text:
On the detection of molecules in electrospray ionisation (ESI)
Of the ESI spectra that I have encountered, positive & negative mode, the detected species always have charge (positive or negative) corresponding to the charge mode that the machine is in. Is it therefore the case that the ESI detector can only detect charged species, or that the overwhelming majority of species are charged?

I feel you have a misconception. ESI is an ionization technique not a detection technique. The sole purpose of the electrospray ionization is to impart a charge on a molecule in the gas phase. The molecule could be neutral, anionic or cationic in the solution phase. The inventor of ESI used to say that ESI makes the elephants fly, i.e., very large molecules could also be ionized in the gas phase.
The important point that you must learn is that if you have an ionized molecule in the solution, it is not necessary that when you bring the molecule in the gas phase, it will have a electrostatic charge. It may become a neutral molecule.
In order to "sort" molecules according to their mass to charge ratios in the gas phase, it is best to ionize them, i.e., remove an electron or add a proton so that the molecule of interest acquires an electrostatic charge in the gas phase. Note that I am emphasizing gas phase again and again. Forget what is the state in the solution. Almost 100 years ago, physicists had worked out how "charges" will move in an electric or magnetic field predictably. If the molecule is electrically neutral it will not interact effectively either with electric or magnetic field. Hence you would not be able to sort them according to their mass/charge rations. This is the job of the mass spectrometer.
Once your molecule has a charge, it is very easy to direct them here and there in the mass spectrometer and even focus all the ions at place  (one of the major advantage of ionized molecules), before exposing them electric fields or magnetic fields.


text:
Current with an ionic solution
For the last assignment of the school year, my science teacher gave the class questions about chem, physics, etc. The third question had me confused and I'll be thankful for all the help given. The question is:

An ionic solution is created and tested to see how much current can go
through it. Two groups are told to create the following solution
consisting of $\pu{1.0 +/- 0.1 L}$ of water and $\pu{2.0 +/- 0.2 g}$ [of what?] . The battery used for the test is $\ce{8 +/- 1 V}$.
Group A measures a current of $\pu{2.1 +/-0.1 mA}$.  Group
B measures a current of $\pu{3.4 +/- 0.1 mA}$.
Describe 3 possible reasons why Group B measures a current greater
than group A.

(The "water" after the 2g was not in the question originally.)

Why are the error margins explicitly mentioned? Aren't they supposed to be implicit?

@AniruddhaDeb What is an "implicit" error margin?

The last digit in any measurement is doubtful, so $1.0 L \equiv 1.0 \pm 0.1 L$. My terminology might be a bit incorrect here, so feel free to correct me.

@Aniruddha Deb Hmm, 1.0 without explicit margins may as well be 1.00 +- 0.05, with errors rounding to 1.0. so 1.0 +- 0.1 is more particular than just 1.0.

@Poutnik you  mean "specific", i guess ;)

@AniruddhaDeb You need to differentiate between +- giving a standard deviation (in which case you might be able to give two significant figures), and a simple assumed measurement error margin.

No offence to you (certainly to the original author of this question), I call such home-work questions such as garbage in-garbage out type questions. No wonder a student will be confused if they see such open ended questions. The reason is that the author is not telling you or us of 2.0 grams of what?
Even if we assume it was 2 grams of salt, several things can affect the measured current
(1) What are the electrode areas?
(2) What are the electrodes made of?
(3) What is the distance between the electrodes?
Last but not the least, the identity of the salt matters a lot. Did both groups use the same salt? Monovalent, divalent, etc

This question (the one by your teacher, not yours) becomes ridiculous if we do not take a few things for granted:

same (unidentified) salt
water from the same tap
identical set of electrodes
identical, clean beakers (or at least they were together in the same washing machine).
identical distance and position of electrodes
same temperature

because the influence of either of them is either arbitrary (distance!) or very hard to ascertain for a high school student (electrode material), or anybody who is given so little information.
Now you can check if the uncertainties given in the text are enough to explain the difference. At 8 V, everything should be linear enough, and yes, it seems that together, the errors could explain the difference.
What the other two explanations could be, I have no idea. Your teacher should probably be ashamed of himself for insulting the students with such a badly devised question. Unless we are grossly misinterpreting his aim and attempt. ;-)

Well ... to be generous, one could think that the teacher is attempting to teach the lesson that all of those things matter, and that if reproduction of an experiment fails, you should consider every possible difference! So the "badly devised" question is designed to make students think outside the box. Possible?

@RossPresser You´re right, in dubio pro reo! ;-)

The question formulation may look strange, but still makes sanse, whatever salt is used.
The difference of results is at the edge to be cause just due uncertainties of volume, mass or voltage, or by (usual ) used electrode material.
I see 3 other potential sources of the difference, as it is unclear if the experiment settings is otherwise identical or not:

Higher electrode surface leads to higher current at the same voltage, as the current density would be about the same.
Shorter electrode distance leads to higher current due higher volatge gradient.
Mixing electrolyte leads to higher current, as it cancels the diffusion limitation.

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sort: votes


text:
Orbital overlap and bond strength
In Peter Sykes' A Guidebook to Mechanism in Organic Chemistry, it says that sp3 orbitals have greater relative overlap compared to sp bonds.
We know that the $\ce{C-H}$  bond is shorter in ethyne compared to ethane. So the intuitive idea that more overlap would lead to stronger bond formation is obviously incorrect.
So what is the relation between relative overlap of two orbitals and their bond strength? Is there any use that we can get from this overlap?

I think there is just one counterpart missing. Greater overlap.... to what?

@Alchimista, I think it would be to another similar orbital. Probably sp3 with sp3 vs sp with sp. But that still implies that the C-C bond in ethane should be stronger then the C-C bond in ethyne with is incorrect. I have added a link to the relevant portion of the text

You see that it mentions one C and four H atoms.... To your note: likely one single C-C bond is stronger in ethane vs ethyne, tough difference might be small and other consideration could be made (eg the relevant region is crowded with electrons) but I do not knoe/forgot. If we touch the yne portion of the molecule, we normally don't go down to total breaking....

@Alchimista I think the text just wants to explain why hybridization happens in the first place, why all the CH bonds are equal in methane, or why O atoms are sp3 hybridized instead of sp.

In that context is straightforward

I suppose the author was talking about the internal overlap which takes place while the formation of sp and sp3 hybrid orbitals.
What I mean is when one s and one p orbital overlap, the extent of overlapping is less than the overlap that happens when one s and three p orbitals overlap.
To understand it more easily, suppose a ball represent a s orbital while a pencil represent p orbital. Think in mind the extent overlap of three pencils and one ball will be more than the extent of overlap of one pencil and one ball.
But the situation changes, when you overlap two sp3 orbitals or two sp orbitals. sp having more s character is relatively stronger then sp3, thus c-c bond is stronger in ethyne than ethane.

The text clearly states that we are talking about overlapping of an atomic orbital from each of the participating atoms

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text:
How to draw the structure and find the oxidation number of Phosphorous in H2P2O7
Find the Oxidation number of Phosphorus in $\ce{H2P2O7^2-}$.
So I thought of drawing the lewis structure to find the oxidation numbers. Because I wasn't sure if Oxygen was bonded to a Hydrogen or not. Then I realised that the two Hydrogens could be ionised, so I'd be left with $\ce{P2O7^4-}$.
But then when I tried drawing this Lewis structure I got really confused, As I tried to draw it with the two Phosphorus as central atoms and I couldn't make it. (I know I can just use general rules to find oxidation numbers but I feel like lewis structures often give more accurate answers). So could you please tell me how to draw the structure and more importantly guide me through the process. Any help is appreciated, thanks!

Please use $\LaTeX$ and also see How to ask a good question

You have O and P, both in their most common valence states. You start connecting them until there is nothing more to connect. What could possibly go wrong?

Take 2 dihydrogenphosphate anions and condense 2 OH groups ( 1 from each ion ) into -O- bridge, eliminating water.

If you didn't understand what  Poutnik has sain in his comment, this is how it should be done:
You have realized your parent compound is $\ce{H4P2O7}$ (adding two $\ce{H+}$). Correct? If so, concentrate on following hypothetical reaction:
$$\ce{2H3PO4 -> H4P2O7 + H2O} \tag1$$
I believe you know what $\ce{H3PO4}$ is looks like. Now, draw two $\ce{H3PO4}$ side by side. Choose two closest $\ce{OH}$ groups (that would be $\ce{-P-\color{red}{OH} \space \space \color{tequish}{HO}-P -}$). Then, remove one $\ce{\color{red}{-OH}}$ and the $\ce{\color{tequish}{-H}}$ from the second $\ce{\color{tequish}{-OH}}$ you chose to eliminate a $\ce{H2O}$ molecule. Finally combine two remaining molecules together such that final structure forms $\ce{-P\color{red}{-}\color{tequish}{O}-P -}$ bond. This is your $\ce{H4P2O7}$. Eliminate any two hydrogens to get $\ce{H2P2O7^2-}$ structure.
Note: Since above hypothetical equation $(1)$ does not require oxidation or reduction to form $\ce{H4P2O7}$, it is safe to assume that oxidation state of $\ce{P}$ in both $\ce{H3PO4}$ and $\ce{H4P2O7}$ is same.


text:
Heat effect on diluting concentrated NaOH solution
I am an engineer in water treatment and have a chemistry question. In water treatment we dose $32\%$ sodium hydroxide solution for $\mathrm{pH}$ adjustment and there is a concern that our dosin lines must be $100\%$ dry before introducing the $\ce{NaOH}$ solution. This is the problem as they have been pressure tested with water and there is water residue in them.
Will $32\%$ sodium hydroxide solution react violently with cold water? It is my understanding that the reaction only occurs when solid sodium hydroxide (lye) is dissolved in water and since my $\ce{NaOH}$ is already in solution then this reaction will not take place. Is this correct?

The solution gets warmer by diluting with extra water, but no violent reaction. I suggest to do a test, if the temperature raise is concerning you, e.g. because of the tubing temperature resistance.
What  can also be done is applying solution with graduating concentration, so warming would be spreaded in sevceral steps. But as hydroxide is already quite diluted, I expect temperature change rather few than many degrees.
Another concern is initial dosage would be affected by water, decreasing concentration of the applied volume.

Thanks for your help, I am not a chemist so I am unable to work out the exact release of heat in the reaction and the amount of residual water present would be impossible to quantify. i can say that the longest dosing line has a volume of 14 litres when full but of course this will have been emptied prior to introduction of NaOH. The worst case possible is that 14 litres of water mixes with 7000 litres of NaOH. This will never happen but at least I would know that my scenario will generate far less heat. Is this something you would be able to quantify at all?

I mean taking a small volume of solution , diluting with water and measure temperature raise. It can be done without lab equipment. But unless it is PVC tubing with low melting point, I would not be concerned.

With such volume ratio, I would be of no concern.

