text:
Why is the enthalpy of a reaction equal to the difference between the enthalpies of combustion of the reactants and the products?
My textbook gives me the following formula for calculating the enthalpy change of any reaction:
$$\Delta H_\mathrm{r}^\circ=\sum(\Delta H_\mathrm{c}^\circ)_\text{reactants}-\sum(\Delta H_\mathrm{c}^\circ)_\text{products}\label{b}\tag{0}$$
where $\Delta H_\mathrm{c}^\circ$ is the standard enthalpy of combustion. Note that the enthalpy of combustion for same elements must be to the same product (i.e. all the elements of nitrogen in whichever compound they are, should be combusted to the same nitrogenous oxide)
My textbook simply assume it as a "fact". However, I have yet not found an intuitive or a mathematical way to prove it.
For starting the proof, I would write this balanced equation involving an arbitrary hydrocarbon:
$$\ce{C_xH_y + H_2 -> C_xH_{y+2}}\label{a}\tag{1}$$
The next step is obviously to write their standard combustion reactions, like this one for the first reactant:
$$\ce{C_xH_y + $\left(x + \frac{y}{4}\right)$O2 -> xCO2 + \frac{y}{2}H2O}$$
and then another two for the other two compounds, and then show that, by Hess's Law, they can be remodeled into the original equation by simple arithmetic.
However, there are certainly a lot of problems with this approach:

What if reaction $\ref{a}$ had more than one reactants, or more than one products? How do I write a proof for the general reaction involving $m$ reactants and $n$ products?
What if the reactant wasn't a hydrocarbon but rather an arbitrary organic compound like $\ce{C_xH_yS_zN_pO_q}$ with many more different elements. How would we write a proof then?

I find that in trying to approach a proof for a general reaction, I have made the situation extremely complex. I wonder if an intuitive or a mathematical proof of such a complex equation is even possible. And if it is not, then how can we even assume equation $\ref{b}$ as true? Or, is it that the reaction $\ref{b}$ does not apply to all organic reactions, but only a subset of them?

PS: While I was experimenting with different equations, I stumbled upon $\ce{N2 + 3H2 -> 2NH3}$. Here, if you write out the individual combustion equations and balance them out, you'll get $\Delta H_\mathrm{r}^\circ=(\Delta H_\mathrm{c}^\circ)_{\ce{N2}} + 3(\Delta H_\mathrm{c}^\circ)_{\ce{H2}}-2(\Delta H_\mathrm{c}^\circ)_{\ce{NH3}}$ as expected from the general formula. However, the interesting thing to note here is that it does not matter whether you combust nitrogen to nitrous oxide, dinitrogenpentoxide, or perhaps any other nitrogenous oxide. In the end, it will all balance out and give you the answer you're looking for. (I call them self-balancing equations.)

Source: KS Verma; Physical Chemistry for JEE (Advanced): Part 1; Chapter - Thermodynamics Illustration 6.56III

I think the heat of reaction should be net "Heat of Formation" ($\ce{$\Delta $H_f} $) of products  minus that of reactants.

@SoumikDas of course you're right. I never contradicted that. In fact, there's also a third way: $\Delta H_r = \text{Bond energy of reactants} - \text{Bond energy of products}$ but I am not asking for its proof in my question, am I? :)

I know the formula seems unconventional, but, apart from that, if there is any other problem in the question, please let me know. Thank you.|| For the people who VTCed, (1) could you please explain what exactly is "unclear"here? (2) if you think the question is broad, why do you think so? Is that because the proof for the formula is long, or that the proof does not exist? In either case, it'd be more helpful if you could please show how the formula is unproveable/difficult to prove, instead of just implicitly  assuming that it is. Thank you.

the simple intuitive way to get the result is to recognise that, if it were not true, there would be some cycle of reactions that could break the law of conservation of energy.

@matt_black Oh that's interesting! But, how does it explain whether $\Delta H_\mathrm{r}^\circ=\sum(\Delta H_\mathrm{c}^\circ)\text{reactants}-\sum(\Delta H\mathrm{c}^\circ)\text{products}$ or $\Delta H\mathrm{r}^\circ=\sum(\Delta H_\mathrm{c}^\circ)\text{products}-\sum(\Delta H\mathrm{c}^\circ)_\text{reactants}$? (the order of the terms)

Second approach, but very abroad assumptions. Assume a reaction
$$\ce{reactants -> products} \tag{1} \label{rxn1}$$
where $\text{A}$ and $\text{B}$ are not necessarily one compound each. Let's analyze element $x$ that appears in the reactants as $x_{a}^{\alpha}$, and appears in products as $x_{b}^{\beta}$, where $\alpha$ and $\beta$ are the oxidation states, and $a,b$ are the number of atoms of the element in each case.
Then, combustion of $x$ can be written as:
$$\ce{ $x_{a}^{\alpha}$ + $\frac{ay}{2c}$ O2 -> $\frac{a}{c}$ $x_{c}^{\gamma}$O_{y}}$$
The enthalpy of this reaction is
$$\Delta H_\mathrm{c}(x_a^\alpha) = \frac{a}{c}\Delta H_\mathrm{f}\left(x_{c}^{\gamma}O_{y}\right) -\Delta H_\mathrm{f}(x_a^\alpha)  $$
Analogously, and assuming the product of combustion is the same:
$$\Delta H_\mathrm{c}(x_b^\beta) = \frac{b}{c}\Delta H_\mathrm{f}\left(x_{c}^{\gamma}O_{y}\right) -\Delta H_\mathrm{f}(x_b^\beta)  $$
If the stoichiometric coefficient of $x$ in the reaction \ref{rxn1} is $m$, the stoichiometric coefficient in the products will be $\frac{am}{b}$
The difference of enthalpies of combustion (taking into account stoichiometric coefficients of \ref{rxn1}), is:
$$
\begin{align}
\Delta H_\text{rxn} &=  m\Delta H_\mathrm{c}(x_a^\alpha) - \frac{am}{b}\Delta H_\mathrm{c}(x_b^\beta)\\
&= \frac{am}{c}\Delta H_\mathrm{f}\left(x_{c}^{\gamma}O_{y}\right) - m\Delta H _\mathrm{f}(x_a^\alpha) - \frac{am}{b}\left(\frac{b}{c}\Delta H_\mathrm{f}\left(x_{c}^{\gamma}O_{y}\right) -\Delta H_\mathrm{f}(x_b^\beta) \right)\\
&= \frac{am}{b}\Delta H_\mathrm{f}(x_b^\beta) - m\Delta H_\mathrm{f}(x_a^\alpha)
\end{align}
$$
Which is the definition we all know. Considering this for all elements in the reaction, then the formula is true. Now, taking this with a grain of salt, since I haven't found a way to prove it for cases where the combustion product is not an oxide (I got a feeling it can be done with charge balance, that's why I was considering the oxidation states, but I haven't gotten to a proof for that case, I left them in case someone else has an idea based on that).

This is an application of Hess' Law which states that the overall enthalpy change of a reaction is independent of the route taken, which essentially comes from the conservation of energy.
It enables the enthalpy change of a reaction, which does not take place directly, to be easily measured.
In many cases, the reactant and product can be burnt and the enthalpy change measured by calorimetry. Because the reactant and product are made up of the same elements, the products of combustion will be the same.
Enthalpies of formation are often difficult to measure directly so you can go through the following steps:

Write down the equation you are interested in.
Underneath, write down the combustion products, which will be the same.
Apply Hess' Law

In general you get:

You can see that, from Hess' Law, the energy change of the green route must be the same as the energy change of the red route, since the arrows start and finish at the same place.
The enthalpy of formation of methane is a good example since carbon does not react with hydrogen directly to give methane. However carbon, hydrogen and methane can be burnt easily and the enthalpy changes measured.
The Hess Cycle looks like this:

Applying Hess' Law:
$\rm \Delta H_f+(-890)=-394+2(-288)$
$\rm \Delta H_f-890=-960$
$\rm \Delta H_f=-960+890$
This where your formula comes from.
$\rm \underline{\Delta H_f=-70\,kJ/mol}$
A graphical method is often better than trying to remember and apply formulae.

Maybe not an answer, but I think the idea may go like this. It is sort of a Born-Haber like cycle. For the reaction of $X$ compound giving products $Y$, we take each element $x_i$ of $X$ and oxidize them to a "high" oxidation state intermediate $I$ (combustion of $X$). Then, we reduce $I$ to the oxidation state $y_i$ of that element in product $Y$ (reverse reaction of the combustion of $Y$). Since $I$ is the same for both reactions, as long as both combustion (or reverse combustion) reactions are balanced by mass, the difference between them will be the same as the difference between $X$ and $Y$, which is the enthalpy of the reaction. Maybe the figure explains it a bit better.

(not my downvote) I think the figure contradicts what you're suggesting. You say that "we reduce $I$ ... in product $Y$" But, in the figure you've shown combustion of $I$ to $Y$. Isn't combustion always an oxidation instead?

Yes, but we're doing the opposite to get to $Y$ (that's why it's the difference of heat of combustions, not the sum).

That's interesting. But how do you know that the moles of the intermediates will cancel out in the end though?

@GaurangTandon see my new answer, it doesn't cover all cases, but I think it covers a few at least.


text:
Single crystal LiF solid state NMR
I haven't been able to find any literature on $\ce{^{19}F}$ solid state NMR studies of $\ce{LiF}$ crystals, not even $\ce{LiF}$ powders. Is there any NMR chemists who is familiar with $\ce{LiF}?$ Any references are greatly appreciated.

You can get this reference free: !http://esl.ecsdl.org/content/14/9/A134.abstract

Here is it: !http://esl.ecsdl.org/content/14/9/A134.full.pdf+html

Another one is here. Hope these are helpful: !https://www.sciencedirect.com/science/article/pii/S0022459608004751

There are quite a few refs for this compound e.g.
J. Phys. Chem. C20091133315018-15023
Rob

This is not a directly responsive answer (mea culpa) but I want to give you any help I can. LiF is a very well known reference standard for X ray powder diffraction. Perhaps parsing this literature would still help you with the problem you are working on. Again mea culpa, not a direct answer...but giving what I can since nothing else has helped so far. Take a look at the powder diffraction literature. May get some comments or insights that help you.

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text:
Hottest region of Bunsen flame

Since region 1 is closer to the source, I presume it to be the hottest as complete combustion takes place there. Also, this is the part where the gas mixture(responsible for flame) reacts with oxygen first(as soon as valve is open). So, the reaction should take place much quicker and more heat should be released. But in my book, the answer is given as region 2. Where am I wrong?
I know there exists as question similar to this in our community : Bunsen burner and hottest part but the answers don’t answer my question.

