What is a virtual state? In talking about Raman spectroscopy, one finds the Stokes line is simply the difference between the energy of an incoming photon and an emitted photon. This energy corresponds to a vibrational transition in terms of energy. Yet, the light sent at the molecule is usually much higher energy than that vibrational transition (or else we would just be doing IR).

The explanation I've heard is that the state which holds all that energy in between the ground vibrational state and decaying back to the first vibrationally excited state is a virtual state.

From doing some reading, I have gathered that a virtual state is called virtual not because it's imaginary, but because it's not an eigenfunction of the hamiltonian of the system.

Where does this state come from? I've heard that it drops out of some perturbation theory. Because Raman is intricately connected with polarizabilities, I imagine this might be perturbations to the time-dependent Schroedinger equation.

So, to put some actual questions out there:

What is meant by "virtual state"? In what sense is it virtual?

Can anyone show me the perturbation theory where this comes from? Even if it is just a reference, that would be greatly appreciated.

I have also seen this is somehow connected with Feshbach Resonance which I know nothing about, so an answer which addresses that point would be a good one.

In most Raman experiments, the incident radiation is not near (or at) an absorbing wavelength, and so you will never access a real, honest-to-God excited state (stationary state). (If it was close to an absorbing wavelength, we would be taking about a resonance Raman experiment.)

Now obviously something is going on between the molecule and the incident light. What you are really doing is preparing a superposition state. A superposition of real excited states, or stationary states, of your Hamiltonian. This is what we call a "virtual state," and it is not an eigenstate of your Hamiltonian, so its energy is undefined. (You are free to compute an expectation value, though.) This virtual state doesn't last forever, and, when it decays, scatters radiation away in some direction (it doesn't matter where, unless you are doing an angle-resolved Raman experiment).

(By the way, if you look closely at the equation porphyrin gave, the sum over $v$ virtual states is actually a sum over real $v$ excited stationary states, such that $E_v$ is defined. Otherwise that expression is nonsense. This equation essentially is the sum of overlaps of the initial state into the states that make up your virtual state, and then multiplied by the sum of overlaps of the excited states that make up the virtual state into the final state. This is the Kramers-Heisenberg-Dirac expression for Raman amplitudes.)

If the superposition collapses back to the ground state, the scattered radiation will have the same wavelength as the incident light, and we have Rayleigh scattering. If the superposition collapses to an excited vibrational states, we have Stokes Raman. If your initial state was vibrationally excited and you collapse to a lower vibrational state, you get anti-Stokes Raman. In the actual experiment you will observe all types and can see a Rayleigh line in the spectra. Most spectrometers will filter this wavelength out, however, leaving just the Raman peaks. Usually the virtual state is dominated by the ground state and most of the scattered light is Rayleigh. This is why Raman spectroscopy was too weak to be useful until we had better optical components.

Obviously this isn't the only way to think about Raman. You can use path integrals and/or QED if you want, which is much closer to Wildcat's answer. In fact, I think you must use QED if you want to work out virtual state lifetimes. Here is a good reference if you want to dig a bit deeper:

  Williams, Stewart O., and Dan G. Imre. "Raman spectroscopy:
  time-dependent pictures." The Journal of Physical Chemistry 92, no. 12
  (1988): 3363-3374.


What is the state of aggregation (gas, liquid) of oxygen in blood? Atmospheric oxygen is in O2 and a gas. Then we inhale the air, our efficient lungs do the magic to filter out the oxygen and push them into the blood stream.

When we say hemo and globin transport the oxygen using the iron ions. In what state oxygen is transported in the blood? as a gas or a liquid or an ion? It is hard for me to conceive of the idea that oxygen would be in gaseous form in the blood. "GAS in blood?" e.g. Arterial Blood Gas Test

Also, how does the lungs convert the gas into something that is compatible to be in blood?

References:
Amount of Oxygen in the Blood

Regarding the state of oxygen in blood: It is in solution in the blood plasma (which mostly consists of water), in the form of single molecules. Think of water which you leave exposed to air: carbon dioxide will be captured and dissolved (along with the other gases in air), but these molecules are not gaseous or liquid, but rather "in solution", which is different from the "classical" states.

Back to oxygen: As your reference already states, most of the oxygen in solution will bind to hemoglobin. The actual state of oxygen in that complex has been debated, but it is believed to be reduced by the hemoglobin iron to the superoxide anion, coordinated to Fe$^{3+}$. See Wikipedia on this.

Also, the lungs do not "convert" the atmospheric oxygen to anything, they rather allow, due to their very large surface area, the quick exchange of oxygen/carbon dioxide in solution and in the air.


Why is Specific heat of Ammonia higher than water? The specific heat of water is lower than that of $\ce{NH3}$, but why?
$\ce{H2O}$ forms more hydrogen bonds if I am right, besides there's more electronegativity for Oxygen.
But still, $\ce{NH3}$ has higher Specific heat. I found this link saying $\ce{NH3}$ has lower molar mass, but its just by one gram so it shouldn't be having much effect right?

The link in the question states that ammonia has a higher specific heat than water. This is true, but it is important to compare the specific heat values under identical (or very similar) conditions.

The isobaric specific heat of ammonia (gas) is 2.1636 kJ/kgK at 1 atmosphere and temperature = 26.9 C. https://www.engineeringtoolbox.com/ammonia-heat-capacity-specific-temperature-pressure-Cp-Cv-d_2016.html At -33.6 C (boiling point of $NH_3$ at 1 atm), the specific heat of $NH_3$ gas is 2.2946 kJ/kgK, not much different from the gas at 26.9 C; however, for liquid $NH_3$ at 1 atm and -33.6 C, the specific heat is 4.4464 kJ/kgK, a large increase from the gas at that temperature.

Liquid water has one of the highest specific heats among common substances, about 4182 J/(K kg) at 20 °C https://en.wikipedia.org/wiki/Specific_heat_capacity At 100 C, liquid water has a specific heat of 4.2157 kJ/kgK (implied pressure = 1 atm). Again, not much difference because of temperature. https://www.engineeringtoolbox.com/specific-heat-capacity-water-d_660.html Now we find water vapor data in a different place: https://www.engineeringtoolbox.com/water-vapor-d_979.html where gaseous $H_2O$ at 216 F (102.2 C) has a specific heat of 1.890 kJ/kgK, a large drop, even bigger than in the ammonia system.