It is PVC pipework

I didn't think so, thanks very much for your help

I'd not concern about the heat increased by the dilution of $32\%$ $\ce{NaOH}$ with negligible amount of water. I'm telling this by my personal experience. In our organic lab program, our student workers have been diluting $50\%$ $\ce{NaOH}$ solution ($\pu{12.5 M}$) with water at room temperature to make $\pu{3 M}$ ($\ce{NaOH:H2O}$ is in $1:3.2$ ratio) and $\pu{6 M}$ ($\ce{NaOH:H2O}$ is in ~$1:1.1$ ratio) solutions in $\pu{20 L}$ capacity. These ratios are worse than the ratio in your worst case scenario ($\ce{NaOH:H2O}$ is in $1:0.002$ ratio), assuming $\ce{NaOH}$ percentage is in $w/v$. None of dilution has reached untouchable (to my skin! :-)) temperatures. This means, the temperature increment would be far below the melting point of PVC tubing in your worst case scenario. Yet, since this is just based on personal experience, I'd like to give you some scientific evidence.
In Caustic Soda Handbook of OxyChem, there is an example saying "the approximate resulting temperature of final solution after diluting $50\%$ $\ce{NaOH}$ at $\pu{120 ^\circ F}$ ($\pu{48.9 ^\circ C}$) to $20\%$ using water at $\pu{80 ^\circ F}$ ($\pu{26.7 ^\circ C}$) is $\pu{143 ^\circ F}$ ($\pu{61.7 ^\circ C}$)." This would result $\pu{63 ^\circ F}$ ($\pu{35 ^\circ C}$) raise with respect to water temperature. The Graph 6: Relative enthalpy of aqueous caustic soda solutions (PDF) in page 36 would provide you with some important insights. The similar scenario you are facing is described here as well.


text:
How many kilograms of graphite are consumed while producing 4 kg of Al through Hall-Heroult process?

How many kilograms of graphite are consumed while producing 4 kg of $\ce{Al}$ through Hall-Heroult process?

This looks like a simple stoichiometry problem:
$$\ce{4Al^3+ + 12e- -> 4Al} \; \; \; \; \; \text{(at cathode)} \\
\ce{3C ->[O2] 3CO2 + 12e^-} \; \; \; \; \; \text{(at anode)}$$
For every $4$ moles of $\ce{Al}$ consumed, $3$ moles of $C$ are consumed. The mass of graphite consumed is thus:
$$M = \frac{4000}{27} \times \frac{3}{4} \times 12 = \pu{1.33 kg}$$
However, my textbook mentions the answer to be $\pu{2 kg}$, along with the following explanation:

for every kilogram of $\ce{Al}$ produced, approximately $\pu{0.5 kg}$ of graphite is consumed.

Why is this so? Could anyone provide a valid reference?

According to Aluminium Production1:

The production of 1 ton of aluminum typically requires 420 kg of carbon, 1920 kg $\ce{Al2O3}$, and 16 kg $\ce{AlF3}$ ....
The theoretical carbon consumption is only $\pu{333 kg C/t} \space \ce{Al}$. The excess carbon consumption is mainly due to air oxidation of the hot anodes in the hooded cells. The reaction occurs preferentially in the pitch binder matrix and leads to the physical loss of coke particles to generate dust that floats on top of the electrolyte.

Theoretically, your answer holds good. However, it fails practically. It's a stupid tricky question by the problem setters, who are apparently testing your knowledge of metallurgy rather than stoichiometry.
References:

Tabereaux, Alton T., and Ray D. Peterson. “Aluminum Production.” Treatise on Process Metallurgy, Elsevier, 2014, pp. 839–917. doi:10.1016/B978-0-08-096988-6.00023-7

Thanks for sharing this. I encountered the exact same problem in my book, and had the same doubt!


text:
Stronger ESI signal for deuterated substances
I observe a constant phenomenon in my ESI measurements, which indicates that deuterated substances give a stronger signal than the same substance in non-deuterated form. Unfortunately I can't make any sense of it, because I can't see any reason why a deuterated substance should be protonated any better. Since I have not seen or read anything about this phenomenon so far, I would be very pleased if someone could clarify whether this is something well known, or whether my samples just happen to show the same artifact over and over again. If it is a known phenomenon, I would also be happy to hear an explanation.
Further bits of information: LC-MS/MS in ESI mode, same acquisition method for both forms of the compound, observed in EPI scans, increase in intensity c.a. 3-fold.
To make this very clear, the two compounds (deuterated and non-deuterated) are measured separately in separate vials with a blank in between, so no co-elution can occur. One measurement is done with a solution of 100 ng/mL estradiol in acetonitrile and one with 100 ng/mL estradiol-d5 in acetonitrile. They are not mixed!

Is your deuterated internal standard co-eluting in LC? How many Ds there?

@M.Farooq Although the deuterated substance is of course intended as IS, it does not fulfil this role in the measurement mentioned above. In fact it is only a reference measurement with one substance each dissolved in 0.1% FA and an acetonitrile gradient. So the two compounds are measured individually. The compound we're talking about is estradiol and estradiol-d5 respectively.

Interesting problem. I believe that the co-elute in your LC. I have seen a paper which talks about ion-suppression by a co-eluting component. This is exactly your case. It is very clear, that estradiol-d5 is suppressing the signal of estradiol. Search for Enke model and ion suppression by deuterated compounds. There must be something out there.

@M.Farooq As I said, it cannot be a related to co-elution as the two compounds are measured separately in separate samples and I even ran a blank in between the two samples to make sure that none of the previous compound was left to be able to suppress the signal. This must be caused by something else. I have to thank you for your reference, though. Even if it's clearly not the case here, it's nevertheless interesting for a later step, when the deuterated estradiol is in fact used as an IS.

Sam: The point is that estradiol and estradiol-d5 are two different molecules. It may not be surprising that their ionization efficiencies are different. What is published in the literature is the opposite effect. The signal of the deuterated compound would bed suppressed by the normal one if they are co-eluting. You may be surprised that one can easily "separate" deuterated compounds in gas chromatography. In order to be 100% sure, I would re-prepare the standards of both. It could be just an Eppendorf or pipetting error.

Great q.  After I wrote my answer I realized maybe you are using an ion-trap instrument.  It does not do beam-type CID, but resonant CID.  I think a similar but slightly different version of isotope effects during fragmentation could happen there.

That's true, I'm using the Sciex QTrap 6500. I'll look into the ion effects on resonant CID.

https://chemistry.stackexchange.com/questions/135401/stonger-esi-signal-for-deuterated-substances

There are a couple reasons this could be happening.
The most interesting possibility: isotope effects during ion fragmentation
You are performing LC-MS/MS, i.e., your mass spectrometer is chemically reacting the estradiol and estradiol-$d_5$ with gas molecules in the collision cell of the instrument.  This process, known as CID or collisionally induced dissociation, will break chemical bonds in the parent molecule and form fragment ions.
On triple-quad and qTOF instruments, the CID is of the "beam" or non-resonant type.  This means a fairly large amount of energy is imparted to parent ions.  In general, this means that a wide variety of fragment ions are formed.  This is true of estradiol; publicly available fragmentation spectra show many fragment ions for this compound (see several GNPS spectra for example).
Deuteration is likely to change the relative propensity of the different fragmentation pathways.  For example, these reactions could represent the fragmentation of a parent ion $\ce{P}$ into two different fragments $\ce{F1}$ and $\ce{F2}$.
$$\ce{P -> F1}$$
$$\ce{P -> F2}$$
Isotopically normal $\ce{P}$ might give an intensity ratio of say 1 to 1 for $\ce{F1}$ and $\ce{F2}$.  But due to deuterium isotope effects, the pathway to $\ce{F1}$ may be much less favored in deuterium-labeled $\ce{P}$.  This could lead to an enhanced signal for $\ce{F2}$ as a much higher fraction of the $\ce{P}$ ion remains available to proceed down that fragmentation pathway.
However, a three-fold increase is pretty large.  I was able to find good measurements of various deuterated tocopherols, and judging from Table 1 in that paper, the response ratio for $d_0$ (unlabeled) tocopherol vs. $d_9$ is 11.8/0.967/10, or 1.22, not 3.  For measurement of trimethylamine N-oxide, authors note "the intensity ratios of the two product ions are different between TMAO and $d_9$-TMAO due to the difference in bond energy among C and H and C and D (deuterium)" but did not quantify this difference.  Eyeballing Figure 1, the intensity ratio of the light and heavy fragment is about 2.5:1 for $d_0$ but 1.4:1 for $d_9$.  None of those values are close to 3-fold.  But maybe estradiol and its site of deuteration and its fragmentation pathway differ and that a value of 3-fold is possible.
Errors in sample preparation
Maybe the concentration of one of your samples is off.  How many times have you independtly prepared the standards and repeated the measurement?
Ways to tell
If you can see a signal with a single quadrupole (i.e. do MS1 instead of MS/MS), do you see the same three-fold difference?  If so, error in sample prep is likely to blame.  If not, then isotope effects during ion fragmentation are to blame.

Thank you very much for your explanations and ideas! I have, in fact, done a triplicate measurement, but I didn't prepare several independent solutions (shame on me!), so a sample prep. error is possible. As I'm pretty confident in the technique I use to dilute the solutions, I am sure though, that the concentrations will not be so massively off that the 3-fold increase could be explained by this. I could imagine, however, that they are in fact off and that this, combined with the isotope effects described by you, adds up to the 3-fold increase. I'll check this! Again, thanks a lot!

Would be very curious to hear how the results turn out.  If you have a chance to prepare independent samples or to try MS1 (if you're not sure how you could monitor the transition of 255 -> 255 for d0 and 258 -> 258 at a collision energy of 0, which would bascially do the same thing).  In fact, you could also pretty easily measure a "survival curve" by monitoring these transitions at a range of different collision energies, and plotting the dropoff of intensity as a function of collision energy.

I'll prepare three independant samples of estradiol, for estradiol-d5 it's a bit trickier. The sample was already dissolved in acetonitrile when I bought it from Sigma, but I suppose that the concentration is correct. I'll combine this with an EPI with a low CE (5 is the minimum on my device). Unfortunately I cannot do an MS1. Estradiol is hardly ionizable and thus the signals are pretty low. In MS1 measurements I'd need a concentration waaayyy higher than I could possibly prepare. Thus I have to use an EPI experiment. I'll report!

Very cool, I'm excited to hear what you find!  You can also make "survival" curves for actual MS2 ions, ramping collision energies, if that gets you better signal.  If you use an actual fragment ion, the maximum won't be at 0 (or 5 CE), but at some other CE.  The great thing about the survival curves is they should be different if isotopic effects are responsible for your issue, and the data is insensitive to concentration.  You normalize each curve to a reference condition.  So you can use samples you already made -- even if there are errors in sample prep, it won't matter.

