text:
Gold Plating on Glass?
I am considering labware options to handle NaOH solutions at ~150 °C. One appealing idea would be to coat my current glassware with gold. I've read different sources mentioning the possibility without further detail.
The most detailed description of a process for this is INSTRUCTIONS FOR GOLD PLATING ON GLASS by Peacock Laboratories, Inc who sells (or used to sell) products for doing so. Unfortunately, this says little as to the chemical mechanisms involved.
Would anyone guess what chemicals are being used in this process and know how to explain the mechanisms at hand, and perhaps describe properties to expect from the result?
Can I expect impermeability from the coating or at least a significantly reduced glass surface exposure to hydroxides? What is the best resistance to expect vs routine cleaning of glassware that would be coated with such a method? Can I expect it to not dissociate upon heat shocks that borosilicate glass can handle? Would there be any chemicals to avoid with such a coating, other than the few that would be expected to react with / dissolve gold metal?

I checked that well known auction site and found lots of titanium cups and such available. Apparently, campers buy titanium bowls, cups, etc. and the cost is lower than I expected. Maybe worthwhile to buy a titanium cup and check it out.

@EdV Thanks for the tip! Using titanium recipients is a possibility I didn't think about. I may have read somewhere that some titanium compounds are amphoteric though. Do you know if NaOH could dissolve traces of it from the cups? The prospect of being able to use my current glassware is appealing because the only work would be the coating part. But if titanium doesn't react I could try to recreate that equipment out of it (I really only need a cup to bring NaOH to high temperature and then a filtration apparatus). Another advantage of titanium would be not having to worry about thermal shocks

@EdV I'm not sure what this translates into in the context of high temperature NaOH for a few dozen minutes, but I found the following statement in a paper:

"Titanium is generally very resistant alkaline media including solutions of sodium hydroxide, potassium hydroxide. calcium hydroxide and ammonium hydroxide. In concentrations of up to approximately 70c for example, titanium exhibits corrosic rates of less than or equal to 5 mpy [(0.12 mm/yr) Table 18.] Near nil corrosion rate are exhibited in boiling calcium hydroxide, magnesium hydroxide, and ammonium hydroxide solutions up to saturation. ...

...
Despite low corrosion rates in alkali solutions, hydrogen pickup and possible embrittlement of titanium can occur at temperatures above 170°F (77°C) when solution pH is greater than or equal to 12" Source: https://www.nrc.gov/docs/ML9932/ML993210187.pdf

I would try a titanium cup, in any event. Also, I just checked and that same auction site lists lots of graphite and SiC crucibles, for not much money. No problem with the temperature, but only you can decide about your NaOH purity requirements.

@EdV Could SiC react with NaOH at high temperature in the presence of O2 (producing sodium carbonate and a sodium silicate)? As for my purity requirements: Should I contaminate the NaOH I would much rather have gold than titanium inside (though I'd also rather have 1-3ppm of titanium and <0.3 ppm of many other elements).

I have no idea if SiC reacts at high temperature with NaOH in the presence of oxygen. But the larger issue is this: you would prefer to have low (or sub) ppm contamination and that means analytical testing somewhere down the road. As an analytical chemist, I can tell you this is where it can start to get expensive: nobody does pro bono trace analysis and nobody wants to possibly contaminate their expensive hyphenated instrument with unsuspected contaminants in a sample from the wild. So, good luck with whatever you decide to do!

How about Sterling silver (92.5% Ag, 7.5% Cu) cups? Lots available fairly inexpensive at that auction site. You could buy a small cup (“child’s cup”) and test it very easily. As well, pure silver is very easily purchased, thanks to all the silver bullion collectors, and could be tested with the NaOH.

@EdV It might be the way to go, though I'm less inspired by copper and the other stuff in there than by silver itself (because the compounds of silver have quite consistently very low solubility). I could get pure silver but then I'd have to get into the absolute mess of casting what I need out of it (though even if I buy sterling silver I'll have to make a filtration apparatus out of it, and most probably plate it with gold to be on the safe side).

Maybe sterling silver > pure silver plating > gold plating. I know that silver slowly diffuses into gold but I don't know if copper and small amounts of heavy metals from the sterling silver could diffuse through the plated pure silver layer.

I am considering labware options to handle NaOH solutions at ~150°C. || AFAIK, Iron is stable in molten NaOH.

Try gold fuming maybe? could take quite a bit to get a consistent layer, it's commonly used in the artistic side of THC consumption but may be worth exploring. good luck!

@Stoner_Scientist. Thanks for the suggestion! I'll look into it!

Gold plating glass could probably be 99.99% complete. It would make a nice mirror and would probably stay unoxidized for a long time.
However, if I wanted to remove the gold, I think I would rinse the glass with HF solution, or perhaps NaOH solution, either of which would find those teeny tiny gaps and undercut the whole gold plate.
I don't think there would be any strong chemical bonding, and no mechanical strength to a gold film. Glass is just a reactive substrate, and protecting it would require a finite thickness, with mechanical strength.
An example of plating not working well is chromium plating (although it can be done very well, it requires much attention to detail). Automobile bumpers used to be chrome plated, but developed rust spots after a few years because of porosity in the chromium film. I believe nickel plating to reduce corrosion tendency (but Ni is not shiny enough) followed by chrome plating works reasonably well. But then there is hexavalent chromium toxicity... So now we just paint the bumpers or make them out of plastic.
Perhaps you could coat the glass with teflon or polypropylene.

I see that polypropylene melts at 160C, and Teflon is only good up to 260C, so I don't think either of those would work for the application described...?

Your second sentence pretty much says it all. And stirring anything in a gold-plated beaker or flask, especially something moderately viscous, would be problem.

@JamesGaidis: Thank you for your answer! I know that is some cases gold compounds (oxides?) can act amphoterically but I'm not sure it would apply here, so if I understand right, the way NaOH would remove the gold layer would be by gradually dissolving the glass underneath it, not interacting with the gold itself in any way? Could I prevent this by coating a thicker layer and somehow try to make it more uniform, maybe by baking the glassware between coating "passes"?

@EdV: Would stirring be a problem because of the friction with the coating layer?

Pure gold is very soft and it is the most malleable metal, so it would be very easy to scratch, which would then expose the glass substrate. Therefore, stirring becomes a bit problematic: no stir bars, no boiling chips, no glass rod and so on. I think the gold layer would have to be unreasonably thick in order to be effective. But a rest piece, e.g., a gold coated microscope slide, or test tube, could be tried.

@EdV: The rest piece is a very smart idea. All I'd need to use (which was prospectively to be in contact with the to-be-coated glass walls) is a very large magnetic stirring rod out of PTFE which I think is softer than metallic gold or a smaller magnetic rod out of glass (which I'd then also coat). It would be smart to have distinctive "wear" pieces where friction is needed (e.g. the magnetic glass rod and a resting plate/disk out of glass), as I could re-coat them separately and give them thicker coating.

The glass rod and resting disk would wear out on each other but the vessels and filtering apparatus which don't require being subjected to much friction would stay unaffected. If PTFE can be used, and if it is indeed softer than the gold coating, then the combination of a soft material vs slightly less soft one might be preferable to having friction between two components of the same material and with the same coating. I'm not exactly sure what that implies but PTFE has a glass temperature at 114.85 °C though, which may affect the suitability of the stirring rod.

As for the resting plate for the stirring rod, I could even stretch a metallic gold piece to fit that purpose, which would leave the stirring rod as the only "wear" part to be re-coated each time (assuming the Au-coated glass magnetic rod is the best option).

Two things: 1) "rest" was actually a typo for "test". Sorry! 2) If a PTFE stirring bar is OK at the specified temperature and putative NaOH concentration, why not use PTFE beakers? I currently have 8 brand new Teflon® beakers left from the dozen I purchased in 1987. Send me an e-mail (you can find my e-mail address pretty easily) and I will send you a beaker, gratis. Then test it! Like Yogi Berra might have said: "In theory, theory and practice are the same. In practice, they aren't."

One more thing: I have some titanium disks, about 2 inch diameter, so I will be happy to send along a few, if you want to test how they hold up to the NaOH. Just let me know.

@EdV That's a very generous offer, and I thank you for it! If possible I would like to avoid abusing your generosity, so I'll first try to determine if that is the best option. I've been thinking for instance about silver as an option also. I think it's amphoteric like gold and titanium in some circumstances, but I'm not sure if it means NaOH could attack the metallic form. The advantage over teflon would be much higher heat resistance, and the advantage over titanium and pyrex is that it forms much less soluble salts which are easier to get rid of later in the process.

I suspect that even if silver does react with NaOH, it could do so much slower than glass/pyrex, which would also make for a more resistant Au gold layer on it compared to glass (if I understood correctly that NaOH would remove the gold layer by reacting not with gold directly but with the glass underneath). Ag diffuses in Au very slowly but if it doesn't react much with NaOH that's fine. Now the bad side is I'd have to work with silver metal and build a filtering apparatus out of it (but probably much easier to work with than titanium)

@Hans - Just a safety related concern with James' answer - The recommendation of using HF to remove the gold if needed is, in my opinion, excessive and a generally unsafe recommendation to make. I would urge y'all to consider using Aqua Regia to remove the gold directly. I only recommend this as HF is scary - like disolves your bones when inhaled kind of scary. Stay safe out there, folks.

Aqua regia is great for gold - HF was just an acid to dissolve glass because it wasn't 100.00% protected. Of course, aqua regia ain't all fun and games either!

@TaylorScott Tanks for the safety recommendation!

I awarded the bounty to this answer. Not that it is a complete answer to the question but it's the only formal answer and contains valuable information. I've also noticed that James frequently makes nice answers to people's questions in general.

How do you coat something with teflon? Are there teflon spray cans?

@Juan Perez: spray-on Teflon powder is available online. Chemical application of Teflon to metal is a commercial process - you could look it up.


text:
Does it sometimes take energy to create a chemical bond?
Wikipedia says so:

Bonds can also be broken apart. Since most bonds require energy to form, they also give off energy when they are broken. But before most bonds break, the molecule has to be heated. Then the atoms start to move, and when they move too much, the bond breaks. Molecules that require less energy to break than they give off when broken are called fuels. For example, a candle will just sit there and nothing happens. But when you use a match to light it, it will burn for a long time. The match brings the energy to break the first bonds, which release enough energy to break the bonds below them, until the candle has burned down.
  But what I have learned is bonds release energy when formed.

Highly misleading and confusing paragraph. "Energy" comes in many forms. Reactions don't occur spontaneously unless the free energy decreases, we know that much.

