Vinegar salt reaction to produce acid; does it matter what type of vinegar I use I am trying to get get hydrogen chloride gas by using the reaction sodium chloride + vinegar = HCl(g) + sodium acetate, does it matter what type of vinegar I use eg apple cider vinegar, malt vinegar; if it does matter where could i find some pure vinegar at a reasonable price

There are two show stoppers :

$\ce{HCl}$ is strong acid, $\ce{CH3COOH}$ is weak acid. The reaction is practically completely shifted toward acetic acid and sodium chloride. Especially if acetic acid is diluted.
$\ce{HCl}$ forms azeotrope with water with the maximal boiling point near 20% $\ce{HCl}$. If you heat the mixture, that would boil water and acetic acid with minimal traces of $\ce{HCl} $ that would dissolve back in the condensate.


What total energy (in kJ) is contained in 1.0 mol of photons, all with a frequency of 2.75 • 10¹⁴ Hz? There is a question, that says:

  What total energy (in $\mathrm{kJ}$) is contained in $1.0~\mathrm{mol}$ of photons, all with a frequency of $2.75 \cdot 10^{14}~\mathrm{Hz}$?

My answer was:
The energy of photon in hydrogen atom is given by the formula $E=h\nu$, where $h$ is Planck constant and $\nu$ is the frequency. After that I got an answer different from the correct answer which is $110~\mathrm{kJ}$.

Could anyone explain why?

The formula $E=h\nu$ is for the energy of one photon. You have a mole of photons. You need to use a slightly modified form:

$$E=Nh\nu$$

where $N$ is the number of photons, in this case

$$\begin{align}
N&=n\cdot N_\mathrm A\\[6pt]
&=1\ \mathrm{mol}\times6.02\times10^{23}\ \mathrm{mol^{-1}}\\[6pt]
&=6.02\times10^{23}
\end{align}$$

Note that you are being asked to report energy in kilojoules not kilojoules per mole.

If I use $E=nN_\mathrm A h \nu$, I get the correct answer you cite of $110~\mathrm{kJ}$. Do you?


Why can't oxalate ion donate two pairs of electrons from the two double-bonded oxygen atoms So, I read that an oxalate ion is a bidentate ligand but since I saw that it has 4 oxygen atoms I was confused as to why it isn't a tetradentate ligand instead.

I searched it on Google and found this:
Oxalate ion dentiticity - why is it bidentate and not tetradentate?

Now, in the link it's implied that the oxalate ion can donate the 2 pairs of electrons either from the 2 oxygen atoms having negative formal charge or from 1 "double-bonded" O atom and 1 negatively charged O atom.

But why can't the two pairs of electrons be donated from Both the "double-bonded" O atoms?
 Why isn't this "third" bonding mode possible?

EDIT:  From the comment below, I infer that because of resonance, there is no double-bonded or negatively charged O atom in the Oxalate ion; rather all of the O atoms have a partial negative formal charge and double-bond character, is that correct?

So, can electrons be donated from any O atom in the ion?

(And so did the answer in the provided link just ignore this for the sake of simplicity?)

Oxalate is a planar molecule. The four oxygen atoms share the pi electron density through resonance, as noted in comments, and thus are all identical. Imagine those four oxygen as four corners of a table. Now you are a Lewis acid, perhaps a transition metal cation, and you walk up to the table. How many of the four corners ligate you? Perhaps you imagine the planar ligand dropping on top of you, so that the four corners are around you. For effective overlap with all four corners of the table, you'd need to be a pretty big cation. In almost all cases you'd achieve better orbital overlap, and a more energetically favorable coordination complex, if you just cozy up to two of the donor oxygens.


Resolution of enantiomers? How to determine chiral compounds apart from each other? How exactly is the interaction between two chiral compounds so they can tell the difference between S or R?

If the compounds have never been made before, then the enantiomers have to be separated prior to analysis. Then analytical and spectroscopic techniques must be used. The R and S configurations cannot even be assigned until the relative stereochemistry had been elucidated. This is usually done by specific NMR experiments. Chiral HPLC can be used to give enantiomeric purity.


Does storing apples alongside potatoes keep potatoes from sprouting? I am looking for a chemical explanation which I wasn't able to find.

I have read in several sources such as this one that, in order to prevent potatoes from sprouting, it is sufficient to store a few apples alongside them. This is due to the fact that the apples produce ethylene.

But this is a bit odd, since the ethylene is typically a ripening agent. This reddit thread also discusses the same doubt, but doesn't dive much into the details. There is also this article and this Quora answer, none of which give a specific chemical reason about why and how this works.

Chemically speaking, do we actually know the exact reason and the necessary conditions to make this work?

Ethylene has been used in storing potatoes for a while, see this article. In it, they address the puzzle of a ripening agent acting to prevent sprouting:

  Sprout suppressant properties   of   apple   ‘vapours’   were   first   reported   in   1932   and   subsequently,  ethylene  and  apple  ‘vapours’  in  1933  (Elmer  and  Huelin,  respectively).    This  was  preceeded,  however,  by  work  demonstrating  a  ‘dormancy  breaking’  effect  of  ethylene  (Rosa,  1925).    These  apparently  conflicting  responses  to  ethylene  were  resolved  with  the  publication of work showing both effects, but under different ethylene exposure conditions.

The mechanism of the different effects are largely unknown. The cited article (by the British Potato Council) gives some references. Potatoes treated with ethylene do have slightly different properties (they mention the color after deep frying), and there is ongoing research to get an overview of all the differences.

  [OP:] Chemically speaking, do we actually know the exact reason and the necessary conditions to make this work?

We don't know the exact reason, but there are companies who figured out how to make this work in a warehouse under controlled conditions. I think on your shelf would be more difficult because you have no way of monitoring the ethylene concentration, so the apples might influence the potatoes in etiher way.


Why does enthalpy change differ in a chemical reaction because of temperature? If enthalpy is a state function, shouldn't the change in enthalpy be the same between a given initial and final state, regardless of how the change took place? Kirchoff's equation describes this variation as
$$\Delta H_{T_2} = \Delta H_{T_1} + \Delta C_p (T_2 - T_1).$$
I'm guessing that this change in enthalpy change is only because of the variation in $C_p$, but I'm not too sure about it. Am I completely wrong about this?

From Hess' law, the heat of reaction at temperatures $T_1$ and $T_2$ are related by $$\Delta H_{T_1}=C_{p,\mathrm r}(T_2-T_1)+\Delta H_{T_2}+C_{p,\mathrm p}(T_1-T_2),$$where $C_{p,\mathrm r}$ is the heat capacity of a stoichiometric molar mixture of reactants and $C_{p,\mathrm p}$ is the heat capacity of a stoichiometric molar mixture of products.


Why does silver plating require undercoats of copper and nickel? I came across a question that was about silver plating (of steel), but I couldn't help but wonder why the sequence was illustrated as this:

step 1 A coating of copper is applied to the object.

step 2 A coating of nickel is applied to the object.

step 3 The coating of silver is applied to the object.

Why are these initial coats of copper and nickel required, and why this seemingly random sequence (since nickel is in between but has the highest reactivity)?

I have read that the nickel plating helps removing oxide layers when working with stainless steel which is beneficial for the silver plating process.

Sadly I do not know why the copper is used but I suspect the context could explain the copper.


What is the effect of temperature during cryotrapping? I'm working on a lab report that has some specific questions that need to be included. The lab work itself dealt with GC–MS.
What is the effect of the temperature selected during cryofocusing/cryotrapping? What is the effect of making this temperature warmer or colder?
My understanding was that the gas based samples would adsorb to the sample loop and then rapidly desorb when the temperature is increased. However, I am not sure what the effect of changing the cold temperature is.

It depends on the temperature regime you of your cryo-trap.  If you are trapping in the temperature range from dry ice to water ice then you must be sure any significant water vapor in you sample has been removed.  If you are using liquid nitrogen, there is some danger of condensing oxygen.  This can be mitigated by using the enough-warmer-that-this-doesn't-happen-with-air liquid argon.


Maximum hydrogen atom bonds per a single carbon atom I was solving a chemical formula which includes a carbon, oxygen, and hydrogen compound. And it turned out to be something like this:

$$\ce{C_nH_{4n}O}$$

when $n$ is a natural number. So that, at least ostensibly, can be one of these:

$$\ce{CH4O, C2H8O, C3H12O, C4H16O, ...}$$

But a solution attached to it said

  The only compound which has $\ce{C_nH_{4n}O}$ as the empirical formula is $\ce{CH4O}$

so the molecular formula is confirmed as above. So simply my question is this: why is that the only one which makes sense in this situation?

I would look at it like recursive problem. Starting with the simplest situation:

Carbon has 4 electrons available to form covalent chemical bonds (so 4 simple bonds for example). Hydrogen has one electron - 1 bond.

If you add one more carbon to the situation, you have to bond it somewhere - you need to use the first carbon. Now with this new carbon you added 3 more possible places to add 3 more elements (like hydrogens) but you canceled one hydrogen place on the first carbon. So every time you add another carbon, you can go up only for 2 hydrogen. It is actually called homologous addition or something like that. (Sorry for my english as well)

I ignored the oxygen whole time, cos as far as he is 2 bonded, he never adds anything he either substracts two hydrogen places, or you can instert it between a carbon and a hydrogen.


How do I call the functional group of this alkane? Given the following chain of carbon atoms (imagine they are all saturated with the correct number of hydrogen atoms), how do I call it? 4-(dimethyl-ethyl)-octane? I'm having trouble naming the short carbon chain extending from the 4th carbon atom of the octane. Can you help me out please?

Under the 2004 provisional recommendations by IUPAC, one possible systematic name is 4-(2-methylpropan-2-yl)octane, which indicates that the longest chain in the substituent is three carbons long, is connected to the main carbon chain via the second carbon (-2-yl), and has a methyl group on the second carbon (2-methyl).

4-(1,1-dimethylethyl)octane is also possible, and indicates that there is an ethyl substituent on the fourth carbon in octane, which is further substituted with two methyl groups on the first carbon where it branches off (see section P-29.4).

However, the preferred IUPAC name is 4-tert-butyloctane; the tert-butyl group is used to describe a $\ce{-C(CH3)3}$ group and is preferred over either of the two above systematic names (see section P-56.2.2.2).

I don't have access to the 2013 recommendations but I believe it should not be different.


What are the microstates of the triplet oxygen molecule? I am trying to understand molecular term symbols of oxygen molecule. Everywhere I looked, it is claimed that the molecule has 6 microstates, three of them are singlet, and three of them are triplet. Now I can draw the diagrams for the singlet states with no problem, but I am struggling to find the tree triplet configurations. According to my microstate matrix there should be a triplet state with a total spin multiplicity of zero - how is this possible?