So although it is often stated that water has one of the highest specific heats among common substances, it seems that either ammonia is not considered common for this issue, or yes, it's true that water has one of the highest specific heats...And that is true of the gas and the liquid states!
But then among all the data, which are truly interesting, there is one other thing that stands out: the heat of vaporization of water is 40.8 kJ/mol, whereas for ammonia, it is only 23.5 kJ/mol. So water can still be considered an extreme molecule.


How to Find Each Element in Nature I tried to google this but 'element' as a programming term complicated the search.

I'm looking for a site or article that covers how one would find each element in nature, in its most abundant form. For example, what's the best natural source/form of Sulfur? Beryllium? Tugsten? etc.

Any help is appreciated.

The Wikipedia page for each element is probably the easiest place to start with. If you want some significantly more in-depth data, the USGS publishes free-access mineral commodity summaries on many (but not all) elements, including information about their origin, abundance and applications. Other good sources are books, especially descriptive inorganic chemistry books, such as a few of John Emsley's publications including Nature's Building Blocks and The Elements, as well as many other college-level inorganic chemistry texts.


Is 4-nitrobenzaldehyde or 4-aminobenzaldehyde more likely to be hydrated? Of 4-nitrobenzaldehyde and 4-aminobenzaldehyde, which has the higher $\ce{K_{eq}}$ of hydration?

 $\qquad \qquad$

Steric factors can likely be ignored in this problem since we are dealing with rather structurally similar compounds.

I suspect the problem comes down to recognizing that a more polar compound is more favorably solvated by water.

I know that the nitro group is always electron-withdrawing, and I drew a few resonance structures as well. The resonance structures for both, however, seem to suggest that both are rather polarized (given the distribution of formal charges).

Resonance forms for 4-nitrobenzaldehyde

I expect the ones with the oxygen possessing the incomplete octet are very minor resonance contributors.
Therefore it seems that the main thing about this molecule is that the nitro group is withdrawing electron density from the ring and the aldehyde group.

Resonance forms for 4-aminobenzaldehyde

The resonance structures here are all somewhat reasonable.

I think you're asking about the following equilibrium

This equilibrium will shift to the left as the aldehyde becomes more stabilized.  So the question becomes, which R group, $\ce{NO2}$ or $\ce{NH2}$, will tend to stabilize the aldehyde the most?  A substituent that further delocalizes the positive charge shown in the dipolar resonance form would tend to be stabilizing.  Just like in electrophilic aromatic substitution, additional stabilizing resonance structures can be drawn involving the amino substituent, but similar structures involving the nitro group are not as significant due to electrostatics (2 adjacent positive charges). Therefore, since the amino will stabilize the aldehyde better than the nitro substituent, the equilibrium involving the amino substituent should lie further to the left, and the aldehyde with the nitro substituent should be more hydrated.


How do you get contents of large flask into a smaller separatory funnel? Separatory flasks that I use have a ground glass joint, but it is not clear how that joint is to be used. Videos I have seen just show people pouring stuff into the throat of the funnel.

Normally the situation I have is that the solution is in a large flask, maybe a 1000ml flask, and I need to separate a thin layer of the solution that is at the bottom of the flask using a 500ml separatory funnel. The question is how to get the bottom of the flask to the funnel.

Saying "oh just use a bigger funnel" is not a solution, because the flasks, reactors and carboys I have tend to be significantly larger than any normal separatory funnel. For example, if I have a 3000ml reactor or 5 gallon carboy with a solution in it, I am not going to go looking around to spend hundreds of dollars on a gigantic separatory funnel.

You should proceed as follows:

Charge the funnel with the mixture until it is almost full.
Let the content settle.
Separate the lower phase.
Separate the upper phase.
Repeat until you have processed all the mixture.

The ground joint is to close the funnel hermetically.
This is necessary when you need to perform an extraction.
A liquid with a mixture of components is charged in the funnel and an immiscible liquid is added.
The funnel is closed with a stopper and it is vigorously shaken.

BE VERY CAREFUL BECAUSE PRESSURE CAN DEVELOP INSIDE THE FUNNEL.

It is better to shake a bit first, take off the pressure, shake again, etc.
To release the pressure, put the funnel upside down and open the valve.
Put the funnel upright and let settle.
When there are two clean phases, release the stopper and take them out separately.
The components will be distributed between the phases in different quantities, so it is a way to purify those substances.


What's the electronic configuration of Nickel and Oxygen in RNiO3 (R=Rare-Earth)? Nickel's electronic ground state is $\mathrm{3d^8 4s^2}$ (there's a dispute on this, but let's not worry about it for now). Oxygen's is $\mathrm{2s^2 2p^4}$. In a Rare-Earth based Nickelate with chemical formula $\mathrm{RNiO_3}$, if we assume simple ionic bonds, then what should the electronic state be?

Literature indicate that the "classical" case is $\mathrm{Ni^{3+} O^{2-}}$ corresponding to $\mathrm{3d^7, 2p^6}$. But this means that Oxygen has gained 2*3=6 electrons. But how can this be possible if Nickel has donated only 3 electrons?

What's the electronic configuration of Nickel and Oxygen in RNiO3
  (R=Rare-Earth)?

As an example, let's look at $\ce{LaNiO3}$.  Each oxygen has an oxidation state of -2, nickel has an oxidation state of +3 and lanthanum has an oxidation state of +3.  Overall the molecule is neutral.
$$\ce{La(+3)~ plus~ Ni(+3)~plus~ 3~oxygens (-6)~=~0}$$


How does one refer to the amino acids in a peptide? Once a part of a peptide an amino acid is no longer an amino acid, so how do we refer to the part of the peptide that was once amino acid x?

A part of the peptide that was previously an amino acid molecule when free can be correctly referred to as an amino acid residue when bound in the peptide.

A source


Which process would separate the mixture of toluene, cyclohexane and water? toluene: $\ce{C6H5-CH3}$
cyclohexane: $\ce{C6H12}$

I think that answer would be:

Toluene and cyclohexane do not dissolve in water (since they are nonpolar), so I would first separate the water from the mixture with a separator funnel.

A mixture of cyclohexane and toluene would be separated by fractional distillation. Cyclohexane, which has a lower boiling point than toluene, would be eliminated from the mixture.

Is there anything wrong in my answer? Would there also be anything to add?

