Pressure exerted by a particle of ideal gas There is a step I fail to grasp in the proof that for a monatomic ideal gas, $U=\frac{3}{2}PV$, where $U$ denotes the internal energy of the gas.
The proof involves considering a particle with mass $m$ travelling with velocity $v$ inside a cube of side $x$. The particle is assumed to travel perpendicularly towards one of the sides before bouncing back off it. The force exerted by the particle at any instant is opposite to the force exerted by the side of the cube on the particle. The latter can be worked out using Newton's second law, prior to which we need to calculate the acceleration of the particle during the collision. Assuming the collision to be perfectly elastic, the change in velocity before and after the collision with the wall is $\Delta v=2v$. To work out average acceleration, according to me, one would need the duration $\Delta t$ of the collision, since assuming it to be instantaneous would imply infinite acceleration. However, the various proofs I've read assign to $\Delta t$ the time that is needed for the particle to bounce off the wall and return after colliding with the opposite wall, in other words $\Delta t=\frac{2x}{v}$. I really can't see how that $\Delta t$ is relevant to the force exerted by the particle on the wall during the collision.

[Edit :] If there are no considerations about duration of the collision, then theoretically, no two particles would ever collide simultaneously with the wall and therefore the wall experiences very frequent collisions with single particles.

Here's your confusion:
You need to consider two different things:

The momentum transfer per particle per collision.  There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that complication is completely unnecessary for introductory kinetic theory, given that there is a simple and direct way to calculate the effect of each collision.]  And the simple and direct way is to use the change in momentum, which is $2mv_x$.

Note: Your statement that "The particle is assumed to travel perpendicularly towards one of the sides before bouncing back off it" is incorrect.  Rather, in determining the momentum transfer per collision, we use the component of the velocity perpendicular to the wall, namely $v_x$.

The number of collisions per unit time.  Here is where the average time it takes for the particle to travel from collision to collision is relevant. This is where ${\Delta t}$ is being used -- not to calculate the momentum transfer/collision, but to determine the frequency of collisions per particle.  And you need the average momentum transfer/collision, and the frequency of collisions per particle, to determine the average force per unit area contributed per particle.

So, in summary: For momentum transfer per collision, we don't consider time, since we can get this directly from $2mv_x$.  For number of collisions per unit time, we need to account for the time between collisions.


What method is used to discover oxidation state of drug binding cysteine residue? I was reading this article and found this sentence:

  KC group found that some cancer cells became resistant to
  Boehringer Ingelheim’s covalent TKI afatinib (Giotrif) due to the
  oxidation of its target cysteine in EGFR

What would be the chemical method to discover that especially the target cysteine is oxidized? Can it be done with NMR or spectroscopy?

NMR could likely be used to characterize a bulk sample of the drug and identify the presence of cysteine residue oxidation products, but would not be of much use for establishing that cancer cells oxidize (inactivate) the drug, nor precisely how it was inactivated, given the tiny amounts of material involved. Maybe LC-MS could be used. Probably worth pulling up the original paper to see how they did the work.


Why is there a max radius for an orbit of an electron around a nucleus? Why is there a max radius for an orbit of an electron around a nucleus ?
I had a course in electromagnetism but I do not get this.

Perhaps you can clarify your question a bit, but if you are referring to the size of the radial distribution of electron density, this increases with increasing principal quantum number, up to the ionisation limit, at which point the spacing between energy levels trends towards 0 and the interaction energy trends towards zero. After this, the electron is for all intents and purposes unbound. This is illustrated for the hydrogen atom in the Lyman series (image borrowed from Wikipedia):

The energy of the largest transition $n=\infty \rightarrow n=1$ is 13.6 eV, or half a Hartree (this is the energy of a photon with a wavelength of 912 Ångström). This is the binding energy of the hydrogen atom. Even the Lyman gamma line - representing a $n=4 \rightarrow n=1$ transition - is quite close to the ionisation limit. As n increases, from a strictly electrostatic point of view the orbital becomes far less stable, and this is reflected in the general negative trend of ionisation energies across the periodic table, however the complete picture is a bit more complex than that because electrons can interact with electrons as well as the nucleus.

The key point is that as n increases, the energy required lose an electron entirely becomes smaller and smaller, and this fixes a practical limit on where we can say the electron is actually bound. This has a caveat though - cold gasses can form unusually long-lived Rydberg states with ridiculous principal quantum numbers with correspondingly enormous orbital radii (up to micrometers!*) and semiclassical electron orbits. The reasons for the stability of these states is outside of my knowledge.

**To paraphrase, Alisa Bokulich points out in 'Re-examining the Quantum-classical Relation' that Rydberg atoms are of comparable size to grains of (fine) sand, which is pretty neat*


Does Bent's Rule only apply to molecules where there is hybridisation? I would just like to ask if the pre-requisite for using Bent's Rule is that the bonding in the molecule involves hybridised orbitals.

I was thinking of why the bond angle in hydrogen sulfide (~ 92 degrees) is larger than the expected 90 degrees. I thought that Bent's Rule would be able to offer a good explanation for this as the s orbital character concentrated towards the electropositive hydrogen atoms results in stronger bond pair-bond pair repulsion, increasing the bond angle. However, I realised that my entire argument may fall apart if Bent's Rule does not work for bonding involving unhybridised orbitals.

Bent's rule is not as much a rule as it is an observation. It was first applied in the context of valence bond theory and therefore used for hybrid orbitals only. At the initial stages, VBT only used s, sp, sp2, sp3, and p type orbitals. Bent proposed to allow also anything in between aiming at more flexibility for the wave function and therefore better accuracy of the description. As such it is only a tool for an already approximate view on bonding.

Most importantly, and in any and all general cases, the hybridisation state of the orbitals is determined by the molecular structure. For any case in which hybrid orbitals are used, there is also a true solution which does not use them.

Though Bent's rule has some predictive character for anything that is close to hybrid orbitals, it is basically just a shorthand for a much deeper rooted phenomenon.

However, Bent's rule is not useful in any case where there is a low s-p mixing, i.e. anything below period 2 is a stretch.


What is the grey film/black sludge from mercury? How to remove it? How to remove it? Any other simple method apart from distillation.

I have a small container of mercury that I’ve collected over the past decade. I opened the container after a few years now and I noticed that there is a grey coating on the surface of mercury. Also there is a black powdery deposit at the bottom. It sticks to the bottom and sides of the plastic container.

First, I would advise against using mercury in a home lab due to the ease which scattered droplets fall into cracks and evaporate.

That said, if distillation is out, then try bulb-pipetting from the center of the mass of mercury to avoid some of the surface contamination. Avoid dropping the mercury when lifting the pipette, since $\ce{Hg}$ is so dense it tends to run out as soon as lifted. Ideally, work in a hood with a large tray with raised sides underneath the apparatus.

If purity is important, electrolysis will yield a higher-grade product... at the risk of working with even more toxic compounds!

Better, get some gallium if you want to play with a liquid metal, and don't need $\ce{Hg}$ to make Grignard reagent or other esoteric substance.


Why vinegar + water to clean stains? The advice for cleaning salt stains (for example from shoes) with a vinegar + water solution is fairly ubiquitous on the internet. Is there a chemical explanation for why the vinegar is important? Why not just water?

Edit: I don't actually know that vinegar helps clean salt stains. There might be actually be no chemical reason for the advice to be valid.

The acid in the vinegar helps to a) dissolve substances or b) makes the substance easier to remove. The water is neutral (more or less) so won't really dissolve any stain or such


Why does stale bread go soft in the microwave? It's fairly well known that stale bread will go soft in the microwave. It makes hard bread edible (albeit a little bit chewy sometimes.)

It's obvious that moisture is being released from somewhere in the bread, and this is what causes the softness.

But the bread doesn't blacken as it would if the carbohydrates were decomposing to carbon, so where (chemically) does this moisture come from?

I believe this has to do with its moisture content. While certainly dry, it still has some moisture, and what little moisture is left gets excited by the microwaves, heats up (the heat helps it soften too) and steams the bread essentially. I think that MaxW's explanation of glutens for the question 'Where does the moisture come from?' is a likely explanation.

Note that if the bread is sufficiently dry enough, this little microwave trick wont help. It is usually advised that you add a few drops of water to the bread before heating up in the microwave in this way. This is to ensure the bread doesn't actually become harder in the microwave, as it boils the water and will drive it off if microwaved in excess.

Also, note that bread WILL become blackened, charred and start smoking if left in the microwave long enough. I did this by typing 30 minutes instead of 30 seconds. It started smoking around 5 minutes.


How does strength of boric acid solution increase in presence of salicylic acid? A while ago, I read that salicylic acid can make boric acid solution strongly acidic when it's added to it.

To my knowledge, I know that boric acid becomes a strong acid in presence of cis-diols (except ethylene glycol). I also know that it happens due to increased stability of conjugate base by chelation using polyhydroxy compounds.

But I can't really make out the case with salicylic acid. I will be really thankful if you could shed some light on this topic.

Equilibrium will be far to the right, as a stable six membered ring is formed. The proton speaks for itself.

References: Queen, A. The kinetics of the reaction of boric acid with salicylic acid. Can. J. Chem. 1977, 55 (16), 3035–3039 DOI: 10.1139/v77-421.


How is heat physically released in an exothermic process? Many non-covalent interactions (e.g. electrostatic interactions) result in decreased entropy and are driven by negative enthalpy. If the heat released in the reaction has its physical basis in an increase in the average kinetic energy of the degrees of freedom contributing to the heat capacity of the solution (i.e. surrounding water molecules move faster), how exactly does binding exert a force on these degrees of freedom? How does binding cause increased thermal motion of solution molecules? Is it simply that the two attractive moieties (e.g. a lysine and a glutamate residue) are accelerated towards each other as they convert potential energy into kinetic, and then release heat via collisions with solution molecules? Is all exothermic heat release the result of potential energy being converted into kinetic energy in reaction products and the subsequent collisions with solution?

I think that you are on the right track. I try to explain how I see things below.

As a bond forms the reactants must absorb some energy from their surroundings so that the activation barrier between reactants and products may be surmounted. (By the Boltzmann distribution there are always a few collisions with far more energy than the average). At the top of the barrier is the transition state (TS), for this to be crossed the excess energy has to be taken away from the TS more rapidly that it may cross back to reactants.

The transition state is effectively a very hot species; it has far more energy than the surrounding medium. This can be thought of as a species having a large number of vibrational and rotational quanta excited. As far as is known$^*$ a transition state lasts for only a few tens of femtoseconds to a few picoseconds because vibrational/rotational energy is lost very rapidly to the surroundings. This excess energy is carried away also as kinetic energy of the products if, say, two molecules of product are formed. As time progresses this energy excites more and more vibrational and rotational modes in the surrounding medium, which may be a solvent or protein's backbone and residues. Eventually this energy diffuses away from the reaction site (via anharmonic vibrational couplings) and the product eventually has vibrational and rotational modes excited with quanta in accord with the prevailing temperature.

$*$ Although transition states are universal there is a paucity of actual, i.e. time resolved spectroscopic data on their properties.