In a $(\pi2p)^2$ configuration the orbital $\lambda$ and spin s quantum numbers have to be tabulated and combinations removed which break the Pauli principle. The total angular momentum quantum number is $\Lambda$ and spin $\Sigma$, for example the combination

$\lambda_1 = 1, \lambda_2 =1 , s_1=1/2, s_2 =-1/2$

generates total angular momentum $\Lambda = 2, \Sigma =0$ and this is part of the $^1\Delta$ state.

There are only 6 valid combinations leading to $^1\Delta, ^3\Sigma^-,^1\Sigma^+ $ states of which $^3\Sigma^-$ is the ground state.

A vector spin diagram shows how a triplet can have spin multiplicity of zero. There is a picture of vector spin for singlets and triplets in my answer to this question; How does spin flipping of triplet carbenes occur?


Does Wurtz reaction follow an ionic mechanism or a free radical mechanism? In a Wurtz reaction, is the intermediate formed a carbanion or a free radical?
And if we can assume both mechanisms to be true, how can we decide the reactivity of a tertiary halide?

I think both mechanisms are valid. As far as I know, both are supported by some experimental evidence. Byproducts, like alkanes (other than the expected one) and alkenes, can be explained by the free radical mechanism. On the other hand, the formation of Organometallic compounds can be verified by the ionic mechanism.

The essential thing to note is that the mechanism involves a $\mathrm{S_N2}$ step. And because of that sterically hindered alkyl halides, like tertiary alkyl halides, instead of giving a substitution product yield an elimination product and you get an alkene.


Bond lengths in naphthalene As I was reading Clayden: Organic Chemistry 2nd Edition, I came across an image of naphthalene showing its varying $\ce{C-C}$ bond lengths, however, I was unable to rationalize the bond lengths by drawing naphthalene's resonance forms. I have searched for a good explanation for the bond lengths shown but I have not found anything remotely convincing.

The 3 resonance forms of naphthalene are shown below however, these resonance forms are inconsistent with the bond lengths displayed by Clayden. From looking at these resonance forms I would not expect the $\ce{C-C}$ bond fusing the benzene rings to be the shortest since it has less double bond character compared to some of the other $\ce{C-C}$ bonds in the molecule.

You are assuming that all resonance structures contribute equally.  The resonance structures on the right and left are equivalent but have one aromatic ring and one non-aromatic ring.  This is in contrast to the center resonance structure, which has two aromatic rings and is more stable, thus giving the bonds between carbons 1-2; 3-4; 5-6; 7-8; and 9-10 more double bond character than the other bonds and making them shorter.


How does one tell if a specific molecule is acidic or basic? Let's take the $\ce{KOH}$ molecule into account. I know it is a base from literature, but how would one go about determining if a molecule is acidic or basic simply based on the structure of the molecule? How about amphoteric?

Also, I understand that there are two complementary systems - Lewis and Brønsted-Lowry theories. How do they work and how do they fit together?

It is sometimes challenging to determine if a molecule is going to be acidic or basic if the system in which it is reacting is not considered.  An important point to consider when dealing with acids and bases is that acid/base strength is inherently tied to the solvent.  For this answer, I'm going to limit the discussion to acids and bases in an aqueous environment.
It is helpful to consider the terms acid and base as a means to classify substances.  This way, Chemists can explain chemical reactivity and structure-function relationships of substances.  Very early classification systems depended on our senses (acids are sour, bases are slippery to the touch) and more recent classification systems utilize structural characterization tools such as NMR or crystallography.  Many classification systems have been proposed over the years, and only a few of them have found sufficiently widespread use to end up in textbooks used in the standard Chemistry curriculum.  Below are a few systems, taken from Miessler & Tarr's Inorganic Chemistry textbook.  The 2nd through 4th entries are those in common use today.

NAME (YEAR) acid definition [example]; base definition [example]
Liebig (~1776) Acid: an oxide of N, P, S [$\ce{SO_3}$]; Base: Reacts with
acid [$\ce{NaOH}$]
Arrhenius (1894) Acid: Forms hydronium ion [$\ce{HNO_3}$]; Base: Forms hydroxide ion [$\ce{NaOH}$]
Brønsted (1923) Acid: Proton donor [$\ce{HCl}$]; Base: Proton acceptor [$\ce{NaOH}$]
Lewis (1923) Acid: Electron-pair acceptor [$\ce{Ag^+}$]; Base: Electron-pair donor [$\ce{NH_3}$]
Ingold-Robinson (1932) Acid: Electrophile [$\ce{BF_3}$]; Base: Nucleophile [$\ce{NH_3}$]
Lux-Flood (1939) Acid: Oxide ion acceptor [$\ce{SiO_2}$]; Base: Oxide ion donor [$\ce{CaO}$]
Usanovich (1939) Acid: Electron acceptor [$\ce{Cl_2}$]; Base: Electron donor [$\ce{Na}$]
Solvent system (1950s) Acid: Solvent cation [$\ce{BrF_2^+}$]; Base: Solvent anion [$\ce{BrF_4^-}$]
Frontier Orbitals (1960s) Acid: LUMO of acceptor [$\ce{BrF_3}$]; Base: HOMO of donor [$\ce{NH_3}$]

I find the various acid/base systems very enlightening.  Note how From Arrhenius through Lewis there was a broadening of the acid/base classification system; Arrhenius can't be used to describe non-aqueous acids and bases and Brønsted can't be used with aprotic substances.  Yet, the Lewis definition incorporates the previous two (a Brønsted acid is also a Lewis acid; an Arrhenius base is also a Lewis base).  One then may ask, what's up with the Lux-Flood definition then?  This definition is counter to the trend of broadening the classification system, and yet it is useful in describing anhydrous solid-state chemistry and is used to describe geochemical reactions as well as the chemistry of high-temperature melts.  My point being: classification of substances into acids and bases is only meaningful if it helps explain chemical phenomena.
Which brings us to the Usanovich definition, which essentially states that every reaction is an acid-base reaction.  Such a broad definition is not overly helpful, and to those of us with an affinity towards electrochemistry (ahem), is somewhat arrogant :-)
I do believe that determining if a substance will behave as an acid or a base requires a bit of chemical intuition (or a Socratic method).  For example, you may have performed an experiment in which $\ce{KOH}$ served as a base; you know that, like potassium, sodium is an alkali metal; therefore you presume that $\ce{NaOH}$ would be a base as well.
Back to the question at hand
So how do I suggest one use this information to predict whether a substance will behave as an acid or a base using its structure alone?  Personally, I find the Lewis theory as the most useful classification system in answering this type of question.  If the structure of a compound is set before me and I am to predict its acid/base chemistry, I will ask two questions:

Are there any lone pair electrons that can be donated?
Are there any electron-deficient atoms that could serve as electron pair acceptors?

If the answer to question 1 is yes, then the molecule can behave as a base.  If the answer to question 2 is yes, then the molecule is an acid.  If both are true, then I have an amphoteric substance.
In proofreading this answer, I realize I said I would restrict myself to aqueous systems, in which case using the Brønsted system may be more helpful.  In this case, the questions become:

Is there a hydrogen that can be donated? (You'll be right more often than you are wrong if you rephrase this question as "Is there a hydrogen that is attached to something other than carbon?").
Is there a lone pair that can accept a proton?


Interaction between metalic nanoparticles and aggregation I can't find any publication to deal with aggregation of colloidal metalic nanoparticles. What type of interaction(s) is/are there between aqeous metalic nanoparticles which give(s) rise to aggregation?

Metallic nanoparticles, considered as colloids, are attracted to each other by van der Waals forces. To prevent their aggregation, we normally use dispersant using electrostatic repulsion or steric hindrance to stabilize the colloids.


How to store bulk gas products and reagents? Suppose I'm doing a batch process where an intermediate (non-hazardous) reagent is a gas, and I need to produce and temporarily store something on the order of a cubic meter of it to later feed back into a reaction.  Assuming the gas can't be condensed at anywhere near STP, isn't there some sort of standard flexible lab gas storage apparatus?

I've been looking for something like a large mylar (or other relatively impermeable) bag with a tube insert I can put on a barb and valve.  Either I don't know what this would be called, or it doesn't exist and I should be using something else.

I realize I could try to pump out a large rigid container, but once we're talking about a few cubic meters, and trying to avoid contamination, that seems like a pain compared to a flexible gas bag.

It probably depends on local regulations and the properties of the intermediates!

Gas bags might indeed be a viable solution.

In the biogas community, see SolarCities for an example, PVC bags from PUXIN, a chinese supplier, are quite popular.


Why does hot oil fry things, but hot water does not? When a slice of potato is dropped into hot oil, it gets fried but the same in hot water, gets ‘cooked’. What is it about oil that gives the slice of potato the crispy feature while water only makes it soft?

Oil can get hotter than the b.p. of water, 373 K.

The browning (decomposition) of starches proceeds very slowly (perhaps years for a noticeable change) at 373 K. In addition, the water in the potato will not evaporate away, since the potato is in water (not the water vapor above the liquid, which can be hotter than 373 K).
Since the cooking oil is hotter (e.g. 440 K), starch decomposes (caramelizes) in a few minutes, and dehydration makes the chip crunchy. Bon appétit!


Which ceramic-like material is used in halogen MR16, GU5.3 and so on bulbs? I hope this question belongs in here; I cannot think of any other better place to ask it.
Which kind of material is used in dichroic halogen bulbs to attach the bulb itself to the reflector and keep it in there? Looks to me like a ceramic or some kind of terracotta, but I wish to know the exact kind and if it is available for home use or, in case it isn't, a good recommendation as a domestic substitute. Of course, I understand it's main property is, appart from its hardness, the ability to be fireproof and/or refractory, due to the high temperatures being reached in there. For this same reason, something like epoxy resin or anything plastic based probably wouldn't work as it would melt or even burn.

Actually, aside from heat resistance, a primary qualification is that the adhesive cement thermal expansion coefficient match that of the quartz (or glass or alumina) envelope over the temperature range of storage and operation, to avoid cracking the envelope (the metal fixture is more flexible and "forgiving").
There are a number of cements in use, such as, "aluminum phosphate refractory cement... a mixture of aluminum phosphate... phosphoric acid [and] alumina". Glassbond sells a variety of adhesives, such as a phenolic/melamine adhesive for glass lamps with operating temperature below 300°C. Sauereisen has a long history making high-temperature adhesives, including those of silicate and of magnesium oxysulfate composition. One could try making these cements at home or in the lab, though they're readily available commercially. You could experiment with refractory furnace cement, though it likely would have a different coefficient of expansion, cracking the envelope.
BTW, I remember, in school, attaching thermocouples to incandescent lamps with a variety of cements... and watching the glass envelope form small star-shaped cracks at each joint, for many of those adhesives, as the lamp warmed. Sorry, but I don't remember which formulations actually succeeded past a few on/off cycles.


