text:
How many ions in 1.00 mol of NaCl
it's my first week of IB chem and I'm already confused. We learned how to find the number of atoms, but not specifically ions... is it the same? Similar? HELP

Is it in an aqueous solution?

@can-ned_food That's is completely irrelevant.

You want to use the stoichiometry of the compound and Avogadro's number.

@Zhe With all respect due, it is not irrelevant if it helps us understand the asker's confusion.  It seems like there at least some here.  This is someone pursuing an educator's curriculum, not a chemist's.

@can-ned_food That specific issue doesn't help the OP answer their question, nor does it change the answer. It would be great if we didn't ask questions that may further confuse the issue by introducing variables that are not relevant to the problem at hand.

$1$ $\ce{NaCl}$ contains $1$ $\ce{Na+}$ and $1$  $\ce{Cl-}$.
$1$ mol $\ce{NaCl}$ contains $1$ mol $\ce{Na+}$ and $1$  mol $\ce{Cl-}$.
$1$ mol ions = $6.02*10^{23}$ ions.
$1$ mol $\ce{NaCl}$ contains $6.02*10^{23}$ $\ce{Na+}$ and $6.02*10^{23}$ $\ce{Cl-}$.


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How do I work out the original concentration of a solution after taking an aliquot and further diluting it for analysis?
I'm trying to work backwards to obtain the original concentration of a 1 litre solution. I feel my calculations have gone wrong and need another set of eyes to have a look.
To start, a tablet containing arginine was dissolved in water and made up to 1 litre solution. A 250 μL aliquot was taken from the solution and diluted to 25mL.
Analysis was done and a concentration value for the 25mL sample was 4.25 x10^-5 mol/L.
When I tried to work backwards from that concentration value and determine what the original concentration is in the 1 litre solution, I got 17 mol/L.
Which seems excessively high for what was suppose to be a tablet.
I then need to give the value of mass of arginine (172.2 g/mol) in the tablet, but using the 17 mol/L, gives a value ~3kg.
What do I need to do to get the concentration value for the 1 litre solution?

You took an aliquot of $0.25 ml$ with unknown concentration.
You diluted 100 times by adding 100(-1) times the volume of your aliquot.
The result was $4.25*10^{-5} mol/L$, or,
$$4.25*10^{-5} mol/L * 172.2 g/mol = 7.38*10^{-3} g/l$$
You diluted 100 times, so going back it becomes 100 times more concentrated:
$ 100*7.38*10^{-3} g/l = 0.738 g/l$.
A good habit, even if you're not stuck, is to write the units corresponding to the values. This way it is very easy to spot when you've for example multiplied by a molar weight you should have divided by.

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What is the pressure point where deviations from ideal gas law occur?
I read a lot on deviations from ideal gas law due to low temperature and high pressure and I understand it, but I am curious to what point (the value) does a pressure is considered HIGH pressure?

Let's just say that high pressure means $P \to ∞$ .

It depends on the gas. If the molecules have strong interactions with each other you can see measurable deviations at pressures around 5 atm. But if you have a "more ideal" gas you may have to pressurize it to much higher pressures. There is no set value.

Calculate the reduced pressure (pressure divided by critical pressure) and reduced temperature (absolute temperature divided by critical temperature), and then see a plot of compressibility factor z as a function of reduced pressure and reduced temperature (in every thermodynamics text book).  If z differs from 1.0 by more than about 10 %, the ideal gas law will not give very accurate results.

When the non ideal gas reaches Boyle temperature it acts like a ideal over a range of pressures. It is at this temperature that the attractive forces and the repulsive forces acting on the gas particles balance out.

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Halogens and Equilibrium

Suppose 1 mol of $\ce{KI (aq)}$ is mixed with $\pu{1 mol}$ of $\ce{I_2^*}$, where $\ce{I_2^*}$ represents a di-iodine molecule which has been radioactively labelled, and the substances are allowed to react for an infinite amount of time. Assume radioactive labeling does not affect reactivity of the atom.
i) Predict the identity of all species in the mixture after an infinite amount of time
ii) Predict the number of moles of all species in the mixture after an infinite amount of time.

For part i), my answer was that after an infinite amount of time, all radioactive tracer decays and hence $\ce{I_2^*}$ becomes $\ce{I_2}$. Then following from the triiodide equilibrium equation ($\ce{I_2 + I^- <=> I_3^-}$), we have that the final species are $\ce{K^+, I_2, I^-, I_3^-}$.
For part ii), after an infinite amount of time, equilibrium is achieved for the triiodide equation, hence the rates of the forward and backward reaction are the same.
Now at the beginning we have $\pu{1 mol}$ of $\ce{I_2}$, $\pu{1 mol}$ of $\ce{I^-}$ and $\pu{0 mol}\ \ce{I_3^-}$, and at the end we have $1-x\ \mathrm{mol}\ \ce{I_2}, 1-x\ \mathrm{mol}\ \ce{I^-}$,  and $x\ \mathrm{mol}\ \ce{I_3^-}$. The number of mols on both sides of the equation are the same (my understanding of equilibrium) and hence $(1-x) + (1-x) = x$ gives us $x = \frac23$. Hence there are $\pu{1 mol}\ \ce{K^+}, \pu{\frac13 mol}\ \ce{I_2}, \pu{\frac13 mol}\ \ce{I^-}, \pu{\frac23 mol}\ \ce{I_3^-}$.
My question is, is my reasoning for the above two questions correct? (especially the equilibrium part, I'm not too certain about that). If it is not, then could I ask what the correct way to approach the above question would be?


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What is the difference between a time of flight mass spectrometer and a spectrometer that uses a magnetic field?
Traditional mass spectrometers used electromagnets to deflect isotopes of different masses which allowed only a certain isotope to reach the detector. But how is the TOF MS different besides the fact that it doesn't use magnets? Does it have a different method of accelerating the ions which leads to them moving with a different velocity? If the methods of acceleration are same wouldn't there initially be no need for magnets to deflect the ions? Or is it the detector that works differently?

Both types of mass spectrometers separate different chemical species based on the mass-to-charge ratio.
The time-of-flight mass spectrometer (TOF MS) is based on the ions being released at a given instant, such as when struck by a laser, accelerating the ions electrostatically and measuring how long it take for them to strike the cathode target. This is ideal for a lightweight interplanetary or cometary mission such as Rosetta.
Magnetic and quadrupole MS work by bending the beam magnetically  or electrostatically so as to spread out the target beam along the width of the target. This is like sending a light-beam through a prism to spread out the spectrum. An advantage of this technique is that it can resolve multiple species simultaneously, but at the expense of fine resolution.
An excellent explanation of TOF and other MS is presented by Young, Waite and Miller.


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When we add soap in low concentrations to water it reduces the surface tension. Why? What is the purpose of this?
I understand how soaps lower the surface tension of water but I don't know what the purpose of this is. How does this help with cleaning, removing stains etc?

The lowering of surface tension is a side effect, though it does relate to the reason why we use soaps when cleaning things.
Many of the things we want to clean are not naturally soluble in water (e.g. things like vegetable oil). The structure of soaps and detergents usually consist of molecules with two parts: one is water-loving and one is oil-loving. When soaps are added to mixtures of oily things and water, they align so the oil-loving parts of the molecules surround the oil and present only a water loving surface to the bulk water. Thus the oil is made much more soluble than it would be otherwise (so it is possible to remove it from your clothes, hands or dishes despite most of the liquid being water).
In reality the situation is quite complicated but the general principles of how soaps and detergents work is captured by this basic view.
The effect on surface tension is related to this. The soap molecules tend to cluster at the surface of the water (to get their oil-loving ends away from the water environment). This breaks up the surface structure of water molecules that normally create strong surface tension. But this effect is basically the same effect that enables soaps to solubilise normally insoluble things in bulk water.

Is the formation of micelle a chemical change?

@Anton Depends what sort of chemist you are: colloid chemists might say "yes"; transition metal chemists might say "no" or possibly "what?" ;-)


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Synthesis Golf V: Denopamine
A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

This fifth round of golf concerns the synthesis of (−)-α-(3,4-Dimethoxyphenethylaminomethyl)-4-hydroxybenzylalcohol (denopamine), a synthetic drug indicated for use in the treatment of angina.

The challenge is to propose a route to denopamine. You may start with anything that has less than 10 carbons, but it must be available in the Sigma-Aldrich catalogue. A method should also be included to set the oxymethine stereocentre (no buying it!).
Given the relative simplicity of the target and the need for large quantities of the drug, consideration to step count and scalability should also be taken.

Sorry (again) for this not being quite to schedule- the last few weeks of summer are always hectic. An easier target  this time so hopefully easier to do on the back of an envelope to get more people involved.

Given that this is the schedule, I’m really happy I got my plaunotol answer in before it was too late.

OK, new route. Retrosynthesis:

Denopamine 1 can be made by a reductive amination between amine 2 and aldehyde 3 (either of the two reductive amination disconnections are fairly obvious choices).
The requisite amine 2 contains 10 carbons and so it becomes necessary to make it using some other means - in this case, reduction of a nitroolefin 4, which can itself be made via a nitroaldol reaction from 3,4-dimethoxybenzaldehyde 5 and nitromethane 6.
The chiral α-hydroxyketone 3 can probably be made in numerous ways. Earlier I proposed organoocatalytic oxygenation with PhNO, but another possibility is to have an addition of "formaldehyde anion" to 4-hydroxybenzaldehyde 7. Nitromethane 6 was chosen as the acyl anion equivalent, as the asymmetric Henry reaction is well-established methodology, e.g. by the Evans group (J. Am. Chem. Soc. 2003, 125 (42), 12692–12693).
The forward synthesis is fairly straightforward:

5 steps from cheap starting materials, with longest linear sequence being 3 steps.

You might want to choose a different reduction for the nitrostyrene, in my experience LiAlH4 reduction is quite low-yuelding. Hydrogenation over Raney Ni in the presence of NH3 might be better

I was going to post something similar, definitely hydrogenation for the nitromethane product. I was going to do a Jacobsen oxidation on 4-acetoxystyrene to give a chiral epoxide. Hydrolysed in last step. BTW, Ti chlorides are a pig to work up on scale.

Sorry for the mistake in the drawing, the synthesis does NOT start from stryene rather the 4-hydroxylated derivative 4-vinylphenol, it does not affect anything throughout the synthesis later on.

5 Steps Total -
Longest Linear Sequence of 5 -
(1) - Asymmetric Sharpless aminohydroxylation:
OsO4, Chloramine, tBuOH, heat, ligand
(2) - Azide synthesis via diazo transfer :
Imidazole-1-sulfonyl azide hydrochloride, CuSO4, K2CO3, MeOH
(3) - Deprotection (not shown)
(4) - Aziridination through nitrene
Ethylene, heat
(5) - Ring opening using aromatic
1,2-dimethoxybenzene

Sharpless aminohydroxylation puts a protecting group on N - the reagent is $\ce{TsNCl^-Na+}$ - so you need one more step to get rid of it.