Yes, a very confused paragraph. There is an activation energy between reactants and products that has to be surmounted before a reaction can occur. If it were not the case all possible reaction would have happened long ago and we would not be here to answer questions :(

OK, it took me a while to realise but this is not the "normal" wikipedia which has a much better articles on molecules at https://en.wikipedia.org/wiki/Molecule Instead this is "simple" wikipedia. A quick sample of this article and links from it found all to be extremely poor indeed, and from what I saw I would recommend against anybody using simple wikipedia at all.

"Molecules that require less energy to break than they give off when broken are called fuels." is the most imprecise sentence in the paragraph, because fuels don't give off energy when their bonds are broken, but when these broken bonds form new bonds (with oxygen) that are much more stable and release more energy than it took to break the original (carbon-hydrogen) bonds.
A candle is stable in air, even tho it does not have complete stability in an oxygen-containing atmosphere. But if you put sufficient energy into a few bonds, the excited atoms will combine with oxygen, which gives off enough energy to excite a few more bonds, which etc., etc. It's this new reaction, which is overlooked in the paragraph, which provides the increased temperature during the reaction.
The energy that "bonds require to form" is an activation energy which destabilizes a molecule enough so that it can reorganize with other reagents (like O2) to form more stable products. This activation energy may be supplied initially from outside sources (the match), but in a continuing reaction, the exotherm from the reaction provides the activation energy and then some.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Parameters by which to distinguish SN1 from E1
I am new to organic chemistry and am struggling with the following very basic question:
Given is the following reaction:
$$\ce{(CH3)3CCl + NH3(aq) →}$$
I am supposed to analyse the end-product (the temperature is not given). However, how do I determine whether this is $\mathrm{S_N1}$ or $\mathrm{E1}?$ I understand $\ce{NH3}$ is a weak nucleophile and that the carbon atom is tertiary, which together eliminate $\mathrm{S_N2}$ and $\mathrm{E2},$ but I was not given any other parameters with which to carry out the evaluation.
Obviously, the end-product will depend on whether the reaction is $\mathrm{S_N1}$ or $\mathrm{E1},$ so how do I establish which one of the two it will be?

Why eliminate the E2 mechanism in basic medium?

@user55119 The way I understood the material, weak nucleophiles and E2 do not go together well. Am I wrong?

tert-Butylamine is not prepared from the chloride. Ammonium hydroxide is basic and will lead to isobutylene via E2 elimination. In a neutral medium such as water there will be an equilibrium mixture of alcohol (SN1) and alkene (E1).

@user55119 I should not be called "Ammonium hydroxide" and is not enough to remove weakly basic hydrogens.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Predicting atomic weight and density of calcium
I have a question from school:

Assuming that the element Ca had not been discovered, predict using the properties of the known element surrounding Ca its own properties such as its atomic weight and density.

How do I go about this? I don't know how to predict this, it all looks like building blocks in my head. I can't connect the dots yet, please help me solve this. I have done some google searching but to no avail.
What I tried: I understand that

atomic mass increase across the period;
density increase across the period;
density decreases down the group;
atomic mass increases down the group.

Why not to average the values across the period and group like Mendeleev did?:)

So, Like I tried that, but I didn't arrive at the same exact value, but very close. I averaged the value of Potassium and Scandium to predict Calcium.

I think this is what you were supposed to do. Don't expect the exact values.

And don't take the density bullets as rules - the various crystal structures play an important role as well. Atomic mass increasing going down is correct. Atomic mass increasing across the period is not always true.

Alright thank you all for your Contributions, I really appreciate @JonCuster.


text:
What does a simple or a complex substituent mean?
I am reading rules for IUPAC Nomenclature. The following statement is from my book.

Prefixes di, tri, tetra, etc., are not considered in deciding alphabetical order for simple substituents but considered for complex substituents.

With respect to the above statement, what does a simple or a complex substituent mean? Could you please give any examples of simple and complex substituents. How to decide whether a substituent is simple or complex? Is there any rule to decide that?

Please give an actual problem you find difficult. Your textbook might explain things correctly or not, we can only put it to the test if we have a test.

@Karl, I am having doubt in determining which groups are called complex. Is that based on the number of carbon atoms, or number of substituents, etc. If it is based on number of substituents, in the group under consideration, what is the minimum number of substituents in the group to make it complex. Since simple and complex are relative terms, I didn't get which groups are simple and which ones are complex.


text:
How to find the resist in pH change when both are same buffers of same concentration?
We are given two buffers [a) 0.1 M phosphate buffer of pH 7.7 and (b) 0.1 M phosphate buffer at pH 6.71 . If acid is to be added to the buffers, which of them,  will resist the pH changes better
I used Henderson Hesselbach equation in both cases but the log part and pka is same, so how to  solve it? Please provide a detailed solution

I am not getting, can you please explain

Also no information has been given about its alpha fractions,  neither of PO43-, nor of HP042- , nor of H2PO4- nor of H3PO4

Why is the log part the same? I think that highlights the root of your problem.

Because both are 0.1M and we do  not know individual conc of P043-, neither of H3PO4, neither of H2PO4 nor of HPO4 and there is no way to find

The pH defines the ratios of the various phosphate species. see https://en.wikipedia.org/wiki/Phosphoric_acid#Acidic_properties

This particular question is testing basic knowledge of buffers. In the most simplest explanation possible, a buffer will resist changes in pH the best when the pH is at the pKa. In the case of phosphoric acid, there are three acidic protons thus making three pKa values. Pulling the following table from Wikipedia
$\hspace{3.25cm}$
We can see $pK_{a2}$ is closest to the values of interest. Buffer solution a) is $|7.20-7.7| = 0.5$ pH units above $pK_{a2}$ while buffer solution b) is $|7.20 - 6.71| = 0.49$ pH units below $pK_{a2}$. Almost equivalent distances from $pK_{a2}$ though in different directions.
Now we must consider what is being added. Since acid is being added to the solution, the solution will decrease in pH. Thus we want to choose the buffer that is more basic as it will be more capable of neutralizing the acid. Therefore buffer a) should be the answer.
As a visual answer, consider the following titration curve for phosphoric acid.
$\hspace{1cm}$
The green smudge is the pH of buffer solution $\color{green}{a)}$ while the red smudge is the pH of buffer solution $\color{red}{b)}$. If acid is added, we clearly stay within the buffer region of the $H_2PO_4^-/HPO_4^{2-}$ equilibrium for $\color{green}{a)}$ while for $\color{red}{b)}$ the pH has the possibility of dropping off to become a very acidic solution. Therefore $\color{green}{a)}$ has the better properties of a buffer.

How is it that buffers of different compositions can have the same pH?for example it is possible to prepare 0.01M phosphate buffer of pH 7.0 and 0.1M phosphate buffer of pH 7.0 How?


text:
Energetic stability in relation to α-hydrogens
I am new to organic chemistry and was asked to rank the following materials according to energetic stability. I think I have understood that the way to do this is to count the amount of α-hydrogens: the more α-hydrogens, the more stable the molecule. However, the definition of an α-hydrogen is "a hydrogen that is attached to an α-carbon", where the α-carbon is the first carbon next to a functional group. In this case, it is not clear to me where the α-carbon is located: the only candidate for a functional group is the CH2/CH3 (depending on the case) at the very end of the chain, but then would the α-carbon be the very first one or the one next to it?
Any further general insight into the ranking of molecules according to energetic stability would also be very much appreciated… Many thanks!

sigh your question is (most probably) about enthalpy of hydrogenation, which in this case approximately depends on number of groups neighbouring the double bond

@Mithoron So what yardstick would you use to rank the energetic stability of three molecules like these? We never studied enthalpy of hydrogenation, so I guess it must be something else...

@mithoron Btw, it says explicitly in my textbook that α-hydrogens stabilize molecules. I am aware I might be looking at the wrong thing, but then please help me find a more correct parameter...

Wow, f*ck alpha-hydrogens. This questions contains five-rings with 0, 1 and 2 sp2 carbon atoms. If your teacher tells you this can be answered by only looking at alpha-hydrogens, that might be correct, but how do you know that ring sterics are similar for all three? How do you know in which alpha-hydrogens actually stabilise the molecule, given that two of them have a rather fixed geometry?

I guess this stupid book of yours was talking about hyperconjugation. One can rationalize that things like change of enthalpy of hydrogenation (which are actual measurables not some  "Energetic stability") are influenced by hyperconjugation depending on amount of α-hydrogens.

Have you looked up the formation/hydrogenation enthalpies of these three isomers? Please do so, and put the numbers in your question. That of course already answers the actual question, but we might all learn sth from it. ;-)

Apparently the intent of the question is to assess the stability of the three alkenes. Evaluation of the heats of formations provides useful information. The NIST site provides heats of formation and hydrogenation all of which are shown in black in the diagram. Computed values appear in red. Because hydrogenation of A, B and C all provide ethylcyclopentane, the greater the heat of hydrogenation the less stable the alkene. Alternatively, the less negative the heat of formation, the less stable. Thus, the order of stability is A < C < B.

You said your problem is finding the $\alpha$-hydrogen. Thus, first find the $\alpha$-carbon (so, hydrogen on it is $\alpha$-hydrogen). To do so you need to know what is the functional group. Here, the functional group of all three examples is the double bond. Following diagram shows you how to find $\alpha$-carbon when the functional group is a double bond or a carbonyl group:

Consequently, structure A has only one $\alpha$-hydrogen (on one $\alpha$-carbon), B has seven $\alpha$-hydrogens (on three $\alpha$-carbons), and C has four $\alpha$-hydrogens (on two $\alpha$-carbons).

How do I know the hydrogens of the methyl group in B count as as three, not 1? Or perhaps, say, 1.4?


text:
Could Calcium Oxide be made from CaCO3 in a bonfire?
I have some chalk and I need to turn it into CaO. Would a big bonfire be hot enough to make it? If anyone has personal experience with this reaction please share, whether succesful or not

Had a look on wikipedia?

I'll split your homework into two questions: What maximum temperature does an ordinary wood fire reach? And At what temperature does calcium carbonate decompose?

Practice trumps theory. According to the internet, temperature of a candle fire is ~ 1000°C throughout most of the flame. I have tried once making a tiny quantity of sodium oxide from sodium carbonate (850°C) in a candle flame, and failed. Heat transfer makes all the difference, and you can't really treat that theoretically. There are some sources on the internet claiming that it was done at first in just campfires, but I haven't seen that done on a video except in kilns.

Lime ( CaO) has been made that way since Roman times; A component of their concrete.

Aah, so you've given this some more thought (and research) than was obvious from your question. ;-)

Needs at least 900C

500 grams of CaO (laboratory grade) comes in under $10. Seems an easier way to do it.

