text:
How does a gas occupy a fixed molar volume?
It is common knowledge that one mole of ideal gas occupies a volume of 22.41 L at standard temperature and pressure (0°C and 1 atm). However, this goes against the claim that gases don't have a fixed volume. So what does it mean for a gas to "occupy" a certain volume under specific temperature and pressure? How is this observed/measured/experimentally found?
Do the gas molecules stay relatively close to each other that the volume in which they are huddled together is said to be the volume they occupy?

Gases do have a fixed volume under specific temperature and pressure.

Related: If a gas always occupies the volume of its container, will its volume always be 22.4 l at STP?

Short answer: Yes! They stay close to each other in a confined volume of 22.4l per mol, on which walls they excert a pressure of 1 bar.

To Karl: Do you mean that if I let 1 mol of some gas into a 100L rigid container, the gas molecules will stay bound to a region in space that has a volume of 22.4 L? I don't think that would be the case.

Pressure.$%please character limit …$

Does this answer your question? If a gas always occupies the volume of its container, will its volume always be 22.4L at STP?

When saying that a gas occupies 22.41 litres of volume at S.T.P. , it means that the gas is under constant pressure even though it's volume isn't kept fixed.
You are right that a gas expands to fill it's container, but that happens only when there is no pressure applied on gas.
Imagine it like this, there is a cylindrical container whose one circular end is fixed and at the other end is a piston which is free to move and is filled with a gas. If you apply a certain fixed force on the piston, gas will compress. As you decrease the force applied, the gas will occupy more and more volume and if you increase the force gas will occupy less volume.
Gas in atmosphere is constantly under pressure from the gas above it and also is under gravitational force, thus doesn't expand freely as a force is acting on it. If you release some gas in very less pressure environment, like in outer space, it will expand freely, if isn't under influence of any other force.
I would also like to point out that gas clouds in space, like nebulae also don't expand freely as their own gravity is stopping them.
Thus, it isn't that gas doesn't have a fix volume, but it doesn't have a fix volume if allowed to expand freely under no pressure.
Volume of gas is measured similarly as you would measure liquid, only the pressure, temperature and number of moles also need to be mentioned.
Besides if you think the volume of liquids(even solids) also vary with temperature and pressure so there also isn't a "perfect" way to measure volume. It's just that their rate change of volume with changing conditions is not as pronounced as that in gases as intermolecular forces  which keep them together or apart is weaker in gases.

Yes gases have no fixed volume,but only in condition when no pressure is applied on them,and when they aren't  under the the influence of any force,but in case of molar volume we are applying a condition of STP,which show that there is a constant pressure of 1atmosphere, so in this case volume can be calculated as a fix volume.

How about adding a reference to $V={nRT \over P}$?


text:
Why does Gibbs free energy correspond solely to non-expansion work?
First of all: I know this question has been asked before (Why does the Gibbs free energy only correspond to non-expansion work?), and although I understand the mathematical derivation given there, I struggle with the concept.
If both $\Delta G$ is negative and the change in enthalpy is negative (exothermic), I don't see why we cannot use that heat to do mechanical work? So how is it that it only cooresponds to non-mechanical work?

E, F, G, and H are all derived from $dQ = T dS = dE +pdV$ based on what two independent variables you want to choose, be it (S,V), (S,p), (T,V), or (T,p). So, if you understand what constraints you have chosen then you should directly be able to get to what the particular free energy means.

As Jon Custer explains that is more or less the definition of $\Delta G$. If you are interested in all of the work look at the Helmholtz free energy instead.

For a closed system in contact with a constant temperature reservoir held at the initial system temperature T, and in contact with a surroundings at an external pressure held at the initial system pressure P, and experiencing a change between two thermodynamic equilibrium states, the change in internal energy is given by:  $$\Delta U=Q-W$$If we express the work W as the sum of expansion work and non-expansion work (the latter, say, obtained by generating electricity or turning a shaft), then $$W=P\Delta V+W_{NE}$$And, because there is a single constant temperature bath comprising the surroundings, the heat added to the system is related to the change in entropy between the initial and final thermodynamic equilibrium states of the system by:
$$\Delta S=\frac{Q}{T}+S_{gen}$$where $S_{gen}$ is the entropy generated as a result of any irreversibility's that may be present.  So, if we combine these equations, we obtain:$$W_{NE}=-\Delta G-TS_{gen}$$where $\Delta G=\Delta U-T\Delta S+P\Delta V$.  So, for any process that goes between the same two thermodynamic equilibrium  states, the maximum non-expansion work that the system can do on its surroundings is $-\Delta G$ (since $S_{gen}$ can only be positive).  The maximum corresponds to a reversible path between the two end states.

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How can I get create complex metal ion solution such that the majority of the central metal ions only have one specific ligand?
I'm looking for a way to have the ligand, Chloride ion, datively bond with Copper Sulphate.
I understand it is as simple as creating a solution of Hydrochloric Acid and dissolving Copper Sulphate inside it will produce tetrachlorocuprate(ii) ions, however there are still the presence of hexaaquacopper(ii) ions in the solution.
Is there any method to create a solution purely consisting of tetrachlorocuprate(ii) or at least make the amount of hexaaquacopper(ii) negligible?

Here two my ideas:

Play by rules, add copper to a solution of $\ce{HCl}$ and $\ce{H_2O_2}$. The basic goal is to "eliminate" copper: $\ce{Cu + H2O2 + 2HCl → CuCl2 + 2H2}$ and then obtain your ion: $\ce{CuCl2 + 2Cl^- <=> [CuCl_4]^2-}$.
Risk. (It might not work). Use $\ce{[Cu(H_2O)6]Cl2 + 2NaOH = [Cu(OH)_2(H_2O)_4]↓ + 2NaCl + 2H_2O}$ (blue precipitate forms). I haven't seen evidence that $\ce{NaOH}$ reacts with $\ce{[CuCl_4]^2-}$, but this idea might be risky. It's always better to play by rules :)

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


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How does critical temperature affect adsorption?
I don't really understand how does a high critical temperature favour adsorption?
What I think is that the lower the temperature, more easy is it for the gas to liquefy and therefore adsorption is favoured

It's exactly opposite.

Higher critical temperature = more adsorbed.
So, what is critical temperature? When gas is heated above it, it can't become liquid, even if you'll increase pressure. Why? Because it's large kinetic energy makes intermolecular forces kinda "irrelevant". And adsorption is all about those forces.
High critical temperature means that gas particles are good in intermolecular forces - "tenacious". This allows them to adsorb (cling to surfaces) more easily than those that won't hold on (low critical temperature).


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Does coffee have the ability to dissolve less sugar into it than water?
Like water can absorb such and such an amount of sugar, can coffee absorb less as it already has coffee in it.

This is not something you can deduce through logic alone.

I will agree with Ivan Neretin. That being said, just because other components are dissolved in solution does not mean that solubility of other components will be affected. (i.e. if you dissolve sodium chloride in water, the solubility of sucrose may or may not be affected). If you have some spare sugar, a scale, coffee, and a little extra time, I'd recommend playing with it a little bit. Look at the literature solubility of sucrose in water, make a known volume of coffee, mass a fair amount of sugar, and see how much you can get to dissolve. Do a few unit conversions and compare the value you get to literature. Could be a good weekend experiment.

Then again, the solubility is far too great to be measured accurately at home.


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Can hydrides react with hydrocarbons? And why?
I have read there exist so called "magic" (super strong) acids which can provide a proton to saturated hydrocarbons, producing cation and hydrogen molecule. Can hydrides act in the same manner (pick a proton to form free hydrogen molecule)? And why?

Depends on hydride.

The pKa of sodium or lithium hydride is (solvent dependent) about 35, the pKa of Butyl lithium is around 50 so clearly H- is not strong enough to tear a proton off an unactivated alkane pKa tables here

Then again, cyclopentadiene is also a hydrocarbon, so...

Not exactly unactivated though as the cyclopentadienyl anion is aromatic

True, but there was no "unactivated" in the original question.

The OP asked about saturated hydrocarbons, which excludes cyclopentadiene,

OK, I stand corrected.


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Wondering about tetrasilicates
I found two papers from which two different formulae for the tetrasilicate (tetrasilicic acid) are found:
1.https://www.sciencedirect.com/science/article/pii/S0022309302015491.
2.https://www.ncbi.nlm.nih.gov/pubmed/26482329.
Of course, the usual formula for polymerization of oxacids suggest the first one is the right one, that is, the tetrasilicate anion being $\ce,{Si_4O_9^{2-}}$, however, how to understand the second formula? Is something missing?

First you mention is cyclic, second linear, probably tetracyclic "tetrahedron of tetrahedra" also exists.

Yeah, but why? I believed the polymerized oxacid tetrasilicic acid had only a single formula...

"tetrasilicic acid" in not a  unique name. SiO4 tetrahedra may be connected in as many ways as possible.

Polymerization: n(oxoacid)-(n-1)$$H_2O$$. That is why I got surprised...

Just read https://en.wikipedia.org/wiki/Silicon%E2%80%93oxygen_tetrahedron

Polycondensation like this occurs for difunctional monomers. Orthosilicic acid has four hydroxyl groups.

Without images this is unclear. Some people are actually sitting behind paywalls. Please [edit] your question to make it more clear.

"Tetrasilicate" is one of special guys. For example, sulfate is $\ce{SO_4^2-}$, while $\ce{SO_3^2-}$ is sulfite. But tetrasilicate is $\ce{Si_4O_n^{(2n-16)-}}$ - thus, it simply has four $\ce{Si}$ atoms (without connotation that they have special number of oxygens near 'em): check for
http://www.biomedsearch.com/nih/Novel-silicate-anion-Si8O2212-Hydrothermal/11487344.html, which is also tetrasilicate (here tetrasilicate is $\ce{Si_4O_11^6-}$).
https://www.researchgate.net/publication/50934696_Preparation_structural_and_luminescent_properties_of_Ba2Gd2Si4O13Eu3_for_white_LEDs (tetrasilicate is $\ce{Si_4O_13^10-}$), etc.
[edited] I've found for you even tetrasilicate with 15 oxygens: https://link.springer.com/article/10.1007/s00710-016-0447-1
Also check for amicite (zeolite): it has approximation $\ce{K2Na2Al4Si4O16·5(H2O)}$. It seems for me like 16 oxygens is maximum for tetrasilicates, but maybe someone would disagree - interesting to know that other tetrasilicates exist.

How many special "guys" similar to tetrasilicates are out there? Wonderful!

Pretty many.$,$

Examples out there?


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Metal Evaporation in GFAAS?
Our lab is doing aqua regia digestion of an ore matrix for the determination of Platinum Group Elements (PGE). We're getting white smoke when we run it through a Graphite Furnace assisted Atomic Absorbance Spectroscopy (GFAAS). Smoke forms above 1000 C, so we think it's some metal evaporating. Any thoughts or ideas? Thanks in advance!

Please give us a little chance to understand your question and [edit] in what those abbreviations mean.

@Martin-マーチン GFASS relates to Graphite Furnace assisted Atomic Absorption Spectroscopy, i.e. a sample is quickly atomised in a graphite tube.  To ensure the sample won't condense at the inner walls of the tube, often, a L'vov plate is added supporting the sample -- then, if the sample atomizes, the walls of the tube is even hotter than the tube.  And graphite, to mute potential ASS fingerprints by metals.  PGE -- speculating -- may refer to platin group elements (Ru, Rh, Pd; Os, Ir, Pt).

