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What happens if we continuously hit an atom with photons
My sir told me that Energy required to remove an electron from one orbit to another depends on hf * n. Where n means the no of photons that will strike on a metal surface and hf is energy of one photon.(Means energy of each photon is going to be 2eV)[Like the example below I have written ].
My question is that if I continuously hit many photons on a metal surface.Why doesn’t the electron come out then since the  energy also increases or hf’s.
If work function(energy required  of an electron ) is 3eV.
My hf value is 2eV and n = 6.Now wouldn’t the total energy value increase and electron should come out (Like from metal surface).?This as I know means that there are 6 photons of energy 2eV.

You have it all backwards. Roughly speaking, the number of photons is always 1. If your electron needs 3 eV, then no number of photons having 2eV will do the job.

Please explain more clearly. Your first sentence mentions the energy necessary to "remove" an electron. Remove ? Are you speaking of absorption of light, or emission of light ? In both cases, the absorbed or the emitted energy does not depend on the number of photons. Later on, you say that "you hit" many photons... What do you mean ? Photons cannot be hit ! They can hit a target. But they cannot be hit.

Have you heard about the photoelectric effect, explained by A. Einstein in 1905 ? In a shortcut. if photons have lower then the threshold energy, it does not matter how many of them is colliding. They will not cause the photoemission. ( Well, then can cause thermal electron emission, but that is other case. ).

@Maurice photoelectric effect

@IvanNeretin - multiphoton ionization is definitely a thing...

@JonCuster True, it is. But before learning that it is, one must learn that it isn't.

Though the explanation of the photoelectric effect is clear, and you should see how Albert Einstein won his 1921 Nobel prize for this, to some extent, it is possible for multiple photons to be absorbed, either simultaneously (or in sequence, if an intermediate metastable state allows it), to cause phenomena that would seem to be beyond their energy.
In "multiphoton fluorescence process, two or more photons are simultaneously absorbed by a molecule, which then emits a single photon at a different wavelength." That is, the energy of the photon emitted in fluorescence may be greater than that of either incident photon.
The Max-Planck-Institut für Kernphysik has an excellent description of the photoelectric effect and multiphonton effects.
One can consider these effects through classical physics, too: as more photons are "squeezed" in time and space, the electric field increases. In the extreme, high intensity fields produced by femto- and attosecond lasers not only tear off inner electrons, they can even cause nuclear particles to be ejected!
So your idea that perhaps the energy of photons could be summed is not entirely wrong... but it requires the photons to interact simultaneously with matter. This only takes place where the radiant flux is immense. Unless you're working with multiphoton microscopy or laser-induced nuclear fusion, the photoelectric effect is the accepted answer.

What does it mean by photon to interact simultaneously with matter. Like 2 photons join each other ?

To some extent, photons might interact with each other: https://www.sciencenews.org/article/lhc-atlas-photons-interact-physics . However, "simultaneously," here, means that within quantum limits, the wave functions supplement each other. This is the Planck on which quantum mechanic stands.

The photoelectric effect is explained in such a way that some people get it right away, but others (including physicists before Einstein) don't/didn't.
Try this: take a photon gun that shoots one photon at a time, with a selectable energy. Fire it at the metal target. Here's what will happen: At low energies (reddish light), the photon is absorbed (let's ignore reflection!) and shakes up the sea of electrons - makes it warmer, but does not make a splash. Turn up the energy of the photon gun (toward the blue end of the spectrum) and shoot again. Now the photon is energetic enough so that when it hits the metal - the sea of electrons in the metal - it actually bangs an electron so hard that it pops out.
The photoelectric effect simply says that electrons are bound into a metal and low energy photons are absorbed by the whole sea of electrons because no one electron is excited enough to leave the metal. But a high energy photon can knock an electron so hard - so fast - that it gets knocked out of the metal before it can distribute its energy to other electrons in the sea of electrons.
Consider the interaction of a photon with the metal as having 3 possible outcomes: a) reflection (which we have ignored), b) absorption with temperature increase (very familiar to people who go out into sunlight with black clothing), and c) electron ejection because the photon energy is enough to excite an electron so much that it is ejected. This third possibility is measured by physicists with specialized equipment that captures the ejected electron.
But we normal human beings can't observe that. Or can we? Sunlight with its considerable blue and UV components can bleach fabric and cause sunburn by generating free radicals (the residue of an ejected electron!). But an infrared lamp will not bleach fabric, and won't cause sunburn (although it can cause thermal burns - which are somewhat different, but I suppose equally painful, so don't experiment on yourself).

Amazing answer.Thank you very much.Just some more clarity form what you wrote.What happens if we do not ignore reflection ? And By black clothing , does it mean that the work function of black is so high that it absorbs a lot of photons and heat up.

While the answer is explanatory and more visual, we should avoid making macroworld analogies which are not correct, such as a photon banging an electron hard that it pops out or a high energy photon knocking the electron hard and fast. Modern physicists will perhaps frown upon these macroworld similarities.

@srijan nahar: I don't have numbers on work function of black material, but since anything will eject electrons if you hit it hard enough, it would seem that you have to hit black stuff harder.

@M. Farooq: You are quite right in insisting on the best analogies for introducing people to an understanding of complex phenomena. Modern physicists (and mathematicians) have a unique mental ability to condense complicated processes into simple symbols which are unfamiliar to beginners. Sometimes a picture (worth 1000 words) is more explanatory than a symbol worth 1e^6 words.


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Why does sodium citrate in water turn black when blueberries are added?
I recently started making oral rehydration therapy for someone I'm taking care of who developed POTS after COVID-19 infection. (ORT was recommended by their doctor.) I use the standard ingredients of ORT: sodium chloride, sodium citrate, potassium chloride and dextrose/glucose. To add some flavor, I tried blending in some frozen blueberries, but when I did the solution instantly turned black. The water became cloudy and opaque as well. By trying to recreate the reaction methodically, making four different solutions of blueberries and one of the other chemicals in water, I determined that this was a reaction between the sodium citrate and something in the blueberries.
Despite its off-putting color, the solution still tastes fine. However, I'm hesitant to serve it to the person I'm caring for without knowing for sure if it's safe to drink! What causes this reaction, and is it safe?

1.) yes, obviously harmless, 2.) we don't give medical advice here, 3.) find a new doctor, ORT is a method to make sure diarrhea patients (particulary cholera patients) get enough fluid.

As your solution contains sodium citrate, it is a little bit basic. As the blueberry juice contains an anthocyane dyestuff that changes color in basic medium, that is what you have made without knowing it. The anthocyanes takes a more bluish color if the solution is basic. That is the same phenomena that happens when cooking red cabbages : they turn blue when boiled in calcareous (or hard) water. This color change is safe and does not modify the taste.

It is "blueberry that turns black", or "a solution of sodium citrate in water when..."

@Karl you say 2) we do not give medical advice and 3) you just gave one.

@Maurice, that's very helpful! Thanks!

@Karl, off topic but the reason was that ORT can apparently address hypovolemia by causing the body to retain more water in the bloodstream. Not sure if that's actually true but that's what they told us! Then again POTS can have several different causes so hypovolemia might not even be the issue.

source: Stack Exchange

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Is there an "off the shelf" way to make magnesium ascorbate?
I make my own sodium ascorbate with ascorbic acid and sodium bicarb. It's easy, I get the vitamin c off ebay and the baking soda from the store.
Is there a similar way to make magnesium ascorbate: ingredients easily available, room temp, etc?

Except as a grassroots science experiment (grow crystals?), what is this good for?

I assume a milk of magnesia, what is essentially suspension of Mg(OH)2 is commercially available for per oral usage as antacid or laxative.

@Karl its a base for making liposomal C for personal use. Mg is typically more needed than Na in humans. Me especially ;)

@Poutnik interesting. I will try it. I suppose the reaction should be identical except swap a little H2 for the CO2 off-gassing? Anyway, make this comment an answer, I'll accept it. Thanks.

@rickopps What about H2 ??

@Poutnik just asking if Milk of Magnesia + Ascorbic Acid would off-gas H2? I did take Chem 101 and 102, haha, but that's all, and that was long ago ;)

I would recommend starting to eat wholesome food instead of potentially poisoning yourself with trendy supplementals, but then of course I'm no authority on nutrition. Good luck.

@Karl Yes, I agree wholeheartedly, however the liposomal C is amazing stuff. I am a baker by trade, so I love "soft" chemistry... making liposomal C is easy and fun to me, and it is crazy effective.

@RickOpps Thanks, but ... how is it "effective"? Effective in doing what? (Btw., baking is great applied chemistry. +1)

@Karl I mean as a Vitamin C in general, liposomal is more effective. Any other vitamin C is thought to be 10% to 20% available through the digestive system. It's easy to take enough C to give yourself diarrhea. But liposomal C is thought to be about 80% available. This is also my personal experience. It is very difficult to take liposomal C to "bowl tolerance". So, if you are into C, a Linus Pauling reader, etc. Liposomal C is a beautiful thing.

Linus Pauling had become a senile idiot when he was promoting vitamin C. His earlier scientific work was excellent and got him a nobel price, his taking a political stand against nuclear armament was admirable and got him the second one, but after that he turned into a pitiful pseudoscientist, which, out of respect for his achievements, I put down to senility. Can happen to anyone.

@Karl, well, lol, okay then. For others who happen upon this rabbit trail and are interested in this idiocy, please also look up Dr. Rhonda Patrick. She might be a senile idiot too? Who knows.

I assume a milk of magnesia, what is essentially suspension of Mg(OH)2 is commercially available for per oral usage as antacid or laxative.
Mixing with ascorbic acid solution will produce water and magnesium ascorbate.

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Why does enthalpy change differ in a chemical reaction because of temperature?
If enthalpy is a state function, shouldn't the change in enthalpy be the same between a given initial and final state, regardless of how the change took place? Kirchoff's equation describes this variation as
$$\Delta H_{T_2} = \Delta H_{T_1} + \Delta C_p (T_2 - T_1).$$
I'm guessing that this change in enthalpy change is only because of the variation in $C_p$, but I'm not too sure about it. Am I completely wrong about this?

The equation does look wrong to me. Shouldn't it be $$H(T_2) = H(T_1) + C_p(T_2 - T_1)?$$ Enthalpy is a state function.

Yes, enthalpy is a state function, and the rest is also true. Now what is your question?

You are right there. IF the total heat capacities of reagents and products  were by chance equal, the reaction enthalpy would be temperature independent.
But as these total heat capacities are generally different, the reaction enthalpy is temperature dependent. As the net heat for changing temperature of reactants and changing it back for products is nonzero.

