes, it is often considered still as approximately pure solvent, e.g water in context of diluted water solutions.
The aqueous (aq) compounds definitely are not pure liquid. They are just dispersed among molecules of water and their chemical activity is approximately proportional to concentration. Therefore, increasing their concentration pushes equilibrium to the other side of the reaction and vice versa.
There is a big touch difference between $\ce{NaCl(aq)}$ and  $\ce{NaCl(l)}$. The former serves for sea bathing/swimming, the latter would make from your body a piece of carbon within seconds.
See e.g. Solubility equilibrium.

I was thinking of a practical example where such an equation could be applied. It is not easy. A solid should react with a liquid, both being insoluble in water. And this reaction should produce two substances both soluble in water. I don't see many possibilities for this reaction. A may be a metal (maybe iron $\ce{Fe},$ or even gold $\ce{Au}),$ and $\ce{B}$ some bromine $\ce{Br2}.$ So the reactions could look like:
$$\begin{align}
\ce{Fe  +  Br_2 &-> Fe^{2+} + 2 Br^-}\\
\ce{2 Au + 3 Br_2 &-> 2 Au^{3+} + 6 Br^-}
\end{align}
$$
Would you believe that adding some $\ce{NaBr}$ in a $\ce{FeBr2}$ or $\ce{AuCl3}$ solution will produce more metal and more $\ce{Br2}?$ Is my example poorly chosen? I would be glad to read your reaction and your interpretation of such an example.

While I think this is an ingenious answer which I happily upvoted, I would recommend not to ask direct non-rhetorical questions. Forums are great for that, but the Q&A sites are not: it's the wrong format. Here, all supplementary questions and discussion with OP should be sorted out in the comment section, and the answer should be complete and not open-ended.

The  more intuitive reason that substances labelled with (aq) are taken into account for shift in the state of equilibrium:
Let us consider activity as concentration of the substance for the sake of this question.
The concentration of solids and liquids is taken as unity while studying equilibrium because they are  incompressible in most cases. So if you add say 100mL  of NaCl (l) to 1000mL of NaCl (l) present already, then that 1000mL still takes up the same volume in the reaction mixture while the new 100mL takes up new space(volume). Also note that 100mL and 1000mL are comparable here.
On the other hand substances labelled as (aq) are taken in small quantities (for example: 5gm) and spread throughout the vast quantity of water (100mL - 2L). Thus if you add more of the former substance it will enter the same water and the population of this substance would become more in the same volume of water. (Eg. 5g in 1000L water is 0.005 g/L ratio to which if again 5g of that substance is added it becomes 0.01 g/L ratio)
Hence in case of the liquid the concentration remains same and for the substance it increases.
Shift in the equilibrium is shown by reaction quotient
Qc =[D] (aq) [C] (aq) / [A] (s) [B] (l)
As explained above solid and liquids get concentration unity while aqueous substance has variability. Thus [A]=[B]=1.
Therefore reaction quotient (which is a projection of shift in the equilibrium) takes the expression
Qc =[D] (aq) * [C] (aq)

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For an electron arrangement diagram, is it necessary to show the inner electron shells or is it customary to only show the valency electrons?

Which electron arrangement diagram is correct? The one that shows all electron shells or the one that only shown the valence electron shell?

Both are correct: for $\ce{Cl-}$, the inner set of electrons would be implicit, as $\ce{Cl}$ has 17 electrons. The point of drawing these orbital diagrams, especially in a molecular context, is to show the nature of bonds and to show how the octet/duplet rules are being followed.
If you have difficulties in using the second diagram (finding it harder to read or are confused by the similarities in $\ce{H}$ and $\ce{Cl}$), you should use the first diagram.

First paragraph is good. I disagree with the second. If you try to draw all the electrons for iodide you have a mess.

@MaxW I assumed OP was just beginning to study chemistry and would find it easier for covalent compounds made from the first 20 elements.


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Does orientation matters to this extent in reactions?

In the above question I know that the three reactions involved are $\mathrm{S_N2}$ reactions and therefore there shall thrice inversion of the substitute.
So, I got the solution for the MCQ as:

However, the solution to the question says that the first structure is only correct and the second one is wrong.
As far I have learnt I know that the orientation of the whole compound doesn't matter because we can rotate the compound in our mind. Is this wrong?

Whilst you're not wrong in the fact that "we can rotate the compound in our mind", do you reach the exact same molecule when doing the transformations on the second one? Do you see the relationship between the two potential starting materials?

i'm a bit confused still, the end product we get on rotating the compound not the same?

Yes you can rotate the compound in your mind, but if you do it accurately, you'll see they are different.

The 5 member ring is planar. The solid arrows mean the substituent is above the plane, the dashed arrorws on  the $\ce{N3}$ functional group mean that the group is below the plane. The reactant only has one configuration since you can flip the molecule over. So you could draw the methyl group down and the  $\ce{N3}$ group up and it would be the same molecule.

@porphyrin It would be 108°, not 72. So after all, the ring is relatively close to being planar. In other words, it is not because its non-planarity that the two structures differ.

@Ivan Neretin oops: yes, as you say it is not the planarity or otherwise that matters. My comment was in response to MaxW's comment that the ring was planar; sp3 makes the ring bent as I mention.  I have deleted my comment anyway.


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Antoine law and ideal gas law
I have the following question:

A $100\ \mathrm{ml}$ sealed flask with $60\ \mathrm{ml}$ acetonitrile is heated to $140\ \mathrm{^\circ C}$. The boiling point of acetonitrile is $82\ \mathrm{^\circ C}$. What will be the pressure in the flask?

1) I understand I can use the Antoine equation, $\log{p} = A-B/(C+T)$, to calculate the vapor pressure of acetonitrile, but that equation only applies if there is some liquid acetonitrile left in the flask. How do I know if this is the case, or instead all acetonitrile has boiled and is in the vapor phase?
2) The total pressure is the sum of the partial pressures of acetonitrile and air. I can calculate the air pressure inside the flask with the ideal gas law, PV=nRT. However, I'm not sure what is the volume? Is it the total volume minus the acetonitrile volume ($100\ \mathrm{ml}-60\ \mathrm{ml}=40\ \mathrm{ml}$), or is it $40\ \mathrm{ml}+x$, where $x$ is the additional volume due to the conversion of liquid acetonitrile into vapor?
Any ideas are welcome.

To know the partial pressure of the air in the flask following heating you need to know how many moles of air you have, which you can calculate from the partial pressure of air at the start.But to determine this, you need the initial temp. of the flask, and the conditions under which it was sealed (e.g., was it sealed such that tthere was 60 mL acetonitrile after saturating the air with it, or was it sealed before the acetonitrile vaporized, in which case the partial pressure of the air would be 1 atm)?  Or did it have any air to start?  You need this info. to begin answering the question.

Ad 1) Remember the pressurized cooking. The water is filling the comparable fraction of the volume, and temperature is raised above the standard boiling point by comparable amount. No cook is afraid of total vaporization.

Thank you for your comments. I will consider your points.

Assume that the initial temperature is room temperature and neglect the initial amount of acetonitrile in the gas phase.  Calculate the equilibrium vapor pressure of acetonitrile at 140 C and compare it with the partial pressure you would calculate from the ideal gas law if all the acetonitrile had evaporated (so that its volume is 100 cc).  If the latter is greater than the former, then there will still be some acetonitrile remaining as liquid.  The partial pressure of the acetonitrile in the gas phase will then be equal to the equilibrium vapor pressure.  Assuming that x gm of acetonitrile have evaporated, you can then calculate, in terms of x, the new mass and volume of the liquid and the new volume of the head space.  You can then determine x from the requirement that the total amount of acetonitrile has not changed.

Thank you, that was very helpful. As the partial pressure calculated with the ideal gas law is greater, I can use the Antoine equation and solve the problem.

EDIT I fouled up my original answer. So as a matter of pride I have fixed it. The OP was right on his idea #1. The answer to the question hinges on the calculated vapor pressure assuming all of the acetonitrile evaporates. I have no idea what coefficients the OP might have had for the Antoine equation.

Let's go down the rabbit hole! Given no other information you have to make several assumptions.

Does the sealed flask initially contain $40\ \mathrm{ml}$ of air or not?

For instance the apparatus could contain two chambers. One $60\ \mathrm{ml}$ to be completely filled with liquid acetonitrile and a $40\ \mathrm{ml}$ chamber which is evacuated. Then when the experiment starts the chambers are first connected and then heated.

Alternatively the apparatus could contain one $100\ \mathrm{ml}$ chamber into which 60 ml of liquid acetonitrile is poured, the other $40\ \mathrm{ml}$ being part air and part acetonitrile vapors.

What is the starting temperature?

This is more important if the flask initially contains $40\ \mathrm{ml}$ of air.

Can the partial pressure of acetonitrile at the starting temperature be ignored?

If yes, then just state the assumption.

If no, then you'll need the partial pressure of acetonitrile at the starting temperature, and you'll have to assume that the $40\ \mathrm{ml}$ gaseous phase is in equilibrium with the liquid phase.

What is the density of liquid(?) acetonitrile at the starting temperature?

Since the problem states "$60\ \mathrm{ml}$ acetonitrile" you'd be led to assume that the acetonitrile was initially a liquid.

I'd assume whatever temperature I could find a density for liquid acetonitrile.

Since $140\ \mathrm{^\circ C}$ is above the atmospheric boiling point, which does the acetonitrile do?

Totally vaporize?

Form gas and a liquid phases?

Turn into a supercritical fluid?

Does gaseous acetonitrile act as if it is a perfect gas?

If yes, then $pV=nRT$ will work.

If no, then some more "advanced" model needs to be chosen and the additional constants that it needs to better fit the gas phase behavior.