Martin-マーチン Sorry for the confusion! Thank you for suggesting edits so I can ask questions better next time on this forum. Buttonwood, thank you for clarifying my remarks!

No, you do not evaporate the metals in question.  The use of aqua regia in the analysis aims to dissolve them, to form salts, which in turn are significantly more volatile than the metals or will decompose in the gaseous state.  (Just have a look alone at the melting points of the pure metals, which still are considerably lower than the boiling points, too.)
Without further detail about the colour of the "smoke", it remains a speculation that the sample placed on the L'vov plate still might have been a bit wet (from the acidic digestion).  If so, it were not good for the instrument, as it may lead to corrosion to metal parts of the housing around the graphite tube.

In the good old days of wet chemical  analysis ; It was typical to dissolve a sample in acids and then heat it to dryness producing " copious" smoke.

@blacksmith37 The good-old silicate analysis of concrete/cement and steel samples I learned in France equally consists of crushing the sample with $\ce{NH4Cl}$, than gentle cooking for half an hour or so with concentrated HCl.  Not too seldom that eventually the sample was boiled (Teflon beaker) with aqua regia for quite some time, either.  Well, the hood had to be cleaned anyway from time to time from the inner, too.

@Buttonwood thank you very much for answering this query of mine! We are planning on switch methods that will involve tellurium precipitation, like how this paper explains link. We think there must be something in the soil matrix that is interfering with our GFAAS method, because even when we did a nitric acid digestion to measure cobalt and palladium levels, white smoke was still emerging. Thanks again for your insight and willingess to answer this question!

@ScienceEnthusiast Interestingly and quite unexpected for me, your reference approaching petrography is indexed by the freely accessible database PubMed (https://www.ncbi.nlm.nih.gov/pubmed/18964774). As so often, at the side there is the column "similar articles" (here, 251 entries) which might contribute to screen literature further. Good luck!


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How to write a cube file?
This is a technical question. I have a real-space representation of a potential in a grid of points, however I have the grid in the format:
x  y  z  w  v
meaning: the coordinates of each point, the weight and the value of the potential.
I would like to visualize this potential but I need to re-write this potential in a cube file. I haven't found any particular software or script to do that.
I know how to get the header of the cube file. However I don't know how to re-write the points.
Is there a specific format to write the points?
Thank you

I'm voting to close this question as off-topic because it's not about chemistry.

The cube files by Gaussian are documented (e.g., http://paulbourke.net/dataformats/cube/, or https://h5cube-spec.readthedocs.io/en/latest/cubeformat.html).  Or, is it a question how to fill them with data? Then it were more on the side of writing a script which reads your data from one source and injects them into this format.  The first source has an example that may serve as comparison when setting up a looped script them, e.g. by Python.

@Buttonwood. Thank you, yes my question is more related in how to fill them with data. I am not sure how to rearrange the points that I have.

This is very related to (technical aspects of) computational chemistry.  It's been lightly discussed in the past and turning our noses up to these types of inquiries would only serve to lower the quality of our site and limit our audience.

I would recommend using the functional PySCF cubegen script as a starting point for your own script. (https://sunqm.github.io/pyscf/_modules/pyscf/tools/cubegen.html)
If we have a grid defined with nx ny and nz points along the three axes, generating the cube file is just a nested loop over z, y, and x. (In that order) You should end up with nz lines of output each with nx*ny entries.


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What is PPM in terms of air concentration?
When they say the $\ce{CO2}$ concentration is $\pu{350 ppm}$ what does this actually mean? Does this mean that if we took a sample of air say $\pu{1 m3}$ then $\pu{350 mg}$ of that cubic meter would be $\ce{CO2}$? Is it simply $\pu{mg//m3}$?

It means there are 350 CO2 molecules for every 1,000,000 molecules in the air (N2,O2,...).

@kelly There is a macro for units you can use, it makes things a bit easier. It is \pu{ ... }.

Please see the criticism section of the Wikipedia page for details why this question is a bit hard to answer.

The unit parts per million refers to the unit molecules. Therefore, 350 ppm CO2 means that for every million molecules of a fluid, there are 350 molecules of CO2.
Concentrations of chemicals in air are typically measured
in units of the mass of chemical (milligrams, micrograms,
nanograms, or picograms) per volume of air (cubic meter
or cubic feet).
However, concentrations may also be
expressed as parts per million (ppm) or parts per billion
(ppb) by using a conversion factor. The conversion factor
is based on the molecular weight of the chemical and is
different for each chemical. Also, atmospheric temperature and pressure affect the calculation.
Typically, conversions for chemicals in air are made assuming a pressure of 1 atmosphere and a temperature of
25 degrees Celsius.
For these conditions, the equation to
convert from concentration in parts per million to concentration in milligrams per cubic meter (mg/m3
) is as
follows:
Concentration (mg/m3
) = 0.0409 x concentration (ppm) x
molecular weight
To convert from mg/m3
 to ppm, the equation is as follows:
Concentration (ppm) = 24.45 x concentration (mg/m3
) ÷
molecular weight
Source: https://cfpub.epa.gov/ncer_abstracts/index.cfm/fuseaction/display.files/fileid/14285

Ok great, Why do they sometimes express it as mg/m^3?

See updated answer (feel free to upvote and accept the answer by clicking the check mark)

Upvoted. Thank you! So if we say there are 300ppm of C02 that is somewhat arbitrary or dimensionless ? I guess thats what is confusing me, the units as in 330ppm what are the units?

@user66634 When ppm is based on molar concentration (as it usually is for gas concentration) the result is a unitless ratio. We are comparing molecule counts to molecule counts. This works well for gases as it the same ratio applies to partial pressures and volumes at constant pressure. IF other contaminants like particulates are the topic, though, this doesn't work and a mass/volume ratio might be used. This will have units of weight/volume.


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Confused in comparing the acid strength of these two cyclic compounds
I am having a hard time understanding the more acidic hydrogen in the following two compounds.

I attempted it by looking at two things, finding the possible resonance and the EN of the Sulfur and Oxygen. All these point to me that h1 is more acidic as the carbon anion formed after removal of H shall be more stable than the h2. But, h2 is more acidic. How is it so? What am I missing?

There is a possibility of back bonding of lone pair generated by Proton removal in second case in the vacant 3d orbitals of Sulphur, which is not possible in case of Oxygen. So, in second case, conjugate base will be more stabilised.

@SoumikDas No.

@Mithoron Why not ?

Because outer $d$ orbitals are not really valence orbitals.  Can we construct molecular orbitals that favor the sulfur containing anion?

When comparing acidic strength, compare the stability of conjugate base. One with more stable conjugate base will be a stronger acid.
When $H+$ will leave, a negative charge will be left on the carbon. And remember:

Lone pair and negative charge can't go in resonance due to high repulsion between electrons of both

And

Stabilization due to resonance is greater than that due to inductive effect

While drawing resonance structures, both of them will have resonance with the +ve charge on the $\beta$-ketone, so the only difference will be due to S and O. Sulphur and oxygen both have lone pair, but it can't participate in resonance.
But S has empty d-orbitals which can do back-bonding, and thus participate in response while oxygen cannot do so. And because resonance trumps over inductive effect, it even if Oxygen is more EN, will not stabilize as much as by Sulphur's back-bonding.
Thus $H_2$ is more acidic.


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Is HMgX also a Grignard Reagent like RMgX?
We do know that the Grignard Reagent is $\ce{RMgX}$ where R is some alkyl group & X is any halogen group.
So, my question is that, whether $\ce{HMgX}$ is also a Grignard Reagent like $\ce{RMgX}$?

I have never come across HMgX as a synthetic reagent, and I'm unsure of its existence. This review may offer more detail though https://onlinelibrary.wiley.com/doi/full/10.1002/anie.201708592

According to Ashby and Goel [1,2] $\ce{HMgX}$ compounds are a thing.  You can't make them with magnesium +$\ce{HX}$, of course (what actually happens?), but you can get them from magnesium halide salts plus an "active" form of magnesium hydride in THF.  They act as hydride ion sources for reaction with $\ce{AlH3}$ and $\ce{BH3}$ (forming mixed tetrahydridoaluminate and tetrahydridoborate salts, respectively), and reference [2] also indicates some substrates such as benzophenone are also reduced through accepting hydride ions from the hydromagnesium complex. Such a hydride transfer would be analogous to conventional Grignard reagents transferring their alkyl or aryl anion moieties.
References

E. C. Ashby and A. B. Goel (1977). "Preparation of HMgX compounds and their reactions with alane and borane in tetrahydrofuran". Inorg. Chem. 16, 11, 2941–2944.
https://doi.org/10.1021/ic50177a055

E. C. Ashby and A. B. Goel (1977). "The preparation of HMgX compounds". J. Am. Chem. Soc. 99, 1, 310–311.
https://doi.org/10.1021/ja00443a085

But it is probably better not to call them a Grignard reagent as they will have far more in common with metal hydrides.

Do we call them "Grignard reagents"?  Do we call Pluto or Ceres a "planet"?  That's ultimately not so much a law of nature as our nature.


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Another salt-dough question
In January 2017, someone asked how salt dough worked.
How does salt dough work?.
In the answer it was said that the salt gives toughness.  My question is would the size of the salt crystals matter. Kosher table or ground to a powder?  This has relevance to a craft project.

I'd say doesn't make a big difference, but the processability might be worse with fine powder. Why don't you try it out?


text:
In the Nernst equation, does T refer to the temperature of the standard cell or the cell under non-standard conditions?
I am doing an experiment where I am varying the temperature of the electrolyte of the anode half cell. I am trying to calculate the cell potential using the Nernst equation under standard conditions but I'm not sure what T to put in.

Think of the Nernst equation as the electrochemical analog of the equilibrium expression in terms of Gibbs functions, so it refers to a constant T. The T you put into the equation should be that of the reference conditions. The equation refers to a process at constant T: T and $E^0 $ refer to the same T.

The Nernst equation is $\displaystyle E=E^\text{o}-\frac{RT}{nF}\ln(Q)$ where $Q$ is the ratio of activities, $a$, of products over reactants, e.g. for $aA+B\leftrightharpoons cC+D$ then $\displaystyle Q=\frac{a^c_Ca_D}{a^a_Aa_B}$. The emf of the cell is $E^\text{o}$ which is measured when the emf of the cell is zero, $E=0$, no current flows, and means that the reactants and products are at equilibrium. In this case $\displaystyle E^\text{o} =\frac{RT}{nF}\ln(K_e)$ where $K_e$ is the equilibrium constant.
The measured emf $E$ changes with temperature as indicated by the Nernst equation and this is a good way to measure thermodynamic properties; for example the reaction free energy is  $\Delta G = -nFE$ and because $\partial\Delta G/\partial T|_p=-\Delta S$ then $\displaystyle \Delta S = \frac{\partial E}{\partial T}\bigg|_p nF$ which is a convenient way to determine the entropy change in a reaction by measuring the reaction emf at two temperatures and letting the derivative be $\Delta E/\Delta T$, which will be accurate for a small change in temperature, and give $\Delta S$ at the mean of the two temperatures. Since $\Delta G = \Delta H -T\Delta S$ the change in enthalpy can also be calculated.


text:
Is there a non toxic moisture indicator that can be applied on a wall?
I am looking for a substance that could serve as an indicator for a risk of mold growth, e.g. by a change of its hue. Cobalt chloride is both irritant and carcinogenic so is not an option. I've found this indicator but its threshold (8%) is not relevant for mold growth on a concrete and plaster wall which I guess is probably around 15-20% (I've found data only for wood where the threshold is 16%).


text:
Polyprotic acid - How does one calculate the CONCENTRATIONS of ALL species present in a solution when you start with 0.3M NaH2PO4?
Can anyone give me a push in the right direction, how do I calculate the concentration of all species present in a solution with 0.3M NaH2PO4.
So the concentration of H3PO4, H2PO4 -, HPO4 2-, PO4 3- and H30+.
I already know how to approximate the ph for an amphiprotic, pH=1/2(pKa1+pKa2). Should I use the [H30+] from this calculation in any way?
I know how to set up equilibrium equations and tables, but starting with the deprotonated form of a polyprotic acid confuses me.
I cant figure out in what order i need to calculate things.
For phosphoric acid
Ka1=7.6×10^-3
Ka2=6.2×10^-8
Ka3=2.1×10^-13

Doesnt answer my question, thanks for the help though!