@OneEyedMushroom Yes, the change in enthalpy is independent of path, depending only on the initial and final states. But ΔH_T2 is going to be different from ΔH_T1. That's because the initial and final states for a reaction at T1 are different from a reaction at T2.  Specifically, ΔH_T1 is the change in enthlapy in going from reactants at T1 to products at T1.  ΔH_T2 is the change in enthalpy is going from reactants at T2 to products at T2.  And your eqn is essentially correct, except it makes the common simplifying assuption that the diff. in Cp between reactants and products is T-independent.

@Martin-マーチン  The equation is OK,  because it's referring to the temperature-dependence of ΔH for a reaction, which in turn depends on the difference in heat capacities between the reactants and products (ΔCp).  Its one simpifying assumption is that it assumes ΔCp is T-independent.

If you're looking at the change at two different temperatures, then the initial states and final states are different?

@theorist now that you say it... Thanks for giving it a more thorough look.

@theorist but why DeltaCp it is thought as a difference? It could change differently along the path/Tactual... I don't understand this.

@Alchimista  Sorry, I can't make sense of your question.   Can you please rewrite it in a clear and precise manner so I can understand just what you are asking?

@theorist I don't understand why Cp is considered at the initial and final stages only. Shouldn't be its variation along the reaction important? Or its T dependance?

@Alchimista You'd need to look at a derivation of the eqn. to understand more fully, but Kirchoff's eqn does do that, i.e., it is obtained by integrating delta_Cp from T1 to T2 (the equations you see here are obtained by assuming delta_Cp is T-independent, so the delta_Cp can be taken out of the integral).

@Alchimista  I think it would help looking at my answers here:  https://chemistry.stackexchange.com/questions/142033/calculating-change-in-state-function-following-different-paths/142083#142083  and here:  https://chemistry.stackexchange.com/questions/61749/enthalpy-of-a-reaction-changes-with-temperature-does-this-mean-bond-strengths-c/132470#132470

Temperature is one of the state variables. If you change the temperature, you have a different state.

From Hess' law, the heat of reaction at temperatures $T_1$ and $T_2$ are related by $$\Delta H_{T_1}=C_{p,\mathrm r}(T_2-T_1)+\Delta H_{T_2}+C_{p,\mathrm p}(T_1-T_2),$$where $C_{p,\mathrm r}$ is the heat capacity of a stoichiometric molar mixture of reactants and $C_{p,\mathrm p}$ is the heat capacity of a stoichiometric molar mixture of products.


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Can bis-quaternary aromatic compounds act directly on acetylcholine receptors?
The book Cholinesterases and Anticholinesterase Agents gives examples of bis-quaternary aromatic compounds that are capable of inhibiting acetylcholinesterase. Page 400 gives examples of some such compounds, designated 2842CT, 3443CT, 3116CT, 3152CT, and 3113CT. The first three are neither organophosphates nor carbamates, while the last two are carbamates. The book goes on to list other bis-quaternary aromatic compounds that have anticholinesterase activity, but does not mention whether they can act directly on acetylcholine receptors. Are they capable of acting on ACh receptors directly, whether as direct agonists or antagonists?

Reference:
Koelle, George B. Cholinesterases and Anticholinesterase Agents. Springer Berlin, 2013.


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Why does the molar conductivity plot behave this way?
On the left image.. we see that specific conductivity is increasing at a decreasing rate, eventually plateauing as it reaches a maximum. I am unable to reconcile this with the molar conductivity graph to the right... shouldn't the decreasing molar conductivity be decreasing at an increasing rate... why does it plateau and such? Thanks for the help

have you checked the units in the right graph?

One effect is crowding -- not enough room for ions to be mobile. See https://byjus.com/chemistry/variations-molar-conductivity/ . Also, as @Karl states, one abscissa inlcudes concentration...

I dont follow why molar conductivity is concave up .. shouldnt it be concave down?

@DrMoishePippik It should. Check again. ;)

Yes it should... that was a hint to the bewildered.

A math note: AFAIK, a concave function is the one with y'' < 0, bending down. A convex one is the one with y'' > 0, bending up.

@DrMoishePippik I don't know if the curve is correct, but the ordinate has a unit that does not make sense. It must be $\rm{S cm^2 mol^{-1}}$.

@Karl uhh all the graphs i find online follow this general shape.. i thought it should be concave down though?

@Poutnik based on that definition.. the curve above is convex.. but i thought it should be concave? like since the curve on the left increases at a decreasing rate.. it should follow the curve on the right is bending down?

Well, the concentration range is at quite big numbers. But following the Kohlrauch law for molar conductivities at very low concentrations, it is convex even there : $\Lambda = \Lambda_0 - const \cdot \sqrt {c}$.


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Filling up a table about a chemical reaction
I am filling up this table but I'm not sure about what it means in the mass row, amount of substance, molecule quantity.
In mass do I just have to repeat 16 in the second column or do I need to put the mass of a single molecule?
Amount of substance 1 mole of the molecule or 5 moles of each atom?
I'm confused.

Is this some reaction that you carried out in a lab? Is there any other context to the question? If this is all that you were given, in the absence of any further detail, it is completely unclear what mass you are supposed to be putting in, so it's quite fair that you're confused.

Well not any other context than there is this simulator for chemical reaction balance, and the chemical reaction in the table comes in there, you have to solve all the other data after balancing the reaction. But the simulator does not contain any other data, just the reaction and blank boxes in the coefficients to balance it, like a game. thats it.

I think they just want you understand the various meanings of a balanced equation. For CH4, molar mass is 16 g/mol, mass is 16 grams, Amount of substance =1 mol, molecule quantity =1


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Wittig reagent with metal–oxo bonds
Could you use the Wittig reagent to attack a metal–oxo bond and replace the oxygen with a carbene?
Specifically, is it possible for the Wittig reagent to do σ-bond metathesis with an electrophilic metal attached to an oxygen?
As I'm trying to understand the mechanism, I would really appreciate an explanation for why or why not along with the answer.


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How does iodine react with amylose and amylopectin, and can iodine colorimetry be used to determine the ratio between the two starches?
Based on some of the methods of studying starches in rice I looked at, I found that there are two general distinctions:

ones that focus on the ratio between amylose and amylopectin
ones that focus just on amylose, measuring percentage content

The former included methods such as electrophoresis, amperometric, potentiometric and spectrophotometric procedures. The latter used procedures primarily based on colorimetry using iodine.
Due to technical constraints, I think that colorimetry will be a realistic option for my task.
However, I am unsure about the differences in measurement for the two types:

given that iodine creates different colors with amylose and amylopectin, can this be used to calculate the ratio?
if the sample contains more than just amylose and amylopectin (such as rice flour, rather than starch), can the iodine colorimetry procedure still be used?

Example of a procedure that uses iodine colorimetry:
[C. M. Sowbhagya; K. R. Bhattacharya (1971). A Simplified Colorimetric Method for Determination of Amylose Content in Rice. , 23(2), 53–56. doi:10.1002/star.19710230206]
Example of a procedure using electrophoresis:
[José Manuel Herrero-Martínez; Peter J. Schoenmakers; Wim Th. Kok (2004). Determination of the amylose–amylopectin ratio of starches by iodine-affinity capillary electrophoresis. , 1053(1-2), 227–234. doi:10.1016/j.chroma.2004.06.048 ]
*I have done as much research as I can on this, but I am not fully familiar with the field. Please forgive any misunderstandings.

If there are different colours, there can be analyzed absorbance at 2 or more wavelengths. ( 2 to differentiate them, 3rd and more to subtract the background ) , implying some standards are available.


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Help in understanding first way of changing the state of system
There is a statement in my book:

THERMODYNAMICS
One way: We do some mechanical work, say $\pu{1 kJ},$ by rotating a set of small paddles and thereby churning water. Let the new state be called $\mathrm{В}$ state and its temperature, as $T_\mathrm{B}.$ It is found that $T_\mathrm{B} > T_\mathrm{A}$ and the change in temperature, $\Delta T = T_\mathrm{B} - T_\mathrm{A}.$ Let the internal energy of the system in state $\mathrm{B}$ be $U_\mathrm{B}$ and the change in internal energy, $\Delta U = U_\mathrm{B} - U_\mathrm{A}.$

By rotating a small set of peddles, let us say we do mechanical work of $\pu{1 kJ}.$ Let the new state then be $\mathrm{B}$ and $\mathrm{A}$ as initial state.
It is found that $T_\mathrm{B} > T_\mathrm{A}.$ Let internal energy of system be $U_\mathrm{B}.$
They say that $\Delta U = U_\mathrm{B} - U_\mathrm{A}.$ Do we assume $T_\mathrm{A} = 0?$
We know $\Delta U = q + W.$ So, is $q = T_\mathrm{B} - T_\mathrm{A}?$
Should we say that work, i.e $W_\mathrm{A} = 0$ and same for $W_\mathrm{B} = \pu{1 kJ}?$
What is the difference in saying $\Delta U$ as internal energy and amount of mechanical work?

We never assume $T_A=0$. Indeed, that would make no sense.

If you are not changing the composition, q (the heat flow) depends on ΔT, the heat capacity of the substance, and the amount of substance; it is not equal to ΔT.   It would make no sense for it to be equal to ΔT, since ΔT has units of degrees, while q has units of energy.   I could give you the formula, but the fact that you don't realize it doesn't make sense to equate q to ΔT tells me you haven't yet read the section of your textbook on this subject.  Thus I think it would be more to your benefit if you first read it and then asked questions if something doesn't make sense.

A $T$ in context of thermodynamics always means absolute temperature, it cannot be T=0, as this state is reachable only after infinite number of steps. $q = C \cdot ( T_\mathrm{B} - T_\mathrm{A})$, where $C$ is a heat capacity of the system $C = \frac{đq}{dT}$. What is $W_\mathrm{A}$ ??

@Poutnik $W_A$ is mechanical work . From 1st law of thermodynamics. U = q + w

@user282657  I have not meant $W$ generally, but $W_\mathrm{A}$  ( and $W_\mathrm{B}$ ) particularly. Is not there 1 single value of $W$ to get us from the state A to the state B along a particular path ?

So $W_\mathrm{A \rightarrow B}$

1kJ is given to us

There difference should be 1kJ but am not sure if W_A would be then = 0kJ.@Poutnik


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Best Cheminformatics API for drawing SMARTS reactions?
I have a need to automatically draw SMARTS reactions. Rdkit's drawings are suboptimal. OpenBabel's drawings don't work properly on macOS. Indigo's drawings are cut off half the page.
I would love to use ChemDraw, but I can't seem to find an API for it to draw reactions with SMILES/SMARTS. I would absolutely be willing to write a wrapper to do this if there is a nice API to actually programmatically do drawings headlessly.

