text:
CuSO4 in SO2/SO3 gas sampling in Sulphuric Acid Production

What role does $\ce{CuSO4}$ has in the gas sampling of an $\ce{SO2/SO3}$ gas mix in a catalytic converter for sulphuric acid production?

To give you a little context: We take gas samples from the beds of the catalytic converter to determine the efficiency of the catalyst. We use "cartridges" that we make with glass tubes filled with $\ce{CuSO4}$. The $\ce{CuSO4}$ (lab grade) comes in crystal version as a pentahydrate. We heat it for around $\pu{30 minutes}$ at $\pu{105 ^\circ C}$ and then we use it to make the cartridges. The cartridges are placed between the sampling point and the sampling bag (made of Teflon). This is a process that has been made since many  years but nobody here is able to tell me why. The answer they give me is that the cartridges are used to "remove" the $\ce{SO3}$ from the sample but I can't see how that could happen. My theory is that because anhydrous $\ce{CuSO4}$ is hygroscopic (I've read it's been used as a desiccant) it may be used to avoid the formation of $\ce{H2SO4}$ that could damage the gas analyzer (a SICK S700). I've really googled it but can´t find the answer. Hope someone could enlighten me in the subject.

Where do you install your cartridges? I suppose you are oxidizing a gaseous current of SO2 + O2 over a catalyst (either vanadium oxide or platinum). SO2 is oxidized in SO3, and this gaseous current is dissolved into somewhat diluted sulfuric acid. Where do you put your cartridges in this setup ? Does the gas go through your cartridges ? Before or after catalytic oxidation?

That's exactly the process as you describe it. We use the cartridges on sampling points after and before each bed. The gas passes through the cartridges and ends in the sampling bag. The gas analyzer measures %SO2 and %O2 v/v and with this data we determine the conversion to SO3 un each bead and of the full reactor.

Heating hydrate copper sulfate at 105 degrees does nothing. The chemically bound water takes a higher temperature to remove all water molecules (water is lost in steps as a function of temperature). This low temperature removes only two water molecules out of five. I think the major role of copper sulfate, besides being a gentle dessicant, is to oxidize sulfur dioxide to trioxide. Cu(II) ion is a mild oxidizing agent, and mild enough that it can oxidize sulfur dioxide to the trioxide.

I don't think that would be the case because the gas analyzer only measures SO2 and O2 in %volume. Or do you think this oxidation has another role?

You would have to share the whole protocol with us. By sharing a small section of a sample preparation, nobody can predict what is going on and what is being analyzed.

That's all the sample preparation really, and the copper sulphate it's just prepared that way: heated 30 min at 105 Celsius. The copper sulphate it's poured in the glass tubes (they are about 15 cm long) each and both sides of the tube closed with a small piece of ceramic fiber. The sample is taken before and after each catalyst bed on bags then it's taken to the analyzer and we get %SO2 and %O2 v/v values. With these we calculate %conversion on each bed. Hope it helps. Thanks

Actually, I could not find any reference on the reaction of sulfur trioxide with  semihydrated copper sulfate. One speculation is that sulfur trioxide may react with the remaining water in the crystals. Then the question is why just semi hydrated copper sulfate? SO3 is also a good dessicant. At ambient temperature CuSO4 may not be a strong oxidant to oxidize sulfur dioxide.

Do you think the water in the semihydrated sulphate molecule could react with the SO3 in the sample to form H2SO4 in the cartdrige and thus, eliminate this SO3 of the sample?

Yes, it may do that. If you pour conc. on hydrated CuSO4 i.e. CuSO4.5H2O crystals, it becomes white because the acid dehydrates it.


text:
Relation between rate of adsorption and the critical temperature of a gas
How is the rate of adsorption related to the critical temperature of a gas? My textbook says more the critical temperature, better are they adsorbed. But I cannot really understand why. They threw in another confusing statement that says that easily liquefiable gases are readily adsorbed. I am thoroughly confused.
I did read the question How does critical temperature affect adsorption? before posting this question, and while that question was answered, it did not answer my question fully.
Edit: To clarify my area of doubt, let me say what I find confusing. Gases with lower Ct will 'prefer' the gaseous phase over the liquid phase. Which is what we want. We want gases to be adsorbed.

I think you have it the wrong way around in the last sentence. (roughly:) Higher Tc means stronger intermolecular interactions means more easily condensed means adsorbs more readily.

Buck, I read that article before posting this question, and while that question was answered, it wasn't to my satisfaction.

As for your answer, I am still uncertain as to why we need the gas to condense. It still is in the gaseous phase when it is adsorbed. Or is it that gases with lower Tc are more difficult to tie down, so to speak. I think I am getting a hang of it. Thanks Buck.

You can argue that the effect of the solid is to "seed" formation of the condensed phase, as a set of interactions that lower the chemical potential below that of the gas, leading to "condensation" (well, adsorption). This is all very approximate as the details matter of how (as in, through what interactions) the adsorbent interacts with the gas molecule.
In addition, the magnitude of the critical temperature may be considered as a proxy for the strength of intermolecular interactions. For instance, if you assume the gas follows the van der Waals equation of state, then the critical temperature $$T_\mathrm{c} = \frac{8a}{27bR}$$ where $a$ is a measure of the strength of attractive interactions. You can find other ad hoc justifications for associating higher $T_\mathrm{c}$ with stronger strength of adsorption. For instance, larger, more polarizable molecules will tend to have higher boiling points and critical points. Such molecules will also adsorb more strongly. However the argument that the rate of adsorption increases should be considered carefully, since steric effects may be associated with larger molecules.
Regarding the point that "we want gases to be adsorbed", well, the point is that you want to predict behavior. The book merely arms you with guidelines that allow you to make predictions. So that, for instance, if you have a number of compounds in a gaseous mixture, you can estimate which adsorb more. Or perhaps you can use that ability to select new substances with desirable properties based on the adsorption behavior of other compounds.

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text:
Lewis Structure of OF+

Would the Lewis Structure for OF+ be drawn so that the more electronegative atom has the more negative formal charge, or would it be drawn to satisfy the octet rule?

Here are my thoughts thus far:
For one, we could make the formal charge on the oxygen +1 and the formal charge on the fluorine 0, and give them a single bond, which would break the octet rule, but make sure that the more electronegative atom (fluorine) had a less positive formal charge. Oxygen would have 4 free electrons, and fluorine would have 6 free electrons.
However, we could also make the formal charge on the oxygen 0, and the formal charge on fluorine +1, and give them a double bond, which would be wrong in terms of electronegativity, but would mean full octets. Oxygen would have 4 free electrons, and fluorine would also have 4 free electrons
How do I decide?

Carbon monoxide is commonly rendered with a triple bond that gives the electronegative oxygen a positive formal charge, and its dipole is negative on carbon.

A good way to think about stuff like this, is to imagine a possible synthesis for the given intermediate.
One way of making this species would be to cleave the H-OF bond of HOF against it's electronegativity, that is, heterolytically cleaving the bond to make H- and [OF]+
As we are going against electronegativity, you get the idea that the intermediates are quite unstable, and so your structure must represent a reactive species
Then, we can simply find out the structure by the curly arrow diagram, assuming that making the structure of HOF is easier than [OF]+
Another reason why your double-bonded structure won't work is that due to the close electronegativities of O and F, the MO picture would probably resemble that of O2 and F2, and so the pi bonding orbitals would experience at least some cancellation by filling of the antibonding pi orbitals(and even the O-F bond would be destabilized a little by the so-called 'antibonding effect')
As a final note, if you are not willing to evoke MO theory, then you can simply say that due to small size and close electronegativities of O and F, a pi bond will mostly not be favorable between them.

@JoshuaRonis What I meant was, OF+ has 16 electrons, so it's MO will resemble O2 with just the MOs rearranged by electronegativity(as Karsten pointed out below). Check out MO diagram of O2; since electrons enter into pi* orbitals there, hence the pi bond of O2 weakens, moreover the sigma bond also weakens due to the 'antibonding effect'(check it out as well). Happy to help :)

Explanation for the downvote would be appreciated

what do you mean?

@joshuaronis Oh sorry, I wasn't addressing you, actually, somebody has downvoted my answer some time ago, so I wanted to ask the reason from whomever it was. My bad :)

A strategy different from the one described by Yusuf Hasan would be to look at the neutral isoelectronic molecule. The oxygen atom is isoelectronic to the fluorine cation, so $\ce{OF+}$ is isoelectronic to $\ce{O2}$.
It is difficult to write a nice Lewis structure for $\ce{O2}$, see https://chemistry.stackexchange.com/a/15061. Experiments show that dioxygen is paramagnetic, i.e. it has unpaired electrons. If you try to write this as Lewis structure, you end up with 7 electrons on each oxygen (counting electrons in bonds twice), i.e. no octets either.
For $\ce{OF+}$, you should be able to write the same Lewis structure (and the same MO diagram). Because fluorine has a different number of protons than oxygen, however, the energy levels will be different and "the best" Lewis structure might be different as well.
As Yusuf Hasan mentioned, the struggles to write a "nice" Lewis structure for this species hints at high reactivity, so I think this is the main lesson of trying to write a Lewis structure. Which of the Lewis structure is best at approximating the real situation becomes secondary, and it makes sense to move on from a Lewis structure model to more comprehensive models of bonding for species such as this one.

Thanks for mentioning about the MO diagram,I was a little skeptical that I could draw it exactly parallel to O2. However,you would still admit that the π* antibonding orbitals are going to destabilize the π bonding orbitals,and so a π bond in the Lewis structure is going to be a little less favored?

@YUSUFHASAN Yes, I think there are at least two strikes against the double bond.

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text:
How to obtain sodium oxide from sodium chloride?
Under what conditions can $\ce{Na2O}$ be made from $\ce{NaCl}?$
I know $\ce{NaCl}$ doesn't oxidise under normal ambient conditions, but in the presence of what temperature and pressure ranges is this reaction possible?

Maybe look at how sodium oxide is made in practice?

Sodium oxide is usually made from oxidation of pure sodium, however I am looking for a direct formation of sodium oxide from sodium chloride.

Methinks this ain't happening, or the method would already be in use.  There must be some reason they trouble to use purified sodium.