That is fair, but alternatively I could do something like azidohydroxylation which is 2 steps in place of the aminohydroxylation + diazo transfer, I picked the latter due to the scalability aspect but disregarded the point you brought up, when I first thought of it.

A synthesis using Sharpless dihydroxylation for stereoinduction:

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Choice of preferred IUPAC names
When looking up names for simple organic molecules, the idea of a preferred IUPAC name (PIN) often comes up.
My understanding was that the PIN was meant to provide a unique and unambiguous name for a given molecule based upon strict applications of the IUPAC guidelines for nomenclature.
Whilst some PINs are clearly applications of these formal IUPAC guidelines, other PINs are trivial names. To give some examples:

Acetic acid (PIN, but trivial name) rather than ethanoic acid (systematic
IUPAC name)
Propan-2-one (PIN and systematic IUPAC name) rather than acetone (accepted but trivial)
Phenol (PIN, but trivial name) rather than benzene (systematic IUPAC name)

Clearly most people still refer to these molecules as acetic acid, acetone, and phenol; but it seems arbitrary that some trivial names are also the PIN whilst others are not (that is not to say that the other names are forbidden, as pointed out in the comments, with reference to P-10 of the recommendations).

Why do we need the PIN, and why does it not always conform to the strict application of the IUPAC guidelines?

Writing this as the idea of a PIN comes up in every nomenclature question and it seems redundant to keep on explaining it every time. Couldn't find a duplicate so possibly hasn't been explicitly asked previously

IMO it should be mentioned that the PIN is not necessarily the only correct name for a compound, cf. P-10: "Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a 'general IUPAC name' [...]" As stated in P-64.2.1.2, "acetone" is acceptable as a general IUPAC name, but is not the PIN.

I think I have something more useful to say. I might have hinted at it in my answer, but I was not very clear (to be fair, my thinking wasn't very clear either). The "misconception" is the idea that the PIN is meant to be obtained systematically, which isn't true. The process is more like: (1) develop a systematic name (2) check if there is a trivial name (3) if there is more than one choice, arbitrarily pick one as the PIN. Does that answer the question better?

@orthocresol- I thought your current answer was pretty well explained (arbitrary is fine, even if it might not be the answer we are looking for").  The steps in your comment are logical, and certainly answer the question, but its hard to verify if that is indeed the case (I'm not sure to what extent they took historical usage into account / chose where to fight their battles and left certain things untouched).

Yeah, that's why I sort of hedged in the answer. I'm unsure if it's possible to ascertain what their motives were. Hopefully somebody can shed some new light on the matter, but if there isn't anything else, then maybe Occam's razor should just be invoked...

Quoting the Introduction (P-10) of the IUPAC Blue Book (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013):

[...] We have developed rules for assigning 'preferred IUPAC names', while continuing to allow alternative names in order to preserve the diversity and adaptability of the nomenclature to daily activities in chemistry [...]
Thus the existence of preferred IUPAC names does not prevent the use of other names to take into account a specific context or to emphasize structural features common to a series of compounds. Preferred IUPAC names (PINs) belong to a 'preferred IUPAC nomenclature'. Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a 'general IUPAC name', in the context of a 'general IUPAC nomenclature'.

Therefore, there may be multiple names for a compound which are allowable in IUPAC nomenclature, one or more of which may be a trivial name (or derived from one). Note that there is no stipulation that the PIN must be the systematically obtained name. So, for a compound with a trivial name, either the systematic name or the trivial name may be chosen as the PIN. How does IUPAC choose which one is the PIN?
It might not be the answer we are looking for, but the end of the day, I strongly suspect it is simply arbitrary. The examples you have cited are very illustrative. Here is one more inconsistency which, to the best of my knowledge, is arbitrary.
"Acetic acid" is a PIN that may be modified by substitution (P-65-1.1.1), but only to a limited degree (P-15.1.8.2.1). Therefore "bromoacetic acid" is also a PIN. On the other hand, acetylene is also a PIN; however, substitution of any kind of not allowed (P-31.1.2.1). So, instead of bromoacetylene, the PIN is bromoethyne. (Bromoacetylene, however, remains acceptable within general use. All the names listed below that aren't followed by "(PIN)" are acceptable within general use.)

What makes IUPAC decide that - for PIN purposes - one trivial name is substitutable, whereas another is not? Or in the examples you have cited, why is one trivial name the PIN whereas another isn't? Of course, the rules state that it must be so, but to the best of my knowledge, there isn't any explanation of the philosophy behind the choice of PIN. Maybe I'll be proven wrong, though, and I'd be glad to see it.
To be honest, if you could come up with one, I'd be surprised. It's very tough to draw a line. Maybe we can agree that "acetone" should be a PIN, but should "neopentane" be a PIN as well? How about "estrogen"? Someone has to just arbitrarily choose where to draw a line, and it just so happens that IUPAC has gone with something stricter that doesn't use "acetone" as a PIN.
Anyway, just to reiterate - many of these names are perfectly valid IUPAC names. It is just that, for one reason or another, IUPAC has not selected them as preferred IUPAC names.


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Can insect repellents kill insects?
One day, there was a fly in my house. I really hate insects flying around me, but I don't want to catch it with my hands, so I sprayed some insect repellent directly onto the fly to try to get rid of it.
This is the insect repellent that I used.

Sticker reads:

"Wirkstoff basierend auf Zitronen-eucaluptusöl"

"Active ingredient based on lemon-eucalyptus oil"

When the insect repellent hits the fly, it slowed down and stopped flying, but it's still moving slowly on the window, so I applied the insect repellent again.
Eventually, the fly completely stopped moving, so I assume it's dead. Does the insect repellent actually killed the fly, or just stunned it?

I think it's only repellent and isn't for spraying insects directly. You don't have fly swatter?

This Wikipedia page on "Insect repellent"  pretty much says everything in the first two paragraphs [ https://en.wikipedia.org/wiki/Insect_repellent ] ;)

According to the sticker on the cap, the active ingredient is "Lemon-eucalyptus oil". I don't think that's insecticidal (but certainly an insect repellent). Of course, too much of anything would kill anything...which means your fly is probably dead ;) (What's written on the can isn't really important...)

There was some foam in the insect repellent, so there may be some surfactant.

I mean, it probably could disorient it enough

This Wikipedia article on "Insect repellent" pretty much answers your question.

An insect repellent isn't necessarily the same as an insecticide.
The repellent, well, "repels" insects; i.e- it discourages them from approaching you (or whatever you sprayed the repellent on) for as long as the chemical(s) linger. These (particularly those based on essential oils) aren't much of a health hazard to other animals.
An insecticide is actually designed to kill insects ("No sh!t Sherlock") rather than simply keeping them at bay. The term "insecticide" refers to substances that kill insects/arthropods at very low concentrations ( "very low" is subjective... I don't think there's an authoritative, precise definition for "insecticides" anyway).

FYI, any substance (including common stuff like air and  water), supplied in excess (again, "excess" is also highly subjective) can kill. Ever heard of "Too much of anything can kill"?
The opposite effect is something called "Hormesis"; where just the right amount of anything (including radiation and toxins) can have a positive effect on you.

Back to question.
The insect repellent you're using contains (according to the sticker on the cap) lemon-eucalyptus oil, whose active ingredient is p-menthane-3,8-diol (and its geometrical isomers). Also, it's supposed to be used against mosquitoes (Mücken) and ticks (Zecken).
It serves as a deterrent, on account of how strong it smells.
Ectoparasitic insects, such as the said mosquitoes and ticks, rely on (to a large extent) their sense of "smell" to locate hosts. So spraying around strong odorants will do a good job of interfering with the bug's ability to track you down.
Remember my "anything can kill" rant from earlier? Well, the same goes for your repellent. If I were a mosquito or a tick (or whatever insect for that matter), I doubt I'd make it out alive if you decided to smother me with some godforsaken liquid/aerosol. For all I know, your fly could've died because it suffocated, rather than being poisoned.
Heck, you could achieve the same effect if you decided to drown the fly in deodorant or even perfume.
So there isn't really much behind the mysterious death of your fly; in all likelihood it suffocated (You murderer!). Take @Mithoron's advice; a fly swatter works better.
If you want to know what biochemical processes the repellent would tamper with in a bug, then that's a question for the Biology.SE. ;)

I heard that surfactants can block the pores that insects use for breath. Maybe that's how it killed the fly.

@Zen Maybe, maybe not. Surfactants (if added) are present in minute quantities. The repellent (as a whole) is a liquid... there's really no point in trying to specify what blocked the spiracles (the "pores" insects breathe through) since you're dealing with a solution (homogeneous mixture of multiple liquids). If this answer helped, you can up-vote it. Also, you can consider ticking the "Accept answer" button. Cheers!

Even if it didn't actually killed the fly, it did certainly stunned the fly because it completely stopped moving after few more sprays.

@Zen Yup, it's clear that death wasn't instant (the first few sprays incapacitated/"stunned" it)...and  I'm also pretty sure the fly didn't get back up afterwards ;P

Today, another fly has appeared, so I had another chance to test my insect repellent. It appeared at afternoon. I applied the spray to the fly. At first, it has convulsions, but then it completely stopped quickly. It's night now, and it's still there, so I am sure it's dead.

@Zen Your zeal for conducting field experiments is laudable (well, you probably might not think of it as an "experiment", but it is). Just a pointer though, using repellent to kill flies is a waste, you should settle for either a swatter or (commercially available) insecticide sprays ;P

it does kills the fly, so I don't think it's complete waste


text:
Can you literally separate air from soil?
My teacher poured water over some soil and said that the air bubbles prove that soil contains air. However I was wondering if there is a more direct approach to proving that soil contains air, like explicitly separating one from the other. Is there an experiment like this? Also, how would the air look like? I've never exactly "seen" air, so how does one know when the air has been separated? Maybe by making it a liquid (liquid nitrogen)?

Well you see, soil is a porous mixture of a lot of things. Porous substances have small pockets in them. Air fills those pockets. So you can't remove the air from soil, unless you take the soil sample to an airtight box of course.

This is a great question.  Not sure why it has down votes.

Gases spread to fill the container. If a substance has holes, air will fill it.

@CurtF. I have no idea why you'd think so.

Your teacher's method of pouring water over soil is an excellent way to demonstrate that there is air in the soil. If you're still unconvinced, it's possible to directly separate the air out. Here's how:
You fill a vacuum chamber with your soil. All of it. So it's all soil and no free space is left. Now you just pump the air to a different chamber. There are several ways to do it so I will not elaborate on the mechanics. This method extracts the air from the intergranular porosity (i.e. the space between the soil grains) to your other container.

Also, how would the air look like?