Also, what do you mean by "Rdkit's drawings are suboptimal"?

Open Babel works fine on Mac - I'm one of the developers and using a MacBook. The easiest solution is to export to SVG. But I'd recommend RDKit - it currently has the best depiction methods IMHO.

@StanislavBashkyrtsev RDKit's drawings are just ugly. For example, double bonds are not properly centered. Try drawing an aldehyde with an implicit hydrogen. You will see that the double bond does not align properly. Furthermore, the reagents in reactions are seemingly randomly rotated. They do not all have the same 30º, 150º, 270º to the horizontal that you would expect. Some are rotated differently than others for seemingly no reason, and this is not configurable. You need to hack it to not draw weird radicals when drawing SMARTS. Finally, bond lengths etc. are not configurable.

@GeoffHutchison OpenBabel reaction SMILES to PNG or SVG does not properly work on macOS. See more here: https://github.com/openbabel/openbabel/issues/2306

OpenEye is probably the most powerful and fast depiction SDK, but it costs money. Also OpenEye is a low-level, cumbersome, complicated library - so I'd consider it only if you have very sophisticated needs. It's written in C++, but it has Python/Java wrappers.
CDK (Java) has an OS-agnostic depiction API, works great for simple needs, haven't tried anything super complicated.
I tried Indigo several years ago and it worked fine, maybe you misconfigured it? Though it did require an additional OS-level dependency for fonts (Cargo?) on Linux. Don't remember if there were problems with other OS's.

OpenEye is great looking. I may shoot them an email to see if I can get some academic license since I am a university student intending to use this software for notetaking. The example drawings on CDK certainly look very promising. I installed Indigo via Homebrew first, and it did not come with indigo-depict. So, I downloaded the executable directly from their website. It produced quite an ugly depiction which was unusable because the whole drawing was offset to the left and downwards and several carbons of the leftmost reagent are cut off.

Accepting this answer because the CDK drawings are fantastic. They are almost as good as ChemDraw, and certainly much better than any of the other tools I have tried (OpenBabel, Indigo, and RDKit) Thanks!

One thing I will note is that unfortunately it apears that CDK does not handle reaction SMARTS but instead reaction SMILES. It does not properly handle things like degree, for example.

@ThorCorreia, you can raise a ticket for this: https://github.com/cdk/cdk CDK team usually is pretty responsive.


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Expression comparing the Solubility of Ionic compounds
In chemical bonding, I am taught an expression that compares the Solvation Energy and the Lattice Energy of an ionic compound which is being dissolved in a polar solvent (could be water, could be something else).
The expression is:
$$\left\lvert r_+ -r_-\right\rvert$$
where $r_+$ denotes radius of cation and $r_-$ denotes radius of anion making up the compound. When this difference is large, solvation energy dominates and when this difference is small, lattice energy dominates. But I don't the reason behind these deductions. Can someone explain?

If the sum of r+ and r- is about the same, the lattice energy may be about the same for various ratios r+/r-. But if this ratio diverges from 1, the increase of solvation energy of the smaller ion due being smaller is higher than the decrease of solvation energy of the bigger ion due being bigger. But it is my guess, so I do not post it as an answer.  It is like 1/3 + 1/3 < 1/2 + 1/4. All implies the same ion charges of compared ion pairs.


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Why are some weak acid equilibrium approximations invalid in very dilute solutions?
Using two approximations for the dissociation of a weak acid, we can write
$$K_\mathrm{a}= \frac{[\ce{H+(aq)}]^2}{[\ce{HA(aq)}]}$$
One of these two approximations we use is that the concentration of the acid HA has not changed upon dissociation, because the dissociation of a weak acid is so small compared to the starting concentration of the acid that is negligible.
However, I was taught that this approximation is invalid when the acid is stronger, but also when the solution is very dilute.
I don't understand why this approximation wouldn't work if the solution is very dilute. I have some ideas though:

The acid concentration decreases significantly when dissolved in a lot of water, so maybe the H+ would become significant then?

Maybe the H+ from water's dissociation affect the equilibrium of the weak acid's dissociation so that the dissociation becomes significant?

Are any of these right? Could someone explain why this approximation is invalid for very dilute solutions?

There is a general equation for acid base equilibrium, see the answer https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566 from which you can make approximations.

The above simplified formula considers 2 simplifying assumptions:

Concentration of hydrogen ions from water autodissociation can be neglected, so $[\ce{H+}] \approx [\ce{A-}]$ .
Concentration of an acid $c_0$ , in relation to its $\mathrm{p}K_\mathrm{a}$, is high enough for the fraction of ionized acid to be negligible, so $c_0 \approx [\ce{HA}]$ .

Especially the second assumption is broken for too dilute weak acid solutions. For more diluted solutions, higher fraction of an acid is ionized to match the equilibrium equation.
if we consider $$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$$
then if we dilute the acid 4 times, concentration of the conjugate base decreases just 2 times, so the fraction of the ionized acid form increases 2 times as well.
It also means the stronger the acid is, the higher concentration must be . As for the same concentration, stronger acids are more ionized and $c_0 \approx [\ce{HA}]$ causes a bigger error.
The rule of thumb is, for the formula to be acceptable, this inequality should be true: $$ - \log {c_0} < \mathrm{p}K_\mathrm{a} - 2$$
so maximally about 1% of an acid is ionized.
The first assumption is broken mainly for too weak acids. E.g. if $\mathrm{p}K_\mathrm{a} = 10$, practically all $\ce{H+}$ comes from water autodissociation and the equation fails.


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A lot of confusion in 1st law of thermodynamics
My sir told me that Total energy of system = K.E + P.E in starting.
Then Change in energy = F(External force on body ) * displacement of walls.
Then from here , change in energy = W+q.
(Don’t understand how we reached till here )
This change in energy is also called change in  internal energy.
Then it should be equal to Kinetic + potential energy of the molecules inside the system that we are considering .
Q1 is that if it is equal to kinetic + potential energy, then in case of ideal gases.P.E = 0.So K.E will be equal to change in energy $\delta U$.
Q2 is that if it change in energy , then it should be change in kinetic energy as well right.
Ok then my sir also tells me that when a reaction happens in non ideal gas scenario , example of 1mole N2 + 3 mole H2 = 2mole NH3.Now my sir says , there will be a lot of potential energy here.So we can’t use the formula kinetic energy = q+w here.
When Na(s) in converted to Na(g) we consider , delta H of sublimation of Na and then potential energy change I.e internal energy change = Na+(g) + e. Is this internal energy change $\delta U$.
All these doubts arrived from this question.
Because here they ask for internal energy change , so does it mean that it is $\delta H_1$ + $\delta H_2$ + Internal energy of 1 + internal energy of 2.
I am sorry if this in not written in good manner but am so much in tension right now that I just gave out everything I knew

you should focus on one general question instead and be precise

Change of energy is not simply F*displacement of the wall, as you write. This formula gives only the work entering or getting out of the system. This work increases the internal energy. But it is not the only way of increasing the internal energy. Heat (or heat transfer) is another way of increasing the internal energy.  And of course changing the chemical nature of the substances present in the container is third way of changing the internal energy.  What is Q1 and Q2 in your question ?

See Wikipedia - Internal energy changes and Enthalpy. Why to write again, what has been written about many times ?

See also Wikipedia - First law of thermodynamics and hyperphysics - First law of thermodynamics. The latter site provies a set of easy, cheat-sheet-like format of short simple pages.

@Poutnik Can we say enthalpy change as heat transfer when pressure is constant but not volume.?

Think longer. Then search a little bit. Then think again. If all that does not help then ask. But ask with results of your thinking.

I'm glad that you have got good concepts but the problem is you're unable to connect the discontinuous dots. See, in order to answer your first two questions I would like to bring attention towards the difference of microscopic energy (energy possesed by the molecules present inside the system) and macroscopic energy (energy you get to see in the system). As a example, a cup of coffee is placed over ground all will say it has zero K.E and 0 P.E. But the molecules present inside the system posses both K.E and P.E. You are right with the fact that change in internal energy for ideal gas means change in K.E but the change in K.E means, difference between the total K.E(summation of K.E of every molecules present in the system) of the sytem before and after the action. In case of non ideal gas, here both P.E and K.E both will contribute.
In case of sublimation of Na, here solid is changing to gas you can't neglect volume change also you have to keep it in mind that change in enthalpy is equal to change in internal energy if the PV-work is zero and the process is isobaric.

Am I right with the Na part I wrote as a whole.

No, you can't consider enthalpy of sublimation with internal energy without proper condition is provided.


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Role of calcium oxide in soda lime decarboxylation
Why do we use calcium oxide $\ce{CaO}$ in soda lime decarboxylation?
$$\ce{\underset{sodium acetate}{CH3COONa} + \underset{sodium hydroxide}{NaOH} ->[\Delta][CaO] \underset{methane}{CH4} + \underset{sodium carbonate}{Na2CO3}}$$
It seems as if it is dehydrating agent to keep NaOH dry), but why is the dehydrating agent required if the heat would be enough to dehydrate the container as the temperature is very high (expected to be over 100 °C)?

If you dehydrate with P205 you get the anhydride. Sodalime first forms the sodium salt. Sodalime is used in preference to sodium hydroxide for ease of handling https://www.chemguide.co.uk/organicprops/acids/decarbox.html

@Waylander I am working with sodium salt already

@Poutnik it feels no difference the reaction i provided is just one step ahead

@Poutnik sorry i didn't catch that

Calcium oxide is the typical catalyst used for this reaction. Caesium salts also catalyze this reaction. wikipedia - sodium acetate

NaOH does not need an extra note it is sodium hydroxide, CH4 is methane etc.

Calcium oxide is not necessary for carrying out this reaction. It only prevents NaOH from touching the glass of the container where the reaction is done. If a solid mixture of sodium acetate and NaOH is heated to high temperature, some NaOH will attack the glass before reacting with acetate, and make the glass melt down. If NaOH is intimately mixed with CaO, this trouble is avoided. The glassware is saved.


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Why does hydrogen bonding require a lone pair?
Why does hydrogen bonding require a lone pair of electrons on the partially negative atom? They would still be partially negative even without the lone pair right? Or does a bond exist, but is much more negligible compared to when a lone pair is there?

In a H-bond electron density is donated to the 'naked' proton, so this electron density has to come from somewhere. There needs to be some orbital overlap, so there must be a lone pair.