Just as a suggestion, how did you prepare the non-labelled estradiol stock solution? I read that your deuterated solution is already provided in acetonitrile by Sigma. Maybe you ran into a solubilization issue in the course of the stock solution. Estradiol is not a very soluble compound. Maybe a quick jump to the spectrophotometer would help and have a look at the absorbance of the stock solution from Sigma and your at a similar concentration in acetonitrile also.


text:
Why does de Broglie wavelength lambda = c / frequency?
For a chemistry problem, I want the velocity of the de Broglie wave to agree with one of these values:

A) $c^2/v$
B) $h\nu/mc$
C) $mc^2/h$
D) $v\lambda$

We know that
$$\lambda = \frac{h} {mv}$$
But I am confused by the next step in my book:
$$\lambda = \frac {c} {\nu}$$
After which:
\begin{align}
\frac{h}{mv} &= \frac{c}{\nu} \\
\implies v &= \frac{h\nu}{mc}
\end{align}
Why is the de Broglie wavelength $\lambda = c/\nu$? In every context I've seen it used, $c$ is the speed of light. But I don't think the velocity of the matter wave could equal the speed of light, because the matter wave is only as fast as the particle – which, having mass, cannot travel at the speed of light.
What am I doing wrong here? Does $c$ moonlight in quantum chemistry as the velocity of the matter wave, separate from its connotations in physics?

Read this: Matter Waves

Welcome to Chemistry! We prefer to not use MathJax in the title field, see here for details.

v is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency. Also. there should be rather p instead of mv, unless v << c.

@Poutnik How is p different from mv? Pardon me if my question is elementary, I have only just started studying quantum chem.

I think $c$ is meant to be the wave speed here, not necessarily the speed of light. This is common usage in physics at least.

$v$ is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency. Also, there should be rather $p$ instead of $mv$, unless $v \ll c$.
As $$ E = \sqrt{(mc^2)^2 + (pc)^2}$$
then
$$p = \sqrt{\left({\frac {E}{c}}\right)^2 - {(mc)}^2}$$
For a massless object like a photon:
$$p = \frac Ec$$.
For an object with mass, there is also $$p  = \frac {mv}{\sqrt{1-(v/c)^2}}= \gamma mv$$
See Matter wave, paragraphs De Broglie relations and Special relativity, where is clear c is the phase speed of light in vacuum:

$$\begin{align}&\lambda =\,\, \frac {h}{\gamma m_0v}\, =\, \frac {h}{m_0v}\,\,\,\, \sqrt{1 - \frac{v^2}{c^2}}\\
& f = \frac{c}{\lambda} = \frac{\gamma\,m_0vc}{h} = \frac {m_0c^2}{h} \bigg/ \sqrt{\frac{c^2}{v^2}-1}
\end{align}$$

"v is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency." How is this different from saying that it is the speed of the electron? What is the nuance here?

It may not be an electron, as wave properties are observable for objects of size/mass up to small molecules.


text:
What is the inter-particle bonding in elemental mercury?
In most pure metals, there is metallic bonding where there exist positive monoatomic ions and a sea of electrons. Is this the same with mercury? If not, then in elemental (liquid) mercury, what kind of ions are present (polyatomic?)? Thanks.

Mercury is as good a metal as any other.

"positive monoatomic ions and a sea of electrons" - that not how metallic bonding works at all.

Metallic liquids are bonded similar to metallic solids, except that Bloch functions don't apply (although an average band structure does, much like amorphous solids).


text:
Stoichiometry and percentage concentration
10 g of Mg(NO3)2 was dissolved in 120 g of water. The result is a solution with a
density of 1.12 g / ml. Determine the percentage concentration
(in %) and the molar concentration (in mol/L) of the resulting
solution

Welcome at chemistry.stackexchange. Please have a look at the How do I ask a good question? article.

As you are new to this forum I want to welcome you but nevertheless I would kindly ask you to first make your own research on a problem you have and then ask your question providing some information on what you do not understand. Your question seems to be a homework which you only copied without any signs of your own thoughts.
That said, here are some explanations regarding your assignment.
Question 1
If you put 10 grams of something into 120 grams of something else, than your resulting mass is 130 grams, right? So, 10 g of those 130 g are how many percent?
$$
\frac{10\,\text{g}}{130\,\text{g}} \times 100\,\text{%} = 7.69 \%
$$
Question 2
Magnesium nitrate ($\ce{Mg(NO3)2}$) has a molecular weight of 148.31 g/mol and you have used 10 g, so what is the amount of magnesium nitrate?
$$
\begin{align}
M &= \frac{m}{n}\\
\Rightarrow n &= \frac{m}{M}\\
&= \frac{10\,\text{g}}{148.31\,\text{g/mol}}\\
&= 6.74 \times 10^{-2}\,\text{mol}
\end{align}
$$
As you know, water has a density of 1 gram per milliliter, thus 120 grams correspond to 120 milliliters, which, again, is equal to 0.12 liters.
To calculate the concentration in mol per liter, just divide the amount in mol by the volume in liter:
$$
\begin{align}
c &= \frac{n}{V}\\
&= \frac{6.74 \times 10^{-2}\,\text{mol}}{0.12\,\text{L}}\\
&= 5.62 \times 10^{-1}\,\frac{\text{mol}}{\text{L}}
\end{align}
$$
Check
Let's check if our results are correct:
If you happen to have 120 milliliters of a 0.562 molar solution, then you can easily calculate the amount of magnesium nitrate by multiplication:
$$
\begin{align}
n &= c \times V\\
\Rightarrow n &= 5.62 \times 10^{-1}\,\frac{\text{mol}}{\text{L}} \times 0.12\,\text{L}\\
&= 6.74 \times 10^{-2}\,\text{mol}
\end{align}
$$
If magnesium nitrate has a molecular mass of 148.31 g/mol, then the mass can, again, be easily calculated by multiplication:
$$
\begin{align}
m &= M \times n\\
\Rightarrow m &= 148.31\,\text{g/mol} \times 6.74 \times 10^{-2}\,\text{mol}\\
&=  10\,\text{g}
\end{align}
$$

While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.


text:
Microscopic image
When I was reading a chapter on solids structure then i thought of how the real structure would be like if seen from microscope.
Hence I want the real microscopic molecular image of NaCl(common salt) and glass
I searched on Google but I don't get the real image

You might look for atomic force microscopy. Do not expect kind of photographs as the technique does not work with light. But the imagery one gets is as close as possible to microscopy picture such as those using visible light or electrons beams. There is a lot of physics behind your Q. Basically at the atomic scale no wavelength can "see". Tough x rays diffraction can also give images of lattices, but those despite based on optics, are even more far from a commonly intended picture of objects.

https://www.doitpoms.ac.uk/tlplib/dislocations/observing.php has a nice overview...

And here is an image with AFM: https://dqmp.unige.ch/renner/2013/11/04/atomic-force-micrograph-of-nacl/

Google image

There are no molecules in solid NaCl. What you are asking about only exists in gas phase.

An optical microscope can magnify the crystals of sodium chloride so you can see their beautiful symmetry:

But you said you want to get down to molecular (atomic) size. At those distances, we generate images with data (positions and bond distances) obtained from x-ray crystallography:

This picture suggests what a tiny crystal of sodium chloride would look like if you could see the individual ions. The colors are not real; these ions are smaller than the wavelength of visible light. But it is interesting that the maker of the picture colored the chlorine ions green.
That picture is the way I think of the crystal stacked: the ions are touching, but then I can only see the surface of the crystal. If I had x-ray vision and could see into the crystal, it might look something like this:

Not really like that. You'll get its diffraction reflected picture.


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Aluminium in homemade liquid castile soap?
I made some liquid castile soap in a slow cooker.  There was a slight scratch in the non-stick surface of the container.  I read that potassium hydroxide when coming into contact with aluminium creates hydrogen gas and hoped my container is stainless steel.  However, the container has corroded along the scratch by about a millimetre deep and wide.   The soap is good, but I am concerned that there is a small amount of alum in the product and this can be absorbed into the body.  Is the soap safe to use?

Al is not all that toxic; besides, you weren't going to eat your soap anyway, were you?

The question is actually is skin absorption of Al3+ a health risk. The answer from a legal perspective varies depending on where you live. In the USA, for example, the answer is no, however, there appears to be a growth industry in selling Aluminum-free products, but not in Europe, where apparently stick deodorants containing the Aluminum ions are banned.
But what is the suspected chemistry possibly suggesting health problems with the Aluminum ion? My prior exploration of this topic points to some recent research. Based upon a review of two articles noting the pro-oxidant property of the aluminum ion, a potential health risk pose by the ability of aqueous Al3+ to recycle transition metals (like Fe, Cu, Mn,..) to lower valence states (via the superoxide radical anion) in Fenton and Fenton-like reactions, contributing to so-called oxidative stress related diseases (including Alzheimer's disease, various cancers,...). More explicitly, with iron ions, for example, recycling between ferric and ferrous ions feeding a fenton reaction via superoxide radical anion (produced in our bodies):
$\ce{ Fe^{3+} + O2^{•−} <=> Fe^{2+} + O2}$
Source: See Eq (1.xxxiii) which can feed a fenton reaction forming the hydroxyl radical:
$\ce{ Fe^{2+} + H2O2  -> Fe^{3+} + •OH + OH-}$
The first alluded to article is Oxidative Stress Gated by Fenton and Haber Weiss Reactions and Its Association with Alzheimer’s Disease by Tushar Kanti Das, et al, published in Archives of  Neuroscience, July 2014 2(3): e20078, DOI: 10.5812/archneurosci.20078.
The authors cite, in Figure 4, “Formation of Aluminum Superoxide Semi reduced Radical Ion and Aluminum Superoxide Complex (43)”, with described reactions proceeding as follows (also adopting notation and water complexing from the second article):
$\ce{ [Al(H2O)4]^{3+} + O2^{•−}   <=> [Al(O2^{•−})(H2O)4]^{3+} }$
$\ce{ [Al(O2^{•−})(H2O)4]^{3+} + Fe^{3+} -> O2 + [Al(H2O)4]^{3+}  + Fe^{2+} }$
And, in the presence of H+ :
$\ce{ [Al(H2O)4]^{3+} + O2^{•−} + H+  <-> [Al(O2^{•−})(H+)H2O)4]^{3+} }$
$\ce{ [Al(O2^{•−})(H+)H2O)4]^{3+} + [Al(O2^{•−})(H+)H2O)4]^{3+} -> 2 [Al(H2O)4]^{3+} + H2O2 + O2 }$
Note per below, the action of hydrogen peroxide on ferric also leads to ferrous:
$\ce{ H2O2 <=> H+ + HO2- }$
$\ce{ Fe^{3+} + HO2- -> Fe^{2+} + HO2• }$
or, pH> 4.88, we have:
$\ce{ H+ + O2^{•−} -> HO2•}$
Another work: Pro-oxidant Activity of Aluminum: Stabilization of the Aluminum Superoxide Radical Ion by J. I. Mujika, F. Ruiperez, I. Infante, J. M. Ugalde, C. Exley, and X. Lopez in J., published in Phys. Chem. A 2011, 115, 6717–6723, American Chemical Society, dx.doi.org/10.1021/jp203290b .
In the Mujika article to quote “In addition, the presence of LMM ligands such as citrate could also have an indirect effect in the oxidation capacity of aluminum by augmenting the bioavailability of Al3+ species, shifting the formation of Al(OH)4- to higher pH’s. However, one should also take into account the effect of citrate chelation itself in the thermodynamic equilibrium of [AlO2•]2+ formation.”
So, my short takeaway, our body produces the short-lived superoxide radical anion capable of recycling problematic transition metals which can engage in fenton (and fenton-like) reactions forming powerful radicals. The latter can attack your DNA leading to associated diseases. The role of Al3+ is that it can form a complex with the superoxide, likely increasing its effective half-life and augmenting destructive radical activity in one's body.
To answer the question: "Is the soap safe to use?", likely that it is a personal decision based on one's understanding of the proposed underlying chemistry, as jurisdictions apparently vary in opinions.

what is "⋅OH"? is "⋅" just a typo?