Number of significant figures for the answer?

It would seem to be 2, so this feeds back into assumptions 2, 3, 4, and 6.

Given the above assumptions, the problem should be easy to solve using $pV=nRT$.

Let's assume:

The sealed flask initially contain $40\ \mathrm{ml}$ of air.

The starting temperature is $\pu{68 ^\circ F = 20 ^\circ C = 293 K}$

The partial pressure of acetonitrile is $\pu{73 mmHg at 68 ^\circ F}$

Density of liquid acetonitrile is $\pu{0.787 g/ml at 68 ^\circ F}$

Since $\pu{140 ^\circ C = 413 K}$ is above the atmospheric boiling point, the acetonitrile totally vaporizes.

The gaseous acetonitrile acts as if it is a perfect gas.

2 significant figures are needed for the answer.

OK, let's check some our our assumptions. For the acetonitrile:
$$m_{\text{acetonitrile}} = \pu{0.787 g/ml}\times\pu{60 ml}=\pu{47.22 g}\tag{1}$$
$$n_{\mathrm{acetonitrile}}\dfrac{\pu{47.22 g}}{\pu{41.053 g/mol}} = \pu{1.150 mol}\tag{2}$$
One mole of a gas at STP occupies $\pu{22.711 l/mol at $0\ \mathrm{^\circ C}$ and $100\ \mathrm{kPa}$. Since the assumption is that only two significant figures are needed for the answer, the approximately $0.040\ \mathrm l$ of the gas phase (air + acetonitrile) can just be ignored. A more careful check confirms the notion.
$$n_{\text{air}}=\frac{pV}{RT}=\frac{\pu{100 kPa}\times\pu{0.040 l}}{\pu{8.314 l\cdot kPa}\cdot\pu{K^{-1}}\times\pu{293 K}}=\pu{0.0016 mol}\tag{3}$$
$$ n_{\text{liquid acetonitrile}}\gg n_{\text{air}}\gt n_{\text{gaseous acetonitrile}}$$
Now assuming that all of the acetonitrile vaporizes, calculate the pressure assuming that the acetonitrile behaves as an ideal gas equation:
$$p=\frac{nRT}V=\frac{\pu{1.150 mol}\times\pu{8.314 l\cdot kPa}\cdot \pu{K^{-1}}\cdot\pu{mol^{-1}}\times\pu{413 K}}{\pu{0.100 l}}\\
=\pu{39 487 kPa}\approx\pu{3.9\times10^4 kPa}=\pu{3.0\times 10^5 Torr}\tag{4}$$
On the Wikipedia data page for acetonitrile gives a formula for the vapor pressure (temperature range from $\pu{229.32 K}$
to $\pu{545.50 K}$) where $p$ is the pressure in $\mathrm{mmHg}$ and $T$ is the temperature in $\mathrm{K}$:
$$\ln p=\ln{\dfrac{760}{101.325}}-3.881710\cdot \ln T-\dfrac{4999.618}T+41.05901+3.155956\cdot10^{-6}\cdot T^2$$
$$\therefore p=\pu{3 379 Torr}\approx\pu{3.4\cdot10^{3} Torr}$$
Since the vapor pressure of acetonitrile in equilibrium with the liquid at $\pu{413 K}$ is less than the vapor pressure that would be created by all of the acetonitrile evaporating, all of the acetonitrile will not evaporate.  The relative volumes of the two phases won't matter for the pressure.

It is one 100 ml chamber. 2. I agree, so I have to assume the starting temp. 3. I don't know. 5+6. That are the problem that concerns me most, because I don't know it.


text:
Adding an essential oil to WHO Handrub Formulation
I'm not a chemist, so I prefer to ask this question to some people, who definitely have knowledge in that area.
WHO Handrub Formulation is available here.
Ingridients: Ethanol 96%, Hydrogen peroxide 3%, Glycerol 98%, Sterile distilled or boiled cold water.
I would like to use it to produce 4l of handrub (I have 92% alcohol, and I re-calculated needed amounts).
Can I also add some essential oil to the formula? Only few drops or more ~10 ml?
For example:

Orange, ingridients: citrus aurantium dulcis peel oil expressed, linalool, citral
Spruce, ingridients: limonane, linalool, geraniol. INCI: oleum picea excelsa
Pine, ingridients/INCI: oleum phus silvestris
Lavender, ingridients/INCI: lavandula angustifolia oil, linalool, limonene

I just want to make sure that the final substance wouldn't be irritating to the hands nor cause any allergies etc. I also found this paper, where they added Piper nigrum Essential Oils to the formula.
Thank you for suggestions.

I don't see why not. The bulk of the solution is aqueous ethanol so the concentration of the oils is pretty low. As for allergic reactions you are far more likely to suffer one from handing the neat oils and getting a drop of your skin than using the solution

This is a great example of a well-formatted question about homemade sanitizers. Well done!


text:
Hybridization theory (orbitals used in different hybridizations)
Why is $d_{x²-y²}$ orbital used in $sp^3d$ (square pyramidal geometry) while $d_{z^2}$ orbital in $sp^3d$ (trigonal bipyramidal geometry)?
I came across this information while reading J.D.Lee Concise inorganic chemistry.


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Open resource for spectroscopic raw data
Is anyone aware of reliable resources (ideally public) which allow the user to download the raw spectroscopic data such as infrared spectrum of small organic molecules, or rotational spectrum of HCl/DCl mixtures or NMR spectra of molecules? I was looking at the NIST database but apparently there is no option for downloading the spectrum as an Excel, csv or txt file. Thanks.
NIST Spectral Database

Excellent question! The last time I needed an IR spectrum, for ethyl acrylate, I had to settles for a JCAMP format file and hack it to get data into a simple .txt file I could use. I hope a good answer appears!

@EdV If working with a Windows computer, try JDXview (cf. my answer).

@EdV: JCAMP-DX is actually pretty self-explanatory and easy (unless it's in one of the weird compressed subformats) - says the maintainer of a number of spectroscopic file format import filters for R... ;-) ... who is usually very wary of .txt or Excel files since the converters usually throw away lots of meta-information that may have been available.

@cbeleitesunhappywithSX Very good point! But the ultimate reason I need ASCII text files is that they are used with my free LightStone simulation software that instantiates both the Jones optical calculus and the Mueller optical calculus. If you are interested at all, please send me an e-mail at evoigtman@icloud.com and I will send along a link to my website and a couple of published papers that explain what I have developed over 30 years.

As a quick fix: To access the IR spectral data of $\ce{HCl}$, I opted for a seach by (Hill) formula  on NIST's web page, which yielded three entries here.  For $\ce{HCl}$ in particular (here) I opted for the first one here, indeed offering the spectrum as image file or as JCAMP-DX.
As in the comment by @EdV, this JCAMP-DX an ASCII file indeed.  Note JDXview by Norbert Haider (Department of Drug Synthesis, University of Vienna, Austria) is a very handy program (only 560 kb, installation free executable) for a quick display of this file by NIST. Via File -> Save, the export of the data in a two-column, tabulator separated .txt file is just a click away.  The program works perfectly well in Windows, and with wine libraries equally in Linux, too.  (The author equally distributes the underlying Delphi / Pascal code.)

Note 1: Because of the tags physical chemistry, and spectroscopy, you may be interested in a python module like jcamp to access either IR, or NMR spectra in the JCAMP-DX formats (there are multiple).
Note 2: Sometimes, all you have indeed is an image (a scanned) spectrum.  If you pay attention and proceed in sections (e.g., $4000 - 2000$, then $2000 - 800\,\pu{cm^{-1}}$) to account for the break on the abscissa in IR spectroscopy, a digitizer like WebPlotDigtizer may be worth to use.  If you want, the program works via a web browser, installation free, with «normal image files» like .png, and since some time with images embedded in a .pdf (think about publications), too.

(+1) Even better than the good answer I hoped for! Many thanks: I just got a new windows laptop yesterday, dumped the “S” kiddie mode and I will download the JDXview program ASAP.

Thanks for a very useful response. Can we use the spectra for publication purposes i.e., NIST data open to re-use? I wanted to try out a new deconvolution algorithm to enhance resolution on rotational spectra or vibrational spectra.

@M.Farooq Re use of NIST data for a publication: I do think it is possible because there is a note how to cite parts / all of it (https://webbook.nist.gov/cgi/cbook.cgi?Contrib=). Wider use probably is a case for Customer support (data@nist.gov, Expiration Date: 06/30/2021, https://www.nist.gov/srd/standard-reference-data-contact-form).  An other source might be the SDBS spectra collection (https://sdbs.db.aist.go.jp): «If more [than 50] spectra are required for some specific purpose or commercial use, you should consult us and describe the intended usage or purpose of our SDBS.»

@M.Farooq: (still SDBS spectra collection): https://sdbs.db.aist.go.jp/sdbs/LINKS/contact_eng.html.  Re IR-data, I speculate sources like Aldrich, Hummel, Sadtler, Biorad/KnowItAll, Specinfo (Wiley, https://application.wiley-vch.de/stmdata/specinfo.php) equally probably require contact prior engaging such a work because they invested time / man power / infra structure into their databases and obtaining (c) clearance from the publishers of primary publication.

@M.Farooq Maybe IR:  https://www.ncbi.nlm.nih.gov/pubmed/27795739 / http://osdb.info/ which seems to be dormant a little in comparison to the Raman sibling COD/ROD (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6557180/, https://solsa.crystallography.net/rod/) which by intent is open domain.

I think the NIST data is digitized version of really old spectra from Coblintz Society. What if one takes a journal permission for a figure and digitize it for testing the resolution algorithm, like as you said WebPlotDigitizer? It is such a grey area.