Distillation
The most effective and easiest way to separate the mixture of toluene and cyclohexane is simple distillation. Cyclohexane has boiling point is about 81°C, and toluene’s boiling point is 111°C.
References:

http://origin-www.thermoscientific.jp/content/dam/tfs/ATG/CAD/CAD%20Documents/Product%20Support%20Bulletins/NMR%20Support%20Documents/pS45-pS80-Simple-Distillation-of-Cyclohexane-and-Toluene.pdf
https://www.sas.upenn.edu/~dbalmer/chem502/Labs/lab2_distillations/balmerexp2.pdf


How to solve this contradiction? I am getting confused about this the more problems I do on it. My understanding was that only strong acids and strong bases will react to produce water and a salt. Something like this:
Molecular Eq: $\ce{HCl(aq) + NaOH(aq) \rightarrow H2O(l) + NaCl}$
Net Ionic Eq: $\ce{H+(aq) +OH- (aq) \rightarrow H_2O(l)}$

Makes perfect sense, both the acid and base break apart and combine with each other. But then I ran across something like this:
Molecular Eq: $\ce{2 CH3CO2H (aq) + Ba(OH)2 (aq) → Ba(CH3CO2)2 (aq) + 2 H2O (l)}$
Net Ionic Eq: $\ce{2CH3CO2H (aq) + 2OH- (aq) → 2CH3CO2- (aq) + 2H2O (l)} $

The net ionic eq tells me that $\ce{CH3CO2H}$ will not break apart, but the molecular equation tells me that they will break apart and form a salt with $\ce{Ba}$.

I'm so confused by this. Why are they giving two opposing pieces of information? Am I missing something here?

The idea behind this is that a weak acid like $\ce{CH_3CO_2H}$ will break apart, but won't break apart completely. If you have some $\ce{CH_3CO_2H}$ in water, most of it will remain as $\ce{CH_3CO_2H}$, so in the net ionic equation it is written whole, but some will separate into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$ ions. This relationship is quantified by the $K_a$ of the acid. Essentially, the dissociation reaction $\ce{CH_3CO_2H -> CH_3CO_2^{-}  + H^+}$goes both ways: $\ce{CH_3CO_2H}$ decomposes into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, and $\ce{CH_3CO_2^-}$ reacts with $\ce{H^+}$ for form $\ce{CH_3CO_2H}$. Since the forward reaction happens at a rate proportional to the concentration of $\ce{CH_3CO_2H}$, and the backwards reaction happens at a rate proportional to the product of the concentrations of $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, eventually the ratio $\frac{[\ce{CH_3CO_2^-}][\ce{H^+}]}{[\ce{CH_3CO_2H}]}$ becomes constant, and we call this constant the $K_a$ of the $\ce{CH_3CO_2H}$. For $\ce{CH_3CO_2H}, K_a=1.7\times 10^{-5}$.

Now its time to connect all this back to your question. When you add $\ce{Ba(OH)_2}$ to a solution of $\ce{CH_3CO_2H}$, the $\ce{OH^-}$ ions present from the dissociation of $\ce{Ba(OH)_2}$ react with the $\ce{H^+}$ ions from the (partial) dissociation of $\ce{CH_3CO_2H}$. This lowers the concentration of $\ce{H^+}$, and so more $\ce{CH_3CO_2H}$ dissociates to keep the ratio of concentrations constant at the $K_a$. Therefore, in the presence of enough $\ce{Ba(OH)_2}$, $\ce{CH_3CO_2H}$ dissociates almost completely, even though, on its own, $\ce{CH_3CO_2H}$ dissociates only to a very small extent.


What does the small, steep curve at the very beginning of a weak acid strong base titration curve come from? I understand that a buffer solution is being made, but I don't understand why the titration curve is different from a strong acid strong base one (apart from the steeper pH change). I understand the half-equivalence point and the rest of the titration curve. If Le Châtelier's applies here, please explain with it, as it makes more intuitive sense for me rather than the Henderson-Hasselbalch equation.

I included this animation if it makes it easier to explain, thanks!

http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons9.htm

This is part of estsblishing the buffer associated with the weak acid. To establish a buffer you need relatively large amounts of acid and conjugate base, and a weak acid can't produce the necessary conjugate base by itself.  You need to push it along with a little strong base to get into the buffering condition.

Say your solution is 0.1 molar acetic acid, with $K_\mathrm{a}=1.8×10^{-5}$. Before titration the pH is given by the Henderson-Hasselbach equation.  The logarithm of the above $K_\mathrm{a}$ value is about -4.74, therefore:

$\mathrm{pH}=4.74-\log_{10}\left(\frac{[\ce{HC2H3O2}]}{[\ce{C2H3O2-}]}\right)$

Now the amount of $\ce{C2H3O2-}$ in the solution is just what you get from dissociation of the acid, but acetic acid is only a weak acid and the actual amount of dissociation is small, in this case only about 1.3% of the acid you dissolved.  Merely 0.01 molar strong base reacting with the acid is enough to make a lot more acetate ion than just the acid dissociation alone, so the ratio in the H-H equation makes a big change and the $\mathrm{pH}$ follows suit.  Only when you have large amounts of both acid and conjugate base does the $\mathrm{pH}$ become buffered against slight additions of base or acid.


Orientation in benzene rings with more than one substituent Why is it that the chlorination of 1-chloro-3-nitrobenzene (18) gives 1,4-dichloro-2-nitrobenzene (19) as the major product, 1,2-dichloro-3-nitrobenzene (20) as a minor product and 1,2-dichloro-4-nitrobenzene (21) is not formed at all?

I know $\ce{-NO2}$ is a deactivating, meta-directing group and hence it decreases the electron density at positions 2, 4, 6 while position 5 is unaffected. On the other hand, $\ce{-Cl}$ is a deactivating, o-p directing group and hence it directs the incoming group to positions 2,4 or 6. Taking into account the fact that $\ce{-NO2}$ is a more powerful deactivating group than $\ce{-Cl}$, the incoming $\ce{Cl+}$ should be attached at position 5 as it has more electron-density than 6,2 or 4 position. But the product with $\ce{Cl}$ at position 5 is not formed. Why so? I am unable to rationalise this.

Source: page 583 of "March's Advanced Organic Chemistry, 7th ed." under the heading "Orientation in benzene rings with more than one substituent".

I quote March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition, page 666.

