text:
Why should I care about the highly radioactive elements after fermium?
After element 100, fermium, there are a bunch of highly radioactive elements like fermium, tennessine, and oganesson, and others. These elements are to radioactive to store, they will instantly disappear, so we cannot use them in any way. If they are to radioactive to use, why should I care about them? I by NO means am saying that we should not care about them, I am asking what kind of potential they have?

Because science! Just because something doesn't appear to immediately have a use does not mean it's useless - in the context of what you are asking, have a look at the so-called Island of Stability.

The potential might not be apparent. The history of science is rife with discoveries that in turn led to other discoveries that changed things profoundly. Maybe the quest for stable elements in or around the island of stability leads to an efficient fuel source for satellites in the future. Or better medical treatments.

@ToddMinehardt I hope a use is discovered.

Thank goodness you didn't think about astronomy. I wonder what applications would you want from discovery of a galaxy 10 billion light-years away... Science is primarily about improving our knowledge and by detecting unknown elements we do just that. What can be done with knowledge is secondary thing.

@ToddMinehardt I am not saying we should not, I am asking if there are any possible applications for them

It's still a matter of opinion, in my opinion. Also, the question would probably be closed as too broad. You might reword it and try to get it reopened.


text:
Polyprotic Acid Titrations
Titrating a mixture of strong and weak acids will yield endpoints such that the volume of base required to reach each endpoint will be different for each acid, e.g. in a mixture of HCl and CH3COOH, the volume of base required to reach the HCl endpoint will be less than that for the CH3COOH endpoint. This is all down to relative acid strengths.
However, the above isn't observed when titrating a polyprotic acid, as each endpoint requires the same volume of base; this is counter-intuitive when considering the above paragraph. For example, H2SO4 also has two endpoints – one for H2SO4 and one for HSO4- – but the volume of base required to reach them remains constant. This doesn't sit right with me, considering that HSO4- is a weaker acid than H2SO4.
So my question is: why does the volume of base required to reach each endpoint in a polyprotic acid titration remain constant, even though each endpoint will result in a weaker acid? Surely the volume of base will increase each time, like is the case in a mixture of stong and weak acids?

** e. in a mixture of HCl and CH3COOH, the volume of base required to reach the HCl endpoint will be less than that for the CH3COOH endpoint **.
kindly illustrate the example by using numbers and calculations

I will start with stating the obvious: in a solution of a polyprotic acid like sulfuric acid, you will ALWAYS have equal amount of each acid (proton). As you say: a molecule of sulfuric acid dissociates in two steps, and the amount of base needed to titrate the protons released at each step must necessarily be the same. Off course, if two different acids are mixed, you can have any ratio.
If you titrate a mixture of HCl and Acetic acid, the first equivalence point reached gives the amount of HCl. As HCl is by far the strongest acid, it will react first with the added base. Thereafter, the base will start to react with Acetic acid until you reach its equivalence point.
The volume of base from zero to first eqv. point tells you how much HCl there is, but it is the amount used BETWEEN the first and second eqv. point that tells you how much Acetic acid you have.

source: Stack Exchange

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

sort: votes


text:
How would one determine the yield of primary vs. secondary hydrohalogenation addition products?
For example, if I was to perform the following reaction:
$\ce{CH3CHCH2 + HBr -> CH3CHBrCH3 + CH3CH2CH2Br}$
How would I quantitatively determine the relative yield of the primary and secondary alkyl halides?
Would IR or UV/Vis spectrscopy be able to differentiate between them, or would I have to utilize reactions such as substitution with hydroxyl then oxidation to form ketones and carboxylic acids to differentiate between the two?

1H NMR would show the relative yield by comparing the integration of the CHBr proton and the CH2Br protons.. IR or UV/Vis will not do this.

Or you isolates them (not sure isolate is the term in English. I mean in the current example you get the individual compounds by distillation, if possible. .).

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


text:
How to calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by some concentration?

Consider the cell described below at $\pu{261 K}$:
$$\ce{Fe | Fe^2+ (\pu{0.925 M}) || Cd^2+ (\pu{0.981 M}) | Cd}$$
Given $E^\circ_\ce{Cd^2+ -> Cd} = \pu{-0.403 V}$, $E^\circ_\ce{Fe^2+ -> Fe} = \pu{-0.441 V}$.
Calculate the cell potential after the reaction has operated long enough for the $\ce{Fe^2+}$ to have changed by $\pu{0.383 mol L-1}$.

I have
$$E_\mathrm{cell} = (0.441 - 0.403) - \frac{0.0591}{2}\ln\left(\frac{0.925}{ 0.981}\right) = 0.03974$$
but I'm not sure how to find the $E_\mathrm{cell}$ when the $\ce{Fe2+}$ concentration has changed by $\pu{0.383 mol L-1}$.

What are the concentrations of the ions at that point? I'm not sure I understand what you mean by "changed by 0.383 mol/L". Is it increasing or decreasing?

I added a screenshot of the problem I'm trying to figure out.

Are you using the natural or the base-10 logarithm?

I'm using the natural log

It's either R T / z F ln(Q) or 0.0591 V / z log(Q). For your formula, the term is too large by a factor 2.303 = ln(10).

As the reaction proceeds towards equilibrium, the cell potential approaches zero. This is one way of figuring out which concentration increases and which decreases. Or you write down the chemical equation of the reaction in the direction it proceeds and work from there.


text:
Nanoparticle Alloys (Gold-Silver) not forming
I have synthesised what I thought were gold-silver alloys at various different compositions.
The UV/vis data showed that nanoparticles alloys with a majority of the composition made from gold had one plasmon band, indicating they were alloys.
However with nanoparticle alloys where the composition is mainly silver, the UV/vis data shows two plasmon bands, indicating the presence of segregated gold and silver nanoparticles which are not alloyed.
Could anyone explain why this has happened?

From a phase diagram perspective there isn’t a good one. I’d contemplate the nanoparticle growth process and how a silver rich vs gold rich particle forms.

Can you add how you made the particles. It would be helpful to know if you did a simultaneous reduction and with what materials.

Yes I carried out a simultaneous reduction using sodium borohydride, the stabilizing agent is PDADMAC

I do intent to answer this just need to find a particular paper first.


text:
Minimizing energy of a molecule
I am minimizing energy of butane on the CHARMM Force Field and I am confused on how to go about getting a uniform unit for each different PE  as in the potential energy for bond, angle, torsion, VDW, ES of butane. For instance if you compute the angle PE you will be using angles vs bond length PE which will be in Ångstrom. I am confused on how to set everything within the same units or parameter.
Also I am programming this in python using numerical methods and will not be using a simulator.

First I only wanted to comment, but then it got too long... But I wanted to pass on this info... this isn't a real answer...
Every simulator I know uses atomic units. I would check out the NAMD and CHARMM manual for their units, I imagine they use nanometer for distance and ultimately kJ/mol for energy. You will need a conversion for the electrostatic since you always input +/- charge, where charge is usually between 0 and 1, but units of charge are actually Coulomb. The conversion factor for this is ~138.9
This is from my own code... qq_convert is the electrostatic conversion I mentioned above. I use nm for all distances. I try to follow the same procedure as Gromacs.
const na = 6.02214129e23      # mol^-1
const R = 8.3144598e-3        # kJ mol^-1 K^-1
const kb = R / na             # kJ K^-1
const e = 1.6021765653e-19     # C
ϵ₀ = 8.854187817e-12 # C²/(J ⋅ m)
ϵ₀ *= 1e-9  # C²/(J ⋅ nm)
ϵ₀ *= 1000   # C²/(kJ ⋅ nm)
const qq_convert = e^2 / ϵ₀ * na / 4.0 / pi # 138.935458  # kJ mol^-1 nm e^-2

after each electrostatic interaction I multiply the energy by qq_convert
For instance the real part contribution of the EWALD summation is
@inline function ERealSingle(diff::MVector{3,Float64},qqCut²::Float64, coord1::SVector{3,Float64},
        coord2::SVector{3,Float64},q1::Float64, q2::Float64, kappa::Float64,
        overflow²::Float64, L::SVector{3,Float64})

    @inbounds for i=1:3
        diff[i] = vector1D(coord1[i],coord2[i],L[i]) # mirror image separation
    end

    rij_sq = diff[1]*diff[1] + diff[2]*diff[2] + diff[3]*diff[3]
    rij = sqrt(rij_sq)

    if rij_sq < qqCut²
        return q1 * q2 * erfc( kappa * rij ) / rij
    else
        return 0.0
    end
end

This is for a single charge interacting with another charge... I loop through all charges etc and then I multiply the summed energy by qq_convert.
I highly recommend checking your code against these SPC/E water benchmarks offered by NIST... really saved my bacon having a test set to duplicate... NIST Test Set. This set allows you to test your Lennard-Jones, all parts of the EWALD Summation and also tail corrections.
Lastly, I recommend using the Julia programming language... it is much faster than Python... Python can almost keep up if you use the Numba JIT decorators, but Julia has other perks.
Edit:
It looks to me like charmm uses Angstrom as you have said, so that is the only units of length you should be using. It would be dangerous for them to deviate.

In my experience with Gromacs, the parameters of each potential energy function will have units such that everything cancels out leaving you with $\ce{kJ/mol}$ (or $\ce{kcal/mol}$ in CHARMM I assume). In Gromacs all I need to do is make sure all units of length are nm. I imagine in CHARMM it is the same, but for Angstrom. Look at the angle parameter units, they do have units and it will work out that the energy is $\ce{kcal/mol}$ or else I say leave CHARMM.


text:
Perturbation theory in Quantum/Computational Chemistry
As we know, the common methods for computation molecular properties in modern Quantum/Computational Chemistry is the Hartree–Fock method and Density Functional Theory.
But from university course of Quantum Mechanic we know also, the perturbation theory method. Why is this method not common in quantum chemistry?

I guess it goes down to what kind of properties/systems you can calculate using a given method and what is the cost. I think it a good question if we view it this way: where perturbation theory can offer significant improvement.

@Greg Can it be argued that perturbation theory requires increasing computational costs, with a slight increase in accuracy?

There are several different perturbation theory methods, but I guess the simple answer to your question is exactly that: in most situations, they don't pay off compared to e.g. DFT.