According to wikipedia, CaCO3 will turn into Calcium if heated above 825 °C (1,517 °F).
As for your fuel
Wood burns in several phases.
In the first phase, which can be referred to as dehydration, the heat produced by the wood's combustion drives of water from it in the form of vapour. It takes a lot of energy to do that which limits the fire's temperature to ~200-250°C.
The second phase, which may be referred to as pyrolysis, kicks in as the wood gradually becomes dehydrated. In that phase the fire's temperature will go from 250°C to topical peaks up to ~800°C.
As the wood further breaks down, you will encounter a third phase which is equivalent to the burning of charcoal. The temperature at the heart of your ambers will be at a minimum of ~700°C and could reach as much as 1300°C if you bring enough oxygen into it (by injecting compressed air into the heart of your ambers through a metal pipe for instance). This temperature would be fully sufficient to produce CaO from CaCO3 in a reasonably short amount of time.
At this point you could also add regular coal into the ambers, which requires a temperature above ~700°C to start burning. In ideal conditions, coal can burn up to ~ 2100°C which would significantly push the limits of what is possible for you.
If you would like to push these limits further, you could add coke on top of your regular coal, which would push the theoretical maximum temperature your fire could reach up to ~ 2500°C.
More practically
As Francis mentioned, there is a wide gap between theory in ideal conditions and practical reality.
The first aspect I already mentioned that will affect the practical temperature your chalk will be exposed to, is the amount of oxidising molecules that is brought to the heart of your fire. Injecting air through a pipe may be the safest option (though it is already significantly less safe than just making a regular bonfire, as it will significantly increase the combustion temperature and is at risk of propulsing fuel around if not adequately set up).
Theoretically, you could achieve a yet better result (at the cost of a lot of danger) by injecting pure compressed O2 into the heart of your ambers (which, for safety reasons, you should NOT do).
The second and very important aspect is insulation. The theoretical maximum temperature that is reached within your fire is of little practical importance. Insulating your fuel combustion zone without compromising the supply of oxidisers into it is key not only to increasing the maximum temperature that is locally reached inside of your fire, but more importantly to just keep calories inside of the combustion zone, which will significantly increase the average temperature inside of it (and tremendously lessen the required fuel expense for achieving any given amount of CaCO3 calcination). You can insulate your fire in numerous ways. Coal itself can act as a (burning) insulator if you place your CaCO3 at the heart of your combustion zone and cover it in a pile of coal/charcoal. Or you could build a structure around your fire: SiO2, Al2O3, CaO and MgO are particularly suitable materials for their resistance to high temperatures and for their refractive properties. You can look up their properties on wikipedia (MgO for instance melts at 2,852 °C, has a thermal Conductivity 45–60 W·m−1·K−1 and a Refractive Index of 1.7355).
Chemistry
My last point is that there might be the possibility for you to make things easier by adding some additional chemistry into the process (if you are inclined to do so). It would seem, for instance that Ca(OH)2 might turn into CaO starting around 512°C, which seems to be lower than CaCO3. So you might be able to lower your requirements on the combustion side by turning your CaCO3 into Ca(OH)2 prior to calcination. There are many ways to do this but one way would be to react CaCO3 with HCl to yield CaCl2 , which you'd subsequently react with NaOH (the very low solubility of Ca(OH)2 would drive the reaction forwards).

Of course,the temperature can be increased rapidly by using the extremely unsafe "Purdue lox grill" treatment: https://www.youtube.com/watch?v=sab2Ltm1WcM . About 3 gallons of lox, 40 pounds of charcoal and about 3 seconds to completion. Sad about the grill ...

@EdV Quite impressive!! I think I wouldn't have the balls to handle lox like that...

Apparently, the trick is to avoid having the lox soak into the charcoal, so the Leidenfrost effect keeps an actual explosion from happening. I used to make lox as a lecture demo (trivial when you have liquid nitrogen available), but never got around to soaking a piece of bagel in lox and setting it off, both for the happy fire and the sheer pun of it!


text:
How can I prepare two phosphate buffer of different concentrations but give same pH?
How is it that buffers of different compositions can have the same pH? For example, it is possible to prepare 0.01 M phosphate buffer of pH 7.0 and 0.1 M phosphate buffer of pH 7.0? How?
I used Henderson-Hasselbalch equation, but still failed. No way is working. Is it something related to volume? Can it be shown mathematically? Can anyone show it mathematically that it is possible?

For a phosphate buffer with $\mathrm{pH} = 7,$ the two dominant species are $\ce{H2PO4-}$ and $\ce{HPO4^2-}$. The relevant $\mathrm{p}K_\mathrm{a}$ is $7.2$ (this is $\mathrm{p}K_\mathrm{a2}$). From the Henderson-Hasselbalch equation, using $\mathrm{pH} = 7,$ the ratio of $\ce{HPO4^2-}$ to $\ce{H2PO4-}$ is about $0.631.$ You can see this in the alpha diagram below:

In this figure, the green trace is the $\ce{H2PO4-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14.$ The blue trace is the $\ce{HPO4^2-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14$. So, at $\mathrm{pH} = 7,$ the blue curve is below the green curve, and the ratio of alpha fractions is $10^{-0.2}$, i.e., about $0.631.$ This is directly from the Henderson-Hasselbalch equation. Since all the phosphate species are in the same buffer volume, this is also the ratio of the respective concentrations.
The difference between the $\pu{0.01 M}$ phosphate buffer and the $\pu{0.1 M}$ phosphate buffer is that the $\pu{0.1 M}$ buffer is ten times higher in both $\ce{HPO4^2-}$ concentration and $\ce{H2PO4-}$ concentration. So the ratio is the same as in the $\pu{0.01 M}$ buffer and the $\mathrm{pH}$ is the same for the two buffers. Of course, the more dilute buffer has lower buffer capacity, but that is another issue.


text:
Formal charge: size vs electronegativity?
My question is why size is not taken into consideration when deciding major contributors of resonance structures, particularly: why is the formal charge always assigned to the most electronegative atom, but not necessarily to the one that can "better hold a negative charge"?
Example: Thiocyanate. Everywhere I've found says the major resonance contributor is that where the
negative charge lies on the nitrogen atom due to the higher electronegativity of $\ce{N}$ vs $\ce{S}$, but when we talk about acids/bases, clearly $\ce{H2S}$ is a stronger acid than $\ce{NH3}$ due to the "better stabilization of the negative charge by the sulfur atom due to its larger size/polarizability" (or so we teach in every OChem course). So why do we disregard this fact when assigning formal charges?
PS. I'm aware formal charges are an artificial construct and do not necessarily represent real-life electron density, I am just curious about these rules from an educator's perspective.

There is a difference between the two cases. In the first, you create (partial) charges in a neutral molecule, while in the second, you have a formal charge on an atom.

@Raphaël I think I see what you're saying, but I fail to see the difference in the end. In the first one you have a formal charge on an atom too.


text:
How to determine pressure and concentration of carbon dioxide in blood plasma?

A man suffering from untreated diabetes mellitus is admitted to a hospital. Doctors fear that his blood $\mathrm{pH}$ may have dropped because of ketoacidosis. Analysis of his blood reveals that $[\ce{HCO3-}] = \pu{16 mM}$ and $p_\ce{CO2} = 30.$ If $\mathrm{p}K_\mathrm{a}$ of $\ce{HCO3-}$ is $6.1,$ determine whether the patient runs a risk of acidotic coma. (Note: in plasma under physiologic conditions, concentration of $\ce{CO2}$ and $p_\ce{CO2}$ are related by the solubility constant for $\ce{CO2}$ in plasma which is $\pu{0.03 mM/mm Hg}.$

We use the Henderson-Hasselbalch equation, then
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log[\ce{HCO3−}][\ce{H2CO3}]$$
Moreover, here $\mathrm{p}K_\mathrm{a}$ of $\ce{HCO3-}$ is given whereas the acid is $\ce{H2CO3},$ and also if we consider second dissociation that is $\mathrm{p}K_\mathrm{a2}$ then also it is wrong because $\ce{HCO3-}$ will dissociate into $\ce{H+}$ and $\ce{CO3^2-}$ so in this dissociation there is no relation of $\ce{CO2}.$  And also why solubility constant is given, how can we find out $\ce{CO2}$ from solubility constant?

Since $p_\ce{CO2}$ is $\pu{30 mmHg}$, and we are given the $0.03$ of $\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}$ ratio, we can calculate the concentration of the $\ce{CO2}$ in plasma.
$$\pu{30mmHg}~\ce{CO2}\times 0.03\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}= \pu{0.9 mM}~\ce{CO2}$$
$$\ce{CO2 + H2O \xrightarrow{\text{carbonic anhydrase}} H2CO3}$$
$$\text{Thus, } \pu{0.9 mM}~\ce{CO2}= \pu{0.9 mM}~\ce{H2CO3}$$

$$\ce{H2CO3 <=> H+ + HCO3-}$$
$$\frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}} = K_{\mathrm{a}_\ce{H2CO3}}$$
$$\log{\frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}}=\log{K_{\mathrm{a}_\ce{H2CO3}}}$$
$$\log{\ce{[H+]}}+\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\log{K_{\mathrm{a}_\ce{H2CO3}}}$$
$$-\left(\log{\ce{[H+]}}+\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}\right)=-\log{K_{\mathrm{a}_\ce{H2CO3}}}$$
$$\mathrm{pH}-\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}$$

$$\ce{[HCO3-]}=\pu{16 mM}$$
$$\ce{[H2CO3]}=\pu{0.9 mM}$$
$$ \mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}=6.1$$
(PS: It is actually $\mathrm{p}K_\mathrm{a}$ of $\ce{H2CO3}$, not that of $\ce{HCO3-}$. Note that $\mathrm{p}K_\mathrm{a}$ of $\ce{HCO3-}$ is $10.3$.)

$$\mathrm{pH}-\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}$$
$$\mathrm{pH}-\log{\frac{16}{0.9}}=6.1$$
$$\mathrm{pH}-1.25=6.1$$
$$\mathrm{pH}=7.35$$

The $\mathrm{pH}$ of blood is maintained within extremely narrow limits of 7.36-7.4 for venous blood and 7.38-7.42 for arterial blood. (Source: Upadhyay - Biophysical Chemistry)

So, we can say the $\mathrm{pH}$ of the blood of the patient is slightly acidic than normal.
I hope I've answered your question.


text:
The heat content of the products is more than that of the reactant in an ............. reaction

The heat content of the products is more than that of the reactant in an ............. reaction.

In this question the answer given is "exothermic"; yet I thought the answer would  be endothermic as the products have more heat content and energy cannot be created nor destroyed. Therefore, from the reactants' side, more energy should have been put in for the heat content of the products to be greater.
Also mathematically endothermic make sense as: $$\Delta_\mathrm{r}H = \sum \Delta_\mathrm{f}H_\mathrm{P}-\sum \Delta_\mathrm{f}H_\mathrm{R}$$
And as: $\sum \Delta_\mathrm{f}H_\mathrm{P} \gt \sum \Delta_\mathrm{f}H_\mathrm{R}$
,$\Delta_\mathrm{r}H \gt 0$
Could someone please explain where am I wrong and why the answer is 'exothermic'?