There is no need (or possibility, really, in terms of standard lab capabilities) to oxidize sodium(I).
In fact, one method relies on sodium(I) reduction to metal as a method of eliminating unwanted chloride.
Method 1

Electrolysis of molten sodium chloride:
$$\ce{2 NaCl(l) -> 2 Na(l) + Cl2(g)}$$
Oxidation of sodium metal to oxide by burning:
$$\ce{4 Na + O2 ->[>\pu{250 °C}] 2 Na2O}$$
Drawback: pure sodium oxide cannot be obtained by direct oxidation of sodium.
Instead, a mixture of sodium peroxide and sodium oxide is formed.
In order to suppress the formation of peroxide, sodium metal or sodium nitrate is added in excess to the mixture in inert atmosphere:
$$\ce{Na2O2(s) + 2 Na(l) ->[\pu{150 °C}] 2 Na2O(s)}$$

Method 2

Convert sodium chloride to sodium bicarbonate using Solvay process:
$$\ce{NaCl(aq, conc) + H2O(l) + NH3(g) + CO2(g) -> NaHCO3(s) + NH4Cl(aq)}$$
Thermal decomposition of bicarbonate first yields in sodium carbonate:
$$\ce{2 NaHCO3(s) ->[\pu{250 - 300 °C}] Na2CO3(s) + CO2(g) + H2O(g)}$$
which is subsequently calcined to form an oxide:
$$\ce{Na2CO3(l) ->[>\pu{1000 °C}] Na2O(s) + CO2(g)}$$
This appears to be a preferred method as it is less energy- and resources-consuming and allows to obtain $\ce{Na2O}$ selectively.

I agree this might be seemingly unnecessary practise, but this would help a bit with my research. Are there any direct ways to obtain the oxide from the chloride, without electrolysis? Possibly with the use of under-pressure or over-pressure with elevated temperatures?

I was wondering if the pressure is reduced from rtp to around ~0.1-0.2 atm this may aid in destabilising the sodium chloride crystal?

@Joeseph123 If you have a muffle furnace, then the second method is probably what you are looking for. Considering both sodium chloride and carbonate are equally very cheap, you may opt out conversion to hydrocarbonate and start from carbonate; this could save some time. And no, to affect the crystal lattice of ionic solids to a noticeable degree you need extreme pressures; reducing pressure a little won't do a thing, I'm afraid.

In my case I am looking for a way that doesn't require the addition of another reagent. By any chance does alumina(Al2O3) react with NaCl? Alumina is of very high abundance, and a lot more accessible in my project.

@Joeseph123 But Al2O3 is another reagent? By smelting NaCl and Al2O3 I suspect you end up with NaAlCl4 or some oxochloride, judging from the related phase diagrams (AlCl3–NaCl; Na2O–Al2O3). Sorry, but it looks like you are over-complicating things that can be done much easier. Thermal decomposition of Na2CO3 is straightforward and there is no unwanted side-products.


text:
Hydrogen reactor, with electrolysis
I was thinking about hydrogen reactors. How they turn hydrogen and oxygen into water.
I just learned about electrolysis, so I was curious how efficient it would be, to fill a cars tank up with hydrogen and oxygen run it for a little bit and then capture the water, send it to an electrolysis chamber and have solar panels to turn it back into hydrogen and oxygen and repeat the cycle.

What you mean is a fuel cell, and you simply discard the water it produces. The electrolysis is done in a large plant.

Yes, that would work.You could also drop a ball from the second floor and then carry it up the stairs again. Or you can just let it roll down the street and take a new ball, like @Karl suggested.

One of the ways of achieving what you have thought about, is a fuel cell, as Karl mentioned in his comment. However, since you have asked about efficiency, I will try to address that question in particular.
What can be done is to produce the hydrogen at some factory, and then use the hydrogen as a fuel (like petrol) that can be loaded into the car to be oxidized in a fuel cell (with oxygen from air) and the resulting water vapor can be released into the air. This does not produce any $\ce{CO2}$, and seems like a green source of energy at first sight.
However, there are many problems that crop up when you try to implement this in the real world. One of the major problems is producing the hydrogen itself.
One source of the hydrogen is water. Electrolysis can produce $\ce{H2}$ but a massive amount of electrical energy is needed, so this is not a solution, it just passes on the problem to somewhere else. Solar cells (Photovoltaic cells) can be used, but they are not very efficient, so the process would not be economic. Nuclear power stations can work, but again , we are held back by the dangers and cost of building too many nuclear plants. Hydroelectric power is not available at all places, and it's overuse also harms the ecosystem. So overall, electrolysis is not economically viable if you want the energy to remain green.
Another source of hydrogen can be hydrocarbons. In fact one of the main sources of industrial hydrogen is the steam reformation of hydrocarbons. Natural gas (which is mainly $\ce{CH4}$) is reacted with steam at around 700-1100$^\circ$C with metal catalyst (e.g. nickel):
$$\ce{CH4 +H2O\leftrightharpoons 3H2 +CO \; \Delta H=+206 kJ/mol}$$
Water gas-shift reaction can produce more hydrogen:
$$\ce{CO +H2O \leftrightharpoons CO2 +4H2 \; \Delta H= -42 kJ/mol}$$
The overall process is endothermic so you have to burn something else to supply that, which creates more $\ce{CO2}$. A simplistic analysis would show you that using hydrogen obtained by this process, as fuel would generate 71% of the $\ce{CO2}$ that you would get from simply using the hydrocarbons as fuel (as we normally do). So, it's not a big difference in terms of carbon footprint.
However, fuel cells and traditional fuel engines work in different ways. The petrol or diesel engine is a heat engine, so its theoretical maximum efficiency is limited by Carnot's theorem. Experiments show that on road, the efficiency of delivering power to the road for a diesel engine is about 22%, which is very low.
A fuel cell is not limited by Carnot's theorem, and the theoretical maximum efficiency for a hydrogen fuel cell is 83%. Tests on road show the efficiency to be around 36%, so not a lot better than traditional engines$\mathbb{^{[1]}}$.
Still if you want to use hydrogen as fuel, a problem is carrying it in your car. Pressurised cylinders can be used to carry more hydrogen in low volume. Another potential way might be to use transition metals that can absorb (or perhaps more accurately, adsorb) large volumes of hydrogen.
And I am afraid fitting cars with solar panels to electrolyze the water is not a very good idea. The solar panel simply won't be big enough to produce the energy required to keep the process going. Remember that you are converting the solar energy to electrical energy, then chemical energy, and finally to electrical energy (fuell cell). In every step, there is some loss, so a far better idea is to simply use the electricity from the solar panel to run the car.
Several countries like Japan and Germany have already started considering vehicles that work on fuel cells. But, it is neither easy nor economic at this stage. So, let's hope that we can discover some source of green energy before all the fossil fuel runs out!

References:
1) von Helmolt, Eberle, Journal of Power Sources, 2007, 165, p.833


text:
Order of magnitude of ions in water
My book says water has approximately $\pu{3.35E28 molecules/m^3}$  and that only a small part of water molecules are dissociated to $\ce{H+}$ and $\ce{OH-}$ ions and that the approximate amount of ions of each sign is $\pu{6E19 ions/m^3}.$
I am trying to prove the last part. I thought I could used the ionic product of water
$$[\ce{H+}][\ce{OH-}] = \pu{1E-14},$$
so that in neutral conditions
$$[\ce{H+}] = \pu{1E-7 ions/L} = \pu{1E-4 ions/m^3}.$$
What am I doing wrong or how should I proceed?

You may know the concentration does not have dimension ion/L, , but mol/L.

You are right: 1 cubic meter of water contains $\pu{1E-4}$ moles of $\ce{H+}$ ions. But one mole is $\pu{6E23}$ atoms or ions.
So, $\pu{1E-4 mol}$ contains $\pu{1E-4}\times\pu{6E23} = \pu{6E19 ions}.$
Here we are.

Regarding typesetting, there are three ways of exponentiation I can think of: 1: use MathJax, $10^{23}$ → $10^{23}$; 2: use MathJax and \pu macro from mhchem, $\pu{E23}$ → $\pu{E23}$ or $\pu{1E23}$ → $\pu{1E23}$; 3: use HTML tags, 1023 → 10²³. Method 1 is preferred and the most versatile; method 2 is handy when physical units (hence \pu macro acronym) are involved, e.g. $\pu{6.022E23 mol-1}$ → $\pu{6.022E23 mol-1}$. I corrected your post accordingly, so you can use it as an example.

For more details please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Thank you for the information. It is exactly what I need. You are an angel

No prob, I'm glad I could help. Sorry, however, I'm anything but an angel.


text:
Ionization of hydrogens in hypophosphorous acid
Does anyone know why the hydrogen atoms bonded with phosphorus in $\ce{H3PO2}$ are not ionizable?
I tried looking in high school level books, but they only describe the process, not the real reason. I would appreciate if someone has the answer.

See, the P-H bond is not all that polar.

Even nitrogen attached hydrogen atom is very difficult to ionize, and phosphorus has almost the same electronegativity as hydrogen. Nonpolar bonds do not get ionized in aqueous solutions.

To begin with, removing another proton from anionic species is not very favorable. For example, the monohydrogen phosphate dianion is not very acidic $(\mathrm{p}K_\mathrm{a}\sim 12)$ despite it having an $\ce{OH}$ group.
In the case of $\ce{H3PO2},$ the second and third protons are bonded directly to $\ce{P},$ which is much less electronegative than $\ce{O},$ and consequently these P-H bonds are much less polarized than an $\ce{O-H}$ bond. So these protons are unable to ionize appreciably in aqueous solution.


text:
Typical reaction yield for combustion of hydrogen
I am writing an essay on the chemistry of rocket fuel, and I was wondering what, if any, would be a commonly accepted value for the actual yield of the reaction between hydrogen, $\ce{H2},$ and oxygen, $\ce{O2}.$ I would also love it if you could cite some sources so I can find more information on this exothermic reaction:
$$\ce{H2 + O2 -> H2O + heat}$$

I would guess about 100% of theoretical yield under ambient conditions, based on the Gibbs energy of reaction. Thermodynamic tables in any General Chemistry textbook give the Gibbs energy of formation for water.

if you people would stop deleting information from others' questions, they wouldn't be so unclear. Just sayin

The title as written by the OP was "Oxygen and Hydrogen actual yield", and now it is "Typical reaction yield for combustion of hydrogen". The actual yield depends on temperature and pressure. I don't think the editing made it less clear. Also, it looks like the OP appreciated the answer.

