text:
Converting pure nylon salt to polymer?
During one of the methods of  polymerization:

Nylon 6,6 is prepared from polycondensation of hexamethylene diamine
  (HMD) and adipic acid. The repeat unit of the polymer is $\ce{–[NH(CH2)6
NHCO(CH2)4CO]}$. The diamine, which melts at 40.87 °C, is normally used
  in the form of a concentrated aqueous solution. The dibasic acid is
  used in its pure solid form (m.p.= 152.1 °C). In the preparation of
  nylon 6,6 polymer, the first step is a preparation of salt formed from
  a mixture of precise stoichiometric quantities of the two monomers.
  The salt can be prepared by mixing the alcoholic solutions of the two
  components, the pure salt precipitates and is dissolved in water to
  give a solution. Alternatively, the salt can be prepared by mixing a
  dispersion of the diacid in water with a solution of the diamine. The
  aqueous salt solution is prepared so that it has a concentration of
  about 50-60%. For the preparation of nylon 6,6 polymer, the salt
  solution is subjected to evaporation at the boil, possibly at elevated
  pressures, until concentration ≥ 60% are achieved. The concentrated
  salt solution together with a small amount (0.5%) of a molecular
  weight stabilizer is then heated in a reactor under a blanket of
  nitrogen so that temperature increases gradually and pressure reaches,
  typically, 1.73 MPa (250 psi). As water evaporates and temperature
  increases from about 212°C to 275°C, the molecular weight of the
  polymer reaches about 4000. The further reaction is achieved by a
  gradual decrease in reactor pressure to atmospheric pressure and then
  holding the polymer under these conditions for about one hour. At this
  point, the polymer is not quite equilibrated but molecular weights are
  in the range of 12000 to 17000. The process is devised to remove all
  the liquid water present in the salt solution as well as almost all
  the potential water, which may get liberated on the reaction of
  carboxyl (-COOH) and amine end groups present in the polymer. High
  molecular weight nylon for industrial applications is obtained by
  conducting the final stages of melt polymerization under reduced
  pressure (to remove water) or by addition of chain coupling agents (to
  help improve stoichiometric balance). The finished polymer is then
  extruded in the form of a ribbon or strand, quenched with water and
  cut to form chips. After drying, the chips are sent for subsequent
  spinning operation. Alternatively, the polymer may be sent directly to
  a spinning machine without prior solidification. In nylon 6,6 the
  polymerization reaction goes almost to completion.

Now my question is in the above method of polymerization involves formation of aqueous nylon salt from pure salt. I want to know why the precipitated pure salt isn't converted to polymer but is made aqueous and then converted to polymer? Isn't there anyway of forming the polymer from pure nylon salt (fully crystal dehydrated salt) instead of the aqueous one from the reactants?
Note: I'm ok with reacting the pure salt with any chemical substance to form the polymer.

From the preparation above: "The salt can be prepared by mixing the alcoholic solutions of the two components, the pure salt precipitates " Collect this precipitate and dry under vacuum without heating. That will give you the pure salt.

@Waylander oops! I had made a mistake in my question. I've corrected it now. My question was actually why the salt is made aqueous?

I presune it is so it can be pumped through the reactor to control the polymerisation and give the desired range of chain lengths. If you just heat the salt you'll have less mixing

The problems with reacting pure nylon salt are as follows:

The molecules of hexamethylene diamine (HMD) and adipic acid will have a very hard time accessing each other's reacting groups, because of limited mobility within the solid structures.
The reaction is a condensation, so it will produce a substantial amount of water anyway, so pretty much any reason for keeping the reactants dry is irrelevant, because the final polymer will end up wet anyway.

It might be possible, but because of (1), your product might be impure, containing a lot of small unreacted adipic acid and HMD particles.
Also, If you don't want to deal with the tedious procedures above, considering that they were probably written for industrial applications, consider using this procedure. It's a lot more lab friendly.


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Different products using different indicators
A question states that 0.1 M solution of compound$$\ce{Na2CO3.NaHCO3.2H2O}$$ is titrated against 0.05M HCL, and that a different amount of HCL is used according to the indicators(phenolphthalein or methyl orange) used.
How does the indicator used affect the titration?

An indicator does not affect a particular reaction. But in your solution, you have two successive reactions when adding HCl to this solution : First $$\ce{CO_3^{2-} + H^+ -> HCO_3^-}$$ and a given indicator must be added to determine the end of this reaction. If you don't, all you see is a colorless solution being transformed into another colorless solution, even though its composition has changed. This added indicator may be phenolphthalein, which goes from red to colorless precisely at the end of this first titration. Then, by adding more HCl, you carry out the following reaction : $$\ce{HCO_3^- + H^+ -> H_2O + CO_2}$$ And you need is another indicator, maybe methyl orange, to determine the end of this second reaction. The color of the methyl orange in solution changes from yellow to reddish exactly after the end of this second reaction.
Phenolphtalein and Methyl orange are molecules which reacts with $\ce{H^+}$ exactly like $\ce{CO_3^{2-}}$ and $\ce{HCO_3^-}$. But they react before or after these two ions. To summarize, $\ce{H^+}$ reacts first with  $\ce{CO_3^{2-}}$ and nothing else, and it takes a couple of milliliters to do it. Then it reacts with Phenolphtalein and nothing else, and it takes less than one drop to do it. Then it reacts with $\ce{HCO_3^-}$ and nothing else, and it also takes a couple of milliters to do it. And last but not least, it will react with methyl orange.

So is there an order of preference according to which H+ ions(and hydroxyls too, for that matter) react with conjugate bases and indicators? Can you tell me where I can read up more about that?

It depends on the pKa's. For the first titration, the pKa of $(HCO_3^-/CO_3^{2-})$ is 10.25. For the second titration, the pKa of $HCO_3^-/(CO_2+H_2O)$ is 6.35. A good indicator for the end of the first titration and the beginning of the second has a pKa between 10.25 and 6.35. The best choice would be an indictor with pKa = 8.35. But there is no such an indicator. We must choose an indicator whose pKa is near to 8.35. Phenolphtalein is acceptable, as it has pKa = 9. For the second titration, we must find an indicator of pKa smaller than 6.35. Methyl orange is OK, with a pKa = 3.9.

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Can I use an electrolyte containing a different cation from the anode when doing an electroplating experiment?
For example, I want to purify copper. So the  anode will be impure copper and cathode will be pure copper. However, instead of using copper sulphate/halides/nitrate solution, can I use magnesium chloride?
My thought process: Magnesium is more reactive than copper, so copper from the anode that are oxidised into Cu2+ ions will still be preferentially discharged, therefore not affecting the reaction(?).
Sorry if there are conceptual errors here and there, Im just really confused about this topic. T_T

There will not be enough chlorides for both magnesium and copper and it will probably end forming insoluble basic copper salts or oxides/hydroxides. Plus, there can be loses of chlorine.

If you use magnesium chloride in solution, you will not get any copper deposited on the cathode ! You will get $H_2$ instead. Is it what you want ?

As others have noted, what you want to do is problematic, but it is also an opportunity! You can purchase copper sulfate quite inexpensively, so why not do the experiments? First, try the magnesium chloride electrolyte and carefully observe what happens at each electrode. Does the copper cathode increase in mass, due to adherent deposited copper? Then do the same experiment with the copper sulfate electrolyte. Again carefully observe what happens at each electrode. Are any gases evolved in either experiment? Just test your intuition!

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text:
Light up Noble gases
I saw you can light up the Noble gases with a Tesla coil.
Does it also work with a plasma globe?
Are there any other simple ways to make the gases glaw like that?
And does it work with all the Noble gases or only when it is law pressure or something?
P.S. I don't really understand in chemistry, just saw some videos on YouTube.

Sure it works with a plasma globe, spark discharges, neon bulbs, and so on. It works for all the noble gases and low pressures just facilitate the electrical excitation process. Neon sign transformers do a great job, although they are rather electrically dangerous and not to be taken lightly (sorry for the pun!).

You can make any gas glow under the right conditions, not just the noble gases.

Many low pressure gases glow when undergoing an electrical discharge
Plasma globes are simple vessels containing low pressure gas mixtures (often noble gases). They are usually driven by something like a Tesla Coil (or by an electronic circuit that produces the same high voltage high frequency output as a tesla coil).
Neon signs and standard fluorescent tubes use the same basic principle: an electrical discharge in a low-pressure gas tube emits light. The colour of the light depends on the gases in the tube and their pressure. Most "neon" tubes, fluorescent tubes and plasma globes use fine-tuned gas mixtures at specific low pressures.
So, yes, it is a general phenomenon, but it usually requires low pressure.

A dinky exception to the low pressure issue is the triboelectric effect observed when sugar crystals (and wintergreen Lifesaver candies) are crushed in air. ;-)


text:
was the octet rule disproven?
so according to my understanding orbitals can hold s can hold 2 electrons and p can hold 8 electron d hold 18 and f can hold 32. wouldn't this violate the octet rule? which say only the first shell can hold 2 electrons and the rest of the shells 8. im also confused why some websites says that each orbital has 2 electrons, s has 1 orbital  (2 electrons) p has 3 orbitals (6 electrons) d has 5 (10 electrons) and f has 7 orbits ( 14 electrons)
im so confused guys plz help me as simple as possible

The octet uses s and p orbitals of the outer shell: 2 + 6 = 8. In the fourth period, you are filling 4s, 3d and 4p; that's why there are 18 elements in the fourth period. See https://madoverchemistry.files.wordpress.com/2017/03/4112.jpg?w=1680

As with many things in chemistry, it isn't so much a Rule as a Suggestion. It helps explain some things, but then it gets complicated really fast.

