text:
Fried potatoes absorbing oil
Suppose we bring a pan and put in it some oil and turn on the heat, before the oil could get very hot and nearly boiling, we put on the potatoes that start frying.
Someone told me that if the potatoes were put in oil before it gets really hot, then they will absorb more oil than if they were put in oil after it gets really hot.
Is this true?? If yes then WHY?
Note:(given that in both cases they stay the same period of time immersed in oil).

How did you try to solve the problem? It will help the answerers to answer your question!

It is an example of the Leidenfrost effect. Ideally the water being expelled from the food would actually keep the oil from contacting the food itself by forming a barrier of steam (vaporizing water) around the food.

Try it first by weighing out similarly shaped potatoes before and after frying using the two methods described in order to calculate the amount on oil absorbed from the difference (do this multiple times and obtain a standard deviation so you can make sure that the difference is not due to experimental errors) and after the real answer is obtained one can start thinking why (or actually have your friend do it so he can prove you that he is correct).
My prediction is that is will make no difference on the basis that oil and water don't mix and considering that potatoes contain significant amount of water they wont let much oil in before the temperature increases (which will evaporate the water and allow the oil to move in the potato). In both cases you ll end up with potatoes saturated in fried oil.

Thank you, but potatoes don't contain only water, but other components as well, so maybe oil would be absorbed by those other components. And maybe as well to water but for a short time, because I realize that after serving the fried potatoes, after some time, the potatoes tend to leak some oil..

Potatoes are 80%  wt water. Try the experiment above and i think you ll see no difference.


text:
What dangers/risks are there in the production of carbon subnitride?
I was reading that carbon subnitride or Dicyanoacetylene has one of the hottest flames when used as a fuel, but it also is unstable and can explode easily into carbon powder (no specifics on cause).  The only thing I know about its production is that it requires extremely hot graphite to have nitrogen blown on it.  What specific risks are there in producing it or storing it, and how does it get produced and stored safely?  For example is it safe to store at room temperature? Can it be produced safely in a normal atmospheric environment?

https://pubs.acs.org/doi/abs/10.1021/jo01266a014 details a convenient synthesis

unfortunately, that website doesn't seem to let me read the article.

You know is is explosive. That sounds like a fairly important and specific risk.

There is no way, dicyanoacetylene can be produced at room temperature. According to the paper "Syntheses of Dicyanoacetylene", decomposing 4,5-dicy ano-1,3-dithiol-2-one at 600-800°C, a 59-76% yield of dicyanoacetylene is obtained along with side-products like carbon disulfide, carbonyl sulfide and sulfur. Another scheme is pyrolyzing solid dicyanodiazomethane at 220°C, which polymerizes. The polymerised product is decomposed at 700°C to get desired product. Further, a note on the storage of dicyanoacetylene is mentioned:

Dicyanoacetylene is a hazardous material. Because of its toxicity and
high vapor pressure at room temperature, it should be handled only in
a hood. It is potentially explosive, both in the pure state and in
concentrated solutions, but its thermal stability in dilute solutions
in inert solvents is greater than previously re-ported. Pure
dicyanoacetylene turns dark slowly at room temperature, but it can be
kept almost indefinitely at Dry Ice temperature.

Dicyanoacetylene can only be kept at very low temperatures, in dry ice or liquid nitrogen, like a dewar, away from light and oxygen, as it may polymerize. Do not store it for long. Diluted solutions of dicyanoacetylene are much more stable, and can be safely kept at standard conditions.
Reference

Syntheses of dicyanoacetylene, Engelbert Ciganek and Carl G. Krespan, The Journal of Organic Chemistry 1968 33 (2), 541-544, DOI: 10.1021/jo01266a014

I suspect that it will be difficult to make, but it will be a very endothermic compound. As a result it has the potential to decompose back into the elements with a release of heat energy.
If we assume that when $\ce{C4N2}$ is a gas, then if it decomposes it will form $\ce{N2}$ and four carbon atoms. As the number of moles of gas does not increase at STP, the explosive power of the substance (product of heat of detonation multipled by the increase of number of moles of gas) will not be very large for a given heat release.
If we look at the literature, (Moureu, C. and Bongrand, J. C., Comptes Rendus Hebdomadaires des Seances de l'Academie des Sciences 1910 vol. 150 p. 225 - 229) it indicates the melting point of the compound is $\pu{21 ^\circ C}$ and the boiling point is about $\pu{76 ^\circ C}$. Thus it will have a high vapour preassure. Thus we can consider it to be a volatile compound.
Sadly I have been unable to find any details of the thermochemistry of this compound.

As far as I know, carbon subnitride is a liquid at room temperature.

We already know it isn't a gas a room temperature. We also know it is explosive and unstable. Looks like practice beats theory.

Wouldn't it also yield carbon dioxide when it explodes in atmosphere?  Or some other compound consisting of carbon and oxygen?

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text:
What are energies in computational methods relative to?
As per my understanding, in molecular mechanics (MM) the computed strain energy is based off the strain inherent in the molecule under study relative to an idealized form. For example, cyclobutane to its acyclic form.
However, what about computed values from other methods like PM6 energies, HF energies or MP2 enthalpies. What are these computed values relative to?

At least in HF and post-HF, energies are absolute.

@pH13-YetanotherPhilipp are they? ADF for example does relative to isolated atoms (or fragments) and Gaussian does relative to isolated nuclei and electrons iirc.

"relative to isolated nuclei and electrons" is the same thing as an absolute (electronic) energy though, isn't it?

Ab initio calculations give the absolute energy of the system of fixed nuclei and moving electrons (all considerations on the type of calculations are implicit such as the Born-Oppenheimer approx.). There isn't any distinction of concepts of absolute energies between softwares when ab initio methods are employed (no ambiguity is allowed as pointed out by Pentavalentcarbon). Maybe the confusion can be illustrated with an example: you can't obtain heats of formation, as an absolute energy, from an ab initio calculation but it serves as a good approximation depending on the method.

'There isn't any distinction of concepts of absolute energies between softwares when ab initio methods are employed" is what I said, but you need to be careful considering that the comment about ADF is correct, and MOPAC (AM1, PM3, PM6, PM7, ...) typically print heats of formation.

"you can't obtain heats of formation, as an absolute energy, from an ab initio calculation' is somewhere between untrue and unclear. It is pretty straightforward to calculate a heat of formation from a series of a calculations. My confusion here is what you mean by "as an absolute energy". It's true that it won't be exact and is only as good as your model, but I feel like you're saying something else.

First of all the concept of "absolute energy" seemed too loose. The concept that I adopted to try to answer the question is that it is not a quantity obtained as the total energy but you need successive calculations to calculate the heat of formation. Moreover, for other properties these additional steps may be approximations on top of the central ab initio method which is an approximation as well.

I agree 100% with your comment. I see what you're trying to do, but ultimately it doesn't answer the question. It isn't about how many steps are required to calculate some kind of energy, it's asking what the energy is that you get, and what the energy is relative to.

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text:
Reporting reaction energies calculated using electronic energies rather than enthalpies
I understand that electronic energy includes a combination of

electron-electron repulsion energy,
nuclear-nuclear repulsion energy, and
electron-nuclear attraction energy

at 0 K, and the energies are relative to a zero with all particles infinitely far apart.
So is this when it's appropriate to use electronic energies rather than enthalpies when determining reaction energies at 0K and when particles are infinitely far apart?

"determining reaction energies at 0K and when particles are infinitely far apart?" - this part of sentences make no sense at all.

In the case of atoms or molecules in a vacuum, at $0~\mathrm{K}$, and infinitely apart, the "reaction energy" of
$$
\ce{A + B -> A \bond{....}\bond{....}\bond{....} B},
$$
where $\ce{A \bond{....}\bond{....}\bond{....} B}$ is effectively at infinite separation, is
\begin{align*}
E_{\text{rxn}} &= \sum_{i}^{\text{products}} E_{i} - \sum_{j}^{\text{reactants}} E_{j} \\
&= E_{\ce{A \bond{....}\bond{....}\bond{....} B}} - (E_{\ce{A}} + E_{\ce{B}}) \\
&= E_{\ce{A}} + E_{\ce{B}} - (E_{\ce{A}} + E_{\ce{B}}) \\
&= 0.
\end{align*}
The reasoning is that energies are extensive, and there is no change in the total energy of the system. The total energy of infinitely separated particles is their sum, as there is no attractive or repulsive interaction between them. They're effectively completely isolated. In that sense, there is no reaction, which is I put reaction energy in scare quotes.
In practice, when you have a finite interaction such as a chemical reaction, the total interaction energy should include electronic, vibrational, rotational, and translational components, of which the latter 3 will grow as temperature increases. Most likely the thermal vibrational contribution is the only significant one.
One missing term that is not dependent on temperature and should be added is the zero-point (vibrational) energy,
$$
E_{\text{ZPVE}} = \frac{1}{2} \hbar \sum_{c}^{\text{normal}\\\text{modes}} \omega_{c},
$$
which is the non-zero amount of non-electronic energy that molecules have even at zero kelvin and is purely quantum mechanical in nature. The ZPVE or ZPE can alter energy barriers non-linearly, since vibrations are not necessarily localized, and is the most significant non-electronic contribution to gas-phase reaction energies.

"Most likely the thermal vibrational contribution is the only significant one." I'm not sure I agree with this. The translational component typically has the largest magnitude unless the species is geometrically constrained (e.g. a species chemisorbed on a surface). Could you clarify?

I should be careful. I am referring the ideal gas kinetic energy, since thermochemical analysis in quantum chemistry is usually within the ideal gas approximation. At room temperature, $(3/2)k_{\mathrm{B}}T\approx\pu{311 cm^{-1}} < \pu{1 kcal/mol}$, which should be dwarfed by ZPE. Can you give a good ref that shows this isn't true in reality? I don't know much about actual thermochemistry.


text:
Why Fe(III) can't coordinate with O2 in methemoglobin
I know methemoglobin's heme group has a change on its shape, what makes oxygen unable to coordinate, but... What causes that change? Does the electronic structure of $\ce{Fe(III)}$ matter or it's only a charge/radius conseqüence?
My solution is the following:
As $\ce{Fe(III)}$ is a hard Lewis acid, it needs hard Lewis bases, such as $\ce{H2O}$ and not $\ce{O2}$.   Furthermore, the His $\ce{F8 }$ proximal's imidazole bond with $\ce{ Fe }$ becomes stronger, as it's relation charge/radius is bigger in $\ce{ Fe(III)}$ than in $\ce{ Fe(II)}$, which causes a torsion of the heme group. Is that right?


text:
How does a d4 high-spin complex have a spin-allowed transition without violating the selection rule for total angular momentum?
1) I am confused by the fact that the spin-allowed transition for a high-spin d4 complex is 5E→T2
Doesn't this violate the spin selection rule for total angular momentum?
2) Also, for a d6 complex, some sources I've consulted with mention a one spin-allowed transition for a high-spin complex, and no spin-allowed transitions for a low-spin complex.
From my understanding of the spin selection rules and the use of the Tanabe-Sugano diagram, I found no spin-allowed transitions for a high-spin complex and 4 spin-allowed transitions for a low-spin complex (i.e. 1A1→1T1; 1A1→1T2; 1A1→1A2; 1A1→1A2)

Can you please describe the selection rules you are using to determine whether a transition is allowed/forbidden? From your description I can’t figure out whether you are applying these correctly and I suspect you aren’t, because there is no spin selection rule forbidding E to T2 transitions (the spin needs to stay the same, which is satisfied as long as you are going from quintet state to quintet state) and as far as I know there is no selection rule for total ang mom either. The only other rule deals with the transition dipole moment but that is always violated for d-d transitions anyway.