There is more than one reaction happening ;)

Of all things, complete combustion is not what happens in region 1.

No, that's irrelevant. Burn pure methane, and the pattern will be the same.

@IvanNeretin Why not? We have an open valve and availability of oxygen.

From Dave’s and MaxW’s comments and answers, my take is that gases near region 1 are just coming out of tube and so they are relatively more cool. The rate of reaction is less than the mixture velocity, so the statement “ No unreacted mixture is present in region 1 “ is false. It is because of this mixture (which still has to react), that region 1 is not enitrely combusted and is relatively more cool than point 2(where there is no unreacted gas mixture). Correct me if I’m wrong and thank you very much for your time.

The question itself is poorly written as given.
The diagram should have "regions" and "points."

So what seems to be labeled "Region 2" and "Region 4" I would name as "Point 2" and  "Point 3" respectively.
So the innermost conical area would be "Region 1", the next conical area would be "Region 2" and the third conical area would be "Region 3."
Region 1 is where the mostly unburned gas and oxygen mixture are pushing above the lip of the Bunsen Burner. Region 1 exists because the gas coming out of the tube is cool. If the air port is open and the gas flow is too low then the gas will start to burn down the tube and you'll get a "strike back" where the flame is either (1) blown out or (2) burns at the jet. If the gas flow is too great you can blow the burning region off contact with the upper tube. If you increase the gas flow even more then you can in fact blow the flame out.
Region 2 would be a reducing region within the flame. This region is hot and burns the fuel and oxygen coming out of the tube.
Region 3 would be a oxidizing region of the flame. Here oxygen from "outside air" (oxygen which didn't come up the tube) is migrating into the flame to burn the excess fuel which is not bunt in region 2.
You'd use the reducing and oxidizing regions when doing bead tests for identification.
Point 2 would be the hottest part of the flame as shown by the composite image of a paperclip in the flame from the YouTube video.

Oh yes, sorry about the labelling and use of language. Same doubt! Sorry if it is stupid,  but why should there be unreacted mixture when the reaction occurs immediately and continues in the same trend as more gas is supplied?

@MollyCool - I'm not sure exactly what your asking. I think you're asking about Region 1. Region 1 exists because  the gas coming out of the tube is cool. If the gas flow is too low the gas will start to burn in the tube and you'll get a sort of "backfire" which then blow the flame out. If the gas flow is too great you can blow the burning region off contact with the upper tube. If you increase the gas flow even more you can in fact blow the flame out. // Does that answer your question?

Yes I am referring to region 1 but you said unreacted mixture exists in that region. I’m confused on why there is unreacted mixture when flame is continuously burning. Entire amount of the gas mixture burns to produce the flame and it is being continuously supplied to keep it lit. So as the gas flows through tube and reaches region 1 by this time it should have reacted to produce flame right?

No, for Region 1 the gas is mostly unburned. Region 1 is cool because it is just coming out of the tube which is cool. Look at the Meker–Fisher burner. Without the metal grid, which is cooled by the cold gas coming up the tube, the very wide top would allow the flame to go down the tube. The wire grid serves as a boundary between the hot gases burning in the flame and the cool mixture coming up the tube. So there is essentially no "region 1" for this type burner, only regions 2 (reducing) and 3 (oxidizing) as shown.

Alright! Yes I get that about this burner. Question is specifically on bunsen burner though. Just saw the video you attached. Seems like even if I put my finger in that region, it wont hurt me much :D. Thank you for your time.

@MollyCooL You seem to be assuming that the gases instantly and completely react as soon as they come out of the tube. They don't.

FWIW I think the original question would have been clearer if they had simply used markers A, B, C and D and not used the word "region" at all.

@David Yes, I was just wondering why shouldn’t they.

Region 1 is coolest... How much temperature does cool here signify? Is it cool enough to be touched with bare hands?

I did some search in internet and in case of candle flame, point 3 will be the hottest region?

Region 1 is much cooler because it contains a mixture of gases that it unreacted.
The gases in the tube are traveling too fast for the flame to enter the tube.
The inner cone is this unreacted gas mixture as it exists the tube.

Oh! But aren’t the reactions too quick That the entirety of the gases react and there is nothing like “unreacted”?

@MollyCooL We know that the reaction does not propagate as quickly as the gases move within the tube, because the flame does not enter the tube.  The mixture in the tube already includes air, so it would react within the tube if the reactions propagated faster than the mixture velocity.  For more detail see: https://nvlpubs.nist.gov/nistpubs/jres/60/jresv60n6p535_A1b.pdf

I get that the reaction rate is slower than mixture velocity in the tube. Is this true even at the inner cone (region 1) where the gas is technically out of the tube?

@MollyCooL yes, because it can not slow instantaneously.  It slows as it interacts with the surrounding, stationary, air.  The center of the gas stream moves the fastest.

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text:
I'm trying to make an optical dissolved oxygen sensor. I need help finding an appropriate oxygen sensitive dye
I have been trying to make a dissolved oxygen sensor, the basics are described on this page.
I have a good understanding of the electronics behind this measurement, but I cannot find a suitable oxygen sensitive dye to use. I have read quite a bit about room temperature phosphorescence, and thought that erythrosin b might but a suitable candidate, it has an excitation wavelength of 525 nm, and radiates in near-IR. After more reading, I found that it needs to be in a sol-gel or similiar substrate. All of the reading I've done seems to require some sort of substrate and this isn't what commercial dissolved oxygen sensors are using. They appear as just a thin, hard disc. Additionally, I don't have any equipment or experience with sol-gels.
I'm looking for any information on what chemical and process commercial systems are using. Or, a simple way to get erythrosin b to work. My tests with it so far have not shown any phosphorescence at all.
Thanks for your time.

This review article might be useful for your purpose:
Title: Optical methods for sensing and imaging oxygen: materials, spectroscopies and applications
Chem. Soc. Rev., 2014, 43, 3666-3761 (DOI: 10.1039/C4CS00039K)
!http://pubs.rsc.org/en/content/articlehtml/2014/cs/c4cs00039k
This is a review article with 694 references (96 pages). The discussion focused on main sections such as direct spectroscopic sensing of oxygen, absorptiometric and luminescent probes, polymeric matrices and supports, additives and related materials, spectroscopic schemes for read-out and imaging, and sensing formats (e.g., waveguide sensing, sensor arrays, multiple sensors and nanosensors). They also discuss future trends and applications, and concluded it with summary of the properties of the most often used indicator probes and polymers.
There are several probes has been discussed. Some of them may be useful to develop oxygen sensors: Those include: a) Hemoglobin and myoglobin as optical probes for oxygen (the affinity of hemoglobin for oxygen is a function of pH and ionic strength); b)  Molecular absorptiometric probes for oxygen (e.g., several cobalt–organic compounds are capable of reversibly binding molecular oxygen); c) Methods based on the oxidative power of oxygen (the reduced or "leuco" form of Methylene Blue (MB) is colorless and quickly converted back
to blue MB by oxygen); etc.
PS: Selected irreversibly acting chromogenic probes for oxygen along with respective polymeric supports and color transitions, and Organic fluorescent probes for oxygen were also given in Table 2 & 3, respectively.

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text:
How difficult is it to synthesize Novichok agents?
In the light of the current UK-Russia events, how difficult is to synthesize Novichok agents (with known structures)? Is it something that can be made at any well-equipped university and/or chemical factory, or does it involve some very specialized or possibly secret knowledge?

"Does it involve some very specialized or possibly secret knowledge..." hmm... making Schedule I chemical weapons of mass destruction in violation of the Chemical Weapons Convention is not illegal only in your nation, but also under international law. As for Novichoks, their specific chemical structure is probably unknown but to the Russian Government and some insiders. All we know is that they have a organophosphate group.

Well, Russians deny that they are the source of the chemical. What I want to know is how plausible is that some other random sources exist. At least some of the structures are publicly known - there is an article on Wikipedia.

Some of the starting materials are controlled substances and are also very toxic.

For well equipped university and labs is, chemically speaking, surely possible. This must be as for any other molecule which synthesis has been reported (explicitly or not). Safety and economical consideration let apart.

It isn't the ease of synthesis that is the problem with nerve agents: it is the expertise required to not kill yourself and your colleagues while making them.

According to Prof. David Collum, it's not that hard so synthesize.

According to Fragmentation pathways and structural characterization of
organophosphorus compounds related to the Chemical Weapons
Convention by electron ionization and electrospray ionization
tandem mass spectrometry Rapid Commun. Mass Spectrom. 2016, 30, 2585–2593:

N-[Bis(dimethylamino)methylidene]-P-methylphosphonamidic
  fluoride (3) was synthesized by the controlled
  addition of N,N,N′,N′-tetramethylguanidine to a solution
  of methylphosphonic difluoride (1). N,N,N′,N′-Tetramethylguanidine (0.60 mmol) and triethylamine (0.65 mmol) in dichloromethane (DCM) (500 μL) were
  added slowly into the solution of methylphosphonic difluoride
  (0.40 mmol) in DCM (500 μL), while stirring at 0–5 °C. After
  30 min, the resulting precipitate was filtered off and the solution was analyzed by GC/MS.
The corresponding O-alkyl N-[bis(dimethylamino)-methylidene]-P-methylphosphonamidates 4 were synthesized by the addition of ROH to a solution of cpd 3. Sodium hydride powder (NaH, 60% in mineral oil, 0.5 mmol) was washed with dry hexane twice. The hexane was removed and DCM (500 μL) was added and the solution was kept stirring at 0–5 °C. The appropriate alcohol (0.5 mmol) was slowly added into the suspension and the reaction solution was stirred at 0–5 °C for 30 min. Compound 3 in DCM (200 μL) was then added dropwise to the solution, while stirring at 0–5 °C for 2 h. Any precipitate was filtered off and the resulting solutions of the end product were analyzed by GC/MS. It should be noted that, due to the extreme toxicity of these materials, the separation and purification of CWC-related chemical are very difficult and therefore should be carried out only by a trained professional in an efficient fume cupboard equipped with an active charcoal filtration system.

Methyl phosphonic difluoride is not something you can just pick up at the corner store

@Waylander you can't readily buy it because it is schedule 1 for CWC https://www.opcw.org/chemical-weapons-convention/annexes/annex-on-chemicals/schedule-1/ But you can buy the dichloride https://www.sigmaaldrich.com/catalog/product/aldrich/228052?lang=en&region=US and react with anhydrous HF.