The octet rule has no theoretical basis. It has been observed to work quite often. But that is all we could say. It cannot be proven. The octet rule works nearly perfectly in organic chemistry,
There are many examples where this rule is not verified. For example the Boron hydrides, the simplest being $B_2H_6$. This compound is commercial, but it does not follow the octet rule. Another example is the familiar molecule of Oxygen $O_2$, which has no double bond as often given in elementary chemistry courses. Other examples : $PCl_5, SF_6$. Another simple example is Nitrogen oxide $NO$. And a lot of compounds made with metals from the transition range.

+1 Good point about organic chemistry.

Even in organic chemistry there are some boners.  Try to apply it to the cyclopropenyl ring system.  The electron count that gives a stable structure is two less than what would satisfy the octet rule.

The octet rule is basically a rule for low atomic number elements like carbon and some other specific elements like the alkali metals and the halide anions. It is typically coupled with the Lewis Dot notation as a first example of chemical bonding. However there are other forms of bonding that involve more than just the $s$ and $p$ electrons.

And even with just $s$ and $p$, molecular orbital structure can override the octet rule.


text:
Does ADP + P => ATP produce water?
ADP + P has one oxygen and two hydrogen more atoms than ATP. Are these atoms released as water in the reaction ADP + P => ATP? To balance the equation, would ADP + P => ATP + H2O be more correct?

Yes. H2O is produced in the reaction ADP + P -->  ATP.  But we usually used not to mention it, because the organic part of the reaction is more important

It is analogical as we often implicitly include water into the equilibrium constant. Consider the above rather as the reaction schema than reaction enumeration.


text:
Comparison of acidity of aqueous solutions of Lewis Acids
What can be the order of acidity of aqueous solutions of following Lewis acids: $\ce{FeCl3, AlCl3, BeCl2, NiCl2}$
The simplified reactions are:
$\ce{FeCl3}$:
$\ce{FeCl3 + 6H2O -> [Fe(H2O)6]^3+(aq) + 3Cl-(aq)}$
$\ce{[Fe(H2O)6]^3+(aq) -> [Fe(H2O)5(OH)]^2+(aq) + H+(aq)}$
$\ce{AlCl3}$:
$\ce{AlCl3 + 6H2O -> [Al(H2O)6]^3+(aq) + 3Cl-(aq)}$
$\ce{[Al(H2O)6]^3+(aq) -> [Al(H2O)5(OH)]^2+(aq) + H+(aq)}$
$\ce{BeCl2}$:
$\ce{BeCl2 + 4H2O -> [Be(H2O)4]^2+(aq) + 2Cl-(aq)}$
$\ce{[Be(H2O)4]^2+(aq) -> [Be(H2O)3(OH)]^+(aq) + H+(aq)}$
$\ce{NiCl2}$:
$\ce{NiCl2 + 4H2O -> [Ni(H2O)4]^2+(aq) + 2Cl-(aq)}$
$\ce{[Ni(H2O)4]^2+(aq) -> [Ni(H2O)3(OH)]^+(aq) + H+(aq)}$
For first step of reaction, the octet of $\ce{Al}$ and $\ce{Be}$ are not complete. Thus, reaction will be faster in case of $\ce{AlCl3}$ and $\ce{BeCl2}$.
For second step of reaction, the charge density of $\ce{Al}$ and $\ce{Fe}$ are more than that in others. Thus, it can break one $\ce{OH}$ bond and the reaction will be faster in case of $\ce{AlCl3}$ and $\ce{FeCl3}$.
But what will be the overall acidity order of these four compounds?

The charge density order would be : Fe3+>Al3+>Be2+>Ni2+. Practically, so should be the order of Lewis acidic strength. However, the point of octet not being complete is also taken into consideration in many answers on the internet and hence AlCl3 has been deducted as better than FeCl3. However, Ka value of FeCl3 is indeed more than that of AlCl3. I know this doesn't exactly answer the question but adds on some necessary information.

@HappyFeetMe Can you provide the reference for values of $K_{\text a}$ for $\ce{AlCl3}$ and $\ce{FeCl3}$?


text:
Solution for Electroplating Selenium?
I need to electroplate selenium onto a conductive metal (for an experiment). The end goal is a thin film of pure, elemental selenium. Unfortunately I can't find anything on the internet: Ideally, I'd like a simple, inexpensive, and easy to make and/or find solution. Does anyone have any suggestions. (Also tips on suitable anodes would be helpful too!)

This might give you some ideas.

I'm not sure of any conventional ways of electroplating selenium, but you could use a technique such as magnetron sputtering to spray a thin film of selenium onto another metal. Heck, you could even use glass or wood if you wanted to do it that way. And if you're going for thin, this technique can get you some of the thinnest, smoothest layers of metal out there. A major downside is that your source metal sample has to be pretty pure selenium. But still, I think It's completely worth it.
However, If you don't have access to one of these devices, They can get expensive. Most major colleges with significant science majors will have one of these or something similar to coat samples for electron microscopy, or other reasons. Be sure to check there before going off to buy your own. Also, there are legitimate YouTube videos out there that show how to make your one device for magnetron sputtering. Check out this one if you're interested.


text:
How does one calculate oxidation states?
For example, how could I calculate the oxidation state of lead in $\ce{PbS}$? Or for that matter how would I calculate it for sulfur? Or I have two unknowns, for example sulfur in $\ce{PbSO4}$? Do I assume that lead has a 2+ charge and therefore a +2 oxidation state and since it has ? Or is there something that I am missing? Additionally, from all the reading that I've done I've only found a rule for bromine, chlorine, and iodine in binary compounds, what happens if they are in a compound with more than two elements? For example $\ce{COCl2}$. How would someone calculate the oxidation state of carbon? This stuff is really confusing for me and I apologize if this seems really basic or I have missed something.
I have read the introduction to oxidation states post, and it was a very well written answer, however it still did not help things click. I still feel like I am missing something

For compounds where you have a non-metal with a metal, there is likely an ionic bond. The oxidation states are just the charges on the ion. PbS has Pb2+ and S2-, so those charges are simply the oxidation states.
For PbSO4, you know sulfate is a 2- ion, so it must be bonded to a Pb2+. Therefore, again, Pb has a +2 oxidation number. When you look at the sulfate ion, you know that the sum of the oxidation states of the elements must equal the ion's overall charge. Oxygens are -2, so set up the equation: (X = oxidation number of sulfur)
X + 4(–2) = -2
X = +6
Calculating the oxidation state of carbon can be more complex. If you draw a dot and cross diagram showing the covalent bonds made by carbon, this will help. Look at the electronegativities of the elements in the bonds. Carbon takes the electron pair if it is the more electronegative, but this is rarely the case. For every electron pair carbon takes, it has a -1 added to the oxidation state, and +1 added if it loses the electron pair. For example, if you draw the methane diagram you find that the oxidation state of carbon is +4. Using this method, you can find the oxidation state of carbon in compounds.


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SARS-CoV-2 virus floating in the air in *aerosols* for up to 3 hours
I have a question related to SARS-CoV-2 virus spreaded in the air.
This Straits Time article from March 18 describes a new study:

The tests show that when the virus is carried by the droplets released when someone coughs or sneezes, it remains viable, or able to still infect people, in aerosols for at least three hours.

Examples of aerosol are fog, dust, particulate air pollutants (e.g. pollen), smoke, etc.
That being said, is essentially saying that the SARS-CoV-2 virus is all over in the air, right?
I am saying that SARS-CoV-2 virus can be everywhere in the world. Why? Because, for example, dust is everywhere in the air and air is not static, i.e. "it flies" from one site to another and hence it can move to a place where no infected people have been at.
Am I understanding this correctly?
Thank you in advance

Sneezing may be considered as one early vector of transmission.  According to https://en.wikipedia.org/wiki/Sneeze, the droplets ejected by a sneeze are in the micrometer range, travel with 4.5 m/s for 0.6 m.  The two later values are much less than I thought, esp. with the current rule to keep at least 1.5 m distance to the next when going for the groceries.  With a known diameter and density of a virus vs. the density of air, it would become a question for fluid mechanics / phys.se to derive how much time is required for a virus to fall to the ground without lateral wind, etc.

This question should be closed due to the medical exclusion. We are chemists not qualified medical professionals.

I'm voting to close this question as off-topic because to derive how long a micrometer-small particle of density $\varrho_A$ may stay afloat in air of density $\varrho_B$ seems more suitable for sibling site phys.se.

@Buttonwood ok.

@Buttonwood sir I just asked it on physics and I have just notice that someone has posted an answer here :O

@America - Thinking you can make this a purely "physical chemistry" problem while mentioning the "SARS-CoV-2 virus" in both the question's title and body is ludicrous. Even trying to ask about some hypothetical virus would do little to hide the intent in the current panic.  Because we are not qualified medical professionals we need to stay out of such "hypothetical" discussions to avoid spreading misinformation.

@MaxW The reason on why I decided to name the virus as "SARS-CoV-2 virus" it's because on another SE site  I was told to name the virus with the proper name. In general I had wrote Covid-19.

You are not understanding it correctly
There is a significant difference between an aerosol, dust and droplets. Furthermore the intermixing of air is not as you claim it is, it is in fact, quite static. If air just went everywhere immediately there would not be clouds, temperature differences, or indeed any weather effects at all.
A question you could ask yourself is: With my current understanding, covid-19 should be everywhere. There should not exist differences between cities and rural areas, or between countries. But there is.
Listen to the advice of health services and the government. They know their shit.

-1 You can answer a question without being patronising to the OP.

Thanks for your answer. I am not a physician, but afaiu air is not static, by just looking at trees and how their leaves are moving I can tell that there's air movement. And hence spreading virus from one place to another.

-1 We need to stay out of such "hypothetical" discussions to avoid spreading misinformation.This is a medical question which we are not qualified to answer.