Thank you for the reply. I am referring to the following rule:                            "The selection Rule for total angular momentum, J, is Δ J = 0 or ± 1. The transitions such as 2P1/2 → 2D3/2 and 2P3/2 → 2D3/2 are allowed, but transition such as 2P1/2 → 2D5/2 is forbidden since Δ J= 2."  So for the d4 and d6 high-spin complexes, isn't Δ J = 2?

These rules refer to electronic transitions in isolated atoms, not complexes. Observe that the term symbols are entirely different. S/P/D/F term symbols only apply to spherically symmetric atoms and A1/A2/E/T1/T2 only apply to tetrahedral complexes (or octahedral if you also specify the g/u parity).

related: https://chemistry.stackexchange.com/a/64486/102629

The photon has one unit of angular momentum and as angular momentum is conserved this must be taken up by the molecule; in terms of symmetry species this means that the product of the symmetry species of the two states involved and of the transition dipole must be worked out. The electron spin in not affected by the photon angular momentum so the spin selection rule applies $\Delta S=0$ unless there is also some spin - orbit coupling then otherwise spin-forbidden transitions can occur but with generally low probability compared to a spin allowed one.
To work out the symmetry species involved in a transition the value of the integral
$$\int\psi_i\mu\psi_fd\tau$$
between initial $i$ and final state $f$ has to be evaluated.; $\mu$ is the transition dipole. This integral is very hard to evaluate but fortunately symmetry can help.
Using symmetry the integral can be determined to be zero (transition forbidden) or not zero. The equation to use is
$$\Gamma\psi_i \Gamma\mu \Gamma\psi_f = \;'A_1\,'$$
where $\Gamma$ means the symmetry species, $A,\; E$ etc and $'A_1\,'$ means the product must contain the totally symmetric representation in the point group used. This is the label given to the top row of characters in the table ($A_1$ or $A_{1g}$).
The transition dipole symmetry is found in the column of the point group just to the right of the characters and are the symmetry species with x, y, and z. In the $T_d$ and $O$ types of point groups the transition dipoles are either the $T_2$ or $T_1$ symmetry species respectively.
Thus in your $A_1\to T_1$ example the symmetry product is
$$A_1 \otimes T_1 \otimes T_1 = A_1$$
in an $O$ point group. The symbol $\otimes$ means multiply characters in each symmetry species column by column.  Hence you can work out the product  by multiplying the characters together or more easily by looking up the values in direct product tables.
In this case this is easy to do as anything multiplied by $A_1$ is unchanged (as its totally symmetric) and any symmetry species multiplied by itself always contains the totally symmetric representation $A_1$. So the transition
$A_1\to T_1$ is allowed in octadedral point groups.
In the tetrahedral point group $T_d$ where the transition dipoles each transform as $T_2$ the product is
$$A_1 \otimes T_2 \otimes T_1 =  T_2 \otimes T_1$$
and now the characters have to be multiplied and the the irreducible representation found (see link to answer below). Instead looking up the product in direct product tables gives
$$T_2 \otimes T_1 = A_2+E+T_1+T_2$$
and in this case as the totally symmetric representation $A_1$ is absent the transition would not be allowed. This means that any transition will be weak, i.e. has a very low but finite extinction coefficient. There are effects in molecules that 'break' the symmetry, such as vibrations distorting the molecules 'ideal' symmetry and these can make forbidden transitions slightly allowed.
This answer gives some details on point groups.
Understanding group theory easily and quickly

What happens for geometries / point groups where the transition dipole moment is not represented by one symmetry element? In $D_3h$ for example z is represented by $A^"_2$ while x and y have $E'$. Is a transition allowed once it transforms by one of the two?

Molecules always have three  dipoles  in different directions and so belonging to different symmetry species. Transitions involving these will lead to different excited states.


text:
Comparing ionic character of group 1 elements
According to Fajan's rule ionic character should increase down the group as the size of cation increase.
So it must be
$$\ce{LiH < NaH < KH < RbH < CsH}$$
However, the following two sources provide two different opinions. Which one is right?
Source 1

Explanation: As the size of the metal cation increases down the group, the lattice energy of the hydrides decreases down the group. Consequently, the reactivity of the alkali towards hydrogen decreases down the group.
iii. The ionic character of the alkali metal hydrides increases from Li to Cs.
Explanation: Since ionisation enthalpy of alkali metals decreases down the group, tendency to form cations as well as the ionic character of the hydrides increases.

Source 2

But it we are asked to compare the ionic character of hydrides (LiH, NaH, KH, RbH, CsH), we can not use the Fajan's rule. This is because, here the anion is very small ($\ce{H-}$) as compared to the cations (alkali metal cations). The ionic character is this case can thus be compared on the basis of Lattice energy.
Since lithium ion and the hydride ion are approximately of same size, the charge density is high between the two, thus the ionic bond so formed will be very strong and hence more will be the lattice energy.
On coming down the group, the difference in the size of cation and anion increases, thereby making the ionic bond weaker, thus reducing the lattice energy.
Also, lesser will be the lattice energy, lesser will be the ionic character.
Thus, down the group, the ionic character of the hydrides decreases.

Just a side note: it's preferable to post text as, well, text, and not as images; also it's a good habit to provide references to the external data sources.

Well, the second source is certainly more wrong. There is a major difference between ionic character of a bond and the stability (lattice energy) of an ionic lattice. $\ce{LiH}$ is certainly more covalent than $\ce{CsH}$ (meaning $\ce{CsH}$ is more ionic), which can be justified simply by comparing charge densities on $\ce{Li}$, which is very high and draws electron density from the $\ce{H}$ atom, forming a more covalent bond.

It is true that the ionic lattice energy goes down as we move from $\ce{LiH}$ to $\ce{CsH}$.  But what happens to the competing mechanism of covalent bonding?
To make a covalent bond you overlap orbitals of the atoms you're bonding, but in $\ce{CsH}$ you have a huge, fluffy, low effective nuclear charge cesium $6s$ orbital meeting a compact hydrogen $1s$ orbital.  This is not going to go over well; the shoe really does not fit here.  Lithium offers much better covalent overlap woth its $2s$ orbital instead of $6s$, with the other (stable) alkali metals in between.
So yes, ionic bonding energy goes down as we proceed to heavier alkali metal hydrides, but covalent bonding fares even worse.  Heavier alkali metal hydrides have less overall bonding energy and a greater part of what bonding energy they have comes from the ionic interaction.

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text:
Does this image represent a molecule?

PLease help me to define this,
Thank you,
Bruce

No.$\mathstrut$

Thanks, it has been a few years since college chemistry and was wondering if this illustration fit in today's chemical world. I took an electrical and process control engineering path in my career and have seen changes in electrical representation since I first started. This is supposed to be a crop circle but looks to me like someone dabbling with atomic representation. Again, thank you, first time asking for help.

There are several ways of drawing/representing molecules:
Lewis Dot: It shows bonding and non-bonding electrons. 2D
Bond-Line Notation: Like Lewis Dot but is simplified. C and H are shown as lines. It simplifies large structures to highlights important atoms, bonds, and groups that could react. 2D
Ball and Stick: It's a three-dimensional arrangement of atoms and bonds. 3D
Space Fill: The "fat" version of Ball and Stick. Shows three-dimensional arrangement as well as size relationships between atoms. 3D
The picture shown would probably be a 2D imprint of a Ball and Stick model.
However, it is JUST A CROP CIRCLE.
If you want to overextend it try using ChemSpider to draw something similar and see if you get any hits.

Thank you folks! Your input is what I am looking for! It has been too many years since my training; my career in the electrical world makes me aware of differences in schematic representation over just a period of 30+ years! Yes, this picture is just a crop circle but I assume the author of it was playing with his chemical passion. I wouldn't do this in another's field and hope if someone tried this in my field my bull and donkey would chase them out!! I am still interested in if this is a compound representation or just someone playing. Thank you for being helpful to an old guy!


text:
Electrolysis puzzle
I created an electrolysis setup to study how different electrolytes, their concentrations, and the electric current affect the "speed" of the electrolysis process. With that in mind, I built a simple  apparatus:

A beaker
two test tubes
PVC-covered copper wires for electrodes.

For each electrolyte and concentration, I then apply different voltages: $\pu{1.5 V, 3 V, 4.5 V, 6 V}$ and $\pu{9 V}$
I think I am missing some detail or not controlling some variable because some of my results were not expected.
For instance, in the electrolysis of aqueous $\ce{NaCl}$ (of all cases!).
I was expecting to have $\ce{H2}$ created at the cathode and $\ce{Cl2}$ at the anode. However, while I experienced $\ce{H2}$ evolution in the cathode (which was collected in the upside-down test tube over), I did not notice any gas evolution at the anode.
I reviewed Jan's  response to a similar question in this forum who explained that, because to elevation of the pH due to the generated $\ce{NaOH}$, the  $\ce{Cl2}$ would react further resulting in the production of $\ce{NaOCl}$, and in no gas being produced at the anode.
However, Jan previously indicated that in the beginning

"...chlorine gas at the anode should bubble after some time".

In my case, I never noticed any bubbling at the anode during the experiment. I varied the concentration of the aqueous $\ce{NaCl}$ solution, from very weak to saturated; tried different voltages, changed the temperature of the solution. Nevertheless, the behavior was  the same in each scenario. I also tested the voltage at different points with a voltmeter (everything checks), made sure I was using pure salt (without Iodine or additives).
One interesting "clue" is that I noticed some "blue powder" mostly at the bottom of the beaker (some suspended too) but none inside of the test tube over the anode. I also noticed some white build up on the copper wire at the anode, which I think to be $\ce{NaOH}$.
Right now, I am at a loss for ideas and would appreciate any suggestion or explanation.