Hey @Steinein How does the linked question not help you? It does seem to be very similar. If you can explain, we can reopen the question.

@Mithoron and etal: The linked question provide a strong idea to answer, but this question seems confusing many chemistry teachers, that they need very detailed completed answer with explaining to remediate misconception. So, the question deserve to reopen.

Try look around for Bjerrum-diagrams for phosphoric acid. These diagrams are illustrations of the distribution between an acid and its conjungate base, at a different pH. Looking at those, you will soon realize that at any given pH only one acid/base system is relevant (unless we are talking about acids with very close pKa).
If you need to calculate the distribution, you must turn to the usual equation for an acid/base-equilibrium. You know [H+] from the pH and [A-]/[HA] can be reduced to one unknown since [HA] + [A-] = Ctotal


text:
How are second and third electron affinities of atoms measured?
Wikipedia data page on electron affinities shows second and third electron affinities. Of course, they are negative, because one tries to attach electron to anion. But how are they measured or calculated in practice?
Electron affinity in Wikipedia

You have an electron rich source and measure the ratios of different charge states.

@JonCuster Thank you. But i still not clearly undestand it in detail. Could you point to some publications or relevant keywords for search?

There are spectroscopic methods, which calculates number of atoms ejected of a salt MX (X has -2 oxidation state) is calculated and equilibrium constant between X2- —>X+2e is calculated and then electron affinity. Here is the pdf https://www.google.co.in/url?sa=t&source=web&rct=j&url=http://www.nrcresearchpress.com/doi/pdf/10.1139/v90-244&ved=2ahUKEwjusJrqzfzeAhUJTI8KHQVkCa0QFjAAegQIAhAB&usg=AOvVaw0zjMhbcHbEDNdiwjvDGVSS

@SauravSingh Thank you! Could you write this as answer?


text:
Why is Diels–Alder adduct exo when we leave it for a while?
Recently we did an experiment of Diels–Alder reaction and we left the adduct overnight. I've read a paper of Woodward that says the adduct of Diels–Alder thought to be endo, whereas in our experiment it turned out to be exo.

The kinetic product is often endo while the thermodynamic product is often exo. Kinetic product is endo due to favourable secondary orbital interactions during reaction while thermodynamic product favoured due to decreased steric hindrance in the exo product.

See this earlier Q&A.

As you probably know, the Diels–Alder (DA) reaction is reversible — the easiest example is the reaction of cyclopentadiene, which polymerizes at room temperature, which — if needed as monomer — has to be cracked just in advance.
In terms of the selectivity you observed, the endo-oriented product of the DA reaction often is the kinetic one, the exo-oriented product the thermodynamic preferred one.  Which is, if the rate of a DA reaction is to be increased while still obtaining endo-preference, an increase of the reaction temperature is skipped in favour a Lewis-catalysis, addition of a perchlorate, or high (external) pressure.


text:
Do molecules with polar bond, but with no dipole moment experiences a greater effect from the london dispersion forces?
Do molecules that have polar bonds and a structure that makes the polarity of those bonds cancel each other out, experience a greater effect from the London dispersion force, than molecules with a similar size, bond length, shape, and mass, but with non-polar bonds?

Yes, an example of a molecule would be CO$_2$, which has no dipole moment but does have two polar bonds. Although CO$_2$ has no dipole moment, its charge distribution gives rise to a quadrupole moment. This quadrupole moment provides a handle for the molecule to interact with other particles (with other quadrupoles, but also with dipoles, induced dipoles or electric charges). As you might know, dipole-dipole interaction scales as $r^{-3}$ where $r$ is the distance between the particles and Van der Waals interaction (induced dipole-induced dipole interaction) scales as $r^{-6}$. It can be shown that quadrupole-quadrupole interaction scales as $r^{-5}$ and therefore represents a stronger interaction than the induced dipole-induced dipole interaction.

What you say about distance is correct but there is also a scaling or proportionality  term that is different for each type of interaction, this can radically change the size of the interaction at a given fixed distance.

Agreed, but for similar sized molecules (as stated in the question) I expect the polarizability to be smaller than the quadrupole moment and, moreover, the quadrupole interaction adds to the the Van der Waals interaction.

Basically, you need dipole polarizability($\alpha$) and first ionization energy (I) of each atom, and distance between them (S). You can approximate dispersion interaction for atoms as $$E_{disp}=-1.5\frac{I_1*I_2}{I_1+I_2}*\frac{\alpha_1*\alpha_2}{S^6}$$
As you can see, I and $\alpha$ are properties of matter, but, for example, double bonds (which are shorter) can change dispersion interaction.
As for molecules, london dispersion forces are "main" forces for nonpolar molecules. Contribution of LDF to intramolecular energy of two methane molecules is 100%; of two water molecules - only 24%. If molecule is not small and highly polar, LDF contribution will be about 90-100%. In order to have more detailed result, use table of electronegativity as well as 3D-structure of given molecule and determine it's polarity. Actually, perfectly polar bonds as well as perfectly cancelled ionic bonds will give the same 100% LDF contribution, but there is nothing truly perfect in our world.
Also, if two molecules have perfectly "similar size, bond length, shape, and mass", they are probably two identical molecules.


text:
Quantity of base needed to neutralise litre of 36% HCl?
Sorry for asking a basic chemistry question. I last did this kind of chemistry at school about 30 years ago and I don't have confidence that I'd get it right from first principles :)
Short version: Hydrochloric acid is used in some kinds of works to remove cement smears from masonry and paving slabs. But on a level surface, if left there, it's also able to degrade the cement that's supposed to be there (cement is highly alkali). Usually you can wash it away with water, but in my situation that's not workable. Instead, I plan to buy some Na2CO3, NaHCO3 or NaOH, all of which are ordinary commodity items at any local shop, dissolve in water, and carefully pour to neutralise any unreacted acid as soon as the surface is clean of cement smears. Excess alkali isn't a problem.
I've used concentrated HCl for many years, and the fumes/reaction heat/safety kit won't be an issue, so I'm comfortable with the safety aspects of this approach - comments not needed on safety.
My question is about making sure I have enough base ready to use, to neutralise the amount of acid I end up using. The acid is already on-hand, it's fuming 36% HCl. I'll dilute it for use, of course, but I can work out how many litres equivalent of 36% ends up being used. The bases (whichever I use) will be dry powder/crystal form. Obviously I pour until no reaction is visible, but if I don't buy enough, I'd have a problem. (Buying a bit more than I need is fine)
So I need a reminder how to work out the mass of each of the above 3 dry/crystalline bases that would be required to neutralise 1 litre of 36% HCl, so I can get an idea what kind of quantity of base to buy, before I start the task.
Given that helping hand with the basic chemical calculations, I can work out the rest easily.

The molarity of 36% HCl is approx 11.7, i.e. there are 11.7 moles per litre. To neutralise the whole litre you need 11.7 moles of NaOH, 11.7 moles of NaHCO3 or 5.85 moles of Na2CO3. Multiply the number of moles by the molecular weight - 40 in the case of NaOH, 106 for Na2CO3, 84 for NaHCO3 - to give you the weight required.


text:
Numbering in a derivative of azulene
I was wondering if anyone can tell me how to number this molecule:

This molecule is derived from azulene by addition of two sulfurs and two oxygens at positions 2, 6, 1, and 5 respectively. I would like to learn the proper procedure of numbering. I tried googling around and reading papers but could not find the correct terminology for this process.

2,6-bis(sulfanyl)azulene-1,5-dione, if you ask ACD Chemsketch (freeware, https://www.acdlabs.com/resources/freeware/chemsketch/), but ChemDraw (http://www.perkinelmer.com/category/chemdraw), or ChemDoodle (https://web.chemdoodle.com/) likely find it, too.  If your aim is to learn the rules, consider Helmchen's pocket guide (https://www.springer.com/gp/book/9783540411383, which partly is in google bookś preview (e.g., page 16), too.

Note that the numbers '9' and '10' which you added in your figure (I removed it) are not correct; fusion atoms are described by '3a' and '8a' as indicated in my answer.

If your aim is to know just the chemical name, you may ask a computer.

If the chemical already is known, chances are you find it in a data base (e.g., reaxys by Elsevier, or Scifinder by The American Chemical Society).  The later even has the option to draw a structure and to request a naming for it, even if there wasn't yet a publication indexed by SciFinder about said molecule.  A resource you typically find in Chemistry departments.
There are standalone programs helping you to find a chemical name for this structure.  2,6-bis(sulfanyl)azulene-1,5-dione, if you ask ACD Chemsketch (here, the freeware was used [free in the naming up to 50 atoms]); but Perkin Elmer's ChemDraw, or ChemDoodle -- often accessible within a campus wide license program by an university -- likely find it / generate it equally well.

If you aim to learn the nomenclature by yourself, the handy pocket guide by Helmchen may be worth to consider.  Portions of it may be seen in googles book preview, like the screen photo taken below.

(credit to Google books, p. 16)

I really Appreciate the fruitful explanation. I have two questions:
1- I understand the meaning of 3a and 8a but when looking for the Vertex-Adjacency Matrix of Weighted Graph, should they not be 9 and 10 instead?
2- Speaking of Vertex-Adjacency Matrix of Weighted Graph, would you be capable of explaining why you have not numbered O and S like the carbons? In other words, would you be able to confirm that numbering is only for carbons?
3- To ask about the Vertex-Adjacency Matrix of Weighted Graph  2,6-dimercaptoazulene-1,5-dione can I ask here ? Or should I post a new question?

The rules for numbering of bicyclic compounds such as azulene are given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).
First, you need to generate a "preferred orientation" of the molecule:

P-25.3.2.3.2 Polycyclic fused ring systems are oriented in accordance with the following criteria considered in order until a decision is reached:
  (a) maximum number of rings in a horizontal row;

Because of this, the two rings need to be horizontally aligned with each other in the preferred orientations. Note that this rule alone does not distinguish between the two orientations on the left.

Then, the numbering proceeds as follows:

P-25.3.3.1.1 The numbering of peripheral items in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. Each fusion carbon atom is given the same number as the immediately preceding nonfusion skeletal atom, modified by a Roman letter 'a', 'b', 'c', 'd', etc.