Blade: "⋅OH" or at times, "OH⋅" and even "HO⋅"  refer to the hydroxyl radical, a very powerful but transient radical.

Blade: FYI, I have since provided background source links.


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Why does 1,4-diiodobenzene have a non-zero dipole moment?
From this answer, I came to know that 1,4-diiodobenzene has a dipole moment of $0.19~\rm D$ in benzene at $20\ ^\circ\rm C$. On the basis of symmetry, I'd expect the bond moments due to the difference in electronegativities of iodine and carbon atoms to cancel each other out, similar to other 1,4-dihalobenzenes. However, it turns out to be non-zero. What is the reason for this observation?
I asked the same question in the comments, and I was suggested that this could be due to the high polarizability of the large iodine atoms. I can understand this reason since this is how books explain the increase in boiling point of alkanes with increase in chain length. And in this case, the boiling point of 1,4-diiodobenzne ($285.0\ ^\circ\rm C$) is higher than 1,4-dibromobenzene ($218.5\ ^\circ\rm C$). But this is the first time I'm seeing that dipole moment is altered by dispersion forces and I'm also confused by the issues described below.
We know that dipole moment $\vec{p}$ of a charge distribution is given by:
$$\vec{p} = \sum_i q_i \vec{r_i}$$
where $q_i$ and $\vec{r_i}$ are the charge and position vector of the $i$th charge about a particular origin. If it's the dispersion forces, arising due to the lack of symmetry of the fluctuating electron cloud, how can we ascertain a particular value of dipole moment as both $q_i$ and $\vec{r_i}$ are indeterminate?

Related: Does 1-bromo-4-chlorobenzene have a permanent dipole moment?

I guess "in benzene" is crucial here. You're treating these molecules as separate, while aromatic compounds have pi-stacking and even CT-complexes.

@Mithoron: If I understood your comment properly, you mean, the bond moments are not only due to the bonding electron but also due to the delocalised pi-electrons from the benzene ring. This is understandable for me. But I think even this case is symmetrical and by logic the net dipole moment must be zero. This is why I thought of those dispersion forces.

No, I mean that molecules bind to each other a that may be source of polarisation.

pi stacking is a good suggestion, and so the dipole moment would then be different, i.e. zero, in other solvents. Of course a small amount of other diiodo impurity  could cause this result.  This compound may be hard to keep pure. It is strange that there is some uncertainty in such a simple measurement.

See my comment here https://physics.stackexchange.com/questions/562140/why-does-1-4-diiodobenzene-have-a-non-zero-dipole-moment#562140. I suspect that its either determined by a method where the "method average" is non planar, or its the heterolytic auto-dissociation. In any case a reference is missing to judge that.

Of course it can be also a simple typo in source you got it from as well. I would be interested in the ionic conductivity of solutions of  p-C$_6$H$_4$I$_2$, possibly the values are already substantial ... (due to the autodisocciation I suspect).

With the method of determination I do not mean the solvent, I mean the physical measurement method. For example one could use microwave spectroscopy to determine the dipole moment, ... . Autodissociation can happen also in benzene ...


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Why does DFT-D finds different optimized geometries for the same structure when using different input geometry files?
When I change c lattice parameter and optimize the structure with DFT-D, every time I get a significantly different geometry with different energy? Why is this happening with DFT-D?

In general, the end point of an optimization depends on the starting geometry. Starting from a different geometry can lead you to wind up in different minimum.

@RagonLoso, how different are these starting structures? Are they minor 0.1Å tweaks? What are the molecules? Finally, what is the magnitude of the difference in the energies?

I personally find this question but very specific. What kind of structure are we talking about? What kind of DFA? Convergence criteria, programs involved, algorithms, etc.. An important question would also be how you reduced the issue down to the dispersion correction (which one?)

Simply put, a geometry optimization finds a local minimum. Many such minima may exist for a given system and it is up to the user to ensure that a reasonable minimum is used for further calculations. Techniques may include using many handmade starting geometries, using a random mutation algorithm or starting from known, experimental structures.


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What could be causing my bath salts containing baking soda to explode out of their containers?
I run a small herbal wellness business and one of the products i sell are bath salts containing baking soda. The base ingredients are Epsom salts, Dead Sea salts, magnesium chloride, baking soda, coconut milk powder, and fractionated coconut oil. I’ve made and sold this recipe for over a year and because i buy my ingredients in bulk i haven’t bought new ingredients recently. All of a sudden, some kind of chemical reaction is happening (which i assume is coming from the baking soda), and after i seal the bags they blow up like a balloon and then eventually explode. I cannot for the life of me figure out what is causing the baking soda to react. I’ve noticed when I’m mixing up batches that there is a fizzing sound that has never been there before, and I’ve wondered if the salts have somehow absorbed water. But water doesn’t really cause baking soda to react like this. Thanks for the help!

Try the other ingredients one by one against the baking soda.

Magnesium chloride will certainly pick up water from the water vapor in air. Perhaps then the coconut milk powder is then producing an acid, maybe lactic acid (?), and then the baking soda and acid react to yield carbon dioxide. This is just my best guess: I never even heard of coconut milk powder.

Coconut milk powder is basically just dried full fat coconut milk. It contains essential fatty acids and lauric  acids (medium chain). Would these be the types of acids to react with baking soda? And why is it just now happening after a year of production? Thanks for the comments y’all.

All acids do react with baking soda. Plenty of substances are transformed into acids by a long standing in contact with the air. The best example is the wine (solution of ethanol) which gets slowly transformed into vinegar by contact with air. And vinegar is a solution of acetic acid.

I think the water pickup is key: dry baking soda and dry citric acid, for example, do nothing. But with moisture, they react to yield carbon dioxide, water and the other stuff. So the reaction gets going faster, etc. I second Ivan’s suggestion: try the coconut powder and baking soda mixture in a baggie, add a little water, and seal the baggie.

Fractionated coconut oil also contains some acids, so I’ll do two experiments- one with the milk powder and one with the oil. Thanks!

I suspect the seasonal introduction of humidity is the root of the problem, which is readily taken up by the magnesium chloride. Then, the following possible reaction:
$\ce{ MgCl2 (aq)  + 2 NaHCO3 (s) -> 2 NaCl (aq)  + Mg(HCO3)2 (aq)}$
Now, aqueous Magnesium bicarbonate exists only in solution (so decomposes on drying), is not stable on standing (per my experience), or warming (whereupon it also decomposes):
$\ce{ Mg(HCO3)2 (aq) ->  MgO (s) + H2O (l) + 2 CO2 (g)}$
where the pressurized buildup of carbon dioxide gas is likely causing a problem.
Solution: Try adding a drying agent or removing the very hygroscopic Magnesium chloride from your formulation.

Thank you so much! Do you have any suggestions just to get the ball rolling in my mind of what to look for in a drying agent?

If the $\ce{MgCl2}$ has already taken up a lot of water, there may not be much you can do---you may just have to buy new $\ce{MgCl2}$. Once you have new material, store it in an airtight container with a bunch of water absorber packets (e.g., silica gel). You should be able to find the absorber packets at various stores... search for things like "desiccant" or "silica gel packet".

I am afraid that this equation has nothing to do with reality. Magnesium bicarbonate would decompose to magnesium carbonate not magnesium oxide. It may happen but only at very high temperatures.
$\ce{ Mg(HCO3)2 (aq) ->  MgO (s) + H2O (l) + 2 CO2 (g)}$

In line with the comment above I do not think that the equations shown are involved. This is as to say that Mg bicarbonate is just more igroscopyc than other ingredients already present in the mix. As stated in other comments it is likely an acid has been formed by the coconut part upon storage. OP You could check the pH of it.

M.Farooq: The more correct (and confusing to those not acquainted with MgCl2) is that on standing in the presence of moisture, it actually creates a basic chloride per my experience. This is nicely outlined in the discussion of 'Basic Chlorides of Magnesium' here http://magnesium.atomistry.com/magnesium_chloride.html.


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Does standard electrode potential depend upon temperature?

Does standard reduction potential depend on temperature?

I am confused because my institute gives the definition for standard reduction potential as the potential that is measured at standard state which defines as measured in atmospheric pressure at $\pu{298 K}$, and when the concentration of all the substances is unity. But as far as my understanding goes standard states can be defined at any temperature and not just $\pu{298 K}$.

Also, as its directly related to $\Delta G^\circ = -nFE_\mathrm{cell} = \Delta H^\circ - T\Delta S^\circ$, this points toward it being dependent on temperature. Also, when someone says standard state, does that put restriction on temperature being $\pu{298 K}$??

Yes, see the Nernst Equation.

@MaxW Thank You, but I still don't get it, how does nerst equation help ?? Like all it does is relate electrode potential to standard electrode potential , it says that the electrode potential will depend on temperature, but nothing about the standard electrode potential

I think I disagree with @MaxW here, but it's really over semantics, not over the substance of his point. The standard potential is standard. If you have a non-standard state, then you have a different value. In other words, the standard value is constant, but the actual electrode potential definitely depends on temperature.

@Zhe Does the standard state mean it's only at 298 K ??

@Zhe As far as I know the standard state does not have any particular temperature show the standard potential at 298 k can be different from that at 273 k

The standard state is 298 K. I would not expect the potential to be the same if you change the temperature...

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

$$ \Delta G = -nFE_\mathrm{cell} = -RTln(K)+RTlnQ $$

@Zhe - Thanks for the correction. I wholeheartedly agree that you can't have a standard potential in a non-standard state.

Yes, the standard reduction potential does depend on temperature. The definition of the standard reduction potential is stated in Ref.1 as:

A standard electrode potential $E^\circ$ is defines as the potential (in Volts, $\pu{V}$) of a half-reaction relative to a reference electrode at a specific temperature, all chemicals being at their standard states at unit activity. These states may be arbitrarily defined as pure crystalline solids, pure liquids, ideal gases at one atmosphere fugacity ($\pu{101325 \times 10^{5} Pa}$), and ideal solutes at unit molality. The most common temperature for the tabulation of standard electrode potentials, as for their thermodynamic data, is $\pu{25 ^\circ C}$ ($\pu{298.15 K}$).