Why is HSO4- shown as an example of a weak acid instead of HSO4? I'm watching some (really awesome) chemistry courses, and I think I have a fairly decent handle on acids and bases (or maybe not, and I'm fooling myself). But the teacher just showed a table of weak/strong acids. $\ce{HSO4-}$  was one of them.

I'm wondering, why hydrogen sulfate ion instead of just hydrogen sulfate? Why is $\ce{HSO4-}$ an acid and $\ce{HSO4}$ apparently isn't? This was the only ion on the table. Or does $\ce{HSO4}$ simply not exist?

The bisulfate molecule doesn't exist in solution. I can only find references to the hydrogen sulfate molecule as part of salts, which implies that hydrogen sulfate is usually found as an ion.

The bisulfate (hydrogen sulfate) ion is indeed an acid. It is a relatively strong weak acid too, with a $\ce{K_a}$ value of $1.2 * 10^{-2}$.

Part of the reason for its acidity has to do with its electronegative oxygens isolationg electron density away from the hydrogen atom bonded to the oxygen. This makes the hydrogen more partially positive and thus more reactive (more likely to be taken by a base).


Which quantum chemistry approximation to use? I'm just getting started with quantum chemistry using some simple test problems.  I have a few of the software packages installed (ASE, Psi4, GPAW etc) and everything is working - no issues with the software itself - but there is a bewildering variety of methods, from semi-empirical (PM7) to DFT (B3LYP) to coupled cluster (CCSD(T)); plus options for each such as basis functions, and nothing in the documentation gives any specific advice on what to use when.
Is there concrete guidance on which method is appropriate for which type of problems?
What I'm mostly interested is biological systems: molecules in water at room temperature, and both covalent and non-covalent interactions; most of the atoms are CHNOSP.  An example would be taking a system of a few hundred atoms and optimizing the coordinates (either locally or trying to find the global minimum).  A slightly more ambitious project would be to optimize the coordinates a few hundred atoms within a much larger system of thousands of atoms (where the other atoms have partial charges and produce electrostatic forces, but are otherwise fixed).

A good guide comes from this series of slides by Jan Řezáč:

It seems that semiempirical methods (of which PM6-D3H4 in MOPAC is probably most accurate for the types of systems encountered in biology) are the only methods able to handle structure optimization of a whole protein (or whole domain of a larger protein).  DFT can handle an order of magnitude fewer atoms, and MP2 or CCSD(T) are mainly used for creating benchmark data sets to test the accuracy of methods, such as S66.


Anion formation in an alpha particle-neutral atom interaction? How can an alpha particle/beta particle/gamma rays interacting with a neutral atom lead to an anion forming? Because i can see how they cause loss of electrons leading to cations forming, but how can anions form?
cheers

If all you have is an atom and an alpha particle whizzing by to kick an electron out of the atom then yes. But how often do you see an atom sitting out on it's own?

If everything happens as before except that now there are other atoms around, when that electron is knocked loose of the first atom, it will likely come to rest in some other atom thereby creating an anion.


Why does tea dissolve better when put in water rather than milk? I sometimes drink tea with or without milk. What I have noticed is that when I dip the bag in water the tea spreads out evenly and I have a cup of tea in like 5 minutes.

But when I do the same with milk in it, the tea takes forever to dissolve into the milk and even after a long time half of the milk is absorbed by the tea bag for some reason.

Could someone explain me why this is happening?

Milk is an emulsion containing small bubbles of oil/fat. Whenever you boil milk or do something drastic, this fat tends to separate. You can do this nicely, and get cream or butter, preserving the bubbles, only part of the water content is separated, or can do more drastically, when these bubbles collapse to a big oily blob.

When you put your filter in the milk, most probably some of the oil/ fat bubbles hit the surface and cover it with an oily film. This film prevents the easy dissolution of caffeine, etc from the tea leaves in water.


Why is the molar entropy of hexane lower than that of butane? Entropy is dependent on the complexity of the molecule.
Due to this, I expected entropy to increase as the complexity of the molecule increased.
According to p 918 of Atkins' Physical Chemistry (9th ed.), I noticed that the standard molar entropy of n-hexane at is $\pu{204.3 J mol-1 K-1}$, whereas that for n-butane is $\pu{310.23 J mol-1 K-1}$.
How can butane have a higher standard molar entropy, even though it is a smaller molecule?

There are two factors to consider. Certainly, the dominant factor is that the degrees of freedom available for the molecule to disperse energy into increases as the carbon chain extends. Thinking along these lines, one would expect a monotonic increase of entropy with carbon chain length.
However, we must also consider phase. As the carbon chain increases in length, the equilibrium phase at RTP changes from gas to liquid. This occurs between butane and pentane. At this point, there is a drop in the standard entropy, which then proceeds to increase monotonically from this point (pentane). The cause for this drop in entropy upon phase-change is due to the motional degrees of freedom being more constrained by proximity to other molecules in liquids compared to gasses.
See this link, where you can see some nice plots of standard entropy vs. carbon chain length.


Wavelength of iPhone 4S camera light for visible spectroscopy calibration I ordered a DIY spectroscopy kit from public-lab, however don't have access to a halogen lamp to calibrate the wavelengths.

I'm wondering if it would somehow be possible to calibrate the wavelength using the iPhone camera light?

According to the site, wavelength calibration is meant to be done with a compact fluorescent lamp (for the mercury lines), but at the bottom there's a tool for calibrating using any two known wavelengths. If the LED on the phone has two distinct lines that you can calibrate against, it may work, but something like a pair of monochromatic LEDs or laser pointers (say one blue and one red) or something will probably work better.


Is the E in Arrhenius equation the change of enthalpy? I'd like to know whether the activation energy in Arrhenius equation is the enthalpy difference between the transition state and the reactant. If it's not enthalpy, is it Gibbs energy or something else?

The "E" is the height of the thermodynamic potential energy barrier. Which potential that is (enthalpy, Gibbs energy, Helmholtz, or internal energy) depends on the conditions under which the reaction takes place - which thermodynamic state variables are held constant.

For example, in a system where the number of moles, the pressure, and the temperature are held constant (called NPT), then the thermodynamic potential energy (and thus the activation energy) is measured using Gibbs free energy.


What is the usage of orbitals more complex than f orbitals? Every high school learner, in each corner of the world, faces the lesson History of Atom during his courses, just as I did. We learned about s, p, d and f orbitals, though there were no signs of orbitals in molecules.

Then I wondered, are there any other orbitals, simpler or more complex, than the four mentioned? Surprisingly, I learned that there are also usages for orbitals g, h, i and even k and l.

Yes, I use the word "usages". Because I believe, unless something is useful, it will never enter the domain of science. Anyway, I read in Wikipedia (though not much of it I did understand) that these orbitals are used when describing and doing the measurements of molecular orbitals.

Since no element in the periodic table has enough electrons to fill even orbital g (in its base state), in cases of molecular orbitals that have a g defined in themselves, atoms must have been excited. Excitation needs energy, doesn't it? Where does this energy come from? Isn't the formation of new bonds usually exothermic?

Surprisingly, I learned that there are also usages for orbitals g,h,i
  and even j.

Actually, the letter "j" is not used, so it is s, p, d, f, g, h, i, k, l, etc.

The higher angular momentum orbitals do enter the domain of science, due to excited states of atoms. Transitions to and from excited states are observable through atomic spectroscopy.

For example there is the article Microwave spectroscopy of Al I atoms in Rydberg states: D and G terms


How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes? How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes?

For example, how to determine the oxidation states of $\ce{Pt}$ in the following compound?:

$[\ce{Pt(NH3)4}][\ce{PtCl6}]$

There is something called an average oxidation state, which is typically not very meaningful chemically, but can (usually) easily be calculated by just knowing the structural formula. In your case, since chlorine is the most electronegative element (save nitrogen), you can assume it to be in $\mathrm{-I}$ state while nitrogen of course would have $\mathrm{-III}$ and hydrogen $\mathrm{+I}$. Thus to get platinum’s average oxidation state just sum up to the compound’s charge:

$$\begin{align}2 x + 4 \cdot (-3) + 3\cdot 4 \cdot (+1) + 6 \cdot (-1) = 0\\
2x - 12 + 12 - 6 = 0\\
2x -6 = 0\\
2x = 6\\
x = 3\end{align}$$

Which gives you an average oxidation state of $\mathrm{+III}$.

But as I said before, average oxidation states do not carry any physical or chemical meaning and can only serve as exercises to students and pupils. Take for example tetrathionate’s ($\ce{S4O6^2-}$) average sulphur oxidation state of $+\frac{5}{2}$.

Instead, one should always attempt to allocate per-atom oxidation states you cannot do that without knowing the charge of each of those complexes, and you need to do it per complex; ideally by drawing it, heterolytically cleaving all bonds (or homolytically for homoatom bonds) and then counting electrons. In the case of your compound — tetraamminplatinum(II) hexachloridoplatinate(IV) — you should arrive at the numbers in brackets in the IUPAC name I gave you, knowing that the cation has $2+$ charge and the anion $2-$.


Is a carbon-fluorine bond stronger than a carbon-chlorine bond? In hydrolysis/nucleophilic substitution of haloalkanes, the bond enthalpy indicates the rate of hydrolysis (e.g. the $\ce {C-I}$ bond is weaker than the $\ce {C-Cl}$ bond so in hydrolysis the alkane with the $\ce {C-I}$ bond will break more quickly and react with the $\ce {OH-}$ ions)

However, in free radical substitution reactions, if you have a molecule like $\ce {CCl2F2}$ , then is the $\ce {C-F}$ bond stronger or weaker than the $\ce {C-Cl}$ bond (so which will require less energy to break)?

For free radical reactions, the most important parameter in assessing bond strength is bond dissociation enthalpy (BDE). Typical values for $\ce {C-F}$ bonds are around $\mathrm {100\ kcal/mol}$, while for $\ce {C-Cl}$ bonds are around $\mathrm {80\ kcal/mol}$; $\ce {C-Cl}$ bonds are therefore weaker.

This can be rationalized by considering the poorer orbital overlap between the $\ce {Cl}$ bonding orbital (3 s/p character, more diffuse than 2s/p) and the $\ce {C}$ $\mathrm{sp^3}$ hybrid (predominantly $\text{2p}$ character) than that between the $\ce {F}$ bonding orbital (2 s/p character) and the $\ce {C}\ \mathrm{sp^3}$ hybrid.

Some data on typical BDE values in $\ce {F/Cl}$ compounds can be found here.


Melting points of alkali metal halides The correct order of melting point of alkali metal halides is:

$\ce{MF}>\ce{MCl}>\ce{MBr}>\ce{MI}$
$\ce{MI}>\ce{MBr}>\ce{MCl}>\ce{MF}$
$\ce{MCl}>\ce{MF}>\ce{MBr}>\ce{MI}$
$\ce{MI}>\ce{MF}>\ce{MCl}>\ce{MBr}$

The correct answer to this is option (1), the reason I could think of is that as the size of anion in ionic compound increases the covalent character increases and hence the melting point decreases.