The equilibrium constant for the reaction $\ce{2H2 + O2 <=> 2H2O}$  is colossal, equal to $\mathrm{2.4 \times 10^{47}}$ at $\mathrm{500\ K}$ (source). This would ensure virtually 100% yield for the combustion of a stoichiometric mixture of oxygen and hydrogen in equilibrium conditions. However, a rocket engine is a reaction vessel that is far from equilibrium, and therefore oxygen could remain unreacted. This is in fact a problem, as it could damage the engine at its operating temperature. This, among other reasons, is why oxygen/hydrogen mixtures for rocket engines are actually very hydrogen-rich, containing up to twice the stoichiometric amount of hydrogen required.
So in practice, taking everything into account, rocket engines running on oxyhydrogen fuel are designed to have a  practically 100% combustion yield based on the limiting reagent, oxygen, while possessing a substantial excess of hydrogen fuel. There is a strong engineering pressure to maximise energy output by ensuring the oxygen is completely consumed; unreacted oxygen is a waste of lift capacity, storage space, materials, cost, etc.


text:
Precipitation of water-insoluble materials from shower or laundry greywater
I am currently working on a project to build a filter that aims to separate or salt out the 'dirt' in shower or laundry greywater. The ideal result from this chemical reaction will be such that the 'dirt' will be in a solid-state (precipitate). Meanwhile, the residual water will be 'clean'. The degree of cleanliness of 'clean' I mentioned is such that this residual water is clean enough to be used for non-potable purposes such as mopping the floor.
Therefore, I would like to ask your help in figuring out how can I turn the soap micelles into precipitate, leaving out only 'clean water'?
Disclaimer: I am ignoring soap scum as I can filter that out easily.

Your ambitious project is not that trivial. Soaps are of several types.
a) If you are talking about salts of fatty acids, the classical soaps, then one can precipitate them out with calcium salts. Many other cations form insoluble salts with soaps. Now if you are trying to purify laundry water, dishwasher water, shampoos, liquid soaps, nobody can remove those surfactants from water by simple means. They (manufacturers) have specifically designed them so that they never precipitate out in water, otherwise your clothes and dishes would always have a layer of scum on them. Potassium ions may form ppts with certain surfactants like sodium dodecylsulfate but forget about ppt'ing out these typical surfactants from water used in ordinary household.
b) Alum can quickly help you in cleaning all the particulate matters in waste water, like magic.
c) Filtration on beach sand beds can remove small amounts of certain surfactants from water.
d) Filtration on activated carbon would be the best to remove surfactants.
Well it turned out to be more complicated than you desired. There is no single cure and certainly this water is not good for mopping floors because it will spread nothing but germs from using laundry water and shower waste. You can think of solar distillation as an alternative. That distilled water would very clean.
Note that in certain countries, household waste water including the sewage water after doing natural bacterial treatments is used for watering gardens and small farms. In one large pit, all wastewater is allowed to stand in open air, far away from residential areas. In the first ou all solids settle down, next pit further purifies via same settling and bacterial oxidation process, finally the supernatant water is transferred to third pit, which is now clean enough that gardens and small farms can be watered with it.  Of course, this water is rich in nitrogen and phosphates (from human and laundry waters) which is good for plants.

Thanks for the detailed feedback, Farooq.
I wish to follow up on your statement that the resulting greywater from using common soap/detergent; nobody can remove those surfactants from water by simple means. Is there a way I can create a new type of soap, which will allow me to remove the surfactants from water in greywater easily?
Looking forward to hearing from you.

I think you can think of natural soaps. I am assuming this is a school project. There are certain plants called $Saponarias$ which have a very good surfactant action along with lather. Check Amazon for "Soap Nuts".  It is not easy to design surfactants.


text:
Formation of intimate ion pair by halogenoalkanes
I was recently doing this organic chemistry question that tested on the concept of the formation of the intimate ion pair in the nucleophilic substitution reaction at an $\ce {sp^3}$ carbon:
$\ce {RX -> R^+||X^- -> R^+ + X-}$
This mechanism is responsible for the slight retention of configuration during an $\ce {S_N1}$ reaction. It was mentioned that the formation of the ion-pair is more likely to occur for a chloroalkane than for a bromoalkane. In fact, it was written that there would be complete racemisation in the case of the bromoalkane while the chloroalkane would have slight preference for retention over inversion.
This seems to make sense since the $\ce {C-Cl}$ bond is stronger than the $\ce {C-Br}$ bond and secondly, the $\ce {Cl^-}$ ion is more charge-dense and is more likely to stay close to the carbocation after it is formed due to favourable electrostatic attraction. However, I have not read about this before from any authorative textbook on organic chemistry so I would just like to verify the validity of this assertion.
Also, can we extend this and assert that the formation of the intimate ion-pair is more likely to occur for bromoalkanes then for iodoalkanes based on a similar line of reasoning?

There's hardly any pure SN1 or SN2, you bring out one of the reasons. I think that something like intimate ion pair is always  made at the start of SN1 and the difference is, if it's broken by solvent before nucleophile attacks or not.


text:
Product of the reaction between hexane‐2,5‐dione and ammonium carbonate

Which of the following statement is INCORRECT?
(A) The product X has a five membered ring.
(B) The product X is aromatic.
(C) The degree of unsaturation of the product X is 3.
(D) The product X has a six membered ring.

Ammonium carbonate will probably lose a proton and one of the carbonyls may be protonated. But from here onwards I can't find any leads. Does the lone pair of other oxygen atom plays a role?
Usually the heat provided may lead to a cyclic structure, but I am not sure.

Yes it is a cyclic compound. First think, what would happen to ammonium nitrate when it is heated?

Ammonia gas , carbon dioxide and water is released...nitrogen lone pair will attack on carbonyl carbon and the oxygen will attack the other carbonyl carbon...is this assumption correct? But what happens next?

Heating $\ce{(NH4)2CO3}$ will generate $\ce{NH3 + CO2 + H2O}.$ Out of these, $\ce{NH3}$ will act as a nucleophile as it's the most nucleophilic of the products. Addition of $\ce{NH3}$ to the two ketones, followed by elimination of water molecules (due to the high temperature provided by ∆) will generate the final product as follows:

The product is a five-membered aromatic ring called 2,5-dimethylpyrrole.

Try to avoid "lassos" in mechanisms. Use arrows. ;)

@user55119 Well,my handmade mechanisms tend to have a lot of lassos,still will keep that in mind later :)


text:
Calculating ΔH with non-integer coefficients in the balanced equation
I am working a practice problem out Ebbing's General Chemistry and for part of the problem I needed to calculate $ΔH.$ The problem considers the burning of ethane in oxygen to yield carbon dioxide and water vapor.
The coefficients used in the solution manual were 1 for ethane 7/2 for oxygen 2 for carbon dioxide and 3 for the water vapor. The coefficients I used were 2, 7, 4, and 6, respectively.
Naturally, it follows the $ΔH$ value I got is twice as large. Can anyone provide clarification here?

There is not much to clarify: you determined dimolar enthalpy of combustion, whereas standard enthalpy of combustion is determined for a complete oxidation of one mole of the substance:
$$
\begin{align}
\ce{C2H6(g) + 7/2 O2(g) &-> 2 CO2(g) + 3 H2O(l)} &\quad &Δ_\mathrm{c}H^\circ\\
\ce{2 C2H6(g) + 7 O2(g) &-> 4 CO2(g) + 6 H2O(l)} &\quad &Δ_\mathrm{c}H^\circ ×2
\end{align}
$$
When composing the corresponding reaction equation, make sure to balance it in such a way that the stoichiometric number $ν$ of the substance the enthalpy is determined for is always $-1.$

Excellent answer, andselisk, very enlightening. Thank you!


text:
Dynamic equilibrium in unsaturated solutions
One standard definition of equilibrium in beginner chemistry is that for a reversible reaction, (dynamic) equilibrium has been achieved when the rate of the forward and backward reactions are equal. We also define the equilibrium constant for a given reaction in terms of a ratio involving the concentrations of products and reactants at equilibrium.
Consider adding a small amount of solid $\mathrm{AgCl}$ to a sample of pure water, where the amount added is small enough that the solution does not become saturated. The associated reaction is
$$\mathrm{AgCl}(\text{s}) \rightleftharpoons \mathrm{Ag^{+}}(\text{aq}) + \mathrm{Cl^{-}}(\text{aq}), $$
with $K_{\mathrm{sp}}\approx 1.77\times 10^{-10}$. After the added amount of salt has dissolved, the concentrations of all species are constant since all of the $\mathrm{AgCl}$ has dissolved and if any solid $\mathrm{AgCl}$ is produced by the reverse reaction, it will be shortlived and will dissociate back into ions almost instantly.
Is it not therefore true that the forward and backward reaction rates are equal at this point? By the simple definition of equilibrium given above, we would then conclude that the system has reached equilibrium. On the other hand, the reaction quotient $Q$ is less than the equilibrium constant $K_{\mathrm{sp}}$ since the saturation point has not yet been reached, and therefore the system has not reached equilibrium.
I am seeking an explanation of why this apparent contradiction arises – I expect it may be due to the basic definition of chemical equilibrium simply being inadequate, and that for equilibrium to be 'properly' achieved (in the sense of the equilibrium constant), the forward and backward reaction rates must approach each other smoothly rather than abruptly as they do in this example due to the $\mathrm{AgCl}$ running out before saturation is achieved. Or perhaps at equilibrium, the reaction rates are necessarily equal, but the converse is not true.  You can assume that I'm familiar with thermodynamics and statistical mechanics.

Edit: The equilibrium constant here is calculated subject to the constraint that both a solid and an aqueous phase must coexist as pointed out in Charlie Crown's answer. This is also alluded to in Buck Thorn's answer where they note that there is no equilibrium with the solid below the solubility limit, and in a comment by Karl below their answer where they state that there is no dynamic equilibrium when the concentration of the solid is zero. Hence the apparent contradiction arises here because the static equilibrium achieved before the solubility limit has been reached is not the same as the dynamic equilibrium which $K_{\mathrm{sp}}$ refers to.

I would say a complex [AgCl(aq)] is partially formed, in equilibrium with ions. No AgCl(s) is formed, as AgCl(aq) concentration is not  high enough.

It doesnt matter why the forward and backward reaction have the same rate. That is the only proper definition of chemical equillibrium, all others are corollaries for certain situations. Obviously the maximum solubility is irrelevant here, as you are well below!

Proper definitions of chemical equilibrium will not involve reaction rates whatsoever. Thermodynamics does not care about time. Chemical potential is the work required to form a molecule in solution, irregardless of the time it takes. Statistical Mechanics says chemical potential is the work required to move a molecule from infinitely far away (ideal gas since there are no interactions at infinity) into the solution.
Also, Free Energy is entirely independent of the equilibrium constant, but the equilibrium constant is not independent from Free Energy. At equilibrium, Free Energy will be at a minimum. Throw out definitions that use equilibrium constants. If you so wish, you can calculate the equilibrium constant from the free energy.
\begin{equation}
    K = \exp^{-\frac{\Delta G^0}{RT}}
\end{equation}
Where $\Delta G^0$ is the ideal gas free energy reaction. If you know the ideal gas free energy change of a reaction, you may calculate K. However, K is a useless thing in reality, or at least, leads to more confusion than it is worth.
For chemical equilibrium, this is the definition at global equilibrium
\begin{equation}
   \left( \frac{\partial G}{\partial \xi} \right)_{T,P} = 0
\end{equation}
Where $\xi$ is the extent of reaction. In your case the free energy was minimized by the solid dissolving rather than staying a solid, so your system came to equilibrium given the constraint that there was no more solid that could be dissolved. There was also no constraint saying that there needed to be solid present.
Equilibrium and solubility limit are two different things. All equilibrium points are constrained by definition. At the very least you are constraining T and P. When you are measuring solubility you are actually implementing an additional constraint: find the minimum free energy of the system with the constraint that a pure solid and an aqueous phase must coexist - this necessitates there initially being enough solid to be able to meet this constraint at equilibrium.
Equilibrium reactions do not require time, and I have no time for equilibrium constants :)

You have to look at two things in terms of equillibrium.