You don't need a lone pair. Water can hydrogen bond to methane, attacking the back side of one of the carbon-hydrogen bonds in the methane. In a molecular orbital the sigma bonding orbitals in the methane are sufficiently delocalized to serve in place of lone pairs, overlapping the antibonding orbitals in the water [1]:
Abstract (from [1])

Quantum chemical calculations were performed at different levels of theory (SCF, DFT, MP2, and CCSD(T)) to determine the geometry and electronic structure of the HOH···CH4 complex formed by water and methane molecules, in which water is a proton donor and methane carbon (sp 3) is an acceptor. The charge distribution on the atoms of the complex was analyzed by the CHelpG method and Hirshfeld population analysis; both methods revealed the transfer of electron charge from methane to water. According to the natural bond orbital (NBO) analysis data, the charge transfer upon complexation is caused by the interaction between the σ orbital of the axial С–H bond of methane directed along the line of the O–H···C hydrogen bridge and the antibonding σ* orbital of the О–H bond of the water molecule. Topological analysis of electron density in the HOH···CH4 complex by the AIM method showed that the parameters of the critical point of the bond between hydrogen and acceptor (carbon atom) for the O–H···C interaction are typical for Н-bonded systems (the magnitude of electron density at the critical point of the bond, the sign and value of the Laplacian). It was concluded that the intermolecular interaction in the complex can be defined as an Н bond of O–H···σ(С–H) type, whose energy was found to be 0.9 kcal/mol in MP2/aug-cc-pVQZ calculations including the basis set superposition error (BSSE).

Reference
1. A. N. Isaev,"O–H···C hydrogen bond in the methane–water complex", Russian Journal of Physical Chemistry A 90(2016), 1978–1985.

That's sure interesting (not being a hard proof, though), but I have a feeling OP simply doesn't get how h-bond works...

The hydrogen bond in $\ce{CH5^+}$ is so weak if compared to $\ce{H3O+}$ that it looks more like van der Wals forces.

Hydrogen bonds occurs when a H atom is bound to a highly electronegative atom like oxygen. In such a bond O-H, the 1s electron of hydrogen is mainly localized between the H and O nuclei. So, at the outer part of the bond, the H nucleus is nearly not surrounded any more by an electron layer. The H nucleus is nearly "naked". Nothing prevents it from touching and attracting electrons from the vicinity, and specially these electrons which are not engaged in a bond, like those in outer layer of the Oxygen atoms. This electrostatic attraction is called a Hydrogen bond.

There's so many errors in this post that, the list could be longer then your answer. See for example https://chemistry.stackexchange.com/a/5506/9961

I'm actually surprised.  With nearly 10k rep I assume your answers can be better than this.

What do you have against my explanation of the Hydrogen bond ? It is made for high schools students.

We should be teaching high school students the truth, not an oversimplified stereotype.

High school students cannot understand Schrödinger equation.

If they understand molecular orbitals, they can understand a proper renderinf of hydrogen bonds.

@Oscar Lanzi. Even so, the explanation would not differ much from what I have written 20 hours ago. Of course the words "naked" and "touching electrons" are nearly a non-sense. But they are well understood by the average student. It is not so easy to be on a level with the average high school student. At least, that is nearly what H. Linkhorn has written in the very beginning of this discussion. And he got points for this answer !


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when is a system of mass-action reactions guaranteed to satisfy detailed balance?
when does a system of mass-action reactions satisfy detailed balance for any non-zero choice of the rate constants?
consider a trivial example:
$X + Y \ce{<-->[k1][k2]} XY$
satisfies detailed balance for any $k_1, k_2 > 0$. Here is a less trivial example:
$A + X \ce{<-->[k1][k2]} AX$
$B + X \ce{<-->[k3][k4]} BX$
we can ask if this system satisfies detailed balance, ie whether for any cycle the product of the rates going clockwise must equal the product of the rates going counter-clockwise (Wegscheider condition)?
It appears that any positive non-zero values of $k_1,k_2,k_3,k_4$ this system satisfies Wegscheider. Is that right and if so is there a simple property of the structure of such reactions that we can check to tell if it will satisfy Wegscheider?

Where do you see a cycle her ?

@Maurice: AX is formed, then BX is formed (path 1) vs. BX is formed, then AX is formed (path 2)

Is it a cycle ?

@Maurice can you point to a definition of a true cycle or argue why this isn't a cycle?

@userip. A cycle is a series of reactions going from A to B, then from B to C, then from C to D, and then from D to A. For example NO plus O2 gives NO2, NO2 plus water gives HNO2 and HNO3, then HNO2 gets NO + HNO3, and NO is already obtained in smaller proportion, to start the cycle again. The final result, after many cycles is that all NO is transformed into HNO3.


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treating PLA (Polylactic acid) plastic to make it biodegradable
I am using PLA for 3D printing, and I know it does not degrade well in the wild. I would like to know if there is a chemical process I could use to treat it at home.
Specifically, I am looking for a cleaner alternative than throwing my 3d printing scraps in the garbage, so I wouldn't be generating toxic components when doing this, ideally. I would like the plastic to degrade (in a few years maybe?) when I put it underground, while limiting the amount of toxic matter propagates in the soil.
I have read that hydrolyzing PLA could help achieving this.
I also found threads on disolving it How to dissolve PLA (polylactic acid)?
But as I am looking for an ecological process, I think I need to stay away from this as the fumes may be very toxic and polluting
Thank you for your help!

You could try leaving your PLA material in a water/vinegar solution for a few days (weeks?). The acidic solution should help hydrolyze the ester bonds, breaking down PLA into lactic acid. Might take a while

That seems like a good idea! Do you know if that might generate fumes, or other toxic byproducts? Or if Heat will help the vinegar act faster? (I have no autonomy in chemistry, maybe there is a resource online that might help me predetermine inputs/outputs of a reaction?)

Assuming your material is pure PLA, the only byproducts of hydrolysis (reaction between PLA and water) will be lactic acid. The vinegar (acetic acid) simply acts as a catalyst. Heat will help. The only safety concern is if you end up boiling off all the water and leave behind concentrated acetic acid. Keep the temp around 60C and do it in a well ventilated area.


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Mole Fraction and Atom Fraction
I'm probably overthinking this but would appreciate a peer check.
Mole fraction of element $i$ in compound: $\frac{\mathcal{M}_i}{\mathcal{M}}$
I used script M so as not to confuse with molar mass (often denoted as capital M).
Atom fraction of element $i$ in compound: $\frac{N_i}{N}$
Now, if Avogadro's number is atoms/mol then isn't the mole fraction equal to the atom fraction?
$$
\frac{N_i}{N}\frac{[\text{atom}]}{[\text{atom}]} \cdot \frac{N_a}{N_a}\frac{[\text{atom}/\text{mol}]}{[\text{atom}/\text{mol}]} = \frac{\mathcal{M}_i}{\mathcal{M}} \frac{[\text{mol}]}{[\text{mol}]}\\
\frac{N_a}{N_a}=1\\
\frac{N_i}{N}=\frac{\mathcal{M}_i}{\mathcal{M}}
$$

Usually the mole fraction is mostly used in studying mixtures of different substances, like the osmotic pressure of sugar solutions, for example, or the partial pressure of Oxygen in air. Atom fraction is mostly used when determining the formula of a pure compound. by elementary analysis. In water the atom fraction of hydrogen is 2/3. Atom fraction can also be used to describe the composition of complex mixtures, like the elements in Earth, or in the human body.

The mole fraction and atom fraction will be the same only if you are clearly talking about atoms. But this is often not what is being talked about in typical chemistry conversations. We might talk about the moles of sodium and the moles of sulfate in sodium sulphate, for example. If you always specify moles of what then you are OK but that if often not done and this leads to ambiguity.

The fallacy lies in the assumption that mole fraction and the number fraction are always the same.
For example, take $\ce{Na2SO4}$
what is the atom fraction of Na: 2/(2+1+4) = 2/7
What about the mole fraction of Na? It 2$N_A$/$N_A$(2+1+4)= 2/7
However, this "simple" formula of yours breaks down in the case of mixtures even if we mix them in 1:1 mol ratio. Calculate the mol fraction of Na ions in a mixture using your formula for $\ce{NaCl}$ and $\ce{Na2SO4}$ and see that you will not get the right answer.


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Diastereotopic methylene protons in achiral molecules
Are the protons in methylene groups of achiral molecules such as propane-1,2,3-triol or 3-hydroxypentanedioic acid diastereotopic?
Consider propane-1,2,3-triol. To me, replacing one of the protons in either -CH2 with an arbitrary X nuclei creates (what I work out as) 2 diastereomers. However, NMR simulations of propane-1,2,3-triol in ChemDraw and on nmrdb.org (https://www.nmrdb.org/new_predictor/) do not flag these hyrdogens as diastereotopic and neither does this "Diastereotopic atoms" computational tool (http://www.cheminfo.org/Spectra/NMR/Tools/Diastereotopic_atoms/index.html).
I have also played around with physical models of each isomer and cannot find a plane of symmetry to make them enantiomers.
What am I missing here?

Well, I can't speak to your NMR simulator and computational tools but from my perspective, and that of ChemDraw, the hydrogens at C1, and C3, in diol 1 are diastereotopic. While diol 1 is meso and achiral, introduction of deuterium in 2a renders both C1 and C2 stereogenic. Consequently, 2a is a diastereomer of 2b where the deuterium has been moved to the geminal C1 position. You may want to look at  this.<b

You are correct, and these protons are indeed diastereotopic: see Magnetic non-equivalence of acetal protons for a similar example. These type of protons often catch people out, so it's good that you spotted it!
As for what you're "missing", it's simply that some of these tools are not particularly advanced. I don't know about the others, but ChemDraw's NMR prediction only uses a limited set of heuristics to predict chemical shifts, and in more complex cases these heuristics won't suffice. I highly doubt that any of them use symmetry information to determine which protons are equivalent.
For another example of ChemDraw NMR prediction fail, see:

Even though this is an achiral molecule, it should be incredibly clear that all of the $\ce{CH2}$ groups here are diastereotopic. ChemDraw isn't smart enough to figure this out. In fact, it isn't even smart enough to figure out that the pairs of protons marked 1.46 and 1.40 should be equivalent (or more precisely, they should be two equivalent pairs of inequivalent protons).

Nice link. I had forgotten about it. ;)

Great answer, thanks. My ChemDraw version does actually recognize all the symmetry of the molecule you show but fails on many others!


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How to determine the source of sub-nm periodic error in NIR spectra
I've run into an issue collecting measurements on deionized water in the $1050-\pu{1200 nm}$ range. To determine the sensitivity of my spectrophotometer in this range, I've acquired measurements on pure deionized water in double-beam mode using two identical quartz cuvettes. I have taken measurements at several time steps near $\pu{1064 nm}$, from $0.1-\pu{100 s}$ per time-point and constant $\pu{0.1 nm}$ per step and $\pu{2 nm}$ bandwidth. I keep finding that the baseline standard deviation (SD) does not improve beyond $1.4 \times 10^{-4}$ Abs.
This value is technically below the acceptable qualification values of $2 \times 10^{-4}$ Abs in that range, but I'm trying to determine if this is the ultimate noise level of the instrument in this range or if these periodic variations are due to a mechanical issue in the monochromator like a lead screw, or due to environmental or electrical line noise of some kind.