@M.Farooq What qualifies as «small molecule» here?  Small molecules (organic synthesis / solution NMR) typically are larger than those inspected for their rotational spectra (diluted gas / low T / low p).  The latter a topic for atmospheric / astrochem (e.g., http://vpl.astro.washington.edu/spectra/, http://vpl.astro.washington.edu/spectra/allmoleculeslist.htm). | With IR spectroscopy in chemometrics (e.g., PCA), there is much of work re (IR spectroscopy) .AND. ((machine learning) .OR. (neuronal network)), too (e.g., retired https://www2.chemie.uni-erlangen.de/services/telespec/ by 1990s.).

By small molecules, I meant rotational spectra (< 300 formula weight). I just wanted to test the deconvolution process on simple spectra HCl/DCl mixture which is partially resolved. BTW thanks for pointing out open access Raman spectra.

Wrt digitizing, years ago the Analytical Chemistry department at TU Dresden had a student preparing scans and digital band tables for the spectra in the Schrader-Meier Atlas. I have no idea whether/under what conditions these are available, but the Schrader-Meier Atlas itself now falls under the "orphaned book" exceptions of German copyright law.

This is "such as":

For Raman spectra of minerals, there's the RRUFF database
They use a JCAMP-DX-like ASCII format
Raman spectra of a number of small molecules important in biochemistry (amino acids, sugars, some fatty acids):
De Gelder, J.; De Gussem, K.; Vandenabeele, P. & Moens, L.: Reference database of Raman spectra of biological molecules J Raman Spectrosc, 2007, 38, 1133-1147. They send the spectra upon asking, IIRC as Matlab file.

Thank you. I sent a request to the author. This article is from 2007, let us if the email is still valid or if they are willing to respond.

The authors did not bother to respond.


text:
I am seeing Doublet of Doublets peaks when I should see just a doublet
The following is a tutorial question I need to do:

I initially thought the answer was isobutyl formate (CAS:542-55-2) but what confuses me is the dd peak at $\pu{3.9-4.0 ppm}$. To my knowledge, isobutyl formate should give just a doublet at that region. What am I missing out?

The peak at 4 ppm is due to the 2 H atoms from CH2 which is bond to CH(CH3)2 on one side and to O-CHO on the other side. The molecule is HCOOCH2CH(CH3)2

@Maurice Yes but shouldn't that CH2 give a doublet? Why dd?

@Maurice It can't be anything else but isobutyl formate, it's just I want to know the reasoning behind the dd peak. Can the CH2 couple with the aldehyde proton as well? How?

Your IR and 13C-NMR matches exactly with the known spectra of isobutyl formate, so your guess of the compound is most probably correct. However, I could not find its 1H NMR. Is it possible that your NMR was recorded with a chiral reagent? The two CH2 protons are enantiotopic and can couple to each other in presence of a chiral reagent. That's the only thing I can think of right now.

@ShoubhikRMaiti We should not forget that the compound in question is distributed commercially, e.g. by SigmaAldrich (e.g., https://www.sigmaaldrich.com/catalog/substance/isobutylformate1021354255211?lang=en&region=US) and that this company has a tradition in publishing spectra catalogues (incl. IR and 1H/13C-NMR), too.  Showing only a doublet around 4 ppm what could be a 300 MHz recording, see: https://www.sigmaaldrich.com/spectra/fnmr/FNMR001772.PDF

Your question is how can the CH2 be split into a dd when the -OCHO shows no triplett, right?
Well, the second splitting at the $\ce{CH2}$ is very small, and the $\ce{CHO}$ looks like it is very nearly split. That´s not just a broad peak, it has a rather clear shape imo. If you do just a bit of zero filling and apodisation, the triplett might well (should) appear.
But generally, you are right. Every splitting must have its counterpart.

Zero filling: Adding blanks to the end of your FID. Artificially increases the resolution, i.e. point density of your spectrum.
Apodisation: Changing the gain of your FID towards the end, by multiplying it with a decreasing or increasing function. The former gives you a less noisy spectrum, the latter a better resolution (sharper peaks actually), each at the cost of the other. The mathematical term for that is "convolution".

yes, that's what I meant. Maybe I should have worded my question better. The singlet at 8.07 ppm does indeed look nearly split, after reading your reply. Thanks.

NOESY experiment would clarify this splitting as well.

Drying the sample might make the splitting a lot clearer already. Or measureing the pure sample, with external lock.


text:
How can a β,γ-unsaturated ketone be converted to 1-pyrroline?
I am working towards a natural product. Presented are three intermediate compounds. After oxidation of the hydroxyl to a carbonyl, I want to introduce an imino-nitrogen atom into the structure. This way, an imino bond creates a five-membered ring, and the double bond disappears.
What reagent(s) should I choose for this type of reaction, and what does the reaction mechanism look like?

This is a tricky. You can't go with something like AcONH4/AcOH because you have other centres that can react, including a second enone (though presumably less reactive due to hindrance). I would be inclined to consider a two step process with 1eq of the anion of trifluoroacetamide or t-butylcarbamate followed by a deprotection to unmask the amine.

You may find the following paper interesting:
Reduction of Oximes and Aliphatic Nitro Compounds to Imines for further in situ Reactions: a Novel Synthesis of Pyrroles and Pyrrolin-2-ones. Barton, D. H. R.; Motherwell, W. B.; Simon, E. S.; Zard, S. Z. J. Chem. Soc., Perkin Trans. I 1986, 2243.

I suggest two possible approaches. Narasaka, et al. have effected similar cyclizations photochemically. I prefer to start with the allylic alcohol 1 to avoid complications with oxime formation. Formation of the bis acetate 2b is not a problem. However, given the greater acidity of oximes relative to alcohols does not preclude the formation of oxime acetate 2a. In the photolysis the naphthalene serves as a sensitizer while 1,4-cyclohexadiene functions as a hydrogen atom donor. The addition of hydrogen is likely from the convex face as shown. If not, after saponification of acetate 3b and Swern oxidation, base equilibration would likely lead to ketone 4.
Alternatively, a Barton-type radical cyclization (5 --> 3a) might be considered.
ADDENDUM: Direct photolysis of a structure similar to 5 without Bu3SnH has been achieved. See ref 2.

1) M. Kitamura, Y. Mori and K Narasaka, Tetrahedron Letters, 2005, 46, 2373.
2) F. Gagosz and S. Z. Zard, Synlett., 1999, 1978.


text:
Generate Surface with CubeFile from Gaussian in VMD or other free programm
I asked a question before, about how to save a mapped surface from GaussView. I don't get an answer, so I get much deeper in my internet research about it again.
Now I want to try to create a mapped surface of the electronic potential with VMD. So I found out, that I must create the cubefile before with gaussian. After that I import my cube file to VMD.
I followed multiple advises and introductions on the internet, but none of them worked for me.
I tried it also on my own, but this works not as well.
What musst I do on VMD or even when I create my cubefile on gaussian? Got somebody a proper link for me or can give me an advise? My target is to create a figure which can be used for publishing a thesis.
Thanks, Andrea!
If you need some files or something just let me know!

I believe this guide answers your question. Archived version just in case it goes offline in the future.

Can you be a little clearer what you want? Do you want to map the electrostatic potential on a isosurface of another property? Something like this: https://people.chem.ucsb.edu/kahn/kalju/chem162/public/smallmol_visual_elpot.html


text:
Addition of acid and base to buffer system
So I was doing my chemistry assignment and became stuck. Can I get some help?
a) Calculate the pH of a buffer system that contains 0.40 M of NH3(aq) and 0.50 M of NH4Cl(aq) . Note that the Kb value of NH3(aq) is 1.8×10−5.
My ans for delta pH: 9.158362492
b) Determine the change in pH if 2.50mL of 0.100 M HCl is added to 0.040 L of the buffer system described in part a).
**My ans for delta pH: **
c) Determine the change in pH if 2.50mL of 0.100 M NaOH is added to 0.040 L of the buffer system described in part a).
My ans for delta pH:

In (b) notice that the problem asks for the change in pH, not the pH itself. $$\Delta \pu{pH} = 4.65 - 4.17 = - 0.48$$ better yet to use two extra significant figures in intermediate calculations and just round the final answer. $$\Delta \pu{pH} = 4.6478 - 4.1734 = - 0.4744 \ce{->[round]} -0.47$$

You can find the way to solve this question in the accepted answer here.

Ok, let's beat this to death with ICE tables.

(a) Calculate the pH of a buffer system that contains 0.40 M of $\ce{NH3(aq)}$ and 0.50 M of $\ce{NH4Cl(aq)}$ . Note that the $K_\beta$ value of $\ce{NH3(aq)}$ is $1.8\times10^{−5}$.

Some observations:

We want the pH, not the pOH.
For $\ce{NH4^+}$, $K_\alpha = \dfrac{K_\mathrm{w}}{K_\beta} = \dfrac{1.00\times10^{-14}}{1.8\times10^{−5}} = 5.556\times10^{-10},\quad \mathrm{p}K_\alpha = 9.2553 $
Since $\ce{NH4+(aq) > NH3(aq)}$ the solution will be slightly more acidic than the $\mathrm{p}K_\alpha$
We'll assume that the equilibrium between $\ce{NH4+(aq)}$ and $\ce{NH3(aq)}$ doesn't shift so that the Henderson-Hasselbalch approximation can be used.