Don't blame yourself.  This is a badly-worded question. Products and reactants don't have "heat content", they have thermal energy. I.e.,  heat is not a property of substances—they don't 'contain heat.'  Instead, heat is a property of processes; specifically, it is the flow of thermal energy across a boundary. So, for instance, in an exothermic reaction in a system with diathermal walls (walls that allow free flow of heat across them), thermal energy flows out of the system, and thus the heat (q) for that reaction is negative.
Having said that, it seems what they mean by "heat content" is the thermal energy of the system after a reaction if the system is adiabatic (no heat flow in or out, i.e., perfectly insulating walls).  Thus, if you have an exothermic reaction in an insulated container, the products you get will be hotter than the reactants you started with.
A much better way of asking the question would be:  Suppose a reaction takes place at constant pressure in an adiabatic (insulated) container, and the temperature increases.  Is the reaction exothermic or endothermic?  Answer: exothermic.

" Suppose a reaction takes place at constant pressure in an adiabatic (insulated) container, and the temperature increases. Is the reaction exothermic or endothermic? Answer: exothermic." - what? That doesn't make sense. First, you're leaving open the possibility of energy entering or leaving the system in the form of work (such as the container walls expanding). Second, if you require no energy entering or leaving through either heat transfer or work, then you can't also hold the pressure constant.

The IUPAC Gold Book definition for exothermic is that the std. enthalpy change ($\Delta H^o$) ("$^o$" =1 bar) is negative (goldbook.iupac.org/terms/view/E02269). That means if you did the reaction at 1 bar in a diathermal container, q would be negative (heat flow out of the system). Now suppose you did the same reaction at const. P (1 bar), but adiabatically. The latter describes a constant-pressure calorimeter (look it up). The walls are insulated, but P is kept constant. In that case, the energy released by the rxn causes an increase in T (even if some of that energy also goes towards work).


text:
Does 1,3-butadiene in bromine/water do addition twice?
Considering 1,3-butadiene (below) added to bromine water (considering it as $\ce{Br2(aq)}$):

I've seen online that the products are 1,2-dibromobutene and 1,4

but what I don't understand is why wouldn't the other double bond undergo substitution again? They're on a carbon chain like normal so shouldn't they react with the bromine water too? Is something stopping it?

Given enough time and Br2, it would.

Please don't use MathJax in titles.

If reaction is carried out in 1:1 ratio of Br2 and 1,3 Butadiene, then the above two products will be formed. If more Br2 is taken, then surely 4 positions will be brominated.

Why answer a question that has already been answered by @Ivan Neretin in a comment?

@user55119 because a comment is not an answer.


text:
How did Thomson calculate the momentum of electron when he didn't know the mass of electron?

This was proven in 1903 by J. J. Thomson who calculated that the momentum of the electrons hitting the paddle wheel would only be sufficient to turn the wheel one revolution per minute. source

J. J. Thomson measured the charge to mass ratio of electron in year 1897, and R. K. Millikan measured charge of electron in year nearly about 1910, i.e., Thomson knew only the charge to mass ratio of electron in 1903.
Then how did he was able to determine the momentum of electron when he didn't knew the mass of electron?

http://physics.princeton.edu/~mcdonald/examples/thomson.pdf

Another way to put it is that the computed radiation pressure is too small: https://en.wikipedia.org/wiki/Radiation_pressure

Recall that momentum is the product of mass and velocity. You don't need to know the mass or velocity of the objects hitting something and transferring momentum to something else - you just need to know the mass and velocity of what is being hit.

Might be a good question to ask at physics SE.

R. K. Millikan? Robert Andrews Millikan?

Thomson described the mechanical effect produced by cathodic rays on the first edition of its book Conduction of Electricity Through Gases (1903, pp. 501–502) and again later on the second edition (1906, pp. 629-630).
As you mentioned it refers to the Crookes experiment with the paddle wheel in which the electron beam collides with the mill coausing it to move (fig.below).

The electron beam carries a current ($I$) that can be measured and is equal to the number of electrons hitting the paddle every second (N) multiplied by the charge of the electron (e).
$$ I =N\cdot  e$$
Considering that the electrons are much smaller than the paddle you can consider that they bounce back with the same velocity (like when you hit a wall with a ball) and therfore the momentum transferred to the paddle per unit time ($p_{tot}$) is twice the momentum of the electron beam.
The momentum of the electron beam per unit time can be calculated with the well known formula $m\cdot v\cdot N$ with N to account for the number of collisions per unit time.
Therefore the total momentum per unit time transferred to the mill is:
$$ p_{tot} = 2m\cdot v\cdot N$$
This is equivalent to (multiplied and divided by the same quantity $e$):
$$ p_{tot} = 2\cdot \frac{m}{e}\cdot v\cdot (N\cdot e)= 2\cdot (\frac{m}{e})\cdot v\cdot I$$
Thompson considered an hypothetical best case scenario with a very high current of $\pu{10^{-5}A}$ carried by the electrons and a very high velocity of $\pu{10^8 m/s}$.
These are the highest reasonable values he could think could be applied in the Crookes' experiments (he even mentioned that the actual current measured was only $\pu{10^{-7}A}$.
The current can be easily measured while the velocity can be estimated from various methods using magnetic and electric fields (i.e. see section Charge-to-mass ratio of the Wikipedia where v is derived). And $\frac{m}{e}$ was already determined (around $\pu{6e-12Kg/C}$.
With those data, you can estimate that in the best case scenario (very high velocity and many particles hitting the paddle per unit time) the momentum transferred to the paddle per unit time is:
$$p_{tot} = 2\cdot (\frac{m}{e})\cdot v\cdot I = 2\cdot 6\cdot 10^{-12}\cdot 10^8\cdot 10^{-5}= \pu{1.2e-8 N} $$
This corresponds to $\pu{1.2e-3 Dyne}$ in the cgs units (units used by Thompson).
If you consider a paddle of $\pu{1cm2}$ then the pressure on the paddle is $\pu{1.2e-4 Pa}$, that is extremely small (an atmosphere is around $\pu{101 325 Pa}$!). Because of this it is reasonable to  attribute the movement to heating transfer.
As you see here, Thompson couldn't calculate the momentum defined as $m\cdot v$ but instead he calculated the  momentum per unit time  (that is a force)! He also considered the momentum carried by many electrons so that he could use the current in the formula of $p_{tot}$.


text:
Confused with 'discharge potential',preferential discharge theory
My book says that the ions which have least value of discharge potential are deposited first in the concerned electrode  because the 'require' the least energy to do so.I am confused by 'require' , do ions upon discharging gain energy or lose energy , are ions in general not less stable than neutral counterparts?further,suppose there are two contenders for the same electrode, say sodium ion and hydrogen ion during alkaline electrolysis of water , what is the need of adding the base , what actually is it doing and how does it improve electrolysis ?

Add some context. This is not the standard terminology and your question is not clear.

What context do you need ? It's a basic question and terminology used is what is given in my book , please elaborate

I can answer your first two questions.

I am confused by 'require', do ions upon discharging gain energy or lose energy?

Ions lose free energy during discharging, as you are referring to the spontaneous process of transferring charge at the electrodes with conversion of some of the ions into neutral species (in general you are speaking about the process described in the next paragraph).
For a process to proceed spontaneously, the free energy change for the system (combined half-cells, say) has to be negative. The Nernst equation tells us that the free energy difference is given by $$\Delta G = -nFE$$ where E is the total cell potential. This value must be negative for the cell to discharge, which means E is required to be positive. This is what is meant by "require" (language is used in funny ways, ions don't really require anything). You have more choices than just discharging the cell: you can charge it (creating more of some of the ions). Unless the potential is such to cause discharge you may actually produce more of the ions that are disfavored under spontaneous (discharging) conditions.

are ions in general not less stable than neutral counterparts?

You should not attempt to generalize, as this depends on the context (details of the reaction in question). An ion may be the most stable state of an atom in one context (an oxidizing environment, say) whereas it will be unstable in another (a corresponding reducing environment). Moreover, you can apply a charging potential and thereby invert the sense of stability, rendering one species more stable than the other.

So why is it written in my book that ions 'require' discharge potential , they should release energy and not require it , isn't it?

Because you have more choices than just discharging the ions: you can create more ions. Unless the potential is such to cause discharge you may actually produce more ions.


text:
What is the origin of the Quadratic factor in calculation of XRD peaks?
Whenever we are to calculate the d-spacing(for cubic system) in XRD peaks with reference to Quadratic Factors we do that with the help of the formula $$\frac{1}{d_{hkl}^2}=\frac{h^2+k^2+l^2}{a^2}.$$
Please tell me how we are arriving at this form and where from are we getting ${h^2+k^2+l^2}$ as a factor. I know that things are different for hexagonal. Would be happy if both are explained with more emphasis on the cubic one.

Should your RHS not have a square root? That is, $d=\frac{a}{\sqrt{N}}$ where $N=h^2+k^2+l^2$.

@jacob1729 Thanks for the clarification. Just edited.

We know that the planes (h k l) are the planes which are perpendicular to the vector (h, k, l). Thus, their equations are:
$$hx+ky+lz=D$$, in which D is a real number.
But these planes have to pass through atoms of the crystal. So, $$D=na$$, where a is the lattice parameter and n is a natural number (e.g., 1, 2, 3, ...). So, the planes have the equation:
$$hx+ky+lz=na$$
The distance between two adjacent planes is, thus, the difference between its distances from the origin of the Cartesian coordinate system.
$$d=\frac{na}{\sqrt{h^2+k^2+l^2}}-\frac{(n-1)a}{\sqrt{h^2+k^2+l^2}}$$
$$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$$
And thus your formula is correct.

Nice answer! Mathworld has a concise explanation of the distance of a plane from the origin: http://mathworld.wolfram.com/HessianNormalForm.html. Basically, the first equation is the plane in Hessian normal form when you divide by the length of the (h k l) vector.

@KarstenTheis:Your answer is has way more depth but this is more comprehendable for beginners.

@Lemoine: Tell me something while you substituted D=na. Tell me whether it is necessary that the diffraction will take place only at planes which are separated by one lattice parameter?Can the diffraction not take place at other planes?

@user586228: It only makes sense to talk about planes with atoms in it. We know that the coordinates x, y and z of an atoms are integer multiples of a (in cubic systems). If D is different from na, the plane does not contain any atom.