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Do the spectra of compound materials combine linearly with respect to the ratios of the constituent compound spectra?
My question stems from reading this article on the FTIR spectrum of oil that has been mixed with various contaminants.
If one has the spectrum for the original uncontaminated oil and the spectrum for water, will the spectrum of a sample that has been contaminated with 10% water (by weight) be equivalent to the addition of original oil sample spectrum and 0.1 of the water spectrum intensities?
(Also, should the unit of contamination be the weight or the volume?)

Yes, you are right. If you have $n$ components in a mixture each with a spectrum $S_1$, $S_2$,...,$S_n$, the observed spectrum $S_{obsvd}$ would be the sum of $S_1$+ $S_2$+...+$S_n$.
In FTIR you either measure the transmittance or the absorbance. The term intensity is more or less used when you deal with molecular or atomic emission of light (fluorescence etc.) As long as there is no chemical reaction among the constituents, the absorbance should be linear with the concentration.

Thanks for your answer, can you point me towards any documentation to support this?

The summation part or the theory of FTIR?

Search, Beer's law is additive. These are paid but highly reliable services 1. "In defence of Beer's law" https://www.osapublishing.org/josa/abstract.cfm?uri=josa-38-10-854 2. Beer's law and its use in analysis. A review https://pubs.rsc.org/en/content/articlelanding/1963/an/an9638800678/unauth#!divAbstract

More specific to Beer's law in the field of FT-IR spectrometry is the 1998 sequel «Extending the Range of Beer's Law in FT-IR Spectrometry.» with part I https://www.osapublishing.org/as/abstract.cfm?uri=as-52-11-1403 and part II https://journals.sagepub.com/doi/10.1366/0003702981943031.  Both sources accessible by subscription.

Provided intermolecular interaction foes not cause shift in the transition frequencies. Strongly interacting molecules (hydrogen bonding)  show such effects.  in these cases,  the observed spectra can be different ( although small)


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Is it correct to say that active carbon is an enhancer of either acid or base?
As a non chemist I tried to read to understand if active carbon acts in solutions as an acid or a base but I understood that it can act as both, depends on the material which used to make the solution.
For example, mixing a sodium bicarbonate with active carbon in a solution will make that solution with higher PH (hence a stronger base, per my understanding).
Is it correct to say that active carbon is an enhancer of either acid or base?

Could you please provide evidence, i.e., measured pH values, for the assertion in the second sentence?

@EdV did you mean the second passage For example, mixing a sodium bicarbonate with active carbon in a solution will make that solution with higher PH (hence a stronger base, per my understanding).?

Yes. I see no reason the pH would change unless adsorption was taking place. But then why? So if measured pH values, with and without the active carbon, displayed a significant difference, then an explanation would be warranted. But a small difference might just be experimental error.

@EdV I don't know where to start - I don't have the equipment to measure the PH if I mix (the amount of) one capsule of Active carbon with say half a teaspoon of baking soda which I bought both in the supermarket; I can do that but don't have the equipment and will not purchase it for this question - if the question cannot be answered regarding this or a similar solution than I guess the question should be closed.

Understood. Best to just wait and see how the answers shake down.

Active carbon is neither an acid, nor a base. It cannot. It does not react directly in water and in acidic or basic solutions.  But it may have a catalytic action on the solutes. It may accelerate the rate of chemical reactions. That is probably what you have observed when mixing sodium bicarbonate in water. Without knowing it, you must have increased the rate of the forward reaction : $$\ce{HCO_3^- <=> CO_2 + OH^{-}}$$ and this reaction is favored if the air in the room contains a low concentration of $\ce{CO_2}$, according to Le Chatelier's principle. As a consequence, the pH of the solution has slightly increased, due to this supplement of $\ce{OH^-}$ ions.

If one is already employing microwave (MW) assisted preparations, in this regard, per this 2007 source, 'Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances', by Xie Quan, et al, then one may, indeed, be able to increase the respective power of acidic reactions.
As to mechanics with bicarbonate at low pH in the presence of any created hydroxyl radicals (from MW irradiation of aqueous activated carbon):
$\ce{HCO3(-) + .OH = .HCO3 + H2O}$
the powerful hydrogen carbonate radical.
Similarly, working with a weak acid (for example, dilute acetic acid) and H2O2, accelerated activity via the formation of the more acidic hydroperoxyl radical from the action of the hydroxyl radical acting again on hydrogen peroxide:
$\ce{H2O2 + .OH -> .HO2 + H2O}$
Source: 'Radical-Enhanced Acidity: Why Bicarbonate, Carboxyl, Hydroperoxyl, and Related Radicals Are So Acidic’, referring in particular, to the hydroperoxyl radical, •HO2. To quote from the abstract:

Comparison of accepted pKa values of bicarbonate, carboxyl, and hydroperoxyl radicals, with those of models having the unpaired electron replaced by H atoms, implied the acidity of the radicals was greatly increased. A Density Functional Theory computational method of estimating pKas was developed and applied to a set of radicals designed to probe the phenomenon of radical-enhanced deprotonation (RED-shift) and its underlying causes. Comparison of the computed pKa values of 12 acid radicals to those of the corresponding model acids confirmed the intensified acidity of the title radicals..

More generally, in the realm of electrosynthesis (which relates to the chemistry of chemical synthesis of compounds in an electrochemical cell) and, in particular to the alkaline 'bleach battery', replacing the copper metal with more noble carbon, will also accelerate the reaction. Source: see comments and reference cited here.


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Are there Exchange integrals in Semiempirical methods?
One of the major features of semiempirical methods is that they significantly reduce the cost of evaluating 2-electron integrals.
In general, they follow the formula for HF, but, introduce many simplifying assumptions. In HF for same spin electrons you need to calculate both coulomb and exchange integrals.
When applying the simplification in, for instance, CNDO and INDO, that $(\mu \nu | \lambda \sigma ) = \delta_{\mu \nu} \delta_{\lambda \sigma}(\mu \mu | \lambda \lambda )$, this seems to de facto rule out calculating an exchange integral.
Is there a scenario where the exchange integral is computed in semiempirical methods (for simplicity, in the case of CNDO/INDO methods)?
Textbooks do not discuss the exchange integral when going over semiempirical, so the answer must be obvious... I just don't know if it is obvious that they cannot be performed, or if they can, and I am missing something.

As far as I can tell from Jensen, CNDO and INDO neglect all exchange integrals, but, NDDO does not

If I remember correctly, the exchange integrals become parameters based on the atomic orbital energies - but it's been a long while and my copy of Pople's Approximate Molecular Orbital Theory is on campus. If you can find a copy, the answer's in there - he explains the derivations of both CNDO and INDO in detail.

I will check it out. As the saying goes... Answer someones stack exchange question and you nourish their knowledge for a day, give them a good book and you nourish their knowledge for a lifetime

Semiempirical quantum chemistry models do carry out proper Hartree-Fock calculations, including both Hartree (electron-electron electrostatics) and Fock exchange contributions to the total electronic energy. As noted in the question, the tensor of 4-center Coulomb integrals is heavily sparsified, approximated, and parameterized to simplify and accelerate semiempirical calculations. Even in the most approximate models (CNDO), the exchange interaction between molecular orbitals does not vanish in general. For example, the exchange integral between a pair of molecular orbitals $i$ and $j$ with real atomic-orbital coefficients $\phi_{i \mu}$ is simplified by the zero-differential-overlap (ZDO) approximation stated in the question as
$(ij|ji) = \sum_{\mu,\nu,\lambda,\sigma} \phi_{i \mu} \phi_{j \nu} \phi_{j \lambda} \phi_{i \sigma} (\mu \nu | \lambda \sigma) = \sum_{\mu,\lambda} \phi_{i \mu} \phi_{j \mu} \phi_{j \lambda} \phi_{i \lambda} (\mu \mu | \lambda \lambda)$.
This answer might be confusing precisely what the original question means by "exchange integral". For example, no explicit integrals are involved in calculating the Coulomb matrix elements used by most semiempirical models, and there may be some class of Coulomb matrix elements in the atomic-orbital basis that are considered to be "exchange-like" that are indeed neglected in most semiempirical models (I'm not familiar with all variants of this nomenclature). The neglect of diatomic differential overlap (NDDO) approximation that eventually superseded ZDO does allow for nonzero $(\mu \nu | \lambda \sigma)$ values when $\mu$ and $\nu$ (similarly, $\lambda$ and $\sigma$) are different atomic orbitals on the same atom.


text:
Why is the boiling point of ethyl fluoride lower than that of hydrogen fluoride?
The book, Solomons' Organic Chemistry (for JEE Mains and Advance), contains the following question:

Hydrogen fluoride has a dipole moment of $\pu{1.82 D}$; its boiling point is $\pu{19.34 ^{\circ} C}$. Ethyl fluoride ($\ce{CH3CH2F}$) has an almost identical dipole moment and has a larger molecular weight, yet its boiling point is $\pu{-37.7 ^{\circ} C}$. Explain.

Now while explaining the (very)  concept of boiling point and why it differs for different substances the book considers:

Dipole-dipole forces
London dispersion forces
Molecular weight

But when I tried to explain the phenomenon stated in the question, I found it hard to explain the lower melting point. Following is my approach on the explanation:

Since both have the same (similar) dipoles therefore there would be very negligible difference in their boiling points.
London dispersion forces increase in magnitude as the surface area of the molecule increases. So contrary to what was said $\ce{CH3CH2F}$ should have stronger dispersion force interaction (due to larger surface area)  and hence higher boiling point as compared to $\ce{HF}$ (given other properties were same).
Given that $\ce{CH3CH2F}$ has higher mass than $\ce{HF}$, therefore it should lead to higher boiling point (as is stated in the question itself).