You can, and it is a good test of your handling skills as it is seriously toxic in its own right

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text:
Difference between Covalent and non-Covalent interactions
It is a general concept that non-covalent interactions are a result of electrostatic interaction while covalent interactions arise due to sharing of electron density. This, however, is not true since I have myself performed one study (link) proving that hydrogen bonding does not consist purely of energy from electrostatic sources.
Another popular method to look at the same is to study the energetics, however, the energy of strong non-covalent interactions (like strong hydrogen bond) does fall in the range of covalent interactions and is sometimes enough to break some low-energy covalent interactions.
Hence, I am willing to know about how to distinguish between a covalent and non-covalent interaction?
I would prefer some answer in terms of wave-function and electron densities with proper reference.

As far as your first paragraph goes, it's a matter of nomenclature. Hydrogen bonds are partially covalent so they aren't really "non-covalent" - your categorisation isn't good.

For the gray areas between covalent and non-covalent interactions, see e.g. https://science.sciencemag.org/content/371/6525/160

This paper is also useful...
https://doi.org/10.1039/B704781A


text:
The confusion of entropy change
The entropy is defined as randomness of a substance or a system.
The randomness increases as we go from solids to liquids to gases. Similarly from hard diamond to soft graphite. Since graphite is soft and slippery, the orderness would be less than diamond.
But acc. to the formula for entropy change during phase transition,
$\Delta S=Q\ \text{(at const. pressure)}/T$
$\Delta S=\Delta H/T\ \text{(at const. pressure)}$
But $\Delta H$ of reaction/process
C graphite --> C diamond is positive
Meaning $\Delta S$ is positive. But it should be negative due to less entropy of diamond.
Where is the problem?

I'm not sure the numbers are right in this quiz I found, but if you run the numbers on the entropy at different temperatures, the entropy does change sign, i.e. at phase transition temperatures, the entropy of diamond is higher than that of graphite (for some reason).

Why do you think hardness should be associated with entropy? Graphite actually has an interested crystal structure, its not just amorphous: https://en.wikipedia.org/wiki/Graphite#Structure

Even if graphite has interesting structure, doesn't it will still be less ordered than diamond. Considering the tight packing of 4 bonded C atoms. I am confused.

Maybe the standard states are not equilibrium states with one another.  I would think that the corresponding equilibrium states would be at a much higher pressure than the standard state of 1 atm.

If $\Delta$S had been negative and as $\Delta$H is positive, the reaction wouldn’t occur, as $\Delta$G would be positive.

Yes, well the reaction will become non spontaneous right?

The standard free energy of formation of diamond is 2.9 kJ/mol and the standard enthalpy of formation of diamond is 2.0 kJ/mol.  So the standard entropy of formation of diamond (relative to graphite in the standard state is -0.003 kJ/mol.K.  So graphite and diamond are not in equilibrium in the standard state.  The thermodynamics favors graphite.  So, if diamond is present in the standard state, it is converting to graphite.  But the kinetics of the conversion is extremely slow.  This is why $\Delta S$ is not equal to $\Delta H/T$.  See my answer below.

The standard free energy of formation of graphite is $0\ \mathrm{kJ/mol}$, and the standard free energy of formation of diamond is $2.9\ \mathrm{kJ/mol}$.  So, at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{bar}$, graphite is favored thermodynamically over diamond.  Let's see what pressure would be required for $\Delta G$ to be zero at $25\ \mathrm{^\circ C}$.  Since the temperature is constant, the change in $G$ is described by $\mathrm dG=v\,\mathrm dP$, where $v$ is the molar volume.  The density of graphite is $2.26\ \mathrm{g/cm^3}$ and the density of diamond is $3.51\ \mathrm{g/cm^3}$.  So the molar volumes are $5.31\times 10^{-6}\ \mathrm{m^3/mol}$ for graphite and $3.42\times 10^{-6}\ \mathrm{m^3/mol}$ for diamond.  So, at pressure $P$ (in Pa), the Gibbs free energy for graphite and diamond are:
$$G_\text{Graphite}=0+5.31\times 10^{-6}P$$
$$G_\text{Diamond}=2900+3.42\times 10^{-6}P$$
So, $\Delta G$ at pressure $P$ is $$\Delta G_\mathrm{G\rightarrow D}=2900-1.89\times 10^{-6}P$$So the change in free energy is zero at $P=1.53\times 10^{9}\ \mathrm{Pa}=15\,300\ \mathrm{bar}$
Diamonds are formed naturally at even higher pressures and higher temperatures.  But this is the approximate order of magnitude.
So at $25\ \mathrm{^\circ C}$, graphite is highly thermodynamically favored over diamond.  The actual presence of diamonds at $25\ \mathrm{^\circ C}$ must therefore be a matter of kinetics, rather than thermodynamics.

2.9 kJ isn't particularly high value.

@Mithoron true but takes even less to form graphite


text:
Which is more acidic?

Which among these hydrocarbons is more acidic?

Here, I thought about hyperconjugation. For this, I counted the alpha H i.e. the number of hydrogens attached to the carbon attached to a sp2 hybridized carbon. What I get is, (C) has most no. of those Hydrogens. So, my assumption is that  it should be more acidic. But that's not the case. Where am I going wrong?

Interestingly, why do you think more hyperconjugation means the compound is more acidic? Where did you even pick that thought from? I earnestly suggest you drop that thought immediately, it has more problems than we can discuss. Acidity is always directly proportional to stability of the conjugate base (barring some very rare cases).

Acidity is directly proportional to the stability of conjugate base. Aromatic compounds are very stable.
In essence only one compound after deprotonating is aromatic - which is the conjugate base of (A).

The conjugate base of compound A becomes aromatic post deprotonation:

This is essentially a cyclopentadienyl anion derivative, that has 6 $\pi$ electrons that are delocalised over the ring. Thus this compound will be by far the most acidic.
As for the other cyclopentadiene derivative, it cannot be deprotonated to make it aromatic: it is localised on an $sp^2$ hybridised carbon atom, and the electron pair will thus be in the plane of the ring itself.
A similar logic can be extended to the methylated tropylium derivative. As for the tropylium derivative that can be deprotonated to give a delocalised electron system, it will have $8\pi$ electrons and thus will be antiaromatic. The molecule will thus prefer to be non-planar instead of being destabilised by the antiaromatic effect.


text:
Salt Bridges/Porous Disks
I am slightly confused about the concept of a salt bridge/porous disk. I understand that these are necessary in a galvanic cell since there is a charge buildup and the salt bridge and porous disks allow ions to flow in order to neutralize this charge and allow for the electricity to continue flowing.
What I am confused about is what ions are flowing. As an example I'm going to talk about a zinc-copper galvanic cell in which the zinc is oxidized at the anode and the copper is reduced at the cathode. What ions flow through the porous disk? Is it some of the $\ce{Zn^{2+}}$ ions and the anion from the copper solution, or is it separate ions? Is this the same in a salt bridge or is the salt used in the bridge the one that moves?
For example, if a salt bridge is made with $\ce{KCl},$ does the $\ce{K+}$ go to the cathode and the $\ce{Cl-}$ go to the anode or do the $\ce{Zn^{2+}}$ ions move through the salt bridge?
In other words, what ions flow through a salt bridge? Is it the ions from the separate solutions in the two half cells or the ions inside the salt bridge? What about a porous disk that has no ions inside of it?

It depends on the geometry/construction of the bridge. In a typical glass-tube bridge, some small amount of the ions from the cell may eventually make their way all the way across the salt bridge, but over reasonable amounts of time, it's mostly the ions from the salt bridge that move into the half-cells (and the ions from the half-cells moving into the ends of the bridge). For a small filter paper strip being used as a bridge, the time it takes should be shorter, as there is much less solution in the salt bridge and the distance is likely shorter. Sometimes, glass salt bridges are filled with gels to limit diffusion further (at the expense of increasing the resistance of the bridge).
The goal of using a salt bridge and separating the half-cells is to prevent the solutions of each from directly mixing. If significant quantities of the ions from the half-cells are crossing the bridge, then they can react directly, bypassing the electrode connections. e.g. if we have a simple galvanic cell of $\ce{Zn∣Zn^2+∥Cu^2+∣Cu}$ and $\ce{Cu^2+}$ makes it across the separator to the zinc size, it can be directly reduced onto the zinc electrode, instead of by the electrons flowing through the electrode connections. Having a single beaker with both $\ce{Cu^2+}$ and $\ce{Zn^2+}$ would thus make a poor galvanic cell as most of the reagents would be wasted reacting directly, rather than doing work in the outside circuit.
With a simple porous separator, there are no other ions than the ones in the half-cells, so naturally, these are the ions that flow across, with the goal simply to prevent bulk mixing of the two solutions.

The motor is being powered by a simple galvanic cell that uses a porous cup rather than a salt bridge to maintain the separation of copper and zinc ions. The separation of ions in crucial to the reaction as a mix between the two would cause an immediate reaction and would not require a lengthy movement of electrons across the wire which is what powers the battery. In this lab, the copper pennies are acting as the cathode where they in a Cu(II)SO4 solution making this the reduction part of the reaction. The opposite is true for the zinc washers, acting as the anode, which are placed in the ZnSO4 solution making this the oxidation part of the reaction. The anions are flowing towards the anode (zinc washers) to balance the out the positive ions (cations) that are being released by the anode and into the wire, into the motor, and into the cathode.


text:
Transformation of an epoxido-carboxylic acid to an aldehyde

So I am battling with the last part of this transformation and was hoping someone could explain this mechanism to me so that I can confirm my answer.
Treating an alcohol with a strong hydrohalic acid results in the formation of alkyl halide and proceeds through an SN1 mechanism. (Protonation of alcohol on the carboxylic acid, then loss of $\ce{H2O}$ to form a carbocation) But $\ce{HCl}$ is supposed to give a substitution reaction and not an elimination reaction. So I am unsure how the final product is formed.

Looking at the product, we can see there's a decarboxylation here due to the "heat". I bet the $\ce{HCl}$ is only to perform acidic cleavage of the epoxide. Though I am unsure how to knit both of them together.

Previously addressed. https://chemistry.stackexchange.com/questions/86726/epoxide-ring-opening-via-decarboxylation/86729#86729


text:
Safest and easiest method to neutralize nitric acid
I would like to know the safest method for neutralizing 70% nitric acid that is used in small amounts of 1-5 ml, for an electronics process in which there will be small traces of silicone and other similar trace materials dissolved into it, to safely contain for proper disposal to not be a risk for acid accidents or fire/explosion risks. Should a calcium carbonate-water mixture be used to increase $\mathrm{pH}$ slowly or is this method still unstable from higher purity of $\ce{HNO3}$ being involved or?

Dilute it with water then cautiously add carbonate until no further foaming occurs.

When diluted with at least 10 times its volume of water, nitric acid is not a fire/explosions risk any more, and it is not a risk for acid accidents.  Trace materials have no more effect on such diluted solutions.