Accordingly, tabulated data is chosen to be at $\pu{298.15 K}$ for the convenience. This is also beneficial that the most common reference electrode is the Standard Hydrogen Electrode (SHE) for the solvent water:
$$\ce{2H+_{(aq)} + 2e- <=> H2_{(g)}} \tag1$$
$E^\circ_{(\ce{H+/H2})}$ for the above half-reaction is been assigned as zero volts at all temperatures. Extended work over $\pu{298.15 K}$ by De Béthune and coworkers shown that the temperature dependence of $E^\circ$ is approximately linear between $\pu{273.15 K}$ and $\pu{373.15 K}$ according to the following equation (Ref.2-Ref.4):
$$E_T = E^\circ_{298} + (T - 298.15)\left(\frac{dE^\circ}{dT}\right)_{298} \tag2$$
References:

Steven G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).
A. J. De Béthune, T. S. Licht, N. Swendeman, “The Temperature Coefficients of Electrode Potentials: The Isothermal and Thermal Coefficients—The Standard Ionic Entropy of Electrochemical Transport of the Hydrogen Ion,” Journal of The Electrochemical Society 1959, 106(7), 616 (doi:10.1149/1.2427448).
G. R. Salvi, A. J. De Béthune, “The Temperature Coefficients of Electrode Potentials: II. The Second Isothermal Temperature Coefficient,” Journal of The Electrochemical Society 1961, 108(7), 672 (doi: 10.1149/1.2428187).
André Jacques De Béthune, Nancy A. Swendeman Loud, In Standard Aqueous Electrode Potentials and Temperature Coefficients at $\pu{25 ^\circ C}$; C. A. Hampel: Skokie, ILL, 1964.

So basically, the Nernst equation will only be valid if I use the same T (Temperature) value at which my E∘ (used in that specific Nernst eq) is defined? i.e. The Temperature parameter of a Nernst equation is a constant, only concentration can be changed?


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Heat capacity of a calorimeter
Why do we mix water with different temperatures to calculated the heat capacity of the calorimeter?
Why can't we calculate it directly using just warm water?
Asked differently, would the following be wrong? (With $T_E$ the equilibrium temperature and $T_C$ and $T_W$ the initial temperatures of the caloriemeter and water)
$$
c_{C}=\frac{\Delta Q}{m_{C} \Delta T}=\frac{c_{W} m_{W}\left(T_{E}-T_{W}\right)}{m_{C}\left(T_{C}-T_{E}\right)}
$$

You need a temperature range to calculate heat exchange (Recall $Q_\text{lost or gain} = m \times c \times  \Delta T$ where $c$ is specific heat of the solution).

Wouldn't it be sufficient to know the initial temperature of the caloriemeter, the water and the resulting temperature of both?

In a thought experiment, a lot of things are possible. To do it for real, you need a reproducible  experiment with a step-by-step list of actions.

@klugreuter: Where is the specific heat capacity of the caloriemeter (assuming $m_C$ is the mass of caloriemeter)? And, now you realize their should be a temperature range.

The specific heat capacity of the caloriemeter is the thing im trying to determine, named $c_C$ above.
Sorry, my knowledge about thermodynamics and chemistry is extremely limited, I'm just trying to figure out why - when we want to get some information about how much energy is lost to the system -  we cant just put in some energy and see how much is lost afterwards.

It’s hard to measure the temperature of a solid. It is easier to equilibrate it with water, and measure the water temperature.

So is it because of practical rather than theoretical reasons? Also wouldn't the initial temperature of the calorimeter be the room temperature?

Karl's comment (see above) prompt me to suggest the following rough experiment which can be done in the classroom with minimum equipment:
Suppose you are required to find out the specific heat capacity of water. Weigh $m_1 \ \pu{g}$ of warm water at $T_1 \ \pu{^\circ C}$ to a Styrofoam cup with a lid. Also weigh $m_2 \ \pu{g}$ of ice at $T_2 \ \pu{^\circ C}$ to another Styrofoam cup. Then add ice to warm water (Note: Measure the temperature of warm water just before addition to make sure it is still at $T_1 \ \pu{^\circ C}$). Stir the mixture carefully while the coup is covered by the lid. Measure the temperature of the mixture until reading is steady. Suppose the steady temperature is $T_3 \ \pu{^\circ C}$. Ensure this time no ice is visible (meaning all dissolved). Now we are ready for the calculations:
Suppose the specific heat capacity of water is $c_{(\ell)}$ and the specific heat capacity of water is $c_{(s)}$. For novices, we can assume that $c_{(\ell)} \approx c_{(s)} = c$. Thus, heat lost by warm water at equilibrium:
$$Q_\text{lost} = m_1 \times c \times (T_1 - T_3) \tag1$$
For ice, if $T_2 \lt 0$, ice would gain heat first to become ice at $\pu{0 ^\circ C}$, then $\pu{0 ^\circ C}$ ice melts to $\pu{0 ^\circ C}$ water, and warm up to $T_3 \ \pu{^\circ C}$ water. If heat of fusion of ice is $L \ \pu{Jg-1}$, total heat gained by ice at equilibrium:
$$Q_\text{gained} = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) \tag2$$
At equilibrium, $Q_\text{gained} =  Q_\text{lost}$. Therefore, equations $(1)$ and $(2)$:
$$m_1 \times c \times (T_1 - T_3) = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) tag3$$
Only unknown in the equation $(3)$ is $c$, you can solve the equation $(3)$ for $c$.
Also note that you can avoid $m_2 \times c \times (0 - T_2)$ part by choosing $\pu{0 ^\circ C}$ ice, because it is easy to get $\pu{-20 ^\circ C}$ from freezer, it it kept in constant temperature.


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Can we fill potato chips bags with a gas other than nitrogen?
I understand that we fill potato chips bag with nitrogen to prevent oxidation. But why do we use nitrogen, instead of neon or hydrogen or something else?
My first guess is that nitrogen is lighter than neon/argon but what about hydrogen or helium?

Because nitrogen is cheap.

And doesn't explode.

Hydrogen? So you want chips bag floating around like balloons and have to be hold down?

Nothing is better than chips freshly baked in hydrogen flame.

From personal experience (traveling to a number of different customers in behalf of Lako Tool  to install heated bag crimpers ("sealing jaws") on vertical form, fill, and seal wrapping machines), I have never seen any gas nozzles that would fill the bags. In fact, that could inflate the bags and therefore interfere with the sealing process.

As Nilay Ghosh said, nitrogen is cheap. Very cheap. Neon is expensive. Argon is cheaper than neon, but considerably more expensive than nitrogen. Helium is also expensive and needs to be used wisely, for important things, e.g., cryogenics. And hydrogen! I can just see the ads: “Buy our chips: they are lighter than air! But avoid open flames and sparks unless you want to be Hindenburged to a crisp (no pun intended)”
Another fill gas to avoid is sulfur hexafluoride. A tennis ball manufacturer once decided to fill tennis balls with sulfur hexafluoride, assuming this would prevent the balls from going flat as a consequence of the high molar mass of sulfur hexafluoride. But the tennis balls exploded on the shelves because air diffused in.
Thankfully, no one has ever tried using nitrous oxide as the fill gas in potato chip bags!

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

Also, SF6 is the most potent known greenhouse gas, with a global warming potential of 23900.

Funny as it may seem, if the bags were filled with nitrous oxide I'm guessing a lot of folks would buy them for the gas.

Nitrous oxide would be a bit of a laugh.

Nitrous oxide is an oxidiser and would cause the oils to go rancid faster.

Exploding tennis balls sounds like a perfect example of planned obsolescence.

Since people use the Nitrous Oxide propellant in whip cream dispensers as an inhalant, and these are referred to as "whippets," would doing a similar thing for chips be referred to as "chippets"?

Could be! I would still like to see ice cream made (safely, outdoors) using liquid nitrous oxide instead of the liquid nitrogen used in the ubiquitous demos. Not that that does not taste good, but laughing ice cream would be a gas! (Sorry for the pun: I could not resist!)

Not mentioned yet: nitrogen is entirely non-toxic, environmentally friendly, does not contribute to global warming or ozone depletion.
In very good approximation, nitrogen is just air with the oxygen removed that would oxidize the contents. The usual production process consists of liquefying air, distilling it, and then selling the gases separately. This makes food-quality nitrogen very cheap. It's not a huge problem if there's 1% oxygen left in the bag.

Not sure about the case of potato chips, but there is often used mixture 80% nitrogen + 20% carbon dioxide, probably to actively suppress eventual growth of anaerobic microflora. That applies probably more on animal products like meat.

I'm recalling when I was working in research for the Air Force, we would get liquid nitrogen for our equipment that was a by-product of the production of liquid oxygen for aircraft.

oxygen and nitrogen separation by fractional liquidation is done in large scale in iron/steel processing industry,  as using pure oxygen is reportedly cheaper than air, reaching high temperatures easier and avoiding useless warming up of nitrogen.

@Poutnik: The process also yields argon, which is a valuable gas for welders. At welding temperatures, nitrogen is no longer inert, but argon is a noble gas. Nitrogen is really what remains when the valuable stuff is sold.

Substances that are gaseous down to, say, -10 are finite (and not very large) number.
Of these, we want:
Safe against ignition (so no hydrogen or gaseous hydrocarbons)
Not poisonous (phosphine, arsine, carbon monoxide - out!)
Not corrosive (chlorine, fluorine - sorry)
Without unpleasant smell (hydrogen sulfide is no-go)
Not related to processes of food degradation (oxygen)
Low diffusion in everything, in order to stay into the package (helium, go home - but sometimes used in cans)
Environmentally friendly (CFCs and friends - great if it wasn't for ozone layer)
Cheap (krypton, xenon - stay away from our mass products)
A lot of them fail more than one of the above.
And those that pass, have to be approved by local food quality regulator.
What we have left:
Nitrogen - numero uno for inert packaging. Cheap as hell, inert for all practical purposes.
Carbon dioxide - a bit more expensive, easier logistics (lower pressure tanks), less suffocating potential (heavier than air and has some smell), alters the taste a bit (but not always bad - see soda drinks).
Argon - adds a touch of luxury, otherwise just like nitrogen. Sometimes just not separated from nitrogen. I have seen it used in an occasion when nitrogen is off.
Air - cheaper than nitrogen and still contains ~80% nitrogen. Sometimes mixed with nitrogen in order to get the oxygen down to acceptable level.
p.s. other gases can be food-grade as well, including, but not limited, to hydrogen, nitric oxide, sulfur dioxide, oxygen, ammonia. They are used for other purposes in the food industry, but not for packaging.

Nitrites do it more efficiently.

@EdV yes, but CO is related to processing and not to packaging.