As with all the periodic trends concerning differences inside a single group, it will have something to do with the size of the atoms involved.

You already noted that fluorine is the smallest of the halogen atoms and iodine is the largest. This is due to there being only two core electrons in fluorine while iodine has 46. While adding an electron to any of these will increase their size, the size differences between them will remain the same due to said core electrons; hence, fluoride will be the smallest and iodide will be the largest.

This means that fluoride has a high charge-per-volume ratio while iodide’s is much lower (compare that to Pearson’s HSAB concept — it’s the same idea). A higher negative charge density means that it draws cations towards it better and thus stabilises the crystal structure. A secondary effect is that the smaller size of fluoride means that it must have a lower coordination number in the crystal structure, thus having less positively charged neighbours, thus attracting them more.

Together, these influences mean that it is just harder to break up the high ordering of the crystal structure to melt the salts the higher up you are in the group.

A higher degree of covalent bonding for iodine may play a role, but I would deem it neglectable.


Does a smell imply particles or chemicals in the air? (I know little science so please help me if I'm in the wrong place or if my question needs editing. I could use help with tagging, for example.)

If I can smell something then does that mean there are particles or chemicals in the air?

My question is motivated by some new window blinds that have been installed in my home. The blinds have a smell, which is most noticeable when the sun is shining on them. I am trying to work out if the blinds are emitting some chemical into the air. This question is my first step.

[This question is not the same as "How does the smell of a compound come about, and is it possible to define a smell?". This is clear from the title.]

Smell works because your nose detects substances in the air.

Everything you can smell is the result of a molecule carried by air interacting with complex proteins in the nose/mouth (smell and taste are related but taste copes with solids and liquids as well as volatile airborne stuff). We broadly understand how smell works but the exact details are often a complete mystery.

The key is that smell-sensing nerves contain some proteins with very specific sensor sites that can detect particular groups of substances when they "fit" into the site. Exactly how the sites translate the fit into nerve signals is often a mystery. But the point that matters for this question is that the stuff being detected has to be carried into the nose via the air (which means it has to be a gas or, possibly, a very small particle that is easily carried by air).

So, yes, smell implies there are chemicals or particles in the air. However, the nose is extremely sensitive to some chemicals and being able to detect the smell does not imply a problem (even nasties like hydrogen sulphide are detectable by smell at below parts per billion concentration, hundreds of times below the level causing harm). Your blinds probably are emitting something but it is most likely both harmless and will probably fade with time like the smell characteristic of a new car.


Equilibrium :. Why does a system at equilibrium try to undo the effect of external stress applied on it? " Le Chatelier principle states that a system at equilibrium when subjected to external stress tries to undo the effect of that stress"

In short, why does system at equilibrium follow Le Chatelier principle?

It is not that a system follows Le Chatelier's Principle, as it is not a law such as Newton's laws of motion,  but rather that this is a shorthand way of working out what happens; a 'rule of thumb' if you wish.

If a system is in stable equilibrium it must have the lowest energy possible under the prevailing conditions. If now a perturbation is made, say by changing temperature or pressure, the system is no longer at equilibrium because some energy has been added or removed. It will now change to lower its energy and again reach equilibrium under the new conditions. Le Chatelier's Principle is then a statement of this.

For example, suppose that as a result of molecular motion, a small part, A, of a larger volume of a gas spontaneously increases by  $\Delta V$. Thus the pressure  decreases, $\Delta p < 0$, and so the net force on A exerted by the surroundings is in a direction that tends to restore equilibrium.


Is the relative natural abundance of isotopes of an element the same everywhere? Wherever you find potassium, the isotopes are present in a set percentage that exists the same everywhere in nature, but why is that? Does it have something to do with how the element is produced?

The relative natural abundance of isotopes is not the same everywhere.  Depending upon what you mean by "everywhere", there are two cases to consider.

Extraterrestial

Dust from before the sun was formed (stardust, presolar grains) has a very different elemental and isotopic composition than that found on earth.  Depending where a star is in its life cycle different elements and isotopes will be created and destroyed, so it is not surprising that dust from different stars has a very different composition than what is found on earth. As Wikipedia notes,

  In these materials, deviations from "natural abundance" are sometimes
  measured in factors of 100

Terrestial

When our solar system was forming, elements and isotopes were not uniformly distributed - close to uniform, but not exactly uniform.  This is due to processes like diffusion in which a mass-dependent fractionation occurs.  A heavier isotope won't travel as far as a lighter isotope in a given amount of time, so the resulting distribution of the two isotopes will not be exactly equal.  While these mass-dependent processes account for most of the isotopic variation in our solar system, mass-independent processes can play a smaller role.  It is also suggested that catastrophic events in nearby stars may have also influenced local elemental and isotopic homogeneity.  These "original" inhomogeneities will be further altered by a continuation of the mass-dependent (diffusion, bond-breaking and making, etc.) and -independent processes. Natural radioactive decay processes can also lead to changes in local isotope ratios.  For example, $\ce{^238U}$ ultimately decays to $\ce{^206Pb}$, so it wouldn't be surprising to find an altered lead isotope ratio (e.g. enriched in $\ce{^206Pb}$) around large uranium deposits.

This paper presents a study of the natural isotopic variation found on earth.  Here is a table from the paper that gives an idea of the range in variation.  As expected it is much smaller than the extraterrestial variation, but certainly not insignificant.  For boron and copper, the isotopic variation is on the order of parts per hundred, but more typically isotopic variation is on the order of parts per thousand.


How does the 1.5 bond order of ozone contribute to its instability? I'm confused as to why ozone is so unstable? I've been told it's because it has a fractional bond order. However, this doesn't make sense to me. Don't fractional bond orders mean resonance and doesn't resonance stabilize ozone?

It's important to bear in mind stability is always a relative judgment.  Things are stable or not only with respect to that into which they could turn. So we only say ozone is "unstable" because it can and will easily decompose into oxygen (O2) if given a chance.  In some sense it might even be more correct to say what's important here is the stability of O2 (which derives from its strong double bond).  If O2 weren't possible, or were significantly higher in energy itself, then we would say O3 is more "stable" just because it wouldn't have anything into which it could as easily turn, so it would stick around longer.  So "stability" is not just a property of one molecule, but a property of that molecule plus all the other molecules into which it could turn.

A famous example of this is the high stability of N2 (with its very strong triple bond), which means certain compounds (like TNT or ammonium nitrate) with nitrogen atoms in higher oxidation states (e.g. bonded to O atoms) can readily decompose to products including N2 and O2, producing a lot of energy.  That is, they are good explosives, and you can consider that as coming from the strong stability of the O2 and N2 molecules.  (This is somewhat akin to the argument in organic chemistry that the pKa of an acid depends strongly on the stability of its conjugate base -- that into which it can turn.  The more stable the base, the higher the pKa, i.e. less "stable" the acid.)

All that said, fractional bond order does indeed suggest resonance, because it suggests the failure of a single Lewis structure to describe the molecule accurately.  And indeed O3 needs two resonance structures to describe it. It's also correct to say resonance stabilizes a molecule, but, again -- with respect to what?  In general, the answer is "with respect to a similar (perhaps hypothetical) molecule that does not have resonance."  So O3 is more stable (because of resonance) than some hypothetical O3 molecule trapped in one of the O3 resonance structures.  It can still be "unstable" for most purposes, it's just less unstable than this hypothetical non-resonant O3.


What would make tracing paper clear? This isn't much of a "chemistry" question, but this is the only category where I can get answer based on science. I already tried soaking it in cooking oil, which did not work (it only made the tracing paper even more slippery.)

I can tell you how to find something that will make tracing paper transparent, but I don't have an example to hand.

Tracing paper is partially opaque because it is "rough" and contains many small fibres of cellulose that scatter light. These scatter light effectively because of their size and because their refractive index is very different from air. If you can find a liquid which has a similar refractive index, that will make the paper much more transparent than in its natural state.

I'm sure their are some, but I don't know any right now.


Is there an alternative wet chemical method or electrochemical method to obtain precipitation of iron, nickel and silicon powder? There are many methods available to obtain powdered metals and alloys. Wikipedia show some of them. Another form is grinding with posterior removal of wheel particles. All these methods produce metal powder with the help of specific machinery. I am asking for an alternative wet chemical method (water solution) or electrochemical method (water or ionic liquid electrolyte) to obtain metal powder (specifically iron, nickel and silicon, in pure or controlled alloy form).

Both metal powders can be preapred electrochemically.

Iron dust is prepared using solution containing $\ce{FeCl3}$ with addition of $\ce{AlCl3}$ to make the solution more conductive. The process occurs at elevated temperature 60°C. Unfortunately exact process parameters are hard to find.  But following graph could help.

Nickel could be prepared electrochemically too and can be prepared by dissolving of Ni-Al alloy in hydroxide -Raney_nickel. The description of patent is here.

This is link to a study book PRÁŠKOVÁ METALURGIE, K. Skotnicová, M. Kursa
, unfortunately only in czech.


Can reversible reactions proceed by different mechanisms in the forward and reverse directions? It seems odd to me that this would be the case, but it also seems odd that it wouldn't.
Imagine a mechanism for a (forward) reaction that includes a slow reversible step and a fast irreversible step:

$$
\begin{align}
\ce{A &<=> B} &\quad &\text{(slow)} \\
\ce{B &-> C} &\quad &\text{(fast)}
\end{align}
$$

If the second step is irreversible, it's presumably going to be highly unfavorable to run in the reverse direction, and so there ought to be some alternate mechanism with a lower $E_\mathrm{a}.$ Is it possible that the reverse reaction is more favorable by a different mechanism? If not, why not?

Yes, you can. An ideal Carnot cycle (admittedly not a reaction) illustrates how this is possible. In such a cycle each step is reversible. To return to a state you can either reverse the direction of the process (running say a heat engine as a refrigerator) or follow the full path of the cycle until the state is again encountered.


silicon nitride and sCO2 I want to use silicon nitride ball bearings with a reciprocating piston that will be powered by $\ce{CO2}$ liquid heated to  $\mathrm{288^oC}$ becoming super critical. That temperature is the maximum and the cylinder will not exceed that staying well below the $\mathrm{768^oC}$ rating of the bearings.

I can not use lubrication and the 1/4 bearings will be stacked 3 high with 4 of them surrounding the rod for a total of 12 bearings in direct contact with the rod that could have traces of supercritical $\ce{CO2}$ upon it. The bearings will be in contact with each other to provide both a seal and to align the piston and if there is any contamination it will spread.

