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Finding region of electromagnetic spectrum with wavenumber
A line with wave number $\pu{1.028\times10^{-2} nm^{-1}}$ is emitted in the spectrum of atomic hydrogen. In what region of the electromagnetic spectrum does this line occur?
Do we find this by referring to the range of wavelength for UV, IR, etc., or is there a method for knowing which series (Lyman, Balmer, Paschen, Bracket, Pfund) this wavelength belongs to?

Raman spectroscopy typically uses wavenumbers, most other types of spectroscopy use the wavelength itself. Yes, knowing the wavelength you can determine if it is UV, Vis, or IR ect. Also the precision is great enough to identify the particular hydrogne line using the Rydberg formula for hydrogen. So what range? What line?

The region of the electromagnetic spectrum is all that the question is asking for. We'll give you extra credit for identifying the actual hydrogen line.

Wavenumbers are a unit of energy. Given that $1\mathrm{eV}$=$8100\mathrm{cm^{-1}}$, and the relations $E=h\nu$ and $c=\lambda\nu$, where $h$ is Planck’s constant, $c$ is the speed of light, and frequency $\nu$ and wavelength $\lambda$, you should be able to solve this with some simple algebra. Once you calculate the wavelength, try to find where it falls on the electromagnetic spectrum$^{[1]}$.

 $^{[1]}$ EM Spectrum - cyberphysics

True. What I'm trying to ask is that, is it possible to  identify which hydrogen spectrum line it belongs to so that even without knowing the range of wavelength for UV, gamma etc. we can identify where it lies in the electromagnetic spectrum. Because we already know that lines belonging to the Lyman series are UV, Balmer series are Visible light, Paschen series are Near IR and so on. What do you think?

Sure; convert the wavelength ranges to energy ranges and see where it falls. It’s all the same.


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Why the etching profile of HF and HCl on titanium surface is different?

Why the profile of etched titanium with HF-HNO3 looks like a hill, while the profile of etched titanium with HCl is U-shaped? Can you please put your source if you know it? Thank you
Note: the picture above is the profile of etched titanium with HF

What are your thoughts on this?

Really unclear what you are asking. Have you considered the sample curvature before etching? The picture above shows it quite clearly.

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Are such high oxidation states as reported by Vojovodic et al. possible?
I am a student in theoretical chemistry and I am confused about the paper: Trends in adsorption of electrocatalytic water splitting intermediates on cubic $\ce{ABO3}$ oxides (Montoya, J. H.; Doyle, A. D.; Nørskov, J. K.; Vojvodic, A.; Phys. Chem. Chem. Phys. 2018, 20 (5), 3813–3818. DOI: 10.1039/C7CP06539F), where the authors report DFT calculations on the oxides:

$\ce{MgBaO3}$ ($\ce{Mg^3+}$ and $\ce{Ba^3+}$)
$\ce{NaLaO3}$ ($\ce{La^5+}$)
$\ce{CaBO3}$ and $\ce{SrBO3}$ ($\ce{B^4+}$)
$\ce{ZrNaO3}$ ($\ce{Zr^5+}$)
$\ce{ScSrO3}$ ($\ce{Sc^4+}$)
$\ce{CaCuO3}$ ($\ce{Cu^4+}$)

... and many more. Are these oxidation states even possible?

You can DFT pretty much everything and publish a paper about it. Some people do that once a month or so. Then again, who said the oxidation states in these compounds are what you think they are?

If oxygen is formally 2-, in total 6-, then you can guess the oxidation states of the metals.

I agree that you can DFT everything, and still your calculations will tell you that the system is unstable, if you compare them with their single oxides.

Who says oxygen is 2-?

I agree; oxygen is probably not fully -2.

Just to note Cu4+ while rare is known - e.g. https://en.wikipedia.org/wiki/Caesium_hexafluorocuprate(IV)

Oxidation states are just numbers, a bookkeeping tool for chemistry. These hardly ever correspond to anything observable. Point in case: hypofluorous acid $\ce{HOF}$, see for example the oxidation state of oxygen and of fluorine in the confines of the definition. Maybe also of interst in that regard is the introduction to "oxidation state"/"oxidation number" in general.
Therefore instead of asking whether the oxidation states are stable, one should ask if the bulk/ molecular structure is stable.
When it comes to computationally aided catalyst design, you can pretty much use the entire periodic table to play with. It is a matter of interpreting the results that count, and obviously if experiment can reproduce such predictions. Where you start becomes a matter of taste and starting with perovskites is as good a guess as any.
The authors of the publication make quite an effort, calculating hundreds of potential catalysts. However, they also indicate problems within their methodology. Quoting from the supporting information:

3 Oxygen evolution data
  [...]
  \begin{array}{lcl}
\text{formula} & \text{values of }\Delta G, \eta, \dots & \text{warnings}\\
\hline
\ce{MgBaO3} & [\cdots] & \text{a,b}\\
\ce{NaLaO3} & [\cdots] & \text{a,b}\\
\ce{CaBO3}  & [\cdots] & \text{b}\\
&\vdots&\\
\hline
\end{array}
  a: One or more runs did not converge or failed in Quantum Espresso, missing adsorbates are reconstructed from successful runs using scaling
b: Changes in atomic positions of greater than 5.5 angstrom found in the slab or adsorbate during optimization, typically indicates structural instability
  [...]

And there are a lot more unstable species.
TL;DR: Basically, in these extreme cases, where you would observe abnormal oxidation states, the calculation also predicted that the bulk/ molecular structure is not stable.

Martin has answered well that the "abnormal oxidation state" compounds appear unstable, but what if they were to form?  The sticking point in the oxidation state calculation is that oxygen in combination with most other elements shall have an oxidation state of $-2$. But at least in theory, that need not be so.  Maybe some of the oxygen is in the form of the radical anion $\ce{O^-}$ rather than $\ce{O^{2-}}$, with the charges thereby balanced with the other elements in their normal oxidation states. In a proposed compound such as $\ce{CaBO_3}$ (if it were to form), for instance, the proposed $\ce{B^{4+}}$ would be so powerfully oxidizing that it would react with any $\ce{O^{2-}}$ that contacts it, forming said oxygen radical anion, until the boron gets down to the ordinary $+3$ oxidation state.

I agree with Oscar Lanzi, half of the proposed Catalysts will immediately oxide to their ordinary oxidation state. The proposal perovskite structures are unrealistic as the adsorption energies reported. They might have calculated hundreds of hypothetical oxides, but most will never be an actual catalyst.

In addition, thanks to Martin I re-read the support information and about half of the structures show computational warnings. As a student, I am surprised about another fact, for Mn, Fe and Co you need to consider the strong on-site Coulomb interaction?

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What is Hydrogen bond Auto-correlation function
Can somebody explain the Hydrogen bond Auto-correlation function in Computational Chemistry (Molecular Dynamics and Simulations) and what does Intermittent and Continuous Hydrogen bond means?
Why the graph decays over time? and it's applications?
This is with reference to the paper: Chandra, A. Effects of Ion Atmosphere on Hydrogen-Bond Dynamics in Aqueous Electrolyte Solutions. Phys. Rev. Lett. 2000, 85 (4), 768–771. DOI: 10.1103/PhysRevLett.85.768.

@D.H.N in general, you should try to edit your prior question to get it reopened rather than posting essentially the same question again.

@Tyberius i am unable to edit mt prior question since it has been locked and hence i posted the question again.

@D.H.N As far as I can see, your question was put on hold, which is not the same as it being locked. With it put on hold, when you edit the question, it gets sent to other users to vote on if the question has been improved enough to reopen.

@Tyberius if i go to the edit option of my previous post which was put on hold, it shows a pop-up which states that the "Post is locked" and also it shows "This post is locked and cannot be edited". Hence i am unable to edit that.

In eqn 1 and 2 the correlation functions are written in the form $<A B >/<A>$ , the denominator  is the average of $A$ and the numerator the correlation function. This is the the integral product $U(t_d)=\int A(t) B(t-t_d)dt$ where $t_d$ is some time delay. When $A$ and $B$ are the same thing then this is the auto-correlation. It measure how similar the signal A is to B at different times. If $A$ and $B$ are random signals (such as white noise, or from a random number generator) then $U$ is 1 at zero time delay (both signals exactly superimposed) and zero at all other delays and this is because the correlation has no 'memory'; any point in a random signal is not related to any other. If $A$ and $B$ are constant then $U$ is one at all times, i.e perfect 'memory'. In your example the H bonds have a certain lifetime because are being broken due to molecular motion (as explained in the paragraph under eqns 1 & 2) and so the correlation function decays away. A plot of the correlation gives the lifetime of the H bonds.

Since porphyrin covered the general case of a correlation function, I'll try to give a little more detail on what they were specifically modelling in this paper.
The correlations functions they define in the paper are correlations of indicator variables (a value of 1 when some criteria is true, and zero otherwise). The continuous correlation function gives the probability that a given hydrogen bond at time 0 remains all the way until a time t. The intermittent correlation function gives the probability that a given hydrogen bond at time 0 is also present at time t, even if the bond breaks and reforms in the time in between.
That these functions decay over time is a consequence of how these bonds fluctuate over time. If t=0, the correlation function is at a maximum, in this case because we know the hydrogen bond is still there. Shortly after, the function is still fairly because the bond hasn't had a lot of time to separate. At very long times, it is unlikely that particular hydrogen bond has remained in place or even reformed.


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Does sodium ethoxide (sodium ethanolate) have a boiling point, or does it decompose?
According to Wikipedia, $\ce{C2H5ONa}$ melts at 260°C, but no boiling point is given. What will happen if $\ce{C2H5ONa}$ is heated further?

I don't know but my instinct is that it decomposes perhaps with loss of ethene

Yeah, it decomposes.

Thanks. Do you have a reference, and do you know at what temperature it decomposes?