  Some substituents have a pair of electrons (usually unshared) that may be contributed
  toward the ring. The three arenium ions would then look like this:

  The stability of these two ions is increased by the extra form not only because it is another canonical form, but because it is more stable than the others and makes a greater contribution to the hybrid. Every atom (except of course hydrogen) in these forms (C and D) has a complete octet, while all the other forms have one carbon atom with a sextet. No corresponding form can be drawn for the meta isomer.The inclusion of this form in the hybrid lowers the energy , but also because it spreads the positive charge over a larger area—out onto the group Z. Groups with a pair of electrons (e.g., as the halogens) to contribute would be expected, then, in the absence of field effects, not only to direct ortho and para, but also to activate these positions for electrophilic attack. (my emphasis)

In your example, if chlorine electrophile attacks at C-4 or C-6, the resulting intermediate  is stabilized by an extra resonating structure, lowering the energy of this intermediate.Hence electrophillic aromatic substitution takes place.

If chlorine electrophile attacks at C-5 , the resulting intermediate is not stabilized compared to attack at C-6 .Hence the resulting electrophillic aromatic substitution is not seen.

This is the possible reason  that  the product with Cl at position 5 is not formed.

Quoting from chemistry.msu.edu:

  The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects:

  The first is the inductive effect of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring.

  The second effect is the result of conjugation of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance).

  The second factor that becomes important in reactions of substituted benzenes concerns the site at which electrophilic substitution occurs.

  The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites.

Therefore chlorine is an ortho para directing group with minor meta product.

References:

Advanced Organic Chemistry 5th edition Part A: Structure and Mechanisms, Francis A. Carey and Richard J. Sundberg
March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition


Do electrons have some volume, area or shape? I am in 8th grade now and when I was in 6th grade, my science book had diagrams of the electronic configuration of atoms. The electrons were round like spherical balls. Is it true that the electrons have some volume, area or shape and they revolve around the nucleus?

Quoting from the Nobel lecture of Hans G. Dehmelt (1989):

  With the rise of Dirac’s
  theory of the electron in the late twenties their size shrunk to mathematically
  zero. Everybody “knew” then that electron and proton were indivisible
  Dirac point particles with radius R = 0 and gyromagnetic ratio g = 2.00. The
  first hint of cuttability or at least compositeness of the proton came from
  Stern’s 1933 measurement of proton magnetism in a Stern-Gerlach molecular
  beam apparatus. However this was not realized at the time. He found for
  its normalized dimensionless gyromagnetic ratio not g = 2

...

  Similar later work at still higher energies found 3 quarks
  inside the “indivisible” proton. Today everybody “knows” the electron is an
  indivisible atomon, a Dirac point particle with radius R = 0 and g = 2.00....
  But is it? Like the proton, it could be a composite object. History may well
  repeat itself.

So there is no evidence that the electron is anything but a point particle, yet a non-zero size has not been ruled out either.

New Measurement of the Electron Magnetic Moment Using a One-Electron Quantum Cyclotron Phys. Rev. Lett. 97, 030801 (2006) (more-official link) shows that any radius of the electron is less than $10^{-18}$ meters.

See also the related webpage Limit on Electron Substructure


Calculation of the lowest energy conformation with a webbased solution or freeware I like to draw a molecule in different conformations in a) website or b) freeware and

Calculate its relative energy difference (for example to the lowest global energy conformation)
Display the lowest energy conformation

I'm particularly interested in the calculation of diaxial interactions, gauche interactions, double gauche pentane interactions (and less in ring strain, like chair / half-chair / twisted rings).

I know there are plenty of tools for modelling conformational and electrostacic properties. Most of those are not freely available. Do you an easy solution?
It does not must to be the most precise one, relative energy differences in steps of 0.1 kcal/mol between two drawn conformations would be fine for me.

May be this question is off-topic, because it's a software request, but I didn't know where else I could ask this question.

Avogadro can do a lot of this for you. It might not be precisely what you want but the molecular mechanics from Open Babel integrated into it can do at least some of the tasks you mentioned. It's free and has a very intuitive and smooth user interface. A real pleasure. The downside is that it's poorly documented.

To play around with conformations, go to Extensions -> Molecular Mechanics. To view the generated conformers (from the Conformer Search option) go to View -> Properties -> Conformer Properties...


Confusion on enantiomers and meso compounds Question: Are these compounds enantiomers?

Attempt: The compounds given are mirror images. But if I rotated the three bonds in the third carbon (configuration will not change), I will get a meso compound (plane of symmetry) :

So are these compounds enantiomers or meso compounds?

You are right, the 2 structures are equivalent and the substance has a plane of symmetry and is a meso compound. When in doubt, make a 3D model if possible.


Reaction between alumina and aqueous sodium hydroxide, which one's right? My chemistry textbook has a rather annoying irregularity.

It cites two different reactions for the same set of reactants (aluminium oxide, sodium hydroxide and water) without the slightest mention of a particular temperature or concentration. Here they are:

In a chapter on the p-block elements:

  $$\ce{Al2O3 (s) + 6 NaOH (aq) + 3 H2O (l) -> 2 Na3[Al(OH)6] (aq)}$$

In a chapter on metallurgy, under the topic of 'Extraction of aluminium from bauxite':

  $$\ce{Al2O3 (s) + 2 NaOH (aq) + 3 H2O (l) -> 2 Na[Al(OH)4] (aq)}$$

Which one of these is correct? Or are they both correct, depending on the situation? If so, under what conditions are they correct?

Quoting from chemguide (emphasis mine):

  With hot concentrated sodium hydroxide solution, aluminium oxide reacts to give a solution of sodium tetrahydroxoaluminate.

  $$\ce{Al2O3(s) + 2NaOH(aq) + 3H2O(l) -> 2NaAl(OH)4}$$

  Note: You may find all sorts of other formulae given for the product from this reaction. These range from $\ce{NaAlO2}$ (which is a dehydrated form of the one in the equation) to $\ce{Na3Al(OH)6}$ (which is a different product altogether).

  What you actually get will depend on things like the temperature and the concentration of the sodium hydroxide solution. In any case, the truth is almost certainly a lot more complicated than any of these.

  The version I am using is perfectly acceptable and is consistent with the aluminium chemistry you will find elsewhere on the site.

You can see that products of the reaction depends on the concentration of sodium hydroxide (Basic pH) but  other factors like temperature and pressure also come into play.