Note, however, that HF and DFT are by themselves often not sufficient. Unless there is a straight-forward operator for the property you are interested in, linear-response theory or higher-order derivatives may need to come into play. Their application is not free, either.

@Sergio I would argue Hartree-Fock on its own is hardly used in computational research these days. Usually it is used as a reference in post-HF methods like Coupled Cluster or MP2 (a perturbation method). I think MP2 does get used a decent amount both as a cheaper alternative to CCSD to include correlation and more recently as a piece of double hybrid DFT methods. However it does seem to strike a bad balance of accuracy/cost, with DFT being much faster while still decently accurate and CCSD being more accurate and only somewhat slower.

What exactly do you mean with "common"?
While HF and DFT are the basis for many methods, I think that hardly anyone does pure HF or DFT calculations.
Perturbation based models like MP2, CASPT2 are used quite often.

@Fl.pf.  "often" and "common" are both rather vague terms, but we can all agree that there is are significant cases where MP2 or CASPT2 are necessary. CCSD(T) is also perturbational and plays an important role, too. Let's not fall to the other direction: while pure HF is indeed rarely used nowadays, DFT methods (both pure and hybrid) are still standard in many fields outside of the "small, closed shell molecule" realms.

Many people have already said this in the comments, but I will put it into an answer. Also, it would not be too hard to find a bunch of references in favor of how common any particular method is in quantum chemistry, so I won't bother to list a bunch of examples of perturbation theory being used.
That being said, if I had to guess, I would bet that second to DFT, second-order Moller-Plesset perturbation theory (MP2) is the most common method in quantum chemistry which includes electron correlation. Also, there are large communities in computational chemistry which virtually ignore the existence of DFT because they are willing to sacrifice computer-time (and possibly some accuracy) to be more certain that their answer is reliable. So, MP2 is a good example of perturbation theory being using in QM.
Also, you will often see people refer to coupled-cluster with singles and doubles excitations and a perturbative inclusion of the triples, CCSD(T), as the "gold-standard" in quantum chemistry. This is a bit more complicated application of perturbation theory, but I would say that CCSD(T) is used in virtually any problem where one can afford to do the calculation. There might be times where one could feel more confident about using DFT with a particular functional over MP2, but you would be hard-pressed to find a good argument to not do a CCSD(T) calculation if you can afford to do so. This is especially true if the alternative is to only use DFT because many functionals are parameterized against CCSD(T) calculations, so getting similar results from CCSD(T) helps validate the use of that functional for the class of problem.
There are also many other electronic structure methods which use perturbation theory such as CASPT2, one can also include perturbation excitations into truncated configuration interaction (CI) calculations, and probably lots of other but less-commonly-used methods. Also, one approach to including spin-orbit coupling in electronic structure is via perturbation theory.
One of the reasons there is such a vast array of methods which use perturbation theory is that they are diagrammatic theories, which means there is a rather simple way of understanding all of the terms physically and the connection between these terms and an algebraic representation of them is quite straightforward within the framework of second-quantization (basically using special creation and annihilation operators).
Also, I know you're mostly asking about electronic structure, but it should be noted that perturbation theory is often used to add anharmonicity onto a harmonic reference calculation of vibrational frequencies. Probably the most common method of doing this is via VPT2 which is basically the vibrational analogue of MP2.
In summary, perturbation theory is an extremely powerful tool which will almost certainly be applied to any problem where a simpler reference answer can be found. On this note, these days I don't think you could get away with doing only a Hartree-Fock calculation and expect people to believe your conclusions about the properties of the physical system. Also, there are such good, parallel codes for MP2 that if you can do a HF calculation, you can probably afford an MP2 calculation.

As a response to one of the comments, I would argue that CCSD by itself is not much more accurate than MP2 for the increased cost. Both methods can be understood as including double-excitations out of a reference determinant, and my experience is just doing CCSD does not improve much over MP2. Although MP2 and CCSD sometimes seem to be wrong in different ways, so they aren't strictly comparable. This is why people insist on CCSD(T) being the gold-standard and not just CCSD.

CCSD yields much better densities than MP2, and in cases of degeneracy, then MP2 fails miserably. Additionally, even if neither are variational (except in the full CI limit of CC), there is a clear way to improve CC (CCSDT > CCSD(T) > CCSD > CCD), whereas MP3 can frequently yield worse results than MP2.


text:
Mechanism for removal of pesticides from fruit by sodium bicarbonate
When Googling around for methods to remove pesticides from fruits and vegetables, a study commonly referred to from some news sites (e.g. here and here ) is one where soaking apples in a sodium bicarbonate solution was shown to effectively remove pesticides (the study was conducted using the pesticides phosmet and thiabendazole), more effectively then a certain method where just water was used.
My question is what would be the likely mechanism for a sodium bicarbonate solution to remove pesticides better than plain water (which was also tested in the mentioned study) and what that says about the applicability of the study to other pesticides (is the result likely to be applicable just to the two pesticides tested, or to be quite general?)
Here is my guess; a guess from just dabbling in chemistry here and there (I'm a computer programmer and by no means an expert on chemistry). The guess is that the additional hydroxide ions in the solution, created due to the sodium bicarbonate having been added, make hydrolysis of the pesticides more likely because the hydroxide can do a "nucleophilic attack".

Hydrolysis is certainly a possibility. It will depend on the structure of the pesticide molecule as to how suspectible it is to this process. Phosmet - quite possibly so. Thiabendazole - not very much, more likely removed by a mild acid (vinegar?) wash as it is basic.

The $\ce{P-S}$ is ~half as strong as the  $\ce{P-O}$ bond so it looks like $\ce{OH-}$  catalyzed hydrolysis would readily occur in phosmet.


text:
Raman Spectrum of Graphene
I recently took a raman spectrum for a sample of graphene-pla. The result had the characteristic G and D bands associated with Graphene, but not the 2D. But there is a broad peak at around 3100, which goes up from 2600. The sample was treated before being characterized. Does anyone know why this might have happened?


text:
Solving Polyester but no chromed steel
In short
I'm looking for an acid that can solve a polyester, fiberglass composite. However it cannot damage steel.
First of all i like to say that i'm not a chemist. I work in a plastic moulding company. Let me sketch a background.
We make bars out of a polyester strengthened by fiberglas. To make this we have the two materials beeing pulled through a heated steel pipe (the mould). The polyester melts around the fiberglass and forms a strong lightweight pole. However sometimes the pole breaks mid process and then there is a part that's stuck in the mould. Like in the picture

It's very hard to remove the polyester/fiberglass bar from the mould, dare I say near impossible.
I was wondering if there is an acid that solves the polyester but makes no damage at all to the chrome steel mould.

Look using the terms " solvent for. ..". It seems a hard task because you have a rod trapped in the mould. If the mould cannot be open it might takes years. ..

Is it not possible to heat the mold above the melting point of the polyester, then PUSH out the stuck bar with a rod?

we tried drilling, melting, ... if there is an acid or solvent of polymer that doesn't damage the steel, i would make a system that circulates, the acid, maybe even heats it to speed up the process. If it works after weeks, i'll be happy with it, a year, or years might be a bit too long

@jeffB we tried pushing it hydraulically with a force of 800kg, hammering, and the rod is usually just too long, and has too much friction, we're talking about a mould of a length of 1m50,

Can't you pretreat the mould by coating it, say with PTFE? Or would this not help?

It's a constant cycle, there's constantly plastic and glass fibers beeing pulled through the mould, so it's not an option.

The thermal exansion of polymers is sufficiently high that if you cooled the entire thing then the polyester might break free.

Cooling to what temperature

I'd cool the whole thing to as low as possible and let it sit for a while. if that doesn't work then I would try heating the metal while the plastic is still cold.

Unless you mean cooling it below 0°C we already tried it

I repeat look for polyester solubility or even better "incompatibility" and surely you find solvents either as pure chemicals or commercial formulations. Then you can try circulating it though your system in they will be no too flammable etc...

Second to Alchimista - Google for solvents. If your polyester is PET, for example, here is a reference: https://onlinelibrary.wiley.com/doi/pdf/10.1002/app.1417. Also a discussion at Researchgate:https://www.researchgate.net/post/How_do_you_dissolve_polyethyelene_terephthalate_PET. The small available surface for contact with the plastic may mean it takes a really long time, though.


text:
Defining sensitivity of TOF Mass Spec - Does the detector play a role?
Looking at TOF MS systems, my current understanding is that temporal resolution results in mass resolution - so by controlling the flight time (with distance) and sampling rate, you can start to distinguish smaller and smaller changes in mass. Hopefully, this is correct.
Where I'm confused is when people start talking about sensitivity.
I heard a talk where they mentioned femtogram and attogram sensitivity.

Does this refer to the amount of the sample they are using?
OR is this referring to the minimum amount of the analyte they can detect?
Does that translate to needing a certain number of ions to hit the detector in a single scan to have a signal significantly above the noise?
Finally, what types of detectors are capable of measuring so low and are there any issues with them?

Thanks for your help!
Cheers

RE: TOF Mass Spec - Does the detector play a role? -- Without a detector how do you get a signal?

@MaxW Apologies if I did not word my question correctly. There would be a detector - I'm asking which detectors would you use for high sensitivity applications (femto and attogram) and are there issues with detectors for this sensitivity.

The term sensitivity has been misused in analytical chemistry a lot (even in German scientific literature). Be careful if you read articles especially the older ones. Sensitivity, in a true sense, refers to the slope of the calibration curve. Thankfully IUPAC sorted this out and defined it as "The slope of the calibration curve. If the curve is in fact a 'curve', rather than a straight line, then of course sensitivity will be a function of analyte concentration or amount. If sensitivity is to be a unique performance characteristic, it must depend only on the chemical measurement process, not upon scale factors."
When we wish to refer to the minimum amount that can be detected, it is called the limit of detection (LOD). Hope this clarifies the terminology. Note that sensitivity and limit of detection are related. Higher the sensitivity (= slope of calibration curve), lower the detection limit, provided that noise does not increase.
Coming to the detectors: The limit of detection of all detectors which are based on counting (basically we count ions indirectly in mass spectrometry) is limited by statistics rather not by electronics! This is the fundamental limit put by mother Nature. It is called shot noise. Whenever you count discrete random events, in case of a mass spec. random impinging of ions, there is always an error associated with it which is proportional to square root of N, where N is the number of events.
You can read the article on shot noise (for educational purposes).The Ultimate Limit in Measurements by Instrumental Analysis: An Interesting Account of Schroteffekt and Shot Noise, https://pubs.acs.org/doi/abs/10.1021/acs.jchemed.8b00202


text:
When finding the liquid density which gives the most accurate results, a 25 ml Pycnometer or a graduated 10 ml cylinder?
Its 10 ml graduated cylinder with 25 ml of pycnometer, and we are measuring the density of the liquid(pure liquid) in each of the materials and we are suppose to compare the densities found , in terms of accuracy
Excuse me guys am doing a a research project on density, these are the two methods i used to find the accuracy, been searching but i have no progress. Feel free to give suggestions.