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Why do bicarbonates tend to be more soluble in water than carbonates, for Group 1 and Group 2 elements?
In Group 1 and Group 2, it is usually found that bicarbonates tend to be more soluble in water than carbonates. For example: $\ce{NaHCO3}$ has higher solubility than $\ce{Na2CO3}$
As most of the trends in s block elements are explained by lattice enthalpies, I think it has something to do with lattice enthalpy, but I’m not sure how we can say lattice enthalpy of $\ce{Na2CO3}$ is higher than $\ce{NaHCO3}$.

Grouo I left out. You will likely reverse the question but bicarbonate is less soluble than carbonate, molarity accounted too

@Alchimista Sorry but what do you mean by "Grouo I left out"?

@Gaurang Tandon.  Group 1 let out.  NaHCO3 is less soluble than sodium carbonate. That is also why the only answer provided until now is not correct.

To increase the chance to get an answer you could pose a completely right question. ..


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What anions does the loss of a proton from HOCN and HNCO produce? Are they same or different?

What anions does the loss of a proton from $\ce{HOCN}$ and $\ce{HNCO}$ produce? Are they same or different?

Once, I had asked my teacher if $\ce{HCN}$ and $\ce{ HNC }$ are different or same compunds, he said they were different and yes, they had different structures. Applying the same logic here, $\ce{OCN-}$ and $\ce{NCO-}$ are different anions because they'll have different structures.

Slightly related: https://chemistry.stackexchange.com/questions/53568/what-is-the-structural-formula-of-alkali-hypohalite-mox-or-mxo

When they are bonded to hydrogen that is $\ce{HOCN}$ $\ce{HNCO}$. The two compounds will be different:

From Wikipedia

Organic cyanates are called isocyanates when there is a $\ce{C−NCO}$ bond and cyanates when there is a $\ce{C−OCN}$ bond.

Due to resonance after deprotonation the two anions will be the same.


text:
Why does salty water make hard boiled egg shells more fragile?
Boiling eggs in salty water rather than regular water makes the salt react with egg shells making them more fragile. My guess is that it's either sodium or chloride reacting with the calcium in eggs. Can someone please explain what is going on in there, and why the reaction makes the egg shell more fragile?
Here's an example YouTube video this effect is mentioned.

The eggshell is made of calcium carbonate mostly, but the crystal structure is held together by matrix proteins (more references in Wikipedia). Salt creates ionic interactions with the protein molecules, breaking protein-protein interactions, and in this case, probably protein-mineral interactions as well, debilitating the eggshell structure.

Thanks! Would you be kind enough to elaborate why salt does that?

Proteins can interact with proteins or other molecules via ionic bonds between the protein (which has some charged - ionic - atoms), and calcium carbonate, which is made up of charged entities (it is an ionic compound). If there is enough salt, then salt - which is also ionic, will displace the calcium carbonate from the protein, debilitating the interaction.

That fully explained the process to me. Thank you.


text:
Do electrons have some volume, area or shape?
I am in 8th grade now and when I was in 6th grade, my science book had diagrams of the electronic configuration of atoms. The electrons were round like spherical balls. Is it true that the electrons have some volume, area or shape and they revolve around the nucleus?

Pictures are simplifications and should not be taken as real representations of the properties of subatomic properties that are best described by quantum mathematics.

Not too say that even point particle are meaningless out of the exact domain where they are modelled in such a mathematical way. Within a molecule or atom, perhaps an electron itself has no size no shape but is surely better modeled by its wave function or the square of the latter.  These latter are quantum mechanical as well , have shapes and volumes, and are of relevance when chemistry is involved.

Related on Physics: Do electrons have shape?

Atoms are composed of a positively charged nucleus and an outer shell of negatively charged electrons. When two atoms come into close proximity, their electron shells repel, preventing the atoms from sharing the same space. The "volume" of an object can generally be understood as the total measure of space that is unavailable for other objects to occupy, as the electron shells of the atoms of the two objects would repel, preventing the intruding object from entering that space.
This macroscopic conception does not apply to an electron. However, there are other measures of an electron that can be considered analogous to the dimensions of macroscopic objects. For instance, quantum mechanics states that all objects exhibit both wave-like and particle-like behavior (although, for larger objects, the wave-like behavior is insignificant). When an electron is analyzed as a wave, that wave has a wavelength, which is called the electron's "de Broglie wavelength". There is also a phenomenon called Compton scattering for which there is a characteristic length that is used in modeling the interaction between a photon and an electron. One could also ask "For what distance would putting two electrons at that distance require more energy than the mass-energy of the electrons?"
All of these, as well as other quantities, can in various sense be considered the "size" of an electron. To what extent these quantities "actually are" the electron's size is a matter of opinion. Some people consider one or more of these quantities to be the size, without any qualification, some consider them to be merely analogous to the concept of size, and others consider them to simply be quantities that are completely distinct from the macroscopic concept of size.
One thing to realize about science is that we use various words in everyday language without bothering to come up with a precise definition of what those words mean. We just have an intuitive sense of the general meaning of words. So when we start doing science, we build off of everyday language, taking everyday words as inspiration for precise terms, but we often have definitions that can be more or less precise than everyday meanings, or only loosely connected. For instance, "energy" in everyday meaning can refer to a wide array of concepts, such as "optimism", "enthusiasm", "friendliness", etc., but in physics it's much more precise. "Salt", on the other hand, in everyday meaning refers specifically to table salt, but in chemistry it is used more generally for ionic compounds. Words can have different definitions that are equivalent in everyday use, but they are distinct when we try to apply them to more exotic situations.
So while "Do electron have volume" may sound like an objective question, if we really look at it, it has to be understood as "Do electrons have a property that corresponds to the 'normal' concept of 'volume'?", and that question depends on how exactly one understand the "normal" definition of volume, how one defines that concept precisely, how one extends that concept to the microscopic realm, and what one considers to be a valid "correspondence".
It is generally accepted that physical laws are the same regardless of direction, and other than spin and some narrow symmetry breaking, electrons haven't been found to act different depending on direction. Which implies that electrons don't have a "shape" in the sense of "looking different" from different direction. So, if they are modeled as having a shape, it makes sense to have that shape be a sphere, possibly pictured with a vector representing spin.
Also, saying that they "revolve" around the nucleus is a simplification. Position and velocity are, somewhat like volume, concepts for which there are rough analogues in quantum mechanics, but they aren't quite the same. In a spherical orbital, an electron has a certain probability of being observed at each location, and this probability is the same for all points at the same distance from the nucleus. So even though the electron has a quantity that is referred to as "velocity", it is some sense not moving (in the sense that the probability of it being found in a particular location is not changing over time). Even though the electron's wavefunction is "moving", the magnitude of the wavefunction is not changing, and so it is both moving in one sense, and not moving in another. Actually explaining this clearly requires getting into quantum mechanics.

In fact in the spherical shells, the most likely place for the electron to be found is in the middle of the nucleus. Where the Bohr radius for that shell (the place where the simplified models pretend that the electron is orbiting at) is just the spherical shell that contains 50% of the distribution mass. That is, the chance of interaction with that electron inside or outside that shell is 50/50.

Reading this, I don't see an actual, clear, literal answer to the question, for an 8th grader. Would you mind adding "no, by our current knowledge, eletrons do not have a volume, area or shape in the sense you are probably thinking about" at the top, or something like this?

@Shufflepants : Er...  The maximum likelihood location of the electron is in the shell, slightly inside the Bohr radius.  The average location of the electron is at the center.  The average need not be particularly likely.

@EricTowers Hmmm, yeah, I guess I don't understand the difference between the distribution function and the density function. http://www.umich.edu/~chem461/QMChap7.pdf

Quoting from the Nobel lecture of Hans G. Dehmelt (1989):

With the rise of Dirac’s
  theory of the electron in the late twenties their size shrunk to mathematically
  zero. Everybody “knew” then that electron and proton were indivisible
  Dirac point particles with radius R = 0 and gyromagnetic ratio g = 2.00. The
  first hint of cuttability or at least compositeness of the proton came from
  Stern’s 1933 measurement of proton magnetism in a Stern-Gerlach molecular
  beam apparatus. However this was not realized at the time. He found for
  its normalized dimensionless gyromagnetic ratio not g = 2

...

Similar later work at still higher energies found 3 quarks
  inside the “indivisible” proton. Today everybody “knows” the electron is an
  indivisible atomon, a Dirac point particle with radius R = 0 and g = 2.00....
  But is it? Like the proton, it could be a composite object. History may well
  repeat itself.

So there is no evidence that the electron is anything but a point particle, yet a non-zero size has not been ruled out either.
New Measurement of the Electron Magnetic Moment Using a One-Electron Quantum Cyclotron Phys. Rev. Lett. 97, 030801 (2006) (more-official link) shows that any radius of the electron is less than $10^{-18}$ meters.
See also the related webpage Limit on Electron Substructure

But from a QFT perspective, the electron isn't a point particle because it isn't in any one place. It's a wave that's necessarily spread out over some volume depending on its state.

@Shufflepants Strictly speaking there is a theoretical question about the effective volume of an idealized free electron, truly unaffected by interaction with the universe. But this is probably only an asymptotic scenario in the real universe.

@Ian I believe the wave function of a free electron truly unaffected by interaction with the universe slowly spreads out over time. That is as time goes on, the size of the spherical region that contains >50% of the distribution "mass" grows without bound over time. My point is that from what we know now about QFT, all indications point to them being spread out excitations in a quantum field, not little balls or little points.

@Shufflepants Well, particle/wave dualism hasn't gone anywhere. You can't say it's not a particle at all.

Missing from detailed answers above, in wikipedia page on electron
https://en.wikipedia.org/wiki/Electron ,

Observation of a single electron in a Penning trap suggests the upper limit of the particle's radius to be $10^{−22}$ meters.

As a generic classical physics observation, if an electron does not have volume but has mass, isn't that implying that it has infinite density?


text:
Why is AC current used rather than DC current for calculating the conductance of a solution(see body)?
I was reading the answer of the above question from a book. it said that while using DC current, electrolysis occurs. Hence it affects the conductance of the solution. But if any gaseous product is created then polarization happens. Then why is polarization not furthermore discussed. So I request you answer what is it and how it affects while trying to get the conductance ?

I cannot answer your question. However, I speak as a moderator. Here on SE we operate on a basis of respect for each other. By repeatedly editing your post, especially without any addition to the content, you are bumping your post to the top of the front page, which seems to be what you want. However, this also pushes everybody else’s questions lower, which is disrespectful. I will warn you that there is no purpose in trying to bargain; we have the ability to lock questions to prevent them from being edited.

I hence apologize to all for my behavior.