If you do not add water and only change the concentration of sodium hydroxide, product will also be different:

$$
\begin{align}
\ce{Al2O3 + 6 NaOH &-> 2 Al(OH)3 + 3 Na2O}\\
\ce{Al2O3 + 2NaOH &->[\pu{900 .. 1100 ^\circ C}] 2 NaAlO2 + H2O}
\end{align}
$$

Sodium aluminate is generally $\ce{NaAlO2}$ but its aqueous solution contains many other species like $\ce{[Al(OH)4]−}$, $\ce{[Al(OH)6]^3-}$ or perhaps the $\ce{[Al(H2O)2(OH)4]−}$. They are all considered to be aluminate species. So, if you are getting sodium tetrahydroxoaluminate or sodium hexahydroxoaluminate, they are no different compounds but are all hydrated form of sodium aluminate (addition of water + change in the concentration of sodium hydroxide).


Hydrogenation of pyridine compared to other heteroaromatic rings Why does pyridine require the lowest temperature for hydrogenation, even though it has it has the highest resonance energy amongst pyridine, pyrrole and furan? Is it just because of the increase in the absolute pressure or does the HCl also effect the reaction somehow?

When we reduce pyridine using the $\ce{H2/Pt/HCl}$ reagent-catalyst combination, we require a significantly lower temperature and pressure compared to other similar reductions (but without $\ce{HCl}$), because $\ce{H+}$ (and other Lewis acids for that matter) will go and add to the nitrogen atom, forming pyridinium salts.

This makes the compound more reactive towards reduction (as a partial positive charge will develop in the ring).


How to separate two liquids with both having the same properties? I was curious if it was possible to separate such liquids e.g. two vegetable oils that happened to have different ingredients, so it is a different liquid, but the same density etc.

Typical vegetable oils are not just one chemical compound, but a mixture (after removing everything that is not) of glyceride triesters. The chance that there is some overlap between your two oils is big, so the answer to your immediate questions likely is no, completely impossible.

Even if there is no overlap in the chemical composition, you would probably still have to separate the oil mixture into all its components, and then know how to mix them together again to get the original two oils.

The latter is possible, of course, with a reasonable analytical effort, the former is close to impossible to perform on a macroscopic sample. A GC(-MS) can easily separate (and identify) the compounds, but that's a few billion molecules or so. Doing liquid column chromatography would be very tricky, with any reasonable apparative effort would give you a few milligrams of each compound, and perhaps there are still a few you can't separate.

Distillation is impossible (boiling points are high and too similar), fractionation by crystallisation would only work well for some of the triglycerides, etc. All in all it's a bad mess. Don't mix two oils and try to separate them afterwards. ;-)


Lewis Acid Catalysts I am studying for the International Chemistry Olympiad and some of questions involve synthesis of complicated organic compounds and i have no idea how to do them.

The setting of the questions is usually like this: they give you a rough idea of the end product however some of the groups are replaced with R. They replace all the other intermediates with A, B, C ... They also give the reactants and catalysts for each step.

Most of the time the catalysts are lewis acids. I wondering if you were able to provide mechanisms and general resulting reactions of these catalyses so i could predict what the product will be.

Also, is $\ce{NaBH4}$ a lewis acid?

Firstly, $\ce{NaBH4}$ is not a Lewis acid because it cannot accept any more electrons. Aqueous $\ce{NaBH4}$ is a commonly used reducing agent often for reduction of carbonyls to their respective alcohols.

The mechanism of Lewis acid catalysis depends on the reaction but often Lewis acids are used to generate cationic electrophiles such as in the Friedel-Crafts reaction.


Bonding in d10 metals such as Zn I recently learned about the MO bonding model in metals, where many metal atoms form lots of orbitals close in energy that form bands. However, I can't understand the bonding in d10 metals. It seems to me that if they form MOs, then the bonding and antibonding orbitals will both be filled, resulting in net zero bonding. The d10 metals obviously bond anyway, so what is wrong with my reasoning?

You miss the fact that the $p$ orbitals lying immediately above the occupied orbitals are also valence orbitals.  These unoccupied orbitals then allow the formation of unoccupied antibonding molecular orbitals whereas the corresponding bonding MOs are occupied.  Such a $p$ orbital contribution is then responsible for both the partially covalent bonding of these elements in compounds and the metallic bonding of the pure elements in their condensed phases.
In the case of metallic bonding in the elements, we see this effect in the alkaline earth metals as well as the Group 12 metals.  In the case of alkaline earth metals, if these were bonded only through the $s$ orbitals they would not be bonded at all; but the real metallic bond structure mixes in the otherwise unoccupied $p$ orbitals and this gives more net bonding (as measured by hardness and by melting and boiling points) than that in the alkali metals that precede them.


Chemical compatibility of polycarbonate and carbonate esters I am designing a battery box to house lithium ion batteries in an electric vehicle. These batteries use an electrolyte which probably contains carbonate esters (eg. ethylene carbonate, dimethyl carbonate, diethyl carbonate).

If I use polycarbonate as the battery box material, should I be worried about an interaction with the electrolyte in the event of a battery leak?

Yes, you need to verify that the housing of the battery is made of a material that is unaffected by its contents. Polycarbonate is a good material for rigidity and transparency, but may be dissolved or weakened over time by organic solvents (carbonate esters or otherwise) and also by pH extremes and high temperatures that could come into play when the battery is charged and discharged. Thermosetting resins might be a better bet for this type of application.


Rearrangement of carbocation During rearrangement of carbocation, we have three choices: methyl shift, hydride shift and phenyl shift. If we have a situation where all the three are possible, then what will be the preference order?

In all cases we study the stability of the carbocation resulting from the hydride/methyl/phenyl shift. That shift will be more favored whose resulting carbocation is the most stable.


Reaction of acyl chloride with excess Grignard reagent I have a reaction to which I must find the majority organic product.

I think I found the correct product but the fact that the reaction takes two moles of propylmagnesiumbromide confuses me a bit. I added an image of the product I think would be the result.

I would like to know if I found the right product, and if the two moles of propylmagnesiumbromide means that there is a second part to this reaction.

The following paragraph from chem.libretext.org1 sums up the answer to your question:

Acid chlorides react with Grignard reagents to produce tertiary alcohols. Two equivalents of the Grignard reagent are needed because the first equivalent reacts to form a ketone which then reacts with the second equivalent. Because of the high reactivity of the Grigard reagent, the reaction can NOT be stopped at the ketone.

Coming to your specific case the reaction would proceed with the following mechanism:

Reference:
(1) Acid Halide Chemistry https://chem.libretexts.org/@go/page/183124 (accessed May 24, 2021).


Slight FTIR band intensity changes across several different copper salt spectra: relevant for characterization? I have several spectra of different samples which are probably all copper (neutral and/or basic) acetate(s) and which are very similar. I present them overlayed and (some of them) stacked:

My question is: can I assume all the samples are qualitatively identical or do subtle/slight intensity differences in the same bands mean a qualitative difference?