As is often the case with chemistry, you can't ignore the fact that your instruments are also made of atoms and have a chemistry of their own.
Naturally, in this setup you just dissolve copper from the wires:
$$\ce{Cu -> Cu^2+ + 2e-}$$
You may check the table of electrode potentials and verify this with numbers. Being more energetically favorable than the other options, this reaction occurs first. Chlorine never had any chance to appear.
That's pretty much the size of it. Voltage is irrelevant (as long as it is high enough for the reaction to occur, which is true even for 1.5V). Salt concentration is irrelevant. Additives are irrelevant. Temperature is irrelevant, as long as the solution is liquid. Time is irrelevant; wait longer, and you'll get all copper eaten up, which will break the circuit and thus force the process to stop.
Due to the lack of acidity, much of the dissolved copper hydrolyzes and finally ends up in the form of hydroxide down at the bottom, which explains the mentioned blue powder.
With electrodes of unobtanium or other inert substance (graphite being a moderately good approximation) you'd get your chlorine all right.


text:
Is there a point where further dilution of ethanol is not exothermic?
Diluting concentrated ethanol with water is exothermic. Diluting 95% refined spirit into 40% vodka makes the resulting product definitely warm. Dilution of 60% absinthe with water that is cool, but warmer than "nearly freezing" results in a lukewarm, unappetizing drink. I haven't noticed the effect when diluting weaker alcohols though.
Is there a threshold of concentration where further dilution of water-ethanol mix doesn't produce heat? Or is it merely an asymptotic relation, where at lower concentrations the effect becomes hard to notice?

Mixing ethanol and water is exothermic all over the range at about room temperature. Data are available, e.g., here excess enthalpy ethanol-water
However, at relatively high temperature and at least around azeotropic composition,  the mixing should become endothermic otherwise the ethanol-water azeotrope should boil at higher temperature instead of having Bp of 78.1 °C.


text:
Can hydrogen plasma reduce carbon dioxide?
There are several studies describing usage of hydrogen plasma to reduce metal oxides. This is not yet widely adopted in industry, but it is of interest for environmental reasons (such as described in Sabat et al. 2013).
I was wondering if the same could be done with carbon dioxide. If not, why not? Is there a way to calculate which oxides can be reduced this way?

The big question isn't whether it is possible but whether it cost a small enough amount of energy to be worth doing.

Not really, because if it isn't possible, then the question of energy efficiency isn't relevant.

Of course it's possible, but why would one want to do so?


text:
How many resonance structures does the oxalate ion have?
I have drawn the Lewis structure of the oxalate ion, and I see that two oxygens are double bonded to the carbons. When finding the resonance structures, I said there are 2, but the exam solutions state 4, which was confirmed by internet images such as this:

However, I see no real difference btween structures 1 and 2, and 3 and 4. Are they considered different resonance structures?

Yes they are. Each member of the "couples" just "weights" as its counterpart.

Resonance structures are not real things. As also indicated by rbw, they have weight on the actual compound. Simpler example is ozone.
http://www.wou.edu/las/physci/poston/ch222/pdf/Ch08-s12-part2.pdf
From the first slide second image there;

you can see that the actual molecule is none of the "resonance" structures, and both bonds between leftmost and rightmost oxygen atoms with central oxygen atoms have equal lenght, and it is between 1 bond length and 2 bond length.
The general idea of drawing resonance structures is, after predicting the most appropriate ones, when you superimpose them and average the bond lengths you will be very near to see the real chemical bonds lengths.
This is straightforward with ozone, oxalate or carbonate etc. but things get complex with increasing atoms, though.
And wrapping up and also answering the question you posed, you should look atoms as if they are all unique, like oxygen 1 oxygen 2 oxygen 3 and oxygen 4. Flip the molecule and you will NOT get the other one, this is not like isomerism. It is just to predict the electron cloud density around atoms of a molecule.
Hope this also helps.

If you begin from the skeleton of the oxalate ion and consider the formal charges than you will find these four structures which are different as pointed out by Alchimista. However, you have structures 1 and 2 contributing equally to the whole system and as well as structures 3 and 4. This is supported by a simple symmetry analysis where structure 2 and 4 can be obtained by a 180 deg. rotation along the C-C axis of structures 1 and 3, respectively.


text:
Trivial name of a compound vs its chemical formula
I had a question like below and I've had trouble with similar problems before and how to go about solving for the chemical formula.

What is the chemical formula for lithium aluminum hydride?

The first thing that comes to mind, we have one lithium, one aluminum and one hydride. Then why is the chemical formula $\ce{LiAlH4}$? Where did the four $\ce{H}$ atoms come from? Earlier on in the course we had stuff like

What is the chemical formula for di-arsenic trioxide?

And that one is easy, I have 2 arsenic and 3 oxygen. Given the full name I can easily deduce the formula (or the other way around). But what do you do in situations where you don't get "all the information". Do I go about thinking up how the structure works and see that it somehow logically follows that it needs to be four $\ce{H}$?
I know most basic compounds, e.g. sulphate is $\ce{SO4^2-}$, but I feel like I can't memorize some of the things. There has to be a more general way to do this. It feels like and I must have missed it. Essentially, the full systematic names are no problem, but the trivial names are giving me a rough time and I'm scared if they come up during my finals next week. Any tips regarding that?

Well, that's by design. Systematic names were designed so that you could easily derive formula from them; trivial names were not. Then again, what is the formula for lithium hydride? Can you tell that? And now what is the formula for aluminum hydride?

No you are absolutely right here, but even sulphate or nitrate is wrong. If you take a closer look at what a sulphate essentially is, you will see the following: It is a separate unit, it has a central atom and ligands and the whole unit is negatively charged. If this was a metal we would say it's a complex and name it with some sort of -ate in the name. And this is exactly what you see here. So tetraoxosulfate in short is sulfate. Whit this shorter notation you are just not able to guess it unless it is so common that you should know about it. For lithium aluminium hydride it's the same thing, it should be called lithium tetrahydroaluminate.
Now you may ask yourself how are you supposed to work with these problems? In all scientific cases just google it. That's what we do as well, if this is class or homework related, then your teacher will only give you certain examples. If he doesn't, talk to him about it.

In many cases, the number of each element in a compound can be fairly easily inferred by the normal oxidation states of the constituent elements. In the case of lithium aluminum hydride, we have lithium, which is almost always at a $\ce{+1}$ oxidation state ($\ce{Li^+}$). Aluminum is typically $\ce{Al^3+}$. Thus, we need $\ce{4}$ hydrogen atoms at a $\ce{-1}$ oxidation state, $\ce{H^1-}$ (i.e. "hydride"). We get $\ce{LiAlH_4}$.
Similarly, magnesium chloride is $\ce{MgCl_2}$ because it can't be anything else. No need to call it magnesium dichloride. (This also gets into the difference in conventions for ionic compounds versus covalent compounds.)
Of course, in other cases it may not be as obvious. What about iron oxide? Well, that's a bit more ambiguous, it could either be $\ce{FeO}$ or $\ce{Fe_2O_3}$. For that reason you're more likely to see iron(II) oxide for the former or iron(III) oxide. Occasionally you'll still see ferrous oxide and ferric oxide, respectively, but that's becoming less common.
As far as the trivial naming schemes, there is a bit of a learning curve, but it really does make things easier. Calling $\ce{H_2O}$ "dihydrogen monoxide" is only useful for trolling people who get scared at anything that sounds "chemical". It's "water", everyone knows water. Even for your original example, $\ce{LiAlH_4}$, Wikipedia says "the preferred IUPAC name" is Lithium tetrahydridoaluminate(III) and the "systematic IUPAC name" is Lithium alumanuide ($\ce{AlH_4^-}$ is apparently alumanuide, I personally did not know that one)
Full systematic names aren't supposed to be "easier". They are supposed to be completely unambiguous. The problem is that in many compounds, especially in organic chemistry, it's not just the number of each elements that are present, it's how those elements are connected, which can get rather complicated. Consider aluminum sulfate: $\ce{Al_2(SO_4)_3}$. Calling it dialuminum disulfide dodecaoxide loses the information on the individual ionic character, never mind the fact that "disulfide" means something else. Meanwhile, "aluminum sulfate" is clear.
For the most part, it really does come done to a matter of practice and exposure. There are a lot of names that exist with a fair bit of historical inertia (bicarbonate comes to mind), but eventually you get comfortable with it. Of course, Google and Wikipedia can come in handy if you see something new, but you'll find yourself resorting to that less and less.

There is a useful rule of thumb when looking at trivial names. Look at the ending of the anion part of the name.
If the name ends in "ide" then the compound only includes the elements indicated in the name. For example sodium chloride.
If the name ends in "ite" then the compound includes the elements indicated in the name plus some oxygen. For example sodium chlorite.
If the name ends in "ate" then the compound includes the elements indicated in the name plus plenty of oxygen. For example sodium chlorate.
If the anion name starts with "per" and ends with "ate" then it normally means more oxygen than is found in the "ate" anion. Such as sodium perchlorate.


text:
Is it wrong to write a coefficient of 1 in a chemical equation?
Take $$\ce{2H2 + O2 -> 2H2O}$$
Is it wrong to write $$\ce{2H2 + 1O2 -> 2 H2O}$$
By wrong I mean that it is against a widely accepted guideline (those of the IUPAC?), not just "your teacher said it's wrong, therefore it's wrong". The only information I could find, is that it is assumed there is a coefficient of 1 when there is none written.

It is not wrong, per se, but it goes against conventions.
These rules are in place to simplify how we write formulas and equations, etc.
If we did write the $1$ all the time, then perhaps we would write it for a chemical formula, and water would be $\ce{H2O1}$, for instance.
If you want to write it as a reminder and placeholder that is totally fine.
A teacher might say that they have a rule, or that they want you to follow standard convention. You should do this if requested. When I taught, I would offer to students that it is OK if they want to do it to get in good habit of always writing coefficient.
All that being said, IUPAC guidelines say to write the species with no coefficient if the coefficient is one. So, technically it is not to the standard, but does not make it less clear, so use it in your work, but if you are submitting as classwork follow the protocol of the class.
Below is cited a brief note to this from the The ACS Style Guide[1].