Now it can be clearly seen that the above stated arguments are against the phenomenon, so I believe that I am missing something in my explanation.
So

What is (if any) wrong/missing in my explanation?
Why is the boiling point of $\ce{CH3CH2F}$ lower than that of $\ce{HF}$?

Hint: Recheck how the dipole moment is defined.

@Karl the said book only talked about the surface area while explaining such phenomenon. It says "and because the
surface areas of larger molecules can be much greater, intermolecular dispersion attractions can also be much larger".

they are in aqueous condition, then hydrogen bonding can be one of the reason in HF both hydrogen and fluorine are able to form hydrogen bond, where in ethyl fluoride only F which can form intermolecular bond.

The three reasons that you present are not sufficient to explain the high boiling point of pure $\ce{HF}$. There is a fourth reason, that you have not mentioned : the Hydrogen bond. Molecules $\ce{HF}$ are interacting with one another through $\ce{H}$-bonds, exactly like $\ce{H_2O}$ molecules attract one another because of $\ce{H}$-bonds. In the liquid state, $\ce{HF}$ behaves as if it is made of heavier molecules, something like $\ce{(HF)_n}$, with $\ce{n}$ being between about $4$ and $7$. The boiling point of $\ce{HF}$ and of $\ce{H_2O}$ is extremely high when comparing to substances of nearly the same molar weight, like $\ce{CH_4}$ or $\ce{Ne}$. The boiling point of $\ce{H_2O}$ and $\ce{HF}$ does not depend significantly on the van der Wals forces. $\ce{H}$-bonds are much more important than van der Wals forces.
And there are practically no $\ce{H}$-bonds in ethyl fluoride, because $\ce{H}$-bonds happens only  when an $\ce{H}$ atom is attached to a strongly electronegative atom.

The last sentence is of course completely and utterly wrong. Just look up ; but these are at least an order of magnitude weaker. Therefore they hardly play a role in the comparison of OP.

@ Martin. Please Martin ! Keep cool ! I perfectly know that some extremely weak H-bonds exist with C-H hydrogen bonds, and even between a lot of other X-H bonds. They can even be measured. But they are so weak that they do not have any visible effect on the boiling point.

Maurice, a technical note, adding a space between @ and the username like you did will not notify the person you're pinging. Just type @user and not @ user.


text:
Change in temperature if salt is dissolved in water with different initial temperatures
I will mix a salt with room-temperature water and 8 degree Celsius water. Then I will compare the heat of solution with each other.
Does the initial temperature of the water affect how much the heat of solution and change in temperature will be?

Welcome to our site!

I think the OP means to ask "Will the $\Delta T$ be different for a salt dissolved in water at two different temperatures?" which I think might be a helpful question.

You will find the answer of your question in the following article :
Nathalie Hubert, Roland Solimando, Alexandre Pere, Dissolution of NaCl at 25°C, 45°C, and 60°C. It was published in Thermodynamic Acta 294, No.2, Pages 157 - 161, July 1997, or under : https://doi.org/10.1016/S0040-6031(97)00058-0 But this article is protected by a copyright.


text:
Separating lanthanides using ion-exchange method
Following statements are on Separation of rare earth metals (lanthanides) via in-exchange method

The tripositive ions are strongly sorbed by cation exchange resin: $\ce{La^{3+}}$, the largest is most strongly held; $\ce{Lu^{3+}}$, the smallest is least strongly held. Elution of $\ce{HCl}$, therefore removes $\ce{Lu^{3+}}$ first and $\ce{La^{3+}}$ last.

Generally, the smallest cation forms the strongest bond, so shouldn't $\ce{Lu^{3+}}$ be most strongly held? The most common cation exchangers, such as $\ce{RSO3^-Na+}$, where $\ce{R}$ is a cross-linked matrix are hard bases, so shouldn't according to HSAB principle also, $\ce{Lu^{3+}}$ be most strongly held?
Follow-on excerpt,

The separation is however not good, particularly between the heavier lanthanides, but is markedly improved by the addition of complexing agents such as $\ce{EDTA}$, or ammonium citrate/citric acid buffer to the eluting solution. The citrate ion complexes most strongly with the smaller elements, i.e. with the heavier elements.

Now, the two compounds (cation-exchanger and complex) helds strongly  different elements, one smaller in size and other larger in size. How will presence of two enhance separating property? Also here ammonium phosphate is mono-, di- or tribasic?
Following graphs are also given

Please explain the meaning and significance of the graphs.

What do you want us to do, exactly ? We are not used to solve all your problems without you having at least tried to answer. We may help you. But we will not all the job, and you nothing. So try to solve some part of your questions. We will help you

@Zenix Yes there are ion exchange resins with sulfonate groups to soften water.  This is plausible to interact with hard cations (HSAB).  An ion exchange resin may be altered, though, for an exchange of any combination of (hard / soft) (metal cation, small anion) may use other groups (e.g., carboxylic acids, amino acids, alkylated ammonium groups) instead to purify water; solution of salts, sugars, proteins, etc. pp. See a list of resins available in bulk (e.g. by Rohm & Haas, https://www.lenntech.com/products/resins/rohm-haas/rohm-haas-ion-exchange-resins.htm), which is a selection of a few.

@Maurice these aren't my homework problem, my queries that came when I was reading about separation of lanthanides using ion exchange resin. If I could figure them out, why would've I asked you? It would we gr8, if u could answer any part..

You must be reading this from really old ion exchange references. It is always good to mention the resources and references for other readers.
Ion-exchange retention can be very roughly approximated with Coulomb's law, which says that the force of interaction between two "ions" is directly proportional. The force of interaction is inversely proportional to the square of the hydrated radius of the ions. Here the word interaction means attraction or repulsion.
So you have to look up the hydrated radii of the lanthanides and try to see a trend on a simple sulfonated cation exchanger.
You second point is a misconception. The "trick" to add complexing agents works when you wish to alter the sign of the species you wish to separate. For example Cu(II) has a +2 charge, if I complex it with EDTA, the [Cu(EDTA)] has a charge of -2 Think about the Coulomb's law now. Can we use a cation exchanger now? This time we must try an anion exchanger.
One can still use a cation exchangers, in that case the complexed metallic anions will be repelled out and those ions which did not complex well will stay on the resin.

This was shared by my chemistry (school) teacher, (lockdown due to COVID-19) can't say if he took it from really old reference. Also I didn't get the graphs, please explain them too. What about the citrates? If we take ammonium citrate, polarity (sign of charge) on complex won't change..

I cannot explain anything from the graphs without seeing the whole figure captions. Post full pictures and all the context.

Caption is just "Elution of tripositive lanthanide ions". It was given at the end, without any reference :(

Sorry, then nobody can say anything. You will have to search. Is it an academic assignment or a practical problem? Please provide the context.

No, these are chapter notes sent by our teacher, since all schools are closed (due to COVID-19). Lol, I guess, our school ain't rich enough to buy lanthanides.

Okay, makes sense now. I think the teacher just needs you to learn the concept of ion-exchange process rather than going into the nitty-gritty details. Just keep the facts I told you in your mind. You would probably need a semester of PhD level separations course to go the depth.


text:
Why does a low concentration of soap in water behave as a strong electrolytes?
According to my text, "a colloid in which the dispersed phase consists of micelles is called an association colloid." An example would be soap in water. They use the term 'associated colloid' and 'association colloid' interchangeably.
They state that, "associated colloids at low concentrations, behave as normal, strong electrolytes."
I can't provide further context as they haven't provided further information.

Could you define associated colloids as it appears in your text? This is not a modern colloid chemistry term. A quick search of Google Books shows that this terminology is popular in Indian test preparation books. Micelle is a modern term. IUPAC defines micelle as "Particle of colloidal dimensions that exists in equilibrium with the molecules or ions in solution from which it is formed" Above a critical concentration, called the critical micellar concentration (CMC), the conductivity of solution drops very quickly. I will let you read more about micelle and conductivity of micellar solutions from the web.
You can watch this video for starting purposes: https://www.youtube.com/watch?v=M6LS4XZswNg. Basically micelle is formed by molecules which have a polar (ionizable end) and a long alkyl chain. Just like any electrolyte they conduct electricity in water. However above CMC, these molecules rather suddenly form a structure called a micelle as shown below (from Google Images). For these reasons, there is sudden change in lot of properties such as surface tension, conductivity and so on. It is not only conductivity.

Here is a reference  https://www.journal.csj.jp/doi/abs/10.1246/bcsj.67.2057 that interestingly notes, to quote: "The ionic conductivity of micelle was found to be inconsistent not only with the micellar size of the identical charge but also with the conductivity of the dynamic radius of micelle of the identical charge. The inconsistencies could be solved by the hopping reaction of counterions between micelles through the overlapping of ionic diffuse double layers around micelle that takes place when two micelles come close together."

My text doesn't provide much details and googling didn't help as it was clustered by Indian websites repeating my text. But I edited my question to clarify what my text said.

@ArnabChowdhury, Watch the lecture, with that you will understand the background.


text:
Can I use an acid other than sulfuric for hydrogen peroxide determination?
I would like to perform a permanganometric titration of hydrogen peroxide at home following a procedure similar to http://www.titrations.info/permanganate-titration-hydrogen-peroxide, i.e. the reaction:
2MnO4- + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O
I do not have sulfuric acid or a licence. Can I use any other acids? I think HCl would release chlorine, and other than that I only have weak household acids such as acetic, ascorbic, citric and tartaric to hand. I can probably get hold of others, but sulfuric is regulated as an explosives precursor here.

If you use an organic acid, you will probably oxidize this acid with permanganate, with or without $\ce{H_2O_2}$.
If you are not allowed to use sulfuric acid, why don't you use phosphoric acid $\ce{H_3PO_4}$?

Thanks. That's exactly the sort of thing I'm looking for. Would there be any extra interactions between phosphoric acid and the H2O2?