This leads to two possible numbering systems for the two orientations above. Which one is preferred is given by a further rule

P-25.3.3.1.2 If alternative numberings for a ring system remain [...] the following criteria are applied in order until a decision is reached: [...]
  (c) low locants are assigned to fusion carbon atoms;

The ketones in your molecule have greater seniority over the thiols (P-41), and hence the ketones are expressed as the suffix (leading to a name ending in "dione") instead of the thiols (a name ending in "dithiol"). Note that if the thiols are deprotonated this seniority ranking would be reversed, as anions have greater seniority over ketones (hence you would have a name ending in "bis(thiolate)").
The quinone derived from azulene is named as azulene-x,y-dione, where x and y are the locants of the two carbonyl carbons (P-64.2.2.2.3). The asymmetry in the quinone leads to ambiguity in the numbering, which is resolved by the next rule:

P-14.4 When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority: [...]
  (c) principal characteristic groups and free valences (suffixes);

Finally, the –SH groups are inserted as prefixes into the name, as they can no longer be suffixes. The appropriate prefix for –SH is 'sulfanyl'. The multiplier used is "bis" and not "di" to avoid ambiguity:

P-16.3.6 The prefixes 'bis', 'tris', 'tetrakis', etc. are also used to avoid ambiguity:
  (a) before a mononuclear subset of a polynuclear acyclic structure;
  Examples: bis(sulfanyl) (preferred prefix; defines two –SH groups, see P-63.1.5; whereas disulfanyl defines the –SSH group; see P-63.3.1)

As you can see, the manual generation of such names is in no way trivial. Most people use software to automatically generate names, which is generally sufficiently clear for the chemical community to understand them. However, it should be mentioned that these automatically generated names are not always the preferred IUPAC names: for example, ChemDraw 16 gives "2,6-dimercaptoazulene-1,5-dione" as the name, even though the prefix "mercapto" is no longer recommended (P-63.1.5).

I really Appreciate the fruitful explanation. I have two questions: 1- I understand the meaning of 3a and 8a but when looking for the Vertex-Adjacency Matrix of Weighted Graph, should they not be 9 and 10 instead? 2- Speaking of Vertex-Adjacency Matrix of Weighted Graph, would you be capable of explaining why you have not numbered O and S like the carbons? In other words, would you be able to confirm that numbering is only for carbons? 3- To ask about the Vertex-Adjacency Matrix of Weighted Graph 2,6-dimercaptoazulene-1,5-dione can I ask here ? Or should I post a new question?

I’m afraid I have no idea what that graph is, and if you have a different question please ask a new question.

My last question would be why as a suffix SH and O are not numbered and only included? Are they because Hetero atoms?

@abedhaidar The numbering system is not applied to the molecule as a whole, but rather to the "senior parent structure", which in this case is azulene itself.


text:
Enantioselective alkylation of Aldehydes by Dialkylzinc reagents
Enantioselective catalysis is one of the important reactions in Organic chemistry which has huge applications in practical  basis for synthesis of optically pure natural compounds. In this respect, I came across the following reaction of Aldehydes aimed towards converting them into Chiral alcohol with large enantiomeric excess with the help of (-)-3-exo-(dimethylamino)isoborneol
[( -)-DAIB] catalyst.
The structure of (-)-DAIB is given as below,

The reaction is given as,

If we slightly rotate the molecule and visualise, we will see that here the methyl group has come from the top side of the molecule,i.e. a si-face attack by methyl group on benzaldehyde has occured.  From the picture of the transition state also this picture become clear,( According to our reaction $\ce{R''}$ is methyl and $\ce{R'}$ is Phenyl in the picture)

But my question comes here, Why the orientation of the bigger group (here Phenyl) in the transition state is preferably axial? Shouldn't it be the opposite direction,i.e. shouldn't the preferred orientation be that transition state where the $\ce{R'}$ and $\ce{H}$ are just swapped with this picture ? In that case $\ce{R'}$ is farthest from $\ce{R''}$ and logically should be the most stable transition state. So, the major product will just be the enantiomer of the originally given product. I am unable to find any flaw in this logical approach. Then, why is the transition state is the given one not the proposed one ?
References: -

Soai, K.; Niwa, S. Enantioselective addition of organozinc reagents to
aldehydes. Chem. Rev. 1992, 92, 833-856.;
Noyori, R.; Kitamura, M. Enantioselective Addition of  Organometallic  Reagents to Carbonyl  Compounds: Chirality Transfer, Multiplication, and Amplification.Angew. Chem. Ini. Ed. Engl. 30 (1991) 49-69
Itsuno, S; and Frechet, J.M.J.  Enantioselective addition of diethylzinc to aldehydes catalyzed by polymer-supported chiral amino alcohols. Evidence for a two zinc species mechanism. J. Org. Chem. 1987 52 (18), 4140-4142

R'' is equally close to R' and H. I'm more concerned about the proximity of R' and H to the proximal methyl group on the zinc.

I guess you are misunderstanding the diagram. R'' is perpendicular to the plane of phenyl ring, hence doesn't make huge difference if it is flipped as far as R'' is concerned. But the interaction with the Methyl groups, Zn atom on the other side, if you have flipped would give rise to much more unfavourable interaction.


text:
Connection of term symbols with specific microstates for atomic carbon
I'm currently studying atomic term symbols. I wanted to try it on a simple atomic carbon with the electron configuration $1s^22s^22p^2$.
I know, that only open-shell electrons are involved in the term symbol classification, so that leaves us working with solely 2 electrons in $2p$ sub-shell.
First of all, we have ${6}\choose{2}$$= \frac{6!}{2!(6-2)!}=15$ possible microstates for the $1s^22s^22p^2$ configuration.
The microstates are listed in the following table:

And, if I understand it correctly, we can categorize them into several "subsets" labeled with the corresponding term symbols.
The atomic term symbol is defined as ${}^{2S+1}L_J$.
In our case, the possible values of S, L and J are following:
$$\max(M_L) = \max(L) = +2 \Rightarrow L = 0,1,2$$
$$\max(M_S) = \max(S) = +1 \Rightarrow S = 0, 1$$
$$J = L+S, L+S-1, \ldots, |L-S| = 3,2,1,0$$
Considering possible values of $S$, multiplicity can be $1$ or $3$, i.e. singlet or triplet. Values of $L$ enable symbols $S,P,D$.
So, our presumed set of states is ${}^3D, {}^1D, {}^3P, {}^1P, {}^3S, {}^1S$.
${}^3D$ "contains" $S=1$ and $L=2$. In that situation, both electrons would have to be spin-up and positioned together in the orbital with $m_l = +1$. That is impossible, as it contradicts Pauli exlusion principle. I.e. the state ${}^3D$ does not exist.
${}^1D$ is possible ($L=2,1,0, S=0$), so we can compute the number of corresponding microstates $N$  with the formula $N = (2L+1)(2S+1) = 5\cdot 1 = 5$.
Using the same formula, ${}^3P$ is going to contain 9 microstates.
With the "substraction method" described in this video, we can arrive to the conclusion, that both ${}^1P$ and ${}^3S$ won't exist.
And finally, ${}^1S$ will contain the last one state.

Question
Now I know the possible term symbols, but I'm not sure, which microstates belong to them specifically.
It's clear, that microstates 13 and 15 will belong to ${}^1D$, but how can I determine it for the other states? ${}^1D$ should contain 3 more microstates with $S=0$ and $m_L$ being equal to -1, 0 and +1, but there are multiple candidates for every configuration.
So how could I distinguish between, e.g. microstates 7 and 9 or 8, 11 and 14?
I don't understand this point, as even $J=2$ for ${}^1D$, so it can't be used to distinguish among the "similar" microstates and to assign them properly.

There is no direct assignment. Please take a look at the very last part of my answer here https://chemistry.stackexchange.com/questions/100413/distinguish-between-two-possible-configurations-for-angular-momentum-in-carbon-a/102973#102973

@Feodoran So, the sets of states belonging to different term symbols are basically overlapping?

Yes, because it is a basis set expansion.

@Feodoran Great! Would you write a short asnwer, so I could accept it?

To add on to what Feodoran has mentioned, you might be interested in looking at the Clebsch–Gordan coefficients, which are essentially the coefficients of the microstates in the term symbols. You can look these up in e.g. the appendices of Atkins Molecular Quantum Mechanics, 5th ed. Note that each term symbol is a collection of states, so a $D$ term (for example) which has $L  = 2$ would comprise five states: one each with $M_L = -2, -1, 0, +1, +2$. These five states can in general be denoted as $|L, M_L\rangle$.
$L$, which is a total angular momentum, is obtained by a coupling of two individual angular momenta. For two electrons in p-orbitals, these individual angular momenta are $l_1 = 1$ (with $m_{l1} = -1, 0, +1$) and $l_2 = 1$ (with $m_{l2} = -1, 0, +1$). The microstates in the table specify both values of $m_{l1}$ and $m_{l2}$, so they are simultaneous eigenstates of $m_{l1}$ and $m_{l2}$, i.e. they can be denoted by $|m_{l1},m_{l2}\rangle$.
Now, in general each coupled representation $|L, M_L\rangle$ will be a linear combination of the microstates, or uncoupled representations $|m_{l1},m_{l2}\rangle$. It is in general not possible to construct a state which is simultaneously an eigenstate of $L, M_L, m_{l1}$, and $m_{l2}$, because the relevant operators do not commute (refer to Atkins or any other QM text on coupling of angular momentum). There are examples of one-to-one correspondence between the coupled and uncoupled representations – for example, microstate 13 perfectly corresponds to $|L = 2, M_L = +2\rangle$, but it is not general, so:
$$ |L, M_L\rangle = \sum_{m_{l1},m_{l2}} c^{L, M_L}_{m_{l1},m_{l2}} |m_{l1},m_{l2}\rangle$$
The Clebsch–Gordan coefficients are simply the $c^{L, M_L}_{m_{l1},m_{l2}}$ in the expansion above. Usually the notation is slightly different – so instead of $(L, M_L, m_{l1}, m_{l2})$, it is usually $(j, m_j, m_{j1}, m_{j2})$ – but the concept is entirely the same.
Using the coefficients for $j_1 = j_2 = 1$ in Atkins (which is relevant to the $\mathrm{p^2}$ configuration of atomic carbon), we find that the $^1S$ term is given by
$$|L = 0, M_L = 0\rangle = \underbrace{\frac{1}{\sqrt 3} |m_{l1}= 1, m_{l2} = -1\rangle + \frac{1}{\sqrt 3} |m_{l1}= -1, m_{l2} = 1\rangle}_{\text{a combination of microstates 8 and 11}} - \underbrace{\frac{1}{\sqrt 3} |m_{l1}= 0, m_{l2} = 0\rangle}_{\text{microstate 14}}$$
It's slightly more confusing because there is a spin component as well, which the coefficients do not cover, and this is why I've given a handwavy "combination of microstates 8 and 11". This "combination" is needed to generate the singlet spin wavefunction $2^{-1/2}(|\alpha\beta\rangle - |\beta\alpha\rangle)$.