➤ Specify the number of each species (molecules, atoms, ions, etc.) of reactants and products by a numeral written on the line and closed up to the symbol.
$$\ce{2Al + 6NaOH -> 2Na3AlO3 + 3H2 }$$
➤ To indicate the aqueous, solid, liquid, or gas state, use the appropriate abbreviations on the line, in parentheses, and with no space preceding them.
$$\ce{Ag(s) + H+(aq) + Cl–(aq) -> AgCl(s) + ½H2(g)}\\
  \ce{4FeS(s) + 7O2(g) -> 2Fe2O3(s) + 4SO2(g)}$$

Within these few lines, it is stated to write a numeral, but then it is omitted when it is $1$.
House rules prevail, as this is commonly accepted practice.
Reference:
[1]: The ACS Style Guide: Effective Communication of Scientific Information; Coghill, A. M., Garson, L. R., Eds.; Oxford University Press: New York, NY, 2006(pdf).


text:
Coins in an Electrolytic Cell
In attempting to reveal the copper under a quarter, I used alligator clips to attach the quarter and a clump of copper wire to opposite sides of the battery. I put the electrodes in a solution of tap water and NaCl.
As the reaction occurred, the solution quickly turned into an icky (for lack of a better word) green, the water became very unclear and some cruddy particles rose to the top. Also, gas evolved from the copper electrode (which I suspect is H2). The coin blackened in this process and I took the quarter out and washed it.
Next, I reversed the polarity and the coin became the more expected copper color. As this happened, gas rose from the coin instead.
While I don't understand what occurred, my concern is the green sludge that was produced. I read solid nickel surrounds the mostly copper quarter coin. If the gas is really H2, then there must be hydroxide ions produced. If this is true, then I suspect the sludge could contain nickel(II) hydroxide. Is this possible? The "sludge" was a dark dollar green and was flaky, unlike dry nickel(II) hydroxide. It really looked like sewage more than anything else. I did not get a picture because I dumped it out before I thought of taking one (I suspected it to be dangerous). Are the conditions I described enough to get an idea of what it was?


text:
Indigo synthesis from anthranilic acid
I am going to perform the synthesis of indigo via Heumann's methods which uses anthranilic or aniline as a starting materials.
I am looking for the indigo synthesis procedure which includes all chemicals and the amounts needed but hopeless to find one.
Can anyone help me ?

https://pubs.acs.org/doi/abs/10.1021/ja01422a030 may help

Synthesis of indigo: In place of anthranilic acid or aniline, you can simply use 2-nitrobenzaldehyde (see: Baeyer–Drewson indigo synthesis). For the mechanism of the reaction see: WIKIPEDIA.
Method: Place about $\pu{0.1 g (\pu{100 mg})}$ of 2-nitrobenzaldehyde solid into a test tube and, using a $10$-$\mathrm{mL}$ graduated cylinder, add $\pu{2 mL}$ of reagent acetone to the test tube. Swirl the tube to dissolve the solid completely. Now, add $25$ drops of distilled water to the acetone solution, and swirl again gently. Then, slowly add $15$ drops of $\pu{3.0 M \ce {NaOH}}$ solution. Your solution should rapidly darken and a deep purple solid (indigo) should precipitate from the solution. Place the test tube in a rack to stand until precipitation completed. Filter the solution using a Büchner funnel ($30$-$\mathrm{mL}$) and a filter flask ($125$-$\mathrm{mL}$). Wash the solid with distilled water several times until no more color washes from the solid, and finally by two portions of $95\%$ ethanol.
Also see: http://slb.faculty.arizona.edu/sites/slb.faculty.arizona.edu/files/152/Indigo1.pdf


text:
How thick is a snowflake?
The x, y dimensions of snowflakes get all the attention for obvious reasons, but the z dimension must be interesting too, in its own humble way.  In fact, I have lived more than half a century and never seen anyone refer to snowflakes' third dimension at all!
So I'd like to know what a snowflake's cross-section looks like, and what forces/phenomena make it that way.
There are numerous possibilities:

that snowflakes are a single molecule thick
single crystal thick then grown over with hoarfrost
thousands or millions of nearly identical snowflakes glued together like slices in a loaf of bread
something else

No doubt, the ratio between x and y dimensions is extremely similar across all snowflakes, but an interesting part of the phenomena involved would be that which defines the ratio between x and z dimensions.

My observations of individual lumps of snow suggest the shapes may a lot in all dimensions. Most lumps consist of more than one crystal, though. Even so, the crystals vary a lot in size.

There are many different types of crystals and their fromation is far too difficult for me to understand. This answer simply deals with the thickness part and that too merely gives a rough approximation due to the lack of data.
 From CalTech

We have two formulae for calculating the width:

For Dendrite
$$\mathrm{h =a_2 D^f} $$
D: crystal diameter (cm) (major dimension)
h: crystal thickness (cm) (minor dimension)
For Needles
$$\mathrm{d = a_2 L^f} $$
D: crystal length (cm) (major dimension)
d: crystal thickness (cm) (minor dimension)

This is a table for $\mathrm{a_2, f}$

For d and l   (right most column )

I calculated the thickness of a $\pu{3 mm}$ Dendrite which is coming around $\pu{24  \mu m}$.

Source
https://web.archive.org/web/20180416083128/https://www.nsstc.uah.edu/data/uahsevere/owles/parsivel/references/Snowflake_Size_Distributions_Lecture.pdf

As a Canadian I have seen some of those possible forms, but by no means all. Unfortunately, I've never seen a snowflake anywhere close  to 2.4 mm thick.  For a 0.5 cm dendrite, using their formula h = aDf, I get 0.009 × 0.5^0.377=0.0069cm or 0.07mm -- so something less than a tenth of millimetre to make it easy to remember.  I do appreciate your digging that out though.  Thanks very much!

Just a word of caution regarding your source: Google drive links frequently become invalid; you did the right thing to include the relevant portions here. (It would have been nice if the author of the slides would have included actual references, but I guess those are on a different handout...)

@ Martin - マーチン I changed the link but wasn't able to find the original document.


text:
How to measure ozone output of an ozone generator using chemical methods
I have the need to optimize an corona discharge based ozone generator. There are certain variables like frequency and voltage which I can change.
The easiest way would be to use an ozone analyzer – however these things are quite expensive and over my budget currently.
I have tried certain ways to get a hold on numbers but failed so far.
Example: I used activated charcoal to convert the $\ce{O3}$ to $\ce{CO2}$ and tried to measure the $\ce{CO2}$ ppm, however it turned out that the activated charcoal did absorb the $\ce{CO2}$, so this measurement did not work..
Is there a easy to do chemical way to react the $\ce{O3}$ with something, that can be measured/done by a hobby chemist? My aim is to find at least a setup with the maximum ozone output. If it is possible to calculate how much grams ozone were produced it would be even better.

For optimization purposes, perhaps something as simple as a solution of KI with added starch (e.g., laundry spray starch). More ozone means more intense blue color from the starch-iodine complex. And a blank is easy.

Back titrate the iodine with standardized thiosulfate solution.

You can do the quantitative estimation of ozone ($\ce{O_3}$) by two reactions. First one is the reaction between $\ce{O_3}$ and potassium iodide ($\ce{KI}$), and the second one is an iodometric titration.
First, ozone is reacted with excess $\ce{KI}$ in presence of borate buffer ($\mathrm{pH} =9.2$), and due to this reaction, iodine ($\ce{I_2}$) and oxygen ($\ce{O_2}$) are liberated as follows, $$\ce{O_3 + 2KI + H_2O -> I_2(^) + O_2(^) + 2KOH}$$
Now this mixture of gases is passed through potassium pyrogallete solution, which absorbs $\ce{O_2}$, and we are left with $\ce{I_2}$. Next, this liberated iodine is dissolved in a solution to make a particular volume, and then titrated with sodium thiosulfate ($\ce{Na_2S_2O_3)}$ of known strength in the presence of starch indicator.
$$\ce{I_2 +_ 2Na_2S_2O_3 -> Na_2S_4O_6 (sodium tetrathionate) + 2NaI}$$
At the end point of this titration, colour of the solution turns colourless from blue ($\ce{I_2}$ + starch gives blue colour complex  which is lost at the end of this reaction).
So, thiosulfate reacts with iodine in a quantitative manner, and from there you can determine equivalents of $\ce{I_2}$ = equivalents of $\ce{Na_2S_2O_3}$. Thus, you can calculate what equivalents of ozone had produced that much of $\ce{I_2}$, and also the grams of ozone present.

seems like if you have a pH meter you could just monitor pH of the borate buffered solution (KOH is formed).  How critical is the pH?  Maybe you could also use unbuffered calcium iodide and just measure the precipitate as $\ce{Ca(OH)2}$ is formed.

Any redox reaction should do for this. Since ozone is a powerful oxidizer.  I would have to agree with Soumik Das on his method.  It would be the easiest.  Most of those chemicals (assuming you don’t have access to a lab like me) are easily purchased.


text:
Steric hindrance during Hydroboration
In response to this diagram (Organic Chemistry By David Klein 3rd Edition Page 364):

The textbook stated

A comparison of the transition states for hydroboration via
  Markovnikov addition or anti-Markovnikov addition. The latter
  will be lower in energy because of decreased steric crowding.

Is it supposed to mean 'former', as the diagram obviously shows higher crowding in the latter transition state?

Yes, you are correct. This would also be more helpful to future readers if you were to cite the textbook and a page number

@orthocresol Done

I would argue that bond polarity has more to do with it. In Markovnikov's studies, he dealt with HX addition where H was always more positive than X. In the B-H bond, boron is more positive than H. Thus, I would classify the lefthand TS as Markovnikov-type addition. After the alkylborane oxidation, where O replaces B, the net result is anti-Markovnikov addition of water.


text:
Is benzene harmful to human health?
I have heard that benzene is a dangerous chemical for humans; it causes cancer and so it's use is restricted/regulated. But many organic compounds that we use contain benzene in some form such as oils, foods, even medicines (paracetomol). So is benzene really harmful to people's health?

Chlorine is toxic, sodium is flammable; how can you eat NaCl?

You should learn to understand the difference between a chemical compound and a mixture.

Yeah, and while drinking a lot of H2O is considered very healthy, drinking a lot of H2O2 is not considered healthier.

@rackandboneman - And, of course, in the US about 3800 people die every year due to overdosing on H2O.

Before we discovered the toxicity of benzene, it was a widely used cheap solvent in labs. Once the toxicity was discovered benzene as a solvent was mostly replaced by toluene, which is benzene with a single methyl group attached. This apparently small change reduces toxicity by several orders of magnitude. You should not assume that single features of a molecule (like a benzene ring) determine its properties or toxicity.

@aventurin your comment is not in focus.  What is the importance of mixtures here?

@rackandboneman - Yeah

It's like with Thimerosal: The ethylmercury ion doesn't make it deadly, even though the ethylmercury on its own is

As Ivan Neretin commented, parts of a compound can never dictate its own properties to the entire molecule's properties. Additionally benzene is not a very effective functional group by itself.
In short, a restricted use is only applied when that certain chemical is exposed by itself. In paracetamol or any other thing including benzene, you will not see any benzene vapor emanating from them or something. It would also be meaningful if benzene containing compounds degradation products included benzene, like in environmental degradation of chlorobenzenes, though this is not a usual case as I see.