The reason to employ nitrogen in place of air, in addition to a cost advantage, is the fact that air contains oxygen, possibly water vapor, dust particles (containing trace amounts of transition metals) and even microbes (bacteria, mold spores,...) as well.
Upon warming with time, I would not expect that such a mix is a good inert medium to store/preserve food.
While using hydrogen may provide some marketing hype, my experience is that the gas is very adept at escaping and any mixing with air followed by accidental ignition may result in burns. So just adding pure N2 from liquid nitrogen is probably a better idea.

There is a cost advantage for using nitrogen over air?

@hkBst depends. Are "customer returns" treated as an externality?

@hkBst Nitrogen will cost more than air unless it happens to be in excess as a by-product in close proximity with transport cheaper than compressing/pumping air. However the reason to use Nitrogen is to avoid the oxygen in the air.


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Mechanism for addition of hydrogen halides in Alkynes
There is a question in my book where one is supposed to add 2 moles of HCl to But-2-yne.
I've attempted this and got 2,3- Chlorobutane instead of 2,2-chlorobutane which is given as the answer.
I'm unable to attach a photo of the same.
My doubt is, after one chlorine has been attached and the triple bond has been changed to double bond, why is the second Chlorine also attached to the same position as the first one.
One explanation, I think -
Since due to inductive effect of first chlorine, there should be delta plus charge near the chlorine attached atom, so H+ from second Hcl molecule goes to the adjacent atom. Therefore,  same C atom has both Chlorine.
But if I think other wise, as in stability of Carbocation and carbanion, adjacent C atoms 2,3 are getting Chlorine.
Please help me clear the doubt

@Guru I checked out the site and mechanism given but that is exactly what my doubt is, why is it a geminal chlorine. I mean on second addition, why is Carbocation formed on the C having Chlorine. Wouldn't -I effect of Cl cause high instability?

@Mithoron Yes I've understood the concept now!


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Why does 'softening' of lead oxidise only elements other than lead?
In a flowchart describing the steps to refine lead obtained after the reduction step, my book1 gives a short note on the process 'Softening of lead' where the crude lead is melted in the presence of air. It's stated that this removes scum of oxides of copper, tin, arsenic, antimony and bismuth which constitute the hard lead. After this soft lead with silver and gold is obtained.
Earlier, under the section which explained about refining of gold, the book described about the process cupellation where crude gold is melted in the presence of air. This method was meant to remove lead and the success behind this method was attributed to the high oxygen affinity of lead. The lead oxide being volatile gets easily separated from the system.
I understood the above two scenarios independently. However, when I thought about the lead refining process with the volatile nature of lead oxide in mind, it seems contradictory on why in the first case lead remains in the molten state whereas in the second it gets oxidised and escapes into the atmosphere.
My question is, why does 'softening' of lead oxidise only elements other than lead? Why doesn't the lead get oxidised and escape into atmosphere as volatile lead oxide? Compared to the impurities, as per the Pauling scale of electronegativity, lead is more electropositive and this seems to suggest it must get oxidised before the impurities which is opposite of what must be inferred in order for the lead softening process to be successful.
References

Guha, Sudarsan. Concise Inorganic Chemistry for JEE (Main and Advanced). 4th ed., Wiley India Pvt. Ltd., 2017.

Apparently, alloys have their unique ratings on a scale of electronegativity as well. So, I suspect it is the latter alloy properties, not pure Pb, that is the explanation in the heterogeneous environments in question.


text:
Common chemcials that produce phosphorescence
Are there any common lab chemicals or household stuff that produce phosphorescence?
I am aware of ZnS, CaS, Fluorescein, Erythrosin B, Vanillin Benzaldehyde, Vitamin B2, Tryptophan etc.
I just need something for sanity check on a fluorescence/phosphorecence lifetime measurement system. I am getting results with fluorescence, but not with phosphorescence.
My samples are biological plant material. With TCSPC in nano second timescale, I am getting fluorescence decay. But with micro and milli second timescale, I am getting only unifom constant counts (looking like background counts). The sanity check is to see if the instrument is able to detect phosphorescence decay from any material known to have phosphorescence, hence the question.
Counts with Nano second timescale

Counts with Milli second timescale (Red = IRF, Black = Decay)

There are plenty of “glow in the dark” phosphorescent things you could try, e.g., the phosphorescent stars some people put on their kid’s bedroom ceilings. You can easily find phosphorescent tape online and likewise various colors of phosphorescent paints.

It is a commercial system -  Horiba DeltaPro. Excitation is with DeltaDiode Pulsed laser 375 nm. Detection is with PPD-650 (230-700 nm range). Timing electronics is DeltaHub-HT (25 ps to 1s lifetimes). Using a sample holder with front face orientation at an off-45 degree angle. In addition there is a long pass filter with 395 nm cut-on to negate scatter.

Thanks, very nice system from what you say. So it is puzzling that you are “not getting results” with phosphorescence. The list of chemicals you gave does not suggest room temperature phosphorescence will be observed with them, particularly if oxygen, even at very low concentration, is present. Freeze-pump-thaw deoxygenation might do the trick (this is standard practice), but I would personally just use a phosphorescent tape or sheet.

Phosphorescence in organic materials is hard to measure unless the solvent has had all the oxygen removed as this is a very efficient quencher, as are heavy atoms/ions such as iodide. Thus you need to  freeze-pump-thaw on a vacuum line with liq Nitrogen, or purge with O2 free nitrogen or argon and seal the cell completely, vacuum taps etc.   Fluorescence lifetimes are best measured by time correlated single photon counting on nanosecond timescale  (many commercial systems exist),  phosphorescence is very  slow (microsecond and longer) that a laser and scope are easily fast enough.

My samples are biological plant material. With TCSPC in nano second timescale, I am getting fluorescence decay. But with micro and milli second timescale, I am getting only unifom constant counts (looking like background counts). The sanity check is to see if the instrument is able to detect phosphorescence decay from any material known to have phosphorescence, hence the question.

@Crops, I copied your main question from comments to the main question, otherwise random people will keep down voting an otherwise useful query.

@Crops, Could you also add the TCSPC of the blank solution? Secondly, why are you assuming that you would have phosphorescence at room temperature in the presence of air and in solution? It is a rare phenomenon.

It is not a solution. These are solid samples placed at front-face orientation with an off 45 degree angle.

Instead of TCSPC of blank solution, I have added the one from the Ludox solution, which is used as the IRF. 10000 is the preset peak, ie. the counts are read until the no. of counts in the largest bin reaches the preset value. I will try again with the maximum timescale (1s) possible with the machine.

I'm not surprised that you are getting a constant background on such a long time scale, just thermal noise will give this even with a cooled detector. Are you running the source at only 100/sec otherwise a bit of scattered light and /or or fluorescence  from this might be giving your signal. Was the long time signal with a blank of with a sample? As you use plant material this article  which shows weak, time independent, chlorophyll chemiluminescence (observed by tcspc) may be interesting  Beddard et al. Nature, vol 267, p373,1977.

@porphyrin, How come the background reading is constant at 1000? The fluorescence decay also starts from 1000.

@M Farooq There could be lots of reasons, we do not know what the experimental conditions are. If the light source reprate is MHz then this could just be fluorescence or scattering.

Try a white LED; some phosphors have half-lives in the multi-millisecond range.

Uranyl ion has a fluorescence lifetime of about 1.5 microseconds: Catalysis of ionic reactions by polyelectrolytes. III. Quenching of uranyl ion fluorescence by iron(II) ions in poly(vinylsulfonic acid) solution, Herbert MorawetzIssam A. I. Taha, J. Am. Chem. Soc. 1971, 93, 4, 829-833 Publication Date:February 1, 1971 doi.org/10.1021/ja00733a005. Uranium glass, sometimes called ‘vaseline glass’, is easily available in university glass shops and on eBay. Not phosphorescence, but relative long lived fluorescence and very convenient to use. Also, Amazon and Home Depot sell phosphorescent paint.


text:
Why is an atom with excess neutrons unstable?
In this chart, I can see that stable nuclides (other than hydrogen) have a neutron count greater than or equal to their proton count, and that the neutron:proton ratio for stable nuclides increases with atomic number.
I can also see that in many cases, radioactive nuclides which have excess neutrons undergo beta minus decay, converting one of the excess neutrons to a proton, and moving toward that stable ratio.
I can see why having excess protons might make an atom unstable: they have positive charge and push each other apart. But it seems to me that excess neutrons would make it more stable, adding to the strong force between the nucleons.
Why is an atom with excess neutrons unstable?

Let's put it this way: why are free neutrons unstable?

OR, why are not  there any stable neutron only kernels ? Even 3H should be by your idea more stable than 3He, but the opposite is true.

Cross-site detailed Q&A: https://physics.stackexchange.com/questions/78107/what-do-we-know-about-the-interactions-between-the-protons-and-neutrons-in-a-nuc

https://chemistry.stackexchange.com/questions/21848/stability-variation-in-heavy-nuclei

Pulling out Theoretical Nuclear Physics by Blatt and Weisskopf (published in 1952, so a bit dated in areas but still a good introduction), one finds (in Chapter VI.1 The Systematics of Stable Nuclei) that the stability of isobaric nuclei (all the same $A$) is determined by:

(1) The symmetry effect: the charge independence and the exchange character of nuclear forces, together with the Pauli exclusion principle, strongly depresses the energy of nuclei with equal or nearly equal numbers of protons and neutrons.

(2) The charge effect: the effect of the Coulomb repulsion of the protons favors nuclei with fewer protons than neutrons. The neutron-proton mass difference also enters the charge effect, but only as a small correction. The charge effect increases in importance with increasing nuclear charge.

(3) In a few special cases a third effect is important: the spin dependence of nuclear forces favors parallel spin over anti-parallel sin of a pair of (extra) nucleons.

So, yes, you picked up on the Coulomb bit (number 2 above), but it is the symmetry effects of the strong force that are more important, particularly for light nuclei. Until one gets to some nuclear physics, the characteristics of the weak and strong forces just aren't familiar to us humans.

Neutrons are fermions; they obey the Pauli exclusion principle. With only a few protons, most baryons in a nucleus would be neutrons, and there would be just a few low-energy states for those neutrons. Hence, most would be in higher-energy states.


text:
How to compare three or more molecular structures?
Let's say you have three or more different geometry files (xyz) for the same structure and you want to compare them. What would be best way for that?

RMSD seems like the first obvious choice. Although, of course, it depends on exactly what you are trying to compare them for...

Isn't it just for two geometry files?

Somebody else will probably have a better idea, but I guess what you're saying isn't entirely clear to me. If you want to sort a list of three or more items, it appears to me that you need some kind of pairwise comparison...

You mean I need a geometry file also for reference?

No, I think my point is just that comparison functions are written for two inputs. For example, you can't compare the numbers 3, 4, and 5 all at once. You can sort them by saying that 3 < 4 < 5, but implicitly you're using a pairwise comparison 3 < 4, 3 < 5, and 4 < 5. So what exactly are you trying to achieve with this? It might be helpful if you added some more context to your question.