Yes, sodium ethoxide starts to decompose on heating and thus its boiling point cannot be determined. According to [1]:

The decomposition of sodium ethoxide is observed to start above 573
  K agreeing well with that reported by Blanchard et al. The gaseous
  products formed on decomposition were mainly methane and ethylene with
  minor quantities of ethane, propylene and butylene. Solid residue was
  found to be a mixture of sodium carbonate, sodium hydroxide and
  amorphous carbon. However, in an earlier study on decomposition of
  sodium ethoxide, by collecting and condensing the evolved gases in
  liquid nitrogen and analysis by gas chromatography, the authors have
  reported the gaseous products to be mainly of ethylene and minor
  quantities of propane, heptane, ethanol and water. The authors have
  also stated that the non-condensable gas at liquid nitrogen
  temperature could be hydrogen or methane. The TG trace for
  decomposition of sodium ethoxide in their study indicated a slow and
  steady decomposition right from 273 K with a sharp weight change above
  573 K which implied that the sodium ethoxide could contain some
  amount of ethanol and moisture that evaporated on heating.[...]

Reference:

Chandran, K.; Kamruddin, M.; Ajikumar, P.; Gopalan, A.; Ganesan, V. Kinetics of thermal decomposition of sodium methoxide and ethoxide. J. Nucl. Mater. 2006, 358 (2-3), 111–128. DOI: 10.1016/j.jnucmat.2006.07.003.


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ABC arrangement of FCC/CCP
I know that FCC shows ABC type but I am confused about it. For simple cubic and body centered, it is pretty easy to work out the packing arrangement but FCC is giving me a headache.
What is this Wikipedia article trying to tell in the diagram and how did choose A, B and C? What's up with the different orientations?

Can somebody tell me these in a simple way as I can't solve FCC problems based on structure.

Perhaps the simple way to look at this is think of the sphere structures as three layers. Think of the (x,y) coordinates of the spheres, neglecting the z coordinate. The first (white balls) and second layers (black balls) are the same in HCP and FCC. So let's label the first layer the A layer and the second layer the B layer.
Now we add the third layer.

In HCP, which is shown on the left in the above image, the center of spheres in the third layer would be placed directly over the center of a sphere in the A (first) layer. So the third layer (white ball)is another A layer.
In FCC, which is shown on the right in the above image, the spheres in the third layer (green ball) would be offset by 1/2 the radius of the sphere, above a sphere in the first layer. (Note that the dashed circle denotes a position directly above the A level circle.) So the third layer is neither an A layer nor a B layer. Let's call it a C layer. Now the fourth layer would again have spheres whose center are directly above the center of the spheres in the first layer. So it would be another A layer.

So HCP has (AB)(AB)(AB)... layers but FCC has (ABC)(ABC)(ABC)... layers. Where A, B and C denote different sets of (x,y) coordinates. Again the z coordinate being the layer of the ball coming out of the 2D image.

Sorry for the late activity.I am stuck at B layer......isn't the black sphere B placed like in three dimensional.I thought the B layer would be 2nd white layer from the bottom of your picture or is the entire white layer A

And how are B layers placed w.r.t A layer...if I can understand that then that will help a lot

I edited this a bit again. The (x,y) coordinates are in the plane of the paper (monitor). The z coordinated would be due to balls stacking. Does that help?

Yeah I understand but I am confused about the arrangement of B layer....I know they fill the voids of layer A but isn't it physically impossible to cover all gaps with theme......does that mean we can't predict stacking of B?

And how come the number of B layer atoms is different in these two structures?

@HydrousCaperilla - It is just a diagram. The artist only showed enough atoms to make the point. In a "real" crystal there would be a "sheet" of atoms of course.

Okay I understand this now


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Citric acid rust removal and neutralization
I'm currently removing rust from steel with citric acid dissolved in water. I generally use a wire brush as the final treatment to remove any surface rust that the acid loosened up. The pieces under treatment end up shiny before being painted over for renewed protection.
I haven't found any authoritative answer on this, but online search results seem to recommend soaking the parts in a solution of baking soda as the final treatment after the citric acid. The idea being to neutralize the acid.
What is the purpose of the neutralization step in this scenario? Does it effectively reduce how prone the acid-treated steel is to rusting again?
And assuming the acid actuates on the surface of the steel, would the wire brush treatment be a substitute for the baking soda, if  it effectively removes some of the acid-treated metal from the outer layer?
Background: Electrical Engineering. But my Chemistry is indeed very rusty (no pun intended).
This question is about why and if it is necessary to neutralize the acid after treatment, whereas Acetic(vinegar) and citric acid as the rust remover is about how to mix acids and which other acids to use. In my view, the questions are completely different, despite of the fact that they both mention the usage of citric acid.

The standard chemical treatment for rust ( when a good mechanical treatment can not be done ) is phosphoric acid with manganese. These are marketed as "conversion coatings" ; Navel jelly is a common brand.

@blacksmith37 Thanks for the reply. My question is on the acid citric treatment, but it's also good to know about the alternatives. So what does 'standard treatment' mean in this context? Does it mean it's better than acid citric, and if so, in which regard?

Is the steel stainless steel? Phosphoric acid is the usual anti-rust treatment for regular steel, but citric acid is recommended as a safe non-passivating cleaner for flash rust on stainless steel. (The wire brush must then also be stainless steel.)

@JamesGaidis: it's regular steel. Citric acid has worked well for me so far in removing rust (not just flash or surface rust) from regular steel. I'm aware of other methods (I have in fact tried phosphoric acid too). The question is rather on the neutralization step when citric acid is used, when that is the method of choice.

I believe the phosphoric acid cleans and leaves some iron phosphate which is a good substrate for paint (  over simplification , I am sure) . The citric just removes oxides ; we used it to clean steel fracture faces for examination.

Links in the comments to the following post might be helpful: https://chemistry.stackexchange.com/questions/39129/rust-removal-with-phosphoric-acid-created-more-rust#124263

See also: https://chemistry.stackexchange.com/questions/68994/aceticvinegar-and-citric-acid-as-the-rust-remover?rq=1

After removing a part from an acid solution bath, it should have no trace of the acid left on it, otherwise, it will rust quickly and aggressively. Depending on the strength of the acid, it can happen right before your eyes.
Flushing the part with water often suffices, especially for weak acids. To be on the safe side, however, it is best to soak the part for a minute or two in a weak base solution, such as baking soda and water.
This works due to basic chemistry. Back in chemistry class, we learned acids and bases neutralize each other, resulting in a chemical that's less reactive than either the acid or the base.  It's the same principle when you pour vinegar on baking soda and it fizzes violently. The vinegar is an acid (acetic acid) while the baking soda (sodium bicarbonate) is a base. The final result of the reaction is water, carbon dioxide, and sodium acetate.


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Reaction of alpha hydroxyketone with KOH
What happens when 2-hydroxycyclohexan-1-one (an alpha hydroxyketone) reacts with aqueous $\ce{KOH}$?

I tried something like Favorskii rearrangement, but it's wrong because OH isn't a good enough leaving group.

Deprotonation, ring expansion to CSID:9972, http://www.chemspider.com/Chemical-Structure.9972.html (accessed 15:03, Jul 4, 2018)? Maybe?

Is that all? Are you sure?

Oh no, not sure at all; hence a comment, hence question marks and the word maybe. There was no sarcasm meant by any of this, just offering the first thing that came to my mind until someone more versed in synthetic chemistry comes along.

Who says anything happens? Aq KOH isn't really strong enough to deprotonate the alcohol.

Well alpha hydroxy ketones in the presence of a Base or acid will tautomerise and thus  have their OH converted to aldehyde /ketone and vice versa. Though I am not sure how that would help since in this case you get the same compound anyways

All you get is tautomerism with base through the enediol, which allows the hydroxyl and ketone to switch positions. Net result: nothing!


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Conjugate addition vs Electrophilic addition
I was just doing an organic chemistry problem from an online source recently, on the topic of conjugate additions (or 1,4-additions). Below is an image illustrating the problem.

I chose the upper route while the solution manual claims that the answer is the lower route. My rationale for choosing the upper route was that perhaps the alkene is not nucleophilic enough to react feasibly in an electrophilic addition with $\ce {HCl}$. This is due to the presence of the electron-withdrawing carbonyl. Thus, it is important that we first get rid of the carbonyl by using the ethyl Grignard. Then, we can easily add the $\ce {HCl}$ for the electrophilic addition.
Upon reconsidering my reasoning, perhaps I was too narrow-minded in saying that the $\ce {HCl}$ must add in the fashion of electrophilic addition. Perhaps, conjugate addition could also take place. And also, by choosing the upper route, I cannot assure that my chlorine atom would end up in the desired position due considerations of the stability of the carbocation intermediate. Thus, I agree that my choice is incorrect.
The correct choice is the lower route and this does ensure that the $\ce {Cl}$ is in the right place in the final product, as it adds via a conjugate fashion. But how is the writer of the solution so sure that the $\ce {HCl}$ would add in a conjugate fashion, instead of via an electrophilic addition?

Grignards unless modified with Cu(I) salts add 1,2 not 1,4 so the EtMgBr will add to the ketone to give the t-alcohol and leaving the double bond intact.
Consider then the HCl step - if it is strong enough to protonate the double bond leading to Cl addition then it will protonate the t-alcohol in preference leading to elimination or t-Cl formation. So the answer is to do the reaction on the double bond first then the Grignard addition. Protonation will occcur at the carbonyl oxygen but you have to remember that the double bond is in conjugation with the carbonyl system. This polarises the double bond, leading to Cl- attack at the end of the system.
The Grignard is more nucleophile than base so you will get majority addition to the ketone though some deprotonation/elimination may occur depending on the reaction conditions.

Ok thanks for the insights! But I would like to know approximately how much of it would be electrophilic addition of HCl across the double bond and how much of it would be conjugation addition with Cl-

I doubt there would be any HCl addition across the double bond as that is not the site of protonation, the carbonyl oxygen is. You simply cannot think of the double bond in isolation, its reactivity is completely altered by the carbonyl in conjugation. If you are happy with the answer please accept it


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Why does chlorination of toluene stop at two chlorine atoms?
My instructor mentioned a reaction for preparation of benzaldehyde by chlorination of toluene in presence of sunlight to form benzal dichloride, which on hydrolysis gives benzaldehyde.
However, I'm not sure why the chlorination of side chain should stop after two chlorine atoms are added? At what step does the side chain chlorination of toluene stop?