Can I expect any type of reaction between the supercritical $\ce{CO2}$ and silicon nitride bearings?

Each of these materials is largely inert and there is no reason to expect that they would react with each other. Regarding supercritical $\ce{CO2}$, this Wikipedia article states the following:

  Supercritical $\ce{CO2}$ is becoming an important commercial and industrial solvent due to its role in chemical extraction in addition to its low toxicity and environmental impact. The relatively low temperature of the process and the stability of $\ce{CO2}$ also allows most compounds to be extracted with little damage or denaturing.

And regarding silicon nitride:

  Silicon nitride is a chemical compound of the elements silicon and nitrogen, with the formula $\ce{Si3N4}$. It is a white, high-melting-point solid that is relatively chemically inert, being attacked by dilute $\ce{HF}$ and hot $\ce{H2SO4}$. It is very hard (8.5 on the mohs scale). It is the most thermodynamically stable of the silicon nitrides. Hence, $\ce{Si3N4}$ is the most commercially important of the silicon nitrides and is generally understood as what is being referred to where the term "silicon nitride" is used.

Additionally, silicone nitride is used in the most critical, high temperature, high stress of applications, such as the bearings for NASA's (now defunct) Space Shuttle engine bearings.


Entropy and Enthalpy ∆S (Entropy change) is given by ∆H/T (∆H = Enthalpy change , T= absolute temperature). Then how can the Entropy change for a reaction be negative if the Enthalpy change is positive, and vice versa?

You are looking at two different entropy changes. The one computed directly from the enthalpy change is the entropy change of the universe. The entropy change of the system is measured by computing the component due to heat transfer. But internal changes in the density of states can also result in a change in entropy, and this can take place without any transfer of heat.


Diagonalization of Hessian H= PkP$^{-1}$: what do the elements of P matrix physically mean? This is quoted from Computational Chemistry by Errol J. Lewars' chapter 2's Stationary Points & Normal-Mode
Vibrations: ZPE:

  \begin{align}\mathbf H &=\begin{bmatrix}
    \dfrac{\partial ^2 E}{\partial q_1 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_1 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_1 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_1 \partial q_9} \\
    \dfrac{\partial ^2 E}{\partial q_2 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_2 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_2 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_2 \partial q_9}\\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    \dfrac{\partial ^2 E}{\partial q_9 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_9 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_9 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_9 \partial q_9}
\end{bmatrix}\\ \\ &=\underset{\mathbf P}{\begin{bmatrix}
    q_{11} & q_{12} & q_{13} & \dots  & q_{19} \\
    q_{21} & q_{22} & q_{23} & \dots  & q_{29} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    q_{91} & q_{92} & q_{93} & \dots  & q_{99}
\end{bmatrix}} \underset{\mathbf k}{\begin{bmatrix}
    k_{1} & 0 & 0 & \dots  & 0 \\
    0 & k_{2} & 0 & \dots  & 0 \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & 0 & \dots  & k_{9}
\end{bmatrix}}\mathbf P^{-1}\end{align}

  The 9 × 9 Hessian for a tri-atomic molecule
  (three Cartesian coordinates for each atom) is decomposed by diagonalization into a P
  matrix whose columns are “direction vectors” for the vibrations whose force constants
  are given by the $\bf k$ matrix. Actually, columns 1, 2 and 3 of $\bf P$ and the corresponding $k_1,k_2,k_3$ of $\bf k$ refer to translational motion of the molecule (motion of the whole
  molecule from one place to another in space); these three “force constants” are nearly
  zero. Columns 4, 5 and 6 of $\bf P$ and the corresponding $k_4,k_5,k_6$ of $\bf k$ refer to
  rotational motion about the three principal axes of rotation, and are also nearly zero.
  Columns 7, 8 and 9 of $\bf P$ and the corresponding $k_7,k_8,k_9$ of $\bf k$ are the direction
  vectors and force constants, respectively, for the normal-mode vibrations: and
  refer to vibrational modes 1, 2 and 3, while the 7th, 8th, and 9th columns of $\bf P$ are
  composed of the x, y and z components of vectors for motion of the three atoms in mode
  1 (column 7), mode 2 (column 8), and mode 3 (column 9).

I couldn't conceive what actually $\bf P$ implies; it is, of course, the eigenvector matrix of the Hessian $\bf H $ but what does it represent?

Lewars wrote that columns of $\bf P$ are actually direction vectors; now what did he mean by direction vector?

I thought $q$ represents the geometric coordinate; I got $k$ is the force-constant; but what are those series of $q$ s in $\mathbf P\;?$

Are the $q$ s in $\bf P$ coordinates of the atoms in molecules? If so, aren't they always changing since they are vibrating, rotating...?

Can anyone please explain this to me what those $q$ s actually represent? What did Lewars want to mean by direction vectors?

Are the $q$s in $\mathbf{P}$ coordinates of the atoms in molecules?

Close, but not quite. The $q_{i,j}$ represent the displacements from the equilibrium geometry of the appropriate atom/coordinate $i$, for each mode j. Thus, for example, $q_{5,9}$ corresponds to the displacement from equilibrium of atom $2$ along axis $2$ (typically the $y$-axis) when the molecule is undergoing vibration in mode $9$.

  If so, aren't they always changing since they are vibrating, rotating...?

There are no changes due to translation or rotation. It is customary to "project out" the translational and rotational motion from the Hessian and the geometry.  Qualitatively, this means that we simply ignore the translation and rotation and consider the molecule as "fixed" in its equilibrium configuration (allowing for vibrational 'wiggling'), with that configuration always oriented in the same way relative to the coordinate system.

As for the vibrational motion, though: yes, the $q_{i,j}$ values do technically also 'wiggle' as the molecule vibrates. However, the magnitudes of the perturbations $\delta q_{i,j}$ are in many cases very small, and can be neglected. In those cases where they cannot, the harmonic oscillator approximation to the vibrational motion breaks down and is no longer a useful model for vibration along that mode.

See my answers here and here for some links to literature on situations where this occurs (emphasis on the 'internal rotation' topics, though some of the 'anharmonic' and 'Coriolis' references might be somewhat relevant). Here is a Chem.SE question about nitrogen inversion that might be of interest.  More generally, a Google search for "low-frequency vibrational modes" should turn up lots of interesting reading.


Hard coating for soft surface Say I have a soft, possibly porous object. I want to coat it with a material,

Resistant to boiling water and some scraping
Not hazardous, at least after the coating
Largely transparent

The idea is to take, say, a fruit with the insides removed and to make a cup or a vase out of it.

Is there something I could buy or make at home for this?

You could vacuform heated thin PET film over it.  That would not stop biology.


How to get 3D molecular structures for 270 compounds? I need to calculate 10 properties for 270 molecules using Gaussian 09. Could you please suggest me what is the best and shortest way to have the structure of 270 molecules (in the format .com or .xyz)?

Is there any website which collected all the molecular structures?

There are a few resources that can get you what you want, but you'll probably want to consider some level of scripting automation. A search for 270 molecules isn't enormous, but doing it by hand would still be tedious.

PubChem: Run by the NIH, it offers many types of searching, and 3D coordinates for most compounds.
ChemSpider: Run by the RSC, it allows searching by name and has 3D structures for many (if not most) compounds. I'm less sure on quality, since they don't clarify how they generate their 3D coordinates.
NIH Chemical Resolver: Run by the NIH as well, it converts between different representations, including name to 3D structure.

Each offers automation tools, including Python interfaces called PubChemPy, ChemSpiPy, and CIRpy.

My choice would probably be the chemical resolver, partly because the access is very simple to automate:

https://cactus.nci.nih.gov/chemical/structure/"structure identifier"/"representation"

So you'd set up a URL like: https://cactus.nci.nih.gov/chemical/structure/aspirin/sdf?get3d=true

The result will be in SD format, but it's easy to convert that to XYZ or Gaussian files with Open Babel.


Buckminsterfullerene double bonds I am not now nor have I ever been a chemist.

I'm practicing by drawing molecules in the Chemspider structure search tool. Today I drew buckminsterfullerene, in order also to test my 3D spatial awareness. I chose to draw the double bonds on the hexagonal rings only - the resulting symmetry making this the intuitive choice, I thought.

Chemspider recognised my sketch as C60 right enough, but in its own 3D rendering has the double bonds placed differently, with some in the pentagonal rings, so of course I run to Google to see what is up with that, and learn the following from Wikipedia :

  C60 tends to avoid having double bonds in the pentagonal rings, which
  makes electron delocalization poor, and results in C60 not being
  "superaromatic".

So according to this my intuition was correct, but I guess Chemspider just has a different version stored for some reason?

So:

Q0: Any guess as to WHY Chemspider has different configuration?

Q1: Is the extract above actually true? (If you can imagine for a moment a world where Wikipedia might not be 100% accurate and authoritative)

Q2: If so, why? As I said, the more maximally symmetric arrangement seems intuitive. I guess the charge of the mutually repelling (repulsive?) electrons would be more evenly distributed this way, and so this is how it would self-organise, like a soap bubble.

Q2: In the context of the extract, does electron delocalization being 'poor' mean that there is a lot of it (this is my instinct), or not so much?

Q3: Either way, why is the degree of delocalization contingent upon where the double bonds are? My first guess would be that the positively charged nuclei in the hexagons keep the electrons in a less vice-like grip than in the more tightly convex pentagons.

Q4: Are the bonds static? Once you have a C60 molecule with its bonds a certain way, do they stay that way, or do they jump about arbitrarily?

Q5: Does it matter? Does the geometry of the bonds affect the chemistry, or indeed physics, of C60 that much?

If your answer is 'go read a book/paper on C60' then that's fine, if you could recommend a good one for a beginner. There's a lot out there.

Cheers MD.

All carbon atoms in fullerene are part of one "double bond", as you draw it. In quantum chemical reality, the structure of the fullerene is made up by fixed "single" bonds between those 60 sp2 hybridised carbon atoms, with the residual thirty pairs on pi electrons being somewhat delocalised.

In a sheet of graphene, the pi electrons are delocalised, and every bond lying between two six-rings gets the same share in the electron cloud.

In fullerene, they are as delocalised, but not equally distributed. The bonds between two six-rings are "more equal" (more aromatic) than those between five- and six-rings, and grab themselves a larger part of the electron density.

Problem of the five-rings is that the bond angles are not 120°, as they should be for a proper sp2 hybridised carbon atom. They become more sp2.4 or something, which let's the pi electrons move out of there a bit.


Why does an increase in Kinetic Energy increase the likelihood of successful collisions? As far as I understand bond formation, bonds must be broken between reactant particles to allow reactant particles to form new bonds and create product particles. This involves increasing the potential energy of the reactant particles, such that the attractive forces between say, reactant A and reactant B > the attractive forces between reactant A particles.