Please improve your question with more details. I know what a pycnometer is, and how it is used. But what do you mean by a cylinder, and how do you use it? Also are you trying to determine the density of solids or liquids?

And what are you aiming for in this (educational?) project?

@MaxW, yes its 10 ml graduated cylinder with 25 ml of pycnometer, and we are measuring the density of the liquid in each of the materials and we are suppose to compare the densities found , in terms of accuracy. the graduated cylinder was weighed and 1 ml of water is added respectively till the 10 ml mark is reached, its masses were recorded.

@Karl, the aim of this project is to find the liquid density of the pure liquid using a 25 ml pycnometer and graduated 10 ml cylinder and compare their densities interms of accuracy

Cylinder is always a "secondary standard" in volumetric measurements. It is never used in quantitative transfers. A pycnometer is designed to hold known volumes and it should always be used for density measurements at known temperature. Good quality pycnometers come with a certificate of quality assurance.
If you are interested in a deeper investigation, analytical glasswares come in two grades, Class A and Class B defined by so-called tolerances by various organizations. If you look closely you will see these labels on quality glassware. Similarly, look for TD and TC label on cylinders and other volume measuring glasswares. TD means to deliver and TC means to contain. If you don't see any A or B label, chances are that this glassware just good for educational purposes (or economical version). European union and other countries have set limits on volumetric accuracy.

A bit less accurate than a pycnometer, but much cheaper, are https://en.wikipedia.org/wiki/Volumetric_pipette s. I'm happy to say my university never insulted us with cheap, lower grade glassware. ;-)


text:
How to report result to the nearest 0.01 mg/L?
what is meant by reporting result to the nearest 0.01 mg/L?
The test result in certificate show 7 mg/L showing no decimal point.

Reporting to the nearest quantity means that you report as many decimal places of the quantity as suggested. This is a common analytical chemistry expression given in procedures or analytical protocols. For example, if we say weigh  $\pu{5 g} \; \ce{KOH}$ to the nearest $\pu{0.1 g}$, one would record the weight to one decimal place, e.g., $\pu{5.1 g}$ or $\pu{5.3 g}$, to one decimal place but you will not report $\pu{5.1176 g} \; \ce{KOH}$ even if your balance is capable of doing so.
Your test report should include at least two decimal places, the report should appear as $\pu{7.24 mg/L}$ or something like that number.

If the quoted value is 7 mg/l and there is no other information given, this means that we must assume that the value has been rounded and is thus between 6.5 and 7.5 mg/l, i.e. $7\pm 0.5$ mg/l.
If a result $X$ is given to nearest 0.01 mg/l or 1% this means that we must assume that the value was rounded up from $X\pm 0.005$  mg/l.


text:
Are the vibrational modes of CO2 active at room temperature?
I was doing an experiment whereby I had to measure the specific heat ratios of certain gases. Carbon dioxide came out to be around 1.3, and checking with the accepted values this is close. My question is, the only way I can justify this answer is if $\ce{CO2}$ had 7 active degrees of freedom at room temperature. This implies that the molecule is vibrating.
I am not a chemist and I thought that molecules only store energy in this degree of freedom at high temperatures.

Your results implies just rotational degree of freedom. Why you see it as to require activated vibrations? 7/5 is 1.3

Its 1.4, the tables in my textbook have the specific heat of C02 listed has 1.28, which is closer to 9/7.

I see.  I was astray by you saying " the only way I can justify this answer is if C02 had 7 active degrees of freedom at room temp".

Yes, the vibrational modes are present at any temperature, including absolute zero where the lowest vibrational energy levels only are populated; the zero-point levels. The molecule is linear ( OCO ) so it has $3N-5=4$ vibrational modes for $N=3$ atoms. The symmetric stretch  corresponds to both CO bonds stretching in phase, and asymmetric stretch to one CO stretching while the other compresses (this has highest frequency), and there are two degenerate bends where the OCO angle changes (lowest frequency). If the z-axis is along the OCO bonds, bends are in the zx and in the zy planes. Wikipedia has lots of diagrams showing these vibrational normal modes.
I have calculated the heat capacity as shown in the figure. The translational value is 3R/2 at all but the lowest temperatures (a few K) and similarly for rotational levels, which have a value of R, again above about 50 Kelvin. (The rotational constant is 0.3902 wavenumbers). The vibrational heat capacity from the modes (1388, 667,667, and 2349 wavenumbers) gives the rising curve on the figure. At 300 K the value is about 3.4 R or 28 J/mol/K. It is clear that only a few vibrational quanta are populated at room temperature. The high temperature value of the heat capacity adds another 4R (each vibration counts R) and is not reached until about 4000 K. Clearly never obtained in practice.

Total heat capacity for CO2 vs temperature. The blue line is the total heat capacity, the lower horizontal line has a value R, the upper one R+3R/2. The top horizontal line has no particular importance and is the value at 300 K and is approximately 3.4R as mentioned in the comments.

Wait, are you saying that there are 4 vibrational modes present at room temperature?

The point is not how many modes are there or how many are vibrational. The point is that the four modes (that of course exist) may or not be excited by heat. At room temperature the answer is no, and we are let with 5. Discrepancies might come from experiment, from the fact that the gas is real, from the fact that a small percentage might be populated statically. I would use for gamma the ratio 7/5, but perhaps I am wrong.@Vishal Jain

As would I, its just slightly weird that N2, another linear molecule, which also has 5 degrees of freedom, has a specific heat ratio of almost exactly 1.4 at room temperature, whereas C02 has a lower value of gamma with the exact same number of degrees of freedom available to it.

Yes 4 vibrational modes at any temperature. As @Alchimista notes if you are measuring heat capacity the population of these vibrations really does matter, also the number of (whole body) rotational levels excited and the translational energy both need to be calculated. All these can be obtained via statistical mechanics using the Partition Function. You need only the pressure, mass, vibrational frequencies and bond lengths and the temperature of course.

@Alchimista The reality is that the constant volume molar heat capacity of CO2 at room temperature is 3.4R, not 2.5R.  So using 7/5 for gamma would be incorrect.

@Chester Miller how to arrive at almost 7 for the accessible DOG?

@Alchimista I have no idea.  And, as an engineer, I don't really care.

@Alchimista I don't think "almost 7 for accessible DOG" is the right wording. There are 5 total translation + rotation degrees of freedom. The additional R worth of contribution to Cv at RT is not strictly 2 additional DOG, but a sum of contributions from the four vibrational modes (mostly the two bending ones), each of which has a small fraction of excited states such that the contribution is not the expected 1R for each one, but just a fraction of that. Coincidentally, the sum of the four fractions turns out to be about R at room temp, a misleadingly round number.

Yes  @Andrew. I just quickly formulated it to be intended as you have resumed. Also plus one for the answer.

Thank you all for the extended discussion, but I was just wondering if anyone could explain why n2's heat capacity is exactly 1.4. Surely the same statistical arguments would lead to some of the vibrational modes being active, contributing to the specific heat, meaning like c02 it ought to have a lower gamma. Not sure if this counts as a separate question.

First as you can see from the figure heat capacity depends on the temperature.  In nitrogen molecules the vibrational contribution at T=300 K is negligible as it has a large vibrational energy compared to RT so only the lowest level is populated. Thus the heat capacity is 5R/2 J/mol/K where R is the gas constant. This comprises 3R/2 from translational motion and R from rotational motion.


text:
Why do we have only one DNA?
I've been crossed a very interesting thought..
Why is it that our body has only single form of DNA, and what would happen if we had multiple forms of DNA?

What do you mean with "multiple forms of DNA"? In what way would/should they differ from the current version?

We have just one type of DNA, because humans develop from a single cell. This single cell contains only type of DNA and cell division makes exact copies of it. Since humans (and basically all other organisms) form by cell division, they therefore have to have the same one type of DNA as the first cell.
The question then is, why should the first cell have only one type of DNA. This is probably because the molecules of DNA "float" freely in the nucleus and having two different "types" of DNA could lead to some weird interactions between the different types.
It would be possible, theoretically, to have more nuclei, each with a different type of DNA, but I don't really see why any organism would want to do that. It would make genetic errors much more probable and it seems that all organisms live all right with just one type of DNA.
*I write "a different type of DNA", but what I really mean (and what I think you also meant) is "a different way of storing genetic information". In a way, humans have two different ways of doing that - RNA and DNA, but only DNA is used for long-term storage, since it is more stable.
A different way could be to e. g. use a different molecule than DNA, maybe different bases than A, C, G, T; or use a different sugar than (deoxy-)ribose or maybe a different link than the phosphate groups. This makes for a very exciting field of study (and I think the part with more bases has already been done somewhere.)


text:
What is a good source for large tables on the properties of water?
I'm currently doing an investigation for school on finding the Clausius-Clapeyron relation/gradient for  boiling water and comparing it to the gradients along my line of best fit. It is given by the equation,
$$
\frac{\text{d}P}{\text{d}T} = \frac{L}{T\Delta v}.
$$
I'm having trouble finding the specific volume of both water and steam (I found this for water but I'm not sure of the credibility) for $\Delta v$. I was hoping someone could reference me to a large data table for finding the specific volume, or a calculator. Thank you.

For every constant related to chemistry and physics, I would advise the "CRC Handbook of Chemistry and Physics". At least it was everywhere before the internet changed the whole game. But it does not make it useless.

Thank you I'll check it out, I believe we have one at school but since I have a large table of data I was hoping I could just copy paste. If worst comes to worst this will be fine.

@JohnMiller When I had vaguely-similar lab assignments in university, they directed us to an online (pay-walled) version of the CRC Handbook. Tables were mostly copy-able, though not always with the best formatting.