$\begin{array}{|c|c|c|} \hline
\ & \ce{NH3} & \ce{NH4^+} \\ \hline
\text{I} & \pu{0.400 M} & \pu{0.500 M} \\
\text{C} & 0 & 0 \\
\text{E} & \pu{0.400 M} & \pu{0.500 M} \\ \hline
\end{array}\\$
The Henderson-Hasselbalch approximation gives us a method to approximate the pH of the weakly acidic buffer solution as follows:
$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\ce{[NH3]}}{\ce{[NH4+]}}$$
$$ = 9.2553 + \log\dfrac{0.40}{0.50} = 9.1584 \ce{->[Round] = 9.16}$$

(b) Determine the change in pH if 2.50 mL of 0.100 M HCl is added to 0.040 L of the buffer system described in part (a).

Some observations:

Again we need to determine the pH, which we can subtract from the pH in part (a) to get the change in pH.
Let's just work in millimoles, mM. Since molarity is moles/volume, and the volume for $\ce{NH4+(aq)}$ and $\ce{NH3(aq)}$ is the same, the volume term just cancels. This saves some work calculating the dilutions.
HCl is a strong acid and shifts the equilibrium according to the reaction:

$$\ce{NH3 + H+ -> NH4+}$$

We'll assume that the equilibrium between $\ce{NH4+(aq)}$ and $\ce{NH3(aq)}$ doesn't shift further so that the Henderson-Hasselbalch approximation can be used.

$\begin{array}{|c|c|c|} \hline
\ & \ce{NH3} & \ce{NH4^+} \\ \hline
\text{I} & \pu{16 mM} & \pu{20 mM} \\
\text{C} & \pu{-0.25 mM} & \pu{+0.25 mM} \\
\text{E} & \pu{15.75 mM} & \pu{20.25 mM} \\ \hline
\end{array}\\$
$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\pu{mM \ce{NH3}}}{\pu{mM \ce{NH4^+}}}$$
$$ = 9.2553 + \log\dfrac{15.75}{20.25} = 9.1462$$
Therefore $\Delta\mathrm{pH} = 9.1462 - 9.1584 = -0.0122 \ce{->[round]} -0.01$

(c) Determine the change in pH if 2.50mL of 0.100 M NaOH is added to 0.040 L of the buffer system described in part (a).

Some observations:

Again we need to determine the pH, which we can subtract from the pH in part (a) to get the change in pH.
Again, let's just work in millimoles, mM.
NaOH is a strong base and shifts the equilibrium according to the reaction:

$$\ce{NH4+ + OH- -> NH3 + H2O}$$

We'll assume that the equilibrium between $\ce{NH4+(aq)}$ and $\ce{NH3(aq)}$ doesn't shift further so that the Henderson-Hasselbalch approximation can be used.

$\begin{array}{|c|c|c|} \hline
\ & \ce{NH3} & \ce{NH4^+} \\ \hline
\text{I} & \pu{16 mM} & \pu{20 mM} \\
\text{C} & \pu{+0.25 mM} & \pu{-0.25 mM} \\
\text{E} & \pu{16.25 mM} & \pu{19.75 mM} \\ \hline
\end{array}\\$
$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\pu{mM \ce{NH3}}}{\pu{mM \ce{NH4^+}}}$$
$$ = 9.2553 + \log\dfrac{16.25}{19.75} = 9.1706$$
Therefore $\Delta\mathrm{pH} = 9.1706 - 9.1584 = +0.0122 \ce{->[round]} +0.01$

Your first result is OK.
But :
First,
it does not make any sense to give a result with $\ce{10}$ significant figures when the initial data has only $\ce{2}$ significant figures. When a data is given like here $\ce{1.8 x 10^{-5}}$, it means that the author cannot be more precise, and may admit that the exact value of $\ce{K_b}$ is somewhere between $\ce{1.75 10^{-5}}$ and $\ce{1.85 10^{-5}}$. So if you repeat the same calculation with say $\ce{1.75 10^{-5}}$, you find a still "acceptable" value of $\ce{pH = 4.6569}$. This means that you final result can be between $\ce{4.66}$ and $\ce{4.62}$. It can be written $\ce{pH = 4.64 ± 0.02}$. The six next figures do not have any meaning.
For the second problem, you add $\ce{0.00025}$ mole $\ce{HCl}$ to a mixture containing $\ce{0.016}$ mol $\ce{NH_3}$ and $\ce{0.02}$ mol $\ce{NH_4^+}$. So the final concentration of $\ce{NH_3}$ and $\ce{NH_4^+}$ are respectively $\ce{0.391 M}$ and $\ce{0.503 M}$. The log of ratio of these concentration is $\ce{0.251}$. So that the final pH is $\ce{4.49 ± 0.02}$.
Try to do the third calculation by the same approach.

LOL - In the olden days you couldn't make such a horrendous mistake using 4 place logarithm tables or a  slide rule!

Thank you. But I am still confused about the last one. Wouldn't it calculate to the same answer as b.

@Samajm - If you add an strong acid to a buffer the pH has to go down. If you add a strong base to a buffer the pH has to go up.

I did all the working but I assume my first answer is wrong.


text:
Confused regarding aromaticity

I can understand the aromaticity in the second but can't in the first structure.
The second structure is aromatic since it has a continuous conjugation in the rings, but the first structure has no continuous conjugation, still it's aromatic,  can anyone help me out ? Where I'm going wrong?

I edited the picture. Hope I got 1 and 2 right...

What is your question?

In 1, ring (a) is aromatic, but (b) is not. However the double bond which is in (b) alone will share electrons with ring (a). // In 2, rings (a), (b), and (c) are all aromatic.

@MaxW I guess I have heard from some source that if one ring is aromatic,then the entire molecule is aromatic. Is that the case?

@YusufHasan - Well wouldn't the opposite be true too? If some part of the molecule is not aromatic, then the whole molecule is not aromatic? // The point I'm trying to make is that chemistry if often ambiguous. You could say that "Looking at the NMR for  compound 1, it is aromatic." What is meant is that it has an aromatic group, not that all carbon-carbon bonds in the compound are aromatic.

@MaxW Ok, I see..No, the thing I had in mind was that, maybe if you have atleast one aromatic ring and all other non-aromatic or aromatic ones, then you can call the entire molecule aromatic. But,if the molecule has even one anti-aromatic ring, then we can't term the entire thing aromatic. Does this make any sense?

@YusufHasan - Sorry I made my point poorly. Let me try again. // When we say that a compound is aliphatic then it has no aromatic bonding. // When we say that a compound is aromatic, then there is some aromatic bonding. But this is ambiguous. It isn't clear if all the C-C bonds are aromatic, or just if some of them are.

Compound 2 is a polycyclic aromatic hydrocarbon named phenanthrene.

Dear OP, I have tried to fix your question best as I could, but you really need to work on your English grammar. And vocabulary.

I still didn't get the answer despite going on the similar question asked previously, the thing is it neither have 4n+2π e and also i can't see a continuous conjugation ,sorry ,but I'm the basic level student still learning to grasp organic chemistry ! What yusuf had said is if one of the ring  is aromatic whole structure would posses some sort of aromatic character ,is it true? If so then MaxW also got a good point. What reason(s) should i state the  for this compound to be aromatic?

"Aromatic" is a qualitative term. "Aromatic" is also ambiguous. Mitheron's reference to the question about "Is 1,2-dihydronaphthalene aromatic?" lists some of the rules that are applied, but the final judge is your teacher: what would the teacher say?
Long ago, my professor said that aromaticity was the measure of surprise that an experimenter had when he determined the heat of combustion of a compound and found that it was considerably less than he expected from his calculations. He would attribute this deficiency to aromaticity. The units of "surprise" have not been established, and have not been established for aromaticity, either.

Okay so I'll consider this as a very rare exception of compound showing aromaticity, well do you know any other such organic compounds having this strange behavior, it will surely come in handy in future.

I've also heard of Friends saying if any of the ring posses a aromatic nature there's some sort of aromaticity developed in whole structure ! Is it? Wouldn't then opposite be true?

Well, the whole point of "aromaticity" is that the system is considerably more stable than the sum of the individual parts, and in addition, contains a ring, like the first-known aromatic compounds, such as toluene and xylene. It's a discussion topic: there is a ring, some kind of symmetry, a neat molecular orbital system, with lower bonding and higher anti-bonding orbitals, all kinds of minutiae that good chemists like to bond over (or anti-bond over). And rules to discuss.


text:
Recycle unreacted ethene
The question is this:

You want to make oxirane from ethene and oxygen gas:
  $$\ce{2 C2H4 + O2 -> 2 C2H4O}$$
  The flow into the reactor is constant at $\pu{200kmol/h}$ of $\ce{C2H4}$ and $\pu{90kmol/h}$ of $\ce{O2}$. Only 6% of the ethene reacts in the reactor. 98% of the unreacted ethene flowing out of the reactor is filtrated and recycled (sent back into the reactor). Oxygen and oxirane passes through the filtrator.
Calculate the flow of gas out of the filtrator:

with no recycling, and
with recycling.

There has to be something wrong with the question (or my translation of it, sorry) because given this information the net amount of ethene in this system is just going to increase.

Why was this demand not closed ? There has been no effort from Hans to try to start solving his problem. Usually when such a situation happens, some moderator closes the discussion, arguing that we are not here to solve the problems. We help those who try to do something and are stopped on a difficulty...

@Maurice Moderators only close unilaterally very clear cases (this isn't that clear to me), other times it takes 5 users with close/reopen privileges to close a question. I don't know why this time no-one put a flag on it, maybe because the OP only wants to clarify the question itself and whether it is possible to solve it. I agree that it would be much better if the OP includes the calculations which lead to the conclusion that the ethene in the system is just going to increase.

@ Martin. OK ! You are really nice.