It doesn't stop, it just gets a lot slower the more chlorine atoms you add. From the abstract of an article titled Photochemical side-chain chlorination of toluene[1]:

Where $k_1$, $k_2$, and $k_3$ are the rate consts. for the successive reactions: $\ce{PhMe}$ → $\ce{PhCH2Cl}$ → $\ce{PhCHCl2}$ → $\ce{PhCCl3}$, resp.; $r = \frac{k_1}{k_2}$ and $s = \frac{k_2}{k_3}$, values of $r$ and $s$, resp., were 6.0 and 5.7 at 100° and 8.4 and 8.5 at 40°, for batch operation, with light varying from ultraviolet to yellow.

[1] Haring, I. H. G.; Knol, H. W. Chemical and Process Engineering (London) 1964, 45 (10), 560–567.

It will not stop at the dichloro compound, it is a free radical halogenation reaction.
The benzyl radical is very stable so it is easy for a chlorine atom to abstract a hydrogen from the methyl grup thus forming a benzyl radical. The benzyl radical then reacts with a chlorine molecule to form a chlorine atom and a molecule of benzyl chloride.
This process can be repeated again to form $\ce{C6H5CHCl2}$ and then $\ce{C6H5CCl3}$. It would be a good idea if you were making benzadehyde on the industrial scale to distill the chlorinated mixture to separate toluene, benzyl chloride, the dichloro compound (dichloromethylbenzene) and the trichloro compound (trichloromethylbenzene).
You can then react the dichloro compound with water to get the target molecule.


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Cyclisation reaction through hydrazine
The following reaction confuses me - I don't understand how and why it happened (couldn't figure out the mechanism):

I thought this is some Wolff Kishner sort of reduction, but the reagents don't match (no alkaline medium). I am not sure how hydrazine lead to cyclisation - for starters, I guess it added to the carbonyl oxygen (aldehyde, as it is more reactive) as an ammonia derivative to form a hydrazone. What's next?

What's next in a further nucleophilic attack by the NH2 at the other end of the  hydrazone on the ketone

I did that, what happens then?

Elimination of water to form the product drawn

Could you please add the mechanism as an answer? I'm not able to figure out how dehydration would occur after internal nucleophilic attack by the nitrogen on carbonyl oxygen. Thanks.

It's essentially just the same reaction once and then over again. The basic medium of the Wolf-Kishner reduction is to remove the hydrazone hydrogen, releasing the $\ce{C=N}$ electrons onto the carbon. In this reaction, the hydrazone instead further reacts with the other carbonyl to form an aromatic 6-membered ring.


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What would be the reaction equation between ammonium nitrate and a powdered metal?
The SDS for ammonium nitrate says it will react with powdered metals spontaneously. What would be the reaction equation, say with iron or copper?
This mixture is used in various pyrotechnics, and it is intentionally allowed to combust with heat and shock, but how does the reaction proceed spontaneously as described in various SDSs in the absence of any ignition source? Is this due to the mix attracting small amounts of water, allowing the nitrate to attack a metal instead?
Would the theoretical products always just be various oxides of the metal involved?

How about your own attempt at an equation?

Iron or copper. I'm assuming they end as oxides? But does the reaction proceed differently when it starts spontaneously?

I'm betting they mean that you can inadvertently make an explosive (technically a blasting agent) or highly flammable mixture by mixing powdered metal and ammonium nitrate.  I used to be a hard rock miner many centuries ago and was well read on explosives.  Aluminum powder and ammonium nitrate is a classic solid rocket fuel and industrial blasting agent/military explosive, for instance ammonal in bombs.  Ammonium nitrate/aluminum powder mixes (magnesium works, too) have better explosive properties than the usual mining ANFO (ammonium nitrate/fuel oil) mixtures, but the latter are preferred usually since ANFO is less expensive.
While blasting agents technically need a booster explosive to detonate, they will burn ferociously.  I'm guessing when a writer of a SDS sheet talks about powdered metal and AN, they are thinking about the potential of creating an extremely flammable mixture.
I looked at some MSDS for ammonium nitrate, and I see powdered metal listed along with reducing agents as incompatible mixtures.  Could you post a link to the MSDS you are reading?  I'm interpreting 'incompatible' as referring to explosive potential and not to a slow spontaneous reaction.
EDIT:  Well.  The usual metal powders that react with AN used in mining and explosives are aluminum and magnesium, so that is all I considered.  However, a bit of research shows that powdered zinc does react with ammonium nitrate, as in this video (note that it didn't actually ignite until set afire, but it definitely reacted).

http://www.sciencelab.com/msds.php?msdsId=9927336

That SDS has a section saying "can self-ignite/detonate when in contact with powdered metals". I'm assuming that when this particular metal powder and AmNitrate mixture is used intentionally as a pyrotechnic, it's the igniter that sets the whole mix off by decomposing the AmNitrate releasing O2, which sets off the reaction by rapid oxidization of the metal. So I'm a bit confused as to how it could spontaneously occur as described by the SDS, and whether that reaction would be any different to start. My guess is that there would need to be moisture involved?

What would be the reaction equation, say with iron or copper?

It is straightforward stoichiometry to do an equation for a simplified reaction that only considers the main products. For your copper example:
$$\ce{NH4NO3 + Cu -> N2 + 2 H2O + CuO}$$
Copper hydroxides will not form, because they are unstable at the high temperatures of this reaction.
However this oversimplifies things quite a bit for three main reasons.
Firstly, in an accidental fire it is unlikely to be a balanced mixture so one ingredient or other will be in gross excess, which increases the likelihood of side reactions.
Secondly, reactions of this type are often violent, which means they tend not to reach equilibrium. Hence there can be small traces of all sorts of weird things, such as even some free hydrogen from:
$$\ce{H2O_{(g)} + Cu_{(l)} -> H2_{(g)} + CuO}$$
Why doesn't the hydrogen react with more nitrate, or oxygen from the air? Mostly it does, but in a fast, violent reaction a small amount escapes unreacted.
Thirdly, ammonium nitrate disproportionates when it is heated. This releases a complex mixture of nitrogen oxides, water, and nitrogen oxyacids. The reaction starts at fairly mild temperatures but both the rate and the mixture of products depends on the temperature.
Some of these volatiles will react with the hot metal powder, but some will escape the reaction zone. The amount that will escape will depend on how intimately the powders are mixed, the granule size distribution, and the thickness of the pile.

This mixture is used in various pyrotechnics,

Is it? So far as I know, even though ammonium nitrate is a fairly good oxidiser it is avoided in pyrotechnics because it is too hygroscopic. That gives unpredictable performance and makes storage difficult.

how does the reaction proceed spontaneously as described in various SDSs in the absence of any ignition source? Is this due to the mix attracting small amounts of water, allowing the nitrate to attack a metal instead?

Yes, very probably. If fine powders are intimately mixed, quite small amounts of moisture may allow the reaction to start in the liquid phase even without energy input. However there are other ways it could start: direct application of heat; electrostatic discharge; local heating by friction or impact. Depending how sensitive the mixture is, some of these stimuli might be small enough that the event appears spontaneous. This will depend on the particle size distribution (smaller is more sensitive) and the reactivity of the metal: fine zinc powder is likely to be very much more dangerous than coarse copper turnings.
Either way, under appropriate conditions the reaction will accelerate:

If a sufficiently large zone is reacting, then heat is being generated (proportional to volume) faster than it escapes (proportional to area);
This causes the temperature to rise;
This accelerates the reaction rate, increasing rate of heat production;
Feedback loop occurs until the temperature is so high that heat loss and heat production is balanced (or the fuel is consumed, whichever occurs first.) For metal/oxidiser reactions, this can be thousands of degrees.
At some point, the metal may melt (most of them) or even boil (zinc), which will change things from a gas:solid reaction to gas:liquid or gas:gas, which is much faster.

Would the theoretical products always just be various oxides of the metal involved?

I guess you mean the solid products, since gases are also certainly released. In general oxides are likely to be the main product since they are usually thermodynamically favoured at the high temperatures of metal/oxidiser reactions. However it depends on the metal. Other possibilities include peroxides and hydroxides. In a few cases, there may also be some nitride formed.


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Are detergents more efficient if poured into more water?
Let's say, I am washing my cloths "by hand". My steps would be:

I take a basin of about 20L
I pour into it some water
I pour my detergent (the required amount, about a spoon) and a bit of bleach
I put my cloths (for a volume of about 3L) and stir a bit.
Then I let everything sit for 6 hours. After that, I rinse my cloths with fresh water.

My question is: will my cloths be cleaner, if, in step 2, I pour 15L of water rather than 5L (enough to "sink under water" all the cloths for about 2 to 3 cm) ?
Arguments in favor of 5L: the detergent and the bleach will be less diluted (but does it really mater for the bleach and for the detergent?)
Arguments in favor of 15L: the detergent will have more water to react with, to remove the stains (but does it really need 15L rather than 5L?)

The detergent does not need anywhere near that much water to react with.

@IvanNeretin Are you saying not anywhere near 15L? Or 5L ? If yes, how much water does the detergent need?

Just enough to dissolve it, if it is solid, or none at all, if it is liquid. The problem is that you'd have hard time getting all your clothes in contact with that tiny amount of liquid, hence we need to add some more, even though it would make the detergent somewhat less active. 5L is a reasonable optimum.

How is this question opinion based? OP is looking for "answers based on facts, references, or specific expertise.", which is contrary to what the closure notice says.