Any thoughts are appreciated.

I had that happen with a UV spec once. If I recall correctly, the culprit was the chopper motor. Check if the frequency matches up.

The periodic variation appears to show 10 steps per nm and your step size is 0.1 nm per step. Perhaps try a different step size and see what happens to the periodicity.

Interesting problem. You can try a couple of things. I assume that you already have the raw data exported in Excel or MATLAB. Something which we traditionally neglect is the error in the x-axis itself. For each export wavelength, plot the difference of wavelength vs. data number. Do you see a pattern or not? If it is a true and constant sampling rate, it should be a perfect straight line. However, I recently had problems with microwave spectroscopy data and the x-axis "spacing" had a lot of jitter as they call it. It can be periodic too.

If your x-axis spacing is not "perfectly" identical, you might see this artifact as a function of integration time.

By “chopper motor” I meant the stepper motor which usually has a chopper drive.

What is also weird is that you have periodic noise as well.

Have you considered the possibility that you are observing an optical etalon effect from reflections from the parallel faces of the cuvettes and the introduction of phase shifts that manifest themselves as the sinusoidal variations that you see in the wavelength space?


text:
Is there a notion of generalized acidity for lithium ions?
From my understanding, the basic notions of acidity can be explained by considering a cup of pure water, and then adding a substance to it which creates either free $\ce{H+}$, or free $\ce{H3O+}$ and $\ce{H5O2+}$ etc...
And these ions are free to react and give rise to what we consider acidic behavior. The classic example is hydrogen chloride characterized with the following reaction
$$\ce{HCl +H2O = H3O+ + Cl-} $$
Whereas $\ce{H3O+}$ is our famous hydrochloric acid.
I wonder if it is possible for an analogous reaction to happen with other polar molecules. For example liquid $\ce{Li2O}$ (I believe it melts around 1438 Celsius) should be a polar liquid for the same reasons that $\ce{H2O}$ is (though maybe not as strongly polar), and therefore: $\ce{LiCl}$ should dissolve in molten $\ce{Li2O}$ to create positive charged lithium ion species such as $\ce{Li3O+}$. Which ought to function as an "acid" of sorts, in that they react with most thing which are dissolved/reacted by hydrochloric acid by acting as a positive charge donor (clearly not a proton donor since we are using lithium and not hydrogen to do the deed here)
Is my line of reasoning sound and has this been experimentally detected before?

I'm not sure if you want more then that Li cation is obviously Lewis acid

(so I can get the terminology right) our $\text{Li}_2 \text{O}$ fluid with dissolved $\text{Li}\text{Cl}$ is a Lewis acid?

$\ce{Li+}$ is a Lewis acid that might coordinate to $\ce{Li2O}$ base forming an adduct. I highly doubt you can get much $\ce{Li2O}$ molecules outside of vapor phase, thogh.

A complex of lithium ion with lithium hudroxide is known as the salt $\ce{[Li2OH]^+[ClO4]^-}$ [1].  As with the hydronium salt, this requires an essentially nonbasic counterion; with a chloride counterion a double salt is obtained with no evidence of a complex cation.  From the abstract (some evident faults in translation from Russian have been corrected):

An effort is made to obtain salts of Li2OH+ and Li3O+ cations, belonging to a series of simple oxonium compound cations H3O+, LiOH2+, Li2OH+, Li3O+, where proton and lithium replace each other. Binary compounds of lithium hydroxide with lithium chloride and perchlorate are obtained: LiOH·LiClO4, LiOD·LiClO4, LiOH·LiCl, LiOD·LiCl and LiClO4·2LiOH. A conclusion is made by IR absorption spectra, that an adduct with perchlorate can be built of ClO4 anions and complex [Li2OH]+ cations. For adducts with chloride, equivalence of hydroxide and chloride anions with regular alternation of the anions in the crystal lattice is most probable.

Reference
1.
Nikitina, Z. K., Nikitina, N. I., Rosolovskii, V. Y., "Complexing Lithium Salts and Lithium Oxide or Hydroxide: Synthesis and Characterization of a New Cation-Lithiumhydroxonium Li~2OH^+", Russian Journal of Inorganic Chemistry 42(9), 1295-1298  (1997).


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How can water below a pH of 7 still be alkaline?
Alkalinity is a measure of the water's ability to neutralize acidity and is expressed in levels of bicarbonates, carbonates, and hydroxides.
How can water that is slightly acidic, e.g. 5.5-6.5, still boast any alkalinity? Or put another way, is there a maximum level of alkalinity for  water at any given acidic pH?
This is based on my understanding that acidity is, put simply, when there are more H+ than OH- ions. So why don't all carbonates react with the surplus H+ ions at pH ranges 5.5-6.5 until either alkalinity is 0 or the pH has reached a neutral 7?

From Wikipedia: "Alkalinity … should not be confused with basicity which is an absolute measurement on the pH scale". Looks like that's exactly what you are trying to do here.

Thanks you two!
So, alkalinity has nothing directly to do with basicity. Instead, it has to do with compounds that will react with (buffer) acids, correct? What confuses me is perhaps what an acid is. I thought an acid is simply a solution with surplus H+ that, subsequently, just raises the H+ surplus when added to water. Which will subsequently "agitate" the buffer.

For your purpose, an acid is a compound able to  release H+, a base is a compound able to either release OH- or to bound H+. Note that a free base can exist in acidic solutions, if it is a weak enough base for given pH. Similarly for acids. A very weak acid can exist in its acidic form even in a concentrated solution of a hydroxide.

A free base can exist in acidic solutions if it is sufficiently weak. Would you expect that to change if the solution was agitated, e.g. stirred and or heated?

No, it is an equilibrium condition. If e.g. pKa of acid is 13 and pH=11, then about 99% of the acid is at the equilibrium conditions in its acidic form and 1% in its conjugate base form ( being quite a strong base ).

Classically, alkalinity is determined by $\ce{HCl}$ titration with methyl orange as indicator ( $\mathrm{pH}$ transition range $3.1 - 4.4$ ). IF some $\ce{HCl}$ is spent by lowering $\mathrm{pH}$ from $5.5-6.6$ to $\mathrm{pH}$ of noticing red colour, there is some nonzero alkalinity.
Note that at $\mathrm{pH} \approx 6.3$, the ratio $\dfrac{[\ce{CO2}]}{[\ce{HCO3-}]}$ is 1 : 1. (*)
Generaly, there is no direct relation between $\ce{pH}$ and maximal alkalinity, similarly as there is no direct relation between $\ce{pH}$ and $\mathrm{pH}$ buffer capacity.
But for typical water composition, containing $\ce{CO2/HCO3-}$ $\mathrm{pH}$ buffer, the alkalinity can be estimated from $\ce{pH}$ and expected $\ce{CO2(aq)}$ concentration, which is supposed to be in equilibrium with gaseous $\ce{CO2}$.
$$[\ce{HCO3-(aq)}] = K_{a1,\ce{CO2}}^{*} \cdot \frac {[\ce{CO2(aq)}]}{[\ce{H+(aq)}]}$$

(*) The  Wikipedia page for $\ce{H2CO3}$ has been recently misedited,  confusing $K_\mathrm{a2}$ and $K_\mathrm{a1}^*$(assuming 100% conversion of $\ce{CO2(aq)}$ to $\ce{H2CO3}$.


text:
As electron energy increases, does ionisation energy decrease?
"Electrons are negatively charged, therefore, two electrons repel each other. When electrons are paired in a 3p orbital there is electron pair repulsion which causes the energy of the electrons to increase. Increased energy makes the electron easier to remove and therefore the ionisation energy decreases."
I found this statement when studying ionisation energy. Is it correct to say that as electron energy increases, ionisation energy (EI) decreases? I understand the factors affecting EI (atomic radius, nuclear charge, shielding and electron pair repulsion) but have not seen this directly compared to electron energy before. Do each of the factors increase/decrease electron energy or is it irrelevant to think about it in that way?

It is very analogous to gravity. If the satellite mechanical energy ( PE + KE ) increases, is the energy needed to leave Earth gravity smaller ? Note that energy of bound electrons is referred to a free electron in rest as zero reference value, so electron energy + electron ionization energy = 0.

The ionisation energy is fixed, so increasing the energy of an orbital means that less energy is now needed to remove an electron from that orbital.

Kinda.
An atom or ion has multiple configurations that can be (very roughly) described as electrons (arrows directed up or down) put into boxes * (representing individual orbitals each capable to accomodate one up-electron and one down-electron). This description ignores many subtleties and configurations that can be tranformed into each other by repositioning 'boxes' belonging to one shell should be considered one state, but as a very rough description it goes.
Assuming this description, you can attribute several ionization energies to particular state, depending on which one electron you removes. Those different energies may be sometimes directly observed in proper experiments.
In this case, yes, removing an electron from doubly occupied cell takes less energy than from single-occupied cell of the same subshell, assuming all other things equal. The last part is important, because there are other effects affecting relative energy of states.
This, however, doesn't mean it is easier in common sense. In quantum physics processes are characterized by their probability and event of particular type, say a particular channel of ionization, may be be more or less probably than event of another type, say a different channel of ionization.
For this reason using word 'easier' is a bad idea. Use more precise words whenever possible.
* Like in this question Electron orbital diagram of vanadium

First of all, the energy of any electron is decided by the energy level in which it stays. So if electron is very close to nucleus that means it is in high energy level which then corresponds to it's high energy and if it's quite away or far from nucleus then it's energy is low. So question says if energy of electron is high , then it clearly understood its close to nucleus in high energy level hence energy required to remove it, that is Ionisation energy is also .

So you say electrons in the 6s orbital have lower energy than those in the 1s orbital, and the orbitals are filled from those with the highest energy. Does not it sound suspicious to you ?


text:
Difficulty in understanding ionisation of sulphuric acid
$\ce{H2SO4}$ when added to water.
$\ce{H2SO4 + 2 H2O -> SO4^2- + 2 H3O+}$
My sir explained me these reaction as $\ce{H}$ will give $\ce{2 H^+}$ ions to the base $\ce{H2O}$.So why does it not become $\ce{H4O^+}$ ?
To try to make myself understand this , I checked my textbook.
$\ce{H2SO4 + H2O <=> H3O+ + HSO4^−}$
$\ce{HSO4^− + H2O <=> H3O+ + SO4^2−}$
I got even more confused that why is H2O is used twice in both the reactions?Why not use the products of 1st equation and then do the same reaction again.The one difference was that we substituted H20 instead of H3O there.
$\ce{HSO4^−  H3O+ <=>} $ why not this ?