The pot with the solution, once all AgCl is dissolved and you have stirred it a bit more, is in equillibrium. Obviously. There is no chemical potential gradient, and you have only one phase. By itself, nothing will ever happen again in it.
The reaction (dissolution of AgCl in water) isn`t: If you add a bit more solid AgCl, the solution comes to rest at a different concentration!

I see, so the reason why the reaction isn't at equilibrium is that the concentration would ultimately settle on a different value if more reactants were available. It seems the basic definition doesn't pick up on this important point. Would a more precise definition be that equilibrium is achieved for a given chemical reaction when the forward and reverse reaction rates are equal and the concentrations of both reactants and products are non-zero?

You are trying to split a hair here. ;) If one side has zero concentration, I say you have no reaction. Why worry about the equillibrium? No solid precipitate, ergo dissolution rate zero, the solution is in equillibrium, ergo the precipitation rate must also be zero, so this cant be a dynamic equillibrium, but is just static. Which is indeed so: The rate of formation of crystallites can only become finite in supersaturation. Small crystallites are less stable thermodynamically. They cannot grow below saturation at all.

Indeed, this is what I am trying to do! Perhaps then my issue is that the equilibrium constant for a reaction is defined at dynamic equilibrium, and has nothing to do with any of the static equilibria that may be attained for the same reaction due to a reactant running out before dynamic equilibrium has been achieved. I'm now satisfied that there is enough information between the three answers provided to resolve my initial question. Thanks!

@Zac Exactly. A dynamic equillibrium constant only makes sense if there is indeed a dynamic equillibrium with two (opposite) equal, constant, and finite reaction rates.

I expect it may be due to the basic definition of chemical equilibrium simply being inadequate

This is the answer (sort of). In essence, when you are below the solubility limit, the chemical potential of the solid lies above that of the solubilized salt, and there is no equilibrium with the solid (because no solid can form). This is illustrated in the following figure:

You might use this to formulate a microscopic argument such as "there is no thermodynamic force driving accumulation of the solid".
On the other hand, once the solubility limit is reached, the chemical potential of solid and solute become equal (this happens at the intercept of the two lines). Once this salt concentration is exceeded, the concentration of solute remains constant when more salt is added. Any added salt does not lead to increased solute since that would increase the solute potential above that of the solid.

So before the solution becomes saturated there is no solid at all? But this seems contrary to the fact that in any reaction as much as high is the equilibrium constant there is some amount of reactants.

@adosar Ksp as written conceals the third "reagent", the solid, which has constant chemical potential.  It's better to think of this "reaction" as a phase transition, given that you form a solid with constant chemical potential (at constant p, T). The same would be true of water freezing or boiling at fixed pressure, say.

Does the same happen when we have a binary mixture of two liquids? If we add more and more from the first component then there would be a point where we should expect two phases, one with both components and another with only the first component (like when the salt precipitates)?

@adosar Yes, this should apply generally. It should be true when for cases of limited solubility or miscibility.

"It's better to think of this "reaction" as a phase transition". Strictly speaking is the formation of the solution a chemical reaction? I was searching to see if mixing compounds is considered as a physical or chemical change but there is no clear answer.

@adosar My advice: follow the definitions provided in your study material. A chemical change alters chemical properties through a reaction. A chemical reaction is a process that breaks or makes new bonds. But what you call a bond is somewhat arbitrary. We usually mean covalent, metallic or ionic, a strong bond in other words, and not intermolecular interactions such as vdW or H-bonding (with partial covalent/ionic character). In that sense precipitation is a chemical reaction and mixing two liquids is not. But what about ice freezing?


text:
The boiling points of Alkali metal group
As I know, the melting points of Alkali metals decrease down to the group due to decreasing of the energy of the metallic bond.

But according to my understanding, after it has become to liquid the metallic bonds are not affecting furthermore because of the lattice has destroyed, so the boiling point affects only the atomic mass. Then according to my logic the boiling points should be increase down to the group. But the above graph shows the opposite of that.
What is the wrong in my logic. Please help me.

A molten metal is still a metal, with (more or less) the same metallic interaction between the constituent atoms.What gets destroyed upon melting is the crystalline lattice.

You can explain the trend in terms of increased charge density, and ensuing charge repulsion between nuclei and between electrons as you go down the group.
An alternative explanation is provided in a libretext, which draws information from a popular educational chemistry site:

When any of the Group 1 metals is melted, the metallic bond is weakened enough for the atoms to move more freely, and is broken completely when the boiling point is reached. The decrease in melting and boiling points reflects the decrease in the strength of each metallic bond.
The atoms in a metal are held together by the attraction of the nuclei to electrons which are delocalized over the whole metal mass. As the atoms increase in size, the distance between the nuclei and these delocalized electrons increases; therefore, attractions fall. The atoms are more easily pulled apart to form a liquid, and then a gas. As previously discussed, each atom exhibits a net pull from the nuclei of +1. The increased charge on the nucleus down the group is offset by additional levels of screening electrons. As before, the trend is determined by the distance between the nucleus and the bonding electrons.

Trends in nuclear screening can be used to explain bonding but there are other ways to explain the trend, and such generalizations should in any case be understood to be simplified ways of describing what requires a more complex answer.

Then why does the mp increase down the group in case of transition elements?@Buck_Thorn

@Shub I am not sure what would explain trends in transition elements. There might be a different explanation there. Also, my own interpretation of the trend (the first paragraph here) is not from a textbook, admittedly, as should be evident, but then in the later part of the answer I also present the textbook explanation.

See also: https://chemistry.stackexchange.com/questions/156254/the-melting-and-boiling-point-decreases-down-the-group-up-to-group-14-not-inclu?rq=1


text:
Meaning of the term "vertex" in the topic of skeletal formulae
From the Wikipedia article on skeletal formula:

Hydrogen atoms attached to carbon are implied. An unlabeled vertex is understood to represent a carbon attached to the number of hydrogens required to satisfy the octet rule, while a vertex labeled with a formal charge and/or nonbonding electron(s) is understood to have the number of hydrogen atoms required to give the carbon atom these indicated properties. Optionally, acetylenic and formyl hydrogens can be shown explicitly for the sake of clarity.

This is the first use of the word "vertex" in the article. Is a vertex

only the intersection of two line segments, or
the end of each line segment?

I assume the latter is true, but on many places I saw cyclopentyl being represented like this:

which should be methylcyclopentane instead, a completely different thing.
So, it looks like (mostly when describing aryls or cycloalkyls) some people assume that "vertex" means only the intersection of two line segments, otherwise they would have added the $\ce{CH3}$ to the end of the line to indicate methylcyclopentane (in this case). It's certainly inconsistent, but is this justified in any system of structural formulae?

The term vertex (plural vertices) comes from the nomenclature used in graph theory, a branch of mathematics which describes how points in a given space are connected to each other. This turns out to be quite fundamental to chemical nomenclature, as the vertices of a graph can (generally) be chemically interpreted as an atom, and the lines coming out of a vertex can (generally) be interpreted as covalent bonds. Two graphs which contain the same vertices, but where the connectivity between the vertices is different, corresponds in general to different chemical compounds.
To be clear, the term vertex applies to any point in a skeletal formula which is not "in the middle" of a connecting line. This means that every explicitly labelled atom, every "kink" in a chain (which implicitly represents carbon atoms) and every termination (also implicitly a carbon atom) is a vertex. Hydrogen atoms are often not vertices in a skeletal formula, though this can vary.
Regarding to your example, with no other context, chemists would call that a skeletal drawing of methylcyclopentane, which contains six vertices in its skeletal formula, one for each non-hydrogen atom. For it to be understood as a methylcyclopentyl radical/substituent, either the context should clearly state it, or preferably we would use other standardised graphical notation, shown below:


text:
Why Does Adding Catalyst Slow Down My Reaction?
I currently did a lab where I was meant to identify factors that increase the rate of the reaction. I reacted Glucose with Potassium Permanganate and used Sulfuric Acid as a catalyst. But in my recordings I found that the more catalyst I added to my mixture, the slower the reaction took to reach its endpoint. Is there any reason why a reaction may take longer to reach endpoint when more catalyst is added?

If there is no reason why this may occur, could you also let me know if I did my experiment wrong?

Too many chefs in the kitchen could spoil the mix.


text:
Troubles with Gibbs free energy
The book I'm reading writes that for a system at constant pressure and temperature it holds:
$$dG=-TdS_{tot}$$
where $dG=dH-TdS$.
So if you want to see if a certain process is spontaneus, you just have to look at the sign of $dG$, since it depends on the sign of $dS_{tot}$.
But I have a lot of doubts about this formula, indeed read the following scenario.
You consider a certain amount of liquid inside a piston at a fixed pressure given by the weight of the piston. Let's assume the liquid is changing its phase, so also its temperature stays constant at the value $T$. Out there is a heat reservoir at the temperature $T_{res}$, which touches the piston. Since the temperature can't be a discontinous function, you assume that there is a piece of metal which connects the reservoir and the piston. You have:
$$dS_{tot}=dS_{res}+dS+dS_{metal}$$
If you consider a stationary transform, the state of the piece of metal doesn't change, then $dS_{metal}=0$.
$$dS_{tot}=\frac{\delta Q}{T_{r}}-\frac{\delta Q}{T}$$
Since pressure is constant:
$$dS_{tot}=dH[\frac{1}{T_{r}}-\frac{1}{T}] $$
Note that if $T_{r}=T$ then $dS_{tot}=0$ since there are no irreversibility in the universe (no finite differences of temperature), otherwise it must be $dS_{tot}>0$. I would stop at this point but let's introduce the Gibbs free energy.
Definig the Gibbs free energy as:
$$G=H-TS$$
$$dG=dH-dTS-TdS$$
$$dH=dG+dTS+TdS$$
For constant temperature:
$$dH=dG+TdS$$
You finally get:
$$dS_{tot}=(dG+TdS)[\frac{1}{T_{r}}-\frac{1}{T}] $$
Which is not the same thing that writes my book. Also note that the temperature and the pressure of the system are constant, so I have the same hypothesis of the book but I get different results. Someone could clarify this? I'd really appreciate.