Such values for various properties of liquid water and steam can be looked up in so-called steam tables. Software versions also allow calculations (ususally interpolation) of additional values. For example, I usually use REFPROP – NIST Standard Reference Database (however, not always the latest version) or some old and simple proprietary steam table software from Siemens/KWU.
If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha.

I was once looking for these and typed "steam tables" into a search engine.  I got lots of hits for restaurant equipement.

The usual reference for phase change information, for various substances, is Perry's Chemical Engineers' HandBook by Perry and Green. Then there are the CRC HandBooks and sometimes such information can be gleaned from the Chemical Engineering Encyclopedias (I forget their names now). Then IUPAC, NIST and some other governemntal institutions make such information available. At one point some of the publishers were touting dataset API's though these were/are behind pay walls so you'll have to go through your School/Universities/Local libraries subscription to get the data.
Most of the properties reported within Perry's Chemical Engineers' HandBook originally appeared in two papers

"Inorganic Compounds (Vapor Pressure of Pure Substances)" appearing in Industrial and Engineering Chemistry 1947 Vol. 39 No. 4 pages 540-550
"Vapor Pressure of Pure Substances" by Daniel R. Stull appearing in Industrial and Engineering Chemistry 1947 Vol. 39 No. 4 pages 517-540
Specifically for water there are the "Steam Tables". I know of two versions
"Steam Tables : Thermodynamic Properties of Water including vapor, Liquid, and Solid Phases" by Joseph H. Keenan, Frederick G.Keyes, Philip G. Hill and Joan G. Moore (ISBN : 0-4710-4210-2)
"Thermophysical Properties of Water Substance" by Jeffery R. Cooper and Edwin J. Le Fevre (ISBN : 0-7131-3222-1)

Using the latter we have the following information for the triple point :
  T(K)    P(MPa)     Phase  *v*(m^3/Mg) *e*(kJ/kg) *h* (kJ/kg) *s*(kJ/(kg K))
 ------ ---------- -------- ----------- ---------- ----------- --------------
                    Solid         1.091     -333.5      -333.5         -1.221
 273.16  0.0006112  Liquid       1.0002          0   0.0006113              0
                    Gas          206180       2375        2501          9.156

You can use these values for checking the authenticity of other sources. Also good luck fitting Clausius-Clapeyron the equation is quite sensitive to the constants. I think it was easier to fit on a log curve, or by inverting it If I remember correctly. FSolve in Excel should handle the fitting alright though otherwise matlab, wolfram and pandas/numpy/scipy provide algorithms for the task.

Yeah I've just finished compiling the data in MATLAB and the errors become as big as 60%, but it's what I want as for my IA (the piece of work we need to do) it's good to have errors as you can talk about what you could of done better in hindsight. Thanks for the advice.


text:
Spin-forbidden transition iof bromine
In one of the lectures of my spectroscopy course, the professor claimed that there is a strong absorption band in the UV/Vis spectrum of bromine, associated with a singlet-state to triplet-state transition. He asked us to think about why this spin-forbidden transition occurs.
So far I haven't really come up with a satisfactory answer. Nevertheless, here are two of my ideas:

Broadly speaking, spin-forbidden does not necessarily mean that a transition is not observed, but rather that the intensity of the spectral line will be considerably diminished. Hence, a spin-forbidden transition could be observed in a spectrum.
An electronic transition is usually associated with a vibrational transition. Through the vibrational excitation the geometry of the molecule is changed such that the spin selection rules are "relaxed".

Am I on the right track with these ideas?

Spin-orbit coupling is significant with heavy atoms, which relaxes the spin selection rule.

Spin orbit coupling (coupling of electron spin momentum and orbital angular momentum) increases as $Z^4$ so this will be large in Bromine compared to fluorine. This interaction mixes the singlet and triplet states which means that the spin selection rule is 'broken'. Bromine only has one vibrational mode so option (2) does not exist.

Thank you for the responses. @porphyrin what exactly do you mean by singlet and triplet states "mixing"?

It means that there is an operator (in this case spin-orbit coupling) that allows some triplet character to be included in (mixed in with) the singlet state wavefunction. The new state is now not just singlet but has some triplet character, this means that the spin selection rule is no longer a strict rule. In fact it is always the case in molecules, but usually small in the absence of heavy atoms. It manifests itself as  causing singlet triplet (intersystem) crossing  from an electronically excited singlet state to a triplet, for example in aromatic and other molecules such as benzene.

The selection rules are related to the transition dipole moment that determines the intensity of a given transition. If a given selection rule holds perfectly, all that it says is that a forbidden transition has zero intensity and an allowed transition has non-zero intensity.
If the selection rule does not hold exactly, just like the in the case of the spin selection rule for heavy atoms, a forbidden transition can be intense. Some examples: the $\text{B}^3\Pi_u-\text{X}^1\Sigma^+_g$ electronic transition in $\ce{Br2}$ and $\ce{I2}$, that are strong when compared to the same transition in $\ce{F2}$ and $\ce{Cl2}$; and the strong $^3P_1-^1S_0$ transition in $\ce{Hg}$, while the analogous one in $\ce{He}$ is very weak. As $L$ and $S$ are not good quantum numbers when there is strong spin-orbit coupling, the selection rules involving them can't be used.
The second answer you came up with makes sense, but not for spin selection rule. Consider the parity selection rule for molecules with an inversion center. As the molecule vibrates (in an asymmetric mode) it can be distorted into a geometry that does not have an inversion center, so the selection rule does not hold. In this case, this vibronic transition will be allowed.


text:
Highly conjugated compounds and the effects of substiuents
TLDR, what is the effect of electron donating and withdrawing groups on the UV-Vis absorption spectra of pi-conjugated system?
For a conjugated system, my understanding is that as the extent of conjugation increases, the difference in energy between the highest unoccupied molecular orbital and the lowest unoccupied molecular orbital decreases. The delocalization of electrons in a highly conjugated molecule provides the molecule with a greater number of molecular orbitals, decreasing their spacing. Conjugated compounds which absorb visible light do so when the difference in energy of the homo and lumo correspond to a frequency in the visible spectrum.
My question is in regards to the affects of electron donating and withdrawing groups and the absorption of light by these compounds in the visible spectrum.
My intuition is that electron donating groups will cause greater delocalization of electrons within the compound, creating more relevant MO's, decreasing spacing and reducing the HOMO- LUMO energy gap, causing the compound to absorb lower energy (longer wavelength light). Is my intuition correct?

You are talking about electron withdrawing groups (so they have an inductive effect) but it is nothing compared to mesomeric effect, which is known to influence color. Also a withdrawing group or a giving group should have an opposite effect which your intuition does not account for.

@SteffX Would you mind articulating this opposite effect, in addition to explaining the relationship between mesomerism and the reflection of light by a compound?

As itt stands the question is very complicated. I disagree with others as for your wording does not limit the discussion to different electronegativity in an elementary book sense. Of course electrons donating or withdrawing substituents / groups influence colour. But it would be better to focus on absorption of a single molecule. Your link to reflection does indeed complicate the issue. Your intuition is correct but the question reflects misconception or at least is not clearly formulated. But in general yes, there are effects on colour. Mesomeric  effects basically extend the conjug. system.

@Alchimista I suppose I should narrow the context then, suppose the molecule in question is an azo dye, let's say methyl orange, which reflects the 520nm and 460nm in acidic and basic environments respectively. In an acidic environment, if one were to substitute the benzene ring with an electron donating group, such as OMe, would the subsituted molecule emit light at a lower wavelength?

Is the reflection that doesn't fit in here. Forget colour for a while and just look at the molecule Establish what happens to its electronic levels, and specifically to the energy gap. The phenyl ring in your ex Is part of the system, removing it will increase the gap almost no matter of what group is going to replace it. You get a totally different molecule. should better place -OMe on that phenyl. Most likely whatever mesomeric effect on a conjugated system is bathichromic, and enhance absorbivity, like a little extension of the pi-system. You should play with a carotene or so for an idea

Yes, your intuition is correct.
Among others properties, systems with pi-electrons delocalisation are actually chromophore whose UV-Vis absorption spectrum is mainly dictated by the HOMO-LUMO transition.
An extension of the delocalised system results in a narrowing of the gap and thus shifts the spectrum to red. The effect is dramatic when extending, e.g., the length of a polyene chain.
However, although with different impact, the same is true if a parent system is substituted with group leading to additional resonance, both because of electron donating or withdrawing effects.
This is fundamentally no different from considering a C=C double bond, which is itself a chromophore, and functionalize it with the various auxochromes that are commonly listed when introducing UV-Vis electronic absorption spectroscopy.
An auxochrome as listed won't necessarily delocalise over the conjugated system in complicated structure, or it can marginally do so. Thus, don't assume that if there is a methoxy - just to mention an auxochrome - somewhere in the molecule or name, the spectrum of the latter is necessarily red shifted as compared to the parent molecule lacking that auxochrome.
If the auxochrome can indeed partecipate in delocalisation, then its effect is bathochromic (red shift) and generally hyperchromic (higher absorbivity).
Rules for predicting the wavelength in the absorption spectrum of conjugated systems do exist, though they fail quite early with increasing the conjugation length.
Consider that the conjugation length of a system is better described by an effective conjugation length, as for the system is susceptible of various effects, first of all departure from planarity due to hindrance. The same is true for the effect of the substituent. If you think of a very long polyene, an end substitution doesn't necessarily extend its effect all along the formally conjugated system.
Similarly the concomitant presence of electron withdrawing and donating groups can have negligible or no effect, as their influences can somehow cancel each others.
Again, is very important to consider that the effective delocalisation is not only influenced by the size of the system, but by an ensemble of parameters, such as the already mentioned steryc hindrance, structural motifs that might pin electrons (think of a phenylene in the middle of an otherwise polyene chain), solvent, etc.
With a bit of caution, you can certainly take that substitution of a conjugated system with groups having mesomeric effects results in absorption maximum at longer wavelength as well as in a higher absorptivity.


text:
Nuclear chemistry home made
A boy in Memphis built 1 year ago a nuclear fusion reactor at his home . How did he achieve so high temperatures in order for the fusion to take place?The fusion that took place was 2 atoms of hydrogen-2 were fused into helium.What I know is that hydrogen-2 can only be made by fusion of 2 hydrogen-1 atoms and then the unstable radioactive diproton which is produced is converted into hydrogen-2 and 1 positron.Where did he find so much energy? And if we pressure a gas very hard,can  we fuse 2 hydrogen atoms into 1 diproton?