When a current is passed through an electrolyte, a chemical reaction takes place. This alters the composition of the solution over time and you won't actually be measuring the conductivity(or conductance) of the initial solution.
Since in AC current, equal current flows in both directions over a given amount of time(larger enough than time period), reaction takes place in both the directions. On an average, no chemical change in composition would be observed.(Of course, that doesn't mean no reaction has taken place.) Therefore the conductivity(or conductance) you measure will be the conductivity you wanted to measure.

Please see my question body and say what polarization is and how does it affect the conductance?

@user187604 I think the full question in your book maybe helpful to answer your question.


text:
Which is more acidic: p-xylene or 3-methoxytoluene?

Which is more acidic: p-xylene or 3-methoxytoluene?

$\ce{OCH3}$ shows -I and +M effect, and $\ce{CH3}$ shows +I effect. Since, the former shows some -I effect, it should be more acidic. But that's not the case. Any other explanation here?

$\text{Acidic Strength}\propto \text{-I, -M, Negative Hyperconjugation effects} \propto \dfrac{1}{\text{+I, +M, +Hyperconjugation effect}}$
Recall that $\ce{-OCH3}$ is a moderately activating group as its mesomeric effect overrules its inductive electron withdrawing effect.
On the other hand, you must be aware that $\ce{-CH3}$ is only weakly activating because of hyper conjugation and inductive electron releasing effect.
Now, its easy to conclude that a strongly activating group would result in more electron density over and around the benzene ring and hence destabilise the anion formed after deprotonation. Thus, we'd expect p-methoxytoluene to be less acidic than p-xylene.


text:
When did the realistic typesetting of chemical structures start?
In many old chemistry books there were limitations on how chemical structures could be represented driven, presumably, by the desire to use available typesetting methods to save costs.
For example, this is a page from a book on industrial chemicals published in the 1930s:

The distorted hexagons are 6-membered aromatic rings but are typeset from vertical and 45° lines with none of the double bonds explicitly shown.
When and how did chemical typesetting become more "realistic" with, for example, proper hexagons for aromatic rings and the correct description of double bonds where appropriate?

Is this helpful? https://www.researchgate.net/figure/Typeset-chemical-structures-Grundon-1962_fig1_251107238

Your example looks like lead typesetting and apparently was limited by the available type sorts.

@Loong I'm fairly sure you are right. I wonder whether anyone has compiled a typographic history of chemistry?

@user55119 An interesting collection of relatively recent examples. Interesting, but I'd love to see a spread of examples over a century.

In that case, you may be interested in Henry Rzepa's blog (split over multiple posts).

this isn't a definite answer but it might show the direction:
I have an old copy of Namensreaktionen in der Organischen Chemie, 2nd ed. by Helmut Krauch and Werner Kunz, published in 1962 at Alfred Hüthig. First edition must have been published in 1960 or 1961.

Most of the arenes in the sketches have the proper shape. Once in a while they seem to be deliberately stretched to align centres that react.
All through the book there's a consistent notation for saturated centres, hydrogen atoms are mostly not omitted.

Looks like a similar typesetting process to my example but with more effort to get the correct unsaturation. Also an odd mix of elongated and correctly proportioned hexagons.

@matt_black You're right. I copied this figure on purpose. Seem it was their way to indicate that "the carbocation goes there"without using curved arrows which probably weren't available.  Other hexagons in the book are mostly correct.


text:
Flaw in my logic on Henry's Law?
It states mathematically $ p=Kx $ where $x$ is mole fraction of dissolved gas and $p$ is partial pressure of the gas.
Dissolution of gases is exothermic, so with increase in temperature, solubility $x$ decreases.
But increase in temperature also means an increase in partial pressure $p$ which in turn means an increase in solubility $x$. Where am I going wrong?

Temperature only affects pressure if you allow it to: by keeping volume constant. If your system has a freely moving piston holding the gas for example, then temperature will lead to a volume expansion, pressure remaining constant.

Henry's law is:

At a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

If temperature changes, the Henry's law constant itself changes.

But how does it change? As in what is the dependence of K and temperature?

@Cowgirl read the section "Temperature dependence of Henry's law constant" in the linked webpage

Read also: !https://www.sciencedirect.com/science/article/pii/S1387380615002183.


text:
Saturated vs Supersaturated
The way I was told is that a saturated solution contains the maximum amount of solute and a supersaturated solution contains more solute than a solution can hold at a particular temperature. But I was also told that if you add more solute than it can hold, that it will not dissolve and will settle at the bottom as a solid. So how is that supersaturated? Since it didn't dissolve, isn't it still just saturated? It was also mentioned that a supersaturated solution needs to be cooled, but if the temperature dropped, wouldn't the solute just come out of solution and settle at the bottom (like sugar and cold coffee) and remain saturated?

You are right about the impossibility to prepare a super saturated solution by dissolving more solute. But you can prepared an almost saturated solution and cool it fast. In particular cases and ckean environments separation of the solute can be slow enough to get in a supersarurated regime.

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

A saturated solution has the maximal amount of solute at equilibrium.
If equilibrium is not required, solutions can have more solute.  For example, when a bottle of soda is first opened, it is supersaturated in CO2, but it is not at equilibrium and CO2 effervesces.

When would equilibrium not be required? Does it depend on the solute?

Also, the reason I'm asking is we're doing a lab with sodium thiosulfate. We heat it up, adding all but one crystal, and then put it in a ice bath and add the last crystal and observe. Is it supersaturated when it is heated? and then the one last crystal causes it to fall out of solution and crystallize? And if so, why is it supersaturated? Is it because it isn't at equilibrium?

Think about it this way. When the solution is heated and you add all but one crystal, you probably saturated the solution. The solution is then at equilibrium with solute dissolving at the same rate as it crystallizes. When you cool down the solution in general the solubility decreases. Therefore, for a while the dissolved solute is greater than the solubility of the solvent. However, eventually the system returns to equilibrium, with the consequence of having solute crystallize out of solution.

Feel free to take my understanding of it with a grain of salt though. I am in middle school and taught myself chemistry, so I might be wrong.

@Sarah It is not supersaturated until you cool it, which lowers the equilibrium solubility.

A saturated solution has the maximal amount of solute at equilibrium.
If equilibrium is not required, solutions can have more solute. For example, when a bottle of soda is first opened, it is supersaturated in CO2, but it is not at equilibrium and CO2 effervesces.

I'll add a little on to DavePhD's answer.
A change in general has two things you need to worry about: the system, or the place that the reaction takes place, and the surroundings, a.k.a. everything else in the universe. So lets start with that $100\deg\pu{Celsius}$ water that you have (just for example). From the Wikipedia page we can see that the solubility is:

$\pu{70.1 g/100 mL}$ ($20$ °C)
$\pu{231 g/100 mL}$ ($100$ °C)

You then dissolve $23$ grams of thiosulfate in $10$ grams of water (just for example). Then, you cool the solution to $20\deg \pu{C}$ using the ice bath. In fact, you cool the solution of sodium thiosulfate beyond its freezing point, something called supercooling or undercooling.

You can think of it as a ball, rolling down a hill coming to a rest because of not having enough kinetic energy to roll all the way down the slope. Our supersaturated solution is currently at (1). It has to overcome an activation energy (2) to reach crystallization (3). Because it wants to crystallize until the solution is $\pu{70.1 g/mL}$, simply providing a surface for it to crystallize on causes it to crystallize rapidly.
I found a resource here that may help.

You might have some honey in the house. Is it clear? Or has it begun to crystallize? Honey is mostly fructose and glucose with a little sucrose and  about 15% water. It is supersaturated and will eventually crystallize out (but you can reheat and cool to clarity, as long as all the crystals have dissolved and cannot act as seed crystals). Sugars are complex molecules with many hydroxyl groups that interfere with crystallization.
The difficulty of getting all the molecules aligned in the preferred order may permit supersaturation for a time; sometimes intermediate alignments are fairly stable but do not sustain orderly growth. Then the partial crystal must partially redissolve and try again; when everything is in perfect alignment, the crystal grows another layer.


text:
What is the molecular formula of the hydrocarbon?

A hydrocarbon is burnt completely in excess oxygen to produce 0.04 mol of carbon dioxide and 0.72 g of water. What is the molecular formula of the hydrocarbon?

Hi Cheong!  welcome to Chem.SE! We require you to show your efforts on this problem. What formulae/steps did you try? Please add this to your question. Thanks!

There isn't enough information to solve the problem exactly, but it is still possible to get the empiric formula which shows the relation of the atoms.
We have 0.4 mol carbondioxide, there is only one carbon in carbondioxide so we know 0.4 mol carbon reacted with oxygen
We have 0.72 g of Water ( M = 18), we need the number of moles so we can just divide it through the molar mass. The result is 0.4, but we have to take into account that water contains 2 hydrogen, thus the amount of hydrogen that reacted with water is 0.8 mol.
The formula is:  $C_nH_{2n}$

good work but we usually don’t provide complete answers on homework questions with no work, maybe a hint instead. We aren’t a homework site, nor do we like the reputation of a homework site.


text:
Variation of equilibrium constant
In an exothermic reaction, the value of equilibrium constant increases with the rise of temperature because,
$\Delta H < 0$ ( exothermic reactions)
$\Delta G = \Delta H - T\Delta S$
Now if $\Delta H $ and $T\Delta S$ are negative and positive respectively, the overall value of $\Delta G$ is also negative. (For positive entropy)
$\Delta G = -RT\ln K$   (where $K$ is equilibrium constant)
Since $\Delta G$ is negative, the value of $RT\ln K$ becomes positive and as $\Delta G$ increases equilibrium constant increases.
Is this logic right?

First of all, How can you assume that, $\Delta S$ is positive in any exothermic reaction? Secondly, the relation you have written $\Delta G = -RT\ln K_\text{eq}$ is not at all correct. The correct relation is $\Delta G = \Delta G^\circ + RT\ln Q_c$ (or $Q_p$ if the reactants are gaseous). Now at equilibrium $\Delta G= 0$ and $Q = K$. So, the correct relation is
$$\Delta G^\circ = -RT\ln K_\text{eq}$$
And, there are obviously differences between $\Delta G$ and $\Delta G^\circ$. So, the reasoning you have given is not totally correct.
The correct reasoning comes from Van't Hoff Equation, which states $$\frac{\mathrm d}{\mathrm dT}(\ln K_{\text{eq}}) = \frac{\Delta H}{RT^2}$$
Integrating both sides, we have
$$\int_{K_1}^{K_2}\mathrm d(\ln K_{\text{eq}} ) = \int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$
So, we have $$\ln\frac{K_2}{K_1}= \frac{\Delta H}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$ Thus we see that increasing the temperature from $T_1$ to $T_2$, the equilibrium constant changes. Here $\Delta H < 0$ (exothermic reaction) and $\displaystyle\frac{1}{T_1}- \frac{1}{T_2} > 0$ so, $\displaystyle \ln\left(\frac{K_2}{K_1}\right) < 0$, which means $\displaystyle {\frac{K_2}{K_1}< 1}$, which implies $K_2 < K_1$. So, the equilibrium constant decreases.
Note: Without any mathematical equation also, the same result can be deduced. In an exothermic reaction, the products are formed along with heat production, which is the definition of exothermic reaction. That means, temperature of the system will increase due to this heat production. Now if you increase the temperature, the system will obviously want to go in the direction where this temperature minimisation is possible, which is actually the backward reaction. So, the reaction shifts backwards, and thus the equilibrium constant decreases as now the product concentration has become lesser and reactant concentration has become larger.