These are all spectra made with a pellet of KBr and I intend a qualitative characterisation to compare with literature, not a quantitative one.

(Answered after your edit of your question:)

Some variation in the location of the absorption bands (vs. 1/cm) and their intensities (vs. remnant transmittance) is unavoidable due to the interaction of your analyte (here: your copper salt/salts) with the matrix (here: pellet of KBr) compared to recording a neat film of your sample.  The spectra depicted by the trace in magenta and blue seem to be a little bit different from the others, perhaps due to some humidity (KCl exposed to air, or a sample not completely dried, or both, ...).  On first view of the current representation, the other look very similar, compared with each other.

As your data represent infrared spectroscopy, figure out how to change the representation such that your ordinate is in remnant %transmittance instead of absorption.  Let the spectrometer's software do this for you.  For a visual inspection of the data, a split in the abscissa, too, for example around 2000 1/cm eases the visual inspection a lot -- equally something typically offered by the software of the spectrometer. (On good instruments, both changes do not alter the raw data collected by the spectrometer.) For example:

(source, the picture was edited by addition of horizontal lines)

Determine the barycentre of your absorption bands. Likely, as you start with IR spectroscopy, a deviation of about 1/cm is not an issue  (ask your instructor).  In other words, if the peak pic by the software suggests 1837.54 1/cm, a report like 1838 1/cm likely is good enough (ask your instructor).
For each important absorption band, have a look about the shape, note if it is a particular broad absorption band -- in your example, around 3350 1/cm, for example.
Beside location, an other important information provided by IR spectroscopy is the intensity of the absorption band.  In the representation of %T, imagine three equal parts between the baseline on top, and the most intense absorption band on the bottom.  A bit like the three horizontal blue dotted lines added to the picture above. The absorbtion bands in the first strip are then (semiquantitatively) classified as of weak, the ones reaching till the second strip of medium, and the ones reaching down to the third strip of strong intensity.  (As an example, the edited spectrum above contains an absorption band of medium intensity at about 3500 1/cm, and two strong ones at about 1700 and 1500 1/cm.)
Compare localisation and absorption band intensity of the spectral data recorded by you, with data found in a reference (textbook, journal article, data base).  Are all important absorption bands found in your spectrum mentioned in the literature reference?  Do you find all important absorption bands described in the literature reference in your spectrum, too?  Are there some important bands in addition or missing?
Of course you may compare sample #1 with sample #2, with sample #3, etc.,
too.

Only by growing experience you will learn what important absorption bands in infrared spectroscopy are.  Because -- in contrast to other spectroscopic techniques -- it is rather the exception than the general case that you may correlate each absorption band with a structural element in the structure of your compound.

For future reference I suggest to record the IR data either already in the %T mode, or (second best) to convert them into this form.  This allows you to check rapidly if even the most intense absorption band does not hit a transmittance of 0% (such an occurence were a rapid indication of a sample with to high concentration of X in your pellet of KBr.)


Concept of heat of reaction in a isolated system One description of the concept is that heat of reaction (exothermic) is the quantity of heat that would given off to the surrounding by the product of a reaction if we allow the product to cool down.

If this description is valid for all exothermic reaction in a isolated system, would it also be valid for all endothermic reactions??

When chemical bonds are broken potential energy increases.  When they are made potential energy decreases. A loss in PE raises kinetic energy and a gain in PE lowers KE.  Since all reactions involve bond breaking and bond making the reaction will involve an increase or decrease in PE and a decrease or increase in KE.  When KE of the system increases, a system in contact with the surroundings will lose energy as heat (exothermic).  When KE of the system decreases a system in contact with the surroundings will gain energy as heat (endothermic).  In an isolated system "exothermic" reactions just raise temperature and "endothermic" reactions just lower temperature.


Reduction of copper oxide by hydrogen in the lab Why does the hydrogen which is passing continuously through the tube have to be burnt when it is passing out from the tube? Why can it not be allowed to just escape simply?

Burning the hydrogen gas at the exit stream of the copper oxide reduction tube is simply one way to assure that it does not accumulate to explosive levels in the lab. Although hydrogen gas is not toxic, it is flammable in air at concentrations as low as $\pu{4\%}$. As the combustion product of hydrogen and oxygen is mainly water vapor, this technique makes for a simple, safe means for disposing of the excess hydrogen from the copper oxide reduction process.


Measuring heat effect of reaction I did an experiment to determine enthalpy change when Zinc is added to Copper(II) sulphate solution.

I recorded values and plotted a temperature-time graph to obtain values of the temperature change to calculate heat transfer .

The trend line that I obtained is when temperature of the mixture started to drop.

I never knew why I needed to extend that trend line and draw a vertical line at the point when Zinc is added and thus never understood why the difference between the interception point and the point when Zinc is added is $\Delta T$

To summarise, I don’t understand why I need to extend the trend line and the interception point gives the highest ‘T’ value...

In an ideal world you would not have to do that. You'd simply wait until the reaction completes and then make a single measurement of temperature.

Too bad our world isn't ideal. The moment your reaction vessel becomes slightly hotter than the environment, it starts losing heat. If you wait for too long, it will cool all the way down to room temperature, so there will be nothing to measure in the first place. If you make your measurement in the first seconds so as not to let any heat escape, that's too early; the reaction has barely started. If you wait just until completion (which is a few minutes into the process, judging by your graph), that's too late, because by then some amount of heat is already lost.

So what do we do? We account for the loss. We assume that the temperature drop is linear (which of course is just an approximation, but still, it is much better than nothing), make a dozen or so measurements, build that trendline, and extend it to the origin. That's where your temperature would stop if there were no drop.

So it goes.


What is the smell of a glass of water in contact with fresh air? When you pour water into a clean glass in a small confined room with no or little fresh air circulation, the only smells you perceive are those of the glass and the water (or the impurities from both of them).

However, expose that very glass to a bit of fresh air, and voilà, it starts to smell different, a bit like the (wet) shells of mollusks in the sea or from a sweetwater source (it is not a "fishy" smell in the sense that is smells like fish but rather one reminding of the sea, lakes or rivers).

I found this smell with wet glass, ceramics, porcelain, but apparently not really or not at all with plastic or metal (maybe the smell of metal somehow obstructs perception, or maybe it alters or destroys the chemical compounds).