Well, it doesn't look like anything, because it is after all air. It is possible to liquefy it by cooling or compressing. It's also possible to measure the composition of the gas you extract out of it and see that it matches the composition of air. It will not be exact, depending on the soil. Some soils have decomposing organic matter (dead bacteria, dead plants, dead other things) so they might have slightly less oxygen and slightly more carbon dioxide, methane, sulfur dioxide, hydrogen sulfide, and a bunch of other things that usually also smell.

It would probably be better to have water on top of the soil as you evacuate. Then you can evacuate the air, and open the chamber again, no air in the soil. Otherwise, air will just enter the pores again at the same rate it left. A bonus is also that you can visually see when the air has been evacuated - "boiling" the water. Of course, if you have a really hard vacuum you need something else than water, since water will actually boil at room temperature somewhere down in the single mbars.

As I said, I didn't want to go into the mechanics of it because there are many ways to do it, and it wasn't the point I was trying to make.


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Diazonium ion positive charge
I am confused about the nature of the positive charge on the nitrogen atom in the diazonium ion. Where does it come from? That nitrogen atom has 4 bonds, but I cannot deduce whether one of them is dative, because the other nitrogen forms 3 normal bonds, and the carbon in the aromatic ring has no empty orbitals or lone pair of electrons. So how does that nitrogen form 4 bonds and how does it have a positive charge? Thanks

In a diazonium, or rather any compound  where nitrogen forms four bonds, the nitrogen has a positive formal charge.
The reason is simple, nitrogen used it's lone pair of electrons to make the fourth bond. This gives nitrogen a positive charge.
A more simple example is the ammonium ion. While ammonia has 3 bonds and 1 lone pair, the ammonium ion has 4 bonds, the lone pair from ammonia was shared with a proton to make the ammonium ion.

Thanks! I didn't read much about the concept of formal charges so I'll have a look at any page about it.

@FahdAl-Seba'ey Sure thing. If you think my answer is satisfactory, click the accept button. This lets everyone know you are satisfied and you got the answer you needed.

Note that formal positive is not the real positive. The entire species is positive, but the real positive charge is delocalized over the whole thing.


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Possibility of π–π stacking in non-aromatic systems
Is it possible to have π–π stacking for non-conjugated double bond systems?
For example, can the carbonyl groups of two urea molecules participate in π–π stacking? Although urea prefers to stay in hydrogen bonded form.
I would be really happy if you cite some resources (research paper, book, etc.) while answering.

There's quite a conjugation in urea.

What you describe would be something very different than ordinary pi-stacking, just like hydrogen bonding is not a simple polar interaction.

https://en.wikipedia.org/wiki/Stacking_(chemistry)#Requirement_for_aromaticity

@Mithoron I did not want some other type of interaction. I just wanted to know if it possible to have $\pi$ - $\pi$ stacking in non-conjugated systems.

It is possible in non-aromatic molecules , but p orbitals are needed. See the answer to this question. https://chemistry.stackexchange.com/questions/60055/pi-electron-stacking-how-does-it-work/60095#60095

There is a paper titled "Amide-pi interactions between formamide and benzene",  doi: 10.1002/jcc.21212. It has been cited by a paper on a crystal structure on a urea transporter (the interaction is between a peptide bond of a ligand and aromatic side chains of the protein).
Technically, the interaction between the arginine side chain and aromatic groups could also fall under your heading, but it is usually described as a cation pi interaction.


text:
Disproportionation Reaction
This is the problem I am trying to balance for redox:
$$\ce{HXeO4 -> XeO6^{4-} + Xe + O2}$$
It is supposed to be in a basic solution. However, my answer does not contain $\ce{OH-}$ on either side:
$$\ce{4HXeO4 -> XeO6^{4-} + 3Xe + 4O2 + 2H2O}$$
I am wondering if I had made a mistake or if it's ok for the reaction to not contain $\ce{OH-}$ ions.

The reaction is not balanced.  When there are charges you have to balance them too.  Try adding hydroxide ions ("basic solution") to the reactants and use them to balance the charges.  You'll need to rebalance the elements of course.

I thought this thing looked weird. The answer confirms that there is no neutral $\ce{HXeO4}$.

The disproportionation reaction of xenates to perxenates in alkaline medium is given in the Wikipedia article of xenon tetroxide. Xenon tetroxide is made on the basis of disproportionatation reaction of xenate salts and further reaction of perxenate salts being  acidified forming unstable perxenic acid, hydrolysing to form xenon tetroxide.

All syntheses(of xenon tetroxide) start from the perxenates, which are accessible from
  the xenates through two methods. One is the disproportionation of
  xenates to perxenates and xenon:
$$\ce{2 HXeO4^− + 2 OH^− → XeO6^{4−} + Xe + O2 + 2 H2O}$$


text:
How to identify where a bond will break and where a bond will form (acetals)
In the following reaction:

The $\ce{H}$ from $\ce{ROH}$  joins the $\ce{O}$ in the carbonyl.
So the Alcohol breaks at the bond between the $\ce{R'O}$ and the $\ce{H}$.
How do I know this occurs, and that the alcohol doesn't break at the $\ce{R'}$ and the $\ce{OH}$ bond (and then the $\ce{OH}$ joining the $\ce{C}$, and the $\ce{R'}$ joining the carbonyl $\ce{O}$)?
Is it because the electronegativity difference between the  $\ce{H}$ and the $\ce{O}$ is greater than the electronegativity difference between the  $\ce{R'}$ and the $\ce{O}$?

I find helpful to look at pKa values. Many of the most common bonds in organic chemistry can be found in a pKa table. Another thing that may help are looking at what constitutes as a good nucleophile, or a good leaving group


text:
How to separate paracetamol and caffeine?
How do you separate paracetamol and caffeine using high-school laboratory equipment?
I was thinking about using the fact that paracetamol has a lower melting point (169 °C) compared to caffeine (around 235 °C): I would heat them up till paracetamol melts while caffeine is still solid. What I can't figure out is how you filter the caffeine from the liquid paracetamol (filter paper would burn at such a high temperature).
I would like to know how you do the filtration at high temperature; or, if you have a better method, please let me know!

Caffeine is basic, paracetamol acidic. Any clues?

@Waylander Would you like to elaborate on your idea as an answer?

The answer that my teachers had in mind was column chromatography. Essentially we use the van der Waals force of silicon gel to do an elution of the caffeine / paracetamol solution dissolved in water.
More details can be found here:
http://www.chemguide.co.uk/analysis/chromatography/column.html

I will mark this as accepted since there are no other answers.


text:
Can iodine be used to disinfect water?
I think it was Nat Geo.. Not sure exactly.. Where I heard, years ago, that a drop of iodine can be used to disinfect water of say, 1 litre.
But isn't chlorine used to kill all bacteria? I mean I don't completely remember that documentary but I'm pretty sure it was  iodine that they used.
Can iodine be used or does my memory fail me?

"Purify" is the wrong word. You don't purify water by adding more impurities to it. Say "disinfect" instead. Then yes, iodine can be used to that end, as well as chlorine.

As @IvanNeretin states, iodine, as well as chlorine and bromine, are effective disinfectants. See https://www.amazon.com/Coghlans-Emergency-Germicidal-Purification-Treatment/dp/B01LXIGV93/ref=sr_1_11. These halogens, as well as ozone, kill microorganisms though oxidation. [Fluorine is too toxic and corrosive to use for this.]

Short answer: Both chlorine and iodine in specific concentrations are used for water disinfection.
Iodine has been used to disinfect water for a long time. It is used in concentrations between 2.5 ppm to 7 ppm, but there is some controversy regarding the maximum safe dose of iodine because excess of iodine consumption in humans can lead to various diseases(thyroid disorders being very common). I believe this is the reason we have moved towards more safer options like chlorine for water disinfection.

What dosage of iodine do you recommend then? You've mentioned it's 2.5ppm to 7 but I regret I'm no pro and don't know into how many drops of iodine it translate into for 1 ltr... Thanks!


text:
How many peaks of HCl are found in a mass spectrum?

In a mass spectrometer some hydrogen chloride molecules will split into atoms. Chlorine has two isotopes. The hydrogen involved here is the isotope $\ce{^1_1H}$ only.
How many peaks does a mass spectrum will show?

My answer was two with $\ce{H^35Cl}$ and $\ce{H^37Cl}$ but I found the answer to be wrong. Where have I been deceived?
I also found the mass spectrum in CIE 9701/02 2003:

Why are there peaks for $m/z = 35$ and $37$? Even if I consider them to be chlorine's, why aren't there any for hydrogen?

You missed the very first sentence in your question.

I didnt missed it, if thats the situation why isnt there any for hydrogen's?

Oh, right, I didn't understand what your last sentence meant... anyway hydrogen would appear at 1, and your graph is cut off between 0 and 35.

Collision between an electron and the molecule during a MS-experiment usually results in two phenomena:
\begin{align}
\begin{cases}
\ce{AB + e &-> [AB]+ + 2e} &\text{electron ionization}\\
\ce{AB + e &-> [AB]-} &\text{resonance capture}
\end{cases}
\end{align}
Resonance capture is less frequent and usually doesn't add much to the spectra. For $\ce{HCl}$ considering electron ionization as the primary process the following process takes place:
$$\ce{HCl + e -> [HCl]+ + 2e}$$
Based on ionization method (e.g. applied energy) several fragmentation processes may accompany the formation of cation/cation-radical ($\ce{[HCl]+}$/$\ce{[HCl]^{+.}}$):
\begin{align}
\ce{[HCl]+ &-> H+ + Cl^. }\\
\ce{[HCl]+ &-> H^. + Cl+ }\\
\ce{[HCl]+ + e &-> H+ + Cl-}
\end{align}
Fragmentation is often followed by recombination acts of different degrees of probability, e.g.:
\begin{align}
\ce{H+ + Cl- &-> HCl}\\
\ce{H+ + HCl &-> [H2Cl]+}\\
\ce{H^. + [H2Cl]+ &-> [H3Cl]^.+}
\end{align}
One can possibly observe peaks of $\ce{[H]+}$ and $\ce{[H2]+}$, but those are usually have negligible intensities due to recombination processes.
Regarding the spectra OP provided, it looks somewhat strange as the base peak @ $m/z = 36$ is not $100\%$ (approx. $90\%$ instead). I would recommend to check for the experimental data in more reliable sources, e.g. in NIST database:

Considering the fragmentation processes and the fact that $\ce{^{35}Cl} : \ce{^{37}Cl} = 3 : 1$, one can perform the following peak assignment:
\begin{array}{rrr}
m/z & \text{r.i.}, \% & \text{fragment}\\
\hline
35 & 18  & \ce{[^{35}Cl]+}\\
36 & 100 & \ce{[^1H^{35}Cl]+}\\
37 & 6   & \ce{[^{37}Cl]+}\\
38 & 33  & \ce{[^1H^{37}Cl]+}\\
\hline
\end{array}
Alternatively, for $m/z = 37$ and $m/z = 38$ fragments $\ce{[^1H2^{35}Cl]+}$ and $\ce{[^1H3^{35}Cl]+}$, respectively, could be proposed, but it wouldn't be correct due to the following:

recombination acts leading to the formation of large molecules as detected end-products are less possible and usually result in much smaller peaks;
observed relative intensities are not in agreement with isotope distribution for chlorine ($\ce{^{35}Cl} : \ce{^{37}Cl} = 3 : 1$, which is in agreement with r.i. ratios for $\ce{[^{35}Cl]+} : \ce{[^{37}Cl]+} = 18 : 6$ and $\ce{[^1H^{35}Cl]+} : \ce{[^1H^{37}Cl]+} = 100 : 33$, but not with alternatively proposed fragments);
the problem itself states that

hydrogen chloride molecules will split into atoms.


text:
Is it possible to calculated the concentration of hydroxide ions from the known concentration of hydronium ions?
Chapter 15 (Acids and Bases) of Holt High School Chemistry:

Neutralizing $\pu{5.00 L}$ of an acid rain sample required $\pu{11.3 mL}$ of $\pu{0.0102 M}\ \ce{KOH}$. Calculate the hydronium ion $[\ce{H3O+}]$ concentration in the rain sample.