The action of a detergent on the water is to lower its surface tension so the solution can wet the oily dirt in the cloth. The action of the surfactant on the oily dirt is to emulsify it by encapsulating it so it can be dispersed in the water. The amount of surfactant needed for this depends on the amount of oily dirt: more dirt requires more surfactant to emulsify it - and intense mixing helps.
Stirring a lot (not a "bit") will be much more effective in emulsifying the dirt and suspending it in the water than standing for 6 hours.
Then rinsing carries away the emulsified dirt. Some of it, anyway. More rinsing will carry away more dirt.
If you are limited by the amount of water available, a starting point would be to use one third of the water to wash, one third for a first rinse and the last third for a second rinse. Squeeze the cloth as dry as possible after wash and after rinses to remove as much of the dirty solution as possible. If you have unlimited water, use a lot of water at first, then a lot more to rinse, and rinse several times, but squeeze dry after washing and after each rinsing.


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Calculate concentrations in a system made up of molecules in order to calculate ionic strength
I understand how to calculate ionic strength where there are concentration data in molality or molarity units available and where the system is made up of only two components.  However, I have been struggling for a few days with what is probably a very simple problem: how to calculate ionic strength where there are more than two components and where the concentrations first have to be calculated because the system is made up of molecules.  I'm not a chemist, but need ionic strength for another calculation.
In one kg of water there are 55 moles (1000 g / 18 g mol-1) of water molecules.  To simplify calculations, my system has 55 water molecules. As an example, let's say it also has one Na atom, three Cl atoms, and one Ga atom.
I think that this (below) is the expression to use, where z is charge and m is the molal concentration of species i.  I want to use a molality-based expression because I want to do calculations at a range of temperatures and expect that it would not be right to use molarities, where temperature can affect volume.
$$\large I_\mathrm{m}=\frac 12 \sum_i z_i^2 m_i$$
So it looks like I only need to know the charges and m for all of my species.  But here's where I'm stuck on something simple.  How do I calculate m?  Is it true that in my system, there would be one mole of Na and Ga and 3 moles of Cl?  So m for Na (for example) would be 3 moles/1 kg water?
Another thing I'd like to check that I haven't been able to find out online is whether the "1/2" in the expression is only valid for two-component systems.  Is this expression valid for my system?  Thank you for any guidance.

Update:
I have realized that I have been confused.  I want to find out which reactions happen in my system at different temperatures.  As part of that, I want to find out the activity coefficients of the various species involved and the standard state free energies.  This requires knowing the ionic strength.
I was confused by something I read online and basically thought that my ionic strength calculation for each temperature would only involve the species participating in the reactions happening at each temperature.  But I now realize (hope) that actually, the ionic strength is the same at ALL of the temperatures I'm dealing with because calculating ionic strength means accounting for all species present in the system.
So if I'm now on the right track, my ionic strength would be 6.5 mol kg-1 based on using the equation above with concentrations of Ga, Na, and Cl of 1, 1, and 3 mol kg-1, respectively, and charges of 3, 1, and -1.

(1) Yes you are right. $m_i$ is the molarity (or molality) of ion species $i$. (2) The 1/2 is because there are cations and anions irregardless of the relative numbers. (3) You do realize that this only partially works. The gist is that the moles of solute should be small compared to the moles of solvent. (4) The $I_m$ means that you are calculating the ionic strength of compound $m$. You can have solutions with more than one solute too.

Thank you, Max W; that is helpful.  Please forgive my ignorance of your point (3).  I don't understand what you mean.  Are you saying that in a system with more Na, Ga, and Cl, there would be interactions like repulsion (for example) that would complicate things?  Do you have any thoughts on whether the expression would lead to valid ionic strength results for my system?

The equation defines ionic strength. The problem comes in trying to use that to predict some behavior like the solubility product of a salt. The "reactivity" of a ion doesn't depend on its molarity (or molality) but rather on its chemical activity. So the overall ionic strength of the solution gives a means to correct for the activity of an ion from a particular salt that is dissolved or in a mixture of salts. See Debye–Hückel theory. But is doesn't work for concentrated solutions. So AgCl has a $K_{sp}$ but NaCl doesn't.

Thank you for that. I'm going to have to take time to read what's at your link in order to understand your final sentence.  I do need to use the ionic strength to calculate activity coefficients, so it sounds like I need to consider whether my solution is too concentrated to do that.  If anyone has any further advice/simplification, I'm all ears.  Thanks a lot, Max.

The point that I was trying to make is that you could use ionic strength to make corrections for the $\ce{a_{Ag+}}$ and $\ce{a_{Cl-}}$ so that the $K_{sp}$ was still useful in a KCl solution, or a CaCl2 solution. However trying to theoretically figure out how much NaCl will dissolve in a KCl solution, or a CaCl2 solution is hopeless.

OK, that's interesting.  Thanks.  I still feel like I'm missing something.  There must be a reason why it's OK to use ionic strength to calculate activity coefficients, but not figure out the amounts of something that will dissolve so I'll have to read up on that.  I'm a bit afraid that my system is too concentrated to get robust ionic strength results.  Tomorrow I'll go ahead and do my calculations and will post the results here in case anyone is interested in taking a look at whether they're valid.  Really appreciate your help.  Good night.

MaxW, if you would like to add an answer, I can mark it as accepted so you get credit.  Thank you very much for your help.


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What is the proper way to prepare a solution aiming to minimize errors?
I am a undergraduate student in chemistry, and my university is very weak in teaching solution preparation. I need to prepare a 10 mg / L methylene blue solution and now the way I do is, using an analytical balance of 5 decimal places, I tare a 50 mL beaker and simply add the 10 mg of methylene blue, trying be as accurate as possible (the amount is very small due to the large molecular weight). Afterwards, I add deionized water to a certain volume and transfer to a 1 L volumetric flask using a funnel (I wash the small beaker several times to guarantee that I transfer as much as possible of the compound) and complete with deionized water until the meniscus line. And so, I shake it many times.
Is this procedure correct? If not, how can I prepare this solution avoiding errors?

I like the final part about shake many times, years ago when I was a young academic I had a student who put an iron solution in a large volumetric flask. The student then added distilled water and topped it up to the mark. The student then used a small short pipette to take a sample. The student could not understand why there was no iron in the sample they had taken. I then told the student to give the volumetric flask a mighty shake. Then they could find iron in the liquid.

@Macro G. If you are to make a solution of 10 mg/liter, then what does the molecular weight have to do with the mass? Why would you say that the amount is very small due to the large molecular weight? The amount is 10 mg which is not "very small".

@Dr.J. I think I didn't make it clear enough, my mistake. What I was trying to say is that due to the fact that methylene blue has a large molecular weight, the amount of actual solid particles I have to take from the reagent bottle and transfer it to the beaker is so little that it is hard to nail the mark of 10 mg in the balance scale. English is not my native language, sometimes is hard to express what you really is thinking. Thanks.

Methylene blue isn't a primary standard so I'm not sure how much accuracy/precision you need.

Typically when wanting the best accuracy the solid would be dried in
a hot oven, then cooled in a desiccator.
Do not to handle the beaker with your hands. Even if it has been
"left out" in humid air that amount of water would be stable. So use
tongs to pull in and out of balance.
Another option is to weigh more and do a serial dilution. So 1 gram
in liter, then 10 ml of that solution into 1 liter.

Not that much of precision since it's going to be used for plasma degradation tests and the kinectics will be measuring its rate of degradation by time, (Co-Ct)/Co vs. Time. (Co being its initial concentration, Ct the concentration in a given time)


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Something about the Roothaan equation
Why we cannot solve Roothaan equation
$\mathbf{F} \mathbf{C} = \mathbf{S} \mathbf{C} \mathbf{\epsilon}$
by just move S matrix to the left, as
$\mathbf{S}^{-1} \mathbf{F} \mathbf{C} = \mathbf{C} \mathbf{\epsilon}$ as a new matrix $\mathbf{F}'$, and then solve the eigenvalue problem
$\mathbf{F}' \mathbf{C} = \mathbf{C} \mathbf{\epsilon}$
why we cannot do that?

We could. Mathematically there is nothing wrong with that - the overlap matrix is positive definite and therefore must have an inverse, and so your manipulation is perfectly correct mathematically. The problem is that $\mathbf{S}^{-1} \mathbf{F}$ is not a symmetric (or in the more general case Hermitian) matrix. Computationally it is MUCH more difficult to find accurate eigenvalues and eigenvectors of a general non-symmetric matrix than a symmetric one. Thus, when solving the problem on a computer, if one can cast one's algorithm in terms of diagonalisation of a symmetric rather than a non-symmetric matrix you are likely to get more correct answers, quicker. And this is what the standard manipulations achieve.

The biggest problem is, that $\mathbf{F} \mathbf{C} = \mathbf{S} \mathbf{C} \pmb{\varepsilon}$ is not an eigenvalue problem.

It is a generalized eigenvalue problem, see e.g. https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix#Generalized_eigenspaces , http://fourier.eng.hmc.edu/e161/lectures/algebra/node7.html , https://epubs.siam.org/doi/abs/10.1137/1.9780898717808.ch6 . Algorithms do exist to solve it directly, and may be more efficient than reducing it to the standard eigenvalue problem, for instance the QZ algorithm, but in Qchem reduction to the standard form via some factorisation of S is the most common

Sorry, but I don't understand where you are going with this. I know that it looks like an eigenvalue problem, but since the Fock operator, and therefore the Fock matrix, depend on their own solutions, it is not. You will in any case have to solve this equation iterative, a transformation of the matrices involved won't change that.

I'm assuming that what is being asked here is about each step of the SCF procedure, i.e. for a given F how do we generate our next set of eigenvectors, not for the procedure as a whole - which as you say is a non-linear eigenvalue problem and requires an iterative solution

Ah, that clears up quite a bit, and I agree. When I read solving the Roothaan equation, however, I immediately think about finding the self-consistent version of the equation. The question might be a bit unclear in that regard.


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At what temperature can molecules no longer form?
// hyper simplistic terminology in use for the question - apologies.
Molecules are formed by atoms bonding together, and those bonds have an equivalent energy.
At what temperature does the thermal energy of an atom exceed the bond energy, preventing it from forming any molecule?
I think an equivalent question is "what is the temperature needed to spontaneously dissociate any molecule into its component atoms?"