@Lemoine what about planes having miller index {111}?Can they be quantified by a natural multiple of lattice constant?

@user586228: I'm not sure if I understand your question. A plane, e.g., 1x+1y+1z=1a has Miller index (1 1 1) and contains atoms (for instance, the coordinates (1,0,0),(0,1,0) and (0,0,1)). The plane 1x+1y+1z=$\sqrt{3}$ a does not contain any atoms

Reciprocal space
One way to derive the formula is to use reciprocal space coordinates. The reciprocal space lattice has three unit cell vectors, a*, b* and c*. The diffraction vector d* is given by:
$$d^* = h a^* + k b^* + h c^*$$
The d-spacing is given by the reciprocal length of d*. The relationship between direct space and reciprocal space unit cell vectors is
$$a \cdot a^* = 1$$
$$b \cdot a^* = 0$$
$$c \cdot a^* = 0$$
$$a \cdot b^* = 0$$
$$b \cdot b^* = 1$$
$$c \cdot b^* = 0$$
$$a \cdot c^* = 0$$
$$b \cdot c^* = 0$$
$$c \cdot c^* = 1$$
So to get the d-spacing, we have to figure out the length of $d^*$. For a cubic system, the reciprocal space unit vectors all have the lenght of 1/a, and they are all at right angles. To get the length for the cubic system, we use Pythagoras' theorem (or the length of a vector in Cartesian space): It is the square root of the coordinates squared.
So:
$$|d^*| = |a^*| \sqrt{h^2 + k^2 + l^2}$$
To get to your formula, substitute 1/d for $d^*$ and 1/a for $a^*$. For the hexagonal system, it is a bit different because the angles are not all 90 degrees, so there will be a trigonometric factor (sin 120 or cos 120 or such) in the respective formula.
Direct space
One plane will go through the origin, the next through a/h, b/k and c/l. In the cubic case, a vector normal to the plane is (h, k, l). To get the distance of the plane from the origin, you can take any point in the plane and form the dot vector with the normalized vector (h, k, l). For example,
$$ d = \frac{(h, k, l)}{\text{length} (h, k, l)} \cdot (1/h, 0, 0)$$
For a cubic system, this will simplify to the expression for the d-spacing. The term $h^2 + k^2 + l^2$ is from the length of the (h, k, l) vector.
(h k 0), no vectors
Both derivations above use vectors. Here is a derivation for the 2-dimensional case using no vectors.

Let the origin be point A. In a cubic system, the plane hk0 intersects the x-axis at a/h (point B) and the y-axis at b/k (point C). Our goal is to calculate the altitude. We use that the area of the triangle can be calculated two ways, using the perpendicular sides or using the hypothenuse and altitude:
$$\text{Area} = \frac{1}{2} * \frac{a}{h} * \frac{b}{k} = \frac{1}{2} * \text{hypothenuse} * \text{altitude}$$
We can get the length of the hypothenuse from Pythogoras:
$$\text{hypothenuse} = \sqrt{(\frac{a}{h})^2 + (\frac{b}{k})^2}$$
Then, we can solve for the altitude:
$$\text{altitude} = \frac{\frac{a}{h} * \frac{b}{k}}{\text{hypothenuse}}$$
$$ = \frac{\frac{a}{h} * \frac{b}{k}}{\sqrt{(\frac{a}{h})^2 + (\frac{b}{k})^2}}$$
With a = b and simplifying, we get the formula the OP asked for (in two dimensions).


text:
Volume of a gas
The kinetic theory assumes that, for an ideal gas, the volume taken up by the molecules themselves is entirely negligible compared with the volume of the container. For a real gas, that assumption isn't true. The molecules themselves do take up a proportion of the space in the container. The space in the container available for things to move around in is less than the measured volume of the container.

Source

After reading the above, my interpretation of volume of gas is that the space for the molecules to move around. When we consider the volume of molecules, then the space for it to move around reduces.
But according to my interpretation the volume of real gas must be lower than the volume of ideal gas which is not true.
When we talk about volume of gas, is it the sum of volume of molecules and space between them?
If that is the case then,

When we talk about volume of ideal gas, we are considering the volume of the container which is the sum of volume of molecules and the space in between them. Then what do we take into account when we say volume of molecule is negligible as compared to volume of container?

It might be a basic question, but I didn't get any answer to it till now(from my teachers and other websites). Hope i'll get it cleared in CSE.

Molar volume...

Your picture has somewhat exaggarated gas particles, and they still only take up $(1/6)^3 = 1/198$th of the volume. The ideal gas law ignores this. Only the the van der Waals approximations claims to make an explicit correction for it, but in the end that's also just an empiric parameter, which needs to be changed for different p/T ranges. vdW is not really a good and useful gas law.

$6^3=180+36=216$

Consider a substance gaseous and liquid phases just below the critical point, when the density  of the former is just slightly lower than the latter. ( for water, it is about 0.2 g/cm3 AFAIK ) . Would you say the volume of liquid is what we are used to be, but gas volume is just the volume of molecules ?

Consider a balloon filled with a gas. The volume we mean is the volume of the balloon, whatever is inside. The perfect gas hypothesis will change the properties of a gas. For example, for the same balloon, we would expect a real gas to exert a stronger pressure outward than a perfect gas, since it has less space to move around, like you said.
Assuming the balloon is flexible, the inward and outward pressure will equilibrate by an expansion of the balloon. Thus, the volume of the real gas (molecule + space around) will be greater than the volume of a perfect gas.
You do put the finger on a limitation of the definition of the volume of a gas. A gas will spread to occupy as much volume as it can, unless restricted. For example, think of a molecule in space; what is its volume?
Volume is a macroscopic property which only makes sense in certain conditions.
When we say that the volume of the molecules is negligible, we mean that the total equilibrium volume of a gas is roughly the volume of free space the molecules have to move around. This is a good approximation at low pressures, since the volume of the molecules is a very small percentage of the total volume, and intermolecular collisions are essentially impossible. However, at high pressures, the molecules are rather close together, do occupy a non-negligible fraction of the total volume, and do collide together often.

How will we define volume of a real gas?

I'd agree with most of this, but you can't really separate the volume the gas molecules occupy and the attractive force between molecules at high pressure. Thus the observed behavior at high pressure depends on both factors.

@Kaushik - The volume of a gas is the volume of the container.

@MaxW Doesn't that volume include the volume of the molecules?

Molecules aren't big balls either, they're mostly electron clouds and aren't "solid" in any way.


text:
Emulsifier for water, been oil
I am trying to blend neem oil and water and I'm wondering what the best emulsifier would be 10% neem oil 90% water thank you

been oil, neem oil ? now which of them? What is "blend", what properties should that emulsion have? What's it supposed to be good for? Also you say you are "trying". Now what have you tried?

There's two ways to do this. The easy way, and the right way.
Traditionally people will use dish soap.
1 tsp neem oil to 1/2 tsp dish soap to 1 quart water
(https://www.greenhousemegastore.com/supplies/insect-prevention/pure-neem-oil)
Ten percent oil (actual oil, not oil phase) is actually pretty high. Just letting you know this in advance. You'll be making a greasy feeling lotion.
Without getting into emulsions (as I don't know what kind of access you have to more "advanced" cosmetic emulsifiers) too much, I'd say to simply melt 2.5% Emulsifying Wax NF (I like Polawax but NF means national formulary, means it's all standardized to be the same. Polawax is different from NF) with the 10% neem oil to $\pu{70 ^\circ C}$, bring 87.5% distilled water up to $\pu{70 ^\circ C}$ and pour oil into water while stirring fast (you can use handmixer, etc.).
If 2.5% emulsifying wax doesn't hold the neem oil for some reason, add a percent or two more wax and heat everything to $\pu{70 ^\circ C}$ again and mix.
For reference, neem oil has an HLB of 9.
(http://www.saffireblue.ca/blog/required-hlb-of-oils-and-lipids.htm)
You don't need to know this if you're going with emulsifying wax, but you will if you want another nonionic emulsifier. Explaining how to use the HLB system is beyond the scope of this answer.
Cheers


text:
Lactone acylation
Can anyone propose/draw a mechanism for Method A, below, from this paper? Almost all other literature use method B and reagents such as SOCl2.

Welcome to ChemSE; Unless you offer a solution to this question it will be voted down. We do not do homework but we will help you with some effort on your part. Get back to us.

This is not a homework question. I do not know where to start since there is limited literature on this.

What is the role of AlCl3? What is its reactivity? How about the lactone?

AlCl3 is a Lewis acid and activates the carbonyl oxygen.

But there is literature that I will share if you put forth your best shot.

My best shot is: the ring breaks in order to stabilize the activated oxygen. I am unsure whether it is at the 2 or 4 carbon... I am then unsure as to whether there is a hydride shift, whether an acylium or carboxylic acid group forms, how to remove the hydroxyl group, and how the benzene attacks...

Additionally I have found several papers that show that this reaction does not proceed eg https://reader.elsevier.com/reader/sd/pii/S0040403903008554?token=F148342816E4F6486CD5E79B1D727C422C4D3B931373B500965B9B20D502E5395EC2FF53B5F24E0FEFE410FF81F6F131

I find this last statement somewhat difficult to believe considering that this has been reported in Organic Syntheses, whose sole purpose is to publish procedures that are supposed to reproducible, to the extent where the published procedure should have been independently verified for being reproducible...

I agree with you.

@JBG you can include all the information of yours in the comments into your question to make it more contextual rather than a mere homework question.

The reaction of benzene and γ-butyrolactone with AlCl3 was reported by Truce and Olson1 in 1952. The authors found that the ratio of γ-phenylbutyric acid to α-tetralone was dependent on the  AlCl3/lactone ratio. The larger the latter ratio; the more α-tetralone. (See Table 1 from the paper). The ratio employed in the Organic Syntheses preparation is 3.7. Table 2 from Truce's paper provides an explanation for the role of excess AlCl3 but it doesn't explain the alkylation of benzene.
The lactone-AlCl3 complex 1 can act as an electrophile aromatic alkylation (1-->2-->3). Subsequent intramolecular acylation of 4 leads to α-tetralone 6. Bear in mind that γ-butyrolactone reacts with HBr to form γ-bromobutyric acid.

1) W. E. Truce and C. E. Olson, J. Am. Chem. Soc., 1952, 74, 4721.


text:
Coffee cup lids
So this morning I purchased a coffee. As a not uncommon event the lid on the coffee cup didn't have a hole completely punched through (a gas inflow hole). This makes it difficult to drink the coffee because I'm essentially creating a vacuum when I suck out the coffee.
I was wondering then. Is it atmospheric pressure or gravity that causes the coffee to drain from the drinking hole?

More of a physics question, this, but it brings up some concepts pertinent to physical chemistry.