This needs a reliable source.

They say, they milk hens beyond the ocean.

Yes, this really happened, probably (it's not a peer-reviewed effort).
The reactor is a Farnsworth fusor, a very low-density, low-rate inertial-confinement reactor. They're surprisingly easy to build and get working; many people have done so, and verified their function by measuring neutron output. An accelerating voltage of 10 or 20 kV produces particle energies equivalent to over 100 million degrees C.
There's active debate about whether a fusor of this design could ever produce more power than it consumes, but there's been some interest in using them as compact neutron sources (that you can actually switch on and off).
You asked about fusing hydrogen-1 to hydrogen-2. These experimenters don't attempt that; they just buy deuterium, which is also readily available.

A common pulsed neutron source is an very small LINAC (linear accelerator) which fires ionized deuterium (deuterons) at a target made of titanium which has tritium impregnated into it. In the target the D + T reaction occurs to form helium-4 and a neutron. Strictly speaking this is a fusion reaction, this is used in a range of applications including neutron activation analysis and the neutron source needed to power up a modern atomic (fission) bomb. While it is a fusion device it is never going to break even in terms of energy in / energy out


text:
HCl rota-vibrational lab: ab initio simulation of rotation constants
I'm trying to set up the simulation portion of the classic undergraduate lab studying the rotavibrational spectrum of HCl and DCl. In terms of simulation (using Gaussian 16), the equilibrium geometry and vibrational energy calculations are trivial. Including anharmonicity is also simple through the use of the freq=anharm keyword. I'm running into trouble with the inclusion of vibrational-rotational parameters, which the program is supposed to support, but I can't find appropriate output information. I'm either missing the output data because I don't know what I'm looking for, or something is wrong in the calculation. Specifically, I'd like the students to get ab initio values for $D_e$, $\widetilde{\nu}_0$, $B_e$, and $\alpha_e$ in the following equation ($m$ is a substitution for $-J''$ or $J''+1$ for the $P$ and $R$ branches, respectively):
$$\widetilde{\nu}(m) = \widetilde{\nu}_0 + (2 B_e - 2 \alpha_e) m - \alpha_e m^2 - 4 D_e m^3$$
I'm currently able to find $D_e$ and $\widetilde{\nu}_0$.
The only clue I can find in the calculation is the following lines:
WARNING: Anharmonic treatment of linear tops is experimental.
         Moreover, an hybrid treatment is used to simulate spectra:
         - Energy: equations including degenerate modes are used.
         - Intensity: summation done on N' modes, considering only one mode
           per couple of degenerate modes. No variational correction done.

    ==================================================
                  Coriolis Couplings
    ==================================================
[No Coriolis Coupling for diatomic molecules

Later, in the entry for Vibro-Rot alpha Matrix (where I'd expect to see entries for $\alpha_e$), I get:
Vibro-Rot alpha Matrix (in cm^-1)
---------------------------------
            A(z)        B(x)        C(y)
Q(    1)    -0.00000        NaN        NaN

I'm assuming the message for Coriolis Couplings is connected to the zero-valued alpha matrix. Is there a way to pull out these parameters in a Gaussian simulation? If not, is there another ab initio package that will perform these calculations? I've tried (without luck) to look through the NWChem documentation.
The input might be useful to reproduce the calculation. Unfortunately, the output is larger than SE.com will allow. Let me know if you require additional lines from it (or a full copy).
Input file:
%nproc=4
%mem=400MB
# MP2 def2tzvp
# opt freq=(anharm,vibrot,noraman) scf=tight freq=ReadIsotopes

Vibrational analysis for D35Cl

0   1
 Cl 0.0 0.0 0.0
 H  0.0 0.0 1.27

298.15  1.0
35
2

Update:
After more reading through the Gaussian references, I've found the rotational constants (Ba(x), Ca(y)) under the following section:
     ==================================================
          Vibrational Energies at Anharmonic Level
     ==================================================

 Units: Vibrational energies and rotational constants in cm^-1.
 NOTE: Transition energies are given with respect to the ground state.

 Reference Data
 --------------
                              E(harm)  E(anharm)     Ba(x)      Ca(y)
 Equilibrium Geometry                             10.643785  10.643785
 Ground State                 1525.714  1515.023        NaN        NaN

 Fundamental Bands
 -----------------
     Mode(n)      Status      E(harm)  E(anharm)     Ba(x)      Ca(y)
        1(1)      active      3051.429  2951.452        NaN        NaN

 Overtones
 ---------
     Mode(n)                  E(harm)  E(anharm)     Ba(x)      Ca(y)
        1(2)                  6102.858  5802.928        NaN        NaN

 WARNING: Anharmonic transition moments for symmetric and linear tops
          are not yet fully implemented.

However, I still can't find an entry for $\alpha_e$, the coupling constant between vibrational and rotational states. Is Gaussian capable of putting out this constant? If so, am I looking for it under the wrong name?

Just a few hints (and I don't know if any of that is actually helpful to the problem at hand): Specify #P to get more output, there might be more stuff hidden between the defaults. Use checkpoint files (%Chk). Split the calculation into one optimisation, and a frequency calculation (you can use %OldChk). I have made the experience that specifying keywords twice may cause problems for g16, so combine the option stacks. I try look into this deeper, but i can't promise

I tried running a separate file, but that seems to mess up the optimized anharmonic frequencies without fixing the missing value for $\alpha_e$. #P doesn't seem to give me any additional relevant information. I did go ahead and remove the freq=ReadIsotopes and just specified directly in the atom specifications H(Iso=2).

Disclaimer:
I never used Gaussian, so I am just guessing about the program.
Coriolis coupling:
Coriolis coupling couples bending modes with rotation and there is no bending in diatomic molecules.
There is no Coriolis coupling in Gaussian for HCl, because there can't be one. The relevant output could also be NaN, it is simply not defined.
Alpha "matrix":
For diatomic molecules you have two degenerate rotations and one stretching mode. Hence your alpha "matrix" is actually well represented by a scalar and used as such in your equations. For this reason one would expect only one alpha value.
That there is written -0.000 and not 0.000 is an indication for a truncated negative number. In addition other values are properly set to NaN. I think that your one reported alpha value is simply close to zero. You can try a verbose output to verify this.
Solution:
If you want to get a non zero alpha
I would recommend a molecule with a shallower potential hyper surface, i.e. a lower force constant. Or you could use this example to discuss floating point arithmetics and truncation with your students.

Thanks for the clarification, this was a concept with which I was unfamiliar. Nice to eliminate one red herring. Any insight into the rot-vib coupling constant, $\alpha_e$?


text:
Nanoparticle tracking analysis versus transmission electron microscopy
I have various bimetallic nanoparticles which have formed alloys.
I then used NTA and TEM to characterize the size of my nanoparticles.
The TEM shows that the size of nanoparticles is considerably less than the NTA.
Can anybody help explain the difference in size?

How confident are you with the viscosity value of your liquid?

not very confident but by looking at the Stokes-einstein equation would i be correct to say that if the viscosity is too low this could lead to a larger particles being reported?

Do you use similar controls in both (as in, particles of known size)?

TEM is a direct method. So, what assumptions come with the other technique?

You don't mention how homogeneous your solutions are or how much "considerably less" is. As summarized in this paper, "Under ideal conditions, particle size analysis by DLS and NTA both can be accurate to 2%, but heterogeneous samples can have errors an order of magnitude larger." (See references in that paper for details).
It's also important to keep in mind that you are comparing techniques which measure samples in very different ways and under very different conditions. TEM measures individual particles in a dried (and often treated) state in a vacuum, while NTA measures an ensemble of particles dissolved in solution. In each case, the size distribution is often just as important as the so-called "size". If the size-distribution of your measurements overlap, perhaps there's no conflict after all.
Also, if your particles have some sort of coating which Au nanoparticles, for example, often do, then you might expect differences in the apparent size measured in these two different states.
Using controls is a good way to get around some (but not all) of these problems, but then the same questions need to be raised about the control particles and any difference in their behavior from the particles under study.
So, as is usually the case in well-done science: "It's complicated..."


text:
Internal standard with low T1 for NMR quantification of ammonia
I am trying to quantify low levels of ammonia with NMR. The good news is that ammonia has a very low T1 (<0.5) which shortens d1 a lot. However, I cannot find any internal standard with a T1 that is as low as ammonia so that I can use absolute qNMR for quantification. For example maleic acid would require around 10 times longer relaxation, because the T1 is around 3-5s which reduces the LOD of cause.
Regards,
Martin

Can you say more about your sample? You are willing to add an internal qNMR standard, what about a relaxation agent ?

Basically it's just ammonia in water. We tried iron as relaxation agent with maleic acid as internal standard. It decreases the T1 but also the sensitivity for ammonia due to line broadening.

You might want to try other relaxation agents and pick your concentration carefully. The combination maleic acid + iron sounds bad because they may form a complex (for quantitation that may not matter depending how the sample is prepared).

I take that back: the formation of a complex may be very bad for quantitation if it is a tight complex, resulting in very rapid relaxation of the complexed maleate. There are other choices, however N-14 relaxation will be a problem with ammonia.

Interesting. It makes sense that maleic acid and iron from a complex, because our data was not quantitative (no linearity). However, we found that the LOD for ammonia becomes much worse due to line broadening which is a problem, because we need high sensitivity. Do you know any relaxation agents that don't cause line broadening for ammonia? I must admit I don't understand the theory behind relaxation agents so this might be a stupid question, haha

There is a vast literature on the subject of relaxation agents. Among the common ones in nmr are GdCl3 and Cr(acac)3 and some nickel salts. Their effect may vary but typically insofar relaxation is concerned they shorten T1/2 for small molecules to the same degree (for small molecules T1~=T2).

It is difficult to give further opinion for your particular case. The chemistry of combining all of the ingredients you mention may be complex. I am not sure changing relaxation agent would make much of a difference insofar as ammonia versus maleic acid is concerned. Feel free to post a more specific question or even data if you need an opinion.