(A more detailed approach has been given by Kramers (see M. Daune ' Molecular Biophysics for a discussion). In the condensed phase 'friction' or 'drag' on the reaction coordinate slows the reaction (compare to gas phase) and the reactants 'diffuse' towards and across the transition state. 'Friction' here means vibrational and rotational energy affecting the reaction coordinate, by impeding or promoting, by turns, the progress of the reaction. The Kramers approach allows a consistent approach to  gas phase reactions from isolated molecules, then in the presence various pressures of inert gas right through to reactions in viscous solution.)


Isopropyl alcohol to clean Nylon plastic I have a hardwood wooden frame, possibly lacquered, containing some very dirty nylon plastic pieces, a bit like the teeth of a comb, where the frame goes along top and bottom.  I know it's nylon because the manufacturers say so, ditto they mention lacquer.
I would normally just scrub the nylon section with soapy water, but the wooden frame MUST NOT warp!
I have some isopropyl alcohol, 99 or 97%, i can't remember, and propose to wipe the nylon clean.  It's 120cm long and has lots of slots in it, so this may take quite a lot of propanol.
My question is:  Is solid nylon 'plastic' affected by isopropyl alcohol of this %?
If I am better off diluting it a bit, would it then warp the wood?  I know it will take the lacquer off - it's not good with furniture finishes generally!
I do know to do this job outside :-)
Thank you

Your nylon plastic should be thoroughly resistant to concentrated isopropyl alcohol.

According to this encyclopedia.com entry, nearly all of the nylon produced (in the US anyway) is one of two types: type 66 and type 64. Each of these is stated in the article to be insoluble in water and most organic solvents, but soluble in strong acids.

As some anecdotal evidence of nylon's resistance to common solvents, this Quora discussion referenced a frustrated biker's need to find a suitable solvent for removing some nylon that had melted onto a motorbike exhaust pipe. There was really no good consumer-type of solvent listed (and nothing remotely chemically similar to isopropanol), and it actually included a link to a patent for a compound specially designed for dissolving nylon-based materials, just because they are so chemically resistant.

That all said, it wouldn't be a bad idea to do a "spot test" of the isopropyl alcohol on a small, inconspicuous as possible piece of the plastic. There's always an outside chance that it's not exactly what you think it is or that for some reason your particular nylon-based plastic is not as immune to a solvent like isopropanol as the above arguments imply that it should be.


Reverse Cannizzaro reaction to make aldehydes? Is it possible (and practicable) to reverse Cannizzaro reaction (disproportionation of aldehydes to carboxylic acids and alcohols) to produce aldehydes from carboxylic acid and alcohol without need to use neither chromate or $\ce{LiAlH4}$?
For example: 1,4 butanediol + 1,4 butanodioic acid - > 1,4 butanedial

No, this is not practicable because aldehydes are higher in free energy than either alcohols or acids.  Thus, the equilibrium constant of the Cannizaro reaction lies strongly against the aldehydes.

The case of formaldehyde may be even more extreme than the example reaction you gave, but it is what I could find data for.

The eQuilibrator database lists the equilibrium constant for the reaction of

$$\ce{2 HCHO + H2O <=> CH3OH + HCOOH}$$

as $$K'_{eq} = 3.4 × 10^{13}$$

This is for each reactant at 1 mM in concentration, at pH 7 and an ionic strength of 0.1 mol / L.  Thus, you'd need a teramole or two of formaldehyde to get one mole of acid/alcohol.  The thermodynamics for other aldehydes may be slightly more "favorable" but you aren't going to get 13 orders of magnitude of favorability for any "regular" pair of aldehyde and acid/alcohol.


Are these two reactions equivalent? I'm supposed to balance a redox equation. I arrive at
$$
\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-}
$$

The "official" solution is
$$
4 \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} \rightarrow 4\ce{NH3} + 2\ce{S} + 2\ce{H2O}
$$

There is no information about if this is occuring in basic or acidic medium. But IMHO in the first equation the charges and all atoms are balanced, so are these equations equivalent?

Let's start with the question: How do we get from your answer to the official answer?  We can transform your answer into the official answer by first adding 3 ammonium ions to each side:

$\begin{aligned}\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-} \\
+ 3 \ce{NH3}+3\ce{H^+} & \rightarrow + 3 \ce{NH3}+3\ce{H^+}\\
\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow 4\ce{NH3} +3\ce{H^+} + 2\ce{S} + 3\ce{OH-}
\end{aligned}$

Then noting that the $\ce{H+}$ and $\ce{OH-}$ will 'neutralize' one another on the product side, producing $\ce{3H2O}$, one of which will 'cancel' the water molecule on the reagent side:

$\begin{aligned}
\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow 4\ce{NH3} + 2\ce{S} + 3\ce{H2O} \\
-\ce{H2O} & \rightarrow - \ce{H2O}\\
4\ce{NH4+}+2\ce{S^2-}+\ce{O2} & \rightarrow 4\ce{NH3}+2\ce{S}+2\ce{H2O}
\end{aligned}$

Producing what is the 'official' result.  Now we can ask: are the two answers equivalent?  Because $\ce{NH4+}$ is a weak acid in water, it will dissociate and we can write your answer as:

$\ce{H2O + NH3 + H+ +2S^2- + O2} \rightarrow \ce{NH3 + 2S + 3OH-}$

Your answer is a balanced redox reaction somewhere in between acidic and basic conditions, which is a rather unusual formulation.  I'm not going to go so far as to say, this can't happen, but it is unlikely, and certainly not how we typically describe balanced redox reactions.


Can a boat conformer of cyclohexane be most stable? It is well-known that the "chair" conformation of cyclohexane is the most stable of its conformers. But is there any cyclohexane derivative where other notable conformers (boat, half-chair) are the most stable?

My own intuition is that if there are 4 very bulky groups in the 1, 2, 4, 5 positions then the boat conformer would be preferred.

TL;DR:

It is highly unlikely that there is a non-multicyclic derivative with a stable boat conformation.
It needs a lot of convincing and quite bulky groups to destabilise the chair conformation and stable another conformation like the twist-boat.

Conformations of cyclohexane

There are quite a few imaginable conformations,[1] but for cyclohexane itself there are only two stable: the chair and the twist-boat conformation. The other ones are transition states, which are given in the comments of the linked question.

A related question is Where is bifurcation situated on cyclohexane isomerisation PES? Unfortunately it is not answered. However, more insight is given by the articles of H. Rzepa on his blog and the pdf of some lecture notes.[2]

Below is an animation of the boat conformation's imaginary frequency, which will yield the twist-boat conformation upon relaxation.

Why is this actually the case? There is a very intuitive answer: There are eclipsed conformations where they do not have to be, see the image below. This stress can be easily relieved in this twist-boat conformation, in which all bonds are staggered (and the chair conformation, too, as it is the global minimum).

You can even do a little bit more, for example looking at non-covalent interactions. I have optimised all structures on the RI-BP86/def2-SVP level of theory using ORCA 4.2.0. Then I have used Multiwfn to analyse NCI with promolecular densities.[3,4] You will see that the isosurface extends further to the hydrogens in the boat and twist boat conformations. Note that this is more a visualisation of these properties than actual evidence for their existence. It should help in understanding, but it shouldn't be used as a general argument. There are plenty of sources explaining why staggered conformations are generally lower in energy than eclipsed ones.

Locking in a boat conformation

You have probably heard of a conformational anchor. Adding groups so that they will lock into a specific conformation. Like when you attach a tert-butyl group in one of the positions, it will rise the energy of the conformation when it is in axial position.

If you do not want to consider extended ring systems to lock the conformation into position, like the earlier mentioned norbornane and derivatives, you'll probably get structures to prefer a twist-boat conformation, but probably not one that will result in an actual boat conformation. I have tried some conformational sampling with CREST on the semiempirical GFN2-xTB level of theory and multiple attempts have failed so far.[5] In my attempts the closest I have gotten was with (1R,2S,3S,4R,5S,6r)‐1,2,3,4,5,6‐hexamethylcyclohexane, where the twist boat conformation is only about $\pu{1 kJ mol-1}$ higher in energy, see the images below (hydrogen omitted for clarity).

I encourage you to get a molecular viewer and look at the structures yourself.[6] You will quickly see how crowded the ring actually is and that there is not much space to put much larger groups anyway. I believe this illustrates the problem well.

However, I was able to push it further and have found a beast that is predominantly available in the twist-boat conformation. Therefore I have added tert-butyl moieties in 1, 2, 4, and 5 position just as you have proposed, (1R,2S,4R,5S)‐1,2,4,5‐tetra‐tert‐butylcyclohexane. The CREST program found two (very, very similar) conformers within a $\approx\pu{25 kcal mol-1}$ window.[7] There was no chair conformation in this window, and the second structure is barely in the given interval, therefore I will not include it here.[8]

Notes & References:

For example, see what Geoff demonstrates for pyranose sugars in Conformations of Cyclohexane. There is also a more general article about ring puckering: https://doi.org/10.1021/ja00839a011, pdf via the Internet Archive
The pdf via the Internet Archive
Johnson, E. R.; Keinan, S.; Mori-Sánchez, P.; Contreras-García, J.; Cohen, A. J.; Yang, W. Revealing Noncovalent Interactions. J. Am. Chem. Soc. 2010, 132 (18), 6498–6506. DOI: 10.1021/ja100936w.
I have previously written about it: How to identify hydrogen bonds and other non-covalent interactions from structure considerations?
(a) Mulliken Center for Theoretical Chemistry, xTB on Github, xTB documentation, CREST: Phys. Chem. Chem. Phys. 2020, xTB: J. Chem. Theory Comput. 2017, 13 (5), 1989–2009 & J. Chem. Theory Comput. 2019, 15 (3), 1652–1671
A good open source choice to view the following Cartesian coordinates is Avogadro.