I basically want to compare atom positions of these files with each other

Do you know the system will be orientated the same in each case

TL;DR There is no generic best way. Develop a meaningful comparison for your system first, then the question of how to compare multiple structures will become less convoluted. Pick a reference structure to draw your conclusions.
Compare only two structures
The whole issue depends on finding a meaningful way to compare structures in the first place. Naively one would assume that an root-mean-square deviation of atomic positions (RMSD)[1] is a good way to compare two structures, but for most practical purposes it won't. With this approach you'll condense two sets of information (quite a lot of information) into a single number. If you don't have a scale, then what does it mean.
A simple example, take the structures of butadiene:[2]

For this example, the calculated RMSD is 1.4513261239839574.[0] In a strict sense, if you swap atoms 9 and 10, because a rotation happens, the RMSD should be different. A well implemented algorithm will know that these atoms are indistinguishable and produce the same RMSD. However, order matters sometimes, so be aware of that, too.
To the more elementary question: What does this number actually mean. So we would need some kind of scale. From a visual inspection, we would say they are quite different structures; from an (implied) chemical understanding we'LL immediately pick up that there is (in first approximation) a single dihedral different. In this case a more meaningful comparison would be structure parameters like bond distances, angles, dihedrals, or out-of-plane angles.
A counter example, now just considering trans-butadiene as structure, but comparing the methods RI-BP86/def2-SVP and RI-PBE/def2-SVP.[3] Now the RMSD is 0.0019595403851554454. If the question is how well these methods agree with each other, then the temptative answer is quite well. But again, if it were not for experience, that value could mean anything.
A counter counter example, would be on how reliably one method of calculation will reproduce itself. Now using repeated calculations of trans-butadiene at the RI-PBE/def2-SVP level of theory, the RMSD is 2.0844418808250816e-16. That is basically zero, isn't it. But without knowing what we have compared and what to expect, this value could again mean anything.
With everything considered this far, we only know how well (in an unknown fashion though) two structures compare. This is like comparing 2 and 3; thus finding out the RMSD is 1. That doesn't really help.
However, if you want to know how well PBE will reproduce the structure of BP86, then the RMSD suddenly becomes a lot more clear. While it doesn't really matter with a comparison of two sets, picking a reference case for your comparison matters to the question.
In conclusion calculating the RMSD for two structures might or might not be the way to go. What you'll always need is a reference, otherwise your question might be meaningless.
Multiple structures
Now that we have established that comparing two structures depends heavily on the question that is being asked, we know that this will apply here, too.
For example, if you consider a conformational space, consider the low energy structure as reference. Then you can actually sort your structures according to the RMSD to that structure. Programs which determine conformational space will actually use such an approach.
If you want to know how well a computational method reproduces a crystal structure, you have pretty much already picked the reference. What you now need to figure out is whether the RMSD is helpful. Going from condensed phase to gas phase will lead to different problems. Angles and dihedrals are influenced by other molecules you might not be considering. However, if you just look at the bond lengths, you can also draw conclusion which method reproduces a parameter best.
In any case, you might want to consider omitting hydrogen atoms from the structure.
This is quite frequently done when benchmarking a methodology for investigations of the reaction mechanism. It is also a good way to use to 'calibrate Density Functional Approximations' (DFA).
If you want to justify that your chosen method reproduces the structures well enough as an approximation, then the obvious choice would be the method you want to use. This has often be done to show that using a simple DFA using less time will produce reasonably well converged structures thus skipping expensive reoptimisations.
There certainly are more things to consider, but I think this illustrates the problem with the whole approach. If you want to compare three things, you need to know how to compare the first, then what to compare them against. If you cannot answer this question, your comparison might not be useful.
As always, look at the literature what others are doing for similar systems, then try to approach yours the same way.
Notes & References

I will be omitting the unit from all values; angstrom is therefore implies. This is simply because the programs used will also imply the units.

There is a nice implementation of RMSD in python available on GitHub: https://github.com/charnley/rmsd. It will be used here, too.

Optimised with xtb, see https://github.com/grimme-lab/xtb
trans-butadiene:
C        0.989655924      0.075383602     -0.024713443
C        2.317729081      0.061870073      0.001720395
C        3.092825332     -0.523680990      1.074346659
C        4.420898600     -0.536792302      1.100972837
H        0.437167570      0.519522240     -0.834969799
H        0.405781945     -0.358847639      0.771074562
H        2.884926249      0.505132896     -0.807127933
H        2.525654828     -0.966802276      1.883291521
H        5.004795037     -0.101504125      0.305779782
H        4.973395433     -0.980161479      1.911645418

cis-butadiene:
C        0.960901519      0.102797367     -0.074150724
C        2.285553694      0.054182217      0.015302774
C        3.054624452     -0.537267624      1.098841827
C        2.562107526     -1.129034473      2.181574196
H        0.470193589      0.563081554     -0.914527335
H        0.316113842     -0.309782226      0.682610316
H        2.879283339      0.488966088     -0.779420603
H        4.129461610     -0.473089986      0.981620967
H        1.504218325     -1.224871958      2.355773288
H        3.205852103     -1.543149959      2.938534295

Calculations are performed with Orca 4.2.1.
RI-BP86/def2-SVP:
C   0.95575989884130      0.08100590986283     -0.03334466433621
C   2.30910835995513      0.06320383319263      0.00080877611734
C   3.10145823671758     -0.52437495050823      1.07559644898752
C   4.45480621149184     -0.54218133779601      1.10974928378639
H   0.40481134955806      0.53782521459098     -0.86973010000421
H   0.35678816482669     -0.36408211138767      0.77899770211471
H   2.87466566256150      0.51941793049988     -0.83303737235663
H   2.53590053819549     -0.98058344364338      1.90944605848559
H   5.05377847073995     -0.09710562028587      0.29740171404077
H   5.00575310711246     -0.99900542452517      1.94613215316474

RI-PBE/def2-SVP:
C   0.95730901184150      0.08075188309749     -0.03290025865850
C   2.30976982144587      0.06285884282725      0.00142432803956
C   3.10079629620100     -0.52403054214147      1.07498080335691
C   4.45325694810625     -0.54192705246080      1.10930497169718
H   0.40637317799734      0.53746037139229     -0.86906027083378
H   0.35957667735107     -0.36458957651717      0.77995109642634
H   2.87598438022296      0.51876441714347     -0.83182376939508
H   2.53458184326634     -0.97993147006835      1.90823171834507
H   5.05099013560010     -0.09659642313629      0.29644859687702
H   5.00419170796757     -0.99864045013644      1.94546278414529

I think when I suggested RMSD I was thinking also about the possibility of clustering similar geometries together. It's not a comparison per se, but a fairly common thing to do, afaik. I refrained from posting an answer, because I really can't tell what OP wants to do...

@orthocresol I agree, it really isn't clear what OP wants or needs. RMSD also was my first impulse, just adding to use a reference structure. When it came down to answering, I realised the problems that come along with that approach. I hope I gave a basic overview about what can be done and what to consider before using such an approach.

I looked at this and decided anything beyond trying RMSD required a lot more depth, and even then wasn't sure I had a good answer - so didn't reply. I was thinking along the lines of the COM (after translation) comparison being like comparing dipole moments and Moments of Inertia like quadrupoles, so in some sense we are generating a multipole expansion of the mass od the structure, but using mass instead of charge. But while I can give recipes that generate number as @Martin-マーチン points out interpreting what those numbers mean is not straightforward.

First I would calculate the centre of mass (com) then go on and calculate the moment of inertia axis vectors for each structure and use these to align them via a rotation matrix. Make the centre of mass the same for both first. Now start to measure the difference in position of each atom; find the mean or centre of the same atom's position between structures and make a sphere of the max radius of each group and plot these to see where your structures differ. You can then home in on parts that are most dissimilar.
A measure of the global difference may not mean much as some parts may be the same and some different.

If the C.O.M. are different it might just be a simple translation of the whole structure, which is otherwise identical

@Ian Bush ah yes, good point. I had assumed structure was about same xyz; finding the inertial axes should still work. I will edit the answer.


text:
How to predict which complex is formed and which are soluble?

How does one come to know (in this table) which metal cations form ammonia complexes and which ones form hydroxy complexes on reaction with $\ce{NH4OH}$
Reference:

J.D Lee - Concise Inorganic Chemistry by Sudarshan Guha

You can't really know specific properties without experimentation, also you can make rules and generalisations for similar behaivour. I think a good starting point would be the Wikipedia article for the 'HSAB theory'. It mainly fulfills your need for the complex formation prediction. The theory even assigns values for acids and bases based on the valence state, ionisation energy and electron affinity called hardness.

Can you please illustrate an example of this using HSAB principle as this has been extremely confusing

You may need more time to understand the topic, assuming it is new to you. There is a picture of the periodic table (f block elements excluded) on the mentioned article found at the Theory paragraph. The picture associates strenght with color, and the bonds tend do be more stable when hard and hard or soft and soft are pairing. As your source states Zn and Cd do form amine complexes, while Al doesnt.

Al is considered a hard acid while Zn is intermediate, Cd is soft. O donors are somewhat harder bases than N so based on the aforementioned assumption the softer Zn and Cd pair better with the softer N donor while Al pairs better with the harder O donor.

I have come across HSAB principle before particularly in the context of conjugate addition.However it is a bit new for me to apply this in inorganic chemistry but I think I get the idea. How would you apply this for determining solubility ?

Predicting solubility reliable would be surely a Nobel-prize canidate, as for now it is mostly impossible to calculate it. The Wikipedia article mentions that generally soft-soft interactions are more covalent wich tend to be insoluble in water, while hard-hard bonds have ionic character that are expected to be soluble in water. While this is a nice statement we immediately face an exception namely Al(OH)3 wich is a hard-hard interaction between Al3+ and OH-, is insoluble in water and has a covalent nature. Overall there are rules you can apply, but there will be always exceptions.

What can be done to make this question better?

Please be mindful, in my opinion, that your source material may not be of the highest quality.
For example, NH4OH is more accurately represented as aqueous ammonia (that is, NH3(aq), see, for example, this source's comment).
Second, Al(OH)3 does not form a complex with ammonia or, as your reference claims, is 'partially soluble' with aqueous ammonia to any significant extent, and if at all, likely due to the presence of OH-. See, for example, this ebook, Basic Analytical Chemistry
 discussing various metal compounds interactions with aqueous ammonia.
Third, a more accurate depiction of the action of ammonia on Cu(OH)2, for example, is the formation of a complex of the form, $\ce{[Cu(NH3)_x(H2O)y](+)}$, where x + y = 6, with also a possible charge of 2+ for cupric, where the concentration of ammonia influences the value of 'x' (supplying guidance on your question as to which possible complex may form).
Lastly, high strength ammonia is needed to react with some of the listed metal hydroxides and is characteristic of the particular metal. See my suggested source, that may address your question on which salts are (or, are not) soluble.

Do you have another link for "Ammonium hydroxide doesn't exist"? The one above requires to be purchased.