My textbook states that reactant particles must have sufficient kinetic energy so the collisions are energetic enough to break reactant bonds. What I don't understand is how the KE of the particles is related to breaking bonds. KE describes the energy needed to shift a body from rest to a given velocity, not the magnitude of a distance between particles (potential energy). So my question is: how is KE related to collisions, potential energy and product formation?

The kinetic energy of a particle is an alternative way of measuring how fast it moves (the mass of the particle also matters but for a given type of particle of a given mass the only thing that matters is how fast it is moving).

Potential energy has nothing to do with how far particles are apart (unless they are very close) and is pretty irrelevant in this situation.

In a gas or liquid, molecules are moving around a lot and banging into each other a lot. The speeds of the molecules are distributed statistically but their average speed is related to temperature. Whether a reaction happens when they collide depends on whether the molecules hit each other with enough energy to cause a reaction to happen (the orientation of the molecules might also matter but there are usually enough collisions in many different orientations that we don't think about this explicitly other than as part of the rate of the reaction).

In general, more energy in the collisions will lead to more reactions as more molecules will have enough energy to make the reaction happen (rather than just bounce off each other). More energy means more molecules have faster speeds. So, in general, more kinetic energy leads to a higher probability that any given collision will lead to a reaction.


Phases in iron carbon diagram The definition of phase :
Uniform chemical and physical composition.

So Alpha iron (0.001% carbon) and alpha iron (0.002% carbon)  are two different phases??

Let me opt for a paradoxical answer. They are not different phases, but you can't call them one phase either. There is no "them".

That is, you can't have a system in equilibrium which contains two different areas with the said compositions, either separated by a phase border or not. If you try, they will mix and form a single phase with averaged composition. It will be uniform all right.

Sure, you may have such compositions in different systems, and indeed many would call them "the same phase" then, based on the very close properties, similar structure, and the fact that they will mix if you let them. But this approach may quickly lead to nonsense. Say, water and acetone will mix in any ratio; are they one phase? Suppose yes. What about acetone and hexane? The same answer applies for the same reason. Now what about water and hexane? Er, well...

So let's not compare phases from different systems in terms of "sameness". Keep your apples and oranges apart.


Is it possible to freeze water by dissolving a salt? Theoretically, by dissolving a salt in water the melting point lowers, approximately 1.86 K kg/mol, making it more difficult to freeze water. However, the process of dissolution of certain salts is endothermic, lowing the temperature of the water. Is it possible that the water freezes due to that change of temperature although the melting point also lowers, or is it impossible?

For instance, consider the next case. In 1 kg of liquid water at 0 °C, we put a mol of KI. The melting point of water should be now about −1.86 °C. However, the enthalpy of dissolution of the KI is about 20 kJ/mol, meaning that the temperature of water should lower till −4.8 °C. So, theoretically, it should  freeze, but does this happen in reality?

It is possible, in a significantly different way that you envision. Freezing point depression via the cryoscopic constant is an example of a colligative property, which holds only for relatively dilute solutions. Once the solutions get very concentrated, intermolecular interactions become more complicated and do not generalize, meaning different compounds will affect the solvent differently.

Therefore, with enough of the right solute, it is possible to actually increase the freezing point of water. Oscar Lanzi's example of clathrates is interesting, but if you want to use salts, then I present you tetrabutylammonium hydroxide. It is known to form stable hydrates with a defined composition and containing a large amount of water, such as $\ce{(C4H9)4N+OH^-.30 H2O}$. The 30-hydrate is a solid which melts at approximately 30 °C, containing 67.6% water by weight.

This specific compound, at this specific concentration, forms a particularly stable network of solvating water molecules. You can think of it as stabilizing the structure of ice, allowing it to occur above the normal melting point of pure water. Thus, in principle, if you add enough of the anhydrous salt to water, when you reach the right ratio, the solution will "freeze".

Many compounds actually display this behaviour of forming solid hydrates at certain defined compositions, but tetrabutylammonium hydroxide is unusual in that it forms solid hydrates with a very large amount of water.


Optical isomerism regarding diastereomers Are diasteriomers always optically active?
If not then why?
Though they are chiral then also.

Some definitions (by IUPAC):

Diastereomer: A stereoisomer that isn't an enantiomer.
Stereoisomer: An isomer that has the same bond connectivity, but arises due to differences in the spatial arrangements of atoms.
Enantiomer: Non-superimposable mirror image.

So, diastereomers do not necessarily have to be optically active. Consider cis- and trans-1,2-difluoroethene.

They are stereoisomers (connectivity is the same, but the fluorines are on opposite sides) which are not enantiomers (they aren't mirror images of each other), which makes them diastereomers.

Neither are chiral because they are superimposable on their respective mirror images. Equivalently, the presence of the plane of symmetry (which is the plane of the molecule) makes both isomers achiral.


Question during radius of gyration calculation from XYZ trajectoris of atoms I have molecular dynamic simulation result of 100 polymer molecules (liquid), and I'm trying to calculate the radius of gyration using atomistic mass, molecular mass, and xyz coordinate of atoms. I'm trying to follow what LAMMPS is doing to calculate the radius of gyration $R_\text{g}$:

$$R_\text{g}^2 = \frac{1}{M} \sum_i m_i (r_i - r_\text{cm})^2$$

Here is my step:

1) Calculate distance from each atom to COM of molecule, for X, Y, and Z direction.

2) Square the distances (for each X, Y, and Z)

3) Multiply distances (for each X, Y, and Z) by atomic mass for each atom

4) Repeat 1~3 for all atoms of corresponding molecule and sum up the values.

5) Divide the value from 4) by molecular mass

6) Now we have the square of the radius of gyration, but it has 3 components (X,Y,Z). Sum the three components together (they are already squared) and then square root. This would be a Rg of single molecule.

7) Perform 1~6 for all other molecules. (and average in the future)

My calculation result is different from what I get from LAMMPS calculation result. Which part of my script is wrong? I'm not sure what is wrong with my calculation process. Do I need to exclude mass, only consider Ri-Rcom?

You don't say exactly what the difference is between your calculation and the LAMMPS calculation, but I'm going to make a guess that you have neglected to account for the periodic boundary conditions properly. The LAMMPS documentation for the compute gyration command says

  The coordinates of an atom contribute to Rg in “unwrapped” form, by using the image flags associated with each atom. See the dump custom command for a discussion of “unwrapped” coordinates. See the Atoms section of the read_data command for a discussion of image flags and how they are set for each atom. You can reset the image flags (e.g. to 0) before invoking this compute by using the set image command.

If you are trying to duplicate this calculation "by hand", you need to "unwrap" the atomic coordinates. This means that, for every molecule, all the atoms in the molecule need to belong to the same "copy" of the molecule, i.e. not to different periodic images of the molecule. You need to ensure this before you even calculate the position of the centre of mass of each molecule.

I'll illustrate one way to do this for a simple linear chain molecule of $N$ atoms. Something similar should work for a more complicated molecule, but you'll need to think about it yourself. I'm going to assume that the atoms are labelled in order $1, 2, 3, \ldots N$, and that $1$ is bonded to $2$, $2$ to $3$, and in general $j$ bonded to $j+1$. Let the atomic positions be $\mathbf{r}_j$. You may want to copy these to a separate working array, to avoid changing the originals.

We'll take the position of atom $1$ as a reference point. In the following loop, start by setting $j=2$.

Compute $\mathbf{d}_{j}=\mathbf{r}_{j}-\mathbf{r}_{j-1}$.
Apply the minimum image correction to $\mathbf{d}_{j}$.
Redefine $\mathbf{r}_{j}=\mathbf{r}_{j-1}+\mathbf{d}_{j}$.
Increment $j\rightarrow j+1$ and return to step 1, unless $j=N$ in which case stop.

Now you should be safe to compute the centre of mass of that molecule, and work out the radius of gyration, in the way you describe in your question. You can repeat the calculation for every molecule independently.

A further point about the LAMMPS compute gyration command is that it not only returns the scalar value $R_g$ (which should correspond to what you describe calculating), it also returns the separate $xx$, $yy$, and $zz$ terms, and also the off-diagonal $xy$, $yz$ and $zx$ terms. This is because the gyration tensor is, like the inertia tensor, a second-rank symmetric tensor. This may be of interest if your polymer configurations are significantly non-spherical in shape. Using these values, you can determine the principal axes of the gyration tensor for each molecule, and the corresponding three principal radii of gyration. It's basically a question of diagonalizing the $3\times3$ symmetric matrix whose components are returned by the LAMMPS command, along with $R_g$. I'm assuming, though, that this is not what you are worried about.
[EDIT: I've just noticed that @porphyrin made this same point in a comment while I was typing in my answer].

EDIT following OP comment.

It seems that "unwrapped" coordinates are already being used for this calculation, in post-processing. So it is possible that my answer does not  explain the difference between the LAMMPS calculation and the one implemented by the OP.

However, different definitions of "unwrapped" may be used for different purposes. In calculating atomic mean-squared displacements as a function of time, to get the diffusion coefficient, it is sufficient to make sure that periodic boundary corrections are never applied to the atomic positions as they evolve in time. However, for that purpose, there is no need to consider which molecule an atom belongs to. In order to compute the molecular centre-of-mass, and from that the radius of gyration $R_g$, the atoms must be grouped together by molecule. This means that, at some stage, a calculation similar to the one described above must be carried out. Maybe this has been done, in producing the "unwrapped" trajectories, maybe not.

So, it is still worth either carrying out the correction described above, or at least checking that bonds between atoms have reasonable lengths in the "unwrapped" trajectories. For example, for a linear chain of $N$ atoms, compute $d_j^2=|\mathbf{r}_j-\mathbf{r}_{j-1}|^2$ for $j=2,3,\ldots N$, and just store the maximum $d_j^2$ in the molecule. Extend this calculation to cover every polymer molecule, and print out the maximum value of $d_j^2$ discovered for the whole set of molecules. How does this compare with the square of the bond length between adjacent atoms, defined by the molecular force field? If there has been no error with the periodic boundaries, the two numbers should at least be similar.


Dissolving metal reductions Could reduction of an organic compound by reacting it with sodium dissolved in $\ce{NH3}$ still proceed if a metal such as aluminium were used instead of sodium? I'm not sure about the aluminium amine or amide structure that would be formed.

Is there a goal here? If you wanted to react aluminium with concentrated aq ammonia, you would get hydrogen but the rate would be proportional to temperature.  However, this would be quite a poor way to create a reductive environment and as someone has already alluded, aluminium and sodium hydroxide or a strong mineral acid is better. From what I've seen, if you use alkali, a catalyst is needed.
Ethylamine can be more conveniently used in place of ammonia.
Calcium can also be used in reductions.I think I've also seen Indium, but that seems a bit expensive.