We can derive the required concentration by substituting into:
$$c_1V_1 = c_2V_2,$$
Can we, after getting the concentration of $\ce{H3O+}$) use the equation with the ionization constant of water
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]$$
to also get the $\ce{OH-}$ concentration?

Of course we can.

Yes, that works. In fact, some textbooks and teachers ask this type of question (asking for the lesser ion of the autodeprotonation of water) to test which students actually read the question and which gloss over details. I fell into the trap of calculating the wrong one and stopping too early, too.
So if your question had asked for the hydroxide concentration, you would indeed calculate the hydronium concentration first (because that is actually accessible) and then invert and multiply $10^{-14}$ to get the desired one as you suggest.

" you would indeed calculate the hydronium concentration first (because that is actually" Which balanced equation we can use in order to calculate $[\ce{H3O+}]$?

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


text:
How can it be determined that a liquid is saturated with a gas?
How can it be determined if a liquid is saturated with a gas?
For example, if I am introducing carbon dioxide to water at 40 PSI at 35F in a sealed container, how would I know at what point the water is saturated with the gas?

If the pressure of the gas above the liquid stays constant over time, the liquid is saturated with gas.

@aventurin That is true, but since the gas is being held at that pressure by a regulator, it is not useful. In other words, if we want to maintain a fixed pressure (which is 40 PSI in the example of the question), then there will be a regulator connected to the container which supplies the gas. The regulator ensures that the pressure remains at 40 PSI, so the pressure will not change.

OK. The duration of the saturation process will e.g. depend on the area of the surface between liquid and gas, if the gas is injected into the liquid or not, if you shake the container, if you stir the liquid, and many other factors. Measuring the concentration of $\ce{CO2}$ in water would be a little more challenging than just waiting. Standard methods include titration or gas sensing electrodes.

You could monitor a spectroscopic peak as the gas is added to the solution.


text:
Colour change of fluorescein in acidic and basic medium
I prepared fluorescein for my chemistry project in school and I realised that this molecule and its preparation steps are very closely related to that of phenolphthalein. I have studied briefly about the colour change in phenolphthalein in different mediums.
What I have thought is that the colour change is due to increase or decrease in density of quantised states due to increase or decrease in pi conjugation (Not sure about that though). I tried to observe colour change of fluorescein in strongly acidic and basic medium.
What I noticed was that in acidic medium, addition of small amount of fluorescein gave red solution whereas when the same amount was added in basic medium, it gave fluorescent green solution.

One form of fluorescein is shown below (1); in basic media it cyclises to form a lactone (2):

But I couldn't figure out how this relates to the colour. I thought that the acidic form 1 would have greater π-conjugation, the energy gap between the electronic states would be smaller, and light of a lower frequency (longer wavelength) would be absorbed.
However, the opposite is observed. What is actually going on?

https://en.wikipedia.org/wiki/Fluorescein - you drew wrong structure

I think it is one of its structure. I am sure that the structure in basic medium is the one i drew. Look http://sites.uci.edu/chem52labs/files/2014/03/M52LC_Experiment1S13.pdf

I really want to understand what is happening and if my concepts are wrong or not.  Any help will be appreciated. @Mithoron

The lactone form is not stable in basic medium, so the cycle opens and gives conjugation to flow and unite all four rings. The problem is you really swithched the acidic and basic forms of fluorescein, so the left one exists at basic pHs (except for the H atom at carboxylate - it goes away), and acid medium favors the existence of the right form.
You may check the forms of phenolphthaleine - they are pretty much the same.

Well you might be correct for the basic medium(i checked those phenolphthalein structures) but in the acidic medium it should be the one which is given because I used concentrated acid and when the same amount was added in a test tube containing phenolphthalein, it gave the red colour which has the similar structure

https://en.m.wikipedia.org/wiki/Phenolphthalein

well, that changes everything. Then we don't talk about form #2 in your question at all. I believe if you want the structures for this case, you'll need to look at structures of phenolphthalein that represent orange and pink colors.

Yes you are right. I will have to edit it again


text:
How to measure the amount of lactose in milk/dairy using high school lab equipment?
I am currently a high school student doing their Biology IB IA. For those who don't know, an IA is, in simple terms, an experiment that you design and execute yourself. My topic is "The effect of temperature on lactase activity in milk".
However, to determine the effectiveness of lactase, I need a reliable method to measure how much lactose will remain in a sample of milk (or some other dairy product if milk doesn't work). So far, what I've seen online calls for some highly advanced equipment that a high school lab does not exist. I am open to all methods (titration, measuring mass, chemical reactions, etc).
What can I do to measure this?

This might help: https://www.youtube.com/watch?v=BV8aFRII1M4


text:
House-hold chemistry experiments for a high-school student
I want to experiment the things which we learn in chemistry as a twelfth grader.I want to do them at my home but I'm not supplied with enough reactants.What can I do by using such reactants.
I tried to carry out paper chromatography at home and that was the separation of a mixture of inks. I didn't have  filter paper so I used a tracing paper.
My solvent was a common organic solvent. But to my dismay,solvent front didn't rise so high. So my experiment didn't succeeded but I'm hopeful I can do other things too which'd be successful.
It was first experiment
Can someone recommend a few projects/experiments (or sources that do the same) for a school student who's looking to improve her understanding of basic chemistry?

Doing simple chemical experiments won't help you advance much; doing advanced chemistry on your own without guidance can be tedious, pricey and sometimes dangerous. I would recommend to find a chemistry workshop. It can be a university section which offers practical lab work with elements of teaching for school students (which is IMO ideal), or some sort of chemistry club/open space where people of the same interest gather together and do lab work under supervision. You can do theoretical chemistry on your own, but for practical synthetical/analytical work you need a community anyway.

I am basically a villager and a college student.We have a lab in our college but I don't think that my Teacher will allow me to spend my time there making reactions)

Well, I would ask the teacher and explain what you've explained to us. I would expect that he/she will try to help you, if not resource-wise, but at least you can get better suggestions of what to do and were to start; also, asking costs nothing)

Certainly yes but the problem is the presence of chemicals in short amount.One of my class mates tried to make paint by using food colours,flour,heating them,and adding various other common things.The thing obtained was,obviosly, not paint,it was something else.But she was appreciated at her attempt.At night of thet day Teacher say some bubbles popping out of that paint and when he opened that all the so called paint burst open and his clothes were no longer clean.)

In short, don't try. The reactions that you learn are overly simplified, and you have no idea what are the complications involved when actually doing it.  Try reading a few research papers about how such reactions are actually conducted.

@andselisk can you guide me about any on line chemistry club or any thing else like that?

@SaraAsfar I believe you need a real lab in the first place and watching/reading experiments online won't help you much and is in my opinion a waste of time. Maybe ask around in social networks, post a few tweets to get in touch with people who work in an actual lab. If you insist on online videos, I can suggest YT channels NurdRage, Doug's Lab and TPM Chem (in order from fun to more advanced).

When I was younger than you currently are I got for Christmas a box exactly called il Piccolo Chimico (the little chemist :). Perhaps you can find some scientific toy box like that. There is quite a lot to do. And with your age and high school education you might be able to integrate it with more chemicals and apparatus. And remember Safety First

From where can I get such scientific toy box @Alchimista.I live in a village.No RESOURCES in my area.How.....?????

@andselisk +1 For Nurdrage! I was going to recommend sciencemadnes.org... and then I remembered that thing's taboo here on Chem.SE XD Also, mind if I include those links (and a few others) in my answer? O:)

@paracetamol Sure thing, go for it! Shame on me, I've never used sciencemadnes.org; is there a good reason it's forbidden on CSE?o_O

@andselisk Well, I believe it's largely looked down upon as a "recipe" site (at least from the tone of a Chem.SE meta post a while ago...I think it was ortho's)...whereas Chem.SE has considerably more "quality". Personally, I can vouch for ortho's (or whoever said that) point: I first stumbled upon sciencemadness looking for fireworks/explosives recipes (pyromaniac ---> ) O:)

A better idea IMO would be to study hard and get into a reputable University w/ a chemistry course you're interested in. You'll not only get to access the lab, but you'll understand the reactions you're performing.

My favorite "textbook" on home-chemistry is the Golden Book of Chemistry Experiments (Robert Brent).
It's a neat little book, and since it was published so long ago (1960s) it's not really available as a (literal) book these days. But fear not, because the PDF version can be easily found online!
A reliable place to download it from is archives.org.

Small warning:
I wouldn't recommend this to young children because quite a few experiments/projects mentioned there make use of an open flame (that's pretty much the only reason...). So be sure to conduct them in an open space far from anything inflammable.
Maybe this is the book/resource you're looking for? ;)

(A spot of light reading)
However, as the gents in the comments section have pointed out, conducting home-experiments won't (necessarily) make you a "Chemist". But I highly recommend you experiment (safety is important too) as frequent as you can, because over time it can firmly root a scientific bent of mind in you.
Intelligence, inquisitiveness and experience (and empathy) are highly prized values in the scientific world.
Besides, as you experiment at home (while you are still a high-school student) you'll soon stumble upon an extremely important idea:
No real experiment can be perfectly replicated
And you can only come to truly appreciate this fact when you experiment yourself...a lot.
No matter what experiment you do, no matter how perfect your set-up or procedure is, every time you conduct the same experiment you will never get the exact same result. Right now this "idea" may not seem like much (as a high-schooler, you might initially find this unbelievable), but its implications are profound and far-reaching.
That the fact that replicating experiments perfectly is impossible will soon lead you to another piece of insight: Probablity & Statistics rule science.
(End of light reading session)

Stemming from @andselisk's idea (in the comments under your post), you might find the following Youtube channels interesting (I'll try to add more when I get the time):

Nurdrage   <--- Highly recommended
NileRed
Cody'sLab
Doug's lab

The next two are largely DIY channels, but some of their videos include projects that count as "Chemistry":

NightHawkinLight
The King of Random

 Thus armed with this knowledge, you shall sally forth and become a chemist (of sorts...)!