I'm sorry. I actually meant delta G^0 only.To answer your question, I am assuming Entropy to be positive.So the variation of the equilibrium constant with Delta G cannot be sought from Delta G^0 = -RTlnK?

Actually if you look into the equation  $\ce {$\Delta $G ^0 = -RTln K_{eq}}$, The LHS is constant for a reaction and independent of Temperature as well. So if you increase the temperature, the RHS says the equilibrium constant should always decrease,irrespective of reaction being  exothermic or endothermic, which is not the case of always. So, better to avoid that logic.

The left hand side is not independent of temperature.  The Van't Hopf equation is an indication of this.

$\Delta$$G^0$ is independent of temperature. Note that it is $G^0$ , not $G$. See this https://chemistry.stackexchange.com/a/91256/58497


text:
What does CkC mean in chemical bonding?
I'm working on a chemistry question where I'm supposed to use average bond energies to estimate enthalpy changes for a reaction. It gives me these bond energies to work with:
$$\begin{align}
\ce{C-H}&=&413\\
\ce{C-C}&=&348\\
\ce{C-Br}&=&276\\
\ce{H-Br}&=&366\\
\ce{CkC}&=&614\\
\end{align}$$
I know how to do the question, but I don't what $\ce{CkC}$ is. I can't find it online or in my textbook. What does the "k" mean?

Which text book did you find this notation in? The info will help future visitors having similar problem.

$\ce{C=C}$ average bond energy is somewhere around $\pu{614 kJ/mol}$ so "k" probably indicates a double bond


text:
Clarification on the term: coordination number
Based on what I've been reading the coordination number of a complex is defined as the number of atoms/molecules that the central atom is bonded to. However, it is simulaneously said that [FeEDTA] has a coordination number of 6. So which is it. Is the coordination number the number of atoms/molecules, or the number of dative covalent bonds?

Coordination number, also called ligancy is the number of atoms that a central atom or ion holds as its nearest neighbours in a complex or coordination compound or in a crystal.
In EDTA, there are two nitrogen atoms and four oxygen atoms, which act as neighbors and bind by sharing their electrons. So, even though EDTA is itself one single molecule, there are six atoms in EDTA which are sharing their electrons. So, the iron has the coordination number of 6.
The complex looks like this:

So, basically, coordination number depends on how many atoms are directly participating in the complexing.


text:
Why is pyrrole more reactive than pyridine and benzene for an electrophilic substitution reaction?
According to me, pyridine should be the most reactive as it has got an extra pair of electrons on the nitrogen atom which is further localised and readily available for electrophile.

Related: https://chemistry.stackexchange.com/q/91322/5026

From Peter Sykes:

Pyridine (62), like benzene, has six at electrons (one being supplied bynitrogen) in delocalised it orbitals but, unlike benzene, the orbitals will be
  deformed by being attracted towards the nitrogen atom because of the latter's
  being more electronegative than carbon. This is reflected in the dipole of pyridine, which has the negative end on N and the positive end on the nucleus:

  Pyridine is thus referred to as a $\pi$-deficient heterocycle and, by analogywith a benzene ring that carries an electron-withdrawing substituent, e.g.$\ce{NO2}$ (p. 6.22), one would expect it to be deactivated towards electrophilic
  attack. Substitution takes place, with difficulty, at the 3-position because this leads to the most stable Wheland intermediate (63); the intermediates
  for 2- and 4-attack (64 and 65, respectively) each has a canonical state in which the charge is located on divalent N a highly unstable,i.e. high energy, state.

  There are certain formal analogies here to m- attack or
  p. 6.24), byt pyridine is very much more difficult to substi
  Thus nitration, chlorination, bromination and FriedelCraf
  really be made to take place usefully, and sulphonation on
  ing with oleum for 24 hours at 230°C, with an $\ce{Hg^2+}$ catalyst attack is due partly to the fact that pyridine has an available nitrogen, and can thus protonate (66), or interact with an electrophile (67):

The positive charge will clearly further destabilise any of the $\sigma$-com
  plexes for electrophilic substitution, as did a substituent such as $\ce{-NR3^+}$, on the
  benzene nucleus (p. 6.24), but the destabilisation will be much more marked
  than with $\ce{-NR3^+}$; as the charge is now on an atom of the ring itself and not merely on a substituent.
  Pyrrole (68) also has 6 electrons in delocalised orbitals, but here the
  nitrogen atom has to contribute two electrons to make up the six (thus be
  coming essentially non-basic in the process, cf. p. 3.26), and the dipole of
  pyrrole is found to be in the opposite direction to that of pyridine, i.e. with
  the positive end on nitrogen and the negative end on the nucleus:

  Pyrrole is thus referred to as a π- excessive heterocycle and behaves rather like a reactive benzene derivative, e.g. aniline (p. 6.25), undergoing very ready electrophilic attack. This may be complicated by the fact that in
  strongly acid solution protonation (69) is forced even on the weakly basic
  pyrrole (it takes place on the 2-carbon atom rather than on N. cf. p. 3.26)

TLDR

Pyridine is a $\pi$-deficient heterocycle owing to nitrogen's more electronegative nature than carbon. Pyrrole on the other hand is $\pi$ excessive since nitrogen donates its lone pair to the ring.
The non-delocalized lone pair on pyridine tends to interact with the electrophile/$\ce{H+}$. This further destabilizes any of the $\sigma$ complexes formed in electrophilic aromatic substitution (EAS).

Well, not really. You didn't rewrite anything. And it's not like we don't know how pyridine looks. If you need pictures badly, you can use some from Wikipedia (with proper attribution).

This can be explained by presence of the aromatic lone pair owing to nitrogen in pyrrole makes the aromatic system more electron rich, and you can get a hint of it by looking at the electron pushing mechanism.
The electron pushing from pyridine fails,
Initially, it was due to the lone pair of electron is fixed at an orthogonal position away from the aromatic ring.
Secondly,  is that particular intermediate cannot be obtained because of structural geometry problems, (=N=)+ should be linear and cannot be formed while maintaining a ring.

One way to awaken electrophilic substitution in pyridine is to N-oxidize it.  See here, about halfway down the page, for an example with nitration (predominantly opposite the N-oxide function), and this answer for an explanation of what happens to the electrons as a result of this oxidation.  In effect N-oxidizing pyridine works by making that molecule more like ... pyrrole.


text:
Does attachment of phenyl to an alkene increases its reactivity towards electrophilic addition reaction?
According to me phenyl should decrease the reactivity as its electron withdrawing nature destabilises the carbocation.

Not always, and it can depend upon the groups attached to a phenyl ring, like $\ce{-OMe, -COOH, -NH2, -NO2, -X}$, etc.
The attachment of a phenyl ring to a carbon-containing double bond can affect both the stability of carbocation and the possibility of attacking with the help of the compound's $\pi$-electron cloud.
If you have only the phenyl ring with no substitution, first of all the $\pi$-electron cloud of the double bond will be in high conjugation with the phenyl ring, and thus the $\pi$-electron density will be reduced on the $\ce{C=C}$ double bond, which can reduce the possibility of attack of that double bond to a electrophile. Though the carbocation formed can be stabilised through resonance with the phenyl ring (as it is a benzyl carbocation), the step of carbocation formation can have a very high activation energy ($E_\mathrm a$) if you don't have enough stabilisation of the carbocation. But, if you have the carbocation which can stabilise through resonance with multiple benzene rings, the actual potential energy of the carbocation can be reduced to a great extent. In that case, the rate of reaction will increase.
So, actually this situation can be thought as a competition between the stability of carbocation and the decreased electron density. Generally, the stability of carbocation is the dominating effect if it is stabilised to a large extent.
The rate will be decreased even more if an electron withdrawing group is attached to the phenyl ring due to even more delocalisation of $\pi$-electron cloud, and less stabilisation of the carbocation.

If you have a strong electron donating group attached to a phenyl ring and the double bonded carbon is ortho/para w.r.t that group, that can itself exhibit the $+R$ effect, and increase electron density near the double bond through delocalisation of its own electron density through the $\pi$-bonds. In that case, the attacking of $\ce{C=C}$ double bond to a electrophile will be much easier, and the carbocation formed will be even more stable. Then, the rate of reactivity of the double bond in the electrophilic addition can increase for attachment of a substituted phenyl ring.

In but-2-ene  one sp3 carbon and a hydrogen atom attached to a sp2 carbon when replaced by two phenyl cations then the resulting molecule is more reactive towards electrophilic addition reaction than but-2-ene. Can you explain it?

answer has been edited. Please check it out.

But then why toluene is more reactive towards chlorination than an alkene for eg. propene?

This case and the case in your question are different. Here the rich $\pi$-electron cloud of benzene itself donates, but in your question, it was not so.


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Are there any drawbacks of the equivalent concept?
I wanted to know because in certain reaction(disproportionation or Comproportionation) equivalent concept does not help much. In many question it is solved by balancing a reaction,but then what's the use of Equivalent method when we are just balancing it.
  In short I want to know that in some case is it neccessary to balance the redox reaction ??
Please help by giving examples.

One drawback is that an "equivalent" may not be a well defined quantity.  Take phosphoric acid as an example.  The first deprotonation is fairly strong but the next two rapidly become weaker.  Therefore you need a rather strongly basic solution to get all three protons off; under more nearly neutral conditions you can remove only one or two protons.  So with respect to Bronsted-Lowry reactions does phosphoric acid have three equivalents per mole, or fewer?  You have to define the reaction conditions to pin it down.


text:
What's a good chemistry book for a practical man
Yesterday I picked up my old organic chemistry book and started to read it from the first chapter. After a couple of minutes I realized that I don't understand the concepts at all.
The first chapter mentions the reason behind polarity of covalant bonds because of difference in electronegativity of different elements, flourine being the highest. Now chemistry is a practical science and all these concepts have been theorized by observing reactions. Is there a chemistry book which gives a detailed explanation of the experiments which led to the formulation of theories we read in school?