Welcome to ChemSE! Thank you for posting your question here! I worked out your problem like this:
Flowing into the reactor:

200 kmol $\ce{C2H4}$
90 kmol $\ce{O2}$

Inside the reactor:

12 kmol (6%200) $\ce{C2H4}$ + 6 kmol $\ce{O2}$ -> 12 kmol $\ce{C2H4O}$

Reactor output (answer, part 1):

188 kmol $\ce{C2H4}$ (200 - 12)
84 kmol $\ce{O2}$ (90 - 6)
12 kmol $\ce{C2H4O}$ (from reaction)

If the ethene is not recycled, (i.e. part 1) then this is the output and therefore the answer. If the ethene is recycled...
Filtrator output (answer, part 2):

3.76 kmol $\ce{C2H4}$ (100% - 98% = 2%, 2%188)(the 2% not captured by filtrator)
84 kmol $\ce{O2}$ (from reactor output; not filtered nor recycled)
12 kmol $\ce{C2H4O}$ (also from reactor output)

The remaining 184.24 kmol of ethene is recycled and becomes part of the 200 kmol pumped into the reactor in the next hour. So, after the first run, only 200 - 184.24 = 15.76 kmol of fresh ethene is pumped into the system every hour, the rest coming from the previous run.
Note: this occurs once per hour. Don't forget that.

But! I have interpreted the question as the flow of fresh ethene is always 200kmol/h. If that is true then this question becomes harder. Or?

@Hans I am very confused. Are you speaking Australian English or Google Translate? Because it doesn't sound american. (I Don't Understand the Comment)

Oh... I think I got it.
They probably wouldn't give you a question like that because there wouldn't be a real answer; The answer would always change over time. I think that by saying "The flow into the reactor is constant" They are just telling you that there will always be 200 flowing into the reactor, regardless of recycling status, so that you don't try to calculate what the total output would be if no new ethene was fed in after the original 200 (which would not run forever because some is consumed in the reactor every hour)

Yea sorry, yes you got me correctly. But the original question really imples that there is constantly a flow of 200kmol/h fresh ethene into the reactor. I will ask my teacher about this. If i can assume what you have said, that the TOTAL ethane into the reactor is 200, then the question becomes much easier.
Im actually a math nerd so i did do some reserach about iterative linear functions. Turns out this system will reach an equilibrium, but when it does, it probably contains so much ethene a realistic reaction vessel couldnt contain it.

Yeah, I think it's just a badly worded question.

It was as i thought, it is 200 fresh ethene excluding the recirkulating ethene. I have solved this problem. Thank you.

As much as I understand, there is nothing wrong with the question, if your translation is correct. Let's see the reaction:
$$\ce{2 C2H4 (g) + O2 (g) -> 2 C2H4O (g)}$$
The flow reactor was fed with constant flow rates of $\ce{C2H4}$ at $\pu{200 kmol\:h-1}$ and $\ce{O2}$ at $\pu{90 kmol\:h-1}$. Only $6\%$ of the ethene reacts in the reactor. That means the reaction rate of the reactor is $\pu{12 kmol\:h-1}$ per $\ce{C2H4}$. Therefore, assuming a perfect world, $\pu{6 kmol}$ of $\ce{O2}$ would be consumed in each hour to produce $\pu{12 kmol}$ of $\ce{C2H4O}$ as well (see the chemical equation).
a) Thus, the gas outlet of the reactor would release 98% of unreacted $\ce{C2H4}$ ($0.98 \times \pu{188 kmol}$) and $\ce{O2}$ ($\pu{84 kmol}$), and $\ce{C2H4O}$ ($\pu{12 kmol}$) in each hour as a mixture. The sum ($188 \times 0.98+84+12 \approx \pu{280.2 kmol\:h-1}$) is the rate of gas flow out of filtrator of the flow reactor. Of course, this is assuming filtrator does not filtered off any of these materials (it wasn't clear on the translation). If filtrator separates the product and oxygen from the mixture (e.g., by gas partition by difference in melting points; melting points of oxirane and ethene are ~$\pu{161 K}$ and ~$\pu{104 K}$, respectively, and melting point of oxygen is ~$\pu{54.4 K}$), the outlet rate of $\ce{C2H4}$ out of the filtrator would be $\pu{188 \times 0.98 kmol\:h-1} \approx \pu{184.2 kmol\:h-1}$ (with no recycling of $\ce{C2H4}$). If  the flow of gas out of the filtrator means only the gases pass through the filtrator, that should be only oxygen and oxirane. Thus, the outlet rate of gases out of the filtrator would be $\pu{(84 + 12) kmol\:h-1} = \pu{96 kmol\:h-1}$.
b) I believe, the recycling of $\ce{C2H4}$ means unreacted $\ce{C2H4}$ is adding to original rate of $\pu{200 kmol\:h-1}$. Thus, rate of $\ce{C2H4}$ fed into reactor after first hour is: $\pu{(200+184.2) kmol\:h-1} = \pu{384.2 kmol\:h-1}$. Only 6% of it ($\pu{23.05 kmol\:h-1}$) reacts with $\pu{11.53 kmol\:h-1}$ of oxygen to give $\pu{23.05 kmol\:h-1}$ of oxirane. Therefore, outlet rate is:
$$\left((384.2 - 23.05) + (90 - 11.53) + 23.05\right) \ \pu{kmol \: h-1} = \pu{462.67 kmol \: h-1}$$
This would be the outlet rate without filtration. If you have to consider only 98% of $\ce{C2H4}$, then the rate would be:
$$\left((384.2 - 23.05)\times 0.98 + (90 - 11.53) + 23.05\right) \ \pu{kmol \: h-1} = \pu{455.45 kmol \: h-1}.$$

But the product is never filtrated; it says so in the question.

Let R represent the recycle rate.  Then the ethene rate fed to the reactor is 200 + R.  If 6% of this reacts, there is 188+0.94R kmol/h ethene exiting the reactor.  98% of this is recycled, or (0.98)(188+0.94R).  But this is equal to the recycle rate R.  So we have, $$R=(0.98)(188+0.94R)$$Solving this equation for R gives:  $$R=2338\ kmol/h$$ The oxygen reacting is now (200 + 2338)(0.03)=76.14 kmol/hr.  So the oxygen in the filtrate is now only 14 mol/h; and the oxirane  exiting in the filtrate is 12 + 0.06R = 152 kmol/h.  So the yield of the reactor is greatly increased by employing the recycle stream.  The amount of ethene lost is now only (0.02)(188+0.94R)= 48 kmol/hr


text:
How does one dilute sulphuric acid (1.28sg@25ºc 460g/L) to 3 M Sulphuric Acid?
Thanks for looking.
The extent of my research around the issue so far.
SA sg table but wrong temp 60f not 77f/25c
http://www.sschemical.com/wp-content/uploads/2013/05/Conversion_Table.pdf
From
http://www.csgnetwork.com/specificgravliqtable.html
Liquid                                 Temp     kg/cu.m
Sulphuric Acid 95% conc.    20 C    1839
Water, pure                               4 C   1000.00
SG curve water, if per chance useful
https://www.engineeringtoolbox.com/water-temperature-specific-gravity-d_1179.html
%(w/w) to molarity and molality question. On which basis I bought battery sulfuric as another supplier had other for same purpose which was 34.5% and in below link 40% came out to 5.9~ M
https://answers.yahoo.com/question/index?qid=20091019225041AASpTj4
Any help appreciated

To ask the first obviously question, how good does the dilution need to be? $\pm 10 %, \pm 1 %, \pm 0.1 %$ and so on... // 1.28sg@25% is really an odd specification. Is the 25 supposed to be some temperature? Perhaps $\pu{^\circ C}$ ?

@MaxW Sorry, yes the 25 was meant to be the temp in C, have fixed now. Regards the accuracy I would like achieve a result suitable for testing for Vitamin C by iodine titration as described here
https://www.thoughtco.com/vitamin-c-determination-by-iodine-titration-606322
Really appreciate your response.

For that $\pm 10 %$ would be fine. Just make enough to do standard and samples with same batch of diluted sulfuric acid. // See this table that shows density doesn't change very much with temperature. So use $\pu{66 ^\circ F}$ densities and assume that they are good for $\pu{77 ^\circ F}$ which is waht you seem to want. Does that solve your problem, or do you need more help?

Great Thanks MaxW. Um, armed with that I think i could just use one of the solutions on the answers.yahoo page but which one to use IDK, they seem to come up with different answers. Sorry for the trouble. Thanks again.

@PeteBottom For future reference, I recommend you take a tour on how to type units along the widely consented guidelines by IUPAC and IUPAP.  There are compilations on ChemSE meta (e.g., https://chemistry.meta.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know/3045#3045) showing how to achieve this here as well in either question, answer, or comment.

If the sulfuric acid has $\pu{460 g } \ce{H2SO4}\pu{/L}$ then the molarity of sulfuric acid is:
$\dfrac{\pu{460 g/L}}{\pu{98 g/mole}}= \pu{4.69 molar}$
As Alex pointed out for dilution  to 3 M Sulphuric
$C_1\cdot V_1 = C_2\cdot V_2$
$4.69\cdot V_1 = \pu{3 molar}\cdot \pu{1.00 L}$
$ V_1 = \dfrac{3}{4.69}\pu{ L} = \pu{0.640 L}$
So take $\pu{640 ml}$ of your concentrated sulfuric acid and dilute to 1.00 Liter.

I'm not 100% sure if this will apply but you can probably use the formula C1V1 = C2V2.
V1 being the volume of stock solution needed to make the new solution, C1 The concentration of stock solution, V2 The final volume of new solution, C2 The final concentration of new solution
For example 5 mL of a 0.25 M solution from a 1 M solution, using the formula plug values in => (V1)(1 M) = (5 mL)(0.25 M). Rearrange a bit =>  V1 = [(5 mL)(0.25 M)] / (1 M)V1 = 1.25 mL and after some quick calculations you should get something like this => 1.25 mL of the 1 M solution into V1-V2 = 5 mL – 1.25 mL = 3.75 mL of dilute.
Sorry if it isn't what you need.