I agree that you can learn from the web. However the theory of how one does a titration and the physical skill to do one are two different things. // I also make the point that labs have hoods, safety showers and eyewashes. You really need to think about safety as you start doing various experiments.


text:
Why did my milled powders catch fire?
I milled a small amount of the sample ($75\%$ Al, $15\%$ Cu, $10\%$ Ni) mixed with toluene for 5 hours after which I set it on butter paper to dry. Then after 13 minutes I switched on the fan in the room, at which point my powder caught fire.
Could you please explain why my powder would have caught fire?

Sounds like your sample is pyrophoric; as toluene evaporated, the highly-dispersed metal mixture came in contact with air and ignited itself. Both Al an Ni powders are pyrophoric agents BTW.

They caught fire because partly because they were milled in toluene and partly because they were such fine powders. As you probably know aluminum forms a passivating oxide on the surface. When you mill the powder, you increase the surface area which will oxidize when exposed to air. You milled it in toluene so oxidation did not occur during and after you were milling as long as it was covered in toluene.
$$\require{cancel}\begin{align}\\
\ce{Al (s) + O2 (g) &->  Al2O3 (s)} \qquad \qquad{\mathrm{\Delta H} = \pu{-1675.7 kJ mol-1}}\\
\ce{Al (s, toluene) + O2 (g) &\cancel{\ce{->}} Al2O3 (s)}
\end{align}$$
Once you filtered off the toluene parts and let it sit the toluene started evaporating and the aluminum started oxidizing. Once you turned the fan on you allowed the reaction to happen much faster to the point that it could not cool as it oxidized and the powder ignited. This is similar to smoldering coals, they are not glowing, but hot and still combusting. When you blow air on them the rush of oxygen will make them hot enough to glow again. This is essentially what turning the fan on did to your powder.


text:
Why does phenol react on oxygen in the Gattermann–Koch reaction?
When discussing the Gattermann–Koch reaction on phenol substrates, it has been said that O-acylation is a problematic competing reaction as

in a phenol, there's high electron density on the O atom [...]

However, there are also numerous other reactions where phenol reacts with electrophiles on the benzene ring. Some examples are:

the Houben–Hoesch synthesis (a kind of FC acylation);
the Riemer–Tiemann reaction;
the reaction of phenol with nitrous acid to form 4-nitrosophenol;
the Kolbe–Schmitt reaction; and
the Vilsmeier reaction, amongst others.

Why is the Gatterman–Koch reaction an exception?

Can anyone please explain what's wrong with my question?

There isn't really any literature evidence to support the fact that it does react @ oxygen (though it is quite probable). In the Gattermann-Koch mechanism, the intermediate is highly reactive, the Gattermann modification avoids this, making it amenable to synthesis (Gattermann conditions + phenol are fine and have been used in many syntheses).  The original Gatterman report of phenols not being amenable to G-K conditions is @ DOI: 10.1002/jlac.19073570209 (German language, very hard to decipher),  DOI:10.1002/0471264180.or009.02 also has an overview (English language).

@NotEvans. I've read the abstract of the English version which says that the Gattermann reaction with $\ce{CN}$ as a reagent reacts on benzene, but I am confused about the Gattermann koch version.

Here is your source:https://www.vedantu.com/chemistry/gattermann-koch-reaction


text:
Working with typical glass: What temperatures would I be dealing with?
I want to be able to melt down typical glass (the kind you usually find at home) and mold it into other shapes. What temperatures would achieve this?
For comparison; if I were to take some sand and convert it to glass (as we know it), what temperatures would I have to generate?
Furthermore, would "heat-treating" glass in some way make it tougher? I mean "Tougher" along the lines of "bullet-proof".

Oh and a pointer: "Bulletproof" glass isn't really just glass...it needs a bunch of other stuff as well. You might want to give this a read [https://en.wikipedia.org/wiki/Bulletproof_glass#Construction] ;)

The softening point of glass depends on its composition. For example,

1650°C for fused silica (pure $\ce{SiO2}$)
821°C for Corning Pyrex 7740 borosilicate glass
713°C for Wheaton 900 soda-lime glass

Since you want to use glass found at home, it might be soft lead-glass (e.g. "crystal" tumbler), soda-lime glass (e.g. bottles), or borosilicate glass (e.g. coffee carafe). Melting together different types of glass is usually not successful because unless perfectly mixed, the difference in thermal coefficient of expansion will cause the object to shatter on cooling. One way to test the coefficient of expansion is to combine two different types of glass into a thread to see if it bends on cooling. This can be used to make a glass thermometer, similar to the bimetallic strip in oven thermometers. Calibration is left to the student.
If you start with fairly pure silica sand, then, you need at least 1650°C to get a glass. Although an inexpensive air-propane torch (not oxy-propane) can theoretically reach 1,995 °C, my own experience is that it takes a long time to melt incompletely just a few grams of sand.
Adding other material to silica lowers the softening point. Soda-lime contains alakali metal oxides, $\ce{CaO}$, etc. Rather than using the caustic oxides, cheaper and safer carbonates (or bicarbonates, e.g. $\ce{NaHCO3}$) can be used, though water of hydration and $\ce{CO2}$ must be driven off by heat. You could try a small amount of very fine sand mixed with baking soda, which would eventually melt to make a lower-temperature glass... but be patient, because the glass is so viscous that it takes a long while to dissolve together.
Lead oxide is used to make lead "crystal", with a lower softening point and higher index of refraction than soda-lime glass, but working with lead in a home lab is definitely not advised!

If you want to completely liquefy the glass, you are going to be in the neighborhood of 1400$\ce{^oC}$ to 1700$\ce{^oC}$. A propane torch (not the kitchen sized ones) would allow you to melt small amounts of glass or soften larger pieces for bending, scoring, or cut. Sand to glass conversion is going to occur above 1700$\ce{^oC}$, with higher temperatures needed with the varying composition of the sand.
See here for discussion on heat treated glass vs tempered glass. In essence, the degree of strengthening imparted on the glass that is being treated post manufacture depends on the rate at which the heat is removed after the initial heating. The strength manifests as higher (flat) surface compression strength and edge compression strength. The treated glass may be more brittle however, that's the trade off for making it more rigid (i.e. higher compression strength).

If I want to do this experiment what kind of thing that melts and work with glass without the grass will get burned?Similar like microwave?

I mean what are things that are required for me to do this experiment in house or in garden?

It largely depends on what you want to do specifically. Do you just want to pour glass into new shapes from a melt you make from other glass or are you making things out of something like glass tubing?

Well,I don't know.Ask only to know

It is also possible to make any glass to be solid and to be also bulletproof?And if it is how many degrees I need for it?

I do not think you will be able to make bulletproof glass without industrial machines and processes. I cannot suggest tools to you if you don't know what you want to do.

These look like temp. for quartz glass. "Everyday" glass like sodium glass melts much easier.

@J.Ari If I want to make a window of a front car then what are the things I need for it?And if I want to make just a normal home window then what are the thing I need to make this?

@J.Ari Or like for example I want to make a glass just like IPhone 6 has,but to have the same size as it has,and what do I need for this to make?


text:
Visualizing s and p orbitals from contracted basis set
I was thinking about basis sets used in Quantum Chemistry programs and thought of why not try to visualize them.
I started with GTO from EMSL website for $\ce{H2}$, specifically cc-pVDZ basis
which is the following,
! s functions
H    5    3
     13.0100000            0.0196850              0.0000000              0.0000000
      1.9620000            0.1379770              0.0000000              0.0000000
      0.4446000            0.4781480              0.0000000              0.0000000
      0.1220000            0.0000000              1.0000000              0.0000000
      0.0297400            0.0000000              0.0000000              1.0000000
! p functions
H    2    2
      0.7270000           1.0000000              0.0000000
      0.1410000           0.0000000              1.0000000

I framed the basis functions for $s-$orbital as following,
$$0.019685\times exp(-13.01  r^2) + 0.137977\times exp(-1.962  r^2) + 0.4446\times exp(-0.478148  r^2)$$
$$1\times exp(-0.12200 r^2)$$
$$1\times exp(-0.02974  r^2) $$
This is only the radial part (the associated spherical harmonic is not here), but still I can plot the above against $r$.

The above plot is for the $s-$functions.

I am able to do the same thing for $p-$functions, and I get similar Gaussians (not shown here) with different widths, and centered at zero.
Shouldn't the lobe of $p-$ be centered larger than zero (which implies that the functional form I am using is not correct/incomplete for $p-$function). Or the lack of explicit spherical harmonic functions does not allow the $p-$functions to be similarly visualized. Could someone clarify.

Right now, this is unclear. Is the plot for s or p functions? What does "Shouldn't the lobe of p− be centered larger than zero" mean? Is it that the amplitude at $r=0$ should be zero and $r>0$ should be nonzero?

@pentavalentcarbon : The plot is for $s-$functions. The $p-$ function have almost zero amplitude near $r$ , while non zero probability at certain distance from $r$ (the dumbel shape in one axis, say towards arbitrary direction , described as +ve). I plan to achieve that.

Your suspicion is correct - you need the angular part, that is, a spherical harmonic (sometimes it may be cartesian instead) to get the shapes of the orbitals and thus, actual lobes of $p$-functions.
Also note that projection to $(x)$ from $(x,y,z)$ may not be straightforwardly the same for all angular momenta. Additionally, the angular part also has influence on the normalization of the basis functions, which may or may not be included in the basis set format at hand. The last point, however, can be checked by numerical integration.


text:
Redundant out of plane bends in gaussian not recognised
I'm having an issue with something that shouldn't be that complicated. I'm trying to generate some surfaces with Gaussian to look at amine planarisations, but the modredundant command doesn't seem to be recognising the out-of-plane bend coordinate type, spitting out the error: "Unrecognised coordinate type "O"".
I've tried different coordinate types, they all work, but it seems this one specifically is not working. Does anybody have any ideas to help me? Code below:
%chk=MR-test.chk
# opt=modredundant tda=(nstates=6) cam-b3lyp/6-31++G(d,p) geom=connectivity

Amine planarisation test

0 1
 N                  0.14115899    0.72065377    0.00000000
 C                  0.63114217   -0.66528146    0.00000000
 H                  1.70114039   -0.66529528    0.00195365
 C                  0.62897198    1.41283584    1.20159164
 H                  0.27231899    2.42164635    1.20159076
 H                  0.27070141    0.90787342    2.07426291
 H                  1.69897019    1.41282074    1.20354879
 C                  0.63336050    1.41438736   -1.19890424
 H                  0.27828574    0.91055325   -2.07353165
 H                  0.27670608    2.42319736   -1.19890424
 H                  1.70335872    1.41437379   -1.19694825
 C                  0.12009706   -1.39042562   -1.25880851
 H                  0.78378957   -1.19762146   -2.07565538
 H                  0.08029786   -2.44332208   -1.07246188
 H                 -0.85887294   -1.03416700   -1.50293458
 C                  0.11550541   -1.39204895    1.25599672
 H                  0.72562653   -1.13043658    2.09516963
 H                 -0.89643994   -1.10106296    1.44624332
 H                  0.15906269   -2.44960896    1.09924986

 1 2 1.0 4 1.0 8 1.0
 2 3 1.0 12 1.0 16 1.0
 3
 4 5 1.0 6 1.0 7 1.0
 5
 6
 7
 8 9 1.0 10 1.0 11 1.0
 9
 10
 11
 12 13 1.0 14 1.0 15 1.0
 13
 14
 15
 16 17 1.0 18 1.0 19 1.0
 17
 18
 19

O 1 8 4 2 S 25 2.000000

I've tried generating the input both with GaussView and typing it myself. It really doesn't seem to like out-of-plane bends. Thanks in advance.