No, phosphoric acid is notmoxidisable as P is at its highest oxidation state

You could also use Nitric acid, although concentrated nitric acid is rather corrosive and produces horrible fumes. Plus, if I'd expect any acid to be labeled as an explosives precursor, It'd be nitric.

I would like to accept your comment as the answer, @Maurice.

Don't use the organics!
This will severely mess up your titration because both permanganates as well as H2O2 will oxidize the carboxylic acids into explosive or corrosive things like peroxyacids (-C(O)OOH). Carboxylic acids are also usually weak, and the peroxy versions can be up to 1000 times weaker.
On the bright side, HCl is a strong acid, and you shouldn't have to worry about the production of chlorine gas; the only chlorine product should be MnCl2. However, since so much O2 Is being produced, your MnCl2 might contain a small amount chlorate impurities, so be wary of that if you plan to use the manganese chloride in the future.

I was just wondering whether I should follow up asking whether my assumption about HCl was valid. So, it would be safe? I read it produces Cl2, but this then forms HCl and HOCl rather than being released. Sounds like this could also cause titration errors, but maybe small.

The major products of using the HCl should strictly be Manganese Chloride. However, as I stated, some things like chlorates will be produced, but I'd be very surprised if a measurable amount of chlorine gas formed.


text:
Does halogen addition occur without sunlight
Can this reaction occur;
(2-Bromopentane) + Br2 --(300°C)--> (2,3-Dibromopentane)
As far as I know, halogenation can only occur in UV or sunlight, not with just a high temperature. So how does this work out?
Sorry about the formatting.

Ever heard of radical initiators?

@HarryHolmes The given reaction though would exchange a hydrogen by a bromine, a substitution.  And because radicals are involved, it is a radicial substitution (e.g., https://en.wikipedia.org/wiki/Radical_substitution).

This reaction is not an addition. It is a substitution, and some HBr is simultaneously produced. So the written equation is wong.

There multiple means to generate the radicals you would need for a reaction of
$$\ce{\mbox{2-Bromopentane} + Br2 -> \mbox{2,3-Dibromopentane} + HBr}$$
An alternative to radiation is the combination of heat and some of radical initiator which is typically deployed at gentle $70-90\,\pu{^\circ{}C}$ like AIBN or dibenzoyl peroxide. To moderate the radical reaction, though, elementary bromine may be replaced by NBS, which lowers the accessible concentration of elementary bromine in the reaction mixture.

I think a very interesting way to perform this reaction, absence light or elevated temperature, especially since Br2 is a liquid at room temperature, is via sonolysis of either bromine or an alkyl bromide:
$\ce{Br2 + Energy (sonolysis) -> {*}[Br2] -> .Br + .Br}$
Here is a description of the mechanics from comments in ScienceDirect on sonolysis (sonochemistry) again based on a radical chain mechanism:

Apparently, the mechanisms involved in sonolysis are based on the same radical chain reaction as thermolysis. As the vapor pressure of the solvent decreases, the crushing effect of cavitation (cavitation energy) collapse of a bubbles increases. Although the reaction rate generally rises with higher temperatures, in sonochemistry an increase in temperature does not always contribute to reaction acceleration. Since the vapor pressure of a liquid goes up with a rise in the ambient temperature, cavitation occurs more readily but the cavitation collapse energy decreases. Therefore, depending on the positive and negative effects of the physical properties of the liquid undergoing sonication, such as the vapor pressure and viscosity, there are cases where the effects of ultrasound on a reaction can be somewhat more pronounced as the bulk temperature is lowered. Therefore, one should be mindful that such an apparent anti-Arrhenius effect may happen in sonochemistry [5].
The extent of a solute, reactant, or solvent molecule being incorporated within a cavitation bubble is dependent on the amount in the system, and under equilibrium conditions is controlled by Henry's and Raoult's laws. In general, the higher the vapor pressure of the compound and its solution composition, the higher will be its relative amount within a cavitation bubble. There are of course exceptions to this due to the nonideal effects of mixing, and the specific effects of adsorption of a solute to the cavitation bubble/solution interface. If a reactant cannot enter a cavitation bubble to any significant extent, then its only means of being activated is at the interfacial region of the bubble and/or in the bulk liquid. It has been shown that the activated species of a reaction was produced by ultrasonic cavitation in the study on the participation of various alkyl bromides in the isomerization of 1,2-dichloroethene Eqn (7.6) [6]. The isomerization process was initiated by bromine radicals, produced from the sonolysis of alkyl bromide.

Um, that seems a bit above my level. It's just, is sunlight absolutely necessary?

This will also breaks $\ce{R-Br}$ bond and major product may be a alkene. See that equation (7.6).

@HarryHolmes - no sunlight is not necessary if another source of radicals is available

I would compare this method to photolysis, where you have to know when to stop (for good yield) and some limited temperature adjustment during the treatment may be advantageous (as is the case in photolysis),


text:
Meaning of chemical energy
I am trying to understand the meaning of chemical energy.
I saw in my book that chemical energy is stored in the bonds of chemical compounds and when the bonds are untied the chemical energy increases.
I don't understand, shouldn't the chemical energy decreases? I think this because the chemical energy is not stored anymore (because there are no bonds)

This page describes the chemical energy level.

https://chemistry.stackexchange.com/questions/15250/how-do-breaking-chemical-bonds-turn-into-kinetic-energy

@EdV Thanks, I am wondering: could the chemical energy of two atoms considered as the energy that is needed to unhook two atoms?

This is binding energy of molecule. All this "chemical energy" is energy from reactions leading to compounds with lower enthalpy.

@Mithoron isn't the chemical energy the negative value of the binding energy of a molecule? It would make sense to me. Do you have other sources to suggest that help me understand better? thanks

Formally there's no such thing as "chemical energy". It's just a vague term for getting energy from reaction (or storing it).

@Mithoron Thanks, but why the chemical energy increases when there are no bonds?

https://chemistry.stackexchange.com/questions/63908/why-is-bond-breaking-endothermic https://chemistry.stackexchange.com/questions/13536/is-bond-formation-strictly-exothermic

@Mithoron thanks for your patience, so is chemical energy the potential of a chemical substance to undergo a chemical reaction to transform into other substances? right?

As @Mithoron said, there is no “chemical energy”. What matters are energy changes in chemical reactions. My PTFE beakers are chemically inert to most common reactions, but they would certainly react with molten Na, K, NaK, etc. Powdered and mixed with powdered Mg, they would serve as the oxidizing agent in the highly exothermic combustion of the Mg. What matters is the reactions potential reactants might engage in.

It is always a bit semantic. Besides the comments above, you are correct as well. Energy can be seen as stored in a bound system if it breaks and rearrange into something at lower energy. Energy would be released. If one, instead, break a system, per se, energy is then stored in the fragments, as small as atoms, and become difficult to be pictorially described. This is to say that such sentences are very clear once one has already a clear picture. Otherwise, they can be confusing. The fact that you are asking means you are getting a clear picture.


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Cyclopropane instability to addition reactions
In a question I saw comparing the reactivity (to addition reactions) of cyclopropane, cyclobutane and cyclopentane, it was said that cyclopropane was most prone to addition reactions.
Ring strain does make it prone to reactions, but how can addition reactions make it any more stable?
In that context, how can you compare the three?

A cycloalkane ring has just one degree of unsaturation. An addition reaction would remove that unsaturation, to give a saturated product. A saturated product can never be a ring.

You're absolutely right: cyclopropane's maximum ring strain makes it highly prone to addition reactions. Keep the following definition in mind:

An addition reaction, in organic chemistry, is in its simplest terms an organic reaction where two or more molecules combine to form a larger one (the adduct).

In your case:

Thus, it breaks down the ring into a straight chain, which makes the low stability of the ring a clear contributing factor to the reaction.
References: Cyclopropane, wikipedia||
Addition reactions, wikipedia

So there aren't any cases where addition occurs without the ring being broken, right?

@Harry Holmes Yes, either a ring or a double bond would need to be broken.


text:
Do esters/carboxylic acids/acid anhydride, etc. show Wolff Kishner reaction or not?
Do carboxylic acid and it's derivatives such as esters, amides, acid halides and anhydrides show Wolff-Kishner reductions and why?

Wolf-Kishner reduction of ketones and aldehydes proceeds by nucleophilic attack of hydrazine on the carbonyl compound followed by elimination of water to form the intermediate hydrazone mechanism here
Nucleophilic attack of hydrazine on esters, acyl halides and anhydrides gives the acyl hydrazide which is a stable species. See the second step of the reaction scheme below. If it does get deprotonated under W-K reaction conditions the anion is stable (note the NH next to the carbonyl group is the more acidic).

Image from Beck et.al..[1]
Nucleophilic acid of hydrazine on carboxylic acids does not occur, instead the hydrazine salt of the acid is formed which is stable. If this salt is then exposed to strong base, free hydrazine is formed and the salt of acid.
References:

Beck, P.; Santos, J. M. d.; Kuhn, B. L.; Moreira, D. N.; Flores, A. F. C.; Martins, M. A. P.; D'oca, M. G. M.; Piovesan, L. A. Regiospecific synthesis of new fatty N-acyl trihalomethylated pyrazoline derivatives from fatty acid methyl esters (FAMEs). J. Braz. Chem. Soc. 2012, 23 (11), 2122–2127. DOI: 10.1590/S0103-50532012005000071.


text:
Can someone help me find the limting reagent?
Okay, so for home work i was given this problem.
Suppose that 8.00g of CH4 reacts with 16.00g O2 to form CO2 and H2O, find the grams of CH4, H2o, and O2 left over
Balanced Equation : CH4 + 2O2 = CO2 + 2H2O
My Calculations for limiting Reagent:
8.00g CH4(1 mol CH4/16.04g CH4)(2 mol H2O/1 mol CH4)(18.01g/1 mol H2O) = 18.0g H2O
16.00g O2(1 mol O2/32.00g O2)(2 mol H2O/2 mol O2)(18.01g H2O/1 mol H2O) = 18.0g H20
So, if they both equal the same amount of H2O created, which one would be considered my liming reagent?