Added proper formating and fixed 0 vs O errors. There is no compound like $\ce{H20}$ nor a ion $\ce{H30+}$.

$\ce{H3O+}$  does not accept another proton to become $\ce{H4O^2+}$.

My sir explained me these reaction as H will give 2H+ ions to the base H2O It does not mean to a single molecule, but H2O as a compound.

@Poutnik Ok.Then what does it accept ?

0 instead of O again ?  I have checked the editing history and no, I have written 2 times H2O instead of H3O. I have just replaced zeros by capital O. Or, I may have not got this your question correctly.

Let us continue this discussion in chat.

We are only giving H+?

not H4O2+. you have to edit it to $\ce{\underset{ }{H4O^2+}}$

Please read "The Aqueous Proton Is Hydrated by More Than One Water Molecule: Is the Hydronium Ion a Useful Conceit? by Todd P. Silverstein in the Journal of Chemical Education. J. Chem. Educ. 2014, 91, 4, 608–610. It is freely available from Google Scholar.
The latest thinking, past 2006, is that $\ce{H3O+}$ does not exist in any appreciable amount in aqueous systems. This ion exists in organic solvents. So consider all these protonated versions of water in your textbook as a very simplified version of a complex reality in water. The hydrated "proton" may look like this, where the blue part is the anion. Water chemistry is fascinating and yet to be fully explored.
Looking at the structure, where do you want to insert the second proton?

This image looks awfully flat to me, and I would actually se quite a few places to stick another proton. Anyway, as I usually do with water posts: I link to the Grotthuss Mechanism, as that is  always important to consider.

Yes, Grotthuss mechanism is still remarkable despite its age. What students have to realize is the the H+ is mobile, so there is no rigid structure per se.

The simple answer why $\ce{H4O^++}$ (note that if another proton is added, the charge must be a double positive) doesn't form is that a water molecule has far more affinity for protons than the $\ce{H3O^+}$.
And there are also far more uncharged water molecules in the solution than products of the first protonation step. Don't forget this is taking place in a water solution so, although we don't usually include the solvent in the reaction equation, neutral water molecules are, by far, the most common molecules in the solution. So, even setting aside the fact that the second proton prefers to add to an uncharged water molecule, the likelihood of there being a water molecule available is far, far higher than there being a $\ce{H3O^+}$ andywhere.
Don't forget that reactions like this take place in a solution and that the molecules of the solution may participate in the reaction.

$\ce{\underset{hydronium ion}{H3O+}}$ may not further react because oxygen already has a formal positive charge and it maynot further bear a more positive charge to form $\ce{\underset{ }{H4O^2+}}$.
As there are approximately $10^{21}$ molecules of water in a drop and (you carry a test tube) so very large number of water molecules exist. $\ce{\underset{bisulphate anion}{HSO4-}}$ can react further with water(which is amphoteric in nature)

downvoters please don't downvote anonymously. Use your own identity. By adding a comment

See also https://chemistry.stackexchange.com/questions/112087/why-doesnt-h%E2%82%84o%C2%B2%E2%81%BA-exist

@Poutnik a new thing to learn for me. I will change cannot to maynot. Don't want to extend the answer a bit long

If people bother to comment, they bother to explain their downvote, if they did so. Or ,they downvote as an anonymous user, without any comment.

While oxygen may bear a positive formal charge, it does not bear an actual positive charge. It also would not become significantly more positive adding another proton. After all $\ce{OH4^2+}$ is isoelectronic with methane. Some advice in non-chemical matters: Don't ask for justification of down-votes in comments; those who voted won't see it anyways and it usually puts people off who would have commented. Also you're only asking for identity, not why those people may have chosen to vote this way in the first place, that's certainly going to make people uneasy. Me, for example.


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Are all exothermic reactions necessarily spontaneous?
I was told that the statement "all exothermic reactions have negative free energy change" is false.
But I learnt Gibbs free energy as the "energy stored inside a system capable of doing non-$pV$ work", so I've always thought that exothermic reactions should have a negative change in free energy, i.e. $\Delta H < 0 \implies \Delta G < 0$. Why isn't this true?

Almost a duplicate (perhaps actually a duplicate) of Are all exothermic reactions spontaneous? I will leave it to the community to vote.

If all exothermic reactions were "spontaneous" in any meaningful sense, then people would catch fire in an atmosphere containing oxygen. If byt "spontaneous" you mean something more restrictive than the normal use of the term, then you should specify what you do mean.

There is thermodynamically based spontaneity due $\Delta G \lt 0$ and there is kinetically based spontaneity assuming $\Delta G \lt 0$  AND the process is not kinetically frozen ( Otherwise all diamonds near room conditions would turn into graphite. ).  By other words, for perceived spontaneity, the thermodynamic spontaneity is a mandatory but not sufficient condition.

Separation of solid $\ce{CaCl2 . 6 H2O}$ and water from $\ce{CaCl2}$ solution would be exothermic, but is not thermodynamically spontaneous, as entropy increase of the surrounding is smaller than the entropy decrease of the system and therefore $\Delta G > 0$. Spontaneous is the opposite,  endothermic process.

Does this answer your question? Are all exothermic reactions spontaneous?

Here is a good example of your statement. The reaction $\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$ is endothermic and spontaneous, because $\Delta $S is huge.. When started at $25$°C with a somewhat concentrated solution of HCl, the temperature goes down to $0$°C by adding NaHCO3. So the reverse reaction is exothermic and not spontaneous.

The Gibbs energy $G$ is defined as
$$G=H-TS$$
For a process at constant temperature, the change in Gibbs free energy $\Delta G$ is then defined by:
$$\Delta G^o = \Delta H^o - T\Delta S^o$$
This quantity $\Delta G^o$ should be negative, for a reaction to be spontaneous.
So if a reaction is exothermic then $\Delta H^o$ is negative, but now the conditions depend on the entropy change $\Delta S^o$, as well as temperature:
Case 1: If $\Delta S^o > 0$ then no problem.
Case 2: If $\Delta S^o < 0$, then up to a certain temperature the reaction is favourable by enthalpy, but above that the entropy dominates over the enthalpy making the reaction non-spontaneous.


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Is the ratio between base pairs by chargaff accurate?
I am reading Lehninger's biochemistry textbook.  It mentions that DNA may rarely contain uracil.  Then it mentions that Chargaff found that the ratio of adenosine bases to thymine bases in DNA is 1.
If uracil is present, will this ratio be disturbed?
Is this the A:T ratio being 1 true, or the ratio is general and approximated?

If the uracil is formed by deamination of cytosine after replication, it would not affect the A:T ratio. If some uracil gets replicated, it would pair with A, skewing the A:T ratio. There are repair mechanisms that continuously re move uracil from DNA so it would not accumulate.

It is true that sometimes uracil could be present in DNA.
Uracil is almost exclusive to RNA but in some cases deamination of cytosine can cause uracil to be present in DNA.
I believe what Chargaff proposed was very crucial in understanding the structure of DNA. It was way back in 1950 (Chargaff ratio gave the clue to the structure of DNA) and we've advanced lot more now.
Just like Markovnikov's rule (organic chemistry) gets opposed when a little re-arrangement of carbocation occurs.
Scientists just try to make things simple. Having exceptions to rules is nothing new in chemistry :)
Moreover, the answer to your question that whether Chargaff was accurate? From everything that was available for research that moment, yes.
I think we for now should have gratitude for what he's done.
You find it easy today because of the ratio. Imagine he hadn't proposed it because of this exception?


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Is there geometrical isomerism in case of 2-methylcyclopentanone?
How can 2-methylcyclopentanone show geometrical isomerism?
A ring and a double bond are in different planes, so even after having restricted rotation there will be no change in areal distance.

The 5 C ring isn't flat.

The bonds C=O and C(1) - C(2) defines a reference plane. The methyl substituent may be above or under this horizontal plane. These two possibilities define two optical isomers.

If you look at the structure of 2-methylcyclopentanone, you'll notice that it is asymmetric about the C-2 atom

This gives rise to two isomers; the R isomer on the left and the S isomer on the right

These are definitely optical isomers, more specifically enantiomers of each other. Whether these are 'geometric isomers' or not is a matter of how you define 'geometric' isomerism; the IUPAC Gold Book strongly discourages the usage of this term, instead preferring to use 'cis-trans isomerism'. If it is defined by the latter, then this compound does not display cis-trans isomerism, as there are no groups for the methyl substituent to be trans to.

@NiveditaChowdhury Please don't post a "thank you" as a comment. Instead, upvote or accept answers that were useful to you.

This answer is partially correct. The fact that there is optical isomerism does not cancel the fact that being the ring non planar, the methyl and the oxygen can point both up or to "opposite" directions. This call for an analysis of conformers.


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How to refine manganese metal from manganese(IV) oxide found in batteries?
I opened up some lantern batteries, and inside there's $\ce{MnO2}$ powder, carbon rods, and zinc casings, and other stuff that I don't know what it is exactly. I scraped out $\ce{MnO2}$-containing powder, but I would like elemental manganese from it.
How can I obtain manganese from manganese dioxide? How strong a reducing agent do I need? How hot does it need to get? Can I just add charcoal powder and smoke it with an oxyacetylene torch?

Make sure you use eye protection. I was once cutting up a PP3 since I needed the connector in a hurry and one of the cells- which these days are little steel tubes rather than the old-fashioned wads- went off in my face.

You may heat mixture of $\ce{MnO2}$ with charcoal to higher than $\pu{850 °C},$ in the absence of air. Manganese will be produced in the solid state by reduction of the oxide. But it is sensitive to air oxidation and easily reoxidized. So better use an excess of charcoal. In the industry, rough $\ce{MnO2}$ out of the mine is first reduced to $\ce{MnO}$ by $\ce{CO}$, then leached and reduced into $\ce{Mn}$ metal by a solution of $\ce{Fe^{2+}}$ in the absence of air. Metallic manganese is then dissolved in sulfuric acid and then reduced by electrolysis. See "Manganese metal" over Wikipedia.

Reduction with charcoal would indeed make sense as the manganese(IV) oxide in the batteries is always already containing graphite for better conductivity.

@ andselisk : Excellent idea !

You can also do the manganese dioxide version of the thermite reaction, aka, Goldschmidt reaction. Just use the manganese dioxide in place of iron oxide. I did this when I was a kid with a basement lab. I used the dried black “manganese dioxide” just like you have. But, if you do this, be really careful: this is an exuberant reaction and safety gear is essential. Also do it outside! I never did this a second time because it was too exciting, if you get my drift.