Your error occurs when you write that
$$dS_{tot}=\frac{\delta Q}{T_{r}}-\frac{\delta Q}{T}$$
since this assumes that
$$\delta Q=-TdS$$
which implies reversibility and therefore that $dG=0$ at constant T and p.
The more general statement is that
$$dS_{tot}=\frac{\delta Q}{T_{r}}+dS$$
which means that
$$dS_{tot}=-\frac{dH}{T_{r}}+dS$$
(since positive heat here means exothermic)
Then
$$T_{r}dS_{tot}=-dG+(T_{r}-T)dS$$
or
$$dG=-T_{r}dS_{tot}+(T_{r}-T)dS$$
Now if you let $T=T_{r}$ (thermal equilibrium with surroundings) then
$$dG=-TdS_{tot}$$


text:
Grignard Reaction with Unsaturated Methyl Ester | Prevention of Acid Catalyzed Hydration
I am attempting to perform a Grignard reaction involving methyl 3-butenoate and methyl magnesium iodide in order to yield the unsaturated tertiary alcohol 2-methyl-4-penten-2-ol. The issue in doing this is that conditions are involved in the acidic workup of the bromomagnesium alkoxide the would also result in the acid-catalyzed hydration of the double bond (cold conditions). I am wondering if it is possible to replace the ice/sulfuric acid acidic workup with a solution of 3 M sulphuric acid in methanol, on an ice bath of course? Firstly, would this prevent the acid-catalyzed hydration of my double bond since no water is present? Also, would this prevent the acid-catalyzed dehydration of my newly formed tertiary alcohol since the conditions are cold (however, the fact there is no water, may result in forcing equilibrium concentrations, so would I have to keep the reaction at VERY low temperature like -78 C?). Secondly, If this would reaction would theoretically work, is there any literature available of anyone attempting it, I have searched and am not able to find any, also would methyl alcohol be the best choice in this case since it will be easy to rotovap in future steps? I will include the procedure I am looking to modify below along with a general scheme for my reaction (I have modified the procedure from Vogel's textobook of Practical Organic Chemistry page 540):

Prepare an ether solution of methylmagnesium iodide from 12.687 g (522.0 mmol) of dry magnesium turnings, 74.093g (32.5 mL, 522.0 mmoles) of dry methyl iodide and 104.4 mL of sodium-dried ether; use the apparatus and procedure described in Exp 5.39, cognate preparation nonan-1-ol. Allow to cool (ice bath), and slowly add a solution of 26.66 g (261 mmol) of A in 26.1 mL of anhydrous ether into the vigorously stirred solution of the Grignard reagent. Reflux the mixture on a water bath for 1 hour to complete the reaction. Pour the ethereal solution into a mixture of 52.2 mL of approximately 2 M sulphuric acid and 196 g of crushed ice. Separate the upper ethereal layer and extract the aqueous solution with two 39.2 mL portions of ether. Wash the combined ethereal extracts with dilute sodium hydrogen carbonate solution, followed by a little water, then dry with anhydrous potassium carbonate or anhydrous calcium sulfate, distill off the ether on a water bath and distill the residue through a short fractionating column. Collect at 117-120 C. A further small quantity of the tertiary alcohol may be obtained by redrying the low-boiling distillate, filtering and redistilling. The yield is 90 g (88%).

I usually work up my Grignrds by slow addition of cold aq NH4Cl soln. You do not need to work up with strong acid to get rid of the Mg salts.


text:
How much oxygen is formed in this first order reaction?
We have a reaction
$$\ce{2N2O5 -> 4NO2 + O2},$$
which has a rate constant $k=6,2\cdot10^{-4}/\mathrm{min}$ at temperature $T=318.15\ \mathrm{K}$. How much oxygen is formed in litres when $1.00\ \mathrm{mol}$ of $\ce{N2O5}$ is at $318.15\ \mathrm{K}$ in $104\ \mathrm{kPa}$ pressure for $20$ hours. Correct answer: $6.67\ \mathrm{l}$
My attempt:
For gases we use partial pressures to measure rates. So using equation
$$\ln P_t=-kt+\ln P_0$$
we get $P_t=49421.76426\ \mathrm{Pa}$. This means that oxygen has a partial pressure of $P_{\ce{O2}}= \frac{104000\ \mathrm{Pa}-49421.46426\ \mathrm{Pa}}{5}=10915.64715\ \mathrm{Pa}$.
Now we say $n_{\ce{O2}}=x$ so according to the reaction $n_{\ce{N2O5}}=1-2x$. If the total volume is $V$ we can type two equations:
$$
\begin{align}
P_t&=\frac{n_{\ce{N2O5}}RT}{V}\\
P_{\ce{O2}}&=\frac{n_{\ce{O2}}RT}{V}
\end{align}
$$
Solving for $x$ and $V$ yields $x=0.153195\ \mathrm{mol}$ and $V=0.0371657\ \mathrm{m^3}$. Now calculating the volume for oxygen at $104\ \mathrm{kPa}$ gives
$$
V_{\ce{O2}}=\frac{0.153195\ \mathrm{mol}\cdot 8.3145\ \mathrm{\frac{Pa\ m^3}{K\ mol}\cdot318.15\ \mathrm{K}}}{104000\ \mathrm{Pa}}=3.900828\cdot 10^{-3}\ \mathrm{m^3}\approx3.9\ \mathrm{l}
$$
My answer is somehow incorrect but what I am doing wrong here?

Is this reaction at constant pressure? In that case, the volume would increase over time, and the first order integrated rate law would not apply. Is this a textbook problem?

It was from a exercise sheet. I don't know the origin of the problem and my translation might not be perfect.

If we assume that the reaction takes place in a large amount of inert gas (like nitrogen) and that the reaction is first order in $\ce{N2O5}$, we can reach the numerical answer given by neglecting the volume change (the reaction increases the number of particles, but if there is a large amount of inert gas, the volume change will have little effect on partial pressures). The path given by Maurice is fine (no reason to figure out the partial pressure first). If you convert the first-order rate constant into a half-life (18.6 h), you can quickly estimate that a bit more than half of the reactant will be gone after 20 h.
If we assume, like the OP, that you start with pure $\ce{N2O5}$, the rate law still applies, but the integrated rate law does not (because while the reaction goes on, dilution also happens through the volume increase). You could numerically integrate, or try to find the integrated rate law that includes the dilution effect. Overall, you would expect the reaction to go slower, i.e. produce a smaller amount of oxygen. Because the final volume is larger, the partial pressure will be lower (compared to the inert gas case) by even more.
The value calculated by the OP is incorrect for either case because it assumes an integrated rate law that does not apply for the assumed conditions.


text:
Are monosubstituted cyclopropanes achiral or chiral?

A classmate an I were discussing the chirality of this molecule which is a monosubsituted cyclopropane. I argue this is an achiral molecule because that carbon which is bonded to bromine cannot be a stereocenter. If you try to determine a point of difference between the other two "substituents," there is none. But if you draw a mirror image, it appears to be non-superimposable which is a trademark of chiral molecules. Is my reasoning on the right track, or is there something unique about the chiral or achiral nature of cyclopropanes such as these.

It is achiral. The cyclopropane ring is planar. The substituted C will be $\mathrm{sp^3}$ hybridised (tetrahedral). If you consider two of the C's in one plane and the H and Br in the other plane, these two planes will make 90° angle between them and one can draw a mirror plane along the plane containing H and Br. Thus, it will be achiral.

Note: every 3-atom ring must be planar no matter what else is on it, because 3 points define a plane

Nit-pick - 3 non co-linear points define a plane. That said I once lost marks during my degree in an organic chemistry piece of work for this very explantation of the planarity of the carbons in cyclo-propanes as the tutor didn't understand my comment.

Here is a 3-D conformer from PubChem

As you can clearly see, a plane of symmetry can be sent along the black line perpendicular to the plane of the screen. Hence, the molecule is achiral. If you take a mirror image, you can ultimately super-impose it again on the parent form
Here is an illustrative 3D image(courtesy of andselisk) which clearly shows the plane of symmetry in pink and ring plane in blue.


text:
Steam rises when the heat supply stops
When water boils and you suddenly shut the heat off, a lot of steam arises from it just after you shut the heat off. Why this happens, I thought that maybe water condenzes very fast in small droplets so they get flown up?

This is explained at StackExchange Physics: https://physics.stackexchange.com/q/4168

Close voters: A similar question was very well received on physics (multiple times in fact). While we don't need to rehash the same points here, I don't see how the question is unclear.

Even water below boiling point has high partial vapour tension.
The gas containing mostly vapour, being hot and having low average molecular mass, has significantly lower density than air. Therefore, it raises up.
At contact with air, it quickly cools down and the vapour condensates.
The droplets are small enough to be taken upwards by the strong thermal vertical convection.


text:
Why isn't hydrochloric acid typically made via the photochlorination of hydrogen?
I understand that in large quantities the reaction is explosive but couldn't the flow of gas just be regulated to control the reaction? It seems like producing hydrogen chloride like this then dissolving it would be a very straightforward way to make hydrochloric acid.
To clarify, when I say 'typically' I mean on an industrial scale.

From gaseous hydrogen and chlorine? You already put a reason why it doesn't make sense and that's like top of a iceberg. Chlorine is plain nasty, and hydrogen is also explosive with air. Something can be possible, but completely unreasonable... probably most things are. It doesn't matter if in theory all could work out, if something goes bad and everyone in building dies.

Wait, but at least some of HCl is actually produced this way.

Might it have to do with the cost of the raw materials??

@IvanNeretin as a byproduct of chlorine production from brine electrolysis I would think.

I suspect it's due to thermal runaway and reaction control. Using heat may give more control that photoinitiation. It would have to be done continuously rather than in batch.


text:
An exception to the IUPAC convention for italicisation (typography)
According to IUPAC, only physical quantities and variables must be italicised.
Section 8 Use of Italic and Roman Fonts for Symbols in Scientific Text of ICTNS Guidelines for Drafting IUPAC Technical Reports and Recommendations states that

a good general rule is that quantities, or variables, can be given a value, but labels cannot.

So why is it that [X] is not written as [X] when the concentration of species X is a physical quantity?

Do you refer to [] as the chemical notation of cencentration? Or do you refer to one of the many other notations?