  36
 GFN2-xTB       -37.96990166 (“chair”)
 C          1.2520828623       -0.5447977556        0.4005613406
 C          2.5529154246       -1.2427641817       -0.0035234170
 H          2.5670950259       -1.4609755375       -1.0683723566
 H          2.6812224740       -2.1742430344        0.5412168974
 H          3.4037282124       -0.6057253357        0.2302177479
 H          1.1934851832       -0.5794558732        1.4947912916
 C          1.2932369007        0.9399180068       -0.0046696921
 C          1.6543517514        1.1471873131       -1.4778364887
 H          1.5052577715        2.1882681334       -1.7560128481
 H          2.7000386472        0.9041963949       -1.6490857234
 H          1.0549606283        0.5311695599       -2.1388312879
 C          0.0004268408        1.6743272530        0.3958681428
 C          0.0003369080        1.9795490247        1.8972278766
 H         -0.8803447484        2.5595415091        2.1655307399
 H          0.0002805364        1.0701573015        2.4915141491
 H          0.8810286139        2.5594814678        2.1656262412
 H          0.0007645822        2.6443828331       -0.1162274411
 C         -1.2926832357        0.9406761004       -0.0050185229
 C         -1.6531427367        1.1479070715       -1.4783512638
 H         -2.6988310362        0.9052087273       -1.6499938197
 H         -1.0536713182        0.5316482470       -2.1390476329
 H         -1.5036207617        2.1889143513       -1.7565761588
 H         -2.1018819331        1.4053817317        0.5738089641
 C         -1.2524695951       -0.5439792518        0.4004522605
 C         -2.5537665456       -1.2411162270       -0.0035423382
 H         -2.5679998358       -1.4596918660       -1.0683148885
 H         -2.6827814315       -2.1722962125        0.5415232111
 H         -3.4041581021       -0.6033840583        0.2298854735
 C         -0.0003947478       -1.2538292818       -0.1423844973
 C         -0.0008645772       -2.7329645415        0.2464953252
 H          0.8741384569       -3.2373492283       -0.1525769397
 H         -0.0008947662       -2.8427074989        1.3295318583
 H         -0.8762127488       -3.2367530896       -0.1525580475
 H         -0.0003167563       -1.1995260135       -1.2346248660
 H         -1.1939183223       -0.5785009660        1.4946918825
 H          2.1024953812        1.4040515111        0.5745201353
  36
 GFN2-xTB       -37.96945643 (“boat”)
 C          1.3009843031        0.3904694151        0.6116442813
 C          2.7523842286        0.8697430480        0.5759961739
 H          3.1370164012        0.8924092382       -0.4400705597
 H          3.3851298683        0.2141081553        1.1682244589
 H          2.8236187015        1.8700221417        0.9985166065
 H          1.0589906677        0.1993477976        1.6630540690
 C          0.3343786187        1.4770082561        0.0930304054
 C          0.8370012219        2.1686754632       -1.1755196945
 H          0.0729240573        2.8439373660       -1.5550562687
 H          1.7313073547        2.7544698863       -0.9832626635
 H          1.0594535849        1.4416329460       -1.9531205354
 C         -1.0843113664        0.9157582793       -0.1622022415
 C         -2.1249907625        1.9785105445        0.1901580387
 H         -3.1130958026        1.6979927608       -0.1639981009
 H         -2.1696260162        2.1243458743        1.2679063910
 H         -1.8631471288        2.9288081963       -0.2704747138
 H         -1.1814251431        0.6916654706       -1.2293744660
 C         -1.3009787861       -0.3905116297        0.6116386064
 C         -2.7523814313       -0.8697957935        0.5761213628
 H         -3.1371030728       -0.8924849595       -0.4399103572
 H         -3.3850864608       -0.2141683561        1.1683995711
 H         -2.8235618371       -1.8700681103        0.9986656463
 H         -1.0588901236       -0.1993916529        1.6630242750
 C         -0.3343953584       -1.4770058647        0.0929158564
 C         -0.8370714193       -2.1686218804       -1.1756271655
 H         -0.0730166306       -2.8438887577       -1.5552044908
 H         -1.7314363652       -2.7543627913       -0.9834449686
 H         -1.0594386262       -1.4414807272       -1.9531506454
 C          1.0843324761       -0.9157815011       -0.1622407002
 C          2.1250455546       -1.9784641695        0.1902373202
 H          2.1696137891       -2.1242687584        1.2679947854
 H          1.8631584859       -2.9287608420       -0.2703581429
 H          3.1131889321       -1.6980202130       -0.1638895009
 H          1.1815268495       -0.6916894600       -1.2294091391
 H         -0.2613905266       -2.2395020738        0.8774919294
 H          0.2613017165        2.2394206185        0.8776918141

The default is actually $\pu{6 kcal mol-1}$, but I refuse to give in to these medieval units.
Cartesian coordinates of the lowest structure of (1R,2S,4R,5S)‐1,2,4,5‐tetra‐tert‐butylcyclohexane:

  66
 GFN2-xTB       -69.56788868 (“twist-boat”)
 C          1.2637535413       -0.4535845271       -0.7036170298
 H          0.8234262820       -0.3514280493       -1.7011134458
 C          0.2874611950       -1.2968906916        0.1262691467
 H          0.4862260613       -1.1072837877        1.1760079950
 H          0.4711152326       -2.3509422427       -0.0576548430
 C         -1.2138745760       -1.0160032197       -0.1668918408
 H         -1.5059514690       -1.6533453308       -1.0111946700
 C         -2.0580638654       -1.5542451449        1.0386873052
 C         -3.5291500453       -1.7879146105        0.6673596086
 H         -4.0389307694       -2.2795212021        1.4949415613
 H         -3.6040044082       -2.4413551590       -0.2001570459
 H         -4.0628013225       -0.8693626960        0.4623391177
 C         -1.5380651982       -2.9543377210        1.4269474062
 H         -1.4806005503       -3.6016258516        0.5545295073
 H         -2.2263828398       -3.4107111998        2.1359237707
 H         -0.5631185572       -2.9193311608        1.9023111302
 C         -1.9997742111       -0.6828458212        2.2953624607
 H         -2.6171623926       -1.1292449936        3.0737145663
 H         -2.3680382876        0.3190177054        2.1083809564
 H         -0.9863340848       -0.6190135645        2.6836308403
 C         -1.2637584056        0.4535847074       -0.7036134097
 H         -0.8234213043        0.3514344229       -1.7011060021
 C         -0.2874690577        1.2968787699        0.1262850454
 H         -0.4862254771        1.1072466416        1.1760203260
 H         -0.4711273811        2.3509334181       -0.0576150207
 C          1.2138643519        1.0160021804       -0.1668873426
 H          1.5059281496        1.6533459288       -1.0111931129
 C          2.0580613014        1.5542510848        1.0386835495
 C          3.5291313266        1.7879648142        0.6673219595
 H          4.0628193815        0.8694121253        0.4623879257
 H          4.0388969819        2.2796720426        1.4948537541
 H          3.6039479309        2.4413286270       -0.2002563066
 C          1.5380254724        2.9543167424        1.4269934992
 H          0.5630890040        2.9192624252        1.9023748186
 H          1.4805221407        3.6016312375        0.5545978702
 H          2.2263377642        3.4106933389        2.1359728252
 C          1.9998372135        0.6828176443        2.2953360365
 H          2.3681483571       -0.3190197145        2.1083145750
 H          0.9864074015        0.6189230717        2.6836201429
 H          2.6172202830        1.1292244475        3.0736878514
 C         -2.5734381625        1.2159695783       -1.0649400350
 C         -3.4459088573        0.3435906290       -1.9774479664
 H         -3.6675713251       -0.6208273126       -1.5422795084
 H         -4.3867433184        0.8479997186       -2.1893090423
 H         -2.9342730730        0.1735856167       -2.9232393678
 C         -2.1913064757        2.4425123599       -1.9212170334
 H         -1.7500553845        3.2379982242       -1.3271846647
 H         -3.0851290926        2.8470606980       -2.3920018369
 H         -1.4898086285        2.1738538773       -2.7080018465
 C         -3.3790378509        1.7599694390        0.1164846666
 H         -3.8515396128        0.9790436919        0.6998702684
 H         -2.7421665034        2.3465041110        0.7744832021
 H         -4.1670675305        2.4137947017       -0.2548140206
 C          2.5734370335       -1.2159683519       -1.0649370663
 C          3.3790279628       -1.7599706209        0.1164927971
 H          3.8516735927       -0.9790428294        0.6997637198
 H          2.7421180248       -2.3463565373        0.7745861394
 H          4.1669484076       -2.4139382233       -0.2547885352
 C          3.4459325455       -0.3435930502       -1.9774226323
 H          4.3867849824       -0.8479910147       -2.1892312850
 H          3.6675602100        0.6208308804       -1.5422534239
 H          2.9343365617       -0.1736062497       -2.9232383093
 C          2.1913055807       -2.4425004188       -1.9212296089
 H          3.0851285923       -2.8470461925       -2.3920160549
 H          1.7500476265       -3.2379922732       -1.3272105313
 H          1.4898131110       -2.1738289364       -2.7080150254


What is the liquid used in instant shoe shiner? What is the liquid used in instant shoe shiner products? Ive tried refilling it with WD-40 but I would like to know if there are any cheap alternatives to refill it?

It's silicone oil.

You can use a mineral oil. You can find it as baby oil and if you don't like the odor, also can be found as fleet enema products (eg. http://fleetlabs.com/fleet_enema_products.php).
For larger and cheaper amounts, you can buy it in computer hardware store, farm supply store, or even in vet clinic.


Why aren't Boron and Aluminium assigned to group 3 of periodic table? What determines the group? I've been curious about this 3D representation of the periodic table "Mendeleev's Flower" and was trying to study it, wondering if it reveals any regularities that are not obvious from classic periodic table.

But since it's hard to look at a 3D table on a flat computer screen, I found flat equivalents.

Theodor Benfey's periodic table (1964):

J. F. Hyde's periodic table (1975):

There I found a discrepancy: Hyde assigned Boron and Aluminium to group 3 (IIIA), whereas everywhere else B and Al are in the group 13 (IIIB).

I can totally see why Hyde split the table like this (apart from being obsessed with Silicon): such a split creates an absolutely stunning symmetry.

You can see this symmetry on the Hyde's table with colors and dashed lines that cross Silicon and connect related elements.

This can't be just a stretch/exaggeration/far-fetching/"squeezing the universe to fit into an appealing model, even though the universe resists". It's just too good!

Well, to be honest, he did do some universe-squeezing: B and Al have electron configuration of s²p¹ — just like other elements of group 13 (IIIB), whereas elements of group 3 (IIIA) have electron configuration of d¹s². Since Al and B appear in the d¹s² column, Hyde had to make a note inside B and Al cells that they actually are s²p¹.

But the symmetry is just too good. It reveals stunning periodicity that no other periodic table is capable of.

And then I remembered this short-form periodic table:

Russian Wikipedia says the short form was banned by IUPAC in 1989 for some reason, but it's still used in Russian schools because it makes figuring out valence values easier.

And this table does not have the split. Boron and Aluminium appear in the same column regardless of whether you assign them to 3 (IIIA) or 13 (IIIB). I think this is absolutely beautiful.

I believe this Al/B positioning thing has scientific significance, but I don't understand the meaning of the groups.

My questions:

What does the group actually mean? Is it simply the electron configuration and nothing else?
What determines Al and B being placed in group 3 (IIIA) and not 13 (IIIB), other than the electron configuration?
Why is the short-form periodic table considered obsolete, even though it clearly has something going on that the full-form table is not revealing?
Is the Hyde's table missing any periodical properties that the classic table has?

Back in 1933, John Drury Clark published 1 his "race track" periodic table that subsequently became famous as the cover of Life magazine in 1949:

Looks like B and Al are nicely positioned. Just need to add an outer track for elements 105 - 118, element names and symbols for elements 97 - 103 and element 41 is niobium, Nb.

1 Clark, John D. (November 1933). "A new periodic chart". Journal of Chemical Education. 10 (11): 675–677. Bibcode:1933JChEd..10..675C. doi:10.1021/ed010p675.

Clark is also the author of Ignition! An Informal History of Liquid Rocket Propellants, Rutgers University Press Classics, New Brunswick, NJ, ©1972 and 2017.


Finding the percentage of pure CaCO3 I found this in a textbook; it seems that the question is poorly worded. It goes as follows:

  $1.5~\mathrm{g}$ of chalk were treated with $10~\mathrm{ml}$ of $4~\mathrm{N}$ $\ce{HCl}$. The chalk was dissolved and the solution made to $100~\mathrm{ml}$. $25~\mathrm{ml}$ of this solution required $18.75~\mathrm{ml}$ of $0.2~\mathrm{N}$ $\ce{NaOH}$ solution for complete neutralization. Calculate the percentage of pure $\ce{CaCO3}$ in the sample of chalk.