@Ed V. You are right. And this thermite reaction words also with the oxide saved after complete extinction of the battery : $\ce{Mn2O3}$. It is a useful way of using a waste substance

I already upvoted your answer, by the way: that is how I would do it if I had it to do over. ;-)

Research notes in this ACS article Kinetics of Hydrogen Reduction of Manganese Dioxide that even heating $\ce{MnO2}$ in an atmosphere of hydrogen gas only results in $\ce{MnO}$.
The source further notes that very high temperatures (some 1,600 K) are required for any appreciable formation of the metal.
As such, a more facile approach, simply dissolve the hydrogen created $\ce{MnO}$ (or $\ce{MnO2}$ in concentrated HCl per Wikipedia) in an acid (say sulfuric or possibly acetic) and place the soluble manganese salt in a vessel with a sheet of pure Aluminum (no oxygen exposure).
In time, the $\ce{Mn}$ will be plated out onto the sheet where it can be mechanically removed.
Or, a related alternate path, convert the $\ce{Al}$ sheet into two pieces (electrodes) for electrolysis procedure.


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How to measure chlorine content using ORP sensor?
In my project I need to measure pH, turbidity, and chlorine level of treated drinking water. I found sensors for measuring pH, turbidity, and also chlorine.
The actual problem begins is that the cost of chlorine sensor is too high, so I googled for an alternative solution, which led me to use an oxidation-reduction potential (ORP) sensor.
Can anyone tell me about the relationship between pH, chlorine, and ORP value in millivolts and how to measure chlorine content in treated water using ORP sensor?

Homework. You need to do the literature study for your project.

Can anyone tell me about the relationship between pH, chlorine, and ORP value in millivolts?, There is no relation because the ORP does not know what species it is sensing.

For a school project like yours, you can use iodometric titration. Iodometric titration is one of standard methods of water examination. The amount of free chlorine in tap water is really small on the order of < 1 parts per million therefore you need a lot of sample. One would typically take 500 mL of tap water (or more), add potassium iodide, and titrate with 0.01 M sodium thiosulfate. You can use an ORP and plot a potentiometric titration curve.
Consult your teacher for this option. He/she should explain to you the details.
There are spectrophotometric methods but the reagents may not be available in your school.

Based on the comment by M. Farooq with respect to the oxidation-reduction potential (ORP) sensor, I would suggest a pure chemical test.
In an inverted container containing the subject treated water, to test for dissolved chlorine and hypochlorous acid, simply inject a volume of stabilized hydrogen peroxide.
The following reactions should govern the system:
$\ce{ Cl2 (d) + H2O (l) <=> HCl (aq) + HOCl (aq)}$
$\ce{ HOCl (aq) + H2O2 (aq) <=> HCl (aq) + H2O (l) + O2 (g)}$
The last is a fast reaction and I would rapidly note (by marking) the volume of oxygen rapidly released. Subsequent elevation in released oxygen could be produced by other catalytic agents acting on the H2O2 itself.
I would recommend performing a trial run on distilled water to which is added a known amount of Sodium hypochlorite. Repeat with boiled tap water plus NaOCl and observe any growth in oxygen release with time.

The amount of free chlorine in tap water < 1 ppm. This test will not work at very low concentrations.

Per this table at https://www.google.com/search?q=ppm+to+moles&rlz=1CAHKDC_enUS750US750&oq=ppm+to+mol&aqs=chrome.2.69i57j0i457j0l6.13439j0j15&sourceid=chrome&ie=UTF-8 , assume since my tests would convert  gaseous, dissolved, and chemically converted chlorine (as HOCl), say 3.5  ppm or about 1/1000 of a mole . As 1 mole equates to over 22,000  ml of O2, this here equates to over 20 ml of O2 gas. So employ an inverted container whose shape would facilely allow such a measurement. Agree, however, special equipment required. Hence, my testing recommendation.

Thank you for sharing your knowledge. This is not a school project, its an industrial project, I need to measure ph, turbidity, and chlorine content and display the values using LCD in the wastewater treatment plant. I found ph and turbidity sensors and i wrote code for it, but now am struct on chlorine measurement, there is a sensor for measuring chlorine but its cost too high, ORP sensor measure the chlorine content but I don't know the relationship between chlorine ph and ORP sensor. I got one from google.(shorturl.at/lwQZ3)].can anyone explain the graph

Again prepare test samples (using the existing water of interest) to which is added further known amounts of added say Ca(OCl)2 in acidified (with CO2 or very dilute mineral acid) solution. See how accurate your electronic sensor is and by repeating with varying amounts, proceed to construct your own graph.  Some graphical fitting/extrapolation likely needed. For comparative purposes to assess accuracy, you can test alternate procedures as well (my answer is one such selection).

Be mindful that Cl2, HOCl and even added H2O2 are impacted, in the water of interest, by the presence of organic matter and transition metals. In fact, HOCl is around 1,000 times more chemical reactive in this regard than H2O2 (so,  your existing water ferrous content, and its exposure to say iron or copper pipes, organics,..., is very important).


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Hampson-Linde cycle: why does expansion need to be isenthalpic
If I understand correctly, the fluid in the Hampson-Linde cycle goes through an isenthalpic (basically a free?) expansion. For ideal gases, T would remain the same. But, depending on the substance and the temperature, the fluid may cool down.
My question is: why enthalpic? If the work (of the expanding fluid) could be used, would that not (1) lead to a bigger cooling effect and (2) useful energy output?
(I'm a beginner, so the answer might be obvious...)

You are talking about the Joule-Thomson step in the process, right?

Yes, that's right

Are you familiar with the derivation of the open system (control volume) version of the first law of thermodynamics?

In this case, would that be (all per unit mass): work + enthalpy[out] - enthalpy[in] + 1/2(speed[out]squared-speed[in]squared) = 0. Somehow I still don't get it. If work is done, then outgoing enthalpy goes down, meaning bigger cooling?

In the derivation, work is subdivided into two distinct parts:  Work to force mass into and out of the control volume, and shaft work.  The work to force mass into and out of the control volume (per unit mass) combine with the inlet and outlet internal energies to give the enthalpies in and out.  So these are accounted for.  And, in the flow through the orifice or porous plug situation in JT, there is no shaft work done..

Right, so what I'm thinking is, say, a turbine stage (mounted on a shaft) located at the orifice. Voila - nice shaft work and a cooler fluid. But surely I am making a mistake, I just can't see it?

I don't quite follow.  But it does make sense to me that, if you replaced the orifice with a turbine, you could cool the working fluid even more and, at the same time, extract some work.  If you're asking why they don't do that, I don't know.  I would have to study the details of the cycle more deeply.


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Does π values vary in Hansch-Fujita equations for a given substituent?
I was reading about Hansch-Fujita equations from Hansch-Fujita paramaters and observed that the $π$ values of substituents (analogous to Hammet $σ$ values) change according to the specific environment as depicted in the following table:
I can't understand why is that the case. I mean in Hammet equations the value of $σ$ for a substituent doesn't vary when changing the environment and what varies is the $ρ$ which is a constant specific for each reaction according to Hammet Equation. That is in Hammet equation we have:
$$\log{\frac{K_i}{K}}=ρ\cdotσ$$
But Hansch-Fujita define $π$ (the analogous of $σ$) as:
$$π=\log{K_Χ}-\log{K_Η}$$
So does they have "include" the $ρ$ constant in the value of $π$?


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What's the chemical reaction between coffee and sodium bicarbonate?
Every morning I put sodium bicarbonate into my coffee and I was wondering what chemical reaction this produces and what are the chemicals produced at the end.
I cannot find an answer to this question online. I can only find the reaction between chloridric acid and sodium bicarbonate, which is:
NaHCO3 (aq) + HCl (aq) → NaCl (aq) + CO2 (g) + H2O (l)
This produces co2 that makes me burp and burn my throat. This is the exact reason I started putting my sodium bicarbonate into my coffee, rather than my mouth. But I was wondering what is happening now.

"I started putting my sodium bicarbonate into my coffee, rather than my mouth. — That is, you do not take coffee orally?:) Look, coffee contain alkaloids (and caffeine is one of them) that will affects the secretion of hydrochloric acid in your stomach, which would lead to different rate of reaction with baking soda you are drinking with or without coffee. But mixing soda and caffeine alone won't result in a reaction.

You are not supposed to eat/drink sodium bicarbonate or add it to the coffee? How much do you add? A pinch or a tablespoon?

Furthermore, $\ce{CO2}$ will never "burn your throat" as you say. Never !

The reason I put sodium bicarbonate in the cup of coffee (and not directly into my mouth) is that in this way I don't burp for some reasons. From the answer below it seems to understand that the reason I don't burp is because co2 is released in the coffee cup rather than in my stomach. Also, co2 cannot burn my throat but it's a fact every time I start burping my throat start becoming painful and I tend to lose my voice. I suspect the burping due to co2 is bringing up something else from my stomach to my throat. I hope it's more clear now what I mean

It has only indirect relation to coffee. Coffee is known to trigger reflux of stomach acidic juices ( typically up to 1% HCl ) and bicarbonate acts as antacid. In hot solutions it tends to release half of CO2, forming carbonate. Both react with natural  coffee organic acids and the rest with HCl when they meet. But AFAIK bicarbonate usage as antacid is rather deprecated.

@user1883212, Please consult a doctor, burning sensation after a burp shows acid reflux from the stomach. Still adding bicarb to coffee is a very bad unhealthy idea.

@M. Faroow why is adding sodium bicarbonate to coffee unhealthy per se? To the OP, normally one would enjoy a nice coffee then use an anti acid... The right way should be skip coffee, in your case.

Alchimista, The OP has not shared how much soda bicarb he/she is adding.

Per a source, to quote:

Here are the nine major acids in coffee, listed from highest concentration to lowest: chlorogenic, quinic, citric, acetic, lactic, malic, phosphoric, linoleic, and palmitic (3). The brewing process releases acids from coffee beans, giving this beverage a pH of 4.85 to 5.10, which is considered acidic.

As such, the action of Sodium bicarbonate with these acids are examples of neutralization reactions producing the corresponding Sodium salts and releasing carbon dioxide gas.
You may wish to research the possible salts for their respective properties. In particular, the same reference further notes with respect to possible health issues:

While coffee’s acidity is fine for most people, it may aggravate certain health conditions in others.

These conditions include acid reflux, gastric ulcers, and irritable bowel syndrome (IBS). Coffee’s effects on these conditions are mainly attributed to its acidity and slight laxative effect in some people (6, 7, 8).