I couldn't get this to run either. From looking at the mod redundant coordinate in Gaussview, I notice that it does not give a degree measure for the out of plane bend initially, so it doesn't seem to be measuring anything right from the start.

According to Gaussian's Modredund description, there is no such type as "O" anymore in Gaussian16.
Hence, the input deck is rejected. Gaussian16 provides generalized internal coordinates, for which more complex expressions are possible.

That's a little annoying, but thanks for the link to the generalised internal coordinates. :)

Maybe Gaussian has a canonical solution for you. I'd ask their helpdesk. I agree that it is ... unfortunate.

@Deathbreath I tried "O" in gaussian09 It just finished the job without error BUT it didn't give me any "scan" result. :/


text:
Two reactions which speed each other up
Recently I have started looking for two reactions that would speed each other up. That means that the products of the first reaction accelerate the second one, and in turn, the products formed in the second reaction accelerate the first one. Do you know any examples of such reactions?

Have a look at some non-linear reaction schemes such as gas phase radical chain reactions that may lead to explosion. The propagation steps in the hydrogen/bromine gas phase reaction do what you suggest:
$$\begin{align}
\ce{H^. + Br2 &-> HBr + Br^.} \\
\ce{Br^. +H2 &-> HBr + H^.}
\end{align}$$

I don’t quite see where these reactions speed up each other?

@Jan They are part of a bigger scheme and only for illustration, but the product of the first is the reactant to the second and vice versa.


text:
Batch conversion of Waters mass spectrometry file formats
Are there any free tools for converting raw mass-spectrometry files into open formats? In particular, is there any way to convert Water's .raw MS files into a readable format in a batch-wise fashion (talking about hundreds of mass spectra, so it needs to be at the command line ...)? After a long search, I only found MassWolf which seems very outdated and which also doesn't seem to provide any command line batch processing. Isn't there something like OpenBabel for analytical chemistry file formats?

Long time ago I remember converting some *.raw data files with Bruker CompassXport. I'm not sure though if it can handle your data. Last time I used it, it was a free command-line tool which you can get for free from Bruker after you register on their website. Also, check out the manual beforehand. Or just upload an example of your data so that me and others could test their suggested converters before posting the ideas here:)

@andselisk Thanks for the quick reply : ) I checked the manual, but couldn't really find anything on the Waters .raw format in there (.raw is really a stupid name for a format...). I am already at the point of thinking about using a Hex editor just to reverse engineer that format because a batch solution would be so much better!

In the lab we use MSConvert from ProteoWizard (http://proteowizard.sourceforge.net/downloads.shtml) to convert Waters files to mzXML format. It is not exactly batch mode but you can input a list of files to process. The .raw to interpretable is done through a DLL provided by Waters, so they did not reverse engineer the format, which is not quite trivial (at least from my first attempts at it).

Waters Databridge actually allows the conversion of multiple files at the same time, but somehow it is not very intuitive to use (not a native UI) and also does not really work in batch mode. Still, it is somewhat possible to quickly export Waters .raw files using Databridge, and it is the best solution I found so far. From there, a simple python script allows me to average the mass spectra (that are still chromatograms), so that it at least doesn't take hours to export every single mass spectrum by hand.
For other people in my position, having to convert hundreds of Waters .raw files, I hope that helps at least somewhat.


text:
How to infer formula from diagram

Hello Folks
I am not now nor have I ever been any kind of chemist.
My question is, how is the formula for Cerin, C30H50O2, consistent with this diagram taken from Chemspider?
I can count the 30 Carbon atoms, assuming that each vertex of a hexagon is a single C. I can obviously see the two Oxygen atoms. But when I count the Hydrogens, I can only get to 28. Where are the other 22?
Thank you for your patience.

You also need to assume that each implicit carbon (like the corners of each ring) is saturated, so the topmost ring has 6 implicit hydrogens.

@Raditz_35 so that's an additional 2 H''s for every completely 'unclaimed' corner, ie 2 x 10 = 20, plus another one each for the methyl and hydroxyl groups on the bottom left ring, making up the 22? Is that right? And have I got those group names right?
Thanks also to pentavalentcarbon, I'm only allowed to tag one user per comment, it seems

@Mark I haven't noticed anything wrong, so I'd say you are right!

@Raditz_35 OK, thank you very much! If you turn this into an answer I'd be happy to make it the accepted.

Each carbon atom (you seem to have identified them) needs four of those stick-like thingies (the lines, sometimes called "bonds"). These include the dashed lines and thick triangles as well as the "double bond" you see at the lower oxygen which looks like this ($=$). That one counts twice. Every time you only see 3 or even 2 for one carbon atom, that means 1 (or 2) hydrogen atoms are implied.

As an explanatory addendum for original poster; for large molecules it is extremely normal to not write out the hydrogens, unless their position or orientation is vital. It would be incredibly tedious and not add any value, nor clarity to the picture (the opposite, actually) to write H's all over that structure.


text:
Distillery Transfer Hoses
I was hoping an actual person could offer me some advice. I work in a craft distillery and I'm getting a lot of mixed information on how compatible PVC is with high-proof alcohol. It definitely seems that "Food-Grade" doesn't necessarily mean "Alcohol-Grade".
A. Is PVC hose safe to use for transferring Grain, Water, and most importantly, Ethanol?
B. If not, is there a type of material you would recommend for any and all distillery applications?

Glass pipes may be best, but it's not like you'd die from vodka transferred using PVC hose. Unless you used it too much ;)

Subjectively, I don't think PVC is that bad, unless the hose is transferring boiling hot beverages or is constantly under sunlight. If you want a bullet-proof solution that can withstand solvents, oils and majority of other chemicals, maybe have a look at cross-linked polyethylene (XLPE).


text:
What is the analogue of "molecule" for ionic compounds?
In a system, if we have $\pu{18 g}$ of $\ce{H2O}$ ($M_\mathrm r = 18$), we can say we have a mole of water molecules or $6 \times 10^{23}$ molecules. But in another system, if we have $\pu{342 g}$ of $\ce{Al2(SO4)3}$ ($M_\mathrm r = 342$), what is the best way to say that there is one mole of "aluminum sulfates" or $6 \times 10^{23}$ "aluminum sulfates"?
Since it is a ionic compound, we can't say molecules, but I can't find an analogous form for ionic compounds. I thought about saying "a mole of aluminum sulfate ionic aggregates", but it still doesn't too good, since I've never seen a definition of ionic aggregate as a unit of the ionic compound formula. It only tells the way the ions organize themselves: as an aggregate.
I know it is a silly question, but in terms of language and symbolism, it is important to me as a chemistry teacher.

Note that you might need to give amounts of each ion separately anyway, due to common ion effect.  So having only one salt with equal parts of cat and an will be a special case.

The term you are looking for is formula unit, I think. Wikipedia doesn't really describe it super well, but just to give an example, you could write the sentence

$\pu{58.44 g}$ of $\ce{NaCl}$ contains $6.022 \times 10^{23}$ formula units of sodium chloride

and it would be pretty well understood.
See also: What is the difference between a Chemical Formula and Formula Unit?

Depending on what is meant by a chemical formula, the terms chemical species, and molecular entity might also be what OP is after. In short, chemical species is a collection of $\ce{NaCl}$ formula units (e.g., as a big solid) whereas molecular entity is one particular unit (molecule, atom, radical, etc). Ion pair is also pertinent in this case. See DOI: 10.1351/pac199466051077 for more details. Feel free to add to your answer if you like.

Thanks for the link and for the extra information. I can see "molecular entity" being used too, perhaps? For the others, I can't quite construct a good sentence with them. "Ion pair" is generally used to emphasise close association of ions in solution, in my own experience, at least.

Not a problem. I believe IUPAC permits using ion pair in similar meanings, the more accurate term for what you describe is tight ion pair. There are also loose ion pairs which are, in turn, categorised by the number of solvent molecules in between. Solvent-shared ion pair refers to there being only one solvent molecule whereas they're solvent-separated if more than one molecule intervenes (IUPAC mentions to be careful with the loose distinction due to ambiguity.). Too many solvent molecules in between, or in the gas phase with sufficient separation the term free ions is used.


text:
How to numerically solve a reaction network with fast equilibrium steps?
For example If I have to solve the following reaction numerically
\begin{align}
*\ce{ + A(g) &<=>[K1] [A*]} \tag{1}\\
\ce{[A{*}] &<=>[k1][k-1] [B{*}]} \tag{2}\\
\ce{[B*] &<=>[K2] [B](g) +}* \tag{3}
\end{align}
Where the first and third reactions are fast equilibrium steps.
[* ] represents vacnt site and [A*]  and [B*] represent adsorbed species.
The reaction is carried out in a closed container, therefore the concentrations of A(g) and B(g) are not constant.
Given initial A(g) and B(g) pressure and assuming zero coverage of reactants initially, how do I solve a system like this numerically.

After you specify all the parameters, it's essentially a differential equation problem. I would suggest use matlab (ode45 or other solver). If it fails, use BZZmath or SUNDIALS solver for stiff problem like the one you mentioned.

I understand that its a differential equation problem and have solved problems of this nature before. But the rapid equilibrium steps confuse me, I have solved problems which dont have rapid equilibration. I dont have a reverse and backward constant for the first and third reaction, only an equilibrium constant.