How to calculate the partial pressure of oxygen in this mixture of oxygen and argon? A $\pu{19.3 g}$ mixture of oxygen and argon is found to occupy a volume of $\pu{16.2 L}$ when measured at $\pu{675.9 mmHg}$ and $\pu{43.4 ^\circ{}C}$. What is the partial pressure of oxygen in this mixture?

The answer is $\pu{436 mmHg}$, but I didn't manage to get to it.

I know that it is found by the formula
$$P_i=X_i\cdot P_\text{total},$$
where $P_i$ is the partial pressure and $X_i$ is the mole fraction.

$P_\text{total}$ is given to find $X_i$. I should divide the amount of $\ce{O2}$ by the total amount of substances of the mixture $n_{\ce{O2}}/n_\text{total}$.

$n_\text{total}$ is easy to find with $n=PV/RT$,
but my problem is, that I can't find the amount of substance of $\ce{O2}$. The mass in the problem is given for the mixture, so how to extract only the mass of $\ce{O2}$, or in other words, how to get the amount of substance for $\ce{O2}$?

How is the mixture composed? Find expressions for mass, volume and amount of substance.
\begin{align}
\tag1\label1
  V_\text{total} &= V_{\ce{O2}} + V_{\ce{Ar}}\\
\tag3\label2
  n_\text{total} &= n_{\ce{O2}} + n_{\ce{Ar}}\\
\tag3\label3
  m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}
\end{align}

Transform $V_\text{total}$ with the ideal gas and solve for $n_{\ce{Ar}}$
\begin{align}
  \tag4\label4
  V_\text{total} &= \frac{RT}{p_\text{total}}(n_{\ce{O2}} + n_{\ce{Ar}})\\
  \tag5\label5
  n_{\ce{Ar}}  &= \frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}
\end{align}

Use total mass \eqref{3} and transform it with the with molar mass to amount of substance equation ($m = nM$), then plug in \eqref{5}:
\begin{align}
  \tag3
  m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}\\
  \tag6\label6
  m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + n_{\ce{Ar}}M_{\ce{Ar}}\\
  \tag7\label7
  m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + \left(\frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}\right)M_{\ce{Ar}}
\end{align}

Now there is only one variable left, which you do not know:

\begin{align}
  \tag8\label8
  m_\text{total} &= n_{\ce{O2}}\left(M_{\ce{O2}} - M_{\ce{Ar}}\right) + \frac{p_\text{total}V_\text{total}}{RT}\cdot M_{\ce{Ar}}\\
  \tag9\label9
  n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot  M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}
\end{align}

We need the following data (It would be by far better to only use SI [derived] units):

\begin{align}
M_\ce{Ar} &= \pu{39.9 g//mol}\\
M_\ce{O2} &= \pu{32.0 g//mol}\\
R &= \pu{8.314 L kPa //K mol}\\
\pu{1 mmHg} &= \pu{133.322 Pa}\\
p_\text{total} &\approx \pu{90 kPa}\\
T &= \pu{316.5 K}\\
m_\text{total} &= \pu{19.3 g}\\
V_\text{total} &= \pu{16.2 L}
\end{align}

Plugging this into \eqref{9}:
\begin{align}
  \tag9
  n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot  M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}\\
  n_{\ce{O2}} &= \left(\pu{19.3 g} - \frac{\pu{90 kPa}\cdot\pu{16.2 L}}{\pu{8.314 L kPa //K mol}\cdot\pu{316.5 K}}\pu{39.9 g//mol}\right)\cdot \left(\pu{32.0 g//mol} - \pu{39.9 g//mol}\right)^{-1}\\
  n_{\ce{O2}} &= \pu{0.36 mol}
\end{align}

Using the ideal gas with the data provided yields:
\begin{align}
  n_\text{total} &=\frac{pV}{RT}\\
  n_\text{total} &=\frac{\pu{90 kPa}\times\pu{16.2 L}}{\pu{8.314 L kPa K-1 mol-1}\times\pu{316.5 K}}\\
  n_\text{total} &=\pu{0.55 mol}
\end{align}

Therefore the mole fraction is
$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\text{total}} \approx 0.64.$$

And the partial pressure is therefore $\pu{57.7 kPa}$ (or $\pu{433 mmHg}$ which is close enough to the solution, given that I rounded the medieval witchcraft units).


Potassium Manganate and Acidified Potassium Dichromate colour changes I would be obliged if somebody could explain the colour changes for the reduction of potassium manganate and acidified potassium dichromate. I understand that for potassium manganate, it is being reduced differently and this gives us a colour change. But I can’t understand the acid, base, neutral reactions.

I also keep on seeing this equation:

$$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$$

Unfortunately  I cannot understand it, possibly because it’s an net equations and not the entire reaction.

Although you are asking for the color changes of the reduction of potassium manganate ($\ce{K2MnO4}$; a green-colored salt, (Wikipedia1)) by potassium dichromate ($\ce{K2Cr2O7}$; a red-orange-colored salt, (Wikipedia2)) in acidic medium, the equation you are showing is reduction half reaction of potassium permanganate ($\ce{KMnO4}$; a purple-colored salt, (Wikipedia3)) in acidic medium. According to Wikipedia1, potassium manganate is an intermediate in the industrial synthesis of potassium permanganate. Thus, color change for that specific reaction is green to purple (disregarding other interference such as color change of the oxidizing reagent).

The reduction half reaction of $\ce{K2Cr2O7}$ in acidic medium is:
$$\ce{Cr2O7^2- + 14 H+ + 6e- <=> 2 Cr^3+ + 7 H2O}  \qquad  E^{\circ} = \pu{1.36 V}$$

The oxidation half reaction of $\ce{K2MnO4}$ is:

$$\ce{MnO4^2- <=> MnO4^- + e-}  \qquad  \qquad  E^{\circ} = \pu{-0.558 V}$$

The total redox reaction of $\ce{K2MnO4}$ and $\ce{K2Cr2O7}$ in acidic medium is:
$$\ce{Cr2O7^2- + 14 H+ + 6 MnO4^2- -> 2 Cr^3+ + 6 MnO4^- + 7 H2O}  \qquad  E^{\circ}_{\text{Rxn}} = \pu{0.802 V}$$

Since $E^{\circ}_{\text{Rxn}} \gt 0$, the reaction is spontaneous.


How does hot tub "shock" work? Hot tub shock is potassium peroxymonosulfate. When added to a hot tub, it creates bubbles. It is said to oxidize organic debris in the water. What is the reaction?

There are several properties you can infer from the compound's name.

The compound is ionic, which the structure cation [...]-ate informs us. In this case, we're dealing with the potassium salt of peroxymonosulfuric acid. This should therefore dissolve pretty well in water.
The compound will be an oxidizer, in the conditions present in your bathtub. peroxy means the compound contains an $\ce{O-O}$ bond, which breaks up homolytically (one electron per oxygen) to give radicals (quite reactive species, electronically unhappy).

As per both hints, the compound is expected to be pretty effective at "eating" through organic deposits, by oxidizing them, rendering some soluble. It should also decrease pH, but not so drastically as it is already the conjugate base of sulfuric acid (which can let go of two protons).

About the bubbling: it most probably is $\ce{CO2}$ coming from the oxidation of the organic matter.


Does it matter which end of a chain one sets to be 1 when drawing the structure of organic compounds? For example, if I were to draw the structure of 1,2,4-trimethylcyclohexane, then would one or both of these be wrong?

Or are both right? I'm wondering if it matters where I start to count from 1 in these cases, if there is some convention to it or if it doesn't matter much.

Both of those are the same molecule (flip the one on the left over its horizontal axis, and they are identical).  It does matter where you start counting from, because the numbers chosen for the functional groups should be as low as possible:

Numbering of the chain. This is done by first numbering the chain in both directions (left to right and right to left), and then choosing the numbering which follows these rules, in order of precedence:

Has the lowest-numbered locant (or locants) for the suffix functional group. Locants are the numbers on the carbons to which the substituent is directly attached.
Has the lowest-numbered locants for multiple bonds (The locant of a multiple bond is the number of the adjacent carbon with a lower number).
Has the lowest-numbered locants for double bonds.
Has the lowest-numbered locants for prefixes.

 From IUPAC nomenclature
So, given the choice between 1,3,4-trimethylcyclohexane and 1,2,4-trimethylcyclohexane, you would choose the latter (for both molecules).


Pronunciation of "alkene" vs. "alkyne" Maybe this is better in English.SE, but I think I'm more likely to get better responses here in the "industry" site. This post got me thinking; how do you pronounce "alkyne" (a hydrocarbon with a carbon-carbon triple bond) differently enough from "alkene" (a hydrocarbon with a carbon-carbon double bond) to differentiate the two in spoken conversation? They're obviously spelled differently and so can be separated in the literature, but I can't find a satisfactory pronunciation for either that isn't "al-keen".

English isn't my first language, and in truth I'm always doubtful about the pronunciation of words in chemistry. Sometimes it's possible to find videos where the words are mentioned, for example here or here or here or here (several examples of alkenes and alkynes being pronounced in this last one). That's what I use, al-keen and al-kyne (ky as in kite or sky).


Flavonoid Xanthine Oxydase Inhibitors vs Hydroxychloroquine Regarding COVID-19 and hydroxychloroquine (HCQ), HCQ is known to increase pH in lysosomes by 1-2 orders of magnitude, and also results in lethal intracellular heme overload in parasites responsible for hemolytic diseases.  Question is, if circulating levels of xanthine oxidase (XO) are also known to be increased in hemolytic diseases, if one considers the natural flavonoids Kaempferol, quercetin, or luteolin, which are known xanthine oxidase (XO) inhibitors, antivirals, and anti-oxidants, what would be different about pharmacology of natural XO inhibitors and HCQ?
Is HCQ an XO inhibitor because it results in heme overload in parasites?  What about anti-oxidant properties of HCQ in terms of radical scavenging?  There is no information supporting the notion that HCQ is an antiviral.
Sticking with hemolytic diseases, for which the goal is overload of intracellular heme in parasites, what would happen if a human took high levels of an XO inhibitor that wasn't HCQ during malaria?
XO inhibitors usually result in an overall decrease in uric acid, essentially rendering them as a natural supplement for gout -- i.e., replacement to allopurinol, which has horrendous side effects.