Hi Alex, thank you for your response. I will try to get my head around it anyway but I just thought to mention I had a critical error in the subject/question. I had 25% instead of 25 degrees C. I'm too much of a novice to know whether your response helps straight up, my instinct says, my error was unlikely to let you actually know the real query I had, but I will go over it a bit more. My response to MaxW above mentions the context for my desired result, ie
https://www.thoughtco.com/vitamin-c-determination-by-iodine-titration-606322
Thanks again very much for your help in any case.


text:
Why is 4-t-butyl aci-nitro cyclohexane stable whereas aci-nitro cyclohexane is not?
During my PhD, about fifty years ago, I had reason to use 4-t-butyl aci-nitro cyclohexane and found it to be remarkably stable. Usually aci- nitro compounds revert to nitro compounds very quickly unless they are stabilised by conjugation. There is no conjugation in this compound and the t-butyl group only serves to keep the cyclohexane ring in the chair configuration. There is no excessive steric hindrance in the chair configuration. So what's going on? My guess is that the conversion of the aci-nitro form to the thermodynamically more stable nitro form happens by the transfer of the OH proton to the carbon via the carbanion and that carbanion can be in the boat or the chair configuration. However, only in the boat configuration does the molecular orbital come near enough to the OH proton for the transfer to occur. Is this on the right track? I am not very familiar with molecular orbital theory (it had hardly started 50 years ago) so I would appreciate some input that I might be able to understand.

Good question. I have no idea why it's stable except for equotorial t-Bu group help the planer aci-nitro to exist (nitro group has to be either axial or equotorial). By the way, I tracked your work on phoyochemistry on the compound (https://doi.org/10.1016/S0040-4039(01)82227-9), I think! :-)

Yep, that's me. Thanks for your interest. I didn't think anybody would be interested in such obscure work.


text:
All-Gauche Alkane: Is It Possible?
I've learned about anti and gauche conformations throughout my scholastic journey, and I have heard often that straight-chain alkanes always adopt an all-anti conformation in their lowest energy state. I also know that, in a pure sample of any given straight-chain alkane, (eg. hexane) the anti and gauche conformations are in a certain equilibrium with each other, And that this ratio of gauche to anti shifts towards gauche as the available energy of the sample (i.e. temperature/heat/ pressure) approaches infinity.  My question is as follows: Has anyone in the scientific community to date successfully forced an entire sample of straight-chain alkanes into the all-gauche conformation without significantly raising the conditions above STP? I have scoured the internet but have not surfaced with a definitive answer, so I now turn to the scientific community.
All contributions are greatly appreciated.

No, the ratio will approach a finite value, likely sth in the range of 2:1. If two states have only a marginal energy difference compared to $kT$, they will be equally occupied. That´s one of the most basic principles in QM and statistical thermodynamics.

The best example would be solid isotactic polypropylene, where one in six (or five?) carbon carbon bonds is in gauche. In a linear alkane it is sth like one in twelve or fifteen at RT.

@Karl but isotactic polypropylene is branched, not linear. Sill a cool tip, tho. Thanks!

At Last! Success!
I found this research paper, which describes a "Self-assembling nanocapsule", which is held together by plenty of hydrogen bonding, and can assemble around a linear molecule of all-anti tetradecane, $\ce{(C14H20)}$ using the energy afforded by the hydrogen bonding to squeeze/twist the alkane into a helical all-gauche conformation. Have a look:

Cool, right?
Sources:
Scarso A, Trembleau L, Rebek J Jr. Helical folding of alkanes in a
self-assembled, cylindrical capsule. J Am Chem Soc. 2004 Oct 20;126(41):13512-8.
PubMed PMID: 15479107.
Note:
Other answers are still greatly appreciated!

Gosh, supramolecular chemistry never ceases to amaze!

@NicolauSakerNeto I know, right? It's so fascinating!


text:
Check My Beginner's Guide to Shells, Energy Levels, Subshells, and Orbitals
I am a beginner trying to summarize what I've learned about:

Shells
Energy Levels
Subshells and
Orbitals

all at the high school level.
Unlike related threads, I'm going to try to:

Define each word and state a few key properties of each.
Highlight what is most important for beginners.
Summarize all of that in a table.

Please let me know how I've done!
Thanks!

tl;dr Summary: Electrons move within orbitals which are in subshells which are in shells which are in atoms.

Shell aka Energy Level
Below atoms, shells are the largest grouping of electrons beginner chemists should care about, and among those, the innermost two shells are most important to beginners.
They correspond to periods (rows) on the periodic table of the element.

In the 3rd period, all elements have 3 shells.
In the 5th period, all elements have 5 shells.
In the $n$th period, all elements have $n$ shells.

Generally speaking, the larger $n$ means:

The shells are farther away from the nucleus and thus the electrons in it are less attracted to the nucleus.
The shells can hold more electrons, i.e. an atom of that element requires more valence electrons to be stable.

Subshell
Each shell consists of at least one subshell. Below atoms, this is the second largest grouping of electrons beginner chemists should care about.
There are 4 types of subshells: $s$, $p$, $d$, and $f$. For beginners, the $s$ and $p$ subshells are the most important.
The notation for shells and subshells:

$1s^2$ means the 1st shell's $s$ subshell has 2 electrons in it.
$4d^5$ means the 4th shell's $d$ subshell has 5 electrons in it
$AB^X$ means the $A$th shell's $B$th subshell has $X$ electrons in  it.

Note: This notation does not specify which orbital those electrons are in... but that info does not matter much to beginners.
Orbital
An orbital is an actual path (probability distribution function, PDF) that either 1 or 2 electrons can be whizzing through at a time. This is the smallest grouping of electrons beginner chemists should care about. The math underlying that PDF is not critical for beginners.
Each subshell consists of at least one orbital.
THE BIG TABLE
Feel free to edit the table. :)

This could be a great blog post. However, Chemistry.SE is a Q&A site meaning it'a place where a well-defined question receives a concise answer. The main Chemistry.SE site is not designed for discussions, polling, collecting feedback or peer reviews. If you can re-formulate your post into a question (see Help Center — Asking), it can be re-opened.

All looks good at a school level, but I would suggest or convince you that we should not attach any physical pictures or meaning to all these atomic concepts. If you pursue science in future, you will have to unlearn a lot of these stories ingrained by school or even university teachers. It is okay to acknowledge and say that we do not fully understand everything as yet. Nature is very very complex. Our only effort is to make simple models of the phenomena we observe in the Nature.
The shell model is outdated and obsolete yet it is good for educational purposes. It was inspired by the planets rotating around the Sun, so learning about it from a historical perspective is good. For example, you wrote "The shells are farther away from the nucleus and thus the electrons in it are less attracted to the nucleus." It may not be a correct picture, because in reality there are no circular shells per se, like an onion ring around a nucleus. Brilliant physicists spent their life time in understanding the atom so don't be surprised if you find atomic structure overwhelming. Indeed it is.
More importantly just delete this interpretation from your mind "An orbital is an actual path (probability distribution function, PDF) that either 1 or 2 electrons can be whizzing through at a time."
An orbital is just a mathematical construct with no physical reality as textbooks show them as balloon models and so on. The rigorous and correction version of an orbital is "Wavefunction depending explicitly on the spatial coordinates of only one electron" (IUPAC)
Their shape depends on the coordinate system being used to solve the Schrodinger's equation. Can you believe that Schrodinger never showed or drew a single picture of orbitals in his four famous papers?
The typical shapes of s, p, d, f orbital are shown in Cartesian coordinate system, but physicists and mathematicians have plenty of other coordinate systems to solve differential equations. Do the electrons care what coordinate system a human chose to represent them or calculate their energies? The answer is no. Their energy turns out to be the same regardless of the coordinate system. Nobody teaches that and I also recently learned this from a physicist who has published a lot on orbital shapes. That gentleman is a big opponent to the teaching of orbitals and the myths associated with them in undergraduate curriculum. Not all agree with him including Linus Pauling.
The atomic structures "stories" as taught to chemistry students as facts is one of the worst things in modern educational system. In my humble opinion, chemists should take the help of modern physicists and ask them how to represent the latest views of atomic structure right from beginning classes. It does not need to have crazy mathematics, but good scientists can explain and elaborate things in words as well. As someone said, "If you can't explain it simply, you don't understand it well enough."
Book recommendation: Hydrogen: The essential element by J. Rigden (from Harvard). A thin book which is very illuminating. Chemical educators should adapt this style. The last chapter or so talks about anti-hydrogen.

"chemists should take the help of modern physicists and ask them how to represent the latest views of atomic structure right from beginning classes..." Where might I find such representations that are intelligible to teenagers? If they don't exist, might that suggest nobody has found a way to do it? If they do, I'd love to read them!

The quotes by Einstein, on simplicity, connect our ability to explain theories to their quality, not to how well we as scientists understand the underlying concepts: "All physical theories, their mathematical expressions notwithstanding, ought to lend themselves to so simple a description that even a child could understand them." By contrast, the idea contained within your quotation, which connects our ability to explain theories to our own understanding of them, famously comes from Feynman:  https://kottke.org/17/06/if-you-cant-explain-something-in-simple-terms-you-dont-understand-it

Dear @WeCanLearnAnything, I added the book recommendation. I liked it a lot (perhaps the only book I read from start to end) and you can easily follow it as well.