For A adsorption:
$$r_1=0$$
because the reaction is in equilibrium.So,
$$K_1=\frac{[A]}{P_A \times [*]}$$
From this equation you can relate $[A]$ to $[*]$.Similarly, using the last reaction you can relate $[B]$ to $[*]$. Your [A] and [B] are connected via reaction 2.
$$\frac{d[A]}{dt}=-\frac{d[B]}{dt}=-k_1[A]+k_{-1}[B]$$
This is a first order differential equation, so you need only one boundary condition. You have three unknown and three boundary conditions $([A]_0=[B]_0=0$ and $[*]=1.00)$.

I think that when A * is in rapid equilibrium with A and * you can not write the rate of change of A as you have written. This expression would be valid if A* had not been in equilibrium with A and *.

Yes, you are right. There  should be two solution regime for this kind of problem (I forgot about that). One is for short time scale and another one for long time scale. I'll suggest consulting William Deen's analysis of transport phenomena book (maybe it was discussed in 3rd or 4th chapter)

See chapter 4 of this link (http://www.mat.unimi.it/users/scacchi/didattica_2017/biomat2/asy.pdf). This is exactly what you are looking for.

Thanks this seems useful, I did not think that you could apply the solution matching approach to a problem of this nature. But they solve the problem analytically, what if I want to do this numerically what do I do? This was just an example system that I chose, the real system that I am interested in solving is rather complicated and definitely doesn't have an analytical solution.
Thanks

Practically, I have no idea; however, you can simplify your problem by taking few relevant perturbation order (say up to 3rd order), form your system of differential equation system, solve it using a ODE/DAE/PDE solver. In essence, before you form your system of DE just make sure you have a well defined set, i.e., your number of equations and number of boundary conditions are consistent.


text:
Why are Alcohol Dehydration Reactions Favored by High Temperatures?
If I go by the Gibbs' free energy formula, it makes sense that high temperatures make the dehydration of alcohols more feasible, as elimination reactions usually have a positive value for the change in entropy ($\Delta S$) and the hydration of the sulfuric acid used is highly exothermic ($\Delta H$). This decreases the value for the change in free energy ($\Delta G$).
However, if I go by the modified Vant Hoff's formula (relationship between an initial equilibrium constant at an initial temperature and a final equilibrium constant at a final temperature), the $K_2/K_1$ ratio would be higher at lower temperatures, owing to the product $\Delta H \Delta T$ being positive (rather than negative if the reaction mixture is heated).
These are the formulas I'm referring to:
\begin{align}
\frac{K_2}{K_1} &= \exp{\left(\frac{\Delta T \Delta H}{RT_1T_2}\right)} \tag{1}\\
\Delta G &= \Delta H -T \Delta S \tag{2}
\end{align}

I agree with your first point (Vant Hoff only working when ΔH is constant); however, ΔT is T2−T1. In the link you sent me, the negative sign was assigned to ΔH, but it will cancel out with the other negative sign once we convert the reciprocal temperature difference to T1-T2/T1T2 which is equal to -ΔT/T1T2.

Heating implies starting with a lower temperature (T1) and ending with a higher temperature (T2); in other words, ΔT = T2-T1 > 0. If we say that ΔT = T1-T2, then we would get ΔT < 0, which would correspond to cooling.

I agree. I think it would make sense anyway since the sulfuric acid is acting as a catalyst and therefore does not participate in the reaction itself.

You're calculations are correct, but now I'm doubting the situation with the acid. Even if the acid itself does not suffer any chemical alterations in the reaction, it does give off heat once it dehydrates the alcohol. Wouldn't this heat have to be considered in the enthalpy analysis as well?

Let's take: $\,\,\,\,\ce{ethene(g) + H2O(g) <=> ethanol(g)} $
The standard enthalpy for hydration of ethene is $\Delta H^\circ_r=-45\frac{KJ}{mol}$. So we have $\Delta H^\circ_r=45\frac{KJ}{mol}$ for dehydration, which means that it is endothermic.
The expression:
\begin{align}
\frac{K_2}{K_1} &= \exp{\left(\frac{\Delta T \Delta H}{RT_1T_2}\right)} \tag{1}\\
\end{align}
(This equation is valid when Ts are similar, because $\Delta H$ varies with $T$ and you are considering it constant.)
with $\Delta T= T_2-T_1$ and $\Delta H_r^\circ>0$ (dehydration) equation (1) gives $K_2>K_1$.
As $$K = \frac{[\ce{H2O}]\,[\ce{CH2CH2}]}{[\ce{CH3CH2OH}]}$$
and $K_2>K_1$. In consequence dehydration enthalpy is favoured by temperature-increase.

Yes, but if the reaction is exothermic, that means ΔH<0  which would make the argument negative as ΔT is positive.

Now I'm confused because my book says alcohol dehydration is exothermic. They included an energy chart with the products (alkene) having lower energy than the reagents (alcohol). I will edit my original post to include it.

What you're saying makes sense if in fact hydration is exothermic and dehydration is endothermic, but other sources like my book (Organic Chemistry Wade) say the latter are exothermic.


text:
How to determine the activation energy in this situation?
There is a rule of thumb that says that the rate of many room-temperature chemical reactions approximately doubles when the temperature is increased by 10°C. Calculate the activation energy for a reaction that exactly conforms to this relationship.
The only thing that comes to mind for me it's the Arrhenius equation, but I am not sure if that is the correct way of doing it since I am given so little information.
How can I solve this problem ?. Thanks a lot in advance!

https://en.wikipedia.org/wiki/Arrhenius_equation#Arrhenius_plot

You would use the Arrhenius equation. You are given that the reaction should double in rate when you increase from room temperature (for convenience, I'll say this is 300K) to when the temperature is increased by 10 degrees. So you can equate the Arrhenius equation at $T=310\mathrm{K}$ to double the Arrhenius equation at $T=300\mathrm{K}$ and solve for Ea.

A general formula for the activation energy can be derived but I don't think we can achieve a specific value

@TanYongBoon That's not true for specific temperature.

@Mithoron There isn't a specific temperature in this case, is there? As in the assumption made here of the increase being from T=300 K to T=310 K is not correct. Because if u take it as T=290 K to T=300 K. The value obtained of the activation energy would differ.

@Tan Yong Boon you are correct in that room temperature does not have an exact meaning and can basically describe anything between 293K and 300K. However, I think for this problem, as it seems to be mainly conceptual rather looking for a specific value (ie about what activation energy will be double in rate with a 10 degree temperature increase) using 300K should be fine and works well because it is a nic round number.

Ok then. Since the question asker mentioned "room-temperature chemical reactions", then I guess it should be fine. Just wanted to clarify.


text:
Phases in iron carbon diagram
The definition of phase :
Uniform chemical and physical composition.
So Alpha iron (0.001% carbon) and alpha iron (0.002% carbon)  are two different phases??

Let me opt for a paradoxical answer. They are not different phases, but you can't call them one phase either. There is no "them".
That is, you can't have a system in equilibrium which contains two different areas with the said compositions, either separated by a phase border or not. If you try, they will mix and form a single phase with averaged composition. It will be uniform all right.
Sure, you may have such compositions in different systems, and indeed many would call them "the same phase" then, based on the very close properties, similar structure, and the fact that they will mix if you let them. But this approach may quickly lead to nonsense. Say, water and acetone will mix in any ratio; are they one phase? Suppose yes. What about acetone and hexane? The same answer applies for the same reason. Now what about water and hexane? Er, well...
So let's not compare phases from different systems in terms of "sameness". Keep your apples and oranges apart.

Thank you so much @Ivan .But sometimes when we have metastable states...that is suppose both state exist in solids..the rate at which they will form an averaged composition will be very slow. In such cases,  they might exist like same for longer times. So here, what should I say ? Single phase or two phase or the concept of phase definition doesn't hold true in such cases?...thanks in advance

If you don't have equilibrium, then the concept of phase doesn't fully apply.


text:
Gaseous state: Boyles, Charles and Lusaac's law
We know that pressure is inversely proportional to volume (Boyles law) and volume is directly proportional to temperature ( Charles law). So then won't pressure be inversely proportional to temperature?

Actually, it's a big NO.
Why?
Because you need to consider the statement of the laws carefully. Boyle's law states that pressure is inversely proportional to volume at isothermal conditions. Boyle's law is not valid if temperature isn't constant. The same goes for Charles' law, where you must have isobaric conditions. The conditions required for both laws aren't the same, and so you can't really make a logical conclusion that way.

If you look at all of the chemistry laws related to the state of gas:

Ideal Gas Law:    P•V = n•R•T
Charles Law:    V1•T2 = V2•T1
Boyle's Law:    P1• V1 = P2 •V2
Gay-Lussac Law:      T1 • P2 = T2 • P1
Combined Gas Law:    P•V / T = k

... they show the interplay of pressure, volume and temperature.  Where you see both temperature and pressure, these show that pressure and temperature are not inversely proportional, but generally proportional.  Take Gay-Lussac; if the initial temperature goes up, and initial pressure, and final temperature remain the same, the final pressure also goes up proportionally.

Welcome to our site, Kurt! I would edit your post to include MathJax, but I'm lazy. Take a look at this and see how we format math expressions for yourself.


text:
Can sodium dithionite reduce the nitro group of 2-nitrobenzoic acid?
Can sodium dithionite reduce o-nitrobenzoic acid to anthranilic acid under basic conditions with mild reflux?
If so, how could I separate anthranilic acid from the reaction mixture?

Just to add to Waylander's answer, sodium dithionate has been used successfully to reduce a wide variety of (more substituted) nitrobenzoic acids. Just as one example (there are many), this one used in Overman's synthesis of plectosphaeroic acids B and C (J. Org. Chem. 2013, 78 (17), 8766–8788):

In this case the product was extracted using ethyl acetate.

Yes I believe it can. Dithionite is a reliable reducing agent for aromatic nitro groups. The reaction of dithionite produces an acid solution which will need to be neutralised before isolation. Anthranilic acid is not very water soluble so an extraction with e.g. chloroform should separate the product cleanly. Tetrahedron Lett. 2013, 54 (21), 2600–2603 discusses dithionite reduction of o-nitrobenzamides and refers to other papers that have used dithionite on similar substrates.


text:
What is the origin of 'tenting' or 'roofing' in NMR spectra?
Given an NMR spectrum with two coupled nuclei one can often observe so-called 'roofing' – an asymmetry in the multiplet.
For example, here is a 300 MHz proton spectrum of 1-bromo-2-methoxyethane, where the inset shows the triplets corresponding to the adjacent $\ce{CH2}$ groups:

(Source: Hans Reich, University of Wisconsin)
What is the origin of that effect?

Not just coupled, but strongly coupled; these roofing effects tend to be seen when the coupling constant $J$ is on the order of $\Delta \nu$ (the difference in resonance frequencies between the two spins).