A combination of gravity and the suction you apply with your lips ("suck") help empty the cup, assuming as you do that the coffee in the cup leaves a vacuum as it exits. This is similar to the "problem" of emptying a bottle of liquid when you invert it (the mouth facing downward). The challenge is most evident with honey or ketchup, viscous liquids or those with higher surface tension that are more challenging to pour. Atmospheric pressure works against drainage of the container because there is no force (gas pressure) against the liquid at the inside top of the container. As liquid pours (or drips) from the mouth, bubbles of air travel up into the container, reducing the vacuum and allowing more liquid to flow out. Notice how thin PET water bottles crinkle when you force liquid out by suction? That's atmospheric pressure. Eventually you have to suck pretty hard to get additional liquid out, and can even make the bottle hang from your lips.

Both, since atmospheric pressure does not exist without gravity. Atmospheric pressure is essentially the weight of air above something, which happens because of gravity.

Ultimately true, but not useful in understanding or explaining the immediate issue which is all to do with air pressure.


text:
How to pool fractions after chromatography?
I used chromatography to purify my protein and now I have fractions with different concentrations ranging from 1-16 mg/ml. Should I pool all these fractions or just the ones with high concentration (where under the actual peak)?
Thanks in advice

How pure do you need your product to be?

I never purified proteins. In column chromatography all fractions containing "just" the target compound(s) are mixed together.

You should run gels on the fractions and verify purity, discuss with others in the lab including your supervisor, and then pool as necessary. Also, keep the fractions at the recommended temperature (probably in a cold room) until then.

If all fractions contain the same protein, I'd combined all and concentrated the combined solutions to desired concentration using biochemical techniques such as dialysis. If they contained more than one protein, then I'd combined all fraction collected under the correct peak. However, it is better ask your supervisor (or adviser) before you combined them. The person who wrote the protocol knows which one to collect. Yet, you can look at following reference to get an idea how to concentrate the combined protein solutions:
Sarah M. Andrew, Julie A. Titus, Louis Zumstein, "Dialysis and Concentration of Protein Solutions," Current Protocols in Immunology 1997, 21(1), A.3H.1-A.3H.5 (https://doi.org/10.1002/0471142735.ima03hs21).


text:
Cup with vinegar/salt/copper solution previously in it ran through dishwasher
Someone enjoys making jewelry, and saw an online tutorial saying that you can mix equal parts of vinegar and salt, and leave copper in it for a while to oxidize it, and turn it green.
After doing this, dumps the resulting solution down the drain, and runs the plastic cup it was done in through the dishwasher, presumably with some residue left in it, with dishes (ceramic), glasses, and silverware (not real silver, whatever cheap'ish silverware is made with.)
Then afterwards reads other places saying not to do this, because it makes a toxic solution.  Some say it makes copper acetate and copper chloride.  Some say it creates copper ions, which can cause heavy metal poisoning, and to only dispose of it through a hazardous material disposal.  Some say all of this stuff is really toxic, even up to 5ppm.
What would you, someone who understands this stuff, do, and what's the science/chemistry behind your reasoning?

Nothing, eat on the plates coming right out of the dishwasher.
Run everything through a few more cycles.
Replace everything.

I'm thinking running everything through another cycle, and I'd be surprised if there would be a problem after that.
But, I've never learned any chemistry, and wanted to get opinions from people who know things in this area, before blowing it off and eating off things that came in contact with at least a minute amount of this stuff.

Just run the dishwasher one more cycle and you are good to go.

Run everything one more cycle and you are totally safe. Copper is highly soluble in water, so it will entirely dissolve and go away, not contaminating your plates.

Copper isn't soluble, but some of its compounds are.


text:
How do you find the heat capacity ratio of combustion products?
For example, if gaseous oxygen $(O_2)$ and paraffin wax $(C_{32}H_{66})$ were to stoichiometrically combust at STP, how would one go about finding the heat capacity ratio ($\gamma={c_p \over c_v}$) of the resulting mixture of gases? I'm guessing that it involves finding the adiabatic flame temperature? (but idk how lol)

What is the approximate temperature and pressure that you want this ratio?  Do you know the mole fractions of the constituents in the product mixture?

Maybe around 3160K and 290kPa. As for the mole fraction, the product mixture is comprised of ~20% CO, ~20%$CO_2$, ~30%$H_2O$, ~10%$O_2$, ~10%OH, and trace amounts of other stuff.

At that temperature and relatively low pressure, the gas mixture will exhibit close to ideal gas behavior.  For an ideal gas mixture, each heat capacity is equal to the pure species heat capacities weighted in proportion to their mole fractions.


text:
Do the groups involved in geometrical isomerism have to be in similar planes to show geometrical isomerism?
In my textbook, under the topic "Conditions for geometrical isomerism", the following has been given:

Groups responsible to show geometrical isomerism must be nearly in the same plane.

Is this statement really true? Because, we are well aware that disubstituted cycloalkanes which show geometrical isomerism are not planar. Eg: 1,2-Dimethylcyclopropane
Kindly explain with examples.

There is nothing to explain. People make blunders, and an example you've just found yourself.

It is a poor definition but it is formally correct. What is important is rigidity. Two points always identify plane (s). There must be a plane *on which the groups cannot rotate, ie at the end a bond with rotation.

In my comment above "with" reads "without rotation", of course.

To have geometrical isomers you need a structurally rigid bond. This bond can be double and this is a case where groups are in the same plane:

Or can be a cyclic compound where the  carbon-carbon single bond is restricted(2 examples below):
1st example:

2nd example:

Your textbook definition is incomplete and I wouldn't use the word "must" there.

The captions of both structures in your first example is identical. Yet 2nd set of that example is 1,2-dimethylcyclopropane. It has both cis- and trans-isomers. Please change the captions accordingly.


text:
Conditions for molecular dissymmetry
In my textbook, under the topic "Asymmetric and Dissymmetric Compounds", the following statement is given:

A molecule which does not possess plane of symmetry, centre of symmetry, and alternating axis of symmetry is called dissymmetric.

Why the definition didn't include the "axis of symmetry" as a condition for molecular dissymmetry?

Because axes of symmetry are irrelevant for the question at hand.

https://chemistry.stackexchange.com/questions/104409/use-of-axis-of-symmetry/104411#104411

For the record, over a 10-year career in chemistry, I have never heard anyone use the word "dissymmetric." The proper word is "chiral."

@Zhe, Thank you for your reply. Actually, it is given in my textbook like that. I am also used to the term chiral, but not dissymmetric.

From Wikipedia:

A chiral compound can contain no improper axis of rotation ($S_{n}$), which includes planes of symmetry and inversion center. Chiral molecules are always dissymmetric (lacking $S_{n}$) but not always asymmetric (lacking all symmetry elements except the trivial identity). Asymmetric molecules are always chiral.

That means that a dyssymetric molecule lacks of $S_{n}$ axis(planes of symmetry and inversion centers) but can have $C_{n}$ axis of rotation. For example the following molecule is dissymetric: it does not have any plane of symmetry or inversion center but it has a $C_{2}$ axis of rotation(the "axis of rotation" is a synonyms of the "axis of symmetry" you mentioned in your question).


text:
One mole of solute (NaCl) is dissolved in 1 litre water. The molarity of solution is

One mole of solute ($\ce{NaCl}$) is dissolved in $\pu{1 L}$ of water. The molarity of the solution is:
A) $\pu{<1M}$
B) $\pu{>1M}$
C) $\pu{=1M}$
D) $\pu{=2M}$

My answer: It’s a fairly simple question so, I did what anybody would do.
$\mathrm{Molarity}=\frac{\text{No. of moles}}{\text{volume of solution}}$
$\mathrm{Molarity}=1$
That is the obvious answer, and it’s probably right, but the answer given is $\pu{<1M}$. I just wanted to confirm that the answer given is right or wrong. If it is right, why?

Look up the difference between molar and molal solutions.

The answer $<\pu{1M}$ is correct.
Suppose you have $\pu{1L}$ ($\pu{1dm^3}$) of water and then you add your salt. When you add your salt the volume of the solution increases so now you have $\pu{1 mol}$ of $\ce{NaCl}$ in more than $\pu{1L}$ of solution. That means that you can't have a $\pu{1M}$ solution anymore.
If you want a $\pu{1M}$ solution then:

Add $\pu{1 mol}$ of salt in a flask.
Add water until you reach $\pu{1L}$ of volume.

EDIT: when you add salt, the volume of the solution increases slightly and less than what you expect(volume of water + volume of dry salt).


text:
Heisenberg's uncertainity principle(HUC)
I know the mathematical form of HUC and its statement i.e.It   is   impossible     to   determine   simultaneously. , the exact  position  and  momentum  of  an  electron. i don't know why this principle exists. Moreover, i read that this all is not due to technique inadequancies, but due to the nature of electron. What all is this? I tried to understand derivation for the mathematical form of HUC (just to get the idea why such thing exist), but it used advanced mathematics (I'm in class 11). Please give me the idea about why this principle exists. Why it is so that we are unable to measure velocity and position at the same time to high accuracy. (No usage of earlier established principles and maths please, in the answer.)

When you'll be ready to digest all the math, you'll see. Before that, just take our word for it.

The usual/standard  explanation is that the act of measuring something (at an atomic level ) is such that it inevitably perturbs what is being measured. However, you should really ask the physics StackExchange community for a proper non-mathematical explanation as to what the current thinking on this is.

You might want to take look at this:  https://physics.stackexchange.com/questions/287126/heisenbergs-uncertainty-principle , noting that it's not due to the nature of the electron specifically, but rather due to the wave nature of matter that reveals itself with any particles that are sufficiently small.

@porphyrin  It's probably useful to note that the current understanding is that the uncertainty principle is independent of the observer effect.

Theoris, i looked at the same ans. earlier, but i was not able to understand it. Any simple answer with some analogy from real life would help.

Unfortunately, you can look this in one of two ways. (1) The real life analogy is all around you because it's how the universe works, or (2) there is no real life analogy because your everyday intuition is totally at odds with how quantum mechanics works.

@theorist, thanks, it is surprising is it not that about a hundred years on QM is still not properly understood even though we use it all the time and it never fails.

I can provide some physical intuition, but it's still going to require math and previous work to make it happen (aside on this at the end).
Let's consider a particle in just one dimension. There are two quantities of interest: (1) the position of the particle along this coordinate, (2) the scalar momentum (related to the velocity) along this coordinate.
I'm going to represent this particle as a plot on a graph. The horizontal axis will measure position. The vertical axis represents some notion of how likely it is to find the particle at this position (formally, it would be squared and normalized). Finally, we also stipulate that momentum of the particle is going to be the frequency of what I draw. This may seem strange when we're talking about an arbitrary plot, but it's actually completely reasonable from the perspective of representing functions from as Fourier series. You're just going to have to assume that this is true because a formal analysis does require math (see other answers).
Now, let's make this more concrete with some plots.
Here's a standard sine function. It has a very well defined frequency, so we know the momentum exactly. Where is the particle? Well, the sine wave stretches out to infinity in both directions, so we have no idea where it "is."