So what is this smell actually?
What chemicals are there causing this smell?

Why does it apparently need contact with fresh air (maybe oxygen radicals or something) to emerge? And it seems to be volatile so that the smell weakens with time (or maybe you just get used to it).

I hope at least some here know what I am talking about... if not, maybe try it out.

Some people are apparently more susceptible to this smell than others (a friend of mine is very susceptible to it while others have no idea what this fuss is all about).

It also works when you get the glass hot and fresh from a dishwasher, though it is recommended to try with a non-scented detergent.

And you absolutely have to try it with non-chlorinated water.

Update:
I've come across two substances that may be related to the phenomenon, Trimethylamine and Dimethyl sulfide, however, I don't know how they smell, so I can't really say if any of these are responsible for the odors, but from the Wikipedia articles, trimethylamine sounds more fishy than "mollusky" while dimethyl sulfide is described to be cabbage-like.

I do have a very similar experience! What the OP calls the mollusk smell, I call it slightly rotting "seaweedy" smell. It has been a very unpleasant experience for us in our new home.

Similar to OP's experience, this seems to be more intense with glasses and ceramics right after the newly washed dishes get exposed to air when the windows are open. But I also detect it with plastic cups as noted below.

In order to eliminate other variables such as dishwasher problems and tap water quality, I bought bottled water and disposable plastic cups. I did the following experiment.

WITH CLOSED WINDOWS

Water from bottle poured in a previously unused plastic cup, sitting for 15 minutes : No smell

Water from bottle poured in a cup and couple of sips taken, sitting for for 15 minutes : No smell

WITH OPEN WINDOWS

Water from bottle in a previously unused plastic cup, sitting for 15 minutes : No smell

Water from bottle in a cup and couple of sips taken, sitting for 5 minutes : Seaweedy smell

So I'm about to conclude that something in the outside air combines with the saliva (and/or chemicals/enyzmes from the detergents) to produce the seeweedy smell! The same thing happens with the newly washed dishes out of the dishwasher, and hand washed dishes after they get exposed to air with the windows open.

Yes, I know this doesn't make any sense to people that haven't experienced it before! It sounds like nonsense, but it's quite annoying every time I take sips from water cup.

I'm almost on the verge of calling up the EPA about it. There must be something in the air, but what...


Filling up of orbitals As we know that the orbitals in a sub shell are degenerate, so is it necessary to fill up the electrons starting from x then to y and then z (if in p-subshell) or they can be filled like shown in the photo.

You are drawing very nice $\mathrm{p}_x$, $\mathrm{p}_y$ and $\mathrm{p}_z$ orbitals onto that piece of paper. But if you take an isolated atom, which direction is $x$? Which is $y$? The coordinate system you’re seeing the atom in is just something we planted over its top because humans like things nice and organised. It is entirely arbitrary.

By the way, the direct calculations do not give $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals (strictly speaking not even $\mathrm{p}_z$. They give you $\mathrm{p}_0$, $\mathrm{p}_1$ and $\mathrm{p}_{-1}$ (the indices referring to the value of $m_l$, the ‘magnetical quantum number’). In spherical coordinates, which are used because it makes the calculations easier, only the $z$ axis is semi-defined by values $\theta = 0$. Because $\psi(\mathrm{p}_0) = f(r, \theta)$ (the wavefunction of one of the p-orbitals is independent of $\phi$), we say that this represents the $\mathrm{p}_z$ orbital. $\mathrm{p}_1$ and $\mathrm{p}_{-1}$ (the direct solutions of the Hamiltonian) cannot even be mapped to directions as they involve complex numbers; only via linear combinations can we transform them into two real orbitals pointing in an arbitrary $x$ and an arbitrary $z$ direction.

Things change if you introduce an external magnetic field. The atom’s $z$ axis will now align with the magnetic field in some way, giving your system one preferred direction. One, not three. $x$ and $y$ are still essentially undefined. However, in a magnetic field it would make a difference which orbitals the electrons populate.

You might also be interested in checking out the Jahn-Teller effect which occurs in transition metal chemistry to avoid filling some of all degenerate orbitals. In a nutshell it is energetically better for certain complexes to distort themself from a perfect octahedron to create distinct energy levels that can be fully populated, half populated or unpopulated.


Does the name 4-nitrodiphenylmethane follow the IUPAC rules? Does the name 4-nitrodiphenylmethane follow the IUPAC rules? It doesn't make sense to me. How can you assign a locant of 4 when your stem name is just methane?

The preferred IUPAC name for the compound that is given in the question is 1-benzyl-4-nitrobenzene; however, that does not necessarily mean that other names are not in accordance with IUPAC nomenclature.

The name diphenylmethane actually is in accordance with current IUPAC recommendations (see Subsection P-44.1.2.2 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)), although the preferred name is 1,1'-methylenedibenzene (see also this answer).

The name diphenylmethane treats the phenyl groups alike as substituents of methane.

If now one of the simple phenyl substituents in diphenylmethane is replaced by a compound 4-nitrophenyl substituent, the analogous name of the resulting compound is (4-nitrophenyl)(phenyl)methane and not ‘4-nitrodiphenylmethane’.


Finding mass percentage of components of a solid mixture A workbook question asks:

  $\pu{2.184g}$ of a solid mixture containing only $\ce{K2CO3}$ ($\mathrm{FW = 138.2058}\pu{ g/mol}$) and $\ce{KHCO3}$ ($\mathrm{FW = 100.1154} \pu{g/mol}$) is dissolved in distilled water. $\pu{31.29mL}$ of a $\pu{0.742\!M}\ \ce{HCl}$ standard solution is required to titrate the mixture to a bromocresol green end point. Calculate the weight percent of $\ce{K2CO3}$ and $\ce{KHCO3}$ in the mixture.

I started this problem by finding the amount of moles of $\ce{HCl}$ added to the titration, assuming that the solution was completely neutralized at the endpoint.

In $\pu{0.03129L}$ of a solution of $\pu{0.742\!M}\ \ce{HCl}$, there are
$$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$

Then I set up the balanced equations:

$$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -> 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
$$\ce{HCl_{(aq)} + KHCO_{3(aq)} -> KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$

This should indicate that the $\ce{HCl}$ will combine with the carbonate and bicarbonate in a 2:1 ratio, favoring the carbonate, since two moles of HCl are required to neutralize one mole of carbonate.