First let us make clear that the products you mentioned do not contain benzene. Likely you intended that plenty of products are aromatic compounds and, as such, contain at least a phenyl ring (or contain at least a benzene-like motif, for those less familiar with chemical nomenclature).
After reading the following, that gives some anecdotal notes as well, it will be clear that the presence of a specific part of a molecule does not necessarily dictates its properties (toxicological in the specific discussion) to the whole.
Translated from my old university notes:

.... many aromatic compounds are carcinogenic. After the London fire,
  a law imposing the maintenance of chimneys was promulgated. As a
  result it was noted that sweeper kids were at high risk of developing
  scrotum cancer. This was lately connected to benzo[a]pyrene present in
  the soot, and clearly established in the 1900s.
Benzo[c]anthracene is carcinogenic as well, and benzene is among the
  most powerful agents inducing leukemia.
In general strong carcinogens are strong electrophiles, such as
  alkylants. On the contrary, benzene is electron rich.
Again in England, around 1960, a kind of plague was causing the mass
  death of turkeys due to leaver cancer. It was established that as due
  to to a toxin present in the feeds that, once metabolized, gave
  epoxides as strong electrophiles.
A link to aromatic carcinogenic properties was then made and
  established. Rather than aromatic themselves, the real carcinogens are
  their epoxides formed during metabolism. They act as substrate for the
  nucleophilic attack by DNA bases as we will see in......

Anything altering the stability or geometry of the metabolite epoxides will therefore modulate their reacivity and in turn the hazard posed by that specific aromatic compound.
This is also the case for PAH were, at least, a particular structure (bay) is linked to strong carcinogenicity.
Edit. As this answer received a  comment pointing to the fact that not all aromatics are carcinogenic I invite you to consider that the above

many aromatics does not mean that all aromatic compounds are carcinogenic
anything altering.... means that each molecule must be considered as a whole. Functionalisation of benzene by a methyl group leads to toluene, for instance.  The latter does not qualify as carcinogenic and should replace benzene whenever possible
PAH is of course a subset of arenes. Among PAHs, again, not all compounds are carcinogens. However, in this case, a specific feature regarding the shape rather generally point to carcinogenicity (see link).

yes benzene is very harmful to human health

But many aromatic compounds are not toxic and it is wrong to overgeneralise. Toluene is very much less toxic than benzene, for example. And while PAHs are often bad, this is a bad generalisation for aromatic compounds as a class.

I do not see why you should comment here. "Many is not all'. "Anything altering the stability of the potentially formed epoxides will. .. modulated the hazard posed by a specific ar. compound." "This is also the case for pah...".  @matt_black. You should even remove your comment at least it does not belong to my answer!

The problem is you are answering a question about the general consequences of having a benzene ring in a compounds with a specific set of examples that are nasty. This doesn't help answer the general question (even after you edited the answer with caveats).

I insist that your comment does not belong to my answer as no generalization is made.  Any normal reader understand that a benzene motif itself is not responsible for harmful effects.

@Alchimista I would say the OP was under that impression. I edited the question to it's current form because the original was a little rough to read, but that was how I interpreted it.

@Tyberius yes indeed OP question was dictated by that impression. But all the answers and in particular that of mine should have cleared that. In particular there is not generalization of sort in my A. I will further edit to account for the actual Q.

I see after long time that even a very strange answer was accepted :))

Containing a benzene-like (phenyl) group does not make a compound have the same biological activity as benzene itself. Nearly all proteins in your body contain the amino acids phenylalanine and tyrosine, both of which have a phenyl group. Your blood constantly carries these amino acids in their free forms. However, these amino acids are not metabolized by enzymes the same way as benzene and therefore are not metabolized into the epoxide-bearing molecules (described in other answers) responsible for DNA damage.

Benzene is toxic to humans, it can be oxidised by enzymes in the body (might be a P450 enzyme) to form benzene oxide which is an epoxide. This is 7-oxabicyclo[4.1.0]hepta-2,4-diene which can then react with DNA or other molecules in the body. If the epoxide reacts with water to form cyclohexa-3,5-diene-1,2-diol. Then this can be oxidised again to form 7-oxabicyclo[4.1.0]hept-4-ene-2,3-diol which can react with the DNA.
As a result of this DNA damage benzene is able to cause cancer, it is important to keep in mind that many carcinogens are in fact precursors of the carcinogen. They require some metabolic activation to be able to cause cancer.
Another example is hydrazine which is converted in the liver by formaldehyde and oxidation into diazomethane. As a result the liver is damaged.

It is generally accepted that formation of diolepoxide-derived DNA adducts such as ones depicted  in following Figure1 is the principal metabolic activation pathway leading to cancer initiation in mammals (Courtesy of https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3812823/).

Polycyclic aromatic hydrocarbons (PAHs) are a class of environmental contaminants (Figure2) that has long been suspected for genotoxicity, metabolism and metabolic activation, DNA adduct formation, mutagenesis, and carcinogenesis. Although, PAHs are relatively non-reactive chemicals toward biological macromolecules under physiological conditions,  upon entering the body, our cellular defense system attempts to “remove” these foreign substances by metabolism, which is principally done in our liver, mainly catalyzed by the cytochrome P450 enzymes. Upon metabolism, most PAHs, with multiple hydroxyl groups, become more polar and water-soluble to be excreted out of the body. However, metabolism of some PAHs generates certain reactive intermediates (see benzo[a]pyrene; [Figure 1]) that are capable of forming damaging covalent adducts with nucleic acids (e.g., free $\ce{DNA-NH2}$ groups), eventually leading to genotoxicity and cancer.

PAHs are a very small subset of aromatic compounds in general and are toxic for specific reasons that don't generalise to all aromatics.

@ matt_black: I didn't imply all aromatics in general are harmful to humans. It's too broad  of field to fin point something like that. Yet, PAHs are a good example to represent benzene. For instance, unlike toluene, benzene under go similar detoxication process like a PAH such as benzo[a]pyrene by the cytochrome P450 enzymes.

The trouble with quoting a particularly nasty class of compounds as an answer to a much more general question is that you reinforce the false impression that the presence of a benzene ring in a compound inevitably makes it toxic. Nothing you say is wrong, but the answer is incredibly misleading as an answer to this specific question.


text:
Gold Half-Life and the Longevity of Gold Curiosity
I was curious if the half life of Gold is 186 Days, then why does it 'Last forever' ... Wouldn't the ancient Egyptian gold artifacts be decayed to lighter elements by now like platinum? I know gold was able to withstand corrosion very well, but the relatively short half-life is kinda throwing off my understanding of why it 'last forever'.

186 days, eh? Wow, it must have decayed completely ages ago. Gold is a myth. There is no gold.

@Ivan maybe the gold is constantly replenished by the action of cosmic rays?  Like carbon-14?  Which reminds me:  Always know what isotopes you're dealing with.

@Oscar It would take quite a bit of ingenuity to explain the apparent stability of certain centuries-old golden items. But I'm sure we can come up with something.

Gold has only one stable isotope, $\ce{^197Au}$, which has $100\%$ natural abundance, thus considered to be one of monoisotopic elements (one of 26 chemical elements which have only a single stable isotope). As a matter of facts, it is now considered to be the heaviest monoisotopic element. For more info, read Isotopes of gold.
Aside $\ce{^197Au}$, there are 36 unstable isotopes of gold been known ranging from $\ce{^169Au}$ to $\ce{^205Au}$. All of these are radioisotopes (radioactive nuclides). The $\ce{^195Au}$ isotope is the most stable amongst unstable isopopes, with a half-life ($t_{1/2}$) of $\pu{186 d}$. The heaviest, $\ce{^205Au}$, has $t_{1/2}$ of $\pu{31 s}$ (ref).

To rephrase so it's perfectly clear, $\ce{^195Au}$ , with a half life of 186 days, is the the most stable of the gold isotopes which are unstable. $\ce{^197Au}$ is even more stable than $\ce{^195Au}$, with an effectively infinite half life.

Gold has a number of isotopes. $\ce{^{197}Au}$ is stable and is the only isotope which is. So all the gold that is mined is $\ce{^{197}Au}$.


text:
The Mathematic of finding the Decay Product Compoistion of Pitchblende/Uraninite
Just as the title says.  I’m curious as to how you would go about figuring out the amount of decay products in a given sample of Pitchblende/Uraninite.  I would hazard a guess at using the half life formula over and over but something seems off about doing that.

I don't understand your question. You can analyse a sample and immediately know the composition. Do you want to derive a former composition, e.g. at 10 million or a billion years ago?

He may not immediately know, some decay products will be in trace amounts and some will be very transient.

Karl:  I trying to figure out what a given sample day 10kg or so of Pitchblende would contain in terms of decay products.  I.e. the element composition of Pitchblende as of today assuming it was made when the earth was formed

Well, it is straightforward. You get the amount of radioactive compounds in your mineral. Then find out the decay chain reactions, it generally goes from Uranium to Lead.
http://metadata.berkeley.edu/nuclear-forensics/Decay%20Chains.html

Then you will find half-life constants for each reaction and start to calculate from U-235 to lead one by one.
Well if you think that it is just chore, write a Fortran code and enter radioactive decay constants, reactions and make a user input chain part and everyone can calculate the decay products for a given mineral/time.

Ah.  That was my original method.  But wouldn’t that be incorrect? Taking the half life of Uranium-235 over the existence of the Earth or said material would get you the remaining amount of Uranium-235.  Then using the difference in masses that would be the Thorium generated.  Now doing the half life equation again for thorium makes it sound like it would be like tacting on another existence of Earth or said material.  Am I wrong?

I don't get what is incorrect, and "would be tacting on another existence of Earth or said material".
Radioactive decay rate is independent of the amount of material, find the production of Th and removal of Th in a unit time, then multiply it with (start-end)/unit time.
Maybe that is more proper. Apply it with all remaining chain.

If you found the half life for a kilo of uranium-235.  And then when to find the amount of the thorium that has decayed in said time span, wouldn’t that be like repeating the same timespan but for new material?

Nope, what I said was this. U-> Th -> Pa, let says U halves itself in 1 billion year, Th does it for 500 million.
10 kg U reduces 5 kg in 1 b year, assume nearly 5 kg Th, but in that span Th halves itself twice, and (5kg/2)/2 = 1.25 kg remained.
So it is produced 5 kg and decayed 3.75 kg in 1 b, net 1.25 kg generated. Then you will just multiply it with 4.5 to find what left after 4.5 billions year, 5.625 kg Th is the answer.