As an aside, the familiar singlet and triplet spin wavefunctions
$$\underbrace{|\alpha\alpha\rangle, |\beta\beta\rangle, \frac{1}{\sqrt 2}(|\alpha\beta\rangle + |\beta\alpha\rangle)}_{\text{triplet spin wavefunctions}}, \underbrace{\frac{1}{\sqrt 2}(|\alpha\beta\rangle + |\beta\alpha\rangle)}_{\text{singlet spin wavefunction}}$$
are also coupled representations. These four wavefunctions are, respectively,
$$|S=1, M_S=+1\rangle, |S=1, M_S=-1\rangle, |S=1, M_S=0\rangle, |S=0, M_S=0\rangle$$
(Different notation again, but it is the same concept.) The uncoupled representations are the states in which each electron has been assigned a spin, i.e. $|\alpha\alpha\rangle, |\alpha\beta\rangle, |\beta\alpha\rangle, |\beta\beta\rangle$. For example:
$$|m_{s1} = +1/2, m_{s2} = +1/2\rangle = |\alpha\alpha\rangle$$
and so we can see a simpler example of how the coefficients link the two representations:
$$\begin{align}
|S=1, M_S=+1\rangle &= |m_{s1} = +1/2, m_{s2} = -1/2 \rangle \\
|S=1, M_S=-1\rangle &= |m_{s1} = -1/2, m_{s2} = -1/2 \rangle \\
|S=1, M_S=0\rangle &= \frac{1}{\sqrt 2}|m_{s1} = +1/2, m_{s2} = -1/2 \rangle + \frac{1}{\sqrt 2}|m_{s1} = -1/2, m_{s2} = +1/2 \rangle  \\
|S=0, M_S=0\rangle &= \frac{1}{\sqrt 2}|m_{s1} = +1/2, m_{s2} = -1/2 \rangle - \frac{1}{\sqrt 2}|m_{s1} = -1/2, m_{s2} = +1/2 \rangle \\
\end{align}$$

Part of your confusion is that the table you show is not the best way to show the information. If you make a table as below and fill in the values and remove those due to the Pauli principle then it is easier to partition up values. ( I'll leave you to complete it)
$$\begin{array}{cccc}\\
m_{l1} & m_{l2} & m_{s1} & m_{s2} & M_L & M_s\\
\hline
1 & 1 & + & + &\text{no good Pauli}&\text{no good Pauli} \\
1 & 1 & + & - & 2 & 0\\
1 & 1 & - & + & \text{same as previous one}  \\
1 & 0 & + & + & 1 & 1\\
0 & 1 & + & + &  \text{same as previous one}\\
1 & 0 & + & - & 1 & 0\\
1 & 0 & - & + & 1 & 0\\
etc& &\text{and no more Pauli conflicts}&\\
\end{array}$$
When you are done there will be 15 valid combinations. Arrange these on a 3 by 5 grid. Three for the $M_s$ and five for the $M_L$ and at each intersection put the number of times these values occur. For example there is  1 value with $M_L=1, M_s=-1$ and 3 with both values of zero.
There is only one value with $M_L=2 , M_s=0$ so this must belong to a $^1D$ state so remove one column with $M_S=0$ (as there a 5 values in the series i.e. $M_L=-2,-1,0,1,2,\;M_s=0$) , what remains is a 3 by 3 array with one element in each except at the centre where there are two. One of these center ones must be the $^1S$ as both $M_L=M_s=0$, those remaining are the $^3P$.


text:
Formulation and Nomenclature of Boric and Phosphoric Acids
Why are boric and phosphoric acids are formulated as:

H3BO3
H3PO4

Instead of:

HBO2
HPO3

Because the acids end with the -ic suffix, the valences of the elements between the hydrogens and oxygens are 3, 4 and 5 respectively, since Hydrogen has 1+ valence and the Oxygen 2-.
In the acids H3BO3 and HBO2, both of the borons have a valence of 3 +, (the multiplication of 1x3 plus 3x1 gives us 6 + which is the same result (with a different sign) of the multiplication of the valence of the Oxygen 2- and 3 (- 6) and in the acid HBO2 (1x1+3x1 = 4 +; - 2x2 =  - 4). How can I know which one is the boric acid when the nomenclature is given? Of course I can memorize it but I don't really know why only one of them is the boric acid.

$HBO_2$ is metaboric acid (left pic.), boric is on the right pic.

$HPO_3$ is metaphosphoric acid (on the left pic),...

Actually, blame old nomenclature. Meta-acid has "one less $H_2O$ molecule" than ortho-acid. Thus, $HBO_2$ is metaboric acid, while $H_2BO_3$ is orthoboric (same about any other acid). But as you can see from pics, it's not really "water molecule" that differs meta-acid from ortho-acid. In case of metaboric acid, it's not even 1 H and 2 O's difference! So now ortho-acids are called simply acids (for example, boric), while meta-acids are still called meta-acids. Maybe because they are formed from ortho-acids. So, ortho-acids are considered "true acids", while meta-acids are, like, secondary. But I should mention that sometimes ortho-acids are called ortho-acids. I believe it depends on occurence of these substances in history of chemistry: the most "historically" important of two (or popular) has the right to skip prefix.
[remark] I know that metaphosphoric acid, as well as pic which I've given, is old fossil and now there is no such thing as "monophosphoric" acid (check for cyclo-triphosphoric acid, though). But the question was about this acid.
[edited]For borates, ortho-acid is almost always "boric acid" (rarely - "orthoboric acid"), and meta-acid is always "metaboric acid".  So, (ortho)boric acid = H3BO3, metaboric acid = HBO2.
Phosphoric acid = H3PO4. It's very unlikely to meet name "orthophosphoric", because you have no metaphosphoric pair. Sometimes (in old textbooks/in poor-made school course of chemistry) you can hear "metaphosphoric acid",  which is on the pic above, but the name and pic itself are false. HPO3 is now considered to be cyclo-triphosphoric acid, as I've said.  As for other possible acids - some have "true name" for meta-acid, while ortho-acids are called ortho-acids, some are called ortho- and meta-, some are like phosphoric acids: there is no ortho- or meta-, only "true acid" and cyclo-acid. Specificity of nomenclature.
Googling for any good explanation now gave me this:
https://www.quora.com/What-is-the-meaning-of-ortho-pyro-and-meta-as-in-the-context-of-phosphoric-acids
Good and detailed info about IUPAC nomenclature. Hope this helps.

Oh God, that's a lot of information! Yes, I knew about the meta, ortho and the di prefixes. This creates a bit of a confusion for me. How do I exactly know which prefix to use when the formulation is given to me? Or if the nomenclature is referring to the ortho when it isn't written (without the Lewis structure)? I remember something like: for the valences that are even numbers you would have to add 1 molecule of water to the oxide, or something like that, in the meta prefixes. Could you give me an example on how to do that? (I don't know if I explained myself well enough)

Please, check [edited] part.

I think I get it now, thank you! I will check the link.


text:
If I add apple cider vinegar to a cleanser with a pH level of 7, will the pH level of it drop?
I am thinking about adding apple cider vinegar to my Physiogel Dermo cleanser which has the following ingredients:
Aqua, PEG-75, Cetearyl Alcohol, Disodium Phosphate, Sodium Cocoyl Isethionate, Methylparaben, Citric Acid, Butylparaben, Propylparaben, Parfum.
Let's say I mix 1tablespoon of apple cider vinegar to a 150ml physiogel bottle. Will the ph level of it drop and will it syill be safe to use on facial skin?

If it will work on your skin is a medical question which we won't answer. We're chemists, not medical doctors.

Since your question consists of two parts, I'll reply in two parts, too.

Regarding the cleanser alone:  among the consititutents you list, it already contains both one, that is an acidic component (citric acid), and one that is basic (disodium phosphate).  It is by intent that the two are present simultaneously in the cleanser, as they -- altogether with water (here: named aqua) -- form a buffer solution.  This is a way to tune the acidity, the pH value of the cleanser to a value suitable to the intended application on skin, where both an aqueous solution of pure acetic acid, or an aqueous solution of pure disodium phosphate would attack your skin.  The ratio of the two determines a buffer capacity, i.e., a certain amount of additional acid, or base added will not alter the pH value of the cleanser.  Just naming the ingredients of the cleanser, but without their quantities, renders it impossible to predict if the addition of a tablespoon of vinegar added to 150 mL of cleanser already will exhaust this capacity, or not. If this buffer capacity of the cleanser is exhausted, the pH value will drop.
Regarding the "is applying of the modified skin cleanser on skin safe": No, Do not do apply this to your skin, because you may harm and (chemically) burn yourself.  The skin already is a complex system -- just think about the discern of dry / wet / fatty skin, changes by season (e.g., dry winters, humid summers), age (e.g., puberty), male / female sex in case of facial skin care; all actually under the assumption your skin were healthy.  It is not a simple layer, actually it equally is an ecosystem protecting you. So, do not use your altered cleanser for skin care just because you found a fancy internet page advertising a recipe. ChemSE definitley is not a replacement for a physician to consult in person, but consider just reading the abstract here of one publication, and the publication here (free access) will indicate you not to alter a skin cleanser.

This is a very detailed and helpful response. Thank you very much!


text:
What is the sense of electronic configuration of an atom if an electron can be in any superposition of energy eigenfunctions?
I recently started learning abou the quantum mechanical explanation of hydrogen atom. In the derivation of the wave functions of hydrogen atom we find its energy eigenfunctions. But according to the Schrödinger equation, the electron can exist in any linear combination of the energy eigenfunctions. So what I don’t understand is that why do we say that the electrons in an atom belong to specific orbitals? Why do we always treat electrons in an atom as if their energy is fixed?

The answer that will satisfy you depend on how familiar you are with the mathematics and principles of quantum chemistry.
So, let's start small. Hydrogen atom. Exact expressions for eigenfunctions for the Hydrogen are known, they are the orbitals we know and love(?). Although the electron can be in any combination of eigenfunctions, it can only stay that way when you are not looking. When you do look, that is, take a measurement, it will collapse to a specific eigenfunction. That eigenfunction could be 1s or it could be 6h, but it turns out that 99.999% of the time, when you measure the energy of a Hydrogen atom, it will display the familiar 1s. That is the sense in which we can say that the electron in a Hydrogen atom is is in orbital 1s, what we really mean is that, typically, for the vast majority of time, the electron of the Hydrogen will be occupying the ground eigenstate.
Now that is for Hydrogen. For heavier elements, you no longer have orbitals, strictly speaking, you have more complicated eigenstates, obtained by solving the Schrodinger Equation to each atom separately. Currently, there is no analytic solution to the equation when you include a second electron. We approximate the ground state of each atom as though the electron were occupying different hydrogen-like orbitals (hydrogen-like because they differ from the pure hydrogen orbitals in a few parameters that are added).
Because electrons cannot occupy the same quantum state (considering spin), the way to do that is to assume that each electron is in a spin orbital. The way to distribute the electrons through the infinitely many hydrogen-like orbitals is to distribute them in the way you have learned, following the Pauli diagram. The resulting electron distribution is an approximation to the true ground state, using hydrogen-like orbitals, because the true ground state is not known.
O hope I made myself understood, but if you need clarification, please ask and I shall edit the answer as much as possible. I have fond memories of the time when I was asking the same questions :)
EDIT 1
You asked why is the probability of finding the electron in 1s orbital highest when we measure its energy?
The doubt comes because from your understanding these probability comes from writing the initial state as a linear combination of eigenbasis of our choice (in this case the energy eigenfunctions) and the initial state can be any function (normalisable and continuous)
Now, while in principle this is correct, in practice that is not so. When you have a vast sample of hydrogen atoms and you measure the energy of each one of them, the vast majority of them will return the ground state for most experimental conditions. That is because experimental setups that actually manage a high degree of superposition with excited states are very hard to prepare.
This means that the experimentally expected state for an electron in a Hydrogen atom is 1s for the vast majority of experimental setups. This is the relevant information. The preparation of an arbitrary superimposed state is not trivial, and they do not occur easily in nature.
This leads us to the approximation using hydrogen-like orbitals and the aufbau principle, according to which the ground state (the most experimentally relevant eigenstate) of an atom can be approximated as though each electron occupied one spin orbital around the nucleus, following the Pauli diagram of energies.
It is not so much a theoretical result as much as a way to make theory and practice agree. There is a way to reach the aufbau principle from first principles, but it involves doing statistical thermodynamics using Fermi Statistics. These calculations would tell you how electrons usually distribute through their available energy levels, and it would become clear that the aufbau principle is pretty sensible, from an experimental standpoint.