Depends on the molecule. Some are not stable above -200 C; others seem to be hold it together even at 4,000 C.

$10,000^\circ;\rm C$ is more than enough for everyone.

What Ivan meant, I think, is that you should ask for "what is the minimum temperature needed..."

That depends what's the pressure and what you call molecule.

The bond will break if you can put more than its dissociation energy into it; $\ce{CO}$ has a huge dissociation energy of $\pu{1080 kJ/mol}$, $\ce{I2}$ less than most and is only $\pu{151 kJ/mol}$. This energy can be easily obtained using a laser of the correct wavelength. (You will need to convert units to electron volts to get wavelength needed. $\pu{1 kJ/mol} = \pu{1.036E-2 eV}$, $\pu{1eV} = \pu{8065 cm^-1}$).
However, when we excite using a laser we only use only a small range of energy, using heat at a given temperature is different as temperature is an average measure of a distribution of energies given by the Boltzmann expression $n_i \approx \exp(-E_i/k_\mathrm{B}T)$ where $n_i$ is the number of molecules with energy $E_i$ at temperature $T$, $k_\mathrm{B}$ is the Boltzmann constant. If  dealing with molar units replace $k_\mathrm{B}$ by $R$, the gas constant.
Thus, even at room temperature ($\pu{300 K}$),  there are, in principle, some molecules of $\ce{CO}$ that will be decomposed  even though the dissociation energy is so large; the chance of being decomposed is $\exp(-1080000/(300\times8.314) \approx 10^{-190}$. It is utterly minute even for $6\cdot 10^{23}$ molecules in a mole, but for iodine $\approx 10^{-27}$, still minute so these molecules are stable at room temperature. At $\pu{1000 K}$, 1 in $\approx 10^8$ $\ce{I2}$ molecule decompose and $10^{-4}$ at $\pu{200 K}$ so this is appreciable over a period of time.
(Strictly we should integrate from $T$ to infinity to get the exact answer but the exponential decays so quickly just one value gives us a good idea)

I've plugged in the "10,000C is more than enough for everyone" comment (as 10000K) into the formula exp(-DissociationEnergy/(T * 8.314)) and get 0.2728 for CO. So even at 10,000K, only 27% of CO molecules have dissociated. Is that correct?

Not quite, you need to multiply this by the number of collisions that give molecules  that energy , in other words calculate the rate constant and then the lifetime of the decomposition to work out how quickly they would decompose. The simplest estimate is an Arrhenius expression, much more sophisticated is called RRKM theory.


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How far-reaching is the effect of conjugation?
Due to conjugation, it is possible for 1,4-addition, or conjugate addition, to occur. I was thinking, since conjugation and resonance effects do no diminish with distance, unlike inductive effects, does that mean that we can indefinitely conjugate and even have 1,6-addition? And even 1,8-addition and so on? My chemistry teacher mentions that 1,6-additions are possible but is there really no limit as to the effects of conjugations? Perhaps, there is increasing "resistance" as the carbon chain lengthens, like how a longer wire would have a larger resistance? I wonder if that would be an appropriate analogy.

Roughly speaking, there is no resistance.

1,6-additions are indeed possible (1) but since you have competing 1,4-addition and 1,2-addition it becomes more difficult. To push it to the 1,6-addition bulky organometallic catalysts are used. Reaction pathways are also determined by the electrostatic interactions, and these diminish the further away you get from the carbonyl.
(1) https://onlinelibrary.wiley.com/doi/epdf/10.1002/adsc.201601342, https://www.organic-chemistry.org/abstracts/lit1/353.shtm


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Up to which temperature and pressure can dendrimers (PAMAM) be stable?
I am preparing PAMAM (poly amido-amine) dendrimer and I want to know up to which temperature and pressure can it be stable?


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If all intermolecular forces are electrostatic in nature, why don’t large non polar molecule dissolve in water?
If I have a large non-polar substance A that forms strong dipole-induced dipole force (DIDF) with water, with strength comparable to that of H-bonds between water molecules, will A dissolve in water?

If you have a cockroach with strong hands, comparable to those of a man, then yes, your hypothetical cockroach could probably beat a man.

The point is that non-polar molecules form only much weaker bond with water because they can form only dipole -induced dipole interactions (which are way weaker than hydrogen bonds in water).

Solubility is strongly dependent on entropic concerns.

@Zhe - solubility at low solute concentrations is strongly dependent on entropy. High solute concentrations (such as in solid alloys) have significant enthalpy contributions.

@JonCuster If we're concerned about something being soluble at all, then we're clearly working in the low concentration regime.

Dipoles can be induced, but they are not permanent. You’re basically asking why a spring doesn’t like to stay compressed when you push on it—it's because that’s not its most stable position. The electron density around  hydrogen in a $\ce{C-H}$ bond is much higher than the hydrogen in a polar $\ce{O-H}$ bond, so electrostatic repulsions overwhelm any attractions that may be induced. See the diagram below$[1]$:

$[1]$ Bond polarity diagram - Iverson Group UT Austin

Don't confuse the strength of interactions with the underlying nature of the interactions
Yes, intermolecular forces are all caused by electromagnetic fields (your title says "electrostatic" but this is slightly wrong as some forces are also caused by interactions you might call magnetic. But the underlying forcefield is the same either way.)
But just because the forces are electrostatic doesn't mean they will be strong. Some interactions involve oppositely charged ions interacting  and they are usually strong (this is easily visible in bulk electrostatic experiments like charged balloons sticking to a wall). Next in strength are the interactions where molecules are not charged but have strong dipoles because of (to simplify a little) some degree of charge separation internal to the molecule (these are caused because some elements hog their electrons much more strongly than others). Some of these are so strong and unusual we give them special names: hydrogen bonding (basically a strong and directional dipolar effect with some extra pixie dust because hydrogen is such a light nucleus). Essentially every molecule that is not symmetric will have some dipole, though many will be very weak (asymmetric hydrocarbons have very small dipoles but chlorocarbons have relatively strong ones). But non-polar things (even atoms of noble gases) eventually stick together enough to liquefy. The dominant forces here are van Der Waals forces. These are still electromagnetic attraction based on dipoles but the dipoles are caused by quantum fluctuations in the electron density in the atoms or molecule's bonds. These forces are fairly weak compared to ionic or dipolar interactions.
It is also worth remembering that everything has van der Waals forces but in polar molecules they are just a lot weaker than the dipolar interactions.
So, although the underlying cause of attraction is always something to do with electromagnetic forces, sometimes those forces are very weak. Crudely, they may be too weak to make something soluble.
But solubility is a little more complicated than that. It isn't just about the nature of the force it is also about the detailed mechanism of solubility. For something to dissolve in water you don't just need a strong attraction between the molecule and water, the interaction has to be strong enough to compensate for the inherent attraction of water molecules to each other (which is pretty strong). van der Waals interactions are usually too weak by themselves (even when induced by a strongly dipolar water molecule) to create a strong enough attraction. And, when you take the loss of water-water hydrogen bonds into account if you stick a large non-polar thing into the middle of a bunch of water molecules, the whole equation looks even less favourable for solvation.
If you had something where the induced attraction was as strong as water-water hydrogen bonds, it might well be soluble in water. But that hypothetical compound is extremely unlikely to exist because of the relative strengths of the induced versus the inherent dipolar interactions (which is even worse when you take account of the large number of hydrogen bonds you have to break to fit a large molecule into solution).

Answer edited; In a polar solvent non-polar molecule will have dipole - induced dipole energy as well as some induced-dipole, induced-dipole energy which is smaller. The enthalpy of dissolving a non-dipole molecule in water, such as cyclohexane, is small, some types of molecules have have a small negative or some a small positive value, so the fact that these type of molecules do not dissolve to any great extent is due to entropic effects because $\Delta G = \Delta H-T\Delta S$ depends on entropy change when $\Delta H $ is small.
Mixing is always entropically favourable ($\Delta S$ positive) so something else must be causing the entropy to become negative. This is due to the fact that water has  hydrogen bonds and these are used to bridge any 'intruder' molecule. This limits their geometry and hence make entropy negative and so $-T\Delta S$ positive and free energy positive.


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Competition of H2O and NH3 leaving groups in nucleophilic acyl substitutions
I've learned in several textbooks, that in the nucleophilic acyl substitution $\ce{H2O}$ is a better leaving group than $\ce{NH3}$. This can be explained by the "thumb rule" of the $\mathrm{p}K_\mathrm{a}$. $\ce{NH3}$ is more basic $(\mathrm{p}K_\mathrm{b}(\ce{NH3})= 4.3)$ than $\ce{H2O}$ ($\mathrm{p}K_\mathrm{b}(\ce{H2O})=15.4)$ and therefore a worser leaving group.
But what is the reason for the fact, that the following common reaction happen (and the worser leaving group has left the molecule)?
My idea: Is it more a kind of "equilibrium discussion"? The reaction can happen but the equilibrium lies not really on the "product side"? Compared to another nucleophilic acyl substitution (e.g. water with a acyl chloride) this equilibrium lies far more on the educt side. But nevertheless, this reaction happens and can be exploited for example in the Strecker Synthesis through several methods (Le Chatelier, etc.)?
Additional idea: The excess of water can push the equillibrium to the product.

This reaction only occurs under forcing conditions (conc HCl, heat) so it is not correct to think of it as comparable to acyl chloride + water

@Waylander With my comparision I wanted to outline the difference reactivity. You have to "force" the equibrillium with harsh condition (as you state) to react and undergo this rather "unhappy" Substitution.


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Why is propanediol soluble in water?
I'm a crafter who likes to make her own body oils and butters but I have no background in chemistry. Now, I just read something about propanediol (which I'm considering for modifying the feel of my anhydrous product) and on one site it says that in appearance it is a clear colourless  oil but then it also mentions it's solubility is in water.
Confused.
So, I want to know how it can be an oil and be soluble in water, but more importantly, I want to know if this means it's not soluble in oil and hence I can't use if for my anhydrous product?