Do you know if there are any relaxation agents that don't cause line broadening? I don't feel competent enough to discuss this subject. Can you recommend a nice review paper?

You may want to read eg     P. Caravan
Strategies for increasing the sensitivity of gadolinium based MRI contrast agents,    Chem. Soc. Rev., 35 (2006), pp. 512-523. This article may give you pointers on what you don't want, ironically. You want something that doesnt relax water quickly. The opposite of the usual goal.


text:
Equilibrium constant for complex cobalt ion

What is the equilibrium constant for the $\ce{Co(H2O)_6^{2+}(aq) + 4 Cl-(aq) <=> CoCl_4^{2-}(aq) + 6H2O(l)}$?

I've looked at various sources but none seem to yield the actual equilibrium constant.

Thermodynamic Properties and Ionic Equilibria of Cobalt Chloride Solution at $\pu{298 K}$ was estimated by a Korean group in literature (Ref.1). The abstract states that:

We developed a chemical model to analyze ionic equilibria in a cobalt chloride solution at $\pu{298 K}$. The chemical model consisted of chemical equilibria, mass and charge balance equations. The activity coefficients of solutes and water activity were calculated with Bromley equation. Values of the equilibrium constants for the formation of cobalt chloride complexes at zero ionic strength and of the interaction parameters were estimated by applying Bromley equation to the reported equilibrium constants at different ionic strength. The effect of $\ce{CoCl2}$ and $\ce{HCl}$ concentrations on the distribution of cobalt species was obtained. The predicted pH values for $\ce{CoCl2-HCl-NaOH-H2O}$ system agreed well with those measured at $\pu{298 K}$.

Thermodynamic equilibrium constant ($K$) for the formation of $\ce{CoCl4^2-}$ complex for the reaction $\ce{Co^2+ + 4Cl- <=> CoCl4^2-}$ was reported in $\log$ scale where $\log (K)$ is reported as $-9.06$ (or $K = 8.71 \times 10^{-10}$). Although this is not the exact answer for OP's question, I hope OP can find insight into his/her question.
Furthermore, cobalt ion has a strong tendency to form various complexes with chloride ion such as $\ce{CoCl^+}$ and $\ce{CoCl3^-}$. $\log (K)$ for them are also reported as $0.22$ and $-3.02$, respectively (Ref.1). The $\log(K)$ for the equilibrium of neutral species $\ce{CoCl2}$ was reported elsewhere as $-3.95$ (Ref.2).
Reference:

M.-S. Lee, Y.-J. Oh, “Estimation of Thermodynamic Properties and Ionic Equilibria of Cobalt Chloride Solution at $\pu{298 K}$,” Materials Transactions 2004, 45(4), 1317-1321 (https://www.jim.or.jp/journal/e/45/04/1317.html).
J. Bjerrum, G. Schwarzenbach, L.G. Sillén, In Stability Constants of Metal-ion Complexes, Part II: Inorganic ligands; The Chemical Society: London, United Kingdom, 1957, p. 97.

You can't find the equilibrium constant because that's entirely dependent on the concentration.
To find the equilibrium constant, you take the concentration of the products to the power of their coefficients (so concentration of cobalt (II) chloride to the power 1 multiplied by the concentration of water to the power 6), divided by the concentration of the reactants to the power of their coefficients (same concept for the denominator).
There's no one set equilibrium constant for a reaction.

Equilibrium constants do not depend on concentration.


text:
What happens when HCl is added to acetylene in presence of mercuric ions?
Normal addition of Hydrochloric Acid to acetylene would give me a 1,1 Dichloro product.
But I read that in presence of mercuric ions only one molecule of HCl is added to give Vinyl Chloride.
Why does this happen?

I'm not entirely sure but I'll have a go.
As a strong acid, HCl will be dissociated into Chloride and protons in water. I'd suggest that the mercury will first associate with the acetylene forming an Hg- Pi complex which will weaken the triple bond. Then the chloride ion will attack one side of the acetylene and the electron density one of the triple bonds will form a new single bond, first to the mercury ion and then this bond will hydrolyse and pick up a proton. This is similar to an oxymercuration reaction but with an attack from chloride rather than water.
The resulting vinyl chloride is more electron poor than the starting species so should should undergo further 'chloromercuration?' slower than the starting species. I would suggest that the selectivity is then derived from the milder conditions this reaction must be run at. Probably you have to force acetylene to react with HCl without a catalyst by heating the hell out of it (if it happens at all) therefore the addition happens twice.


text:
ML6 Type Coordinate Complexes - Drawing a Molecular Energy Orbital Diagram
In class we recently discussed how to draw out molecular orbital energy diagrams for compounds such as NH3 (ammonia). Building upon this, we touched on how to construct a molecular energy diagram for ML6 type coordinate complexes, namely [Cr[NH3]6)3+. I am confused as to where to begin our diagram formation. All I know right now is that we consider that NH3 coordinates to Cr through it's a1 HOMO. I don't know how to proceed beyond this.

Have you tried looking in a textbook? Because of the importance of octahedral complexes, all inorganic texts devote a fair amount of space to construction of MO diagrams for ML6 complexes.

Unfortunately our prof doesn't have a textbook for us in this class. We're only left to our handwritten notes.

Look in a library? There is a list here: https://chemistry.stackexchange.com/q/37303/16683

@Henro You can literally open any textbook about inorganic chemistry...


text:
Define, (actually define) the "stability" and "energy" of a compound
When we say a conformer or compound is of "higher energy" than another, are we quantifying energy in terms of the bond strength? Or are we going off of how much energy it takes to break the bond?  I don't think we can just assign energies to each compound... so what is this based off of?
One might answer that its based off of stability. Well, what is "Stability?" Is it how rigid/strong the bonds are? Is it how reactive the compound is?
Often times i find myself reading explanations with circular language on this topic. A higher energy conformer means lower stability or A stable compound is one that has low energy. I'm finding these definitions to be crazy ambiguous, and don't thoroughly define what these are.
If someone can clearly explain this in terms that a first year undergraduate student can understand, you have earned my gratitude.

There's more than one kind of stability...

Energy is generally a free energy.

These terms are well defined in (chemical) thermodynamics, however, chemists like to use them in a more casual way in high-school and early undergrad textbooks, too. It is not necessarily bad, as the intuitive meaning more or less correct, and helps the discussion. As @Zhe said, the energy is generally Gibbs free energy if we are talking about an ensemble, but in certain discussions like comparing conformers and potential energy surfaces it is actual energy (potential energy) of the system.

As a chemist, I would agree that textbooks are not clear about the term stability and energy. It is not your fault and undergraduate organic chemistry books make the situation worse. I cannot recall the text, it was an old book, however it clearly said that stability of a compound does not mean anything. We should always ask, stability with respect to what? Are we talking about thermodynamic stability or kinetic stability? For example H2+O2 mixture is not stable. This statement is meaningless because we are not saying stability with respect to what? This mixture should form water right away but people actually waited for 30 years and did not find much water in a mixture of H2 and O2. Thermodynamically, this mixture is not stable and it will react violently once you provide the right activation energy. When you are reading about stability, ask this question yourself.
In the same vein, the term energy should also be clarified by authors. As a crude analogy is that you can think of energy as the potential energy of the whole molecule when the authors talk about conformers. Have a overall picture of the molecule. Imagine two springs, A) one which is less stretched and the other one (B) is more stretched. You can say spring B has more (potential) energy than (A). An eclipsed conformer has higher potential energy (overall, no need to think locally) than a staggered one.

Finding minimums: a general view
When minimizing energy you are searching (using numerical methods) for the minimum of Potential Energy). You are at the minimum when the derivative(Jacobian) is equal to zero. However, there are a couple unstable places that also have a derivative of zero, namely saddle points and maximums (top of a circle). We are after the minimum (bottom of a circle).
Think of a pendulum. It has two spots where the derivative is zero... when it is all the way up, and all the way down, but only when it is in the all the way down position is it stable. When the Hessian (second derivative) is positive definite, you are at the "bottom of the pendulum swing" and at a minimum. Pendulum up the derivative is zero, but this is a maximum.
There is also the problem that often in energy landscapes it looks like a rollercoaster... At a commenters request, a better image (taken from wikipedia, source in image description)

We need to find not just the a minimum that has a derivative of zero,but the lowest of all minimums present. This is a real tough problem since numerical solvers stop once the derivative is zero. They don't know there are other minimums. There are methods to try to get around this, but it is a very active area of research and a huge prize awaits whoever can guarantee a global minimum is found.
A thought on energy
What is this energy we are minimizing? We are minimizing the energy compared to a reference point. Everything is always with respect to a reference point.
I am not a chemist, and I don't know much Quantum mechanics but I will try to give a chemistry answer...This is my impression of what is done in the scenario you are asking about. I perform minimizations on Free Energy, but that is related but not quite what you are after. In Quantum mechanics the reference point is when all nuclei and valence electrons are infinitely apart. Another common reference point in chemistry, physics and engineering is the ideal gas, which is when all molecules are infinitely apart from each other but atoms (and electrons) are still bonded to each other.
Given a reference point which we define to have "zero" energy, for each conformer we would calculate the energy to form all of the covalent bonds in the molecule, which is to say, calculate the energy to pull the valence electrons and nuclei in from infinitely far apart and put them in the exact geometry you want. Do this for many geometries. The one that took the lowest energy to form will be the stablest of the ones you tried (There may be others you did not try though! you may be in the wrong loop of the rollercoaster still). It takes different amounts of work to put two nuclei different distances apart etc. In practice, QM programs do all of this for us, and they are getting quite fast but it can still take a very long time. It is a hard thing to calculate.
Bond Energies
Generally in molecular mechanics bond energies are calculated using the harmonic spring approximation (hookes law), however the below image shows the QM calculated energies.

Angle Energies
Here is an image from the simualtion software Espresso's website which shows that the energy required to bend an angle is closely approximated by a quadratic function

Torsional Energies
Torsional energies are more complicated than the bond and angle potential energies because a torsional angle needs to be able to be turned 360 degrees and be back where it started. Here is the energy for different conformers as the geometry of the torsion angle is turned. Because of the different geometries, the nuclei and electrons are different distances apart from each other, and it takes a different amount of work (energy) to move them from infinitely far apart to each of those geometries. The image is from here.