I know good sir.  But what I’m saying is when you use the half life equation on thorium, the time used in the equation is it the same time interval as the uranium on or a different one

All the elements have different half-lifes. If you understand previous question you must know this Dalis, and the following equation:
[generation rate(Th)] = [decay constant (U)][time]
[decay rate(Th)] = [decay constant (Th)][time]
[accumulated amount (Th)] = [time]{[decay cont.(U)]-[decay const.(Th)]}
Here all [time] are 4.5 billion years. Is it clearer now?

Ok I will call defeat on answering the question.  I would need a whiteboard to explain it better.  I will use my original method


text:
Do 2-aminoalkanes undergo iodoform test?
I was doing problems in an excercise book and they gave a list of compounds that may undergo iodoform test- which included 2-aminoalkanes.
I dont find any keto methyl gps or methyl carbinol gps that would give iodoform test that normally would give the test. Can someone shed some light here?

Does this answer your question? What is the mechanism for iodoform test given by 2-Aminoalkanes?


text:
Elimination reaction with 1,2-dibromo-4-methylcyclohexane
I was doing a question from a practice exercise on the topic of elimination reactions and alkene synthesis. It is shown below:

I predicted that the product would be 1,2-diiodo-4-methylcyclohexane because I thought the reaction conditions favoured substition with iodide via the $\ce {S_N2}$ mechanism. Firstly, iodide is a rather good nucleophile and secondly, the substrate is a secondary alkyl halide. Even if it was not $\ce {S_N2}$, I would expect perhaps $\ce {S_N1}$ since iodide is a rather weak base and the conditions do not seem to favour elimination. However, the answer provided is very interestingly, 4-methylcyclohexene. Would someone kindly enlighten me on where I went wrong?

I think they should have mentioned heat in the question. 1,2-diiodo when heated undergo elimination to form alkenes.

Iodide is an even better leaving group than bromide. So if anything is to happen, it has to be more complicated than a simple substitution.

Dehalogenation of vicinal dihalides by iodide ion occurs stereospecifically and
  concertedly via an $\mathrm{E2}$ mechanism. An anti-coplanar arrangement of both bromine atoms is required (see diagram below):

Since, the given compound is a 1,2-trans-dibromocyclohexane, two bromo groups  can arrange in 1,2-anti-diaxial positions in one of chair confirmations, fulfilling the requirement. It won't be possible if it'd have been a 1,2-cis-dibromocyclohexane.
Reference: Structure & Synthesis of Alkenes (Chapter 7): Leroy G. Wade, Organic Chemistry; 8th Edn, Pearson Education Limited: Harlow, Essex, UK, 2014.

Here it would also good to redraw important parts yourself.

As you can see from above mechanism the dibromo needs to be in an axial positions for elimination

I see a set of structures and a textual description posted as an image, but not the mechanism itself. Please post text as text, not a screenshot, whenever possible. Also, if you are using a third-party source, also add a proper reference and format the copied part accordingly (e.g. as a block qoute with >).


text:
What kind of mechanism is adopted here?
The relative reactivities of acyl compounds towards nucleophilic substitution is acyl chloride> anhydride>ester>amide.
I would like to know how  this kinetic property is decided. Is it on the basis of better leaving grp or is there another explanation for it?,

Yes. It is often explained by the stability of the leaving group. Chloride is more stable than carboxylate, which is more stable than alkoxide, which is more stable than the amide anion. Thus, the order of reactivity is as such.

@Tan Does stability mean the stability of the leaving grp like Cl- Ion. Cl is more electronegative than the others .Is that why acyl chlorides are more reactive because of the stable leaving grp. Then what about the other 3? How are they stabilised?  How is it 'their' relative orders are decided?

How about arranging all leaving groups according to each of their relative strength as a base? Start with weekest base and end with strongest base. And then, compare with    the order of reactivity (@Tan Yong Boon).

It is likely to be due to the different amounts of partial positive charge which exist on the carbonyl carbon.
We can consider for example the rate of reaction of a given amine (such as aniline) with the acid chloride, acid anhydride and an ester such as the methyl ester. The partial positive charge argument would explain this reactivity series.


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Why atomic mass is taken in comparison?
The definition of atomic mass says that it is the mass of one atom of the element with respect to 1|12 of the mass of carbon-12 atom.
My question is that why there is need to compare the atomic mass with carbon-12 or any other element , why not just we measure the mass of 1 atom of an element and call it as the atomic mass.
Please do not mark this question as duplicate because I have searched the web but didn't find any appropriate answer.
Any help is appreciated.

Please read this reference with historical point of view.

Not sure if I get this question right but if I call the mass of 1 atom hydrogen 1 atomic mass and I also call the mass of one atom helium 1 atomic mass and also 1 atom fluorine is one atomic mass then one atom mass isn't defined at all because every atom weights one atomic mass while in reality they all have a different weight.

@DSVA I think he's just asking why is there's amu at all, and not just in SI unit. OP should put it more clear and not taunt me with supposed unduplicability of the question ;)

https://chemistry.stackexchange.com/questions/77295/why-amu-is-used-as-a-unit-in-relative-atomic-mass https://chemistry.stackexchange.com/questions/29211/what-is-the-mass-of-1-atom-of-carbon

The reason that we compare the mass of atoms to one standard atom is because of the scale of atomic mass.
The mass of one atom is super small. For example, the mass of one carbon atom is $1.99 \times 10^-23$g, which in full is 0.0000000000000000000000199g. This clearly is a very long number, and so is inconvenient in working with.
To address this inconvenience, a standard atom needs to be set to compare others. It was first hydrogen having an atomic weight of 1, though was discontinued since non-integer masses were given for certain elements (e.g. C - 11.9, O - 15.9). Oxygen was then used from its chemical significance, which worked well. However, in 1961 carbon-12 was decided to be the standard between an international agreement of scientists and physicists.


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Detection limits
I am given this question:
A study showed that over 2000 children in China have been poisoned by lead. Blood samples have been taken and show that there is 10 times as much as the acceptable amount of lead in their blood. How can you analys the lead in their blood?
How do I determine which of the following analytical methods is most suitable:
ICP, FAAS, GFAAS and ICP-MS.

Presumably your [$H^+$] really a difference in concentration (also probably multiplied by some negative power of 10) with respect to  some initial condition as a negative concentration does not make sense?

If you have considerably less than 30 ng/ml of Al in your sample, you won't find it with flame atomic absorption spectroscopy. Is that what you want to know? Otherwise please elaborate on your actual problem.

I will edit my question and elaborate on an actual problem where the table has to be used :) @Karl

[...] am I looking for high numbers or low numbers? What do the numbers indicate?

Unless I misunderstood your question, the answer is in the legend on the left side below the periodic table.
The numbers for each element describe the detection limits for particular experimental setups.
You can only use the instruments that you own (or have access to). The periodic table will tell you whether your own machines will reach the intended sensitivity. If not, you might need to send your samples to somebody with a different, more sensitive setup or you have to give up your overly ambitious plans alltogether.

I have edited my question, perhaps it clearer what I am asking about. What does it mean that my machine "will reach the intended sensitivity"? Do I want the sensitivity to be as high as possible or low? Because sometimes our professor chooses a technique with highest sensitivity and sometimes she chooses something in the middle.


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What is the CNC bond angle in methyl isocyanate?

Methyl isocyanate, $\ce{HC3NCO}$, is a toxic liquid which is used in the manufacture of some pesticides. What is the approximate angle between the bonds formed by the nitrogen atom in a molecule of methyl isocyanate?
(A) $104^\circ$ (B) $109^\circ$ (C) $120^\circ$ (D) $180^\circ$

I answered (A), but the correct answer is (C) according to the marking scheme.
I thought it was a bent-shaped, because  the lone pair of $\ce{e-}$ would repel the bond pairs of electron (similarly to $\ce{H2O}$ or $\ce{SO2}$).
My current assumption is that the $\ce{e-}$ in $\ce{N}$ in $\ce{CH3NCO}$ forms $\mathrm{sp^2}$ hybridization (and therefore should forms a shape like ethene).
I'm not sure if my thinking is correct, so hopefully you can help me understand why (C) is correct and why (A) is impossible.

This is not a good exercise question, because none of the options is actually correct. It basically asks you to blindly apply a model which cannot yield the correct result.
Before continuing any deeper into the matter, let's have a look at the actual structure. I was not able to find a crystal structure, so I ran a quick optimisation on the RI-BP86/def2-SVP level of theory. For this purpose, this is accurate enough.

The bond angle $\angle(\ce{CNC}) = 144^\circ$ is right in between $120^\circ \dots 180^\circ$.
The (pure) VSEPR model must fail for this molecule, as it is  only really working for central atoms with distinct and non-delocalised ligands.
For simple cases it performs very well in describing the approximate shape of a molecule. Once you have gotten this down, you can refine predictions with other tools. Please note, the lone pairs take up more space argument is actually a post-rationalisation and should not be used within the framework of VSEPR theory. More reliable predictions can be made with Bent's rule, which says that more electronegative ligands will concentrate p-orbital contributions towards them, decreasing the bond angle (this is a very, very brief summary).
As far as the VSEPR model goes, one should concentrate on simple coordination patterns:

linear; the bond angles will be close to $180^\circ$
trigonal; the bond angles will be close to $120^\circ$
tetrahedral; the bond angles will be close to $109^\circ$
higher coordinations like trigonal bipyramidal, octahedral, etc.

(TL;DR) From that we see that the central nitrogen has three ligands, so the general coordination around it is trigonal and the bond angle should be close to $120^\circ$. This is the answer, which your book will accept as correct, because it blindly applies the VSEPR model.
Applying a bit of Bent's rule won't help much. Carbon is slightly less electronegative than nitrogen, hence one would expect a slight opening up of the angle (more s-character directed towards carbon), but that's not enough to push it to $144^\circ$.
More importantly, the $\ce{-NCO}$ group is highly delocalised. Another very important concept comes into play: resonance. Ignoring the underlying structure for a moment, these are the most important contributors:
$$\ce{Me-N=C=O <-> Me-N^+\bond{3}C-O^-}\tag1\label{eq:resonance}$$
Within the concept of resonance, the positions of the nuclei will not change, and the structures do not exist on its own. A single structure of a set is therefore always an incomplete description. Please read What is resonance, and are resonance structures real?
The left structure in \eqref{eq:resonance} is basically what we have described above with VSEPR. From the depiction on the right, we would expect a $\angle(\ce{CNC}$ of about $180^\circ$, because triple bonds are coordinated linear. Since these are electronic descriptions of the same structure, we can expect the actual (underlying) structure to be somewhere in between these hypothetical forms.
The observed molecular structure is always the result of a delicate balance of effects. So repulsion of nuclei, electronic repulsion, electronic attraction, dispersion, etc. will all have an effect. Many of these interact quite complicated with each other. Sometimes our used models will yield good approximations, sometimes they don't.
In this particular case, we see that VSEPR alone fails; including resonance will refine the prediction. However, without a more rigorous approach, it is impossible to predict the bond angle correctly, i.e. we do not know the contribution of the right structure without doing some more maths. I used a quantum chemical approach, which is obviously nothing you could do in an exam. This is one of the reasons why I dislike this question.