Here's a different graph. Where is the particle? It's very close to $x=0$. What is it's momentum? It's kind of hard to pick out single frequency here, so you cannot easily establish what the momentum is.

And that's basically it. Identifying frequency (momentum) and position for a plot like this are traded-off against one another. You'll probably end up in the intermediate case where you know some information about both. This is wave packet.

This particle is roughly centered around $x = 0$, and we can roughly say what its momentum is.
The uncertainty principle simply formalizes this idea except using the formal mathematical structure of quantum mechanics.
Now, for the aside I promised. Your question was entirely reasonable until you got to the last line, and I quote: "No usage of earlier established principles and maths please."
This is impossible. Science does exist in a vacuum. It is based on the past observations and structure built by previous generations of scientists. If we were to start from scratch, would we arrive at the same science that we have today? Possibly, possibly not. But the understanding we are working with is built upon foundations that we are stuck with, for better or for worse.
You also can't really do this without math. Part of science is to describe relationships, and mathematics allows us to do so unambiguously. Words suck at conveying ideas because words mean different things.
Lastly, it's fine to try to build intuition on a concept by relating it to the macroscopic world. However, you should not assume that this is possible. It may be philosophically satisfying that objects behave in similar ways regardless of size, but that's not the way the world is.

There is a simple way to understand this if you understand the relationship between frequency and wavelength in a vibrating string
But you can't avoid at least some mathematics.
They key to understanding where the position momentum uncertainly relationship comes from is to think about the relationship between the frequencies present in a musical note and the time over which the note is sampled. This can be understood if you know about Fourier transforms but that involves mathematics. But the basic principles can be understood without that detail.
A vibrating string can produce a nearly pure note. Just one dominant frequency is present. If you look at a long time period while the string is producing that note and analyse the frequencies present you will get a very narrow set of frequencies. But, if you only sample the vibration for a very short period of time, you will see other frequencies. To put it another way, if you look at an infinite pure sine wave, it will have a single frequency: the curve is smooth and there are no sharp edges when the vibration is plotted on a chart. But, if you take a small sample (over a short period of time) both the start and end of the chart will show sharp edges where the sample starts and stops. If you now translate that into a chart of the frequencies present, there will be many other frequencies present.
Real musical instruments also show some of this behaviour. A plucked string will emit a wider range of frequencies than a smoothly vibrated one.
The link to Heisenberg is this. If you have infinite time to observe the frequency you can get a pure tone with little uncertainly about the frequencies present. But if you only have a short time to observe the vibration you inevitably have other frequencies and the "certainty" about the frequencies present is much larger. This is very clear in the mathematics of Fourier transforms (which describe the relationship between the worlds of vibrations over time and the frequencies present).
The key takeaway is that you cannot simultaneously have a very precise frequency and a very small localisation in time when observing the vibration. The more time you have, the tighter your mix of frequencies will be: the less time, the wider the range of frequencies you will see.
It isn't a perfect analogy to position and momentum uncertainly in the quantum world but it is a similar type of tradeoff. And it isn't a practical limitation which we could bet with better instruments or techniques: it is baked in to the very physical definitions of the things we observe (in music: the relationship between observed vibrations over time and the frequencies present in the mix).
The position and momentum of particles are two points of view of the same thing just as a vibration over time is one view of a musical note and the frequencies present is another. But the process of translating from one point of view to the other (mathematically a Fourier transform) cannot give us precision in both frequency and time simultaneously. For quantum particles it is the same: we cannot know at the same time how fast something is moving and where it is.
The mathematics would give a more precise picture, but the physical analogy provides some intuition as to why this is so.

This is a nice explanation but surely there is more in it than just a Fourier transform between conjugate variables? If you observe for a long (effectively infinite) time then just one energy will be measured but if for a short time many waves, i.e. a localised wavepacket is observed and hence many energies exist; why? Where do these different energies come from and are there just because we observe for a different length of time?

No, not really. The Heisenberg Uncertainty Principle is just a result of the the variables being related by a Fourier transform. See e.g. https://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle

@porphyrin It isn't a perfect analogy, but the mathematics is very similar. And the analogy does give a non-mathematical picture that people familiar with the basics of sound will understand. Ian Bush has it right.

As you can see from figures in Zhe's answer that it is possible by adding waves to make a wavepacket. The characteristic of a wave is that superpositions are possible and these lead to interference/diffraction phenomena. By analysing the wavepacket's spread in wavelength and position, the expression $\Delta k.\Delta x \ge 1$ is found where $k=2\pi/\lambda$. The exact value of the product$\Delta k.\Delta x $ will depend on the shape of the wavepacket but $1$ is the minimum value. This expression is also found by Fourier analysis between what are called conjugate pairs, such as position $x$ and its reciprocal $k$. Time and frequency are another such pair. The wavepacket at $x$ with width characterised by $\Delta x$ and that of $\Delta k$ (width of wavevector)  are related by Fourier Transforms as described in other answers.
However, the relationship  $\Delta k.\Delta x \ge 1$ is not an inherent property of quantum mechanics but is a property of Fourier Transforms and well known in classical physics. An example is that it is not possible to form a train of electromagnetic  waves for which it is possible to measure, at the same time, the position and wavelength with infinite accuracy.
Quantum mechanics enters when a particle is given wavelike properties as satisfied by the de Broglie relation. A material particle (such as an electron or a molecule) of energy $E$ and momentum $\pmb p$ is now associated with a wave of angular frequency $\omega=2\pi\nu$ and wavevector $\pmb k$ i.e. $E=\hbar\omega; \pmb p=\hbar \pmb k$. The fact that the Schroedinger equation is linear and homogeneous means that for particles a superposition principle applies which gives them wavelike properties.
(Aside: The Schroedinger equation describes what happens to a particle under the influence of a potential (such as in an harmonic oscillator) and allows the calculation of the energies and wavefunctions. The Schroedinger equation has no derivation, it is assumed to be correct, as is the de Broglie relationship, and is justified by experience as it has not yet been found to fail to describe experiments.
The small value of Planck's constant makes the limitations of the uncertainty principle totally negligible for anything macroscopic, i.e. greater than approx micron size. )

It's simply a consequence of wave mechanics and operators. Any observable, i.e., any observable which can be measured in a physical experiment should be associated with a quantum-mechanical operator (like + or - operators, just more complicated). Furthermore, if these operators are intrinsically non-communicative, or if the order in which they are applied matters (like the - sign; 5-3 $\neq$ 3-5), then they can't be known simultaneously. This is important because your measurements effect the quantum mechanical state. If you're measuring 1 then 2 and you get a different answer than if you measured 2 then 1, then you can't get anything meaningful out of that measurement because you don't really know whats what. It gets more complicated with eigenvalues and eigenstates so hopefully we can leave it at that.
So from quantum mechanics we know that the operators for position and momentum are $\hat{x} = x$ and $\hat{p_x} = -i\hbar\frac{\partial}{\partial x}$. If these two operators did commute, then the difference between their resultants would be zero.
However, we can test these operators via a 'trial' function to see if they commute:
$$\hat{p_x} = -i\hbar\frac{\partial}{\partial x} ; \hat{x} = x ; f = f(x)$$
$$\hat{x}\cdot\hat{p_x} \cdot f = \hat{x}(\hat{p_x}\cdot f) = x(-i\hbar\frac{\partial f}{\partial x}) = -i\hbar x \frac{\partial f}{\partial x}$$
$$\hat{p_x} \cdot \hat{x} \cdot f = \hat{p_x} (\hat{x} \cdot f) = -i\hbar\frac{\partial}{\partial x}xf = -i\hbar x \frac{\partial f}{\partial x} - i\hbar f$$
Since those results differ, the operators are non-commutable, and thus you can't know momentum and position precisely and simultaneously.

"It's simply a consequence of wave mechanics and operators." But those would not mean a thing if they did not just happen to describe our universe correctly. So the answer to the question "why?" is more like "We don't know but that is what we observe".

Yes you can ask enough why questions of every domain of science until you bottom out at the answer which is "well, that's just how it is"

This is all fine, but it basically concludes that you need to understand some fairly complex maths to get the answer. And the OP wanted an answer that didn't do that.


text:
Comparison of stability of aromatic compounds
What is the basis of comparison between two aromatic compounds having different ring structures
Like in propyl and heptyl rings

There is no point in comparing the stability of two entirely different things.

Do you mean comparing how much "aromaticity" it has? The degree of aromaticity can be evaluated either through computations or hydrogenations; it represents how much more stable a compound is due to aromaticity (and thus how much more energy it takes to hydrogenate it compared to an "equivalent" non aromatic compound)

To put Ivan's comment differently, how do you compare the stability of an electron and the stability of nothing?

You have to take ring strain into account. If you analyze the energy of an organic cycle as a function of the number of atoms in the ring, you find out that cyclohexane and cyclopentane rings are most stable, because the internal angles of an hexagon (120°) and a pentagon (108°) are closer to the angle in a tetrahedron (109°28').


text:
Does LPG (cooking gas) react with water?
If cooking gas (or LPG) was exposed to some water in the gasline system, what would be the resultant by-product, and would this induce flame into the cylinder for explosion?

LPG is a mixture of hydrocarbon gases (propane and butane) and the mixture will have non-polar character. Water is polar. Many chemistry students learn that non-polar can not react well with polar substances (and vice versa), thus you will not observe like chemical explosion like you wrote. If they are not react, you might want to check the fact about their solubility. From there, we know that the mixtures are slightly soluble in water, it means that there is tendency that the water will evaporate (and the evaporation does depend on the temperature of the system).

@làntèrn While your comment might be useful for OP, it is inaccurate. Even quite low solubility may be enough; not to say that substances can react on interfaces between the phases. Also in specific conditions H2O and  hydrocarbons may react - in vapor phase, at very high temperature, when O2 isn't present.

Everything that comes out of a refinery has been exposed to water and contains residual water in the ppm range or lower. A menial job in a refinery is to occasionally open valves on the storage tanks to drain water ( that has separated from the product) into the refinery sewer system. And corrosion inhibitor is added to products going into pipelines to prevent corrosion of the steel pipe by water continuing to separate from the hydrocarbons

@Mithoron Well, the OP doesn't tell about the temperature and I know the temperature does impact on interfacial phenomena of the mixtures. I do learn it but given the OP information, it is not suffice enough to explain the outcome of the system. Perhaps it is not too good to elevate one thing to some extents.