From your balanced reaction, you see that one mole of methane reacts with two moles of oxygen, so 16 g methane with 64 g oxygen. Therefore, 8 g methane requires 32 g oxygen, but you only have 16 g oxygen.

8g of methane is 0,5 mol.
16g of O2 is 0.5 mol. There are the same number of moles of both, however you need two times as many moles of O2 as you do of CH4 for the reaction (2:1 mole ratio). Which is the limiting reagent?

Please rewrite this expression : 16.00g O2(1 mol O2/32.00g O2)(2 mol H2O/2 mol O2)(18.01g H2O/1 mol H2O) = 18.0g H20. And change the * 2 * by * 1 *

Limiting reagent is the reactant used which produces lesser amount of the product as compared to other reactants if other reactants are used in sufficient amounts. Thus it limits the quantity of the product.
Now, according to your question,
Moles of CH4
= (Mass of CH4 taken)÷(Molecular mass of CH4)
= 8/[12+(4×1)] = 8/16
= 0.5 moles of CH4 are present
Also,
Moles of O2
= (Mass of O2 taken)÷(Molecular mass of O2)
= 16/(16×2) = 16/32
= 0.5 moles of O2 are present
The reaction given is
CH4 + 2O2 => CO2 + 2H2O
Thus,
For 1 mole of CH4
=> Produces 1 mole of CO2 and 2 moles of H2O
This means that
For 0.5 mole of CH4
=> Produces 0.5 mole of CO2 and 1 mole of H20
For 2 moles of O2
=> Produces 1 mole of CO2 and 2 moles of H2O
This means that
For 1 mole of O2
=> Produces 0.5 mole of CO2 and 1 mole of H2O
This means that
For 0.5 mole of O2
=> Produces 0.25 mole of CO2 and 0.5 moles of H2O
Since, O2 produces the lesser amount of products, thus, it is the Limiting Reagent.
Therefore, CH4 will be left as it is the reagent in excess.
Thus, with every 2 moles of O2
1 mole of CH4 is used.
This means that
With every 1 mole of O2
0.5 mole of CH4 are used
With every 0.5 moles of O2
0.25 moles of CH4 are used.

This means that in the reaction, quantity of

CH4 = (0.5–0.25) moles = 0.25 moles
O2 = 0 mole
CO2 = 0.25 moles
H2O = 0.5 moles

Therefore mass of a compound
= (Moles of compound that are present) ÷ (Molecular mass of that compound)

**Mass of

CH4 = 0.25 × 16 = 4 grams
O2 = 0 × 32 = 0 gram
CO2 = 0.25 × 44 = 11 grams
H2O = 0.5 × 18 = 9 grams
  **

Sorry if there's any mistake.
Thank you.

Your final answers were correct. But you have gone awfully long way to prove it.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, check the correctness of the unit symbols and avoid using "textual" formulas. Note that it is illiterate to replace a physical quantity with its units: the amount of substance is the proper term, not moles of X. And I agree with @MathewMahindaratne: the answer is unnecessarily over-complicated.

@MathewMahindaratne, Well, I am a student, although I can do it in no more than a minute but I just wanted the answer to be clear to the one who has asked question, maybe that is why you feel that it was long. You may know the topic elaborately but it may not be the same for the person asking the question. That is why I wrote it in this way. But as I have read about you, you are a chemist and i am a great fan of people who do research (because I too want to do it in future), so I am very sorry that it wasn't according to you, but I'll take your words as advice that I'll follow. Thank you.

@Sanya: Don't get too personal. I already up-voted your answer. My comment is for future perspectives. In future, avoid repeating same argument. :-)

@MathewMahindaratne:It's okay, I didn't took it in a personal way.


text:
Effect of optical isomerism on colour of transition metal complexes
I know that different stereoisomers of a given transition metal complex (cis- and trans- forms) have different colours. But do different optical isomers (non-superimposable mirror images) have different colours, or do both optical isomers have the same colour?

@EdV Do you mean the splitting of the 3d subshell that causes the colour? If so, are you saying that both enantiomers would have the same splitting and so same colour?

If you ask about enantiomers then there's no effect and color is identical.


text:
Why is phosphorus trifluoride the strongest Lewis acid among the phosphorus trihalides?
I had read that for the trihalides of boron, boron trifluoride is the weakest Lewis acid due to backbonding. The order I had seen was -
$\ce{BI3 > BBr3 > BCl3 > BF3}\$
$$\ce{}$
$- order of acidic strength
Why is the same not true for phosphorus trihalides as backbonding exists in them too?

related https://chemistry.stackexchange.com/questions/41556/why-cant-phosphorus-trifluoride-be-used-as-a-lewis-acid-catalyst-in-electrophil

In phosphorus trihalides backbonding effect is not as big as boron trihalides. Firstly boron trihalides are trigonal planar and have an empty-and-small p orbital (violates the octet rule), so it is favourable that the p orbitals of the boron and the halogen atom will overlap; meanwhile phosphorus trihalides are trigonal pyramidal which is less favourable for orbital overlap, and the phosphorus atom already has a complete octet which includes a lone pair. Lone pair-lone pair repulsion would make backbonding even less favourable.
The Lewis acidity should be largest for PF3, since the electronegative fluorine atoms would reduce electron density of the phosphorus atom and make it more hungry for electrons. Moreover the fluorine's small size allows more space for extra bonding.
Hope that helps and forgive me for grammar and spelling errors - I am not a native English speaker.


text:
Entropy Change In Rusting of Iron
I presume that the entropy of the system should decrease during rusting of iron because the atmospheric oxygen gets converted to solid state from gaseous state which should result in the decrease of randomness. So am I correct?

You know very well that you are wrong. Homework: Where do you suspect the entropy goes?

It is necessary to differentiate between the change in system entropy and surrounding entropy.
Rusting involves a decrease in system entropy, but causes an even greater increase in the entropy of the surroundings. This is because it is an exothermic reaction, releasing heat energy to the surrounding molecules, increasing their disorder.
Overall, when you calculate total entropy (system entropy + surroundings entropy) you find an increase. Hence, the process in spontaneous.


text:
Difference between chemical energy and potential energy
I know that chemical energy is the potential energy stored in the arrangement of atoms within molecules. Forming new chemical bonds releases energy.
what does potential energy mean?
Is it the energy released when new chemical bonds are formed?
Or is it the needed energy to break the bonds of the compost?

Potential energy is the energy of attraction/repulsion between "things", they can be atoms/electrons etc.
While Potential energy can be used to calculate Free Energy, which, in your context is the WORK required to form a molecule, Potential energy is not the same as Free energy.  I would start here: understand what Free Energy is, for instance the Gibbs Free energy.
While Potential energy is always used, it is the context in which it is used that makes it useful. On its own, with no guidance, it is not all that useful.

To further this, at T=0K the internal energy = electronic + ZPVE. Also, at T=0K, U=H=G, so from the electronic POTENTIAL energy and zero point vibrational energy, you can calculate the INTERNAL energy, ENTHALPY, and Gibbs Free energy. I do not offhand know what "energy" is.

@CharlieCrown Thanks Charlie, but I am a little bit confused, do you have any sources to suggest that could help me understand better the chemical energy

And so we return here.

Edoardo, I would just Wikipedia Gibbs free energy, and/ or Helmholtz free energy. My feeling is that your requirements are basic enough that Maurice answer will give a strong enough feeling of false security to power you through :)

@Charlie Crown. All my explanation is related to the production and absorption of heat as energy. Enthalpy is a potential energy, and the difference of energy is a certain amount of heat. It is possible to start again the same development with the free enthalpy, or Gibbs energy, and it will introduce a new sort of potential energy, the difference of which does not give the heat, but gives the work or electric energy. Enthalpy and free enthalpy are different concepts. Enthalpy does not change with the concentration. Free enthalpy does change with the concentration. Shall I explain this ?

There exist many different potential energies. Maybe you would prefer an analogy. For example let's consider an object on a table (a ball, a book or a piece of chalk). It has a potential energy which is not chemical for the moment. It is not visible. It is a hidden energy. It can only be shown if you give the object an opportunity to fall down and to go to a place where the potential energy is lower. During the fall, the potential energy is transformed into kinetics energy (its speed increases). When it touches the ground, the kinetics energy is transformed into heat which is another sort of energy. This gravitational potential energy is equal to mgH, where m is the mass of the falling object, H is the beginning height of the table, and g is a universal constant (9.81 m/s2 on Earth). Before the fall, the object has a sort of hidden energy, which we call gravitational potential energy (or mechanical energy), and this energy could only be expressed if you give the object an opportunity to go to a lower place. The object has a higher gravitational potential energy on the table than on the ground.
It is the same in chemistry. H2 and O2 have a similar hidden energy, which is a chemical potential energy. If you break the bonds H-H and O-O, you need energy to do it. You give energy to the atoms. It is similar to the operation of picking up the object on the ground and put it higher, on the table. If you imagine going back to H2 and O2, you would get the energy you had consumed for breaking the bonds. But now, if you prefer recombining the H and O atoms into H2O molecules, you will get much more energy (much more heat).
The object on the ground is similar to H2 and O2. When you give it some energy, you can lift it up to the table, or better to a state of separated H and O atoms. But now if the table is just at the edge of an abyss, you can get more energy by letting the object fall down the cliff. In this case you obtain H2O and not H2 and O2. The chemical potential has decreased to a very low level if the final state is H2O. The fall has produced much more heat than the recombination H2 + O2.
Chemists often use this analogy between height and chemical potential energy (which is usually called enthalpy). They draw short horizontal lines above one another to simulate  different states, as if one line represents the table (H and O atoms), and another one (lower) the level of the ground (H2 and O2). Of course the level of H2O is another horizontal line : much much lower.
Have you followed me ? Is it what you needed to understand the potential energy ?


text:
Heats of combustion of cycloalkane homologues
Heat of combustion increases as the number of carbons increases, but on comparing ethyl cyclopentane, ethyl cyclohexane, andethyl cycloheptane, shouldn't you consider ring stability too?
In that context, it should be 6-membered ring < 7-membered ring < 5-membered ring.
Why isn't this so?