To look like an oil and to be an oil are quite different things. That said, it may or may not be soluble in actual oil; there is no way to tell in advance. Chemistry is an experimental science, they say.

There is no strict definition of what is "an oil". In this case it is merely a descriptor saying that it is a clear mobile liquid

Okay, so, is propanediol oil/fat soluble at all?

Its not an oil but a di-ol in other words has two -OH groups in it , ethanol for example has one -OH group, glycerol has 3.  As it  is a form of an alcohol (a more or less generic name for molecules with OH groups) it is soluble in water but it is also viscous, as a result of its chemical structure, and as it is viscous it feels a bit like an oil.

This question requires some additional specification before it can be answered.
Propanediol is ambiguous as a chemical name. There is 1,2-propanediol and 1,3-propanediol, and these are different compounds (namely, structural isomers).
Though either way, if you check references, you will see that both compounds are miscible with water, so the solubility is infinite.
See: 1,2-propanediol and 1,3-propanediol.
The term "oil" simply refers to the fact that it is a viscous liquid at room temperature.
Unfortunately, generalized solubility is hard to determine. 1,2-propanediol appears to be miscible in several different polar solvents, so you might be able to dissolve in something anhydrous.

Thanks, Zhe. The one I'm referring to is 1,3 propanediol. Here's the site I was checking it out on: www.lotioncrafter.com/propanediol-1-3.html


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Disposal of methanol solution
Can I add a 4 L bottle of methanol solution with 0.1% formic acid into a drum containing non-halogenated waste?

Immediately contact your local ES&H personnel. They should know all appropriate local/regional/country laws on disposal. Note that mixing acids and organics is generally not a good thing, and there may be further issues with whatever else is in the drum. Get professional advice right now, and not from random folks on the internet.

Seems you're asking rather if it's a good idea, which may or may not be the case and we can't know it in general, as it depends on situation. Asking this on internet probably is not good idea. That being said, much worse things go to drain around the world, and you seem to worry too much.

Jon is right, while I know that I am a nice person and a decent chemist. You do not know the people here in person. I would trust your local health, safety and environmental persons more than random folk off the internet.

This is a very fact specific question. Dependent on the contents of the non-halogenated waste and what waste streams your facility has established.

If there is nothing else (bad) in your solution, dilute it in 10x the amount of water and pour it down the drain. I mean definitely ask the responsible people, but that's what I'd say if I was responsible. 0.1% of formic acid is irrelevant, methanol is harmless unless you drink or put your hand in it.

@Karl - that would be illegal where I am.


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P2NP synthesis ammonium acetate catalyst?
I am going to try to prepare some P2NP from benzaldehyde and nitroethane following this procedure:
The aldehyde (5g), 5 ml. of the nitroalkane, and 2g of ammonium acetate were added to 20 ml of glacial acetic acid. The resulting solution was refluxed for two hours and then poured into ice-water. If a solid product was obtained it was collected and recrystallized from methanol, ethanol, or acetic acid. If the product was an oil it was separated and crystallized, if possible, from one of the above solvents.
However, I got a bottle of reagent grade ammonium acetate from a chem supplier and it is very wet. To be fair, they never marketed it as anhydrous, though.
Does that matter when I'm not using a dean-stark trap? If so, how much will it affect the yield, and is there a way accessible for a hobby chemist to dry it?
What hydration state it is in - like is it trihydrate? This is important in order to calculate the amount needed. I guess 2g of anhydrous is needed in the procedure.
Update:
10g Benzaldehyde was poured in 50ml Sodium bicarbonate water solution, stirred and left to settle down for 30 min or so.
3.5g of very wet ammonium acetate was measured - it was foolishly assumed the salt is trihydrate and calculated that 2g salt is present in 3.4g trihydrate. However, that was left for 30min on the open as an experiment and picked up another 500mg of water, so this calculation is more or less incorrect. Some of the water can be removed by pressing against something like a toilet paper btw.
20g of GAA was added to 500ml RBF. The acetate was added to the acid and shook for a few minutes until fully dissolved. 5.5ml Nitroethane and 5ml from the washed Benzaldehyde were added.
This was refluxed for 2 hours with the condenser hardly needed at all. After that, the heat was stopped and the hot solution was poured into 150ml ice-cold water. No crystallization occurred whatsoever. The organic layer was picked up with a syringe and was 5ml of thick deep-yellow oil weighting 5.87g and with PH of approx. 7. This was put in the fridge for two hours but no crystallization occurred, just a hint of some cloudiness like in this photo:  (not as much, though), probably contained some P2NP but frustratingly little, and was not worth the efforts.
This was poured in 50ml of ice cold IPA for fun and it partially dissolved but was poured down the drain.
Obviously, a dean-stark apparatus is needed to monitor when water evolution will stop so the reaction is complete, and to get a reasonable yield.
My question is - what is in that yellow organic layer? Unreacted benzaldehyde and some P2NP, right? So, how can those be separated easily and efficiently? I read somewhere that this can be washed with sodium bicarbonate solution but I don't see why.

Dissolve your wet NH4OAc in MeOH add anhydrous sodium sulfate to remove the water, filter into the reaction flask then vac off the methanol. Use immediately. Or make it in situ by adding ammonium carbonate to glacial acetic

Can anhydrous MgSO4 be used instead?

Yes, I would think so

You can separate and recover benzaldehyde by washing with sodium bisulfite solution   http://chem2all.blogspot.com/2009/03/addition-of-bisulfite-to-aldehydes.html

Thank you, again. Why make it in situ when it won't be anhydrous, though? (NH4)2CO3 + 2CH3COOH → 2CH3COONH4 + H2O + CO2

Correct, but probably less water than is in your ammonium acetate. This reference may assist you http://www.orgsyn.org/demo.aspx?prep=CV1P0413


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Confusion About Acid/Bases That Don’t Follow the Arrhenius Definition
I understand that acidity is merely a reflection of how many free hydrogen ions are in water, and basisicity is just a reflection of the number of hydroxide ions, but I don’t understand how this definition holds true for acids and bases that don’t dissociate into hydrogen ions or hydroxide ions.
For example, how is sodium ethoxide, a very strong base, able to have such a high ph (14) if it doesn’t dissociate into hydroxide ions? I get how it’s called a base due to the Bronstead-Lowry definition, but I don’t get how it has such a high pH.

Sodium ethoxide does not have any pH at all.

When dissolved in water, shouldn’t it have a pH of 14, though?

When dissolved in water, it will cease to exist.

When discussing the strength of an acid/base we consider its dissociation constant, for bases this is its $K_\mathrm{b}$, or for easier comparison its $\mathrm{p}K_\mathrm{b}$. pH is a property specific to aqueous solutions of a specific composition / concentration / temperature, not any molecule itself.
A 1 M solution of sodium ethoxide will have a pH of 14 because:
$$\ce{OEt- +H2O\rightarrow HOEt +OH-}$$
I would strongly recommend reading up on Lewis acid-base theory, since it provides a more general/useful idea of acidity/basicity. Most reactions are in fact not done in water. Also you may be interested in knowing that it is impossible to have a base stronger than $\ce{NaOH}$ in water due to something called the leveling effect.


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Visualization of TLC plates
I'd like to know what exactly happens when various agents are used after developing a TLC plate, during the visualization phase, of sugars with anisaldehyde for example.
I know that those compounds are used in order to make invisible stains, visible. Also, sometimes, application of heat is necessary, I reckon so as to speed the "colouring process". However, I'd like to learn what kind of reactions take place or what kind of products are there formed, since I was not able to reach a conclusion after my research.

Google is your friend! Start here https://www.silicycle.com/faq/thin-layer-chromatography/stains-for-tlc-plates then look up each named stain reagent

I've found similar charts but they don't describe what exactly happens reaction-wise. The only webiste I've found that partially describes the occuring reactions is this one: http://www.reachdevices.com/TLC_stains.html However, even here, some reactions are missing or are simple theories, yet to be corroborated.

authors: Αντώνιος Κελεσίδης


text:
Enthalpy of neutralization of strong acid and strong base differing from enthalpy of formation of water from ions
If the neutralization between strong acid and bases has enthalpy of neutralization of around $\pu{-57.1 kJ mol^-1}$, why is it that when I try to calculate the enthalpy of formation of water from $\ce{H+}$ and $\ce{OH-}$ ions I get approximately $\pu{-55.8 kJ mol^-1}$ using the reference values?
Why is there this difference in enthalpy if the reaction between say $\ce{NaOH}$ and $\ce{HCl}$ is only between $\ce{H+}$ and $\ce{OH-}$ ions, while $\ce{Na}$ and $\ce{Cl}$ remain dissociated?
Calculations: From the Wikipedia page, the standard enthalpies of formation are:
$$\begin{align}
\ce{H2(g) + 1/2 O2(g) &-> H2O(l)} &\quad \Delta_\mathrm{f,1}H &= \pu{-285.8 kJ mol-1}\tag{1}\\
\ce{1/2 H2(g) &-> H+(aq) + e-} &\quad \Delta_\mathrm{f,2}H &= 0 \tag{2}\\
\ce{1/2 H2(g) + 1/2 O2(g) + e- &-> OH-(aq)}&\quad \Delta_\mathrm{f,3}H &= \pu{-230 kJ mol-1} \tag{3}
\end{align}$$
Computing $\Delta_\mathrm{f,1}H - \Delta_\mathrm{f,2}H - \Delta_\mathrm{f,3}H$ gives us the reaction $\ce{H+ (aq) + OH- (aq) -> H2O (l)}$ with the associated enthalpy change $\Delta_\mathrm{f}H(\ce{H2O(l)}) = \pu{-55.8 kJ mol-1}.$

Good first question, there are some things to consider. Firstly there might be the possibility that the different answers are due to differences in the values measured for a given thing by different authors. Secondly it looks like you are using a Hess's law method, the thing is that errors might build up there. Lastly what do you mean by "strong acids and strong bases", do you mean an infinitely dilute solution of both or do you mean something like 4 M nitric acid and 5 M sodium hydroxide ?