No. It produces water after combustion, so it does not react with it.


text:
Why do we use coke instead of coal in order to reduce FeO as basic reaction in ironmaking process?
Why is coke used as a reducing agent to reduce FeO to produce iron instead of coal.I admit that coke is carbonaceous but what is it that compels us to use coke instead of the naturally available coal?

coke burns hotter than coal.

When coke burns, it leaves little ash or smoke, it contains less impurities, and as @MaxW said, it produces a higher temperature.

The single most important factor is strength ( mechanical compressive ); coal is heated to make coke, the resulting coke is stronger than the original coal. Also, coke helps to make the charge of iron oxides and limestone more porous to permit gas flow up and droplets of liquid iron and slag down. The coke oven heating drives off volatiles from the coal, which are a valuable source of chemicals, and contain much of the undesirable $\ce{S}$ and $\ce{P}$, which would otherwise go into the pig iron. Minerals like silica (ash) remain in the coke but they are easily collected by the slag. A bit off topic, but powdered coke and gasses are blown into the bottom of the blast furnace to provide more $\ce{C}$, which makes modern blast furnaces much more productive than 60+ years ago.

"Strength" is a bit dubious quantity. ;-) Coke reacts faster (with CO2 to form CO) because it is more porous, and thus always keeps a large surface area. The gas flow argument isn't really one.

If you are unfamiliar with a blast furnace think of a 20 ft diameter tube 100 ft high filled with dense gravel ( iron oxide, lime stone and coke) at temperatures approaching 2500 F . It is critical that this "gravel" not collapse and maintain porosity for very high gas flow up though the column. So the strength of the coke is critical.

The word "strength" means nothing. Strength of what? Mechanical? If you say "coke is more porous", that describes the individual coke particle (remember you are on a chemistry site here). You mean the bulk density and porosity of the coke/ore mixture, right?

You should edit your answer accordingly if you want my upvote. ;-)

@blacksmith37 Can u give a good synopsys of the BlastFurnace with mentioning prcoess where high pressure,sintering agglomeration etc are required?

USS book - "Making and Shaping of steel" is old but  the best source for all phases of steel production. Today there will be more use of sintered ore pellets but I expect the coke considerations have not changed.


text:
Applying Raoult's Law in volatile solids–liquid solutions
My textbook explains applying Raoult's Law for the two cases:

liquid–liquid solutions, in which the vapour pressure of the solution is due to both the components
non-volatile solids–liquid solutions, in which the vapour pressure of the solution is only due to the liquid component of the solution

I am curious as to what will happen (how things change for the application of Raoult's Law) if the solution consisted of some volatile solid like camphor, in some liquid (possibly some liquid hydrocarbon).
Certainly, the solid now also contribute to the total vapour pressure of the solution, but does it still block the escaping of the solvent molecules into vapour phase as much as the non-volatile solid blocks the liquid solvent molecules?
If yes, then how can I apply Raoult's Law for the case to find the total pressure of the solution?

Raoult's Law is only valid when the solute is non-volatile. Therefore, you cannot apply it.


text:
What instrument is used to detect heavy metals?
If i have blood sample, organ tissue sample or food sample, what analytic technique is used in laboratory to detect the ppm levels of heavy metals? What electronic intrument they use?

The best is probably Atomic Emission Spectroscopy

It depends. There is no real answer to such a general question.


text:
However what is so acidic about CaO and basic(no pun intended) about SiO2 while calculating basicity of slag?
I have seen Basicity to be calculated as $\mathrm{B} = wt\%\:\ce{CaO}/wt\%\:\ce{SiO2}$ particularly in slag bascity/acidity calcuations. Now I do not think that $\ce{CaO}$ is the most basic oxide that we have or $\ce{SiO2}$ is the most acidic oxide then why are they calculated accordingly? On a lighter note in my geology classes, I have been taught to classify minerals as acidic on the basis of Silica content. Are they related? Please explain with close reference to context.

The title question does not make much sense. Aside of that, it is not about being so basic ( high B ) nor so acidic ( low B ), but about being relatively basic or acidic. Extracts of rocks or soil with high(low) B have relatively high(low) pH.

@Poutnik But why $CaO$ and $SiO_2$ are only chosen?

It is an empirical parameter based on the major representatives of basic and acidic items, easy to determine.

What is so acidic...and basic...?

Nothing.
This derives from the historical perception that silica in geological systems and in melts was in the form of silicic acid ($\ce{H2SiO4}$) and the alkali and alkali earth elements were considered as bases.
We now know that in high temperature silicate liquids (that eventually solidifies into slag, or rocks, or glass) there is no acid–base chemistry, at least in the form that we know it from low-temperature aqueous acid–base chemistry. However, the usage of the terms acid and base in relation to rocks and slags still persists, primarily with older professors (who still teach) or in certain parts of the world, such as Russia. Nowadays geologists usually prefer the term "felsic" over the term "acidic" for silica-rich rocks.
Nonetheless, the measure of $\ce{CaO}$ and $\ce{SiO2}$ contents in slags, glasses and rocks is still useful for various reasons. Therefore, the ratio is still calculated and used, and the old name persists. A slightly better parameter, called optical basicity, takes into account the compositional variability other than $\ce{CaO}$ and $\ce{SiO2}$, by adding other oxides such as $\ce{Na2O}$ and $\ce{K2O}$ into the calculation. It is widely used and has many applications.


text:
Is the density of solute same as the density of solution?
This problem came up while solving some problems on mass percentage and mole concept.
Essentially, our teacher said that the $w/w\%$ and the $v/v\%$ for a given solution will be the same, because in order to obtain $v/v\%$ there will be a mutual division by density.
However, the density of solute is different from density of solution, right? For example, the problem we were solving involved $\ce{NaOH}$ dissolved in water. But $\ce{NaOH}$ as a compound has a characteristic density, different from the density of the solution. So $w/w$ and $v/v$ should have different values, right?
I'm not sure if I'm able to clarify what I'm thinking, but I need help with this nonetheless.

Maybe your teacher meant %w/v, otherwise no, it doesn't make sense.

Teacher said something wrong anyway

The density of the solution isn't needed for either wt/wt% nor vol/vol%.
wt% = (grams solute)/(grams solution)
vol% = (vol solute)/(vol solution)
For vol% it only really makes sense if mixing two liquids like ethanol and water.


text:
Why does a NOx container get cold when NOx is released? and What is the relationship of the vapor pressure curve with the liquid-vapor dome?
The temperature at which $\ce{NO_x}$ is in equilibrium with its liquid and vapor phase at 1 atm is about –84 °C. Does that mean $\ce{NO_x}$ exists as a liquid at that temperature? What about it's vapor phase then, because it's vapor and liquid phase should be at equilibrium, right?
Also, it's that the reason why compressed $\ce{NO_x}$ when released from its closed and pressurized container (e.g. 50 bar) at 20 °C becomes a gas while its temperature drops? Therefore, the compressed $\ce{NO_x}$ was in it's liquid state at 50 bar, right?
If that's the case, then what is the relationship between the vapor pressure curve and the liquid-vapor dome? How would you use the saturated liquid-vapor tables to predict the vapor pressure for your substance? And which specific volume would you chose to define your substance: its liquid specific volume or its vapor specific volume?

You should write about particular substances, like NO, NO2, N2O4, equilibrium 2 NO2 <=> N2O4, as considering particular behaviour or properties for NOx does not make much sense.

The vapor pressure curve describes the relationship between temperature and pressure for which the pure substance can exist as liquid and vapor, together in equilibrium.  So it is not necessarily all liquid or all vapor.  Both can be present in different proportions.  As far as specific volume is concerned, your tables should list both the vapor specific volume and the liquid  specific volume.  The specific volume of the combination of vapor and liquid is proportional to the mass fraction of each.  (So you need to know the mass fractions of vapor and liquid).

The temperature at which NOx is in equilibrium with its liquid and vapor phase at 1 atm is about –84 °C. Does that mean NOx exists as a liquid at that temperature? What about it's vapor phase then, because it's vapor and liquid phase should be at equilibrium, right?

Yes, the vapor and liquid can coexist, provided P and T are on the coexistence (!) line (as already explained in another answer).

Also, is that the reason why compressed NOx when released from its closed and pressurized container (e.g. 50 bar) at 20 °C becomes a gas while its temperature drops? Therefore, the compressed NOx was in it's liquid state at 50 bar, right?

No, the reason it cools when depressurized has to do with the adiabatic nature of the expansion. For an adiabatic expansion of an ideal gas, the final and initial pressures and temperatures can be related using the following expression: $$PT^{-C_{pm}/R}=\text{constant}$$
By the way, 50 bar is not quite enough to bring NOx to the coexistence line at $20^\circ C$, at least according to data tables in reference 1, which say 50.6 bar is required.

If that's the case, then what is the relationship between the vapor pressure curve and the liquid-vapor dome?

The region under the dome defines the liquid-vapor coexistence region, and is usually represented as a 2D projection, say in P-V coordinates, of the higher dimensional phase diagram. The coexistence line in a typical 2-D P-T diagram represents a projection of the dome in 3D P-T-V space in which the V-dimension is collapsed.

How would you use the saturated liquid-vapor tables to predict the vapor pressure for your substance?

At a temperature of interest, the (saturation) vapor pressure is given by the saturated vapor pressure indicated in the table.

And which specific volume would you chose to define your substance: its liquid specific volume or its vapor specific volume?

As another answer explains, the observed specific volume (on a point within the dome) is an average of the saturated vapor and saturated liquid $V_g$. The specific volume of the gas is likely more responsive to the position on the coexistence line, since liquids are typically only slightly compressible. According to the ideal gas law, $$V_g=\frac{RT}{MP}$$
References

Thermophysical properties of nitrous oxide (2004). ESDU 91022; ESDU Series on Physical Data, Chemical Engineering and on Physical Data, Mechanical Engineering. ESDU International plc, 27 Corsham Street,London N1 6UA.

Every gas at decompressing cools down (adiabatic expansion).
See for example:
https://physics.stackexchange.com/questions/14140/why-does-the-gas-get-cold-when-i-spray-it

Note entirely true: https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect


text:
How to convert 1,4 pentadiene to 1,3 pentadiene?
Are there any 'conjugating agents' to accomplish the above transformation? If not, then how to proceed?

Deprotonation with nBuLi then quenching of the anion at -75 will produce a mixture that is majority the conjugated diene.

My first instinctive answer would be to use a precious metal catalyst. Palladium usually displays this sort of activity. However, this article should cover most methods from basic and metallic to photochemical.
https://www.thieme-connect.de/media/10.1055-b-003-125756/lookinside/10-1055-b-003-125756_chapter012-1.jpg