Because an additional $\ce{CH2}$ group gives an extra mole of $\ce{CO2}$ on combustion, the energy released in this process trumps the effect of stability of the ring.

Have a look at this table from Organic Chemistry by Wade and Simek [1, p. 134]:

Every time a $\ce{-CH2}$ group is added, the heat of combustion is increased by almost 650 kJ/mol; whereas the maximum relief in ring strain attained by adding that $\ce{-CH2}$ group is not more than 25 kJ/mol.
So there is no competition between increase in heat of combustion due to more number of carbon atoms and decrease in it caused by ring stability. Every time you see two hydrocarbons with different numbers of carbon atoms, be assured that the one with more carbon will release more heat on combustion.
P.S. In case you are unfamiliar with how ring strain is quantitatively measured [1, p. 133]:

The molar heat of combustion of cyclohexane is nearly twice that of cyclopropane, simply because cyclohexane contains twice as many methylene ($\ce{-CH2}$) groups per mole. To compare the relative stabilities of cycloalkanes, we divide the heat of combustion by the number of methylene ($\ce{-CH2}$) groups. The result is the energy per $\ce{-CH2}$ group. These normalized energies allow us to compare the relative amounts of ring strain (per methylene group) in the cycloalkanes. The reference value of 658.6 kJ (157.4 kcal) per mole of $\ce{-CH2}$ groups comes from an unstrained long-chain alkane.

References

Wade, L. G.; Simek, J. W. Organic Chemistry, 9th ed.; Pearson: Glenview, IL, 2017. ISBN 978-0-321-97137-1.


text:
Equation for Galvanization and Electrolysis of Saltwater with Copper and Zinc
I am looking for the chemical equations for both the galvanization and electrolysis of saltwater utilizing copper and zinc. The closest I have gotten was from this site: Explanation for the reactions in a saltwater battery with zinc and copper electrodes
Although, that question and answer was exactly the format I was looking for and was impressive, the final equation was ultimately for galvanization in a two-tank bridged battery. I would like to see what the equation looks like for a single-tank with both copper and zinc sharing the same solution.
In addition, I would very much appreciate the reaction equation for the electrolysis version. If those answering could keep it to a similar format as the answer I linked above, it would be much appreciated.

If you dip a Copper plate and a Zinc plate into a salt solution, you create a galvanic cell, where Zinc is oxidized and is the negative pole. If you want to apply an external tension from another cell, the result depends on the choice of the positive and the negative poles.
If you connect positive poles to the negative poles, both cells produce courant, and no electrolysis occurs.
If you want to have the $\ce{Zn/Cu}$ cell work in electrolysis, you first must connect both positive and both negative poles together, and apply an external tension greater that that of the $\ce{Zn/Cu}$ cell.
If you apply a sufficiently high tension to overcome the tension produced by the $\ce{Zn/Cu}$ cell, you will produce an electrolysis with the following reactions:

at the negative pole (Zn), for lack of any reducible ion in solution, H from water is reduced according to :  $$\ce{H_2O  +  2 e-  ->  H2  +  2 OH^-}$$
at the positive pole (Cu), two oxidation reactions may happen simultaneously :      $$\ce{Cu-> Cu^{2+}} + 2 e-$$ with probably a side reaction :$$\ce{2H2O ->  O2 + H+ + 4 e-}$$The relative importances of these reactions depend on the concentration of $\ce{NaCl}$. The first must be the most important .

Thank you for the response, Maurice. Would you be able to include NaCl into the formulation? My primary goal is I'd like to see the result in equation form with all ingredients involved (Zn, Cu, H2O, NaCl, and there by products) first by galvanic reaction (the electrolyte providing current to something miscellaneous like a bulb) then taking the same exact solution/set up/by-product and then reversing it via electrolysis by introducing an opposite current (probably with a generator), in place of the bulb that was there during galvanization. Please let me know if this makes sense. Thank you!

I'd like to see as much of the details such as if it produces hypochlorite, CuCl, NaOH, etc. Thank you very much!


text:
looking for alternatives to lysol wipes (Alkyl C12-18 dimethyl benzyl ammonium chloride)
Lysol and other recommended disinfectants claiming virucidal properties have been basically impossible to find here for weeks. I did however, find a bottle of clorox all-purpose disinfecting cleaner with the same active ingredient as my lysol wipes (dimethyl benzyl ammonium chloride) with a .3% concentration as opposed to Lysol's .26%.
https://clorox.ca/products/sprays/clorox-all-purpose-disinfecting-cleaner-spray/
I've been doing as much research as a chemistry inept mechanical engineer can to find studies on the active ingredient and viruses, with mixed reviews, and am turning to you for help. I'm wondering if i follow similar instructions to lysol's (wet for 10 minutes before air drying) with this product, am I going to achieve the same effect, or is that not necessarily the case? Is the lack of indication and instructions on the packaging for virucidal disinfectant what's holding it back from making it onto the EPA and canadian equivalent lists, or could there be a legitimate reason they cant make that claim?
I know this is a long-shot, but any insight you might share would be greatly appreciated in these interesting times.

This is medical advice which is off-topic here. Consult your doctor instead.

I'd love to, the main problem is doctors tend to be fairly busy during global pandemics, and had you read the question, you'd know it was specific to finding more information about a specific compound, which i doubt my doctor knows all that much about... My doctor being a neurologist because I have MS which puts me in a high risk group, hence i'm spending so long to see if what i have might actually be useful. But sure I get it, you're a very important person who only has time to be negative rather than just ignore something.

I'm not trying to be negative. Your question was downvoted because you are a new contributor to this Q&A site and do not know (yet) what is considered off-topic. We don't like medical-related questions to be asked here, hence it was my duty as a member to flag it as medical advice. I understand you though.

From a purely standard chemistry perspective, the longer one allows a potentially slow reaction (attack of the outer membranes, for example) to occur, I would expect better results.
Photochemistry is also an option to consider, which may be catalytic with other treatments. In particular, explore safe natural light sun bulbs (I would select the higher output, again intensity matters) and any comments relative to benefits you seek.
There are also reputedly disinfecting air filters available (again check reagents and reviews), which is probably the best idea.

thanks for the response, very good ideas to consider.
I will also keep digging on the subject compound. I did find a handful of studies (and others i cannot access) with varying conclusions as to the effectiveness on enveloped viruses, the consensus being not as effective as you'd hope. it just boggles me, how then, these lysol wipes can make the 99.99% claim. perhaps a mechanical effect of the wipes in addition to the disinfectant, or a non-active ingredient catalyst? Just odd because i've also found some studies about how the wipes may be detrimental to the solution's effectiveness


text:
Concentration of gas above aqueous ammonia
I am a physicist who has run into some chemistry in the lab (and unfortunately, I've completely evaded any formal training in chemistry). I have exposed some samples (I think details of the sample are not important here, but would be happy to provide further information if necessary) to a solution of 30% ammonia and 70% water, and am trying to calculate approximately how much ammonia (in ppm, for example) reached my sample.
My samples are placed in an 3″×3″×1″ (L×W×H) container with the top exposed to the air. Until now, I've just been marking down the volume of the solution that I've placed (about 15 mm) directly below my samples in the container (e.g., 5 μL, 10 μL, …, 200 μL). I think it's good enough to assume room temperature and standard atmospheric pressure for my labs. I've looked into Henry's law and related relationships, but am not sure that these are the appropriate tools. I would greatly appreciate any help or guidance!
Maybe these tables will help (I've stumbled upon them trying to figure this out in the context of Henry's law):

https://www.tannerind.com/PDF/green-aqua-amm.pdf
https://www.atmos-chem-phys.net/15/4399/2015/acp-15-4399-2015.pdf

P.S.
If a closed container is a more tractable problem, I also collected some data with a closed container (same dimensions), and would very much appreciate insight into that problem too (if it is very different, or the only solvable problem). I should also mention that the solution was removed from the container in about 10 seconds in all cases (if that makes a difference).

"Ammonia has a very strong odor that is irritating and that you can smell when it is in the air at a level higher than 5 ppm" - did you smell it? Depending whether your nose or your sample is closer to the solution (and the geometry of the setup), your nose will experience a higher or lower concentration than your sample.

It's not doable even with closed container, unless you were real careful to have equilibrium vapor pressure in there.

It is a good question which relates to liquid-liquid extraction equilibria and head space GC. I would argue that the question should stay open, a person who knows the thermodynamics of liquid-liquid extraction or vapor / liquid systems should address it

If the container is open, as shown below, then you have a total mess.

Obviously eventually all the ammonia and water will evaporate into the room eventually. There is absolutely no way to tell how much ammonia gas would be with the volume of the container at any given time, and the ammonia concentration within the container wouldn't ever be homogeneous.
If the container is closed, then you have a somewhat more tractable problem.

However there are two equilibriums to consider. First the gas phase must be saturated with water vapor, and second the ammonia must be distributed between the gas and liquid phases (if any liquid phase remains).
Just out of curiosity, 3x3x1 inchs has a volume of 9 cubic inches, or about 147 mL. So if the air in the lab was at $\pu{20 ^\circ C}$ and completely dry then about $\pu{2.5 \mu L}$ of water would be needed to saturate the volume to 100% humidity.