Cool. Hopefully my library has it!


text:
Product of the reaction between chalcone and sodium methoxide in methanol

I'm a bit lost on how to do this, for Product C I can see two things happening:
1) The $\ce{-OCH3}$ attacks the alpha hydrogen, but that doesn't really make sense as then you would get two double bonds next to one another
2) The $\ce{-OCH3}$ attacks the electrophilic carbon, but then we know the negatively charged oxygen cannot collapse back because then the $\ce{-OCH3}$ would leave. So, would it just get printed by $\ce{HOCH3}?$ And if it does that how would it react with the second set of reactions?

$\ce{-OCH3}$ will attach on $\beta$ carbon, then keto-enol tautomerization would occur. Finally bromination at $\alpha$ carbon will be there.

The first step is called Michael addition. Have a look at this wikipedia page.

The Michael reaction or Michael addition is the nucleophilic addition of a carbanion or another nucleophile to an α,β-unsaturated carbonyl compound. A newer definition, proposed by Kohler, is the 1,4-addition of a doubly stabilized carbon nucleophile to an α,β-unsaturated carbonyl compound.

Of course, in your case, the nucleophile is $\ce{OCH3-}$. Perform 1,4-addition by adding it to the double bond at the beta position (with respect to the carbonyl), and you'll expect an enolate ion, which looks like this -

Finally protonate the $\ce{O-}$ and tautomerise to get your product as @Zenix already stated.
For the second step, check out this link.

References: Michael reaction, wikipedia || Ketone halogenation, wikipedia || Enol, wikipedia

So for question 6 would I just be getting the same product again?

@user690808 Nope, I'm not sure if you got the idea. For question 6, the answer would be 3-methoxy-1,3-phenyl-propan-1-one.

Im still a bit confused could you draw a mechanism of the first reaction for me.


text:
If you placed silver metal into silver nitrate, would a reaction occur?
thinking about oxidation and reduction, what happens to the silver metal and the silver ions?

How would you tell the difference between "yes, it would occur" and "no, it wouldn't"?

@IvanNeretin use Ag-110m; compare diffusion of gold in gold

For OP: As I have realized from James Gaidis' answer, you were very minimalistic in describing context of your question, what is very bad practice, as much more burden is put on shoulders of responders via the question interpretation and clarification. As I have just now realized, you may, or may not speak about a solution.

Yes, the reaction will occur, and no, the reaction will not occur.
There is ongoing equilibrium reaction $\ce{Ag <=> Ag+ + e-}$ with zero net result.
Initially, when metallic silver is inserted, very tiny net result of the above reaction happens, until there is achieved the equilibrium potential. It is determined by the initial potential before inserting into the solution and by the effective capacitance of the electrode.
If you charged positively ( e.g. electrostatically ) a piece of metallic silver before inserting into the solution, a very small amount of silver would dissolve. If negatively, a very small amount would deposit itself.
As @MaxW has mentioned in his answer, isotope tracing is frequently used to track the reaction mechanisms of complex processes and also kinetics of systems in equilibrium, that seem macroscopically static.

This is a very important and subtle point you made. I wish all teachers like electrochemistry like your teachers. East Europeans were too good in electrochemistry.

:-) Hehe. But the truth is, I liked electrochemistry at university.

Oops... Took to long to type my answer.

@mpprogram6771 That is OK. The quick answers are offen overcome by later thorough answers. :-)

@Poutnik, You have made really important point, "If you charged positively ( e.g. electrostatically ) a piece of metallic silver before inserting into the solution, a very small amount of silver would dissolve. If negatively, a very small amount would deposit itself." Have you seen this described in your textbooks.

Not really seen, but is not it obvious ? . An electrode potential is regulated via charge balance between the electrode and solution. If potential is too high, the summary rate of oxidative processes is boosted and the summary rate of reductive processes is supressed, decreasing the potential toward the equilibrium potential when both summary rates are equal. And vice versa. But it is also obvious the total exchanged charge is very small.

Yes, it is obvious. Such experiments were done by Faraday. I am talking from a typical modern day electrochemist PhD. If you ask them they won't know anything about such a situation.

Between the solution of silver ions and the silver atoms on the electrode there is ongoing exchange of silver ions for silver atoms with a zero net change. Since Ag mined from the earth has no radioactive isotopes, it would possible to detect the change by using a radioisotope of silver say $\ce{^{108\mathrm{m}}Ag}$ in the solution. After immersing the Ag electrode in the radioactive solution it too would become radioactive indicating that a exchange did happen.

Short answer: No
Long answer: Honestly, there's really no good way to tell.
Think about it this way: the only plausible redox reaction between the Silver(I) cations and the Silver metal is the exchange of electrons on the surface of the metal. And if this happened, the $\ce{Ag(s)}$ would donate one of it's valence electrons to $\ce{Ag+(aq)}$, resulting in a $\ce{Ag+(aq)}$ ion and a $\ce{Ag(s)}$ atom. See the change? there isn't one. So as far as we know, the surface of that silver electrode could constantly be exchanging electrons with the solution of ions, but we would never know, and It really has no significance whether it does or not, because it wouldn't affect much.
Credit to @Ivan Neretin for prompting this answer.

I'm going to be a literalist, and then an extremist, and then a speculator.
If you put a silver wire into silver nitrate (which is literally a solid, mp = 210 C), no visible reaction will occur.
However, silver nitrate decomposes when heated:
2 AgNO3(l) → 2 Ag(s) + O2(g) + 2 NO2(g)
Qualitatively, decomposition is negligible below the 210 C melting point, but becomes appreciable around 250 °C and it totally decomposes at 440 °C. (Wikipedia)
So, if you warm the silver nitrate solid until it is extremely melted, it may be enough of an oxidizer to oxidize silver metal to the oxide Ag2O, which melts at 300 C, but begins to decompose over the range 200 - 280.
Now, just speculating, but if AgNO3 decomposes to O2 (+ NO2), it may be thru 2 Ag2O --> 4 Ag + O2, that is, thru an intermediate Ag2O, which may be too difficult to isolate. However, this sort of reaction could be demonstrated by heating/melting AgNO3 with silver metal, and when the AgNO3 is all decomposed, the silver metal may have disintegrated, indicating a reaction where the metal was oxidized, then decomposed back to metal.

Good point about dry chemistry !


text:
Why is the chlorine ion more stable than the chlorine atom?
In heterolytic cleavage, for instance, $\ce{NaCl}$ breaks in two fragments $\ce{Na+}$ and $\ce{Cl-}$.
If chlorine is fully filled because it takes one electron from sodium, then why does chlorine make further bonds to other species?
My question is: Which is more stable, the chlorine atom (17 protons and 17 electrons) or the chlorine ion (17 protons and 18 electrons)?

RE: "If Chlorine is full filled because it takes one electron from sodium ,now chlorine is full filled then why further chlorine makes bonds to other species ?" -- Huh? Yes a $\ce{Cl-}$ ion will react with some ions or compounds. But if you dissolve ethanol and NaCl in water no chlorocarbon compound will form. So a $\ce{Cl-}$ ion doesn't react willy-nilly with everything.

I asked which is more stable ? If chlorine have already full filled shell then why reacts to some other species to make compound.

Species make compound to make more stable but in that case chlorine is stable in ion state!

You are trying to compare apples with oranges on the scale of vegetable. There is no such thing as absolute stability. Especially anions are very seldom the most stable in gas phase.

Chlorine is more stable in ion state than in the neutral atom. But it is only more stable with respect to its own electrons. For the rest of the world it has one negative charge in excess. So, stable or not, it attracts positive charges like Na+ or any other positive ion. You cannot avoid that it will produce neutral assemblies of atoms, like NaCl. It is not a question of stability.


text:
Boundary conditions in a closed system for diffusion (2nd Fick's law) in a porous medium
I would like to calculate the chloride redistribution in concrete (assumption: porous and homogeneous structure, purely diffusion-controlled mass transfer, closed system). For this I have derived a solution of the differential equation of Fick's 2nd law.
Since the chloride ions cannot leave the system at the borders, I have assumed the following boundary condition:
$$\frac{dc(0,t)}{dx} = \frac{dC(L,t)}{dx} =0, ~~~t>0$$
I have taken this boundary condition from another post on this site (When is diffusion steady-state?)
My solution of the differential equation assuming this boundary condition is correct. I have checked it with the help of an FEM simulation. Now to my question:
Does anyone know of a literature source with which I can prove my chosen boundary condition?
Can one generally assume that the derivative of the concentration at the edges is always zero when the system is closed?
Thanks a lot for your help in advance!

There are two classic texts on this well studied topic J. Crank, 'The Mathematics of Diffusion' publ. 1979 OUP,  and Carlsaw & Jaeger  'Conduction of Heat in Solids', publ. 1959, Clarendon Press. Either or both are well worth looking at.


text:
Calculate collision density of air

Caluculate the collision density between $\ce{O2}$ and $\ce{N2}$ in air at $T = \pu{10 ^\circ C}$. Assume that the air is an ideal gas and that the molecules are spheres with diameters $\pu{0.36 nm}$ for $\ce{O2}$ and $\pu{0.37 nm}$ for $\ce{N2}$.

This is what I have done so far, the collision density equation is:
$$ Z_{AB} = \frac{N_A}{V} \frac{N_B}{V} π d^2_{AB} c_A$$
And since we assume that air is an ideal gas, we can use:
$$\frac{N_A}{V} = \frac{N_B}{V} = \frac {p}{kT}$$
$$\frac {p}{kT} = \frac {1.01325 \cdot 10^5}{(1.38 \cdot 10^{-23}) \cdot 283} =2.6 \cdot 10^{25}$$
The collision cross-section can also be calculated:
$$ \sigma = \pi d^2_{AB} = \pi \left((3.6 \cdot 10^{-10})+(3.7 \cdot 10^{-10})\right)^2 = 1.67 \cdot 10^{-18} m^2  $$
But to calculate the average speed I need the mass of the molecules, which I don't have and I am not sure if I can determine it from the information given either. I thought about calculating the volume of each molecule and then looking up the density of oxygen and nitrogen to get the mass, however I am not sure if this is the correct way that the problem should be solved. Is there a specific assumption I need to take in order to determine the masses?