I don't get it when the question says the chalk was dissolved and the solution made to $100~\mathrm{ml}$. Does this refer to the chalk which already reacted with $\ce{HCl}$, or was another $1.5~\mathrm{g}$ sample of $\ce{CaCO3}$ taken?

The question is a bit poorly written. But it also gives a sense to the procedure of dissolution of chalk upon treatment with $\ce{HCl}$ and dilution of the solution.

The reaction that takes place after the treatment with $\ce{HCl}$ is;

$\ce{CaCO3}$+ 2$\ce{HCl}$ → $\ce{CaCl2}$ + $\ce{H2CO3}$

$\ce{H2CO3}$ decomposes into $\ce{H2O}$ and $\ce{CO2}$.
$\ce{CO2}$  is a gas. It bubbles up out of the solution.

$\ce{CaCl2}$ is soluble in water. So it should be dissolved by adding water.Further, it should then be diluted to the required volume.

"The chalk was dissolved" might signify the addition of water to dissolve the water soluble $\ce{CaCl2}$ produced during chemical reation with $\ce{HCl}$.


Molar absorption coefficient of H2SO4(aq) for UV-vis spectrophotometry I am wondering if there is anything in the literature that can tell me what is the molar absorptivity (i.e. molar extinction coefficient) of aqueous sulfuric acid at a given wavelength (in my case, I am interested in $\lambda_\mathrm{max}(\mathrm{ferroin}) = \pu{508 nm}$).

In addition, is there a good resource in the literature to look up these values for common substances? I haven't yet been able to find anything of the sort.

I never ran into uv absorption data for (aqueous) sulfuric acid and I doubt that there is any absorption at $\lambda$ = 508 nm.

There is an old paper by K. F. Palmer and D. Williams in Appl. Optics, 1975,14, 208-219 (DOI) on the optical properties of sulfuric acid droplets with respects to the composition of clouds on Venus and a more recent one from the group of J. B. Burkholder in Phys. Chem. A, 2003, 107, 1112-1118
(DOI) on Vibrational and Electronic Spectroscopy of Sulfuric Acid Vapor. Both, while interesting and therefore mentioned here, are not relevant for your experiment.

My suggestion is simple:

Measure your ferroin sample in aqueous sulfuric acid (or whatever solvent is required) and use the same solvent as a blank. Most spectrometers will automatically perform the subtraction of the blank from your sample!

Concerning a (free) source of UV-vis absorption data, there is the NIST Chemistry WebBook.


Enthalpy of combustion of 1-butanol 1-butanol $\ce{C4H10O}$

So I have absolutely no idea how to find the enthalpy and I did some looking on the internet with no use since I don't understand.

I came here hoping someone generous could do the sample calculations so I can understand and do the rest my calculations and test for my experiment.

That's the data for one the mixture of alcohol.

How would I figure out enthalpy for test 1?

(maybe I could cut out test 2, 3 and average in the picture since its irrelevant in figuring out the enthalpy of test 1, and causes confusion)

For temperature before and after. The solution was used to heat up 200 ml of water. By burning the mixture, igniting it.

Answering your question as it stands is impossible, as we readers do not have enough information about your actual experiment. What do you mean with 'enthalpy of 1-butanol'? You can only measure changes in enthalpy, so you need to specifiy what 'change' you're looking at.

You could mean the enthalpy of evaporation or the reaction enthalpy of some chemical reaction or hundreds of other possibilities. It's essential to clarify what process you're referring to, so one can associate the corresponding starting (e.g. liquid) and end state (-> gas) to this energy change.

However, you stated that 200 ml of solution were heated up by whatever your process was. At constant pressure, the enthalpy required to change the temperature by $\Delta T$ is

$$\Delta H = m c_p \Delta T$$

where $c_p$ is the isobaric heat capacity. Knowing the value of $c_p$, you can calculate this energy change. Now where did that energy come from? Presumably, from your process. At this point, assume that there are no energy losses (f.e. neglect the heat lost to the surrounding air). Then by conservation of energy, the enthalpy increase measured by the temperature increase must be equal in size to the energy released by the process.

Finally, note that by convention we say if a process releases energy, it's enthalpy change is negative (we could've just done it the other way around; think of it as the same idea as defining a sea level).


Teeth friendly alcohol drinks What is the best alcohol for your teeth according to the pH?

Is it true that if the acidity of the drink is higher (and ph is lower) than it's worse for your teeth (enamel erosion + bacteria friendly acidic environment) than if the alcohol drink is less acidic?

Does vodka have the highest pH of all the alcohol drinks out there and therefore is the best for your teeth?

Is quite reductive think that pH is the onlyfactor that determinate the health of your teeth, the content of carbohydrate have a strong effect too.
However in fact to my knowledge clear vodaka have one of the highest pH and very few carbohydrates so indeed is one of the most "teeth friendly alcohol drink".
These is take from Victor R. Preedy, Ronald Ross Watson Academic Press, 24/dic/2004 Comprehensive Handbook of Alcohol Related Pathology with the adding of Vodka and Coca Cola.

\begin{array}{|c|c|c|c|}\hline
Alcolic \space beverage & Carboydrate \space content(\%) & pH \\ \hline
Beer               & 3.0-5.0   &  4.1-4.5    \\ \hline
Wine & 0.0-12.9 &  2.8-3.8     \\ \hline
Sweet \space liquor           & 30.0-31.0 & 3.3-3.9 \\ \hline
Strong \space alcohol & 0.0-1.2 &   6.5-6.9\\ \hline
Vodka                & very \space few       & 6.0-7.0        \\ \hline
Coca Cola & 10.6(?) & 2.8\\ \hline
\end{array}

However keep out  from Vodka lemon!


Dehydration of primary alcohols without a β-hydrogen What happens in dehydrations of primary alcohols (in presence of acid catalyst at high temperatures) which don't have a β-hydrogen?

For example, in the case of neopentyl alcohol, will it dehydrate as it doesn't have a β-hydrogen? Obviously, E2 reaction will not be possible, but can it react via an E1 reaction, with a rearrangement in the carbocation step?

In this case, $\mathrm{E2}$ elimination is impossible regardless of condition used, because of lack of $\beta$-hydrogens. However, in acidic conditions, it is possible to have elimination reaction. Since 2,2-dimethylpropanol is a $1^\circ$-alcohol, initial carbocation formation is difficult. However, this formation of carbocations is accompanied by a structural rearrangement, which is the driving force for the $1^\circ$-carbocation formation, which is otherwise not favored:

Such rearrangements take place by a 1,2-shift of a neighboring alkyl group or aryl group or hydrogen, and are favored when the rearranged carbocation is more stable than the initial cation. In this case, migrating group is methyl group and carbocation changed from $1^\circ$ to $3^\circ$, which is very stable compared to initial carbocation and favored. Once $1^\circ$-carbocation formed, it quickly rearranged to $3^\circ$-carbocation and as a result, the reaction make forward progress. It can also be considered a concomitant elimination of water and methide migration where methide group acting as an intramolecular nucleophile (See the mechanism shown in bottom box of the scheme). Either way, the initial step is very slow because of very high activation energy:


What reactions create a lot of gas really quickly? I'm looking for a chemical reaction that produces a large amount of a gas in a very short amount of time (like less than a second). What are my options?

The reaction used for car airbags involves sodium azide (NaN3). A spark initiates the reaction and it generates gaseous nitrogen really fast.

I don't know much more about it but you can certainly find some good informations on google!


What could be causing noisy pH measurements? I designed and built a hydroponics system with pH logging and I am trying to understand why pH measurements in the nutrient reservoir vary ~ ±0.2 pH while measurements in a separate bottle of probe storage solution only vary ~ ±0.01 pH.  The nutrient solution is in a 3.5 gallon plastic pail, as shown below.  The pH probe is the one with a blue top.  The black probe measures EC and removing it has no effect on the pH noise in the reservoir.

The plot below shows the  logged pH. The abrupt change is when the probe is moved from the nutrient solution to the storage solution.  Note the relatively large noise in the nutrient reservoir and the nearly imperceptible noise in the storage solution.

In the plot,the pH is measured every 5 minutes at the start of a spray cycle, when the pump is off and has not been running for the last ~4.5 minutes.  Note that similar noise occurs when measuring at a rate of 1 Hz.  During the storage solution measurements, the probe tip is in a separate O-ring sealed bottle.  The bottle is sitting at the spot where the pH probe would normally be inserted into the reservoir.

What are potential causes of the pH noise in the nutrient solution?

Several factors may influence your sensor readings:

Electromagnetic noise. However, since EMF should be directly proportional to pH, it seems to be not your case. If it were the main reason, the same amount of noise would show up in both the nutrient and the storage solutions.
Temperature. I've included this one because it is unclear from your post whether you are measuring the actual solution temperature when you calculate pH from EMF, or just assume that the temperature is some constant value. The Nernst equation may be used to estimate the amount of "noise" (i.e. the difference in EMF) one may expect from the temperature fluctuations: $\Delta E=2.303\frac{R\Delta T}{F}pH$. That is, if we can neglect the temperature dependence of pH itself, i.e. when $\Delta T$ is relatively small (when $\Delta T$ is large we can no longer neglect the temperature dependences of equilibrium constants). If, for some reason, the daily temperature change in the nutrient bath is different than that in the storage solution, you may expect to get slightly different readings.
Math. We should not forget that the value measured (pH) is logarithmic, and that it is the concentration (well, activity) of protons that makes the difference. For example, to change pH from 6.45 to 6.05 (±0.2), you need to add just around $5.4\cdot 10^{-7}$ mol/l of protons, but to change it from 4.26 to 4.24 (±0.01) - $2.6\cdot 10^{-6}$, or almost 5 times as much. In other words, the same actual $[\textrm{H}^{+}]$ fluctuations cause different pH changes around different pH values. In turn, these $[\textrm{H}^{+}]$ fluctuations may be due to slight inhomogeneities of the nutrient solution, differences in dissolved carbon dioxide concentration, etc. The more neutral your solution is, the more sensitive, with respect to $\Delta [\textrm{H}^{+}]$, your pH electrode becomes. In this regard, you may want to prepare a few solutions with different pH and try to investigate the noise vs. pH dependence. Or do something even more simple - as a reference system, instead of the storage solution (which, in addition, probably possesses some buffer characteristics), try to use something more neutral, like water, and see if the noise is as large as in the nutrient solution.

Since my analytical chemistry is a bit rusty, no other potential cause comes to mind.


Why do (did) they make trans fat instead of using saturated fat? When making chocolate unsaturated fat is used for a good melting point (and perhaps health reasons.)

To make chocolate edible it has to be a bit more solid, so other fats were required. Therefore they use trans fats, which are unsaturated fats made saturated in cis or trans form.

Why do they do that instead of just mix it with normal saturated fat? Why use trans fats?