Force-Fields
In molecular mechanics we use functions (called force-fields) to approximate the above potential energy contributions. Here is a general set of the equations used. You can find the minimum of energy by changing bond lengths/angles/torsion parameters for a single molecule, and after each change, calculating the energy. The geometry with the lowest energy is the most stable because molecules, like people, always take the path of least resistance. I have copied the equations as an image from here

Could you label the axes? What are you plotting?

It is from wikipedia. What I am plotting doesn't matter, it is a general take on numerical methods and finding a minimum. Since the OP is after Energy, pretend y-axis is Energy. Then pick a thing to minimize it with respect to... Bond length, angle, anything you want to minimize it with respect to can be the x axis. Bond lengths and angles look like quadratics. If OP wants I can put images for the Bond/Angle/Torsion potentials, but that is forcing the question in a specific direction. I tried to keep my answer ... general

Actually I liked your pendulum analogy and I thought the graph represented that!

Since you have a theoretical background, I am an analytical chemist, what do theoreticians mean by energy when they show it on the vertical axis? Is it the sum of kinetic and potential energy of the electrons in a bound state at a given geometry? Your reference point is infinitely separated H and O atoms?

For me, I am only interested in potential energies. My reference state is generally an ideal gas, rather than the separated nuclei/electrons. Kinetic energy is import too in simulations, it keeps the temperature correct, and also, you need to know the total energy to make sure you are conserving energy. I don't want to delve to much into the QM side of things since I am a novice in that area :)

@M.Farooq Which axis? Generally, it is the Born-Oppenheimer solution Schrodinger equation, containing electronic energies as well as nuclear repulsion etc, too.

I was talking about the y-axis of the figure no 1 (top). I haven't solved Born-Oppenheimer equations but I am sure that it includes kinetic and potential energies.

When finding conformers, it is for a single molecule. You gain nothing by letting it move around (have kinetic energy). Generally, especially in Monte Carlo, you would hold the center-of-mass constant and just let its bonds "wiggle" around while you scanned potential energies of each different geometry. I would think it is the same in QM too, you are just using QM to calculate the PE, rather than approximations to QM (force-fields).

The answer is complicated because there are different kinds of stability and when we talk about energy, we also can mean different kinds of energy.
When we speak of energy, we generally refer to a free energy. Your normal laboratory setup is an open-air, constant-pressure system, so we use Gibbs free energy. If you were working with bomb calorimeters,  you'd want to use Helmholtz free energy for constant-volume processes.
Note importantly, that energies are relative. "What is the energy of this compound?" is a meaningless question. And when we speak of "high energy," we mean high relative to some reference point. That's precisely why most examinations of energy in chemistry focus on a change, most commonly $\Delta G$, the change in Gibbs free energy. Gibbs free energy is a tool for measuring changes in internal energy (most commonly bond-breaking and bond-making) while compensating for the entropic costs of those changes and pressure-volume work done by the system.
At standard state, the Gibbs free energy change for a reaction or transformation can describe the relative preference of the "reactants" versus the "products" via an equilibrium constant $K$.
$$\Delta G^{\circ} = -RT \ln K$$
This is a way to measure one kind of stability, thermodynamic stability.
Considering $\ce{A <=> B}$, $\ce{A}$ is thermodynamically stable relative to $\ce{B}$ if the free energy change of this process is positive (i.e., $\Delta G > 0$), and vice versa.
Unfortunately, thermodynamics only tells half the story for a reaction. There's also the kinetic piece, which determines how fast a reaction proceeds, that is, the rate of the reaction. Rates are determined by the relative energy of a fleeting transition state, captured by $\Delta G^{\ddagger}$. For a reaction that is highly spontaneous, if the reactant is kinetically stable (that is $\Delta G^{\ddagger}$ has a large value), then reaction will still proceed very slowly. This is kinetic stability.

Couldn't the stability of a single compound refer to its Gibbs free energy at standard conditions? Why we need to introduce the relative term?

Gibbs free energy at standard conditions is loose. The term is the standard Gibbs free energy of formation, which is relative to the component elements in their standard states. In that case, you are still referenced.

Ok but now we have a measure of stability of the compound which doesn't depend on the type of the reaction unlike reactivity which does.

I don't understand your comment... @adosar

A compound A can react fast with a compound B to give C (slow) and with compound D to give E (fast). The Gibbs free energy of formation of A is the same on both reactions but the reactivity (which is reflected on the activation energy) will depend on which coumpound A is reacting with (B or D) i.e. in the specific reaction.

I don't understand why this is a problem. The thermodynamic stability of A can be measured in terms of its energy relative to C and E. It's kinetic stability can be determined by the rate of conversion to C and E (which determined in part from the relative energy of A versus the transition state). You're forever bound to a relative perspective if you work with energy. @adosar

As folks are saying in the comments, energy is used throughout chemistry and comes in many different flavors, so instead of attempting to answer the broad question "How is energy and stability defined throughout chemistry", I will focus on your two examples.

When we say a conformer is of "higher energy" than another, are we quantifying energy in terms of the bond strength?

Conformers are different shapes (3D structures) a given molecule assumes over time (usually by rotating about single bonds, i.e changing torsion angles). High energy conformers are rare, and low energy conformers are common. Conformational energy is a potential energy. It can be used to figure out the forces on the molecule as you change the torsion angles.
The reason that molecules don't just assume the lowest energy conformation is that there is kinetic (thermal) energy available. Other molecules keep bumping into our molecule of interest, giving it the energy to adopt higher energy conformations. On the other hand, when our molecule of interest goes from a higher energy conformation to a lower one, the potential energy is transformed into kinetic energy.
The reason that different conformers have different energies is because in some conformations, atoms that are not directly bonded come too close to each other ("steric clashes", associated with high conformational energy) or have favorable interactions (hydrogen bonds, dipole-dipole interactions etc). Also, some conformations don't allow optimal bonding. An extreme example would be a double bond, rotation about which would break the pi-bond and increase the potential energy a lot. Even with a single bond, some conformations have lower energy than others because the allow for a slightly stronger bond (this is called a stereo-electronic effect).
So the conformational energy depends on subtle changes in bond strength (stereo-electronic effects) and on non-bonded interactions (steric clashes, non-covalent interactions).

When we say a compound is of "higher energy" than another, are we quantifying energy in terms of the bond strength?

For a clean comparison of the "energy" of two compounds, they have to contain the same set of atoms, i.e. they should be isomers. So you could ask whether silver fulminate (AgCNO) or silver cyanate (AgOCN) has the lower energy. Or you could ask whether acetaldehyde occurs as aldehyde (CH3-CHO) or enol (CH2=CH-OH). Conceptually, we could add up all the bond energies in the two compounds to see which one has the lower potential energy.
In cases where the potential energy difference is very high, we can safely say that one compound will be common and the other rare or non-existent. You would say one is more stable than the other. If the potential energies are similar, we also have to consider how the compounds would interact with surrounding molecules (e.g. solvent), and we have to consider entropy (or, equivalently, we have to compare Gibbs free energy instead of potential energy only), which is beyond the scope of this answer (and some might argue beyond first year undergraduate study).

"When we say a conformer or compound is of "higher energy" than another, are we quantifying energy in terms of the bond strength? Or are we going off of how much energy it takes to break the bond?"
The two options are expressing the same and yes, we are doing that.
But for an analysis is generally convenient to think of the same bonds and consider other potential energies, which can be due to torsion, electrostatic repulsion / steryc hindrance, tension... Entering in the discussion of what is more conceptually appropriate might astray from the question, but often the above partition is surely motivated by the different energies involved.
Suppose to separately burn two conformers of an alkane. The products are totally insensitive to this, still the heat of combustion will be higher when burning what we then call the unstable conformer, right because its energy content was higher. Note that so was, independent of the way we thought to distribute the energy surplus in & around the molecule.
" don't think we can just assign energies to each compound... so what is this based off of?".
We can as we have just did above, assuming a reference of course. In the above example we could have taken the energy of the products to be zero, tough you may look for how standard states are defined.
Once you have standard states, you can order different molecules by their energy of formation, and this gives you an order of stability.
When we consider the free energy instead of entalpy, we can even compare different and differently sizes molecules. The order is done per mole. The fact that burning more matter gives more heat is of course trivial and does not tell itself about the stability of that matter. Enthalpy alone worked above because we had a) conformers and b) the same reaction.
It is not necessary to mention quantum chemistry, if not calculations or deeper insights are required. Thermochemistry is older and deal with this concept as well, as in the given example. While the core quantity in thermochemistry is heat of formation, combustion makes (made) easy to run the experiments and get things by comparison. This is to say that my example must not be taken as a definition of stability, but it was to show that there is no circular language. Thermodynamics quantities are of course real and quantifiable.
"Is it how reactive the compound is?"
As addressed by others reactivity should not be confused with thermodynamical instability. It is just matter of wording dictated by the fact that a system wich does not change is called "stable", both in daily chemistry and daily life. I have difficulties to think of a language which instead of still stones speak about stones in metastable equilibrium. However in chemical physics we have this term, as well the term activation energy. Once the activation energy is available to a system in a metastable state, then the fate of the latter is dictated by the thermodynamics, as above.


text:
if a balloon full of warm air was placed in a refrigerator, what is the result?
If a balloon full of warm air was placed in a refrigerator, What will happen to the Ballon ?

What do expect from Charles law? How is volume related to temperature?

LOL -- Not being exposed to the UV rays of the sun means that the ballon will degrade much more slowly in the dark.

The balloon will shrink and get smaller as the gas inside it gets cooler, the pressure inside the balloon falls. You might also have seen that a bottle partially filled with warm water kept in a refrigerator squeezes a bit because of the air pressure inside the bottle falls. This result directly follows from the gas laws.


text:
antisymmetric wavefunction
Why can't we choose any other antisymmetric function instead of a Slater determinant for a multi-electron system? Why do we choose our wavefunction for a multi-electron atom as a product of single-electron wavefunctions?

Of course we can. Go on, choose any other one, see what happens.

Go for it, mate!

A product of single-electron wavefunctions is, in general, neither symmetric nor antisymmetric with respect to permutation. The second question here seems to be slightly non sequitur.

@orthocresol Sorry , By product i mean linear combination of product wavefunction .What does the taking product signify?