“The overall rule” [1, p. 7] in

The overall rule is that symbols representing physical quantities or variables are italic

a keyword here meaning that not every symbol of a physical quantity should be italicized.
Chemical element symbols are always upright unless they are used to denote a locant in a name of a chemical substance [1, pp. 8–9]:

Symbols for elements in the periodic system shall be roman. […]

Symbols for physical quantities are single, or exceptionally two letters of the Latin or Greek alphabet, but they are frequently supplemented with subscripts, superscripts or information in parentheses to specify further the quantity. Further symbols used in this way are either italic or roman depending on what they represent. […]

[…]

[…] Letter symbols for elements are italic when they are locants in chemical-compound names indicating attachments to heteroatoms, e.g. О-, N-, A-, and P-. The italic symbol H denotes indicated or added hydrogen.

For example, amount concentration of nitrate can be displayed as $c_\ce{NO3-}$, $c(\ce{NO3-})$ or $[\ce{NO3-}]$, where “$c$” is italicized as a symbols for physical quantity, whereas “$\ce{NO3-}$” is upright as it is a specification with its own typography rules.
As for the general use of brackets in written text, according to Wikipedia

In English, typographers mostly prefer not to set brackets in italics, even when the enclosed text is italic.

Further, from Bringhurst' The elements of typographic style [2, p. 85]:

The sloped square brackets usually found on italic fonts are, if anything, even less useful than sloped parentheses. If, perish the thought, there were a book or film entitled The View from My [sic] Bed, sloped brackets might be useful as a way of indicating that the brackets and their contents are actually part of the title.
Otherwise, vertical brackets should be used, no matter whether the text is roman or italic: “The View from My [sic] Bed” and "the view from my [sic] bed."

Similarly, from The Chicago Manual of Style (for earlier edition see this answer on English.SE):

6.5 Parentheses and brackets in relation to surrounding text. Parentheses
and brackets should appear in the same font—roman or italic—as the surrounding text, not in that of the material they enclose. This system, though it may occasionally cause typefitting problems when a slanting italic letter touches a nonslanting roman parenthesis or bracket, has two main virtues: it is easy to use, and it has long been practiced.

Specifically for use in technical documentation and in mathematical notations, brackets could be treated as a notation for operator (like matrix delimiters) [4, p. 316], yet another reason to use roman typeface:

6.2.5.3  Special Bracket Notation for Specific Operators
[…]

Brackets only should be used to indicate chemical concentrations:
$[\ce{S^2-}];$ $[\ce{Na+}];$ and isotopic prefixes: $[\ce{^{32}P}]\text{AMP}.$

The following comparison between the symbols fro amount concentration and specific optical rotatory power illustrates the problems arising from using italicized brackets, namely uneven spacing and clashing symbols:
$$
\begin{align}
&[\ce{^{7}NO3-}] &\quad &\textit{[}\ce{^{7}NO3-}\textit{]} \\
&[α]_λ^θ &\quad &\textit{[}α\textit{]}_λ^θ
\end{align}
$$
References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Bringhurst, R. The Elements of Typographic Style, 3rd ed.; Hartley & Marks, Publishers: Point Roberts, WA, 2004. ISBN 978-0-88179-206-5.
The Chicago Manual of Style, 17th ed.; University of Chicago Press: Chicago; London, 2017. ISBN 978-0-226-28705-8.
The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers, 1st ed.; Rabinowitz, H., Vogel, S., Eds.; Elsevier/Academic Press: Amsterdam; Burlington, MA, 2009. ISBN 978-0-12-373980-3.

Suppose in a reaction, $A$ + $\text B$ +$B$ $\ce{->}$ $AB$ +$\text B \text C$, I want to differentiate the unknown elements $A$ and $B$ from known elements $\text B$, i.e., boron and $\text C$, i.e., carbon.

@Apurvium If this is an isotopic exchange reaction, then put appropriate mass numbers next to the elements' symbols. If the atom exists both in ground and in excited state(s), use asterisk(s) for the latter. If you just want a symbol to pop out, use prime(s). If you are writing a book for kids, then probably use different colors.

But use of italics clearly differentiate unknown $B$ from $\ce{B}$, i.e., boron.

@Apurvium Sure, $B$ differentiates so clearly that the slant implies it is a physical quantity (for example, susceptance). Why one would place a physical quantity as a term in a chemical equation and what this is supposed to signify is an open-ended question. Atomic labels are always upright (except, of course, when they are locants).


text:
Why does saliva clean so well?
I notice that saliva seems to clean well, much better than plain water. For example, tar spots on the car can be difficult to get off. I typically use really powerful solvents to get them off, like 95% acetone and things like that. Yet, saliva works too.
Why is this?

Saliva has proteins which act like mild surfactants (detergent effect) and of course mechanical abrasion helps to physically dislodge the spots.

Saliva has alpha-amylase (ptyalin), which hydrolyzes carbohydrates (so starch starts to taste sweet in just a few seconds), kallikrein, an enzyme that breaks down protein, and lingual lipase, which helps disintegrates fats (lipids).
So next time you clean off a splattered insect from your wind-screen (windshield, should you prefer), think of it as external digestion, as practiced by flies and by starfish. Or perhaps you might not want to consider that thought...


text:
Synthesis of silver nanoparticles
Synthesis of silver nanoparticles  by electrochemical Method

The silver content in these products consists of both silver ions and silver particles. Typically 90% of silver content is in the form of ionic silver and the remaining 10% of the silver content is in the form of silver particles. The silver ions are produced by electrolysis are frequently described as "dissolved silver". Because the majority of the silver in these products is dissolved silver rather than metallic silver particles, it would be more technically accurate to describe these products as silver solutions. In the present work, the electrochemical synthesis of silver nanoparticles is presented.
Tech.Journal,Vol.33,Part(B),No.7,2015
Q.1  What is the scientific meaning of the following sentence?
"The duration used for exchanging the polarity was 4 min."
Q.2 How can we exchange the D.C current polarity regularly?
Q.3 What is the suitable current and any other conditions for preparing silver nanoparticles?
Q.4 How can we extract silver nanoparticles from silver solutions?


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Are there any chemical pathways or catalysts that could produce electricity from fossil fuels without combustion?
While fossil fuels are dense in energy, I have known only one way to retrieve energy from them, namely, to burn them. I wish we could have a way to get energy from them without combustion.
Is there any way we could get energy from oil, gas, diesel or other fossil fuels without burning them? Ideally, so we could making something like a battery out of them.

If you mean, can you get energy out of it without breaking bonds? Not really. I doubt you would get very far trying to take electrons from it either.

Check out fuel cells. Usually they require hydrogen as fuel, and the hydrogen often comes from processing of fossil fuels. However, some kinds of fuel cells have been developed that can directly use hydrocarbons as fuel.

Even if this was easy (it is possible but not easy0 it would not be much of an advantage. A fuel cell could generate electricity from oxidation reactions (and such fuel cells using methane are known) but the products re the same as burning and the overall efficiencies are not good enough to justify the overall cost in most circumstances.


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Mechanism of chlorine disproportionation
It is a common fact that $\ce{Cl2}$ reacts to form $\ce{HCl}$ and $\ce{HOCl}$ when dissolved in water. However, how exactly does the reaction proceed?
The first step could either be that the oxygen atom in $\ce{H2O}$ attacks $\ce{Cl2}$, forming $\ce{Cl-}$, and $\ce{H2OCl+}$, which then loses a proton to become $\ce{HOCl}$.
But it could also be that chlorine attacks the $\ce{H}$ in $\ce{H2O}$, forming $\ce{Cl+}$, $\ce{HCl}$, and $\ce{HO-}$. The first reaction seems more probable (based on the products), but I am not sure.

According to this paper,
the attacking species is $\ce{OH-}$ rather than $\ce{H2O}$ on $\ce{Cl2}$
The paper also says that:

The proposed mechanism also provides a good explanation for the high initial rate of reaction. At the instant the strong chlorine solution is mixed with the distilled water the hydroxyl ion concentration is considerably higher than that calculated from $\ce{kw}$ and the hydrogen ion concentration, for it takes a finite time for the hydrogen and hydroxyl ions present to react to reduce the ion product to the equilibrium value. During this period chlorine molecules also react with hydroxyl ions at approximately the same rate as hydrogen ions are reacting. So for this initial period a higher rate of hydrolysis of chlorine is expected than that found after the hydroxyl ion has reached its equilibrium concentration.

It also says that:

During the course of the reaction the equilibrium concentration of hydroxyl ion is only about 10^(-12) M. Since about 10^-(3) mole per liter of chlorine must react before final equilibrium is reached, new hydroxyl ions must be supplied by the dissociation of water molecules. This could easily complicate the reaction kinetics, if the dissociation rate were not rapid enough. However, calculations assuming that the dissociation of water is a normal bimolecular reaction with an activation energy equal to the endothermicity of the reaction show that the total hydroxyl ion required may be formed in 10^(-5) sec. This is so rapid that the steady state concentration of hydroxyl ion should not be measurably different from the equilibrium value.

What this stuff basically means is that using $\ce{OH-}$ as the nucleophile, as it can explain the initially high rate of reaction(at the back of our minds, we know that $\ce{OH-}$ is a better nucleophile than $\ce{H2O}$). When $\ce{Cl2}$ is added to the distilled water, we find that $\ce{[OH-]}$ is more than that what would be expected at equilibrium value, so $\ce{OH-}$ and $\ce{H+}$ keep combining to reduce the ion product, while at the same time, $\ce{OH-}$ also reacts with $\ce{Cl2}$. So at this time, rate of reaction will be quite high.
Also, the ionization of water to provide the $\ce{OH-}$ and $\ce{H+}$ will be keeping up with the consumption. Also, the reaction rate tells us that collisions in this reaction are very effective, and lead to reaction almost all the time
So, essentially, to explain the initial high reactivity, we take the reaction mechanism to be a coupled equilibria of $\ce{Cl2 + OH- <--> HOCl + Cl-}$ and $\ce{ H2O <--> H+ + OH-}$

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please note that in English, there is a space after the punctuation mark.

On the matter at hand, I doubt there is enough evidence for your proposed mechanism. It is one path in a complex coupled-equilibrium reaction. Neglecting the existence of $\ce{H+(aq)}$ or $\ce{OH-(aq)}$ is probably not doing the system justice.

@Martin - マーチン♦ I appreciate the edit,and the reason I neglected  H+(aq) or OH-(aq) is because the maximum amount of these ions produced by neutral water would be 10^(-7) M,which is not a lot of amount unless the solution is quite dilute.

Dear Yusuf, Organic chemists do make up lot of plausible stories with their arrow notations but that does not mean that the reaction proceeds that way.

@M.Farooq So, is my suggestion correct or incorrect, objectively speaking, and if wrong, why?

I am afraid, there is no proof. I don't know of a primary reference. My "feeling" is that these could be free radical reactions. I may be wrong as well. Please search from Google Scholar. Proposed mechanisms have to be supported by actual experimental findings, which is a difficult task. Organic chemistry texts should be blamed for spreading so much non-sense on arrow pushing mechanisms.