Catalytic hydrogenation tends to make trans fats... unfortunately catalytic hydrogenation is the easiest (read lowest cost) way of hydrogenating fats so it became the industry preferred way of hardening fats before anyone realized there was a health risk to trans fats. Cheap plant oils can be hardened by hydrogenation... and hydrogenated plant oils are generally cheaper than naturally saturated fats.

Because they are unnatural, trans fats also don't go rancid so quickly... so goods made with trans fats can have longer shelf lives. Which cuts costs. Given so much of mass food production is severely cost conscious, it was inevitable that trans fats became widely used and are now having to be removed as ingredients.


Does the shielding effect affect atomic bonding? Electrons within an atom don't experience the full charge of the nucleus due to the shielding effect? Likewise, when creating atomic bonds, does the shielding effect affect the attraction between two bonding atoms?
Thanks for any help offered

Yes, it does. It is due to the poor shielding effect of f-orbital electrons that we see the inert pair effect; that is, $\ce{Pb^2+}$ is more stable than $\ce{Pb^4+}$. Lower oxidation states are more stable as we go down the 13, 14 & 15 group. This affects the number of bonds that an atom can form.


Increase in Viscocity in Gases with Temperature Why does viscocity increase with temperature in gases? When they gain kinetic energy, they should be more free right?

From the basis of the kinetic theory of gases, viscosity in gases increases with temperature because the increase in kinetic energy will increase the frequency of collisions between the molecules. This increased number of collisions decreases the ability of the molecules to diffuse in the desired direction (i.e. the gas doesn't flow down the pipe to a reactor).


How does acetone help to make a liquid cleaner stronger? Acetone, as a solvent, seems to behave somewhat different than other solvents. It seems to solve things other unpolar solvents do not.

For example, I have two cleaner sprays from the same brand, and the one with acetone seems to be "stronger" in some way. I think I have noticed that difference before elsewhere, so I assume at least part of what makes it stronger is the acetone.

Well, acetone is at the same time polar (so it likes water) and lipophilic (so it likes fat). As such, it is a very good solvent.


What makes a ligand stronger than another? A ligand is a neutral molecule or negative ion with at least one lone pair of electrons which forms a dative covalent bond with a complex ion that has a vacant, energetically accessible orbital.

It is said that in terms of ligand strength, or 'binding power', the cyanide ion is stronger than ammonia, which is in turn stronger than water, which is in turn stronger than the chloride ion.

What determines the strength of a ligand?

One definite factor is the ability of the ligand to stabilize its lone pair. You observe that the cyanide ion is a better ligand than the nitrogen ligand. Let's think about this. The lone pair that reacts on the cyanide ion is the lone pair on the carbon atom. Not the lone pair on the nitrogen atom.

Carbon is less electronegative than nitrogen. Thus carbon is less able to stabilize its valence electron density, which in turns enhances the reactivity of its valence electron density through increased negative charge density.

Same with sulfide and oxide anions. Sulfur is less electronegative than oxygen, and thus less able to stabilize valence electron density. Hence the much greater stability of silver sulfide as opposed to silver oxide.

Other competing factors include charge density; the Lewis acid the ligand is coordinating to, medium (solvent), etc. For example the oxide anion is more basic than the sulfide anion in water solvent; i.e. the oxide anion forms a stronger bond with the hydrogen proton in water. But we just saw that silver sulfide is much more stable than silver oxide.


Direction of arrows in electrocyclic reactions We recently started learning about pericyclic reactions and this is a step from a mechanism I'm working on for an assignment. I believe that the arrows on the right are better than the arrows on the left because if I throw in some charges instead of arrows, the outcome I get on the right is somewhat more "stable?" for the reasons stated on the picture. Am I correct in my assumption?
 On the overall reaction,  it's the 6 to 7 step that I'm interested in.

The answer is rather bland and probably not what you want to hear. Electrocyclic reactions are called electrocyclic because they are cyclic. (Please don’t see cyclic circular reasoning in there.) They proceed in a cyclic, i.e. fully symmetric fashion. You simply cannot draw one arrow direction and claim it is the most logical or the only one possible. You could draw these arrows as you drew them, but you could also formulate a radical-based mechanism (each double bond splits into two radicals which combine with the two of the other double bond) at no loss.

Thus, there is no way to say whether left or right. The reaction happens; in a cyclic fashion.


Perplexing claim in Chang and Overby's "General Chemistry" I'm reading the chapter on gases in Chang and Overby's book "General Chemistry: The Essential Concepts" [1, p. 145], and the following passage is causing me some confusion:

The work of the Italian scientist Amedeo Avogadro complemented the studies of Boyle, Charles, and Gay-Lussac. In 1811 he published a hypothesis stating that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules (or atoms if the gas is monatomic). It follows that the volume of any given gas must be proportional to the number of moles of molecules present; that is,
$$V \propto n$$
$$V = k_4 n \tag{5.7}$$

I don't quite understand the emphasized sentence; in particular, I don't understand how the emphasized claim "follows" from the preceding statement. In fact, I don't actually think it does.
For example, suppose instead that the volume of any gas at a given temperature and pressure was inversely proportional to the number of moles of the gas present; that is, suppose the relationship between volume and amount was instead
$$V \stackrel{?!}{=} \frac{k}{n},$$
for some constant $k$. Then it would certainly still be true that equal volumes of different gases contain the same amount of gas. For example, if I had $\pu{5 L}$ of $\ce{H2}$, then, according to the above "false law", I would have $k/5$ moles of $\ce{H2}$. Similarly, if I had $\pu{5 L}$ of $\ce{N2}$, I would have $k/5$ moles of $\ce{N2}$ — the same amount!
So the claim that the proportionality follows from the fact that equal volumes of different gases contain the same number of molecules seems illogical. In fact, it seems to me that the only thing one could logically deduce from the first hypothesis is that there is some relationship between volume and amount which holds for all gases (that is, the relationship doesn't depend on the gas itself).
Am I missing something, or is the statement in the book an error?

Added much later:
To clarify what I'm asking here (though I thought it was pretty clear), consider the following two statements:
Statement A: at a given temperature and pressure, equal volumes of different gases contain the same number of molecules.
Statement B: at a given temperature and pressure, the volume $V$ of a gas is related to the number of molecules by $V = kN$ (where $k$ depends on temperature and pressure).
My point is simply that these two statements are not logically equivalent. B implies A, but not the other way around.
So why is it that the two statements are (seemingly quite often) presented as logically equivalent?
References

Chang, R.; Overby, J. General Chemistry: The Essential Concepts, 6th ed.; McGraw-Hill: New York, NY, 2011. ISBN 978-0-07-337563-2.

The  other answer is, I think excellent.

Your question is basically "but what if there isn't a direct proportion between number of gas molecules and volume?" Instead, that was Avogadro's hypothesis in a nutshell: the volume of the gas is determined exactly proportionate to the number of molecules in it.

You are right in principal that you could have such a "false law" where the number of molecules is inversely proportionate to the volume and still have equal number of molecules for $\ce{H2}$ and $\ce{N2}$.

Let's follow your course of logic. That would mean if I get 10 L of $\ce{H2}$, I would then have $k/10$ molecules. Let's imagine I add more and more volume, to get 100 L then 1000 L of gas. At some point, I must necessarily divide the molecules in fractions in your false law. Certainly a molecule must occupy some indivisible volume, since Avogadro believed in the atomistic theory.

Instead, the obvious solution is that the number of molecules must be a measure of the amount of "stuff." So if we have more "stuff," we need more volume for it to occupy. The expression must be directly proportional between the volume and the number of molecules.


The connection of the "R" in the ideal gas law and osmotic pressure I am studying chemistry and biology at the same time, and have encountered a problem. In the ideal gas law, the $R$ is the ideal gas constant, and its value is $\pu{0.0821 L atm mol-1 K-1}.$

However, in the osmotic pressure formula $\Psi_s = -iCRT,$ the $R$ is called the pressure constant, and its value is $\pu{0.0831 L bar mol-1 K-1}.$ Is it a coincidence that their values are identical (after adjusting for the $\pu{atm}$ and $\pu{bar}$ conversion)?

It's not a coincidence at all!  If you do an online search for "derivation of osmotic pressure", you'll see how $R$ enters into the derivation.

Indeed, that's one of the beauties of the van 't Hoff equation for osmotic pressure – it reveals that (under the modest simplifying assumptions of the van 't Hoff equation derivation) the osmotic pressure created by a dilute solute is equal to the pressure that solute would exert if it instead acted as an ideal gas!:

$$\text{van 't Hoff equation:  } \Pi_i = \frac{n_i}{V} R T = C_i R T$$

$$\text{ideal gas equation:   } p_i = \frac{n_i}{V} R T = C_i R T$$

Here the "$i$" refers to species $i$.


How do I write out the chemical reaction of ammonia with strong acid? Write equations for the reactions of each of: (a) nitric acid (b) sulfuric acid *(c) hydroiodic acid, with each of: (i) sodium hydroxide (ii) zinc oxide (iii) ammonia.

So, the reaction of nitric acid with sodium hydroxide and zinc oxide is simple because I know that acid + base = salt + water. Therefore:

$$
\begin{align}
\ce{HNO3(aq) + NaOH (aq) &-> NaNO3 (aq) + H2O (l)}\\
\ce{2HNO3(aq) + ZnO (s) &-> Zn(NO3)2 (aq) + H2O (l)}
\end{align}
$$

It makes sense. In the second part, for example, $\ce{Zn}$ is 2+ and $\ce{H}$ is -1. So using the criss-cross method, you get $\ce{Zn(NO3)2}$ and water. I understand this.

However, with ammonia:

$$\ce{HNO3(aq) + NH3 -> NH4NO3}$$

How am I supposed to know this equation? It's not metal + acid = salt + water. Is there word equations for everything?

In the second part, for example, $\ce{Zn}$ is 2+ and $\ce{H}$ is -1. So using the criss-cross method, you get $\ce{Zn(NO3)2}$ and water. I understand this.

I don't agree on the -1 for hydrogen, let's look again at that example:

$$\ce{2HNO3(aq) + ZnO(s) -> Zn(NO3)2(aq) + H2O}$$

Metal oxides, such as zinc oxide, are basic anhydrides.

Imagine that

$\ce{ZnO}$ was made by heating $\ce{Zn(OH)2}$ until it loses water:

$$\ce{Zn(OH)2 ->[\Delta] ZnO(s) + H2O}$$

$\ce{ZnO}$ is a salt of $\ce{Zn^2+}$ and $\ce{O^2-}$

The reaction between nitric acid and zinc oxide and just a neutralization reaction between two protons and $\ce{O^2-}$ (which, as such, isn't present in solution).

Now for the reaction with ammonia: For this you may want to remember that there are different acid-base theories: The one that helps here is the Bronsted-Lowry theory:

Acids are proton donors

$\ce{HNO3}$ definitely is that.

Bases are proton acceptors

How can something else other than the $\ce{OH-}$ that you knew and the $\ce{O^2-}$ from $\ce{ZnO}$ be a proton acceptor = form a bond to a (proton)?

Is that possible if this acceptor doesn't even have a negative charge?

(Yes, otherwise I wouldn't buld up the tension like in a suspense story :D)

If the proton can't provide the electrons for the bond, the acceptor has to deliver them! And that can happen in the form of an electron pair, a double filled, non-bonding orbital.