A quick word on hybridisation:
The concept of hybridisation is independent of VSEPR. Many books will throw them together, but it is actually incorrect. You can make a prediction of the shape of a molecule with VSEPR. After that, you can use the molecular structure to choose the closest matching hybridisation model (if any at all). As such, hybridisation is always a result from a particular structure, never a cause. Therefore, your initial assertion is reverse:

My current assumption is that the $\ce{e-}$ in nitrogen in $\ce{CH3NCO}$ forms sp2 hybridization (and therefore should forms a shape like ethene).

Correct is, because the structure around nitrogen is trigonal, the electronic structure can be approximated with sp2 orbitals.

Why $\angle\ce{CNC} \approx 104^\circ$ cannot be correct
In order for this angle to be achieved, the nitrogen first needs to be surrounded by four ligands (or electron pairs), giving us an angle of about $109^\circ$ to start with. Then more p-character needs to go into the bonds, reducing the angle. Hence the ligands must be more electronegative than the other ligands. A popular example is dichloromethane,  or even water (but that is a bit more complicated). See the Wikipedia article on Bent's rule, our Q&A Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?, and multiple other questions with regard to Bent's rule.

Hnmmm. I did not expect 144° at all so TIL... Did you manage to look at hybridisation at N?

@orthocresol It is indeed more extreme than I expected, too, (Spoiler: even more in ~thio~). I don't have NBO or similar program currently available, but with Coulson's theorem, you could approximate this. For $\mathrm{sp}^x$, $x = \sqrt{\left(\frac{1}{-\cos(\angle)}\right)^2}$, ergo $\mathrm{sp}^{1.2}$.

would an x value less than 1.5 indicate that the resonance structure with sp contributes more than sp2?

@Safdar these are quite a crude approximation, so I wouldn't read too much into it. A more involved investigation would be necessary. At first glance though, it appears like you say. However, I don't feel confident enough in these calculations to make that statement. Also, there are probably more effects at play than electronics.

@Martin-マーチン I did not expect to find a crystal structure for methyl isocyanate (after all, a mp of −45 °C contributes to some difficulties [yet orderly crystallization at T lower than RT in a capillary mounted on a diffractometer is possible]), but fond some $\ce{R-N=C=O}$ in the COD to compare with: http://www.crystallography.net/cod/4025444.html, http://www.crystallography.net/cod/1504501.html, and http://www.crystallography.net/cod/1502171.html.

@Buttonwood Thanks. I actually did not think about doing that. Guess I'm too much a computational chemistry to not run a calculation.

The answer is C because double bonds act as only one pair of electrons around the central atom (the Nitrogen) therefore in this case the nitrogen has 2 bonding pairs (one from the double bond and one from the single bond) and one lone pair. As the nitrogen is surrounded by 3 pairs it will take a triangular planar structure around the atom, meaning that is extra repulsion from the lone pair are ignored the best angle approximation for the bond is 120 degrees.

I think that to answer this question, you need to do the process of elimination (A Level multiple choice questions are like this sometimes). Not to mention that the question you supplied looks like it's from an OCR new specification exam (I could be wrong), and hybridisation isn't on the OCR specification so I highly doubt they'd expect students to think about that. My following method of thinking is purely from being taught the OCR specification:

It can't be A, since to achieve a 104° bond angle, you require 2 bonding pairs and 2 lone pairs,
It can't be B, since to acheive a 109° bond angle, you require 4 bonding pairs and 0 lone pairs, and,
It can't be D, since to acheive a 180° bond angle, you require to have a few bond areas and 0 lone pairs (for example, the 2 double bonds in CO2 count as 2 bond areas).

The bonds around nitrogen have 2 bond areas and 1 lone pair, which isn't covered by the OCR specification; this leaves C as the answer due to the fact we eliminated the other answers.

I would vote for C, the C-N-C angle will be slightly smaller than 120 degrees. This is becuase the lone pair is slightly closer to the nucleus than the bonded pairs in sigma bonds. As a result the lone pair occupies a slightly larger solid angle than a bonded pair.
As a result the H-N-H angle of ammonia and the C-N-C angle of methyl isocyanate are slightly smaller than the values which would be predicted by the VSEPR theroy if it is assumed that all electron pairs in sigma bonds / lone pairs are equal.


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Is 3,4-diethyl-3-hexene cis or trans?
I'd really appreciate it if someone could explain this one, please! According to my answers sheet it is not cis nor trans. But it looks like it is cis? Why, according to my answers sheet, is this neither?

Thank you in advance!

It is neither because you have identical groups on either side of the double bond so there is no possibility of isomers

Welcome to Chemistry.SE. We’d like you to take the Tour to familiarize with the site.

As each of the groups surrounding the double bond are all identical there is no optical isomers of this molecule, as no matter which way you rotate or swap groups around you will always have the same molecule. therefore as it has no optical isomers it is neither cis nor trans

@AlkiKaka if you find that this answer satisfies your needs, consider upvoting and eventually accepting this answer so that others can find this also.

your welcome :)


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Utility of proton vs carbon NMR
I've just learnt about both types of NMR, and I haven't ever used NMR in real life, so anything I say may be wrong.
The crux of this question is that, in my own opinion, proton NMR gives you more information than carbon NMR, which I'd imagine to be much more helpful when analysing compounds. Because of this, can anyone tell me whether carbon NMR is ever used instead of proton NMR (which seems to be more 'powerful')? If it is, when?

Good question! The first spectrum an organic chemist will take after isolating the product of a reaction is the 1H NMR. It is quick, routine and informative.  But a synthesized natural product is better .compared with the natural product by the decoupled 13C NMR as well as 1H NMR. The 13C NMR is all singlets and and a field of chemical shifts that is ~20 times that of a 1H NMR spectrum. Chemists are fortunate to have many different tools to identify structures.

TL;DR Yes, sometimes. However, in general 13C data is much less useful than 1H (and 2D) data and the truth is we can often do without it.

I will limit my answer to the context of organic chemistry. Firstly, I directly answer your question: is 13C NMR ever used instead of 1H NMR? The answer is, yes.
However, I can only think of one case where one would do so. This is when there is severe peak overlap in the 1H NMR, such that one cannot reliably use it to ascertain whether one has (for example) a mixture of compound, or just a single compound. In such a case the 13C NMR (which has much more dispersed peaks and hence nearly no overlap) is more informative than the 1H NMR. This is generally not very common.

I want to go beyond a direct answer, though, and talk about the usefulness of 13C data in general. You are correct about the 1H NMR being more helpful than the 13C. This is generally true, and the main reasons for this are twofold:

13C NMR simply does not give as much information as 1H NMR. The only useful thing that one can obtain from the 13C NMR is the list of chemical shifts – but in turn, these are virtually impossible to interpret for all but the simplest molecules. On the other hand, 1H NMR gives you information about proton environments (shifts), number of protons (integrals), and limited information about molecular connectivity (couplings).
As porphyrin wrote, 13C NMR takes significantly longer than 1H NMR, because (a) the natural abundance of 13C is much lower (b) the gyromagnetic ratio of 13C is smaller.

As such, out of all the NMR data that organic chemists collect, the 13C NMR is by far the least useful and most time-consuming spectrum. On its own, its only real use is to provide a "fingerprint" of the molecule, and perhaps some information about functional groups present. So, in most cases, the 13C NMR is indeed less helpful than the 1H NMR.

The question then becomes, "why do people still take 13C NMRs?" Before going on, though, I should make it clear that people do not take 13C NMRs preferentially over the 1H, but rather in conjunction with it.
It primarily comes back to the fingerprint idea – when everybody tabulates 13C shifts of molecules they make, this gives chemist A the information they need to figure out whether they were successful in making what chemist B previously made. You might argue that the 1H is good enough as a fingerprint, but this is not always true, and anyway the consensus is that two is better than one: all major journals require submission of 13C NMR data for novel synthesised compounds.
The only other real case where the absolute 13C shifts are useful are when one needs to resolve very close peaks in 2D spectra (e.g. HSQC, HMBC), or in compounds with very few protons (thus rendering H–H or H–C correlations in 2D spectra less useful). Even then, it is not true to say that the 13C data is preferred over the 1H data.
All in all, your question is perfectly justified. In most cases, the 13C NMR is less useful as the 1H NMR, and only in some limited cases is it as useful as the 1H NMR.
In fact, there is a pretty recent perspective article by Liermann and Schlörer[1] arguing that the 1D 13C NMR should not be recorded, except for in the special cases described in the previous paragraph. The authors also make the point that the 2D spectra serve as much more useful fingerprints and are also much better at validating a proposed structure for a molecule. Even though in crowded spectra the 13C fingerprint may be more useful than the 1H fingerprint, it's still less useful than the 2D data, where peaks are dispersed in two dimensions.

Reference

Liermann, J. C.; Schlörer, N. E. Data handling in NMR facilities and assignment of NMR spectra in synthetic chemistry labs: Why electronic structure validation should become part of the routine. Magn. Reson. Chem. 2017, 1–7. DOI: 10.1002/mrc.4675.

Many thanks for the detailed answer. I asked the question simply because I have to analyse both proton and carbon spectra in exams, and I always end up coming to the conclusion that proton NMR is much 'easier' to analyse due to the information proton spectra give you compared to carbon spectra.

In principle the method is the same for both nuclei but there are practical differences. Proton NMR is approx 4 times more sensitive due to the difference in magnetogyric ratios but more importantly 99.98% of $^1$H are present in a molecule but only 1.1% of $^{13}$C. So unless you synthesise with $^{13}$C, (v expensive etc.) proton nmr is most commonly used.  $^{13}$C can then be used as well if proton NMR is not definitive. Note also that NMR does not generally give you a structure (bond lengths, angles) just the topology, i.e. this atom next to that group etc.

It is possible to get bond angles from proton NMR, if you consider the torsion angle between H-C-C-H then it is possible to use the Karplus equation, it is also possible to use NOESY and other NOE experiments if you deoxygenate the sample before doing the experiment.

I agree, that is why I used the words 'not generally',  some angles using empirical equations and NOESY etc, but its not like doing x-ray diffraction.