Welcome to Chemistry! Take the tour to get familiar with this site. This is far too broad of a question and isn't really appropriate for our site. For starters, I would look at a general chemistry book before an organic book, and go from there. If you don't have one, check our list for some guidance, and try some videos too: https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry


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Are there bonds which share single electron?

Are there bonds which share single electron?

During chemical bonding, I asked our teacher why there are only 2 electrons shared in a bond. He said that because it's that one orbital can accommodate only two electrons. Two electrons make it stable. He said that there are bonds which share a single electron which has some term I don't remember and is out of our understanding for now. So, I am curious whether there are such bonds, what are they called, some examples, and are they unstable due to the reason aforementioned?

Not exactly a single electron in one bond, but there's other examples like 4c-3e in PCl5, or 3center-2electron bond in diborane. There are unpaired non-bonding electrons like in nitrogen dioxide. But, I am unaware of a single electron in one bond.

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

There exists an interesting molecule dicarbon with bond order two and two pi bonds instead of the usual one sigma and one pi. This is the case when 2s-2p mixing is allowed in molecular orbital theory. However when the aforementioned condition is not allowed, dicarbon exists as a molecule with one sigma and two pi bonds of, interestingly, one electron each. So in a way, one electron bonds do exist.


text:
Why does the temperature of a tyre rise when it is pumped up rapidly?

Why does the temperature of a tyre rise when it is pumped up rapidly?

The way my textbook explains this is that $V\propto T$, so, an increase in volume causes an increase in temperature.
However, I think that this justification is flawed, because air is a real gas instead, and follows (approximately) the van der Waals' equation of state: $\left(P+\frac{a}{V^2}\right)(V-b)=RT$. I'm not not sure how I can neglect the $a/V^2$ term in it, considering the volume in a tyre is not that high.
Moreover, I actually had thought the reverse statement to be true. Pumping air "rapidly" implies the conditions are adiabatic i.e. heat exchange is negligible. So, $TV^{\gamma-1}=\text{constant}$ holds, implying an inverse relation between volume and temperature.
So, why actually does pumping a tyre increase its temperature? And what is the fault in my reasoning above?

Source: KS Verma; Physical Chemistry for JEE (Advanced): Part 1; Exercise 5.1 Question 6

Hmm, I'm tempted to downvote all answers.  Probably more important process is loss of energy as heat, which accumulates in fast process.

Isn't $TV^{\gamma-1}=\text{constant}$ and ideal gas property?

@EashaanGodbole Oh you're right. Well it doesn't. But I thought it may hold "approximately" if not exactly. Like the "inverse relation between volume and temperature" may still be true.

Well, around 34 psi and 298K, the molar volume of an ideal gas would be around 10 L/mol. Ideal gas law won't be too far off. (It's usable.)

Once the pressure in the tire has risen a little, the volume of the tire doesn't change much any more.  But, you are increasing the number of moles of air inside the tire when you pump air into it.  The new air pumped into the tire compresses the gas that was already in there previously and raises its pressure.  So the new gas is doing work on the previous gas to compress it  adiabatically and approximately reversibly.  This causes the internal energy of the gas and its temperature to increase.

Oh, nice! This makes sense. Could you please also add a line or two about why my attempted explanation was leading to the wrong result?

It was leading to the wrong result because your answer didn't take into account the fact that the number of moles of air inside the tire was increasing.

Plus, the pressure is building up in the tire, so the new air being injected is getting adiabatically compressed by the pump even before it enters the tire.  This causes the new air to enter at a higher temperature.

Yep, that makes sense.

We know that $\Delta U=Q-W$, $\Delta U$ being change in internal energy. Further, $\Delta U=nC_V\Delta T$.
As process is adiabatic, $Q=0$. Therefore, all work done is gone to change the internal energy which implies its temperature rises rapidly. Also, $TV^{\gamma-1}$ does hold. Consider $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}=\text{constant}$ ($T_1$ and $T_2$ being initial and final temperatures). Clearly, $V_2$ is less than $V_1$, implying $T_2$ is greater than $T_1$ for above equation to hold. Thus it is not reverse but decreasing volume increases temperature in adiabatic process.

" it is not reverse but decreasing volume increases temperature in adiabatic process." Actually that contradicts the given statement, which is that increasing volume increases temperature.

What I am saying is decreasing volume increases temperature.When you do work on gas,you reduce gas volume.By PV^γ=Constant,decreasing volume implies temperature increases for to above eqn to hold true.You must have used pump to inflate tires of bicycles in your childhood.The pump can be seen as piston container system,where when you pump down the volume decreases thereby increasing temp.

Kinetic theory of gases, if your volume is constant, but your pressure is increasing as consequence of adding molecules, they are going to increase the number of hits against the wall, in this case the wire. In turn, the frequency by which each molecule transfer energy to the wall is going to be larger than the previous state. Then, temperature rises.

Give an argument on why -1, don't be mediocre .

(not my downvote) I don't know why there was any need to mention the KTG theory at the beginning of the answer. I guess it only applies to ideal gases and the rest of the answer seems to make sense without it as well. What specific postulate of it were you trying to refer to?

Suppose the tire's (absolute) pressure is increased
by the tire pump from $P_0 \ge P_1$, where $P_1$ = atmospheric pressure, to a final pressure $P$
at the ambient temperature $T_0$ to $T_1$.
Also, the process is adiabatic, as you mentioned it is done rapidly.
$TP^{1/\gamma-1} = \text{constant}$
Therefore, $T_0P_0^{1/\gamma-1} = T_1P_1^{1/\gamma-1}$
Putting $\gamma=7/5$ for air,
$T_0P_0^{-2/7} = T_1P_1^{-2/7}$
$T_1/T_0 = P_1^{2/7}/P_0^{2/7}$
Since,  $P_1>P_0$, $T_1>T_0$.


text:
Why isn't ionization energy called cationization energy?
I'm learning chemistry and ionization energy is "[...] the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation." It seems more intuitive to me to call this cationization energy. Other than historical reasons and convention, are there any reasons why it's not?

What about anionization instead? ;-)

@MaxW It seems to me anionization energy would be the opposite - the amount of energy required to add a valence electron, right?

The cat prefix does not add valuable information. Since language tends to be efficient there's no such prefix.

Should be migrated to History of Sciences.SE

The reason I ask the question is that I've found the concept confusing because "ionization" alone is ambiguous. I think cationization would be valuable information and immediately imply the direction of movement. The implication from the above comments is that my intuition is correct, so I appreciate the responses. Thanks!

@MaxW Maybe what you were implying with your comment is that any time cationization occurs, anionization also occurs; however, is it possible for an electron to be removed from an atom without it being added to another atom and become a free electron? Or maybe what you were implying is that the energy required to add or remove an electron is the same?

Atoms can have electrons added to them or removed from them, which in turn ionises the atoms. However, electron addition and removal aren't the same thing in terms of enthalpy changes. The key here is the difference between the
process and the enthalpy changes during the process - as much as an atom is ionised during the production of a cation or an anion, the enthalpy changes are different. Because of this, there are two different terms used for the enthalpy changes during these processes:

'Ionisation energy' is the enthalpy change when one mole of gaseous atoms form one mole of gaseous positive ions (cations), under standard conditions,
'Electron affinity' is the enthalpy change when one mole of gaseous atoms form one mole of gaseous negative ions (anions), under standard conditions.

As seen from the definition for 'ionisation energy', it's already described that there's a production of cations, so it would be redundant to call it 'cationisaion energy'.  The word 'ionisation' in itself is typically used to describe the removal of electrons from an atom, which could also contribute to the term's name.

The reason why there are two different terms is due to how the electrons are dealt with during ionisation. The removal of electrons is always endothermic, whereas the addition of electrons is only sometimes endothermic.
When removing electrons, more energy is required than what's released to remove the electrons from the atom, hence why it's always endothermic. This is due to the Coulombic attraction between the negatively charged electron(s) in the atom's orbit and the positively charged proton(s) in the atom's nucleus. The required energy will increase with each successive removal, since the electrons will get closer to the nucleus and the attraction will increase because of it. This enthalpy change is explicitly called 'ionisation energy'.
When adding electrons, only the first addition is exothermic whilst the rest are endothermic. The first addition is 'easy'; you're just adding an electron to a neutrally charged atom, and the energy released during this process is greater than the energy required. However, with each successive addition, the energy required will increase. This is due to the Coulombic repulsion between the anion and the electrons, so more energy is required than what's released to 'force' the electrons into the atom's orbit, hence why successive electron additions are endothermic. This enthalpy change is explicitly called 'electron affinity'.

It finally makes sense, thanks so much! By the way, I was just watching a Khan Academy video, and although Sal's no expert in the field, he makes the same remark about "cationization energy" at ~3:00 so this seems to be a common confusion: https://youtu.be/5CBs36jtZxY?t=3m

It's called 'ionisation energy' for a reason. We can't think 'Oh, in this process cations are produced so we should call it cationisation energy...'. Maybe it could be called 'cationisation energy' like Sal said (emphasis on 'maybe'), however that would be redundant and unnecessary since the term 'ionisation energy' isn't ambiguous - we have a term for the opposite thing! No problem, by the way.


text:
Creating water from hydrogen and oxygen
Assuming a scientist has two canisters: One with a liter of pure gas of hydrogen (single atoms gas), and one with half a liter of pure gas of oxygen (single atoms gas).
The scientist desires to conjunct the two single atom gasses inside some instrument somehow, to create water.
I understand that the process of doing so is some kind of explosion.
What are the names of that reaction and of the instrument in which it should take place?
I ask this after I saw the film "The Martian", in which the protagonist "Mark Watney" created water from separating hydrogen from hydrazine, then burning it with oxygen in extremely low stream of gasses, creating very small amount of water in each step of a process.
Here is a video in which a youtuber seemingly creates water from burning hyrdogen and oxygen (legal note: I ask any reader not to try this without being certified as a chemist from, a internationally known academic center).

This question is for general knowledge: To know the name of the reaction and the common instrument for it.

Hydrogen and oxygen are not "single-atom gases".

@user9303970 single hydrogen/oxygen atoms are very unstable and don't occur naturally. Hydrogen and oxygen gases are diatomic gases ($\ce{H2}$ and $\ce{O2}$, respectively).

@user9303970 Unstable here means that the bond dissociation energy in diatomic hydrogen and oxygen is 436 kJ/mol and 498 kJ/mol, respectively. To give you an idea, see also atomic hydrogen welding.

What do you want with an actual apparatus and safety instructions for a hypothetical reaction?!? Total nonsense. Burning hydrogen with oxygen is done with a torch, in the same way as the one used for oxy-acetylene welding.