Thank you for your answer! First of all can you explain why is the probability of finding the electron in 1s orbital highest when we measure its energy? Because from my understanding these probability comes from writing the initial state as a linear combination of eigenbasis of our choice (in this case the energy eigenfunctions) and the initial state can be any function (normalisable and continuous).

My main problem is that I am not able to connect to the prior knowledge of orbitals and how electrons reside in them which brings out all the chemical properties of them to the new knowledge I acquired about wavefunctions and schrodinger equation. I have seen some your answers and I guess you are experienced in the field of quantum chemistry so can you suggest me some ways by which I can learn these things with understanding?

Okay now I think I have understood. Just to confirm, I think you are saying that when we take a large sample of hydrogen atoms and measure its energy, almost all of the times we will find it in 1s state because thermodynamically it is not easy to prepare superposition of higher energy states. And aufbau principle comes as its consequence. Am I right?

@Kartikeya yes, that is pretty much it :D

Your question is in fact quite general in quantum mechanics and does not depend on H atoms or any atoms for that matter.
If a perfect experiment is repeated under the same conditions and in exactly the same way then what will be the outcome? Surely we would expect the same answer?
In general this is not true, and this is not due to experimental uncertainties because we perform an ideal, or perfect, experiment. Thus experiments in general do not have a right or wrong result, i.e in general we cannot predict the outcome, repeating the same  experiment gives a different result.
What we can predict is what the range of results will be, what the the chance (probability) of observing a particular result is but we can always accurately calculate the average result via the expectation formula.
To be more exact we suppose that there is a 'system' that has a wavefunction $\psi$  and we operate on this with an 'operator' and so make a measurement. The 'system' is quite general and might be an H atom or a particle in a box and so forth. The measurement might be that of position or energy etc. If the operator is the Hamiltonian which is the total energy ( kinetic  and potential) operator then the measurement is that of the energy. If the result of applying the operator $\hat A$ to the wavefunction is to produce a non-zero number multiplied by the wavefunction then we have an eigenvalue equation. We call the wavefunction  that has this property the eigenfunction of the operator $A$ and label it $f(x)$, where $x$ is a coordinate, then the eigenvalue equation is $\hat A f(x) = af(x)$ where $a$ is the number we call the eigenvalue.
Returning now to our prediction of an experimental outcome, if the 'state' is in an eigenstate of the observable then the result of an experiment will be only the eigenvalue associated with that state, i.e. the number $a$. Thus if the state is the 1s wavefunction of the H atom then its eigenvalue will be measured every time an experiment is performed. In general if the eigenvalue equals, say, 2.1 then each repetition of the experiment will give the value 2.1 exactly.
If the 'system' is not in an eigenstate of an observable then the result will be one of the eigenstates associated with that observable, i.e. an eigenvalue will be measured but not always just the same one, i.e. eigenvalue 2.1 will still  be measured some of the time but others as well. However, an important distinction now occurs because if we measure the system once and observe the eigenvalue, say, $E_3$, if we do the same experiment again on the same system then only eigenvalue $E_3$ will be observed no matter how many more times the experiment is done. The wavefunction has 'collapsed'. This is an unavoidable consequence of measurement (also called 'observation') of a quantum system, i.e. observation inevitably affects the system. As far as I am aware no one knows why or how wavefunction collapse occurs.
If the system is not in an eigenstate, i.e. is 'messed up' then to calculate the chance of observing each of the eigenvalues depends on the similarity or overlap of the 'messed up' state $M$ to the each eigenfunction $f_n$ belonging to each eigenvalue. The equation for the overlap is $S_n=\int_\text{all space}M^*(x)f_n(x)\mathrm{d}x$ where $x$ is the coordinate, and the probability of observing eigenvalue $a_n$ with eigenfunction $f_n$ is $|S_n|^2$.  (By messed up I mean that the system is in a linear combination of eigenstates).
Notes:
The wavefunctions $\psi$ are functions  that have parameters and also variables that are the coordinates of the systems components. Thus for a particle in a box the wavefunctioon depends on the position (variable) in the box and its length (parameter) and its quantum number (parameter). In a given quantum state the parameter cannot vary. If it does then it has become another state.
The Born interpretaion is that the squared modulus of the wavefunction gives the probability density, i.e. $\rho(x)=|\psi(x)|^2$.
The probability of finding the particle at a particular place is found by integrating the probability density between two limits say $a \to b$. Thus $
P_{a\to B}=\int_a^b|\psi(x)|^2dx$
In an experiment we operate of the wavefunction to produce an 'observable' which is the result of an experiment. Each and every observable has its associated operator. The operator always operates on the wavefunction.  Examples of observables are position, momentum, energy. A position operator $\hat x=x$ and a  momentum operator along  $x$ is $\displaystyle \hat p_x= -i\hbar\frac{\partial}{\partial x}$. The Schroedinger equation has both kinetic $\hat K$ and potential energy $\hat V$ operators, these form the Hamiltonian, which is the total energy operator, $\hat H= \hat K + \hat V$
The average of many measurements gives the expectation value which is just an average. If the operator is $\hat A$ it has the form
$$\langle A\rangle =\int_\text{all space} \psi^*(q)\,\hat A\,\psi(q)\mathrm{d}q$$
where $q$ represents whatever the coordinates are, these are just $x$ for a particle in a 1D box. The * after the wavefunction means take the complex conjugate; if the wavefunction is complex replace $i$ with $-i$.

Thank you for your answer! Your answer explained how observations affect a quantum state. But I was asking that why electrons in an atom are not in the “messed up” state? Why do we always assume that electrons have wavefunctions which are energy eigenstates, not linear combination of them?

These are what we get from a direct solution of the Schroedinger equation, and it makes life a lot easier if we use them rather than some messed up state. In practice any messed up state is going to have more energy than the lowest eigenvalue so it is usually easier experimentally to make a single eigenstate, but not always so and so experimental methods have to work hard to achieve this, such as cooling isolated molecules to a few K in high vacuum.


text:
Neuter Reactants: What do they do?
So imagine a hypothetical chemical equation
$$
A + B \rightarrow C + B
$$
It is obvious from the equation that $B$ is not consumed during the reaction. If $B$ actively take part in the reaction, if it participates in the transition state and isn't just sitting there in the environment, I classify $B$ as a Neuter Reactant.
Under these defitions I would like to know:
Is a Neuter Reactant always a Reaction Inhibitor or Catalyst?
or are there instances of neuter reactants $B$ such that their reactions $A+B \rightarrow C+B$ have a similar rates to the straight reaction $A \rightarrow C$?

Since you named Neuter Reactant, it is up to you to describe the behavior that amkes it special. // More traditionally if B increase the reaction rate it is called a catalyst, and if B causes the reaction rate to be slower it is an inhibitor.

I meant it as a query to the community. I have learned that a Catalyst, for instance, is a substance that take part in the reaction but which is not consumed. Is there any substance that also takes part in a reaction and is not consumed but does not alter the kinetics of the reaction, like a catalyst or an inhibitor do. I don't know if such things exist and I am asking to see if anyone does, that is the question. Maybe I should edit the original question to make it more clear? What do you think?

If "B" is "consumed" it can't be on the right hand side of the equation.

Do you want it to take part in the reaction or not? I mean how can it take part without having any influence? That makes absolutely no sense.

There are reactions similar to the one you show, which is straightforward catalysis, these are autocatalytic reactions such as $\ce{A + B\to 2B}$ or $\ce{A + 3B\to 3B}$ which lead to interesting non-linear behaviour. Of more importance is the Michaelis-Menten scheme  $\ce{E + S\leftrightharpoons ES \to P + E}$ where E is an enzyme , S a substrate, ES a complex of both, and P the product.


text:
Sigma electron movement in resonance forms
I have come across an example in my textbook, which shows resonance structures for the bromonium ion. However, I am confused as to why the sigma bond electrons "move" to get the tertiary carbocation and the secondary carbocation. Are sigma bond electrons not strictly localized and pi electrons delocalized?

My question is: When are we allowed to move the sigma electrons when writing resonance forms?

In the text, which talks about the diagram says this: "In resonance terms, we say that the weighting factor (coefficient) for the more-stable form is greater than that for the less-stable form (c1 > c2, Fig. 11.13)." Are you sure these are not resonance forms?

Yes, I am sure these are rearrangement structures, see Wagner-Meerwein rearrangement in my answer. And yes the factor which decide the major product is the stability of intermediate (here carbocation).

@Mithoron there are multiple descriptions possible, but they would involve sigma type orbitals on MO and VB theory. That doesn't make your point and more value. The bromonium is a very good example for nonclassical cations. // Very approximate rule of thumb: as long as you don't move the nuclei, you can use (for better or worse) the concept of resonance. (Here it's actually for better.)

@Martin-マーチン I think it can be also viewed as an analogue of alkene - metal complex.

@Mithoron yes indeed. The pi orbital would still do a sigma type donation. There was a typo in my last comment (I'm on my phone), it should have been: that doesn't make your point any less valid.


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Why is petrol also called octane?
Why do people commonly use octane as a synonym for petrol (gasoline)? Isn't petrol a mixture of hydrocarbon distillates--not a pure substance?

Octane is a specific molecule with 8 carbons normally found among the many different molecules in gasoline. Pure iso-octane (2,2,4-trimethyl pentane) was chosen as the reference fuel standard for resistance to preignition for gasoline fuels. So the octane number of a fuel is a rating of the preignition potential of that fuel compared to a fuel containing 100 %  iso-octane molecules. Since iso-octane is a mouthfull, the "iso" part was dropped and people just call it octane.  This is why some vendor calls his product petrol or gasoline or gas or octane or high test, etc...

Octanes are components of fuel. There is also such concept as "octane number", used to determine quality of gasoline. Higher octane number = higher activation energies, and higher $E_a$ = lower chance of uncontrolled ignition = higher cost per liter.
"Gas station numbers" for fuels are octane numbers. You can see (I think) numbers from 87 to 97 (depends on your country), but usually even 92 is considered "premium". As we can see, octane number determines main properties and even gives "name" to types of fuel at gas stations. So, maybe it's okay to call petrol "octane", though I prefer to call it "gasoline" :)

Octane number  is a scale that is used to determine the quality of a fuel.
It may be defined the  percentage of isooctane by volume in a mixture of isooctane and n heptane that has the same anti knocking properties as fuel under examination.
Petroleum or crude oil is obtained in complex mixture of hydrocarbons (chiefly aliphatic with small amount of aromatic hydrocarbons) along with  some trace amount of sulphur and nitrogen.