There are also a large number of ascribed benefits of drinking coffee as noted in this source.

This may be a good answer to the question how it is presented, but the real issue behind is treatment of stomach reflux and excessive stomach acidity, for which coffee is kind of contradicted. The bicarbonate  is not here to neutralise coffee acids, but as antacid to neutralise stomach HCl.


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The rusting of steel may be promoted by brushing with HCl followed by brushing with hydrogen peroxide solution. What is the chemistry involved?
A welder showed me a way of promoting rust on steel by brushing on muriatic acid and then a hydrogen peroxide / water solution. What's going on, chemically speaking?

You know that all substrates need to be equation? Also steps are usually written separately.

First muriatic acid is an old name from the 18th century. Forget about it. Today it is hydrochloric acid. Second, an equation must show the same atoms on both sides of the horizontal arrow. So Iron must appear in the left-hand side of the equation.

@Maurice: It is sold as Muriatic Acid in the hardware store. It is HCl diluted with water.

@Mithoron: No, I don't. I am not a chemist.

Why all the hate for the question? Is this forum intended only for professional chemists and chemical engineers?

Why promoting rust on steel ? Usually it is the contrary. Rust is an unwanted phenomena that everybody is trying to fight or to prevent.

@AlLelopath Stack Exchange sites are not forums, and no, Chemistry.SE is not strictly for professionals. Some comments may appear harsh, but they get straight to the point. Please don't take them personally, they are solely targeted at your question, not you. I think this is a good question, but the main reason for downvotes, I suspect, is the chemical reaction scheme which does not really make sense. I would probably exclude it at this point, and go with textual description.

I downvoted because both the title and “chemical equation” are nonsensical and there is an informative article on rust in wikipedia. As for what the welder said, I do not doubt it because the hydrochloric acid just cleans off oxides and such and then the hydrogen peroxide solution provides the two other necessities for the rusting of iron in steel: oxygen and moisture/water. I will remove my downvote if the title is fixed and the “chemical equation” is deleted.

And rust is not Fe2 O3.

@Maurice: "usually" being the operative word there. See for example https://en.wikipedia.org/wiki/Weathering_steel

@EdV: I did not know that the muriatic acid was for cleaning purposes, so thank you. I've removed the equation. What should the title be?

Perhaps something like this: “Apparently, the rusting of steel may be deliberately promoted by brushing with hydrochloric acid followed by brushing with hydrogen peroxide solution. What is the chemistry involved?”

I understand the point of your question, here is a discussion of the associated science.
The process remains a classic electrochemical process where here H2O2 is a replacement for oxygen and the require $\ce{H+}$ is provided by the $\ce{HCl}$ which is also a source of chloride ions noted to promote corrosion as well as forming salts which are excellent electrolytes.
Some references, starting with a good depiction of the underlying electrochemistry, to quote:

Iron (or steel) rusting is an example of corrosion -- an electrochemical process involving an anode (a piece of metal that readily gives up electrons), an electrolyte (a liquid that helps electrons move) and a cathode (a piece of metal that readily accepts electrons). When a piece of metal corrodes, the electrolyte helps provide oxygen to the anode. As oxygen combines with the metal, electrons are liberated. When they flow through the electrolyte to the cathode, the metal of the anode disappears, swept away by the electrical flow or converted into metal cations in a form such as rust.

And some comments from ScienceDirect on the referenced chloride properties:

The chloride ions present in the water promote local corrosion such as crevice corrosion and pitting, and the higher the chloride content, the more corrosive the water.

I hope this helps.

This confuses me in that it says the HCL provides H+ which contrasts the comments above in which I'm told the HCL is just for cleaning.


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How do you connect a battery to a galvanic cell in order to reverse the flow of electrons?
Assume you have a galvanic (voltaic) cell with E = +2V. If you wanted to reverse the current by inserting a 9V battery, which end of the battery should connect to the cathode of the galvanic cell?
The Kaplan answer key states that the "positive end of the battery should connect to the cathode."
Since electrons flow from anode to cathode in the galvanic cell, we want them to flow from cathode to anode to reverse the current. In a battery, electrons move from the negative to positive end. Therefore, why wouldn't the positive end of the battery be on the galvanic cell's anode to attract electrons in order to reverse the current?
Could someone point out the flaw in my reasoning?
Thanks :)

Electrons always flow from anode to cathode. When you reverse the flow, you rename the electrodes. See https://chemistry.stackexchange.com/a/135974/72973

Note that it is not specific for galvanic cells, the answer is general for any voltage sources. So the question belongs rather to electronics or similar than chemistry.

Poutnik, Please don't make chemistry so narrow that the public literally start thinking that chemistry is nothing but titrations. You are good at electrochemistry and this very well belongs to electrochemistry.

The electrochemical signs and labels have confused several generations of students. You have to remember a couple of things about the labels for the electrolytic and galvanic cells.
Anode: The electrode at which oxidation takes place.
Cathode: The electrode at which reduction takes places.
The signs of the electrodes of a galvanic cell are electrostatic signs (positive and negative) of the electrodes. This is an important point to keep in mind, which is often not emphasized during teaching. So (-) sign on the battery terminal indicates excess of electrons and (+) sign indicates deficiency in electrons.
In a galvanic cell, the cathode is positively charged and the anode is negatively charged. Work this out.
In order to reverse the flow of electronic current in the 2V galvanic cell, you need to connect the electron rich electrode of the 9V battery to the electron rich electrode of the galvanic cell i.e., you will connect the negative terminal of the 9 V battery with the negative terminal of the 2V galvanic cell and connect the positive terminal of the battery with the positive end of the galvanic cell.
Kaplan's wordings is quite bad!
Consider the analogy of water supply in your own house: When you open a tap, water spontaneously flows out because the actual source of water is at a higher gravitational potential than your tap. What do we need to do to reverse the direction of water flow in the tap. We need to connect the tap to a powerful water pump which overcomes the gravitational potential energy of the water source and pumps the water against the gravitational potential. This battery connection is exactly the same because the battery generates 9V and the galvanic cell produces 2V.

(+) Well explained!


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Does carbonation permanently increase the acidity of soda water?
If you carbonate water, and then expose the resulting soda water to the atmosphere and wait until it goes completely flat, will the (former) soda water end up more acidic in final equilibrium than water that was never carbonated and is also exposed to the atmosphere?
(Assume that both the water that was and was not carbonated are exposed to the exact same atmosphere, so that (a) they end up at the same temperature, and (b) the tiny addition to the level of atmospheric CO2 released during the carbonation process itself doesn't have any differential impact between the two waters.)
I understand that the soda water will be significantly more acidic than the flat water while it's still bubbly, because of the carbonic acid formed by the water and the injected pressurized CO2. I also believe that both waters will end up slightly acidic (although not as acidic as the soda water while it's still bubbly) because of the (modest) partial pressure of atmospheric CO2 in contact with both waters. My question is whether the high concentration of carbonic acid introduced into the soda water during the carbonation process permanently changes the equilibrium acidity of the water.
I'm also curious whether the answer would be different depending on whether you start with pure distilled water vs. "realistic" tap water, where the presence of impurities might somehow change the story.
My strong physical intuition is that the flattened (former) soda water will eventually "forget" about the carbonation and both waters will end up at the same equilibrium acidity, but some chemist friends disagree with me.

You're right. Henry's law states that at equilibrium the amount of gas dissolved in solution depends on the partial pressure of that gas which is in contact with that solution. So it doesn't matter if you have no carbonation or excess carbonation, when the solutions reach equilibrium then they will have the same amount of carbonation.

Both water samples ( assuming pure water ) would theoretically end up slightly acidic with $\mathrm{pH} \approx 5.6$, but the carbonated one after long time. Tap water would have its baseline $\mathrm{pH}$ determined by its $\ce{CO2(aq)/HCO3-(aq)}$ $\mathrm{pH}$ buffer.
This is valid for an ideal case, as side processes - especially for tap water -  will affect the result. As examples come to mind microbial/algae consumption of carbon dioxide, forming compounds affecting $\mathrm{pH}$, or precipitation of carbonates.

This is an interesting question and environmentally related significant topic.
First, a pertinent related observation on dissolved oxygen content, to quote a question and its answer, which I believe to be factually accurate:

How does salinity affect Dissolved Oxygen solubility?

Dissolved oxygen decreases exponentially as salt levels increase. That is why, at the same pressure and temperature, saltwater holds about 20% less dissolved oxygen than freshwater.

So, likely adding say $\ce{NaCl}$ to carbonated water reduces dissolved carbon dioxide levels.
Per another source  discussing the environmental problem of ocean acidification:

As atmospheric CO2 levels increase due to anthropogenic causes, dissolved CO2 also increases, which in turn decreases the pH of water. When water becomes saturated with CO2, it not only reduces the ocean's pH, but depletes the calcium carbonate sources as well ³⁵

As such, I would expect the impact of adding salt to carbonated distilled water would result in some reduction of dissolved $\ce{CO2}$ leading to an elevation of pH.


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Tendency of formation of complexes in Group-1 elements
My book states the following:

Down the group(Group-1), tendency to form complexes decreases due to decrease in charge density$\bigg(\frac{\text{charge}}{\text{radius}}\bigg).$ This results in decrease in ion-dipole interaction between metal atom and the approaching ligand.

My logic-
There is a sudden availability of inner d-orbital while moving from 3rd period to 4th period. This can help accommodate the incoming lone pairs. So the tendency to form complexes must increase.
So combining the two facts, the trend should be irregular- decrease down till sodium, suddenly increase in potassium, and then slowly decrease till Caesium.
Can you provide any insight into the theory I have given above?

The fallacy is the assumption that "inner $d$ orbitals" become "available".  Generally they are not, in the Group 1 and Group 2 metals.  There are rare occasions where some evidence of $d$-orbital bonding is found for heavier G2 elements, but the impact is small and not widespread.
In Groups 1 and 2, where there are inner $d$ orbitals they are so well shielded that an electron placed there is not tightly bound and thus in general tends not to stay put.  Somewhat ironically, $d$ orbitals become available to accept electrons from ligands just at the point where in neutral atoms they are being filled (i.e., the transition elements).  This is because we add protons to the nucleus as we add electrons to the subshell, and the electrons we are adding to the $d$ orbitals do not fully shield the remaining $d$-orbital states from the increasing protic charge.  It's as if the ligand electrons need to see their doppelgangers in the swimming pool before they figure out the water must be fine.
In truth, without readily available $d$ orbitals and with low charge density (except maybe lithium), the alkali metals are poor at forming complex ions.  That includes complexing with hydroxide ions, thus explaining the unique basic strength of the hydroxides of these metals.

Thank you for your insight! It helped a lot!