HCQ does show antiviral properties in vitro, but, in vivo (in humans), when taken after an initial dose of 800mg for 4 days at 600 mg/day, it was not found to prevent COVID-19 in high-risk exposure settings.  Thus, it didn't stop people from getting SARS-CoV-2 infection followed by onset of symptoms/severity.
I suspect this is exactly what would happen with flavonoids, many of which have shown antiviral properties in vitro (laboratory settings).  The information on flavonoids and SARS-CoV (2003) is very sparse, but several flavonoids did inhibit SARS-CoV in vitro (see last paragraph below).  I do know there was a clinical trial approved in 2020 by NIH in which the investigators received an IND (From FDA)  to use diosmin as a supplement to treat patients with COVID-19 (i.e. hospitalized).  However, I don't know if was completed.  Diosmin is available OTC as a supplement, and used via prescription under the brand names Dafflon® and Vasculera® for chronic venous insufficiency (CVI), which is caused by venous hypertension (VH). VH, in turn, aggravates metabolic imbalances leading to a self perpetuating cycle of further metabolic changes, including venous acidosis. These changes promote further inflammation in vascular tissue leading to edema, skin damage and possible ulceration and deep vein thrombosis (DVT). Varicose veins and hemorrhoids are also part of the spectrum of CVI disorders.
Interestingly, it took 16 years before a group published the most comprehensive paper on in vitro inhibition if SARS-CoV (2003) by many flavonoids that were considered.  I review these in a pre-print on flavonoids that was just accepted for publication.


Does imidazole and hydrochloric acid yield imidazole hydrochloride salt? I'm asking this because I found a link for imidazole-HCl buffer: , but from what I understand, a buffer is a weak acid or base with its corresponding salt. Would imidazole-HCl buffer be a salt solution, or does it have a weak acid or base as well (if it does, what is the weak acid/base)? Also, I found two different pictures for imidazole hydrochloride:

Which one is correct? Is imidazole-HCl a salt or a buffer?

The correct structure of the salt is as shown above. This structure is more stable than the one that has two hydrogens on the same nitrogen - due to electron delocalization.

If you have a water solution of about 50/50 imidazole base and the hydrochloride salt, you get the maximum buffer capacity, due to the fact that both acid and base forms are present, and a small amount of a strong acid or base subsequently added to the mixture will react with these imidazole species, and therefore have little effect on the pH.


How does aluminium amalgam work as a reducing agent? I can't seem to find much information on this topic but how does aluminium amalgam work as a reducing agent?

Could it be used to deoxygenate a carboxylic acid or amino acid to the corresponding alcohol or alkane?

If not , why not?

Could aluminium amalgam be used to deoxygenate a carboxylic acid or amino acid to the corresponding alcohol or alkane?

No. Aluminium amalgam has been used to selectively reduce other groups in molecules containing a carboxylic acid group, while preserving the carboxylic acid group.

See for example in Industrial Organic Nitrogen Compounds at page 139:

  Some pyridine-carboxylic acids have been reduced to the corresponding dihydropyridine-carboxylic acids with aluminium amalgam.

See also the patent Polyene carboxylic acids and esters and manufacture thereof where aluminium amalgam is suggested as a mild reducing agent which preserves the carboxylic acid or ester group.


Why does the specific conductivity against concentration curve for NaCl curve more than KCl? I understand why KCl has a higher specific conductivity than NaCl.. but I dont understand why the curvature is more significant for NaCl (KCl is a straight line). Ive provided a picture of this here, and this is also what I get from my experiment. Does anyone mind explaining why?

For equally charged ions a smaller radius implies a higher charge density, and therefore stronger Coulombic interactions with other charges in solution, including stronger ion-dipole interactions with water molecules, resulting in a more stable hydration sphere, and stronger interactions with counterions (here chloride). Such interactions reduce the mobility of the ion by opposing the effect of the applied external electric field, by either increasing solvent drag or generating an opposing electric field (see for instance this chem libretext for an explanation). This slowing down can be interpreted as an increase in the hydrodynamic or Stokes's radius, and also results in a reduced diffusion coefficient, as shown in the following table (source CRC, $\pu{ 25^\circ C}$):

Ion
$\Lambda_+ \pu{(10^-4 m2Smol^-1)}$
$D \pu{(10^-5 cm^2s^-1)}$

$\ce{Cs+}$
77 .2
2 .056

$\ce{Rb+}$
77 .8
2 .072

$\ce{K+    }$
73 .48
1 .957

$\ce{Na+}$
50 .08
1 .334

$\ce{Li+}$
38 .66
1 .029

$\frac12$ $\ce{Sr^{2+}}$
59 .4
0 .791

$\frac12$ $\ce{Ca^{2+}}$
59 .47
0 .792

$\frac12$ $\ce{Mg^{2+}}$
53 .0
0 .706

$\frac12$  $\ce{Be^{2+}}$
45
0 .599

Therefore the smaller monovalent cation, $\ce{Na^+}$, and divalent cation, $\ce{Mg^{2+}}$, have lower molar conductivity than the species lower in the periodic table ($\ce{K^+}$ and $\ce{Ca^{2+}}$).

Additional comments:
(1) Instead of discussing NaCl or KCl, start discussing individual cations and anions, recognizing that these salts completely dissociate into ions in solution. Notice that the different salts share a common anion (chloride, Cl-). Therefore any differences are due to the choice of cation. Sounds obvious, but then you can look at everything from the perspective of the cation's interactions (even if those of the anion will also change between salts, and affect conductivity).
(2) Distinguish between the size of an ion before and after hydration. Bare (unhydrated) $\ce{Na^+}$ is smaller than $\ce{K^+}$, but hydrated $\ce{Na^+}$ is bigger than $\ce{K^+}$.
(3) Bigger bare (unhydrated) cation means weaker interactions and therefore faster mobility.
(4) What drives the formation of a larger hydrated ionic radius is deep down the same thing that drives interactions with other ions. The hydration shell is not that tight. Interaction with other ions will readily override them.
(5) Stronger interactions with other ions means more deviation from ideality. Ideality here means that individual ions behave the same (their mobility is the same) at all concentrations.


POCP: How many ozone molecules can be created by one ethene molecule? One ethane molecule can create two molecules of ozone, one ethanol molecule can create one molecule of ozone:

In the following table the POCP is twice as high for ethene than for propane:

From this fact I expect four molecules of ozone from ethene, but in my own formulated reaction mechanism I get only three ozone molecules:

How many ozone molecules can be created by one ethene molecule and if my proposed mechanism is wrong what is the correct one?

I think this question cannot be answered so easily - your proposed reaction mechanism, however, looks good. A detailed explanation and "main reaction mechanism" can be found here, page 192 (just search for "ethene" in the book). As you already indicate ethene reacts with a hydroxyl radical, followed by the addition of oxygen. The reaction with NO generates the HOCH$_2$CH$_2$O radical which can either dissociate or react with another oxygen molecule.

Further reactions of the oxidized products occuring during this pathway can also be found in the above reference, see page 372 (unfortunately only some parts of the book are available for free).


Symantics of parts per (million, billion, etc) I have always thought that parts per million meant that in a solution of one million particles, one of those particles is the particle that you are looking for its concentration in ppm. However looking at the formula ppm=(mass solute/mass solution)* 1 million, it seems like it may be meaning one part of mass per mass of solution?
Doing a example calculation for one atom of mercury in 999,999 molecules of water, I get about 11 parts per million and not the 1 ppm. It seems like the ppX changes with molar mass of the X. Am I missing some key concept here or do I have the right idea. Thanks

(1 atom Hg)(mol Hg/Avogardros#)(20.59g/mol Hg)= 3.3309E-22g Hg = variable A

(999,999 molecules H2O)(mole H2O/ Avogardros#)(18.015g/mol H2O)= 2.9915E-17g H2O=variable B

ppm=(moles solute/moles solution)*10^6
ppm=A/(A+B) *10^6 = 11.13ppm

The notation "parts per" is just a dimensionless unit to describe small amounts.
In chemistry it's used for concentrations of dilute liquids.. whereas in physics it can be used for defining change with respect to another quantity and in Six sigma it's used to define "defects per million opportunities".

In chemistry however, $ppm$ is mostly used as concentrations of dilute solutions in water and sometimes used interchangeable with $mg/l$. The notion of "one particle per million particles of solvent" can be used for solid-solid solutions or initial understanding of the term.


Relationship between Gibbs free energy and potential energy (related to reaction coordinate diagrams) For a rate law, the rate constant $k$ is given by the equation

$$\ln k = \ln A - \frac{E_\mathrm{a}}{RT}\tag{1}$$

If $k_\mathrm{fwd}$ is the forward rate constant and $k_\mathrm{rev}$ is the reverse rate constant:

$$\ln k_\mathrm{fwd} = \ln A - \frac{E_\mathrm{a,fwd}}{RT}\tag{2}$$

$$\ln k_\mathrm{rev} = \ln A - \frac{E_\mathrm{a,rev}}{RT}\tag{3}$$

Subtracting the two equations yields

$$\ln\frac{k_\mathrm{fwd}}{k_\mathrm{rev}} = \frac{E_\mathrm{a,rev}}{RT} - \frac{E_\mathrm{a,fwd}}{RT} = \frac{1}{RT} (E_\mathrm{a,rev} - E_\mathrm{a,fwd})\tag{4}$$

If $K$ is the equilibrium constant for the reaction, then $\displaystyle K = \frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$:

$$-RT\ln K = \Delta E\tag{5}$$

where $\Delta E$ is the change in potential energy on a reaction diagram, such as the one shown here.

However, we know that

$$-RT\ln K = \Delta_\mathrm{r} G^\circ\tag{6}$$

Does this mean that potential energy is equal to the standard Gibbs free energy of reaction?

Equation (1) is an empirical one that does not consider entropy. If you substitute it for the Eyring equation (considering free energy of activation $\Delta G^\ddagger$) you end up with equation (6) as expected, not equation (5). Wikipedia has a nice section on the relationship of activation energy and Gibbs energy of activation:

  Although the equations look similar, it is important to note that the Gibbs energy contains an entropic term in addition to the enthalpic one. In the Arrhenius equation, this entropic term is accounted for by the pre-exponential factor A. More specifically, we can write the Gibbs free energy of activation in terms of enthalpy and entropy of activation: $ΔG‡ = ΔH‡ – T ΔS‡$. Then, for a unimolecular, one-step reaction, the approximate relationships $E_a = ΔH‡ + R T$ and $A = (k_B T/h) \exp(1 + ΔS‡/R)$ hold. Note, however, that in Arrhenius theory proper, A is temperature independent, while here, there is a linear dependence on T. For a one-step unimolecular process whose half-life at room temperature is about 2 hours, ΔG‡ is approximately 23 kcal/mol. This is also the roughly the magnitude of Ea for a reaction that proceeds over several hours at room temperature. Due to the relatively small magnitude of TΔS‡ and RT at ordinary temperatures for most reactions, in sloppy discourse, $E_a$, ΔG‡, and ΔH‡ are often conflated and all referred to as the "activation energy".

So in a more rigorous treatment, the pre-exponential factor is temperature-dependent (this is almost impossible to measure because the exponential is "very" temperature-dependent) and has an entropy component. That is why equation (5) in the question is not accurate (equation (6) is correct, and can be derived in a rigorous manner from first principles and definitions).

Related question: Y-axis of the reaction co-ordinate graph