Calling gasoline "octane" would be mistake. There are a few widespread misconceptions about this topic. Most importantly, many people think that gasoline contains large amounts of octane, mainly because of the frequent mentioning of the high octane number of gasoline. Accordingly, many textbook chapters about combustion use the reaction of octane as example for a combustion reaction. This is of course wrong or at least misleading.
In order to clear up some of the misconceptions, I have copied the contents from another answer, which fit even better here:
High-octane fuel does not contain a high concentration of octane.
The composition of gasoline varies widely; anyway, typical gasoline contains less than 1 % octane.
The octane number of gasoline does not describe the composition of gasoline. It refers to the measurement of the knock resistance of gasoline, which is usually expressed by their research octane number (RON) or motor octane number (MON). Both parameters are determined in single-cylinder four-stroke test engines with an adjustable compression ratio according to standardized test procedures, e.g. EN 25164 or ASTM D 2699 for RON, and EN 25163 or ASTM D 2700 for MON. During the measurement, the compression ratio is raised until a certain standardized level of knocking is reached. The corresponding octane number is determined by finding a reference mixture of heptane and “octane” that gives the same knock level. The octane number is the fraction of “octane” (in %) in the “octane”-heptane mixture. For example, gasoline with an octane number of 95 has the same knocking performance as a mixture of 95 % “octane” and 5 % heptane; however, this does not mean that the gasoline contains 95 % “octane”.
Nevertheless, the wrong idea that gasoline would contain a high concentration of octane is widespread. For example, in the first relevant Google hit, Ritter, S. Chem. Eng. News 2005, 83 (8), 37:

Octanes are an important component of gasoline because they help provide smoother combustion in the car’s cylinders and prevent knocking.

Also remarkable is the fact that many textbooks and lectures select the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion:

$$\ce{2C8H18 + 25 O2 -> 16 CO2 + 18 H2O}$$

The reference substance for the octane number is not octane.
The octane number refers to a mixture of heptane and “octane”. However, the “octane” used for this mixture is not the unbranched alkane octane. It is actually a branched isomer of octane with the traditional name “isooctane”. Thus, gasoline with an octane number of 95 actually has the same knocking performance as a mixture of 95 % “isooctane” and 5 % heptane; however, this also does not mean that the gasoline contains 95 % “isooctane”.
The octane number of pure octane is actually very low. Its knock resistance is worse than the performance of pure heptane, which has a octane number of zero by definition.
The “isooctane” that is used as a reference substance for the octane number is not isooctane.
The name isooctane was in accordance with older nomenclature rules. The corresponding systematic name according to current IUPAC recommendations is 2-methylheptane. However, the “isooctane” that is used for the determination of the octane number is not isooctane, i.e. it is not 2-methylheptane. This “isooctane” actually is 2,2,4-trimethylpentane, which is another isomer of octane. Thus, gasoline with an octane number of 95 actually has the same knocking performance as a mixture of 95 % 2,2,4-trimethylpentane and 5 % heptane; and once again, this does not mean that the gasoline contains 95 % 2,2,4-trimethylpentane.
By way of comparison, the octane number of pure isooctane (2-methylheptane) is only about 23.


text:
Why Na/C2H5OH not preferred in the reduction of carboxylic acid, despite being used for the reduction of esters
why is Na/C2H5OH  not preferred in the reduction of carboxylic acid to alcohols, despite being used for esters?

You could get deprotonation as a side reaction, because the reagent generates a strong protic base (ethoxide ion).

Correct. You get the sodium carboxylate which does not reduce.

As the comments point out: the sodium in ethanol generates sodium ethoxide, just as sodium in water generates the hydroxide.  The ethoxide ion, acting as a strong base, deprotonates the acid to a salt and thwarts the reduction.  Get rid of the easily removed proton by using an ester and you unleash the reduction reaction you want.


text:
In an Electrotlytic cell using MgSO4 electrolyte, Cu(OH)2 and Mg(OH)2 are visible products, Where Does SO4 go?
I have an electrolytic cell with $\ce{MgSO4}$in an aqueous solution at full saturation, running at a potential of $3.6v$ with a copper anode because I need some $\ce{Cu(OH)2}$.
Production of $\ce{Cu(OH)2}$, $\ce{CuOH}$ and $\ce{Mg(OH)2}$ are all visible.
There's no gas production at the anode but I expect that the oxygen is remaining in solution due to $\ce{OH}$ production.
There's no sign of $\ce{SO2}$ or $\ce{SO3}$, so
what Happens to $\ce{SO4-}$?
EDIT
$\ce{SO4-}$ eventually rejoins $\ce{Mg}$ because the more reactive base will always oxidize a less reactive neutral or acidic salt, not to mention that $\ce{H2SO4}$ is a species that will form under these conditions as $\ce{Mg}$ and $\ce{Cu}$ both become insoluble (which should actually yield $\ce{H2SO5}$ because $$\ce{Cu + MgSO4 + H2O -> Cu(OH)2 + Mg(OH)2}$$ @cathode, $$\ce{H2O + SO4- -> H2SO5}$$ @anode, which doesn't hang around too long $$\ce{H2SO5 -> 2H2SO4 + O2(g)}$$ and the obvious reformation of $\ce{MgSO4}$ via $\ce{H2SO4}$ + $\ce{Mg(OH)2}$ and $\ce{CuSO4}$ + $\ce{Mg(OH)2}$ so I was eventually left with $\ce{Cu(OH)2}$ as the only insoluble product.
I have since been able to produce $\ce{Mg(OH)2}$, omitting $\ce{Cu}$ entirely of course, but coming up with a process wasn't easy with my 3 months of experience. I however have been able to completely separate $\ce{Mg+}$ from $\ce{SO4-}$, yielding $\ce{H2SO4}$and $\ce{Mg(OH)2}$using perlite.

It forms $CuSO_4$, which is soluble in water. $Cu(OH)_2, CuOH$ and $Mg(OH)_2$ prefer to precipitate, while $CuSO_4$ just floating there, completely invisible. Actually, even when ppl try to obtain $CuSO_4$ in such electrolytic cells, they simply collect $Cu(OH)_2$, add $H_2SO_4$ and crystallize sulfate. So it is there, but there is no way you can obtain it directly.
[remark] You can also try to crystallize it from solution, since it's there. But maybe you won't succeed.

That's what I thought, if not elemental sulfur, which I thought I saw in the top layer. If I centrifuge (I made one) the solution to separate the insoluble Copper and Magnesium would it be possible to plate the copper out of solution to obtain H2SO4?

Check this website. https://www.quora.com/How-can-I-separate-CuSO4-and-H2SO4-they-are-in-the-same-solution
It shows various ideas about this process.

I originally started this process because I thought it would be a cheap source of H2SO4 while being able to produce Cu(OH)2 for a reaction with ammonia to produce Copper Nitrite [Cu(NO2)2] Is that where all of the sulfate anions go?(I've only been playing with chemistry for a month or so.)

It's not really a good idea. I've found youtube video for you, it's like instruction for doing this, but I don't recommend trying anything dangerous at home, without teacher or supervisor who can help you. But at least now you have CuSO4 solution, which is needed (still it's better to use not home-made solution, but as you wish). Promise that you will be careful! Don't work without proper personal protective equipment. https://www.youtube.com/watch?v=t5nHi5_Obd8

Lol, I'm a quick learner. I've already HCl by two methods, H2SO4 + NaCl and via Cl oxidation from NaCl. I'm currently designing the circuit for a high power O3 generator to make pure NaClO for a more efficient source for chloroform synthesis. I'm guessing I can synthesize H2O2 from water if I bubble the ozone through it. I'm trying to synthesize H2O2 electrochemically using carbon fiber electrodes (far superior to regular carbon or lead dioxide, no exfoliation), but I'm having purification issues. How do you take the H2SO4 out and how can I avoid making peroxymonosulfuric acid as a byproduct?

Check for H2O2 + H2SO4 ⇌ H2SO5 + H2O. Adding water shifts equilibrium back to H2SO4, but maybe you don't even need it, since there is trace amount of H2SO5

How is magnesium hydroxide precipitated and copper (II) hydroxide isn't?  Based on solubility products copper (II) hydroxide should be much the less soluble of the two.

@OscarLanzi You are correct, eventually the Mg(OH)2 disappeared and the only insoluble product was Cu(OH)2

@KellyShepphard I was eventually able to obtain  CuSO4 but not like that, I had to separate the reaction to half cells with a few drops of electrolyte at the anode and  the solution at the cathode at saturation resulting in a product with minimal MgSO4 contamination.


text:
How does NH4+ come to being?
There are 5 valence electrons in Nitrogen and in order for it to have a complete outer shell (8 electrons) it needs 3 more electrons. Why is it the case that the Nitrogen needs 4 Hydrogen instead of 3 Hydrogen?
(What is the full name of $\ce{NH4+}$ other than Ammonium?)

Check this picture. Nitrogen has 5 electrons, as you've said: in ammonia, 3 for covalent bonds and 2 as electron pair (dots on picture). That's why ammonia is a base (check for Brønsted–Lowry theory). And bases are capable of donating electron pair - thus, accepting proton. Ammonia then becomes its conjugate acid - ammonium.

p.s. It's common IUPAC name is azanium.
http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf

Does the picture still represent NH4+ then?

No, it's NH3. I choose it to show electron pair. It's ammonia - conjugate base of ammonium.


text:
Manganese (II) oxidation by air
I am studying the reaction of dissolved dioxygen with manganese(II) hydroxide for the Winkler method:
                   2Mn(OH)2(s)+O2(aq)⟶2MnO2(s)+2H2O(l)

The data I have been given is Mn(OH)2's pKs=12.7, as well as the standard potential for MnO2(s)/Mn(OH)2(s), which is 1.23V, and the one of O2/H2O, which is also 1.23V; nothing more. Using Gibbs free energy  for each half reaction, I have managed to calculate the quilibrium constant for the reaction at 298K, which I found to be 10^-25.4, making it dramatically unfavorable to products.
Does anyone find the same thing for this constant, or have I made a mistake? If it is the right constant, how come the reaction takes place?
I am aware that it is not the commonly used reaction when it comes to the Winkler method, but its the only one I can propose when it comes to having the dioxygen react with manganese hydroxide owing to what few data I have. Also sorry for my shaky English, I'm french.
Thanks in advance for your answers!


text:
Getting the pressure of a system by series expansion

An average number, $N$, of bosons of spin $S = 0$ is conned to a two-dimensional domain
  with surface $A$. The gas is ultrarelativistic with a single particle energy $\epsilon = cp$, where $c$ is the speed of light in vacuum and $p$ is the absolute value of the momentum. Compute the pressure, P, of this system as a function of z, A and T. While still being
  in the high-temperature regime, use the result for $N(z, A, T)$ to find $P(N, A, T)$ (keep up to quadratic terms in N). Discuss your results and the relation
  to the ideal gas law.

$N(z, A, T)$ in function of $z$ has been previously calculated:
$$N = \frac{2\pi A}{(h \beta c)^2}(z + \frac{z^2}{4})$$
Where:
$$\beta = \frac{1}{K_B T}$$
$$z = e^{\beta \mu}$$
The following equation also holds:
$$\frac{P}{K_B T} = \frac{2\pi }{(h \beta c)^2}\sum_{n=1}^{\infty} \frac{z^n}{n^3}$$
With this information we should be able to get $P(N,A,T)$ (keeping up to quadratic terms in N), but I am not getting the stated pressure.
The provided solution is:
$$P \approx \frac{NK_BT}{A} (1 - \frac{N(h \beta c)^2}{16 \pi A})$$

I can see one or two areas where clarification might help people to answer your question. Firstly, the $n=2$ term in your sum is $z^2/8$, not $z^2/4$, so this is not consistent with your first equation. Secondly, maybe it would help to define some quantities such as $z$, and explain what system you are considering, and where the first two equations come from. Can you edit your question to explain these points a bit more?