Calculation:
  water vapor pressure at $\pu{20 ^\circ C} = \pu{17.5 torr}$
  atmospheric vapor pressure = 760 torr
  molar volume at $\pu{20 ^\circ C}$ and 760 torr = 24.1 liters
  Moles water = $\dfrac{17.5}{760}\cdot \dfrac{0.147}{24.1} = \pu{1.4\cdot10^{-4} moles}$
  water is $\pu{18 g/mole}$ and $\pu{1 g/ml} = \pu{0.001 g/\mu L}$
$ \pu{\mu L water} = \dfrac{\pu{1.4\cdot10^{-4} moles}\cdot \pu{18 g/mole}}{\pu{0.001 g/\mu L}} = \pu{2.5 \mu L} $

NOTE: If the container is sealed there is still the problem of estimating how long it takes the system to reach equilibrium.

At STP, 147 mL gas contain about 6 mmol particles. Only 2% or so of that are water (at 100% humidity), i.e. 0.12 mmol, or about 2 microliter. Huh? Did you start with vacuum, and the container is pure water vapour?

@KarstenTheis - Missed your comment. Thanks for the correction, which I've made. I've also shown the calculations, which I should have done before.

Careful, if you have samples that contain transition metals and/or their salts, in the presence of ammonia fumes, air (a source of oxygen), water (or even moisture) and an electrolyte (any soluble salts), you may have some electrochemistry afoot as well. In particular, for example, is the leaching of copper ores (removing the copper ions) with NH3/air with a spontaneous (after an inception period) electrochemical reaction mixed with some more standard chemistry.
See this source, as an example of some possible unexpected chemistry: "Kinetics and Mechanism of Copper Dissolution In Aqueous Ammonia", (source link).
Now, if the ammonia is being consumed in a reaction with air exposure, then equilibrium reactions relating to NH3 exposure are going to be invalid.


text:
Shielding effect of d electrons
In transition elements shielding effect is observed . Due to this zinc shows abnormality in atomic size.
But at the same time we also say that gallium has same similar size as  aluminium due to poor shielding effect of d electrons . Why is it that d electrons have good shielding effect in case of zinc but poor shielding effect in case of gallium?

Slater's rule may be used for quantitative reasoning

When we are talking about zinc... We have to think about period trends. So in periods along moving from left to right covalent radius decreases. But in case of transition metal..
Two effects works first is $Z_{eff}$. Another effect is electronic repulsion.At the starting the no of electron in $d-$ orbital are less so $Z_{eff}$ is dominant over electronic repulsion but as we are moving alosng d-orbital electronic effect become dominant factor over $Z_{eff}$, So in the middle redius become almost constant and at the end it increases.
When we are talking about gallium and Aluminium radius comparision thats mean we are talking about group trends ,So in group trend abnormal size is due to poor shielding effect and relativistic effect. Normal trend on going down the size increases from aluminium to gallium size decreases because here a poor shielding of 3-d orbital present so $Z_{eff}$ penetarate and valence shell contract so size decreases.

Why shouldn't electronic repulsions taken into account in case of gallium and aluminium size comparison?

We can compare electronic repulsion in case of zinc and gallium because as both have d10 orbital.  And it more clear when we see radius of gallium  is 187pm and radius of zinc is 139 pm.  But when we are talking  about Aluminum  there is no d orbital and galium has.  So that's can be a possible  explaination because  we don't  have d orbital  in case of aluminum  and have d orbital in case of galium so basically  we can't  compare it.  I can just mention Galium radius 187 pm and zinc radius 139 pm and Aluminum  radius 184 pm. So we can see difference  of radius length of zinc and gallium is to

Yes the values are correct they are as expected . But we know that there are electronic repulsions in gallium and  they aren't in aluminium . So I am asking that why can't gallium atom be bigger than aluminium because of the above reason .

I mean gallium should be far more bigger.

In atomic size [neutral atom] Ga fits perfectly in, to the right of Zn and below Al. When comparing ions Ga loses the s and p electrons so the outer electrons remaining are the 3d10 These do not shield each other so Ga+3 is rather small.


text:
Quantitative measurement of ring strain
While looking for data about ring strain in cycloalkanes, I came across two types of data:

Ring strain per methylene group (let's denote this quantity by X): The difference between heat of combustion of the cycloalkane per methylene group and the reference value of $\pu{658.6 kJ/mol}$.
Total ring strain (let's denote this quantity by Y): The difference between molar heat of combustion of the cycloalkane and the reference value of $\pu{658.6 kJ/mol}$ multiplied by number of carbon atoms in the cycloalkane.

From what I have read till now, it appears that the first quantity is the better representation of ring strain. (I say this on the basis that Y has the same value for cyclopentane as well as cycloheptane but still cycloheptane is considered to be less strained as it has lower value of X).
I want to know why X is a better measure for ring strain than Y. When we talk about a ring, we consider it as a whole; then why do we calculate ring strain per methylene group?

Take a look at this explanation: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/cycloalkanes/cycloalkanes.html

I read it, and I failed to understand why it talked about cycloheptadecane at the end. What was the conclusion of it?

The present answer has given a nice reasoning, but it's more about how the value of X confirms with the experimental findings. If anyone could provide a theoretical reason (using thermochemistry maybe) as to why X is a more suitable measure, it would be really helpful.

Let's look at following table from a Yale website:
$$
\begin{array}{c|ccc}
 & \ce{(CH2)3} & \ce{(CH2)4} & \ce{(CH2)5} & \ce{(CH2)6} & \ce{(CH2)7} \\
\hline
\Delta H_\mathrm{Combusion} \ (\pu{kcal/mol}) & -499.8 & -656 & -793.5 & -944.6 & -1108.3  \\
\Delta H_\mathrm{Combusion} \text{ per } \ce{-CH2}-  \ (\pu{kcal/mol}) & -166.6 & -164 & -158.7 & -157.4 & -158.3 \\
\text{Ring strain per } \ce{-CH2}-  \ (\pu{kcal/mol}) & 9.2 & 6.6 & 1.3 & 0 & 0.9 \\
\text{Total ring strain}  \ (\pu{kcal/mol}) & 27.6 & 26.3 & 6.5 & 0 & 6.3 \\
\hline
\end{array}
$$
Now, just concentrate on ring strain values for cyclopropane and cyclobutane only:

Ring strain per $\ce{-CH2}-$ (X-values) are $9.2$ and $\pu{6.6 kcal/mol}$, respectively.
Total ring strains (Y-values) are $27.6$ and $\pu{26.3 kcal/mol}$, respectively.

According to Y-values, cyclopropane and cyclobutane are equally unstable (relatively speaking), while according to Y-values, cyclopropane is significantly less stable than cyclobutane. Which valuse should be more reliable? Let's look at experimental findings:

Although cyclopropanes are far less reactive than alkenes, they react with chlorine and bromine under polar conditions to form corresponding addition products, 1,3-dihalopropanes, as the predominant product (Ref.1). To my understanding, cyclobutane and higher cycloalkanes do not give this ring opening reaction.
Cyclopropane also reacts with HCl and HBr to form corresponding addition products, 1-halopropanes (Ref.2; Reactions of bromocyclopropane with hydrobromic acid has given mixture of 1,1-, 1,2- and 1,3-dibromopropanes). Again, to my understanding, cyclobutane and higher cycloalkanes do not give this ring opening reaction.

So which one you think should be more reliable and why?
References:

Joseph B. Lambert, Erik C. Chelius, William J. Schulz, Jr., Nancy E. Carpenter, “Polar bromination and chlorination of cyclopropane,”  J. Am. Chem. Soc. 1990, 112(8), 3156–3162. (https://doi.org/10.1021/ja00164a043).
Choi Chuck Lee, Bo-Sup Hahn, Kwok-Ming Wan, D. J. Woodcock, “Reactions of cyclopropane with hydrochloric acid and of bromocyclopropane with hydrobromic acid,” J. Org. Chem. 1969, 34(10), 3210–3211. (https://doi.org/10.1021/jo01262a097).

I get this reasoning. Can you also please address this (maybe silly) question: why does a larger cycloalkane show greater total ring strain than another smaller cycloalkane having similar value of X (for example, cyclooctane and cyclopentane)? Does the quantity 'total ring strain' hold any significance?

The flexibility in the ring release some strain energy. This can be explained by the values of higher ring sizes, some of which gives even negative values for ring strain (See the Yale website given in my answer).

Umm... My question was about increased total ring strain in larger cycloalkanes. It seems you have explained why large cycloalkanes have lower ring strain.


text:
Are centre of symmetry(COS) and alternate axis of symmetry(AAOS) dependent on each other?
COS=Centre of symmetery
AAOS=Alternate axis of symmetry
Lets suppose a molecule has a COS. Does it need to have AAOS?
What about the converse? If a molecule has AAOS does it necessarily have a COS?
With some research and help I can see that having AAOS doesn't mean it has COS. Still problem with the 1st question does COS imply AAOS?

What is your reasoning, e.g. after passing both tutorial and the interactive gallery of Symmetry @ Otterbein, http://faculty.otterbein.edu/djohnston/sym/index.html?  Selecting 18-crown-6 in the section $S_n$ (http://faculty.otterbein.edu/djohnston/sym/gallery/index.html), for example, allows you to (pan, rotate, zoom) the molecules, to toggle on/off the display the symmetry elements, to launch their operation -- and if the sequence is too fast, to use a slider to see the intermediate (marked in yellow) at any stage between $E$ and the operation applied.

@Buttonwood thanks man that helped a lot... Now I understand the difference they cause... Visualising without any add is difficult.