It is a common calculation see this for example https://www.tec-science.com/thermodynamics/kinetic-theory-of-gases/mean-free-path-collision-frequency/

The "in air" means you need more facts than the problem gives:  You need to know the fractions of O2 and N2 in air.  Given that you need to add that to the problem, it's probably OK to also use the (well known) masses of O2 and N2.

The mass of $1$ molecule $\ce{O2}$ is $\pu{32 g/6.022 10^{23} = 5.314 \times 10^{-23} g}$

In case that previous comment isn't obvious enough: atomic-mass-per-mole / Avogadro's number = mass / particle, and in this case each particle has twice the atomic mass of an oxygen atom.


text:
Change in pH when changing the sequence of acidic solution
I have 4 acidic solutions: Barium nitrate, aluminium nitrate monohydrate, cobalt nitrate hexahydrate, and hexachloroplatinic acid. Everything is dissolved in water.
I mixed them in order to create a final acidic solution. Following is the mixing procedure
1)Platinic acid =>Cobalt Nitrate=>Alumnium Nitrate=>Barium Nitrate
(This means Aluminium Nitrate was added to Barium Nitrate , to the mixture of which cobalt nitrate was added, and finally, chloroplatinic acid was added.)
Now, when I add everything to barium nitrate, the final $\mathrm{pH}$ of the mixture is always around 1.5,
2)Barium Nitrate =>Aluminium Nitrate =>Cobalt Nitrate =>Platinic acid
When I add everything to chloroplatinic acid, the $\mathrm{pH}$ of the mixture drops to around 0.3-0.5. Also if I do some other sequence, the $\mathrm{pH}$ is different.
The following are the details of the solutions-
Barium Nitrate- 2.1 g in 40 ml of water (0.008 moles) (0.2 M)
Aluminium Nitrate Nonahydrate-33.1 g in 40 ml of water (0.088 moles) (2.2 M)
Cobalt Nitrate-1.86 g in 40 ml of water (0.0064 moles) (0.16 M)
Hexachloroplatinic acid-0.83 g in 40 ml of water (0.0016 moles) (0.04 M)
I am trying to understand why this is happening because ultimately every time the constituents are the same, but why is the sequence of the addition causing this change?
Since always conceptualizing $\mathrm{pH}$ in terms of $\ce{H+}$ and $\ce{OH-}$, I am having a hard time trying to understand how the nitrates and the chloroplatinic acid are behaving with each other and with the water molecules and how exactly the final $\mathrm{pH}$ is dependent on them.
The $\mathrm{pH}$ of individual solutions when I measured them in our lab are:

barium nitrate: 6.2;
aluminium nitrate: 1.71;
cobalt nitrate: 5.46; and
hexachloroplatinic acid: less than 1.

Looking at these numbers, I can see why adding everything to hexachloroplatinic acid is giving less $\mathrm{pH}$ than adding everything to barium nitrate, but what mechanism is playing part here?
Any help would be greatly appreciated.

Know what the concentration of each solution is? Was anything added to any of the solutions to stabilize it?

It would be interesting to know exactly what you have done. For example, what are the concentrations of your four solutions, and what are the volumes of your solutions  used for preparing your mixture ? Do you use the same volumes for all your different mixtures ? Do you use one drop of the first reagent and 1 liter of the second ? And also why do you carry out this strange mixture ? What for ? Hexachloroplatinic acid is an expensive reagent...

yeah sorry should have mentioned the concentrations-

@MaxW I dont add anything to stabilize them

@Maurice I use it to synthesis metal oxides .Yes ,I am aware of the cost of Hexachloroplatinic acid

Did you boil the distilled(?) water to get rid of $\ce{CO2}$ before making the solutions. Did you measure the pH of that water?

As @Maurice asked, please explain your mixing experiment in detail. Which solutions, the order, and volume of each.

@MaxW  I have edited the question, please take a look

@user86428 - what volumes of each solution?

@user86428 - all in all there is obviously some sort of concentration dependent reaction happening. I'd suspect that the reaction is between the solutions of aluminum nitrate and hexachloroplatinic acid.


text:
Why does a proton prefer to attack a double bond instead of protonating an alcohol in a reaction of unsaturated alcohol with sulphuric acid?
A solved problem in Organic Chemistry by Solomons, Fryhle and Snyder asked to write the mechanism for the following transformation in the presence of sulphuric acid ($\ce{H2SO4}$):

The mechanism provided by the book is as follows:

It can be seen that a proton from sulphuric acid attacks the double bond thereby forming a tertiary carbocation as an intermediate. Later the lone pair on the oxygen atom of the hydroxyl group attacks the carbocation to form a cyclic intermediate which after deprotonation gives the required product. I understood the mechanism given in the book.
However, when I attempted the problem with the given reactant and the reagent, instead of attacking the carbon-carbon double bond, I protonated the alcohol. I attained at a much different product by following the mechanism given below:

The reason I chose to protonate the alcohol was I felt that electron density near oxygen is much higher than that in the double bonds. Moreover, I also end up with a conjugated diene, which is resonance stabilized.
Further, the following statement given as an explanation for a different problem supports my method of protonating over attacking the alkene:

Protonation of an alkene forms a carbocation with an open octet, whereas protonation of methanol does not form an open octet. Therefore, we predict methanol to be a stronger base than alkene. The mechanism starts with the reaction of a stronger base with sulphuric acid.

Now, my question is, why does a proton prefer to attack a double bond rather that protonating an alcohol? Under what conditions, does a proton protonate the alcohol instead of attacking the double bond? Are there any well-defined rules to predict which attack will be preferred over the other in such cases?

Image source: I made them using ChemDoodle.

That would probably depend on situation https://chemistry.stackexchange.com/questions/41836/reactions-of-alcohol-with-sulfuric-acid

The alcohol is protonated but reversibly. It is a unproductive event. Protonating the alkene is a productive process.

Both the reaction mechanisms are equally correct. Reaction takes place at numerous sites in most of the organic reactions, many kinds of products are formed, but the thing which actually give the major product is one of the intermediate, which is relatively stable than its peers.
We know, that 3° carbocation is a more stable intermediate than an allyl carbocation (irrespective of the fact that the latter is going to give a conjugated product), so the former will be present in adequate amount in order to give the major product.
Also, the formation of conjugated alkene takes place through multiple intermediates, and therefore it becomes important to care for the stability of each intermediate. I especially doubt the 3rd step, where along with a 1° carbocation, alkene is present, which acts as an EWG, and hence lowering the stability of the intermediate.
So, taking account of these factors, I would likely go with the given solution.

Edit: As per Yusuf Hasan's points on the stability of THF in comments, the given product, being a THF derivative is likely going to solvate the other intermediates and it is itself stable. That's why, THF is an unreactive compound and hence it is used as a solvent in many organic reactions. (source, under the heading "Reactivity of cyclic ethers - Oxacyclopropanes")

I guess you didn't reflect on the product viability of both the pathways. One of the paths is leading to the formation of a THF derivative, and being a cyclic ether,it would tend to be much more stable than the dialkene. Also, the ether formed as a product would also be much better at solvating the intermediates and also itself than the aqueous medium,so the reaction would again be driven forward by Le Chatlier's principle

Thanks for your answer. I don't understand how you decided that a tertiary carbocation is more stable than an allyl carbocation. In the chapter "Radical Reactions", the same book says that allyl radicals are more stable than tertiary radicals. And since, the stability of carbocation parallels the stability of free radicals, I face a contradiction with the second paragraph of your answer. It would be helpful if you could clarify this.

@GuruVishnu: in general, hyperconjugation (hypcj) is relatively weaker effect than resonance. But, in case of that 3° C+, there are two things: 8 hypcj structures + inductive effect of surrounding carbons, which overrules a single resonance (as you know, hypcj is also a form of resonance).

@YusufHasan: thanks for your comments. I'll edit my answer to include it.

@RahulVerma: I agree with your comment explaining the stability of tertiary carbocation over an allyl carbocation. So does it mean the stability of radicals don't always parallel the stability of carbocations and vice versa? I thought both being electron deficient must be governed by the same stability rules.

@GuruVishnu: Yes. My chemistry teacher says that radicals are stabilized by both, electron-donating and electron-withdrawing effects, unlike carbocations. So, in case of radicals, allyls are more stable.


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Measuring Testosterone Levels Using ELISA
I am new to this topic so please be patient. I am looking to understand ELISA better and how one may attempt to not only detect but measure the total calculated level present in a sample.
I would like to know the process and the materials required to conduct such a test, I know I need a well plate which I have but I'm unsure of the different chemicals required and in which order.
A part I am also very curious about which ties into my background is the sensor(s) used. From my understanding, a laser is commonly used a 450 nm that is shined through the sample that has reacted with colour and measures the absorbance of photons - the intensity of the colour is associated with the desired target that's present in which we're trying to quantify or measure; my question here is the laser shining on ever single individual well to get the mean?
Additionally, I am curious how the calibration sample is obtained and how it's stored for example what signature does 500 ng/ml of testosterone have that can be used as a reference point and where are these obtained.
Again please bear in mind im new to this subject and want to learn more. If you can explain this in simple terms initially and build onto the technicalities that would be great, thanks.

There are lots of protocols available online that explain how one carries out this type of experiment. See for instance: https://www.abcam.com/testosterone-elisa-kit-ab108666.html Click on "protocols" and open the pdf (Protocol Booklet). Good luck!

You can think of separate wells as separate samples.