Therefore a reaction, such as

$$\ce{H+ + NH3 <=> NH4+}$$

is perfectly valid. The rest of the example is just arranging the counter ions.


Why do carbon dioxide and sodium hydroxide not form sodium oxide and carbonic acid? Why do carbon dioxide and sodium hydroxide not form sodium oxide and carbonic acid?

$\ce{CO2 + 2NaOH -> Na2O + H2CO3}$

as opposed to:

$\ce{CO2 + 2NaOH -> Na2CO3 + H2O}$

This is in the context of fractional distillation of liquified air, before which carbon dioxide must be removed using sodium hydroxide.

Sodium oxide is not a base in either the Bronsted-Lowry or Arrhenius senses.  It is technically a base anhydride, meaning that it can be hydrated to yield a base:

$$\ce{Na2O + H2O -> 2NaOH}$$

Carbon dioxide can similarly be viewed as an acid anhydride, meaning it can be hydrated to form an acid.

$$\ce{CO2 + H2O <=> H2CO3}$$

The difference is that the hydration of sodium oxide is essentially irreversible.  One program I used estimated $K_{eq}=10^{39}$ for sodium oxide hydration.  In contrast the hydration of carbon dioxide is reversible, with $K_{eq}\approx1.5\times10^{-3}$.

The reaction that you are wondering about, $\ce{CO2 +2NaOH⟶Na2O +H2CO3}$, is essentially the dehydration of sodium hydroxide by $\ce{CO2}$ to form sodium oxide and carbonic acid, that is, it is reaction 2 minus reaction 1.  But the equilibrium constants I just mentioned show that sodium oxide has an "affinty" for water that is about 42 orders of magnitude greater than carbon dioxide's affinity for water.  So that reaction is very unfavorable.  The reverse reaction would be extremely favored: Sodium oxide would be a powerful reagent for $\ce{CO2}$ removal prior to air liquefaction, at least as powerful as sodium hydroxide.

If you are curious, here is the R code I used to estimate the equilibrium constant of sodium oxide hydration, using the package CHNOSZ:

> subcrt(c("Na2O","H2O","NaOH"),c(-1,-1,2),T=25)
info.character: found H2O(liq), also available in gas
subcrt: 3 species at 298.15 K and 1 bar (wet)
$reaction
     coeff         name formula state ispecies
2031    -1 sodium oxide    Na2O    cr     2031
1       -1        water     H2O   liq        1
1157     2         NaOH    NaOH    aq     1157

$out
      logK         G         H         S         V       Cp
1 39.01742 -53229.29 -57143.24 -13.24728 -36.04979 -40.8745


Is citric acid is a anti foaming agent? I added citric acid in my dish washing liquid. It stopped foaming.  Iwant my liquid again to foam. because I prepared 4 litres of liquid. Please let me know how can i get back my liquid foam. please answer immediately.

Citric acid is not a defoaming agent in itself. It is routinely used for pH adjustment in many commercial hand-dishwashing agents.

But it depends on the surfactant in the product. As Ivan tells, anionics will at some point be protonated by acids, and thereby loose their surface activity –and foaming capability.

Commercial anionic surfactants like ethersulfates will only do so at rather low pH, outside what is relevant for dish-washing products. But if your product is based on saponified oils it’s another story. Many fatty acids are weak acids (pK in the rage of 4-5) and will be significantly protonated at pH even as high as 5-6.

By the way: making good foam is as much an art as a science :-). If you want to dig deeper in the subject, I can recommend “Prof Steven Abbot –Surfactant Science, Principles and practice”. It is freely available on the web.


Decomposition of Gold Hydroxide On heating will gold hydroxide $\ce{Au(OH)3}$ decompose to form pure gold or will a gold oxide (e.g. $\ce{Au2O3}$) be stable? Does anyone know the temperature of these reactions?

There is yet no consensus on what occurs when gold(III) hydroxide is heated. Part of the problem is that the exact structure of $\ce{Au(OH)3}$ is unknown, or dubious at best. It has been inconsistently formulated as "truly" $\ce{Au^{III}-(OH)3}$$^{[0]\ [1]\ [2]\ [\text{a}]}$, oxide-hydroxide hydrate $\ce{AuOOH*H2O}$$^{[3]}$, oxide hydrate $\ce{Au2O3*xH2O}$$^{[0]\ [4]}$, even an acid-like hydrate $\ce{H[AuO2]*H2O}$$^{[5]\ [6]}$. I am going to use $\ce{[AuOOH*H2O]^{*}}$ with an asterisk in equations to keep this controversy in mind.

At temperatures below $130\ ^\circ \mathrm{C}$, the decomposition is probably best described by$^{[3]\ [7]}$

$$\ce{[AuOOH*H2O]^{*} ->[below $130\ ^\circ \mathrm{C}$] AuOOH + H2O}.$$

Around $140\ ^\circ \mathrm{C}$ the oxide-hydroxide decomposes further.$^{[3]\ [7]}$

$$\ce{2AuOOH ->[$140-150\ ^\circ \mathrm{C}$] Au2O3 + H2O}$$

Gold(III) oxide is already considerably unstable from $155\ ^\circ \mathrm{C}$ .$^{[1]\ [3]}$ It is not definitively shown what reaction follows. Some posit $\ce{AuO}$$^{[3]}$, which might not exist or actually  be better described as $\ce{Au[AuO2]}$$^{[3]}$, others propose a combined product of elemental gold $\ce{Au}$ and $\ce{Au2O}$$^{[1]}$. There are studies that rule out these products, at least as isolable intermediates, and instead propose $\ce{-> Au + O2}$$^{[0]\ [2]\ [7]\ [8]}$.

$$\ce{2Au2O3 ->[$150-165\ ^\circ \mathrm{C}$] 4AuO + O2}$$

$$\ce{4Au2O3 ->[$150-165\ ^\circ \mathrm{C}$] 4Au + 2Au2O + 5O2}$$

$$\ce{2Au2O3 ->[$150-165;\ 300\ ^\circ \mathrm{C}$] 4Au + 3O2}$$

The proponents of the bottom reaction, a more direct $\ce{-> Au + O2}$ claim that the oxides were mistaken for a complex mixture of $\ce{Au}$, $\ce{Au2O3}$, and interstitial or dissolved $\ce{O2}$.$^{[9]}$

The authors$^{[9]}$, nevertheless, still note

  It  is  not  yet  clear  how  this  “dissolved”  oxygen  should  be  described.  Is it really dissolved in interstitial metal lattice positions,  are there trapped $\ce{Au2O3}$ oxide  clusters in a densely sintered metal  matrix, or encapsulated oxygen gas in lattice imperfections of the gold?$^{[9]}$

If the intermediates $\ce{AuO}$ and $\ce{Au2O}$ exist, then not for too long.  The decomposition of these oxides would most likely consist of multiple interconversions between $\ce{AuO}$ and $\ce{Au2O}$. For example$^{[1]\ [3]\ [\text{b}]}$,

$$\ce{2Au2O + O2 <=> 4AuO},$$
$$\ce{Au + AuO <=> Au2O},$$
$$\ce{Au2O -> Au + O2}.$$

Over $300\ ^\circ \mathrm{C}$ only gold metal $\ce{Au}$ and dioxygen $\ce{O2}$ remain.$^{[9]}$ Another experiment$^{[10]}$ supports this conclusion but differentiates between surface and subsurface oxides. The latter are found to be stable well over $300\ ^\circ \mathrm{C}$$^{[10]}$. More research is required to shed light into what is actually the main decomposition pathway. An additional way of expressing the process at lower temperatures is$^{[9]}$

$$\ce{6AuOOH -> 4Au + Au2O3 + 3O2 + 3H2O.}$$

$[\text{a}]$ $\ce{Au(OH)3}$ is presumed to assume a square planar coordination with four $\ce{-OH}$-groups.$^{[2]}$ A most recent study provides evidence for a four-coordinated linear polymer.$^{[0]}$

$[\text{b}]$ Equations are written on the analogy with copper.$^{[1]\ [3]}$

$[0]$ Daisuke Kawamoto, Hiroaki Ando, Hironori Ohashi, Yasuhiro Kobayashi, Tetsuo Honma, Tamao Ishida, Makoto Tokunaga, Yoshihiro Okaue, Satoshi Utsunomiya, Takushi Yokoyama. 'Structure of a Gold(III) Hydroxide and Determination of Its Solubility'. Bulletin of the Chemical Society of Japan, ($2016$), 89(11), 1385$-$1390. DOI: 10.1246/bcsj.20160228.

$[1]$ H. Karik, K. Truus. ($2003$). Elementide keemia. (p 313)

$[2]$ C. Klanner, N. Weiher, E. A. Willneff, S. L. M. Schroeder. 'In Situ Quick XAFS Studies of the Thermal Decomposition of Gold(III) Oxide'. Hasyweb Annual Report $2000$, ($2000$), part 1, 46, 2986.

$[3]$ R. Ripan, I. Ceteanu. ($1969$). Chimia metalelor. II volume. (p 768$-$769)

$[4]$ I. V. Mironov. 'Properties of Gold(III) Hydroxide and Aquahydroxogold(III) Complexes in Aqueous Solution'. Russian Journal of Inorganic Chemistry, ($2005$), 50(7), 1115

$[5]$ Κ. V. Astakhov, A. G. Elitsur, K. M. Nickoluev. Zhurnal obshcheĭ khimii, ($1951$), 21, 1753.

$[6]$ A. G. Massey, N. R. Thompson, B. F. G. Johnson, R. Davis. ($1973$). The Chemistry of Copper, Silver, Gold. (1975 reprint)

$[7]$ O. Diaz-Morales, F. Calle-Vallejo, C. de Munck, M. T. M. Koper. 'Electrochemical water splitting by gold: evidence for an oxide decomposition mechanism'. Chemical Science,  ($2013$), 4, 2334$-$2343. DOI: 10.1039/C3SC50301A.

$[8]$ C. Klanner, N. Weiher, E. A. Willneff, S. L. M. Schroeder, C. Figulla-Kroschel, M. Jansen. 'Thermal Stability and Decomposition Kinetics of Crystalline  Gold(III) Oxide and Disordered Gold(III) Oxides/Hydroxides'. Hasyweb Annual Report $2001$, ($2001$), part 1, 46, 5440.

$[9]$ M. Peuckert,  F. P. Coenen, H. P. Bonzel. 'On the Surface  Oxidation of a Gold Electrode in $1\ \mathrm{N} \ce{H2SO4}$ Electrolyte'. Surface  Science, ($1984$), 141, 2$-$3, 515$-$532.

$[10]$ L. K. Ono, B. R. Cuenya. 'Formation and Thermal Stability of $\ce{Au2O3}$ on Gold Nanoparticles: Size and Support Effects'. Journal of Physical Chemistry C, ($2008$), 112(12), 4676$–$4686. DOI: 10.1021/jp711277u.