@M.Farooq I found this paper on Google Scholar,but I can't interpret what it means. Can you please go through this and tell me what is the experimental evidence for this reaction's mechanism?

Good find Yusuf. The summary of their paper is that the Cl2 molecule reacts with a OH(-) ion in water and forms Cl2+ OH(-) <=> HOCl + Cl(-) rather than directly reacting with the water molecule.

@M.Farooq Oh,and where does the HCl come from then?

A solution has to be electrically neutral at all times. So when there is a hydroxide from the ionization of water, there is an equivalent amount of protons.


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Equation of combustion of heavy fuel oil (HFO)
We are researching the use of heavy fuel oil (HFO) on ship vessels, but are having trouble finding the equation of the chemical combustion process of HFO. I understand that it contains long chains of hydrocarbons, however it would be helpful if someone could help us towards finding it.
$$\ce{CH4 + 2O2 + N2 + H2S -> 2H2O + CO2 + NO + NO2 + SO2}$$
So far we have found that the base equation, including pollutants in HFO, comes down to to the above according to www.petroleum.co.uk.

"The chemical composition of HFO is highly variable due to the fact that HFO is often mixed or blended with cleaner fuels, blending streams can include carbon numbers from C20 to greater than C50." There cannot be a single equation and it is highly dependent on the batch you are using. The general case you have for methane would work for many of the hydrocarbons $\ce{C_nH_{2n+2}}$ (and others) just fine, but that's just the resulting stoichiometry. The process might be more complex. Have you done any literature research?

There is nothing to be found. Heavy oil doesn't even have a single formula, much less a single equation of any reaction.

@IvanNeretinso so you suggest HFO doesn't combust at all? I understand that it is not a single formula, but i'm interested in the base equation which can be applied to longer chains of hydrocarbons. And if it really doesn't have any equation, point me towards a source since without a source it does not help me further.

Why, it surely does combust regardless of our inability to write an equation for that. (Likewise, planets used to rotate around the Sun long before Newton wrote his equations.) So what? Well, I guess you may write something along the lines of $$\ce{C_xH_y + \left(x + {y\over4}\right)O2 -> xCO2 + {y\over2}H2O}$$ but this is hardly very useful.

Also ring type hydrocarbons and likely compounds up to 4 + % if you mean "bunker C" . If you mean fuel for a Diesel type engine and not a boiler --- still could have many compounds.

You either need to know how much of each molecule you have and have a combustion equation for each one, or you need to use model compounds... lump all c10-c20 into being c15 and combust X amount of c15, lump c21-c30 into c25 etc.. or even just lump everything into c20 etc... Also, if you are combusting this as actual fuel oil, there better not be H2S in it, that will have been converted to other sulphur containing molecules by that point in the refining process.


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How to clean pocket pH meter before storage?

Hello.
I read about different methods how to clean this specific(attached photo)pocket pH meter before storage and i can't figure out what is the right way.
One method says to stir the pH meter(glass electrode)in distilled water and
replace the protective cap,other says to stir the glass electrode only in tap water before replace the protective cap.One says also to wipe dry with cloth(or filter paper)the measuring electrode,while other says to add few drops of storage fluid in the protective cap before replace the protective cap.Those too many methods for the same type of pH meter made me confused.What should be the right way?

The right way is the one described in the official manual.

Yes,but each manual  gives different method (as i described) for the same type of ph meter and also the manual are not clear at all.

What do you mean by "each manual"? There should be only one official manual and the rest is something else. And thank you for removing the watermark from the image;)

This is a cheap chiness no brand ph meter,that you can't find a normal understandable manual for it and when you read the instruction of use for the same type of ph meter on different places a round the web,you get different methods.I mean,almost the same,but difference in small important things as i mentioned.

I see, now it makes sense, sorry for misunderstanding.

No problem.I guess that by looking at the photo,the chemical expert here may know what should be the right way to clean and store this type of ph meter and help with that issue.

see pH electrode storage

MaxW@I read your link,but it seems that the instruction in the link is refer to a different type of ph meters.For example:the link says to soak the ph meter in electrode storage solution or ph 4 buffer solution for hydration,while this yellow ph meter is instructed to be soaked in distilled water for hydration.Could this yellow ph meter has a special bulb electrode?


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cyclopropane's reaction with lithium aluminium hydride
what is cyclopropane's reaction with lithium aluminium hydride? I think it must lead to ring-opening but I didn't find any article favoring this in my texts.

Cyclopropanes do not react with LiAlH4. Substituents of cyclopropanes rings may be reduced with std LiAlH4 conditions with the cyclopropane ring being untouched
example in this patent

but cyclopropane is an extremely strained structure and hydride could attach its LUMO leading to ring-opening and thus releasing strain. And if the reaction isn't happening what are constraints that oppose it?

LiAlH4 is not really a source of naked H-. The reaction generally runs through Al complexation to the substrate before H- transfer. There's nothing for Al to bind to with a cyclopropane.

shouldn't sn2 reaction of ammonia on methyl chloride increase if we use polar aprotic solvent because otherwise in polar protic solvent ammonia would be h-bonded and additional energy to overcome but I was rated wrong for this my test and the correct answer said that rate would increase if we use polar protic solvent.

That is a separate question

you are right .they said I can post only one question every 40 minutes but this way I won't be able to complete my assignment even in several days.so sir please and thank you.

Polar aprotic eg DMSO or SMF will still H-Bond with NH3 as it is both donor and receiver. SN2 is accelerated by polar solvents.


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Effective nuclear charge and Ionization energy
A common reason given on why 3rd ionization energy > 2nd > 1st is because of increasing effective nuclear charge.
As per my book $Z_\mathrm{eff}$ = Atomic number $-$ Number of inner electrons.
Now let us consider a Carbon atom. C has an Electronic configuration of $\ce{[He] 2s^2 2p^2}$ or $2,4$
Right now the effective nuclear charge is $6 - 2 = 4$
Now if we remove the first electron the first ionization energy will be $=\pu{1086.45 kJ/mol}$
After doing this effective nuclear chrarge is still $6 - 2 = 4$
Now if we remove the second electron the second ionization energy will be $=\pu{2352.62 kJ/mol}$
If the effective nuclear charge in removal of 1st and 2nd electron is same then why is 2nd ionization energy > 1st.
note : these values are taken from here

The website you link to gives a 2nd ionization energy that is more than twice that of the first. Also, you should regard effective nuclear charges as aids in interpretation. As soon as you remove one electron you can expect repulsion between the remaining electrons to be reduced.

Effective nuclear charge $Z_{\text{eff}}$ is given by Slater's Rules.
According to these rules, for $\ce{C}$, effective nuclear charge experienced by an outer shell electron is:
$$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 3 = 3.25$$
For $\ce{C+}$, effective nuclear charge is
$$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 2=3.6$$
Since the effective charge for $\ce{C+}$ is more than that on $\ce{C}$, the second ionisation energy is more than that of the first.


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As phase transitions are considered reversible processes, shouldn't they be non-spontaneous?
On the one hand it is stated that reversible processes cannot occur, as they are infinitely slow.
On the other hand phase transitions are said to be reversible processes?


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Nuclear physics
When considering Fusion,  the matter that gets converted into energy what specific matter is converted exactly like a quark and is most of the energy created from the matter  turning into energy?

Perhaps better a question on Physics SE - but, it all comes down to the relative binding energies of the parent and product nuclei...

Is it possible to convert some of the energy from fusion and convert it back to matter

Yes, of course. Particularly easy if you use fusion to generate an unstable nucleus.

yeah because I have this whole dream of making a business on transmutation

Good luck - you should learn about nuclear cross sections and how hard it will be to transmute elements in an economically viable way. Sure, nuclear powers made lots of plutonium, but it wasn't economically viable in particular.

There is not a specific 'piece' of matter exchanged for energy in nuclear reactions.
The mass difference comes from a more subtle change in the energy of the system: how stable the nuclei are compared to one another.
Mass and energy are a conjoined quantity (In fact, we often measure the masses of particles in energy units), so speaking of them as distinct entities is already questionable in this context.
Famously, the relationship between the rest mass and energy is given by
$$
E = mc^2
$$
This relationship tells us that mass is energy and energy is mass, so something as simple as stretching a rubber band to store elastic potential energy will increase the mass of the rubber band. In fact, any kind of potential energy change will change the mass whether it's forming a chemical bond or walking up a hill.
This is the principle at work in nuclear reactions. A helium nucleus has less mass than the sum of the masses of its component nucleons because they are more stable (i.e. lower energy) together than apart.

oh thank you thank you thank you thank you

so same for fission the protons neutrons and electrons mass dosn't change

you don't realize how much hope you have given me for my dream

You really  made my day thank you so much


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What do X-ray diffraction and NMR tell about the Proteins?
I have gone through a simple idea that proteins have different conformations; it is a dynamic system. Moreover, whenever we do ensemble experiments, so we get an average picture. There are some questions-:
Q.1 Often protein structures are reported in PDB (protein data bank), so what do they signify? The most stable conformation or some averaged picture of all conformation or both?
Q.2 In the case of the NMR, proteins have the freedom to move/vibrate through space as they are in the solution phase, but in XRD, they are rigid. So what will be the difference if I get one structure from XRD and one from NMR?
Q.3  Why structure of some proteins has not been resolved yet? I get it that a large protein cannot be solved using NMR; it will be much complex. However, why we cannot use XRD?
a) Why is the crystallisation of some protein hard?
b) Will there be any other challenges even if we get the crystals of a protein?
Q.4 Investigators often use different buffers to crystalise the given protein, but how they make sure that structure that they are getting in that buffer is the "actual" structure? How they make sure that crystalisation do not affect the "actual" structure of the protein.
From "Actual" Structure, I mean the structure or set of conformations that are retained in the physiological condition (in the working conditions).