The origin is quantum mechanical in nature. I will try to give a simplified overview of what is happening without going into the explicit details. For a pair of coupled spin-1/2 nuclei, labelled 1 and 2, the Hamiltonian in a magnetic field is
$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + 2\pi J(\vec{I}_{\!1}\cdot\vec{I}_{\!2})$$
If the spins are not coupled at all, then
$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z}$$
and the eigenstates of the Hamiltonian are exactly
$$\begin{array}{ccc}
\text{Eigenstate} & \text{Value of }m_1 & \text{Value of }m_2 \\ \hline
|\alpha\alpha\rangle & +1/2 & +1/2 \\
|\alpha\beta\rangle & +1/2 & -1/2 \\
|\beta\alpha\rangle & -1/2 & +1/2 \\
|\beta\beta\rangle & -1/2 & -1/2
\end{array}$$
NMR theory dictates that only transitions with $\Delta M = \pm 1$ are observable, where $M$ is the sum of the quantum numbers $m_1$ and $m_2$. For example, $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ is an observable transition ($\Delta M = \pm 1$), but $|\alpha\alpha\rangle \leftrightarrow |\beta\beta\rangle$ is unobservable ($\Delta M = \pm 2$). From the table above one can see that there are four observable transitions:
$$|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle; |\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle; |\alpha\beta\rangle \leftrightarrow |\beta\beta\rangle; |\beta\alpha\rangle \leftrightarrow |\beta\beta\rangle$$
Each transition corresponds to a fixed energy, and hence each observable transition gives rise to one peak in the NMR spectrum. The area under the peak - or loosely speaking, the height of the peak - is related to how much "$\Delta M = \pm 1$ character" there is in each transition. Of course, there is a way to formalise this, but for the sake of simplicity that will have to be omitted.
For the transition $|\alpha\alpha\rangle \leftrightarrow |\beta\beta\rangle$, this is entirely $\Delta M = \pm 2$. So, the amount of "$\Delta M = \pm 1$ character" is exactly zero, and the transition is entirely unobservable.
On the other hand, for all four of the allowed transitions, the amount of "$\Delta M = \pm 1$ character" is exactly one, and hence all transitions have equal intensity. [Of course, because of the lack of coupling, there will be two degenerate pairs (i.e. there are two pairs of transitions with the same energy), and so you get two singlets in the spectrum.]

Now, as the coupling term $2\pi J(\vec{I}_{\!1}\cdot\vec{I}_{\!2})$
becomes larger and larger, the eigenstates are gradually shifted away from these. These can be expressed in terms of linear combinations of the four original eigenstates. Without proof I will present the results
$$\begin{array}{cc}
\text{New eigenstate} & \text{Expression in terms of old eigenstates} \\ \hline
|1\rangle & |\alpha\alpha\rangle \\
|2\rangle & \cos\theta|\alpha\beta\rangle + \sin\theta|\beta\alpha\rangle \\
|3\rangle & -\sin\theta|\alpha\beta\rangle + \cos\theta|\beta\alpha\rangle \\
|4\rangle & |\beta\beta\rangle
\end{array}$$
where $\theta$ is defined by $\tan(2\theta) \equiv J/\Delta\nu$. As the ratio of $J$ to $\Delta\nu$ goes from zero to infinity, $\theta$ goes from $0$ to $\pi/4$. It is easy to verify that the deviation from the old eigenstates increases as this ratio $J/\Delta\nu$ increases.
Let's look at the transition $|1\rangle \leftrightarrow |2\rangle$. How much "$\Delta M = \pm 1$ character" is there here? Since $|2\rangle$ is a mixture of $|\alpha\beta\rangle$ and $|\beta\alpha\rangle$, the transition $|1\rangle \leftrightarrow |2\rangle$ can be thought of to be a "mixture" of $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ and $|\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle$.
So, there is $\cos\theta$ amount of "$\Delta M = \pm 1$ character" arising from the $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ contribution; and there is also $\sin\theta$ amount of "$\Delta M = \pm 1$ character" from the $|\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle$ contribution, for a total of $\cos\theta + \sin\theta$.
For quantum mechanical reasons it is necessary to take the square of this quantity, i.e.
$$\begin{align}
(\cos\theta + \sin\theta)^2 &= \cos^2\theta + \sin^2\theta + 2\sin\theta\cos\theta \\
&= 1 + \sin(2\theta)
\end{align}$$
and working through the list of transitions, one finds that the amount of "$\Delta M = \pm 1$ character" for each transition is
$$\begin{array}{cc}
\text{Transition} & \text{Amount of “}\Delta M = \pm 1\text{ character"} \\ \hline
|1\rangle \leftrightarrow |2\rangle & 1 + \sin(2\theta) \\
|1\rangle \leftrightarrow |3\rangle & 1 - \sin(2\theta) \\
|2\rangle \leftrightarrow |4\rangle & 1 + \sin(2\theta) \\
|3\rangle \leftrightarrow |4\rangle & 1 - \sin(2\theta)
\end{array}$$
Aha! So, as long as $2\theta = \tan^{-1}(J/\Delta\nu) \neq 0$, the intensities of the peaks will be different. There will be two strong peaks (those with plus signs), and two weaker peaks (the minus signs).
When the ratio $J/\Delta\nu$ is small, i.e. the weak coupling limit, then $\sin(2\theta)$ is small and the difference in intensity is hardly noticeable - not something that you can tell by eye. For example, in a $\pu{400 MHz}$ spectrometer, two protons that have a chemical shift difference of $\pu{1 ppm}$ would have $\Delta \nu = \pu{400 Hz}$. If we take $J = \pu{7 Hz}$, then $2\theta = \tan^{-1}(7/400) = \pu{0.0175 rad}$ and we have $\sin(2\theta) = 0.0175$, a difference which you'd be hard pressed to spot!
But when this ratio increases, the value of $\sin(2\theta)$ will increase, and the difference in intensity will become more and more pronounced. I won't go into the maths, but it is possible to show that the middle two transitions have the higher intensity, and the outer two transitions have the lower intensity - which, of course, gives rise to the "roofing effect" which you speak of.


text:
Evaluating (∂U/∂T)P - (∂U/∂T)V for an ideal and van der Waals gas
What is the value of $$\left(\frac{\partial U}{\partial T}\right)_{\!P} - \left(\frac{\partial U}{\partial T}\right)_{\!V}$$ for an ideal gas and for a van der Waals gas?
($U$ refers to the molar internal energy.)

Here I am going to use $U$ to refer to the total internal energy. If you want the molar internal energy ($U_\mathrm m \equiv U/n$) just divide your final result by $n$.
In another of my answers I have proven (equation $(14)$) that
$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V}$$
The proof up until there is completely general, so this applies for both an ideal and a van der Waals gas.
Furthermore here I have shown that (combining equations $(6)$ and $(8)$)
$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p$$
Putting these together gives you
$$\begin{align}
\left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} &= \left(\frac{\partial V}{\partial T}\right)_{\!p}\left[ T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p\right]
\end{align}$$
All that's left is to evaluate the partial derivatives. The ideal gas case is very easy; since $p = nRT/V$, $(\partial p/\partial T)_V = nR/V$ and the term in square brackets goes to zero.
For the van der Waals gas you need to work at those partial derivatives which are a bit tedious, but should be doable, especially if you employ implicit differentiation. For example, if you take the van der Waals equation of state
$$\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT$$
and differentiate with respect to $T$ throughout, keeping $V$ constant, one gets
$$\begin{align}
\left[\left(\frac{\partial p}{\partial T}\right)_{\!V} + 0\right](V - nb) + \left(p + \frac{an^2}{V^2}\right)(0) &= nR \\
\left(\frac{\partial p}{\partial T}\right)_{\!V} &= \frac{nR}{V - nb}
\end{align}$$
and likewise, barring any careless mistakes, I obtained
$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = nR\left(p - \frac{an^2}{V^2} + \frac{2abn^3}{V^3}\right)^{-1}$$
Maybe somebody has a less tedious way (and I hope so), but this is all I can offer you myself.

For the ideal gas, you can get to the result a little quicker if you know the equipartition theorem. Using the equipartition theorem, one can show that $$U=\frac{\mathrm{d.o.f}}{2}nRT=U(T)$$ where d.o.f is the degrees of freedom of the molecule. This says that for an ideal gas, U is just a function of temperature, so $$\left(\frac{\partial U}{\partial T}\right)_P=\left(\frac{\partial U}{\partial T}\right)_V \to \left(\frac{\partial U}{\partial T}\right)_P-\left(\frac{\partial U}{\partial T}\right)_V=0$$
From statistical mechanics, we can use the partition function of the van der waal gas to show that $$U=\frac{\mathrm{d.o.f}}{2}nRT-\frac{an^2}{V}$$
Now with a direct expression for the energy, we can solve using the relation: $$\left(\frac{\partial U}{\partial T}\right)_P-\left(\frac{\partial U}{\partial T}\right)_V=\left(\frac{\partial U}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P} + \left(\frac{\partial U}{\partial T}\right)_{V}-\left(\frac{\partial U}{\partial T}\right)_V=\left(\frac{\partial U}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}$$
with
$$\left(\frac{\partial U}{\partial V}\right)_{T}=\frac{an^2}{V^2}$$
$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = nR\left(p - \frac{an^2}{V^2} + \frac{2abn^3}{V^3}\right)^{-1}$$
If the energy of the van der waal gas is taken as a known result (or derived as shown by juanrga in this answer on Physics SE), you only have to evaluate two derivatives, though similar to orthocresol's answer, it is a bit tedious to compute $\left(\frac{\partial V}{\partial T}\right)_{\!p}$

What does "d. o. f" mean?

@aventurin Degrees of Freedom of the molecule


text:
Polymerisation of methyl methacrylate - can the C=O group get involved?
When methyl methacrylate undergoes free radical addition polymerisation, the C=C pi bond is broken during the process:

I'm wondering why the C=O pi bond doesn't get broken during the polymerisation. I suspect it may be because the C=O bond is stronger than the C=C bond, but are there any other ways to rationalise this?

Compare radicals involved.

C=O bond is strongly polar, and the cleavage of the double bond would rather involve heterolytic breaking with ions formed than homolytic with radicals.


text:
Program to compute multipoles of water at CCSD(T)-F12 accuracy
I want to compute traceless cartesian electrostatic multipole moment tensors up to fourth order (hexadecapole), at CCSD(T)-F12 accuracy using a large basis set. I'd like to be able to choose the expansion center, and to do this for small molecules like water.
What code (program) can do this?
I know molpro can do it, but the license is quite expensive. I have access to Gaussian16 which gives the multipoles by default, but not traceless and not for this method, and i also can't figure out if there is a way to choose the expansion center coordinates.
It is not easy given a code to figure out if it has the required capabilities, and there are many codes, so some advice would very much appreciated!

My understanding is Gaussian always uses the origin as the expansion center, so you could change the origin to put the center where you want. Also, you might be interested in the new Matter Modeling SE. While your question is on topic for Chem SE, Matter places a greater emphasis on software/computational modeling questions.