@B.Anshuman These previous chem.SE questions(here and here) will be of help.


text:
Compact way to regulate 3000psi (200bar) down to 30psi (2bar)?
This was originally posted on the Lego page, and closed because it's not actually about Lego itself.  I thought I'd give them a chance first, in case someone had done this already and had some application-specific knowledge.  Then my next stop is here, since you guys are probably more familiar with the gear that I'm looking for.

I have this, that I want to power with a HPA, paintball, auxiliary SCUBA, or whatever-someone-decides-to-call-it tank:

I think a 6.1cm diameter, 0.22L or 0.35L tank will fit nicely in place of the 0.5L plastic water bottle.  I can redesign things and move some structure around if I have to, but the overall size is pretty well fixed.
The air inlet is a sports inflator needle pushed into the black hose in the front, with a bicycle pump currently attached to it.  The maximum pressure there is about 35psi (~2.3bar).
I can use a longer hose there if needed too, but that's about as short as it can be.  The straight section with a slight kink is the inflator needle.
So, is there a compact way to regulate the HPA tank from 3000psi (200bar) down to 30psi (2bar) with a schrader-valve type connection at the end?

I see that a fair number of HPA tanks (but not all) come with a factory-attached pre-regulator, that can apparently be adjusted by taking it apart and changing the spring.  But it seems like a stretch to me, to take something that is designed for 3000psi in / 800psi out, and get it all the way down to 30psi out instead.  Maybe it's possible, and if so, that's great!  But it seems like a bit much to me.
It seems more likely to me, to just barely get into the range that a shop air regulator can handle, and use that for the final delivery as well as a complete shutoff.  But there are still adapters to be found, and perhaps some creative folding to make it all fit in the available space.
I also found this, which says it takes 4000psi in and is fully adjustable from 4000psi out to 0psi out.  It looks a bit cumbersome at first glance, compared to the space that it would go in, but no real size reference except for the gauges.  Also, with such a wide range of adjustment, it might be difficult to have it "barely on" and stable at my required 30psi.  Not to mention a rube-goldberg chain of adapters, probably on both the inlet and the outlet sides of it.
Is there an elegant way to get the required 100x drop in pressure, with decent regulation (there's a throttle valve after the final regulator), in the space shown, and with the right connector at the end?  (imitation schrader valve)

:/ Unfortunately I can't see how it would be about chemistry, either.

@Mithoron Do you know a site that would take it?

I'm not sure, perhaps https://engineering.stackexchange.com/ ? You've account there so should know better... Mod can migrate it if there's consensus,

@Mithoron Oh yeah.  I'd forgotten about that one.  I'll have to poke around there again and see if it fits.  Thanks for the reminder!

And it's posted.  Should I delete this one now?  Or just let it close on its own?

I’m voting to close this question because it belongs to engineering and was already cross-posted.

:/ Didn't I just tell you that mods migrate? What you did is cross-posting, and you shouldn't do it in the future.

Google ‘compact pressure regulator 3000 psi’ and look through the results.

@Mithoron So I'm just supposed to guess where it works and wait for a non-migrated close before I try somewhere else?  There were three places I could have posted, the third of which I had forgotten about.  The most obvious one just kicked it off with no mention of migrating at all.  That's not the first time it's happened either.  So I don't trust that system.  Yes, it was a different site with different mods and a different culture, and maybe that's part of the problem, but anyway, I hope you see my side too.

ALL of this is SE, and that's what I was going on.  If there was a consistent trend of migration instead of just closing, then I'd be more inclined to use that.  But if I just get closed and that's it, I'm going to cross-post because they're effectively independent anyway.


text:
Can potassium nitrate and sulfuric acid be used to nitrate organics?
If we are doing a classic nitration to produce glyceryl trinitrate, one uses a mix of concentrated sulfuric and nitric acids. Additionally, one can produce concentrated nitric acid by distilling it from a mixture of potassium nitrate and sulfuric acid.
So, is it possible to do a nitration by mixing the glycerol with sulfuric acid and then adding potassium nitrate?
Disclaimer
In case anyone is wondering, I am not going to try this, and I do know how to make many better and safer explosives since I used to be a member of the alt.engr.explosives group way back.

@NilayGhosh No. I know how to make that. My explosive of choice would be Kinepak. It's a LOT simpler

I'd rather add nitrate first, as H2SO4 would start reacting with glycerine immediately, or at least be quick with adding the salt.

A simple, but dangerous path to nitrating a metal oxide (say $\ce{MgO}$) is to treat with $\ce{NO2}$ gas (caution: toxic and corrosive), as to quote Wikipedia on Nitrogen dioxide:

$\ce{NO2}$ is used to generate anhydrous metal nitrates from the oxides:
$\ce{MO + 3 NO2 -> M(NO3)2 + NO}$

whereas heating a mixture of $\ce{Cu}$, $\ce{KNO3}$ and $\ce{H2SO4}$ will generate $\ce{NO2}$.
A safer path to nitrating, starts with the introduction of the powerful  radicals, like the hydroxyl radical ($\ce{^.OH}$). The latter can transform commonly available nitrate salts  (like $\ce{KNO3}$) to the nitrate radical ($\ce{^.NO3}$) especially in the presence of an acid catalyst per the reaction:
$\ce{^.OH + NO3- <=> OH- + ^.NO3}$
A supporting source reference, Nitrate radicals and biogenic volatile organic compounds: oxidation, mechanisms, and organic aerosol. Per the paper's abstract:

Oxidation of biogenic volatile organic compounds (BVOC) by the nitrate radical (NO3) represents one of the important interactions between anthropogenic emissions related to combustion and natural emissions from the biosphere.

Also, see The Reactivity of the Nitrate Radical ($\ce{^.NO3}$) in Aqueous and Organic Solutions, by
Stephen P. Mezyk, Thomas D. Cullen, Kimberly A. Rickman and Bruce J. Mincher, per the abstract to quote:

ABSTRACT
Rate constants for the nitrate ($\ce{^.NO3}$) radical reaction with alcohols, alkanes, alkenes, and several aromatic compounds were measured in aqueous and tert‐butanol solution for comparison to aqueous and acetonitrile values from the literature. The measured trends provide insight into the reactions of the $\ce{^.NO3}$ radical in various media. The reaction with alcohols primarily consists of hydrogen‐atom abstraction from the $\alpha$‐hydroxy position and is faster in solvents of lower polarity where the diffusivity of the radical is greater. Alkenes react faster than alkanes, and their rate constants are also faster in nonpolar solution. The situation is reversed for the nitrate radical reaction with the aromatic compounds, where the rate constants in tert‐butanol are slower. This is attributed to the need to solvate the $\ce{NO3−}$ anion and corresponding tropylium cation produced by the $\ce{^.NO3}$ radical electron transfer reaction. A linear correlation was found between measured rate constants in water and acetonitrile, which can be used to estimate aqueous nitrate radical rate constants for compounds having low water solubility.

With respect to the hydroxyl radical, the UV photolysis of a photocatalyst like $\ce{TiO2}$ in aqueous $\ce{KNO3}$, could be a path to the powerful, but very transient, hydroxyl and nitrate radicals.

Nitration reactions rely on the absence of water to proceed efficiently, in fact part of the purpose of the sulfuric acid in the mixture is as a drying agent to remove the water produced by the nitration reaction.  Because potassium nitrate (and potassium sulfate for that matter) is not meaningfully soluble in concentrated sulfuric acid, it is impossible to create the dry solution required by this reaction in this way.

In that case, why can one add pure sulfuric acid to potassium nitrate and get nitric acid, in the abscence of water?

This is thoroughly incorrect.

Could you please correct me, or at least offer insight into what exactly is incorrect then?

I realize that by some definitions what's being discussed here isn't even technically nitration, since nitration can specifically refer to aromatic structures, and that the primary role of sulfuric acid in either case is the formation of the nitronium ion, but I wouldn't consider either of those to be strictly in the scope of the question.


text:
How do I figure out the composition of a material given specific densities?
The Problem
A waste material consists of sand (SG 2.7), a chemical salt (SG 2.1), and a dense plastic (SG 1.4). 200 g of a representative sample from the material is found to have a density of 2300 kg/m3. The salt is dissolved out of the sample which, after drying, now weighs 170 g. What is the composition of the waste material?
My Attempt

Based on the given information, we can derive:

densitysand = 2700 kg/m3
densitysalt = 2100 kg/m3
densityplastic = 1400 kg/m3
masssalt = 30 g = 0.03 kg
proportionsalt = 0.03 kg / 0.2 kg = 0.15

To try and calculate the individual mass of sand and plastic in the sample, I used the density = mass / volume equation to first figure out the volume using the density and mass of the total sample.

densitytotal = masstotal / volume =
volume = masstotal / densitytotal = 0.2 kg / 2300 kg/m3 = 0.00008696 m3
masssand = 2700 kg/m3 * 0.0000869 m3 = 0.2348 kg
massplastic = 1400 kg/m3 * 0.0000869 m3 = 0.1217 kg
But I don't think this step was the right choice since the mass of the sand came out to be larger than the mass of the sample. Was I correct in trying to figure out the value of the mass of sand and plastic or was that the wrong approach to determine the material's composition? If it's the first case, how should I find out the mass values?

Your mistake is that you have calculated the mass of a sample with the same volume as ours consisting entirely of sand, and a sample with the same volume as ours consisting entirely of sand.
volume = mass_total / density_total = 0.2 kg / 2300 kg/m^3 = 0.00008696 m^3
gives the initial volume of the sample before the salt is removed, however by multiplying this by the densities of sand and plastic, we calculate the mass of a 0.0000869 m^3 sample of sand and a 0.0000869 m^3 sample of plastic.
First we need to find the new volume and density of the sample after the salt has been removed.  This can be done by calculating the volume of salt removed:
volume_salt=0.03kg_salt/2100 kg/m^3=0.00001429m^3
By subtracting this volume from the initial volume you already calculated:
0.00008696m^3-0.00001429m^3=0.00007267m^3
dividing the new mass by this gives:
0.170kg/0.00007267m^3=2,339kg/m^3
Now we need to use this to calculate the proportion.  We can do this by creating and solving a system of equations.  We know that some mass of sand plus some mass of plastic gives us a 170 gram mixture, and that some volume of sand plus some volume of plastic gives us a mixture with a volume of 0.00007267m^3.
xkg_sand+ykg_plastic=0.170kg
am^3_sand+bm^3_plastic=0.00007267m^3
Now, we know that the volume and mass of each material are related by their densities, and because of this it is possible to replace the terms in one equation with the terms in the other by relating them through density, so if we decide to solve for mass, our volume equation becomes:
(xkg_sand/(2700 kg/m^3))+(ykg_plastic/(1400 kg/m^3))=0.00007267m^3
Because the x and y in these equations are the same, we can solve a system, which yields:
x=0.1418kg
y=0.0282kg
Dividing these by our starting mass gives:
proportion_sand=0.709
proportion_plastic=0.141
These can be rounded to match your answer if needed, however these proportions combined with your calculated salt proportion adds up to 1.