This goes against the "one question per post", and even the sections numbered Q1 through Q4 contain more than one question, and many of these questions are very broad. I recommend reading a review article (such as https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3011226/), and then asking a more specific question (or multiple question, one per post).

https://chemistry.stackexchange.com/questions/48788/is-the-structure-of-a-crystalized-protein-the-native-one

Why don't you ask about femtosecond, or electron microscopy, too ;) https://pdb101.rcsb.org/learn/guide-to-understanding-pdb-data/methods-for-determining-structure

Dynamics can mean many things, and not all proteins undergo extensive conformational changes during their lifetime. Some are fairly rigid in solution.
XRD or NMR data represents an average over the conformations of a protein construct sampled under experimental conditions. A protein construct is a variant of the protein, selected to allow expression and purification, and possibly also to ensure stability and crystallization (as necessary for XRD).
NMR or XRD data is processed to derive an ensemble of geometries that are consistent with the data and satisfy other criteria, usually imposed by a protein force field or prior statistics about the frequency of geometric features, which allows scoring the quality of a geometry. Before being deposited, the ensemble undergoes quality control by the PDB, amongst other things to verify that there are not too many or extreme violations of expected protein geometric parameters (bond lengths, angles, dihedrals, vdw contacts). Deposited NMR geometries are intended as representative, not exhaustive.
Whether experimental geometries are relevant depends both on the protein and what structural (or dynamical) questions are of concern. Approximating physiological conditions improves the likelihood that the data is relevant.
Proteins under XRD have less conformational freedom, they are more "rigid" (having looked up the definition in Meriam-Webster's dictionary) than those in solution. Different XRD samples (crystals) may still capture views of significantly different conformations, and complementary techniques such as NMR can be used to fill in gaps.
Yes, bigger proteins are more difficult generally to study, but progress has been made. Some of the limits are instrumental, others with sample prep. There is no single answer (too broad). Q4 is a standard problem. You get the structure and then worry whether there are inconsistencies with what is known/expected. You can try different media and see if they generate different structures.
Why is crystallizing proteins hard? Crystallization has to happen spontaneously! The protein must be kept stable (folded) as it crystallizes, which means being restricted to a narrow range of solvent and temperature conditions. There's a lot more that can go wrong versus what needs to go right.  Even as you want the protein to associate and form an ordered crystal, you don't want it to form the wrong kind of interactions resulting in a disorderly aggregate.

Sorry for the mistake I meant why structure of "some" protein has not been resolved yet even though they are not very big. They don't crystallize, why?

@AdityaShrivastav Added an explanation to my answer.


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Constructing chemical equations
When constructing chemical equations, how can you work out the subscript of each product. For example:
$$\ce{3Ca  +  N2-> Ca3N2}$$
How do I know that it's $\ce{Ca}$ with a little 3 and N with a little 2.

Welcome to Chemistry.SE. Those subscripts come from the minimum number of calcium and nitride ions needed to product a charge-balanced neutral ionic compound. Here is a related question on nomenclature of binary compounds that covers some of the basics: https://chemistry.stackexchange.com/questions/98159/how-to-name-binary-inorganic-compounds-given-their-chemical-formula-and-vice

For more information on the linear algebra check out the book "Chemical Reaction Equilibrium Analysis: Theory and Algorithms" by Missen & Smith.

When constructing chemical equations, you cannot work out the subscript of each product. The formula cannot be guessed. In your example, you must know the formula $\ce{Ca, N2}$ and $\ce{Ca3N2}$. This being given, your job is to find the good coefficients 3, 1 and 1.

I hope nobody thinks that people study chemistry to make their living after graduation in a job where they balance equations all day as their prime task.


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How does the relative acidity of 2,3-dinitrophenol compare to that of 2,4-dinitrophenol?
I am assuming that the -I inductive effect in 2,3-dinitrophenol is stronger than in 2,4-dinitrophenol since it operates through less bonds, therefore on inductive grounds it should have a lower $\mathrm{p}K_\mathrm{a}$?
However it's unclear to me whether there are any differences in  π-resonance effects. Any help?

Can you draw resonance structures for the conjugate base?

@orthocresol I got 3 resonance forms for the conjugate bases of each. Is this correct?

You seems to forgot to include nitro groups. Try again and see.

@MathewMahindaratne Are 6 resonance structures for the ortho and para nitro substituents, but only 5 for the ortho meta substituents since 3-nitro group cannot delocalize the charge?

Sounds like you've got it correctly. Then, which conjugate base is the relatively more stable?

@MathewMahindaratne the conjugate base of 2,4-nitrophenol would be more stable and therefore it would be more acidic. Thank you for your guidance.

You got it! :-)


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Does the salt bridge in a galvanic cell deplete over time?
I searched for the answer for this question on Wikipedia and Stackexchange  a lot but couldn't find same question or a required answer in some other similar question.
I understand that the function of the salt bridge is to complete the circuit and maintain electrical neutrality in both the half cells of a galvanic cell. This is achieved by movement of cations and anion in the salt bridge. The anions move towards the anodic half cell and cations move towards the cathodic half cell.
My question is: are these cations and anions provided by the salt bridge itself and as such, do they deplete over time as the electrochemical reaction progresses?
OR
Is the salt bridge just a medium that is passing the anions from the solution of the cathodic cell to the anodic cell and passing the cations from the solution of the anodic cell to the cathodic cell?
OR
Do both of these mechanisms take place?
I will be grateful for insight on this topic.

Consider a salt bridge as a part of the cell electrolyte, that is not allowed to mix with the rest of electrolyte.
For a particular ion, there are 2 cases:

The ion is shared with the electrolyte part, from which the ion migrates toward the bridge. Then the ion gets depleted from electrolyte and not from the bridge.
The ion is not shared with the electrolyte part, from which the ion migrates within the bridge toward the other electrolyte part. Then the ion gets depleted from the bridge.

But there is needed to add the bridge maintains ion neutrality, so ions that leave the bridge are replaced at the other bridge end by the same or other ions of the same total charge.
Bridges are not usually used in power cells and cells where significant portion of chemicals gets depleted or react. If they are used, they usually use at least one of ions shared.
Power cells often use a single electrolyte and a diaphragm, soaked with this electrolytes, so ion depletion does not apply.
Bridges are used mostly in cells where accurate potentials matter, to minimize phase potentials, where the current is minimal. Like calomel ( $\ce{Hg2Cl2}$ ) or $\ce{AgCl}$ reference cells, or in laboratory test cells.

@Poutnic: Do the components of the salt bridge change over time on cell operation? How?

The ions migrate in and out, that is the change. Plus, in a particular case, there may be chemical changes, not directly related to being a cell bridge.

I am replying late but thank you @Poutnic for the answer.

Salt Bridge is like a source which provides cations to the the half cell where the negative ions are more than the  positive ones in the electrolyte .
Similarly it gives anions to the half cell where positive ions are more than the negative ones in the electrolyte.
It basically ensures continuity of the flow of electrons from anode to the cathode by neutralising the  ions present in the respective electrolytes by providing an electric contact between the two solutions without allowing them to mix with each other.
One thing which should be taken care of while choosing appropriate salt bridge is that the substance/solution present in it should have equal migratory aptitude of both the cations and the anions so as to ensure that the neutralisation occurs almost simultaneously in both the half cells.eg Potassium chloride, ammonium nitrate.
Usually an electrochemical cell stops functioning when no more electrons are released at the anode.
There can be multiple reasons for this and malfunction  of the salt bridge might be one of the reasons.
Eg.the salt bridge getting depleted i.e. it is not releasing enough counter ions in the electrolyte. Therefore  the solution in the anode compartment would become positively charged and the solution in the cathode compartment would become negatively charged, because of the charge imbalance, the electrode reaction would quickly come to a halt


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how can I measure the pH of a HF:H2O:CH3OH solution?
how can I measure the pH of a HF:H2O:CH3OH solution? more specifically, how can I precisely and accurately measure the H+ ion concentration?

Well, there are various options there, particularly what are the approximate proportions?  Another thing is what you're really trying to achieve there? If there more methanol then water, it's possible there's also more protonated methanol. And if there's a lot of HF you should use Hammett scale of acidity. Also HF reacting with MeOH may be a problem...

The pH is a quantity that is measured in an aqueous solution by dipping a glass electrode in the solution. It is equal to -log[H+] in dilute solutions. For concentrations > 0.01 M, the pH deviates from log [H+], because the concentration [H+] should be calculated by dividing the number of moles H+ by the volume of  the free water (and not the volume of the solution). Free water is the water not attached in the Debye layers around the ions. It is very difficult to calculate the volume of the free water in a given solution. For concentrations > 0.01 M, the volume of the free water is smaller than the volume of the solution.
In mixtures water + methanol, the problem becomes worse and cannot be solved by measuring the pH. The pH value, measured with the glass electrode, gives a value which is not the logarithm of the concentration of $\ce{H+}$. In this case, you have better determine the concentration of HF in a volumetric titration, by adding NaOH solution drop by drop, until the indicator changes its color. The indicator may be thymolphtalein. The volume of free water is smaller than the volume of the solution.


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what wavelength should I use to measure the transmittance of iodine colouring?
I am doing an experiment on how pH buffers may affect amylase's break down of starch into glucose in a starch solution. I am using iodine reagent to identify how much starch is left after amylase is added. I am wondering, to which wavelength should I set the spectrophotometer to in order to measure the absorbance and percent transmittance values of the solution after iodine solution is added to measure the amount of starch?

As the solution is blue, you should choose a wavelength in the red region, around about 700 nm.

You know that in first approximation the absorption recorded is a product of the optical path length $d$, the extinction coefficient $\varepsilon$, and the analyte concentration $c$ as $$\mathrm{abs} = c \cdot \varepsilon \cdot d$$
Now, depending on your specific task, consider the three variables to your advantage.  Because for the wavelength eventually elected to record your data, you want to stay in a linear relationship between analyte concentration and detected absorption as the read-out.

The pathlength $d$ is something typically fixed to $1\,\pu{cm}$, but be aware that there are optical cells offering a shorter path length (e.g., $5\,\pu{mm}$, $1\,\pu{mm}$ as here or even $0.1\,\pu{mm}$) or longer paths (e.g., $50\,\pu{mm}$ (example), or $100\,\pu{mm}$.  It depends on the design of the spectrometer if mounts are interchangeable, or not, to benefit from an intended adjustment to your needs here.
What is the range of concentrations $c$ to record?  As the extinction coefficient depends on the wavelength of recording, it may be beneficial to record at a wavelength maximizing $\varepsilon$ of your amylose iodine complex, provided you want to detect low concentrations of this specis.  (Of couse it shouldn't be in the range other molecules of your solutions absorb intensely.)
But: if you want to determine solutions with a high concentration of your analyte, than it makes sense to record the absorption at a wavelength where $\varepsilon$ of your analyte just is high (not maximal), while «all the rest» of the sample solution does not contribute significantly to the absorption.  This approach will guard you to encounter absorptions too high to be sensible for detectors in the spectrometer.  Look up the manual of the spectrometer or ask colleagues already using the instrument; for some, an absorption greater than $2.0$ equates to «skip it, there is no light on the detector at all».

The iodine-amylose complex is known to absorb intensely.  This equally may justify to record $\mathrm{abs}$ in function of analyte concentration $c$ a little «off» from $\varepsilon_\mathrm{max}$.  By recording the spectra of your blank, and of a typical solution with your complex to quantify, you may justify how many nanometers above or below $\lambda$ yielding $\varepsilon_\mathrm{max}$ of your analyte is fine.  If time and resouces permit, you may record at multiple wavelengths, too; eventually, the readings should lead to the same concentrations per sample.