@NuclearChemist, regarding the Hess's law method, what are the possible errors that can build up in there? The strong acids and base I am referring to are dilute solution of strong HCl and NaOH, e.g. 2M HCl, but not infinitely dilute. (I am not sure what infinitely dilute means)

Well if you have A forming B, and the measured delta H is 100 +/- 1 kJ mol-1. If you measure A to C, C to D, D to E and E to B all with an error of +/- 1 kJ mol-1. Under some conditions the effect of the errors will build up so this Hess's law based method will give a worse answer than a more direct measurement of A to B.

-55.8 is pretty close -57.1 kJ/mol. There could be other reasons, but remember that:
1) The enthalpy of neutralization was measured colorimetrically. Different authors reporting different results. I personally only trust articles from the late 80's and onwards since accurate solution calorimeters were not available prior to that era. It is typical to see +- 2kJ/mol difference.
2) many use a single concentration of acid and bases to be reacted and repeat the experiment multiple times to get and average value. What I do instead, I do multiple experiments in incremental amounts of acid to be titrated on excess of base. This way, you're changing one variable and probing against the change in that variable only (the amount). You will get a slope of J/mol in the best statistical way: linear regression.

2) Many authors are not careful about using only dilute solutions and correcting for the enthalpy of dilution when titrating in a calorimeter
3) Hess's law method combines all the errors in the heats of formation


text:
How to divide the peaks of two anion ions using ion chromatography?
I used an ion chromatography system (Dionex IC-2000) to determine the inorganic ions and acid from the samples collected in the atmosphere.
For measuring $\ce{SO4^2-, NO3, Cl-}$, and low molecular acid together, I used a gradient elution method to divide them and quantify their amount.
However, I found that an artifactual band always overlapped with malonic acid shown like this:

This figure was captured in the situation of the single standard of malonic acid.
I want to divide the overlapped peaks efficiently, and have not found any solutions?
Any useful suggestions or tips would be highly appreciated.
My gradient dissolution method
0.000   Autozero
        Concentration =     1.50 [mM]
        Curve =     5
        Load
        Wait    CycleTimeState
        Inject
        ECD_1.AcqOn
        ECD_Total.AcqOn
        Channel_Pressure.AcqOn
        Concentration =     1.50 [mM]
        Curve =     5

7.000   Concentration =     1.50 [mM]
        Curve =     5

8.000   Concentration =     3.00 [mM]
        Curve =     5

59.000  Concentration =     3.00 [mM]
        Curve =     5

59.500  Concentration =     1.50 [mM]
        Curve =     5

60.000  ECD_1.AcqOff
        ECD_Total.AcqOff
        Channel_Pressure.AcqOff
        Concentration =     1.50 [mM]
        Curve =     5

Never did ion chromatography. But it immediately occurs to me that malonic acid is a dicarboxylic acid. Could it be two different species of malonic acid? What is the pH of the eluent?

See this paper Determination of Anions and Carboxylic
Acids in Urban Fine Particles

@MaxW. Thanks for your reply. But one peak occurred even in the situation of clean water. So, I assumed it was $CO_3^{2-}$. The eluent we used is $KOH$

I'm not quite sure what you want but if you want to analyse them as two peaks just to obtain their areas then there are standard methods with which to do this. The simplest is to assume peak shapes, say gaussian/Voight etc and add them in proportion via least squares until you fit your data.

@porphyrin. Thanks for your reply. One of the peaks was an impurity introduced to the system, and the other one was malonic acid for detection.

A mixture of roughly-symmetric peaks can be achieved by a sum of exponential functions.
$$f(x; \vec a, \vec m, \vec s) \triangleq a_{1}\exp\left(-\frac{1}{2}\left(\frac{x-m_{1}}{s_{1}}\right)^{2}\right)\ +\ a_{2}\exp\left(-\frac{1}{2}\left(\frac{x-m_{2}}{s_{2}}\right)^{2}\right)$$

There are a variety of ways that you might fit $f(x; \vec a, \vec m, \vec s)$ including gradient-based methods or Monte Carlo methods.

Edit: You can of  course add other terms. From the appearance of the plot above, one may wish to include a constant baseline parameter $\beta_0$.
$$f(x; \vec a, \vec m, \vec s, \beta_0) \triangleq a_{1}\exp\left(-\frac{1}{2}\left(\frac{x-m_{1}}{s_{1}}\right)^{2}\right)\ +\ a_{2}\exp\left(-\frac{1}{2}\left(\frac{x-m_{2}}{s_{2}}\right)^{2}\right) + \beta_0$$

Edit: If you want to create skewed peaks I would looks at a couple of options.
Having different standard deviations to the left and right of the mean is one way of getting asymmetric peaks, which you can model as mixtures.
A second option is to look at transforming (hopefully simplifying) existing skewed distributions into non-distribution peaks which can be put into linear combinations. An extreme example is the skewed generalized $t$ distribution.

In this answer, I miss the keyword deconvolution which often is used to introduce the technique of multipeak fit in general and in programs (e.g., fityk).

@Buttonwood I agree, one could use deconvolution. Although the OP did not mention the requisite conditions, sometimes signals can also be separated tensor decomposition. And there are further techniques for separating signals. In every case we must suppose something about in what sense the signals can be separated.


text:
Evaluating Franck-Condon factors
Franck-Condon overlap integrals are given by $\langle\psi^e_{v_e}|\psi^g_{v_g}\rangle$, where $\psi^e_{v_e}$ and $\psi^g_{v_g}$ are the vibrational functions for the $v^{th}$ vibrational states of the excited and ground electronic states, respectively.

Why is it considered difficult to obtain Franck-Condon factors (judged by the fact that there is a lot of literature dedicated to their calculation) – is the problem only in obtaining high-accuracy vibrational wavefunctions for the two electronic states, or there something else at play?
If you can obtain both $\psi^e_{v_e}$ and $\psi^g_{v_g}$, could you calculate the overlap integral in Cartesian coordinates?

As FC factors are the integral of the product of two vibrational (wave) functions their calculation is not particularly hard if you have the functions in the first place and in cartesians, usually, as they are vibrational functions.  Thus obtaining these is the hard part.


text:
Why don't such ions form if half-filled or full-filled configurations are considered stable?
We all know atoms want to stabilize by having an octet or duplet configuration. But if you see nitrogen which has an electron configuration of $\ce{2p^3}$. It has half filled p orbital, so it is technically stable, isn't it? Then why don't elements like oxygen form a monovalent oxygen ion?
By similar logic, sodium has $\ce{3s^1}$ configuration which should also be stable because it has a half filed orbital. But sodium is reactive as we know.
Can you please point out the flaw in my argument?

Nitrogen which has am electron configuration of 2p3. It has half filled p orbital, so it is technically stable right? No, a N atom is not stable in that two such atoms will form a molecule $\ce{N2}$.

Stability is relative. A half-filled orbital configuration is more stable than usual, but not as stable as an octet, etc.

As also pointed out in the comments, there is no such thing as absolute stability. One species is stable only with respect to another species.
You may have come across the fact - that half-filled configurations are "stable" - while studying ionization enthalpy/electron gain enthalpy, or the electronic configuration of $\ce{3d}$ elements. Pay careful attention: the half-filled configuration is "stable", but only with respect to the neighbouring configurations. That is to say, the $\ce{2p^3}$ configuration is much stable than $\ce{2p^4}$ or $\ce{2p^2}$. Yet, $\ce{2p^3}$ is still very less stable than $\ce{2p^6}$. This is why, in nature, nitrogen prefers to exist in diatomic molecules, so as to complete its octet.
Now for the fully-filled $\ce{3s^2}$ configuration (sodium anion, aka "sodide"): again, note that the sodium cation is way more stable than the sodium anion, as the former has a fully filled octet, while the latter doesn't. (Sodide has indeed been formed, but by using special techniques only. See Wikipedia).
Also see: Why are atoms with eight electrons in the outer shell extremely stable? and Why do atoms "want" to have a full outer shell?

Well, I think there may be need to do a bit of debunking of debunking. Inherent instability is quite a thing and you seem to be rather talking about reactivity.

@Mithoron Could you please elaborate a bit more on your statement: " Inherent instability is quite a thing"?

If something is inherently unstable, it breaks down by itself.


text:
Claisen rearrangement in substituted ring

This is an example of Claisen rearrangement, property of allyl vinyl/phenyl ethers.
I believe that the allylic migration should take place to the position ortho to both $\ce{-CH3}$ and $-\ce{OR}$ because:

There's more electron density there due to the presence of electron donating group $\ce{-CH3}$

and I read this in March under Claissen rearrangement:

Since the Claisen rearrangement doen't involve ions, it shouldn't greatly depend on the presence or absence of substituent groups on the ring. This is the case. Electron donating groups increase the rate and electron withdrawing groups decrease it, but the effect is small.

So I expect the product to be 2-[(2E)-but-2-en-1-yl]-3-methylphenol:

However my teacher made us write the product to be 2-[(2Z)-but-2-en-1-yl]-5-methylphenol:

His rationalisation was that the carbanion like species attacking the allyl group is more stable at the place away from methyl group.
But my question is that if this reaction is single step, followed by tautomerism why should it even depend on the carbanion like species which isn't even formed.

I would have expected a mixture of both but hopefully someone with database access can find the exact reaction outcome

I have to call your teacher on the Z-stereoisomer. It will be  an E-double bond as well.

Oh crap. I didn't read your question properly. I would argue that your teacher is wrong, and you are right. I'll submit an answer with justifications in a minute.