1. A pragmatic view
As the comments suggested you can use different functions than slater determinants.
But this would be more or less as bad as an idea
as describing a movement in a circle without using polar coordinates.
If you include the constraints of your system implicitly in the coordinate system or basis of your description, nearly everything gets better.
More elegant equations, faster convergence... .
Determinants are a fast and elegant way to include constraints like the Pauli Principle in your description.
2. A bit more mathematical view
Let's go back to the definition of a Determinant in general.
From Wikipedia:

[...] Another way to define the determinant is expressed in terms of the columns of the matrix.  If we write an $n \times n$ matrix $A$ in terms of its column vectors
$$A = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}$$
where the $a_j$ are vectors of size $n$, then the determinant of $A$
is defined so that
$$
\begin{align}
  \det \begin{bmatrix} a_1 & \cdots & b a_j + c v & \cdots & a_n \end{bmatrix}
          &= b\det(A) + c \det \begin{bmatrix} a_1 & \cdots & v & \cdots & a_n \end{bmatrix} \\
  \det \begin{bmatrix} a_1 & \cdots & a_j & a_{j+1} & \cdots & a_n \end{bmatrix}
          &= -\det \begin{bmatrix} a_1 & \cdots & a_{j+1} & a_j & \cdots & a_n \end{bmatrix} \\
  \det(I) &= 1
\end{align}
$$
[...]

The last constraint comes from the normalization of the atomic orbitals and the second constraint is just the Pauli-Principle.
To the best of my knowledge you don't have to force the linearity in the first constraint, but if you do, you end up with a determinant.


text:
How does acidifying a compound in qualitative analysis make sure any interrupting compounds get removed?
So I've learnt that acidifying a compound for salt analysis is to remove any basic interference and remove interrupting compounds such as carbonates. How does it remove them or what reaction is taking place?

One of the fundamental reactions of chemistry is that acid + carbonate gives carbon dioxide + water plus the salt of the acid. The carbon dioxide bubbles off removing it from the system and you are left with an aqueous solution of the salt of the acid.
$$\ce{XCO3 + HY -> XY + CO2 + H2O}$$

We acidify salts for two reasons, either to remove offending compounds such as carbonates (which can give a false positive for the sulfate test since it too forms a white precipitate with $\ce{BaCl2}$). Additionally $\ce{NaOH}$ (example), if you add $\ce{BaNO3}$ to it (sulphate ions reagent) you’ll get $\ce{Ba(OH)2 + NaNO3}$, barium hydroxide is only slightly soluble at normal temperatures so you see precipitate. If we had added an acid it would’ve been fine, no false positive with the sulphate reagent. Another example would be $\ce{AgNO3 + Na2CO3 -> AgCO3 + NaNO3}$
Also we acidify to remove any basic interference for the halides. The solution needs to be acidic so that $\ce{AgO}$ doesn’t form, silver oxide forms in alkaline solutions. It is brownish in colour and can thus interfere with the precipitation of the halides since that depends on colour.
So if there was any basic interference in the solution of lets say $\ce{NaCl}$.
$\ce{AgNO3 + LiOH -> AgOH + H2O}$
$\ce{2AgOH  -> Ag2O + H2O}$ (Highly favourable energetics)
And we use nitric acid especially since the salts it forms are unoffending, let me give an example. SO for the halides, if we have a solution of $\ce{NaBr}$, and we add $\ce{HCl}$ to the solution and maybe there is basic interference like $\ce{NaOH}$. $\ce{HCl + NaOH}$ would give us a chloride salt which would obviously interefere in the precipitates colour when we add $\ce{AgNO3}$.
But this is simply a precaution to add $\ce{HNO3}$, I myself have done ppt reactions without acidifying (it was a mistake, but it worked).

Was the answer downvoted for not using the markdown formatting?


text:
Potassium Manganate and Acidified Potassium Dichromate colour changes
I would be obliged if somebody could explain the colour changes for the reduction of potassium manganate and acidified potassium dichromate. I understand that for potassium manganate, it is being reduced differently and this gives us a colour change. But I can’t understand the acid, base, neutral reactions.
I also keep on seeing this equation:
$$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$$
Unfortunately  I cannot understand it, possibly because it’s an net equations and not the entire reaction.

Although you are asking for the color changes of the reduction of potassium manganate ($\ce{K2MnO4}$; a green-colored salt, (Wikipedia1)) by potassium dichromate ($\ce{K2Cr2O7}$; a red-orange-colored salt, (Wikipedia2)) in acidic medium, the equation you are showing is reduction half reaction of potassium permanganate ($\ce{KMnO4}$; a purple-colored salt, (Wikipedia3)) in acidic medium. According to Wikipedia1, potassium manganate is an intermediate in the industrial synthesis of potassium permanganate. Thus, color change for that specific reaction is green to purple (disregarding other interference such as color change of the oxidizing reagent).
The reduction half reaction of $\ce{K2Cr2O7}$ in acidic medium is:
$$\ce{Cr2O7^2- + 14 H+ + 6e- <=> 2 Cr^3+ + 7 H2O}  \qquad  E^{\circ} = \pu{1.36 V}$$
The oxidation half reaction of $\ce{K2MnO4}$ is:
$$\ce{MnO4^2- <=> MnO4^- + e-}  \qquad  \qquad  E^{\circ} = \pu{-0.558 V}$$
The total redox reaction of $\ce{K2MnO4}$ and $\ce{K2Cr2O7}$ in acidic medium is:
$$\ce{Cr2O7^2- + 14 H+ + 6 MnO4^2- -> 2 Cr^3+ + 6 MnO4^- + 7 H2O}  \qquad  E^{\circ}_{\text{Rxn}} = \pu{0.802 V}$$
Since $E^{\circ}_{\text{Rxn}} \gt 0$, the reaction is spontaneous.


text:
Help required for the calculation of moles
If 4.5ml of yeast prep + 0.5ml of 0.25 M glucose is separated into 2ml in 2 similar flasks. How many micromoles of glucose are there in one flask?
Using M1V1=M2V2
(0.5ml/1000)*0.25M = M2 * 2ml
= 0.0000625M
Using n=cv
n= 0.0000625*2
= 0.000125moles
=125umoles
Is this correct?

No, V2 should be 5.0 mL

Use of $M_1V_1 = M_2V_2$ is a good start, but you lost at the same place. Identifying three known values out of four is important. You have initial volume ($V_1$) and concentration ($M_1$) of your glucose solution, which are $\pu{0.50 mL}$ and $\pu{0.25 M} = \pu{0.25 mol\:L^{-1}}$, respectively. You also knows the final volume ($V_2$) after dilution, which is $\pu{(4.5+0.5) mL}=\pu{5.0 mL}$ (See A.K.'s comment). Now you see, your unknown is $M_2$. From rearranging the equation $M_1V_1 = M_2V_2$ for $M_2$, you get:
$$M_2 = \frac {M_1V_1}{V_2} = \frac{\pu{0.25 mol\:L^{-1}} \times \pu{0.50 mL}}{\pu{5.0 mL}} = \pu{0.025 mol\:L^{-1}}$$
Now you can use similar approach to  find amount of glucose in  $\pu{2.0 mL}$ of $\pu{0.025 M}$ glucose+yeast solution:
$$\text{amount of glucose in}\; \pu{2.0 mL}= \pu{0.025 mol\:L^{-1}} \times \pu{2.0 mL}\times \frac{\pu{1.0 L}}{\pu{10^3 mL}}\times \frac{\pu{10^6 \mu mol}}{\pu{1.0 mol}}=\pu{50 \mu mol}$$

No.
0.25M = 0.25 mmol/ml
0.5 ml of this contains 0.125 mmol, and diluting it to 5 ml (4.5+0.5ml) gives a concentration of 0.025 mmol/ml.
Transfering 2 ml of this dilution to a flask then gives 0.05 mmol (=50 umol) glucose in the flask.


text:
diamine synthesis
I want to synthesize a new polyimide based on new diamine. In the first step of the synthesis of this diamine, my monomer was added to a mixture of 1-chloro-4-nitrobenzene and potassium carbonate in dimethylacetamide (used as the solvent) and stirred the mixture in $\pu{100 ^{\circ}C}$ for $\pu{24 h}$. The reaction mixture is then cooled to room temperature and precipitated in methanol/water mixture (1:1). At this step, I have 3 questions:

When the solution poured in antisolvents, is there need for agitation?
and if agitation needs, how should be the speed of it?
How can I determine the amount of potassium carbonate that needs for this purpose?

How can I improve the efficiency of this reaction? its yield is very low

It's hard to answer your second question since you have not given complete description of your monomer. Since you need diamine, your monomer must have one amine and a nucleophilc center (which must be more reactive than amine) or just have two nucleophilc centers. Either way, if you are doing your reaction in small scale, stirring (you called it agitation) is important to complete the separation of solvent (dimethylacetamide, which may dissolve in methanol/water) from product and to dissolve extra $\ce{K2CO3}$ so that you have only crude product in remaining solid material. If you are doing your reaction in large scale, then it is a different story.
One needs to know the your monomer structure to determine amount of $\ce{K2CO3}$ needed for the reaction. For example, if you have more acidic functions in your monomer other than reactive nucleophilc centers, you may need additional equivalents for those groups. Since it is not really strong base, use of excess is always better. For example, following is the relevant polyimide preparation with two phenolic reactive centers (Ref.1):

In this case, they have used a little more than two equivalents of $\ce{K2CO3}$ in the reaction. The differences between this and yours are the monomer (may be) and the solvent (dimethylformamide vs dimethylacetamide). So, you may find some insight for your project from this reference.
Reference:

S. Devaraju, M. R. Vengatesan, M. Alagar, “Studies on thermal and dielectric properties of ether linked cyclohexyl diamine (ELCD)-based polyimide POSS nanocomposites (POSS-PI),” High Performance Polymers 2011, 23(2), 99-111 (https://doi.org/10.1177/0954008310391825)


text:
What happen if KMnO4/H+ added to phenylethanoic acid oxidation?
What happen if KMnO4/H+ added to phenylethanoic acid oxidation?

Adding KMnO4/H+ to any compound containing benzene ring oxidizes the aliphatic part to COOH so benzoic acid is formed.
Note:The aliphatic part should not be 3°.

Thanks, what about the the other product? I guess the -COOH ganna be HCOOH..... so benzoic acid + HCOOH?

probably forms CO2