Quantum calculation of free energy change in neat conditions I am slightly baffled by a seemingly simple situation: I want to calculate the free energy of association between a molecule $\ce{A}$ and the solvent $\ce{B}$ using quantum methods (e.g. DFT).
From what I've understood, the basic way to do this in most situation is to optimize the geometry of each compound, then perform frequency calculations to obtain free energy corrections to the SCF. Let the SCF plus the corrections for specie $\ce{X}$ be $\ce{G_X}$. This yields the free energy of each compound at 1 atm, which can be converted into the free energy at 1 M via a factor of 1.89 kcal/mol. Thus, for the complexation $\ce{A + B \rightleftharpoons AB}$, the free energy of complexation would be
\begin{align}
\Delta G_{comp}^0 &= (G_{AB}^0 + \pu{1.89 kcal/mol}) - (G_{A}^0 + \pu{1.89 kcal/mol}) - (G_{B}^0 + \pu{1.89 kcal/mol}) \\
&= G_{AB}^0 - G_{A}^0 - G_{B}^0 - \pu{1.89 kcal/mol}
\end{align}
However, in the case of association between molecule $\ce{A}$ with concentration $\ce{C_A}$ and solvent $\ce{B}$ with concentration $\ce{C_B}$, how could one obtain the free energy of complexation?
There is the equation $\Delta G = \Delta G^0 - RT \ln\left(\frac{C_{AB}}{C_A*C_B}\right)$, but the "initial" concentration of the adduct is null. There seems to be a $RT \ln(C_B)$ term floating around in the literature, but I'm not too sure where it's coming from and how to handle this situation rigorously.

One needs to distinguish between the standard free energy (of reaction) for a process, $\Delta _r G^\circ$, and the free energy in general, $\Delta _r G$. You show how to use quantum methods to compute $\Delta _r G^\circ$. This tells you how much the free energy changes if you take 1 mole of each reagent at 1 M concentration and generate one mole of complex at 1 M concentration. At other concentrations one uses
$$\Delta_r G = \Delta_r G^0 + RT \ln\left(\frac{C_{AB}/C^\circ}{(C_A/C^\circ )(C_B/C^\circ)}\right) \tag{Eq. 1}$$
for the reaction $\ce{A +B<=>AB}$ (assuming all components behave ideally; $\pu{C^\circ= 1 M}$ is the standard concentration). In fact, you use an analogous expression to compute the free energy difference between 1 atm (at $\pu{0^\circ C}$) and 1 M concentrations for each reagent or product:
$$\begin{align}\Delta_r G(\pu{1 atm}\rightarrow\pu{1 M}) &= RT \ln\left(\frac{C_X(\pu{1 M})}{C_X(\pu{1 atm})}\right)\\ &= RT \ln\left(\frac{\pu{1 M}}{(\pu{1e-3 \pu{L/m^3}})\cdot(1/(RT/\pu{1 atm})}\right)\end{align} $$
You use analogous relations (usually in the form of Eq. (1)) if you intend to compute the free energy change for one mole of complex formation when reagents and products are at particular concentrations (information which you then have to provide).
If you are interested in computing the equilibrium concentration of complex given specified concentrations of A and B (assumed constant) then you solve for $C_{AB}$ in Eq. (1) after plugging in values for $C_A$ and $C_B$, remembering again that you are assuming your solution behaves ideally.


How does Combinatorial Chemistry differ from Rational Drug Design? I stumbled on Vertex Pharmaceuticals - Wikipedia.

It was one of the first biotech firms to use an explicit strategy of rational drug design rather than combinatorial chemistry.

Then I looked up Combinatorial Chemistry. What is Combinatorial Chemistry

     Combinatorial chemistry is a laboratory technique in which millions of molecular constructions can be synthesized and tested for biological activity.  It has generated massive numbers of targeted molecules for testing and the developing techniques of high throughput screening has automated the screening process so larger numbers of biological assays can be done.  All this together has reduced the discovery-to-market time from what used to be 10-14 years to 5-8 years.

But this doesn't explain how Combinatorial Chemistry differs from Rational Drug Design? Isn't Combinatorial Chemistry rational too? Mustn't chemists pick reactants rationally? Even if she can't predict the products, she won't picked her reactants randomly.

From a medical point of view the problem in drug discovery is to find a compound that causes a desired outcome, not one that interacts with a specific target. For instance, if you want to reduce mortality due to infection with Covid, you have many strategies that may be pursued at the molecular level, from vaccination to interference with viral propagation to reduction of symptoms or boosting of the immune response.
The difference between a combinatorial strategy and a rational strategy is to a first approximation simple: in a combinatorial strategy compounds are tested for activity without immediate concern for how they bring about an outcome. There is no "a priori" information to guide compound selection other than the common properties of known drugs (as encoded eg by Lipinski's rules) and the ability to procure or synthesize compounds. Therefore the "rational" approach in this case is to create as large a library and to screen as many compounds as is reasonable given cost and time.
Rational drug design takes into consideration a molecular target (such as a protein) and perhaps the mechanism of action of a known lead compound and designs future potential drug compounds around that target and/or lead. That narrows the set of compounds under consideration reducing time and cost.
In general, the strategies of combinatorial and rational design are not mutually exclusive. It is typical to start a rational strategy after early leads have been identified some other way, say through combinatorial chemistry. It is also common to design a set of compounds during a stage in rational drug design using a combinatorial strategy applied to develop next generation leads from a starting compound.


HPLC: peak area vs concentration Many text books say that there is a proportionality between the peak area of the chromatogram and the concentration. However, in my opinion, there should be a proportionality between the concentration and the peak intensity, not the peak area. The peak intensity is increasing proportionally with respect to the absorbance of the substance. Hence, by using the Beer-Lamberant law, we can induce there is a proportionality between peak intenstiy and the concentration.

Is there any mathematical relation between the peak area and the concentration?

Any help will be appreciated.

However, in my opinion, there should be a proportionality between the concentration and the peak intensity, not the peak area.

There is a proportionality between both peak area vs. concentration and peak height vs. concentration.

Peak height is proportional to the instantaneous amount of analyte that is transiting the detector.
Peak area is proportional to the sum of all of analyte moleucles that have transited the detector.
From 1 and 2 you might be able to infer the relationship betweeen peak height $h$ and area $A$: $$A = \int{h(t)\;dt}$$
People are usually interested in the total amount of substance injected into the column.  (If they know the injection volume, they can calculate the concentration from this value.)  The total amount of substance would be calculated from the peak area.
The maximal peak height is proportional to the peak area, but only if the peak "shape" is constant.  Here are some scenarios where peak shape will not be constant:

You want to compare injection #2 you made on your column two years ago to injection #2000 that you made recently.  Due to column degradation, the recent injections have much more noticeable peak tailing.  Because the longer tail leads to wider, asymmetric peaks, there will be less total height at the peak maximum relative to "better", symmetric peaks.
You change the flow rate of the HPLC.  All peaks are narrower, and thus, higher.
The scan rate of your detector changes, and the detector reports "counts" or "intensity" rather than counts per time.  This is actually very common for mass spectrometers, but less common for absorbance detectors.  If you scan twice as fast in MS, your peak heights go down by approximately twofold, but you have data points twice as often, so the peak area is relatively unchanged.
Your detector undersamples the peak.  Say your chromatography is nice and the "real" peak shape is nicely gaussian, but is ~ten seconds wide and your detector only gives you a data point every two seconds.  Say that due to very small random drifts in retention time, on some injections one of the data points coincides exactly with the maximum of the "real" peak, but in others, it is a little bit off.  The peak heights in this scenario will vary considerably more than the peak areas. (If the recorded maximum is off-center from the true maximum, there will be two data points that are higher, i.e. "closer" to the maximum than if the data maximum coincides with the true maximum, so integration will partially correct this error.)  Essentially, the peak maximum is a single-point sample from the gaussian distribution of the peak, while the area is a several-point sample from the gaussian distribution, so it has better sampling properties.

  Is there any mathematical relation between the peak area and the concentration?

Yes, there is, but it depends on the peak shape.  For perfectly gaussian peaks, $$h_{max} = \frac{A}{\sigma \sqrt{\tau}}$$

where $\tau$ is $2 \pi$ and $\sigma$ is the width of the peak, which is related to the full-width at half-maximum of the peak by $\mathrm{FWHM} = 2 \sigma \sqrt{2 \ln 2}$.

However, for non-gaussian peaks, this relationship does not hold.


Is it wise to use ECPs on light atoms for electronic structure calculations? So, I have a big organic molecule consisting of $\ce{H}$, $\ce{C}$, $\ce{O}$, and $\ce{N}$ atoms and the goal is to find its equilibrium geometry and US/Vis spectrum. The molecule is so big that I was thinking about reducing the complexity of the calculations by using effective core potentials (say, Stuttgart RLC) on light atoms ($\ce{C}$, $\ce{O}$, $\ce{N}$). However, I'm not sure how trustworthy the results would be comparing to using full-electron bases, thus, I'm looking for some reviews if there are any.

I think you will find that the time savings are minimal. Modern integral screening will eliminate virtually all inter-atom core-core integrals anyway and if you only have a 1s core then the cost of evaluating intra-atom core-valence integrals will be comparable to evaluating the ECP-core interaction.


Why don't salt crystals attract each other? If salt is made of ions in a giant lattice, surely if I have two crystals they would attract each other and fuse? Or maybe the ions can't get close enough to each other to attract, or the surface is made of a different, neutral material?

Sodium chloride ("salt") is made of a giant lattice of ions: Na+ and Cl-.

The crystals overall do not have any net charge as the the number of chloride ions is equal to the number of sodium ions. So, overall, there will be no electrostatic force between crystals as they are neutral in the bulk compound.

On the surface, crudely and without getting into the detail of the crystal structure, the ions will alternate on the surface at an atomic scale. You might imagine that this creates atom-sized patches where there will be net positive or negative charges. Would these attract another crystal? Would these attract the patches of charge in another crystal if they were adjacent. Maybe, but only if the surfaces were both atomically flat and lined up exactly. In reality the surfaces are rough and there is no net charge so large salt crystals won't stick because of this effect.

So the intuition that some crystal made of ions should have strong electrostatic attraction to another similar crystal is wrong because of electroneutrality and the detail of the physical surface.


Para effects on substituents in aromatic compounds I wanted to know how to compare the following compounds on basis of acidic strength.

I have to find the acidic strength of following compounds with $\mathrm{X}=\ce{CH3, \ OCH3,}$ and $\ce{NO2}$.

For ease of explaining, denote the first compunds as (A), the second as (B), and the third as (C)

Due to steric inhibition of Resonance (SIR) effect, compound (A) will be the most acidic, as its anion will be the least stable.

Now for ordering the other two, I know that there are two things affecting acidic character:

Bond strength; the weaker the bond, the higher its acidic strength
Charge on hydrogen; the more positive charge their is on the hydrogen, the more its acidic strength.

So now if we take $\mathrm{X}=\ce{CH3}$, we should have that $\mathrm{(A)>(C)>(B)}$; as $\ce{CH3}$ is an electron donating group and it would make the hydrogen less positive, and inductive effect is distance dependent. This answer is incorrect as per my textbook, which says $\mathrm{(A)>(B)>(C)}$.

If we take $\mathrm{X}=\ce{OCH3}$ which is an electron withdrawing group, we should have $\mathrm{(A)>(B)>(C)}$. This answer is correct.

If we take $\mathrm{X}=\ce{NO2}$, which is also an electron withdrawing, we should have $\mathrm{(A)>(B)>(C)}$. But my textbook says the answer is $\mathrm{(A)>(C)>(B)}$.