Practical aspects are very important when deciding which experiments to run, and Proton 1Ds are by far the simplest and fastest experiment you can run for common organic molecules. 13C experiments take much more measurement time, that alone is a severe drawback. There might be exceptions, but in almost every case a Proton 1D is the first experiment you measure, and the only question is which experiment you run in addition to that one.
There are cases where the 1H experiments are not all that useful, for example if you molecule has very few Protons. You'd probably still run that experiment, just to make sure that there isn't other stuff in your sample that has more Protons.
But one important aspect is that those experiments give you different kinds of information. And then there are many more advanced NMR experiments that give you much, much more than those simple 1D spectra. What you measure depends on how difficult your problem is, and how sure you need to be that you're right. You can get a lot of information from a 1H 1D spectrum, but a lot of it is somewhat indirect. With 2D experiments like COSY, TOCSY, HSQC or HMBC you can directly measure the connectivity between nuclei in your molecule. Each experiment gives you different information, no single experiments gives you all the information about a molecule.

http://www.chem.ucla.edu/~harding/notes/notes_14C_cnmr.pdf
Take a look at this, for example.
When an organic chemist decide to make an NMR analysis, it is mostly to characterize a new synthesized molecule. And (s)/he will do both H1 NMR, C13 NMR, IR, MS and try to understand by looking all of them.
But if you still want to compare or only select one of them, H1 NMR will be much faster and in many cases gives many information, yet for instance when there is no hydrogen bonded to an atom, for especially bigger molecules, not just for the questions of organic chemistry or NMR courses, you will mostly not be able to see the entire molecule by merely H1 NMR. Well, again noone will guarantee that C13 NMR will make you completely solve the structure.
Funny it is, that even there are professors who say that looking solely on MS, if you know enough you will get the entire molecule in the end. In theory you can even say this to IR spectrum since neighboring atoms really have some effect on place and strength of absorption. However, these are some marginal thoughts, consumes very high amount of time if you try to apply and much prone to err compared to having as many analysis data as possible.

1H is probably the most common one, it is fast and it is quite intense for small quantities. You can use it as a fingerprint to compare results or you can determine at least some information from it. Hence it is preferred. But whenever we make a new compound or whenever we are unsure we also make a 13C. The problem is that you require more substance or a higher concentration or you won't be able to see much later. In 13C you cannot 'hide' your carbons as easily as in 1H. And it is sometimes useful because you can see if something is quarternary like in a substituted benzene ring. I think it is less used because it takes more time to measure and is often done during the night so the NMR will be available for others during the day. But that is just an operational factor so I wouldn't count that here. In general, although many groups do not do it when they publish their results, it is of interest to always mention the retention factor for the column chromatography or the boiling point under vacuum, the 1H NMR, the 13C NMR, the IR and the mass spectrum. 13C is also less used in practical courses at universities so students often believe it would be a problem. Same for 2D NMR. The option is always available and shouldn't be a problem if you ask the operator. But this common myth that it is something you wouldn't do because of costs or difficulty is just not true. If you cannot determine your compound you can always ask for the more advanced option. So in the end it depends, yeah 1H is standart because in most cases you know what you want to do but 13C is quite common as well.


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What is the actual mechanism of Fenton reaction?
Fenton reaction was supposedly discovered by Fenton as Tartaric Acid oxidation by hydrogen peroxide in presence of ferrous cation in 1894. Research Paper link.
I have worked on this reaction for years, not an exaggeration. Whenever I mixed $\ce{H2O2}$ with $\ce{Fe^2+}$ solution, or directly solid ferrous sulphate hepta hydrate, I instantaneously got a dark beer-like color, and a vigorously reacting mixture without adjusting $\mathrm{pH}$. $\pu{10 ml}$ of 35% $\ce{H2O2}$ and $\pu{150 mg}$ of $\ce{Fe2SO4.7H2O}$ is mostly what I used. This much peroxide generally defined the reaction as Catalyzed Hydrogen Peroxide Propagations by RJ Watts group. When mixing with 1g soil in first 12 seconds temperature escalated to $\pu{84^\circ C}$, then slowly calmed. Overall bubbling ceased in 60 seconds, what kind of soil did not make any difference. And for $\mathrm{pH}$ measurement, it is very problematic for many reasons so I simply could not get any meaningful results. This just increased my curiosity.
There are experimental and computational studies over its mechanism of reactive oxygen species generation and organic matter degradation. The second one is a bit easier since somehow catching the end products gives the idea, especially with spin trap compounds.
I have lost all of references so I can only give the categories I have encountered.

Majority went for radical species mechanisms and chain reaction, involving hydroxyl, perhydroxyl, superoxide radicals and uses as a base Haber & Weiss,
Haber, F. & Weiss, J. J. (1934) Proc. R. Soc. London Ser. A 147, 332-351.
Yet these reactions are inherently fast, they are just proposed mechanisms.
There is a considerable number of people stand for ferryl ion pathway, which mostly does not involve radicals but unstable iron (IV) oxidation state compound.
Usually computational studies propose this idea, as hydroxyl radicals lifetime is very short for making it as an efficient reactant.

The reaction of Fenton I am asking is reaction between aqueous hydrogen peroxide and $\ce{Fe^2+}$ ion without any ligand other than water. If it does not help for explanation you may very well ignore more than 30 variations like sonofenton.

The dark beer like color is due to iron(III) being formed as perixide oxidizes the iron(II)

The Fenton reaction
The oxidation of organic substrates by iron(II) and hydrogen peroxide ($\ce{H2O2}$ ) is called the Fenton Reaction. It was first described by Henry John Horstman Fenton in 1894 who first observed the oxidation of tartaric acid by $\ce{H2O2}$ in the presence of $\ce{Fe^{2+}}$ ions [1]. After more than 100 years, the mechanism of the reaction is still unsettled. The recent review by Dunford (2002) stated that:

The mechanism of reaction of hexaquo iron(II) with hydrogen peroxide has been unresolved for 70 years. Most scientists, perhaps by default, have accepted the free radical chain mechanism of Barb et al. (1957)[3]. However an earlier proposal involved formation of the ferryl ion, $\ce{FeO2+}$ [4]. Recent work has favored a mechanism involving $\ce{FeO2+}$ and $\ce{FeOFe^{5+}}$ species. Similarly there are differences of opinion on the mechanism of reaction of iron(III), both hexaquo and chelated, with hydrogen peroxide. These differences have fostered a recent burst of activity, with claims on one hand that hydroxyl radicals play a key role, and on the other, that there is a non-free radical mechanism. In contrast, the mechanism of reaction of the heme-containing peroxidase and catalase enzymes with hydrogen peroxide, orders of magnitude faster than reactions of iron(II)/(III), now appears to be well established.

According to the free radical chain mechanism [3], the formation of radicals $\ce{HO^.}$ and $\ce{HO2^.}$ have been proposed. Although, this mechanism has some flaws, it has still been used to explain result in current research as recently as 2014 [5]. The proposed mechanism was depicted in Figure 1, which is followed by several other well received mechanisms for the reaction discussed in reviews sighted.
Mechanism of the redox reaction involved in the Fenton’s reagent (Figure 1)[5]:

Basic reactions and intermediates involved in the classic Fenton and the metal centered Fenton reactions (Figure 2)[6]:

Proposed non-radical mechanism for the Fenton reaction (Figure 3)[6]:

Possible reaction pathways for the Fenton reaction in absence of organic substrates (Figure 4)[6]:

Requirements of the reaction:
$\mathrm{pH}$ should be adjusted to 3-5: if the $\mathrm{pH}$ is too high the iron precipitate in $\ce{Fe(OH)3}$ and will decompose the $\ce{H2O2}$ to oxygen. Basically, the optimal $\mathrm{pH}$ occurs between $3$ and $6$. It's really important to pay attention to the double $\mathrm{pH}$ drop due to the addition of $\ce{FeSO4}$ ($\ce{Fe^{2+}}$ source) and $\ce{H2O2}$. Indeed, $\ce{FeSO4}$ catalyst which contains residual $\ce{H2SO4}$ and the $\ce{H2O2}$ addition is responsible for the fragmentation of organic material into organic acids. Thus, the iron catalyst should be added as a solution of $\ce{FeSO4}$ with slow addition of $\ce{H2O2}$, you may able to control the suddern variation of $\mathrm{pH}$ and the temperature during the reaction. It's better to complete the reaction step by step with a continuous adjustment of these factors.
References:
[1] LXXIII.—Oxidation of tartaric acid in presence of iron: H. J. H. Fenton, J. Chem. Soc., Trans., 1894, 65, 899–910 (DOI: 10.1039/CT8946500899).
[2] Oxidations of iron(II)/(III) by hydrogen peroxide: from aquo to enzyme: H.Brian Dunford, Coordination Chemistry Reviews, 2002, 233–234, 311-318 (https://doi.org/10.1016/S0010-8545(02)00024-3).
[3] Reactions of ferrous and ferric ions with hydrogen peroxide. Part I.—The ferrous ion reaction: W. G. Barb,  J. H. Baxendale,  P. George,  and  K. R. Hargrave, Trans. Faraday Soc., 1951, 47, 462-500 (http://dx.doi.org/10.1039/TF9514700462).
[4] Ferryl ion, a compound of tetravalent iron: William C. Bray, and M. H. Gorin, Journal of American Chemical Society, 1932, 54(5), 2124-2125 (DOI: 10.1021/ja01344a505).
[5] Design Polysaccharides of Marine Origin: Chemical Modifications to Reach Advanced Versatile Compounds: Nathalie Chopin, Xavier Guillory, Pierre Weiss, Jean Le Bideau, and Sylvia Colliec-Jouault, Current Organic Chemistry, 2014, 18(7), 867-895 (DOI: 10.2174/138527281807140515152334).
[6] Fenton reaction - Controversy concerning the chemistry: Krzysztof Barbusinski, Ecological Chemistry and Engineering S, 2009, 16(3), 347-358.

I think that the reaction of hydrogen peroxide with ferrous sulfate under acidic conditions (Fenton type reaction) is rather similar to radiation chemistry in water. In both cases the $\ce{HO}$ radical is important.
$$\ce{H2O2 + H+ + e- -> H2O + HO}$$
While when water it irradated it forms H atoms and HO radicals.
I used to react pivalic acid ($\ce{Me3CCOOH}$) with ferrous sulfate, sulfuric acid and hydrogen peroxide as the first step of the synthesis of $\ce{CyMe4BTBP}$. This forms $\ce{HOOCCMe2CH2CH2CMe2COOH}$ which is formed by dimerisation of the $\ce{Me2C(CH2)(COOH)}$ radical. This radical in turn is formed by the abstraction of a hydrogen atom from the pivic acid.

I have also found this article regarding your method:
https://www.sciencedirect.com/science/article/pii/1350417795000356
Quite reasonable in any aspect. I also have tendency to look this as a radical reaction. But with it I saw some specific details as well, like how hydrogen peroxide displace a water ligand and when there is an electron transfer occur from what oxygen etc.