@LonelyProf I have added more details. If there is more information you think would be helpful, please let me know.

Thanks. This clarifies what the series in $z$ applies to, and gives some useful background. I'm puzzled by the appearance of $V$ in that equation, though, when it does not feature in the final solution, or in any other equation. Also, do take a moment to review the policy on homework and related questions which may apply here.

@LonelyProf My bad, actually it has to be $A$ because we are dealing with a surface.

Since this is clearly an exercise, I don't think it's appropriate to give the full solution, but I see no harm in pointing you in the right direction.
You are aiming to convert an expansion in $z$ into an expansion in the number density $\rho=N/A$. These are often called virial expansions. Your first equation can be written
$$
\rho = q\left[z+\frac{z^2}{4}\right] \qquad\text{where}\qquad
q = \frac{2\pi}{(h\beta c)^2} .
$$
I have defined $q$ just to cut down on needless clutter.
The brute force approach to this would be to solve this equation, to give $z$ in terms of $\rho$,
and then substitute this expression into your series formula for $P/K_BT$ as a function of $z$. You can do this; it's a bit fiddly, since it is a quadratic equation in $z$, but it could be done.
For your purposes, a slightly more straightforward approach is to assume that you will get a result of the form
$$
z = c_1 \rho + c_2 \rho^2 + \ldots
$$
where you need to determine the coefficients $c_1$, $c_2$ etc.
Hopefully you can see why there is no need for a $\rho$-independent term $c_0$, when you look at the equation for $\rho$.
Also, it should be obvious that you don't need to consider more terms like $c_3$ etc, if you are only interested in the final result having the first couple of powers of $\rho$.
So you determine the coefficients $c_1$, $c_2$ by substituting that expression for $z$ into the right hand side of the equation for $\rho$,
and equating the coefficients of corresponding powers of $\rho$ on each side.
The term linear in $\rho$ will give the coefficient $c_1$ immediately:
$$
[\rho^1]:\qquad 1 = q \, c_1 .
$$
The quadratic term in $\rho$ will give an equation for $c_2$, which
also involves $c_1$,
$$
[\rho^2]:\qquad 0 = \text{I'll leave you to work this out.}
$$
Since you just determined $c_1$, this will be easy to solve to get $c_2$.
So now you have $z\approx c_1 \rho + c_2 \rho^2$ and you can insert this expression into the given series expansion for $P/K_BT$,
only keeping the terms in $\rho$ and $\rho^2$.
If you do all this carefully, you will get the answer provided.

Thank you, your explanation is really helpful! I got $z\approx  \rho -1/4 \rho^2$ which led me to the provided answer :) By the way, for the sake of curiosity, this is the complete exercise and the provided answer: https://imgur.com/a/dnILMPR https://imgur.com/a/hrCNdoK https://imgur.com/a/4i5tei1 What I did not know was how to get EQ 26.


text:
How can a carbohyrate be hydrolysed?
Polysaccharides and disaccharides if they are neutral how do they undergo hydrolysis?
When making a polysaccharide or a dissacharide dehydration synthesis occurs and why does it not get reversed immediately because of hydrolysis needing water. Do only soluble carbohydrates go through hydrolysis and the insoluble ones don't ?

Look at the picture below. You are right, if you add water to dissacharide, it will hydrolyze, that's why condensation builds up by removing solvent. If you remove water, equilibrium shifts to obtain more of it - thus, condensation proceeds. If you add water, equilibrium shifts to lose it - thus, hydrolysis proceeds (check for reversible reactions). All carbohydrates undergo hydrolysis, "insoluble" are those that bulk instead of dissolving when ingested (such as amylopectin or lignin). But they still can be digested - just find a right ferment for it, as microbes do!


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Molpro - Connection between atomic term symbols and system symmetry
Let's say I have a carbon atom with the electron configuration $1s^22s^22p^2$. We know, that using atomic term symbols we can describe the following states:
$${}^3P_0, {}^3P_1, {}^3P_2, {}^1D_2, {}^1S_0$$
Now I'd like to compute their energies with Molpro. But how should I specify those terms using just symmetry point groups?
I saw something similar using state-averaged multi computation from several different symmetries. But, I don't know, how to derive them just from the term symbol?

My attempt
I was thinking about writing the state wavefunction as the linear combination of microstates and specifying each of then separately.
I.e. let's have the state ${}^1S_0$. According to this answer it can be described like this:
$$|L = 0, M_L = 0\rangle = \frac{1}{\sqrt 3} \left( |m_{l1}= 1, m_{l2} = -1\rangle + |m_{l1}= -1, m_{l2} = 1\rangle - |m_{l1}= 0, m_{l2} = 0\rangle \right)$$
I presume, that I can choose any of the available point groups in Molpro for 1 atom. Let's say I'll choose $D_{2h}$ point group.
Then the orbitals' symmetry can be described like this:

$1s$ ($A_g$)
$2s$ ($A_g$)
$2p_x$ ($B_{3u}$)
$2p_y$ ($B_{2u}$)
$2p_z$ ($B_{1u}$)

According to this, the symmetry of the state $\left| m_{l1} = 1, m_{l2} = -1 \right>$ should be
$$(A_g)^2 \otimes (A_g)^2 \otimes (B_{3u})^1 \otimes (B_{1u})^1 = B_{2g}$$
and the corresponding wavefunction specification in Molpro: wf,6,6,0.
The state $\left| m_{l1}=-1, m_{l2} = 1 \right>$ posses the same symmetry, i.e. $B_{2g}$ and so the Molpro wavefunction specification stays the same wf,6,6,0.
Finally, the state $\left| m_{l1} = 0, m_{l2} = 0 \right>$ has the symmetry
$$(A_g)^2 \otimes (A_g)^2 \otimes (B_{2u})^2 = A_g$$, i.e.
the wavefunction will be specified like wf,6,1,0 and multi will be called
with both of the wavefunctions together
{multi;
 occ,2,1,1,0,1,0,0,0;
 core,1,0,0,0,0,0,0,0;
 wf,6,6,0;
 wf,6,1,0;
}

Is this the correct approach?

You are trying to use the wf card to pass individual configurations. This is wrong in 2 ways:

The wf card does not specify which configurations to use, it specifies the symmetry of the wave function (or electronic state).
You don't need to tell Molpro which configurations to use. It will figure this out on its own based on the symmetry of the wave function you are requesting.
All you need is the IRREP of the $^1S_0$ state, which is totally symmetric.

{multi;
  occ,2,1,1,0,1,0,0,0;
  core,1,0,0,0,0,0,0,0;
  wf,6,1,0;
 }
However, this will give you different a state than expected. Add expec2,lxx,lyy,lzz; and check $\langle L^2\rangle$ in the output. It should be 6, telling us the $^1D_1$ state is lower in energy than $^1S_0$. So we need to request more states of that symmetry (use the state card). Also, we should make sure the $^1D_1$ is described appropriately, as it might negatively affect the targeted $^1S_0$ state. So you should add all of the 5 degenerate $^1D_1$ components. I leave that as an exercise for you (see also below).
Some further note about your configuration $(a_g)^2\otimes(a_g)^2\otimes(b_{3u})^1\otimes(b_{1u})^1=B_{2g}$ (Note that the symmetry of orbitals is commonly denoted by small letters, while capital letters are used for electronic states.)

Only IRREPs of same symmetry mix, so this one will not contribute to $^1S_0$.
It is breaking the spherical symmetry of the atom. This can only happen if there is some external perturbation, e.g. a magnetic or electric field.
It is however, one component of a degenerate (different) state. But you need to consider all the degenerate components to preserve symmetry. This is an issue of the symmetry reduction from $O_h$ to $D_{2h}$. For example you can calculate the $^3P$ state with

{multi;
  occ,2,1,1,0,1,0,0,0;
  core,1,0,0,0,0,0,0,0;
  wf,6,4,2;
  wf,6,6,2;
  wf,6,7,2;
 }
Check that you get 3 exactly degenerate $p$ orbitals, and 3 exactly degenerate electronic state energies.
If you would request each state separately, by running multi 3 times with only one wf card each, you will get 3 degenerate state energies, but the $p$ orbitals won't be degenerate.

Thank you! But I'm not sure, how did you derive, that ${}^1S_0$ is totally symmetric? And how did you derive the three degenerate states corresponding to ${}^3P$?

It is written in the Molpro documentation, you even used it yourself (for the orbitals). In principle one should be able to derive this by reducing the full spherical point group to $D_{2h}$, I just can't find its character table. Alternatively you can just look at the spherical harmonics and determine their characters.

In Molpro documentation? I know, how to specify the irreducible representations in Molpro, my problem is much more basic - I don't understand, how to determine the symmetry of the state (e.g. ${}^1S_0$) without its microstates (electron configurations).

Well, how do you know a $1s$ orbital is $A_g$? Same story ...

Of course, I know, how $1s$ "looks" like, so I can determine it easily. On the other hand I don't know this about ${}^1S_0$, as it's a superposition of several electron configurations...

But $S$ still is the angular momentum $L=0$, just as $s$ is $l=0$. If you only occupy $s$ orbitals, the total symmetry is just the same. And if you equally distribute electron occupations between the three $p$ orbitals, they will be spherical symmetric as well.

Actually it is the other way around. The symmetry is fixed be specifying an $S$ state. What configurations are allowed to serve as a basis follows from that.

Would you mind to add your explanation in the answer with maybe a step-by-step derivation for the singlet S state or all three P states? I'm afraid, that I'm still not understanding it completely and I think it would be very helpful even for some other readers.

Somehow I fail to see your problem. A $1s$ orbital is $a_g$, just as you wrote in your question yourself. It is the exact same story.

So, as far as I know, $S$ term doesn't equal correspond with $s$ orbitals in any direct way.

Both are things with a spherical distribution without any total angular momentum.

Maybe make a new question "Why has an $S$ term $A_g$ symmetry?". Then somebody else can try to explain it in a different way.

I'll try that both with $S$ and $P$ states - hopefully it'll fill the missing space in my understanding. Anyway, thank you very much for your responses!


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Bond length in NO3-
I am doing a multiple choice question for which one of the possible answers is, for the Lewis structure of NO3- (one of the oxygens is double bonded to the nitrogen and the other two are single bonded),
"one of the N-O bonds is shorter than the other two".
Apparently this is not the right answer, but my understanding was that a double bond is shorter than a single bond. I just googled it and I'm reading that the more bonds there are the shorter the bond length. So shouldn't that answer be correct?
Thanks!

Resonance, resonance... all three bonds are equal.

Could you explain that? Is it just that the "double bond" moves around equally around the N?

It's not really a duplicate (IMO), but it contains the answer to this question nevertheless.

The main reason for your answer being wrong is the phenomenon of resonance in nitrate ion dur to presence of a conjugated system. The resonance is shown in the figure below. Dur to resonance none of the cannonical structures explain the bond length in nitrate except the resonance hybrid ( which happens to be the last drawn structure). Due to resonance all bond lengths would be the same