I wanted to know where am I going wrong, and if there is some other effect I am not taking into account.

Note: my earlier question was closed due to the similar question: Ortho-effect in substituted aromatic acids and bases.
But that question only deals with the ortho effects of substituents, not the para.

At meta position the M effect is not considered. The reason behind is that during resonance of a six membered cyclic compound charges never reaches at meta position that why it is so try to proof it by drawing resonance structure and if you didn't get us then tell me in the comments below.


Why does the reaction between Na (s) and Cl2 (g) occur spontaneously if the entropy of the system decreases? 1- The reaction is exothermic.
2- My book says that it is because the total entropy change is very positive.
Here is my book's explanation and it is the highlighted part which i dont understand :
"Despite the fact that there is a decrease in the entropy of the system, the energy leaving the system produces a substantial increase in the entropy of the surroundings, because there are more ways of arranging the energy quanta in the rest of the universe than there are of arranging them in the system alone.

As you point out there are two things to consider, the heat released in the reaction and the change in entropy. In forming NaCl from chlorine gas and solid sodium the entropy decreases because the gas is converted into solid NaCl which has a far far smaller total volume. The energy released has the effect of heating the solid NaCl just formed and then the surroundings as energy flows away to eventually reach equilibrium again but now at a higher temperature than that at equilibrium before reaction occurred.

When a quantity of heat $dq$ from a body at $T_2$ transfers to one at $T_1$ the overall entropy change is $dS = dS_1 + dS_2 = dq/T_1 - dq/T_2$. As $T_2 \gt T_1$ the entropy change is positive.

In the reaction the entropy change overall depends on comparing that decreased due to volume decrease vs. that increased due to the energy released on reaction.

Entropy is a measure of the number of ways that the particles can be placed into their energy levels, this is often called the number of configurations $\Omega$, and Boltzmann found that $S=k\ln(\Omega)$ so that increasing the distribution of energy into the energy levels increases $\Omega$ and hence S. As the temperature increases more vibrational and rotational levels are populated and the kinetic energy distribution widens, all this means that there is an increased number of configurations and hence entropy just because energy can be placed into  energy levels and in more ways.

If there was just one level then all 'particles' would have to be in this level and so entropy would be zero, $\Omega=1$.  When there are two levels some can be in the first and some in the second level and now the entropy is increased ($\Omega \gt 1$), and so on for many levels.


Why is 1-bromotriptycene inert to nucleophilic substitution? According to my textbook, 1-bromotriptycene is inert to any nucleophilic substitutions.

Due to steric hinderance we can say SN2 reactions cannot occur, but I think SN1 reactions should be favoured as formed carbocation will be resonance stabilised.

Can someone explain why it doesn’t undergo SN1 reactions?

Sn1 reaction proceed through carbocation intermediate which has planar structure. The structure of 1-bromo triptycene  is cage like structure so the bridgehead carbon cannot assume planarity , hence,the formation of a carbocation at the bridgehead position does not take place. Consequently 1-bromotriptycene is inert to sn1reaction.


Nitrosonium ion in the nitration of phenols Peter Sykes mentions in the book Guidebook to Mechanism in Organic Chemistry (6th ed.) on p. 137 that:

  Highly reactive aromatic compounds, such as phenol, are found to undergo ready nitration even in dilute nitric acid, and at a far more rapid rate than can be explained on the basis of the concentration of $\ce{^⊕NO2}$ that is present in the mixture. This has been shown  to be due to the presence of nitrous acid in the system which nitrosates the reactive nucleus via the nitrosonium ion, $\ce{^⊕NO}$ […]

However, there was no mention as to why the nitration of more reactive aromatic compounds would result in there being nitrous acid in the system. I understand that the conventional explanation for why the nitration of activated benzene derivatives is more rapid is that there is a stronger attraction between the electrophile and the nucleophilic benzene ring due to the greater amount of electron density in the ring. This explanation using the presence of the nitrosonium ion seems interesting but I don't see how nitrous acid could be generated due to there being a more reactive aromatic compound being nitrated. Could someone enlighten me on this?

Possibly the electrophile is molecular $\ce{HNO3}$. Nitric acid is strong but not super-strong ($\mathrm{p}K_\mathrm{a} = -1.4$), so solutions on the order of a mole per liter have a significant amount of molecular $\ce{HNO3}$. As $\ce{NO2(OH)}$ this can transfer its nitryl ($\ce{NO2+}$) ion moiety to the aromatic ring in exchange for a proton to make the nitrated ring and water. Obviously molecular nitric acid is less electrophilic than the straight-out nitryl ion, so an activated aromatic ring is needed for this to work.


Relative atomic mass Is it possible for an element to have a calculated relative atomic mass that is higher than the relative isotopic mass of its most massive component isotope?

The equation is:

$\mathrm{Relative~Atomic~Mass}=\sum[{\mathrm{(Isotopic~Mass) \cdot (Relative~Abundance)]}} \times100\%$

No. Mathematically, that would not make sense.

Consider an imaginary element with only one known isotope: imaginarium-100.

The percent abundance of imaginarium-100 is $100\%$, so its relative atomic mass is: $$100 \cdot 1=100~\mathrm{amu}$$

Now imagine a group of scientists were to discover that in fact imaginarium-99 consists of $0.0001\%$ of all imaginarium samples.

The new relative atomic mass would be:

$$100 \cdot 0.999999+99 \cdot 0.000001=99.999999~\mathrm{amu}$$

The relative atomic mass goes down with the discovery of a less massive isotope. So the closest that the relative atomic mass of an element can be to the mass of its most massive isotope is if the most massive isotope is also the only known isotope, but the relative atomic mass can never go over this.


What is the systematic name of a "double benzene"? Consider the molecule representation below. What is the systematic name of that molecule?

Could it possibly be 3,4-hex-2,6-anebenzene?

As @ron already said in his comment, the systematic name of the naphtalene is Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene or Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene since you can start the carbon numerotation from 2 chemically equivalent carbon.

You can find all the references here and here


Intermolecular interaction Van Der Waals In the empirical demonstration of the Van der Waals equation at Khan Academy, they made a demonstration of the real volume, for example, it was said that it has to be larger than the ideal volume, since the size of the molecules is not disregarded, hence it would be:
$$ V_{\text{ideal}} = V_{\text{real}} + nb $$
Where, $ b $ was said to be related to the size of the molecule.
But why don't I also include intermolecular interaction? Depending on it, the volume may even be smaller.

Intermolecular attraction can reduce the volume, in the same way that increasing the pressure can lower the volume (Boyle's law).
One way to see the effect is to take the derivative of the volume, as given by the van der Waals (vdW) equation, with respect to the attractive parameter $a.$
The vdW equation is
$$\left(p+\frac{a}{\bar{V}^2}\right)(\bar{V}-b)=RT\tag{1}$$
Ignore for simplicity the excluded volume term $b,$ so that
$$p\bar{V}=RT-\frac{a}{\bar{V}}\tag{2}$$
Then the derivative of the volume wrt $a$ is
$$\left(\frac{\partial V}{\partial a} \right)_{p,T} = \frac{\bar{V}}{a-p\bar{V}^2}=-(RT)^{-1}\tag{3}$$
In essence what this result says is that the volume decreases as you increase the parameter $a.$


Colligative properties I have some problems with these properties.
1) Lowering vapor pressure and boiling point.
It happens because the solute molecules take space in the surface of the liquid so a lower number of the liquid molecules change to the gas form. Also the vapor pressure increases when you increase the temperature of the liquid because more molecules get to escape into a form of gas.

So the real question here, What makes boiling point related to vapor pressure?
I have read a lot of articles about that.. The definition of boiling point is the temperature at which the vapor pressure of a liquid is equal to the pressure of the atmosphere on the liquid. Which will rapidly increase the rate of forming gas but why does that happen? It says because of the vapor pressure balanced the atmospheric pressure which makes bubbles form. How does it exactly balance out? Isnt pressure in all directions? How can you be sure that it balances out? Also, Bubbles are created at the bottom of the liquid, How does atmospheric pressure affect it?

2) Same applies to Freezing points.

I need a microscopic explanation. Also without some advanced chemistry complexity.

Here's an answer without complexity which I read in my book. When you heat a liquid in open vessel, its molecules turn to vapours, but the atmospheric pressure above is pushing these vapours down. Some vapours would have enough energy to tackle this atmospheric pressure and would get away from the liquid leading to evaporation. Only some particles at the surface would have enough energy but when vapour pressure reaches atmospheric pressure, all liquid molecules have enough energy to counteract the pressure on them and thus liquid can form vapours from the whole bulk of the liquid leading to boiling.


Why are naturally occuring liquids more rare than solids and gasses on Earth? There are only a few naturally occurring liquids I can think of on earth, such as water, mercury and animal oils.

Why are naturally occuring liquids more rare than solids and gasses on Earth?

To be specific,

By Earth I mean the surface which is inhabited by humans, not the interior core or mantle.
I am referring to naturally occurring substances, i.e. naturally occurring liquids are far more rare than naturally occurring gasses or solids. Why is this?
I am referring to average conditions of temperature and pressure that prevail on our planet's surface, not man-made conditions or extremities.

Current Line of Thought

I looked into inter-molecular forces for an answer. Here's what I've come up with.

At Earth's temperature, substances held by dipole-dipole forces are solid and substances held together by weaker london forces are liquids or gasses, such as oils or noble gasses.

The majority of substances have dipole-dipole interactions, hence are solids. In comparison, london forces are weaker so require only a little energy to change phase, which is why non-polar and noble-gasses are gasses or liquids at the same temperature. However, the majority of substances or not noble gasses or non-polar, which is why solids are more common on Earth.

To sum, dipole-dipole are the most common inter-molecular forces. Substances with dipole-dipole forces are solids at earth's temperature, hence the majority of substances are solids.

I am aware that many exceptions arise to this general trend, such as water, however the fact remains that liquids are more rare than solids and gasses on Earth.

I still have a feeling I am wrong, so if anyone could provide a better answer or critique on my line of thought that would be great.

Updated the question with specifics to make it less broad

Your argument misses two major points.

First, a great number of solids are ionic and metallic substances. Those show a really strong bonding mechanism between elementar particles.

And second, to get your theory to work, you have to take into account the size of molecules. To evaporate or melt anything, you have to break or at least disturb all the bonds each molecule has to its surroundings.

For something as small as H2O, not even hydrogen bonding is strong enough to make it solid. A large molecule like Polyethylene, on the other hand, is only held by weak London forces, but has melting points as high as 145°C, and cannot be evaporated.


Suspension or Solution Tomorrow I have a science test, and in it there are some mixtures, and we have to identify whether they are a solution, a suspension or a colloid. Is there an easy way we can identify this WITHOUT getting the mixtures up and seeing whether they sediment or not?
Note: They are everyday mixtures, like lattes, freshly squeezed orange juice etc.

If the analyte happens to be in a transparent container, shine light on it, see if the light is scattered (Tyndall effect) -- if so, you have either suspension, or colloidal matter.  Instead of a laser pointer, a good pocket light (like the ones with halogen bulb halogen or LEDs) may be used as light source, too.

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