I didn't ask for "safety instructions" anywhere, and even if I would, calling a good question for general knowledge with different possible benefits to those who will read it is redundant.

If you want to have clarification about why your question is still closed, please open an issue on [meta.chemistry.se] and tag it with [meta-tag:specific-question].

I am sad that my edit suggestion was rejected; I myself created this question from the above account that I already abandoned (I no longer use its email and I don't know its password). I think the edit sharpened the question and is plausible.

@JohnDoea Please contact support about merging your accounts; then you can edit your question as you wish. On a side note: element names are not capitalised; don't use subscript for text which is not a subscript; your edit did not address the reason as of why this question was closed. I support the decision of the reviewers to reject the suggested edit. If you have more questions about the procedure, please open a question on our meta site.

Hello @Martin-マーチン; I have a comment one thing you wrote: I don't agree with your suggestion not using subscript: From my long experience with StackExchange it is used for smaller text notes; If you suggest using html inline-styling as in `style="font-size: 10px" instead - with that I would agree.

There is absolutely no reason to change the text size. If it is not important, delete it. If it is important, keep it at the same size. I know this is a practice on other sites in the network, but here we generally discourage this.

Having a single atom hydrogen or oxygen gas means having lots of free radicals just waiting to react with their own species, liberating a huge amount of energy within a fraction of a fraction of a fraction of a second. This is very unstable.
Lets say the hydrogen only reacts with oxygen and vice versa, then such a reaction would be possible. In general it is called radical addition or free radical addition when radicals react with each other.
The bond energy of a O-H bond is about 464 kJ/mol.
If we had 1 mol magic $\ce{O.}$ ( the oxygen radical) gas and 2 moles magic $\ce{H.}$ the result would be 1 mol (normal) $\ce{H2O}$ and 928 kJ energy. This is the energy delivered by 221.8 g of TNT.
This reaction should be slow or taking place in an explosion proof container.
If you did the same with ordinary oxygen and hydrogen, the energy liberated would still be high but not as big as with radicals ( the O-O and H-H bond needs energy to be opened).

So much thanks Legolas ! If I understood you correctly, it will be less energetic, hence a bit less explosive, to create water from dihydrogen+monooxigen? This is the only way to create the water?

There is no stable monooxygen, you have to say dihydrogen reacts with a half dioxygen.

There are many ways to create water, all condensation reactions liberate water ( They are extremely common).

I thought that in the film "Bring Him Home" (with Matt Damon), he actually separated atoms of hydrogen from hydrozine, then mixed with oxygen in low doses, just to make small amount of water each time again.

This is true, but unfeasible.

First of all he needs to make the hydrazine react to nitrogen and hydrogen, which is an explosive reaction and then burn the hydrogen.


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What is the difference between transaldolase and transketolase in the pentose phosphate pathway?
What is the difference between transaldolase and transketolase in the pentose phosphate pathway? From what I understand, they both catalyze the transfer of carbon chains from 1 aldose into 1 ketose to produce 1 aldose and a ketose, though transketolase transfers 2 carbons while transaldolase transfers 3 carbons. Is that the only difference? If so, then why they were named "aldolase" and "ketolase"?

An aldol is a $\beta$-hydroxy ketone or aldehyde, and is the product of an aldol addition.

The enzyme that cleaves the bond formed (red) is known as an aldolase. The transaldolase reaction begins by aldol cleavage that is, cleavage of the bond in red.

Transketolase on the other hand simply cleaves off the ketone group from the substrate. It does not perform an aldol cleavage.

Apart form this, the mechanisms of action of the two enzymes vary considerably. For instance transketolase requires thiamine pyrophosphate (TPP) as a cofactor but transaldolase does not.


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Can you create pure sodium metal by electrolysis of aqueous NaCl rather than molten NaCl?
I was researching how to make pure sodium from salt because I think it would be interesting to do. In all of the articles I read that talked about using $\ce{NaCl}$ they said to melt it and then use electrolysis. Would dissolving the $\ce{NaCl}$ in water then using electrolysis work? Electrolysis of $\ce{NaCl}$ dissolved in water produces $\ce{H}$ gas at one electrode and $\ce{Cl}$ gas at the other, which would leave $\ce{O}$ and $\ce{Na}$. So could you just use electrolysis until a lot of $\ce{Na}$ was produced then boil off the water? Would the $\ce{O}$ bond to the $\ce{Na}$ to produce $\ce{Na2O}$? If it would make sodium oxide then how difficult would it be to remove the $\ce{O}$ to get just the sodium?

In one word, no.

https://chemistry.stackexchange.com/questions/42081/electrolysis-of-sodium-chloride-solution

And if sodium were produced, I suppose it would be stable in water.

Some more ideas at  -  https://chemistry.stackexchange.com/questions/19500/extracting-sodium-from-salt/90123#90123

As someone mentioned in his/her answer last year, you cannot get sodium metal from electrolysis of aqueous sodium chloride. But, contrary to his/her explanation, you do actually get metallic sodium at the cathode in this reaction. Even if the half-cells are separated with Nafion (chloralkali process), the elemental sodium produced at the cathode instantly reacts with the water to produce $\ce{NaOH}$ and hydrogen. So, it appears as though it liberates $\ce{NaOH}$ and $\ce{H2}$ from the outset so most people confuse what's actually happening.  You literally have ZERO time to capture the metal before it reacts. It's instantaneous. Using a MERCURY cathode will actually capture metallic sodium inside the mercury as an amalgam.  But I know of no method of extracting the sodium from the amalgam. I believe that process is also one that produces $\ce{NaOH}$ in the end. You would have to melt the $\ce{NaCl}$ and perform the electrolysis directly on the salt to get free sodium.
I did think about testing a chloralkali cell with the electrodes right up against the Nafion (and holes in the electrodes) and leaving the cathode side dry to see if sodium ions will permeate and give elemental sodium on that side.  I doubt it would work though. Both sides probably have to be wet.  And we know when there's water on the negative electrode, you'll get $\ce{NaOH}$, not $\ce{Na}$.

Filling in:  In the old technology the amalgam was reacted catalytically with water in a separate chamber to form NaOH without contacting the anodic products.  Now they use a membrane instead of mercury to effect the separation.  Pure sodium never was and cannot be extracted from the amalgam, you get intermetallic compounds.

The answer is no.
Electrolysis of a $\ce{NaCl}$ solution produces $\ce{NaOH}$ as a remainder, not $\ce{Na2O}$. Even if Sodium reacted in such a electrolysis, it would react with water to produce $\ce{H2}$ and $\ce{NaOH}$ again.

Actually, you have some time to collect Na. It will float and if you add oil on tge top of the water, Na will collect there. Then you just have to collect the very small pieces and press it togeather into a bigger piece, which you then store in parafine or something simillar.

No, you really don't.


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What are some examples of pure substances with no liquid state?
For example,

A molecular substance that decomposes before melting, and under pressure can only assume a liquid state that doesn't contain the same molecules, for example a mixture of the elements of which it's composed
A polymer that breaks down into monomers when it melts, and doesn't have a liquid state with the same polymeric bonding regardless of the pressure
An ionic solid that sublimates upon heating, and under pressure can only assume a liquid state that clearly does not exhibit ionic bonding

What are some such substances? (Non-pure substances, such as wood, are non-examples.)

Interestingly enough, when considering not just the set of "typical" or "known" substances, but the vastly larger amount of all possible compounds, almost all pure substances do not have a thermodynamically stable liquid state or gaseous state. To become a liquid, a molecule must gain enough energy to set itself loose from the tug of its neighbours in the solid state. Typically intermolecular forces are fairly weak, and so for simple molecules this energy can be provided at relatively low temperatures.
However, there is no upper bound on how strong the intermolecular forces between two molecules can become; just take an enormous molecule. Even with relatively weak intermolecular interactions between each atom, when summed up across a huge molecule, the required energy becomes vast.
Combine this with the fact that, thanks to combinatorics, almost all possible substances are composed of massive molecules (in a chemical analogue of the frivolous theorem of arithmetic). For example, there are only five possible distinct substances with the formula $\ce{C6H14}$, but over a trillion trillion trillion trillion ($10^{48}$) distinct molecules with formula $\ce{C120H242}$.
The conclusion is that almost certainly the molecules would decompose (breakage of covalent bonds) well before getting close to the extremely high temperatures needed to produce a true liquid state.
Therefore, there should be many, many examples to choose from. A heavy-handed example would be maitotoxin (or many other natural products). Though the theoretical melting point would be very high, the presence of all those hydroxyl groups suggests it would decompose by dehydration around ~200 °C, as is typical of polyols (I'm sure it would actually decompose much sooner).
In the same polyol vein, a much more relatable example would be sucrose (sugar). In the strict thermodynamic sense, sugar doesn't actually melt when heated on a stove. The clear colourless syrup that's formed is actually the first step in caramelization, the complicated thermal decomposition of sugar.
The above cases represent your first suggestion of possibilities. Polymers also fit, with the complication that artificial polymers are almost never actually pure substances but mixtures of very similar molecular chains, varying in size. Something like  your last suggestion is interesting, but examples seem hard to provide. Perhaps dinitrogen pentoxide is close, due to its unusual transition between ionic and molecular forms.


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What is the definition of current efficiency?
What is the exact definition of current efficiency of an electrolytic cell? I've looked online and there is no suitable answer to it.
I thought current efficiency would be the amount of electrons that actually participate in a desired reaction. For example, if we are electrolyzing a an aqueous solution of $\ce{FeCl3}$ and we have a $\pu{10A}$ current running for, say, $\pu{100 s}$, at $60\%$ current efficiency, two possible products are $\ce{H2}$ gas and reduced $\ce{Fe}$ metal. Current efficiency of $60\%$  means that half of the electrons that pass through this circuit would reduce the $\ce{Fe}$ instead of reducing $\ce{H2}$.
However, there is another way to think about it, in that it means 60% of the theoretical mass yield of $\ce{Fe}$ comes out of the reaction instead of my original definition, which is different because the reduction of iron (III) requires 3 electrons rather than only 2 electrons for the reduction of $\ce{H2O}$/$\ce{H+}$ ions. So what is the actual definition?

The theoretical definition is the ratio of electrons used to make the useful product out of the total electrons that were supplied. The problem in the chemistry literature with efficiency definitions, however, is that you will more than likely find that different people will use different definitions for one reason or another (easier to calculate, some type of comparison in their work, or simply because it sounds or looks better). The right way to deal with efficiencies, then, would be to define how the efficiency is calculated in order for the reader to know what you are talking about. If this is an academic setting, I would stick to the theoretical definition, but ask your professor as to what their favorite definition would be.