Well, you're both half-right. ;)

Another great example of how it's important to look at actual evidence. There have been various studies on Claisen rearrangement selectivity, here's a combined experimental/theoretical one on meta-substituted ethers specifically: Gozzo, F. C.; Fernandes, S. A.; Rodrigues, D. C.; Eberlin, M. N.; Marsaioli, A. J. Regioselectivity in Aromatic Claisen Rearrangements. J. Org. Chem. 2003, 68 (14), 5493–5499. DOI: 10.1021/jo026385g.
An important screenshot is shown here:

Based on this article at least, it turns out that for your specific reaction, it's likely that the regioisomers would be formed in equal amounts. (See allyl aryl ether 1). We further see that sterics are not as significant of a consideration as electronics; allyl aryl ether 6 actually prefers the "more sterically hindered" product 14 to 1.
The article concludes that "the preferred conformation of the reactants ... with no exception, was found to lead to the major isomeric product.... The regioselectivity for the Claisen rearrangement of meta-substituted allyl aryl ethers is mainly influenced, as expected, by the electronic effects of the meta substituent, which governs the relative stability of conformers, transition states, and intermediate to final products.

I am not sure how electronics are affecting allyl ether 6. Both positions look equally good to me, then why is $P^A$ the major product?

It looks like it is difficult to rationalize the reactivity using simplified approaches.  Here's what the article says. "However, B3LYP/6-31G(d,p) calculations have shown that, owing to the complex electronic structure of the allyl aryl ethers, simple frontier HOMO-LUMO intramolecular orbital interactions fail to mimic transition-state orbital interactions, and therefore, these interactions cannot be used to predict the reactivity or regioselectivity of the aromatic Claisen rearrangement."

So which electronic effects do we see if we get a question?

Like in 6, I would have surely predicted para product with migration to 4. But data says something else

First, consider if this reaction reversible? If so, we'll just form the thermodynamically favorable product. But the temporarily breaking of aromaticity and tautomerization to reform aromaticity suggests that this process is not reversible. So that means we're looking at kinetic factors.
There are a few key concepts I learned from Dave Evans' class that have really helped me understand organic chemistry. One of these is to draw out the transition state, here, for a [3,3]-rearrangement, which is a chair-like transition state.

In this case, there are four transition states you can draw, dictated by the stereocenter at that innocent looking methyl group and the substitution on the phenyl ring. These are drawn below.

I drew the corresponding product for each transition state. Notice that, as drawn, the pseudo-equatorial methyl group in the transition state gives rise to the E-product.
I would think that the methyl substitution on the phenyl ring would push the reaction to the other side. I'm less sure about this, but I would prefer the two products on the right for the aryl substitution. This is based on steric repulsion and on the fact that electronic effects look about the same either way.
Between these two, we should consider whether it's better to have the methyl group in the pseudo-equatorial position or the pseudo-axial. Since it's still a chair-like transition state.
One of the flag pole hydrogen atoms is missing because the phenyl ring isn't substituted like a normal carbon chain, but I would venture to say that you would still favor the pseudo-equatorial substitution because of other hydrogen (which is not fully axial either). A normal methyl group has an A-value of 1.74, but it's probably significantly less here. However, even at half, you can determine that the preference for the methyl group to be pseudo-equatorial is going to be about 4:1. (We're plugging into $e^{-\frac{\Delta G}{RT}}$)
Again, since the reaction is not thermodynamically controlled, we're looking at the transition states to figure out kinetically favored product. The third one from the left looks best, so the corresponding product is probably major.

I like your reasoning, but the calculations in the paper I posted in my answer say the product is both the kinetic and thermodynamic one

@DennisCao The excerpt from the paper does not address E/Z selectivity at all. That is more certain than the regioselectivity, which I expressed some uncertainty about.

Ah of course that's your point, not sure what I was thinking of your answer originally.


text:
How much muriatic acid to dissolve cinder block?
Can a cinder block be dissolved in hydrochloric (muriatic acid) ?
Also if it can then how much hydrochloric acid will be needed ?

I think that the question does relate to some interesting issues in chemistry. So I would suggest we keep it in a modifed version

Keep in mind that a cinder block is likely to contain silicates, these will not dissolve well in hydrochloric acid. The silicates will either not react or they will form silica.
If you want to dissolve silicates then you need to use something which will complex to the silicon such as fluoride (HF). I have seen methods for the digestion of soil samples to get the plutonium out. To obtain the plutonium from a soil sample it is often important to digest down the silicate minerals. To do this you must boil the soil sample to dryness many times with a mixture of hydrofluric acid and nitric acid. This will convert the silicate minerals into silicon tetraflouride which will boil out of the mixture.


text:
Reaction of solid silver sulfide on warming with dilute nitric acid

Solid silver sulfide is warmed with dilute nitric acid.

I thought that the nitric acid would engage with silver sulfide in a double displacement reaction, as follows: $\ce{Ag_2S {(s)} + 2H+ {(aq)} -> H_2S {(g)} + 2Ag^{+} {(aq)}}$. This double displacement reaction would be driven by the production of $\ce{H_2S}$ gas.
However, my textbook says that the answer is: $\ce{Ag {(s)} + NO_3^{-} {(aq)} + 4H^{+} {(aq)} -> 3Ag^{+} {(aq)} + NO {(g)} + 2H2O {(l)}}$. Does this redox reaction occur because $\ce{HNO_3 {(aq)}}$ is a strong oxidizing agent? Why doesn't the textbook answer contain $\ce{AgS {(s)}}$ in the left hand side of the net ionic equation?
In the future, how do I know when nitric acid will engage in a double displacement reaction to liberate gas, or when it will engage in a redox reaction? Furthermore, if the nitric acid was concentrated, then would $\ce{NO_2}$ be produced?

Obvious textbook mistake...

@Mithoron What should the correct answer be?

Nilay's analysis looks correct.

Nitric acid is a strong oxidising agent and it will oxidise both the reactants and products. Thus, a reaction involving nitric acid is not a simple double-displacement reaction but a redox reaction where nitric acid itself get reduced to a couple of nitrogen oxides depending on concentration of the acid and the other reactants gets oxidised. According to [1]:

Nitric acid, when of concentration above 5%, dissolves
  precipitated silver sulphide rapidly. Very strong acid yields silver
  sulphate alone, while acid of lower concentration forms some nitrate
  in proportion to its dilution. 15-20%, acid yields the
  maximum, about 95 per cent, nitrate, greater dilution again resulting
  in a smaller percentage. Heating the acid, or prolonging its time of
  action is unfavorable to the formation of nitrate.

Another paper[2] suggests formation of normal silver(I) sulfate through a series of exotic sulfates which is confirmed by the change in color: -

[...]the acid additions used correspond to 75%, 150% and 200% of the
  amount required stoichiometrically based on the following reaction:
$$\ce{3Ag2S + 8HNO3 <=>> Ag2SO4 + 8NO + 4H2O}$$
[...]
The distinct colors of the various phases involved in this system
  (black silver sulfide ($\ce{Ag2S}$), brown silver sulphide sulphate
  ($\ce{Ag6S3O4}$), white to light yellow silver sulfate ($\ce{Ag2SO4}$)
  and bright yellow sulfur($\ce{S}$)) are a further indicator of this
  behaviour; the leach went from brown ($\ce{Ag6S3O4}$) to white
  ($\ce{Ag2SO4}$) with increased pressure and temperature and some
  residue samples showed evidence of the presence of elemental sulphur.
  This transition from $\ce{Ag2S}$ to $\ce{Ag6S3O4}$ to $\ce{Ag2SO4}$
  shows the progressive oxidation of sulphur in the leach solids as the
  leaching reaction goes to completion.

In conclusion, reaction of silver sulfide and nitric acid will give different products depending on stoichiometry of the reactants, concentration of acid, effect of temperature and pressure etc. As you noted correctly, if the concentration of nitric acid is high enough, it may even produce $\ce{NO2}$ gas.
$$\ce{Ag2S + 10HNO3 → 2AgNO3 + H2SO4 + 8NO2 + 4H2O}$$

References:

Gruener, H. Silver Nitrate Formed by the Action of Nitric Acid on Silver Sulphide. J. Am. Chem. Soc. 1910, 32 (9), 1030–1032. DOI: 10.1021/ja01927a002.
Holloway, P.; Merriam, K.; Etsell, T. Nitric acid leaching of silver sulphide precipitates. Hydrometallurgy 2004, 74 (3-4), 213–220. DOI: 10.1016/j.hydromet.2004.05.003.


text:
Rule to determine the total number of elements in a period

If each orbital can take maximum of 3 electrons, then total number of elements in 2nd and 4th periods respectively are = ?

I am not aware of the rule which helps to determine the total number of elements in a period.

The total number of elements formed due to their last electron entering the $n$-th shell will be equal to the total number of electrons that shell can fill. For each electron added or removed (along with a proton), you'll get a new element, since a unique element is itself defined by a unique count of protons.
Thus, this problem is reduced to finding the total number of electrons in $n$-th shell. And, simply enough, the total number of electrons in $n$-th shell = number of orbitals in $n$-th shell $\times$ number of electrons each orbital can hold.
For example, for $n=3$, we have $\ell=0,1,2$. Thus, the number of orbitals here is $=1+3+5=9$ (recall that s-subshell has 1 orbital, p-orbital has 3 orbitals, and so on[1]). Since $s=\pm 1/2$, each orbital can fit two electrons, thus total count of electrons is $=2\times9=18$, which is also the total number of elements in the third shell.
In your question, the only change is that each orbital can now hold three electrons (by some phenomenon we don't know yet!) So, in the final step, just multiply the total number of orbitals by three instead!

To calculate the number of elements in a particular period, you need to note the individual subshells that belong to that period, and then sum up the number of the elements from those subshells.
In my solution above, I have shown how to get the total number of elements formed due to their last electron entering the $n$-th shell, which is not equivalent to the number of elements in $n$-th period.
For example, in the fourth period of the periodic table, you have the $\ce{4s, 3d}$, and the $\ce{4p}$ subshells. This is the calculation that your solution manual has shown in the second row.

[1]: In general, the total number of orbitals $k$ in shell $n$ is:
$$k = \sum_{\ell=0}^{n-1}(2\ell+1)$$
since: a) each $m$ value stands for a unique orbital b) $-\ell\leq m\leq\ell$, thus $m$ has a total of $2\ell+1$ values for a particular $\ell$ value