Then:

$$\ce{\pu{0.0232mol}\ HCl \times \frac{1\; mol\; CO_3}{2\; mol\; HCl} = 0.0116\; mol\; CO3}$$

$$\ce{0.0232\; mol\; HCl \times \frac{1\; mol\; HCO_3}{1 \pu{mol} HCl} = 0.0232\; mol\; HCO3}
$$

And:

$$\ce{0.0116\; mol\; CO3 \times \frac{1\; mol\; K2CO3}{1\; mol\; CO3} = 0.0116\; mol\; K2CO3}$$

$$\ce{0.0232\; mol\; HCO3 \times \frac{1\; mol\; KHCO3}{1\; mol\; HCO3} = \pu{0.0232\; mol}\; KHCO3}$$

Converting moles to grams:

$$
\ce{0.0116\; mol\; K2CO3 \times \frac{138.2058\; g}{mol\; K2CO3} = 1.60\; g}
\\
\ce{0.0232\; mol\; KHCO3 \times \frac{100.1154\; g}{mol\; KHCO3} = 2.32\; g}
$$

When I add these masses together, I get $\pu{3.92 g}$, which is greater than $\pu{2.184 g}$.

Where did I go wrong?

We first start by writing the system of equations in this reaction, assuming that we don't know the ratios that the elements combine in.

Because we know the mass of the unknown mixture, we'll use $x$ to represent the mass of $\ce{K2CO3}$ and $y$ to represent the mass of $\ce{KHCO3}$, in grams. This is our first equation:
$$x + y = 2.184$$

Next, we know the final amount of moles of $\ce{HCl}$ that reacted. It was
$$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$

Now, let's reconsider the balanced equations for the reaction. We have
$$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -> 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
$$\ce{HCl_{(aq)} + KHCO_{3(aq)} -> KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$

We can see that it takes $\ce{2 HCl}$ to neutralize one $\ce{CO3}$, but it only takes $\ce{1 HCl}$ to neutralize one $\ce{HCO3}$.

We also know the formula weights of the individual components. Now, we can write the second equation as an expression of the sum of the moles of the reactants. Our coefficients for $x$ and $y$ become:
$$
\ce{\frac{2\; mol\; HCl}{1\; mol\; K2CO3} \times \frac{1\; mol\; K2CO3}{138.2058\; g}} \times x
$$

$$
\ce{\frac{1\; mol\; HCl}{1\; mol KHCO3} \times \frac{1\; mol\; KHCO3}{100.1154\; g}} \times y
$$

Now we can cancel out units and write out the second equation:
$$\ce{\frac{2\; mol\; HCl}{138.2058\; g}}x + \ce{\frac{1\; mol\; HCl}{100.1154\; g}}y = \ce{0.0232\; mol\; HCl}$$

After that, this is straightforward linear algebra. This system of equations is simple enough to be solved by basic substitution and elimination. However, you can choose to use a matrix. In that case, we have

$$
\mathrm{
\begin{bmatrix}
    1 & 1 \\
    \frac{2}{138.2058} & \frac{1}{100.1154}
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
}
=
\begin{bmatrix}
2.184 \\
0.0232
\end{bmatrix}
$$

Rearranging gives
$$
\mathrm{
\begin{bmatrix}
    1 & 1 \\
    \frac{2}{138.2058} & \frac{1}{100.1154}
\end{bmatrix}^{-1}
\begin{bmatrix}
2.184 \\
0.0232
\end{bmatrix}
=
\begin{bmatrix}
x \\
y
\end{bmatrix}
}
$$

Using a calculator we find that
$$\mathrm{
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
0.309\\
1.875
\end{bmatrix}
}
$$
$$x = \ce{0.309\; g\; K2CO3}$$
$$y = \ce{1.875\; g\; KHCO3}$$

Finally, the mass percentages are determined to be:
$$\ce{\frac{0.309\; g\; K2CO3}{2.184\; g\; total} \times 100{\%} = 14.1{\%} K2CO3}$$
$$\ce{\frac{1.875\; g\; KHCO3}{2.184\; g\; total} \times 100{\%} = 85.9{\%} KHCO3}
$$


Total entropy change for a spontaneous reaction For the oxidation of iron,
$$\ce{4 Fe(s) + 3 H2O(g) -> 2 Fe2O3(s) + 3 H2(g)}$$
the entropy change is $ \pu{–549.4 J K^{-1} mol^{-1}}$ at $\pu{298 K}.$ In spite of the negative entropy change of this reaction, why is the reaction spontaneous? $(\Delta_\mathrm{r}H^\circ$ for this reaction is $\pu{–1648 kJ mol^{–1}})$

In this question the solution given is that we consider the total entropy change for the reaction, i.e. the entropy change for a system plus the entropy change of surroundings, and that comes out to be positive.
But shouldn't the entropy change for surroundings (calculated by $q_\mathrm{rev}/T)$ be the negative of the entropy change for the system (heat absorbed by the surroundings is same as the heat emitted by system)? Thus, shouldn't the total entropy change be zero?

The initial state of your system consists of 4 moles of $\ce{Fe(s)}$ and 3 moles of water vapor at $\pu{298 K}$ and $\ce{1 bar}$ in separate containers. The final state of your system consists of 2 moles of $\ce{Fe2O3(s)}$ and 3 moles of $\ce{H2(g)}$ at $\pu{298 K}$ and $\pu{1 bar}$ in separate containers.
The $\Delta H^\circ$ and $\Delta S^\circ$ correspond to a process for taking this system reversibly from the initial state to the final state. They don't include the changes in enthalpy and entropy of the surroundings in executing this process.
Do you have an idea of what such a process would be?


Why does being "bonded" decrease atomic size? I have a furiously overwhelming question-

  Why is the atomic size of noble gases larger than the atomic size of
  the previous halogen  when atomic sizes decrease across a group?

I did search it on the net and asked my teacher but everyone says that because noble gases are usually not bonded.

But:

Why does being "bonded" decrease atomic size?

Also we are talking about the atomic sizes of isolated elements?

The first thing you should realize is that there are various definitions of the size of the atom. One distinguishes for example the covalent radius, the Van der Waals radius and the ionic radius. The covalent radius is based on the binding of atoms into molecules and on the resulting bond length. The van der Waals radius is half the minimum distance between two atoms that not form a covalent bond and the ionic radius is derived from crystal structures of ions in salts. The covalent radius cannot be determined for noble gasses as they do not bind under normal conditions. The Van der Waals radius however can be determined and since the noble gasses are not very polarizable, the Van der Waals radius does increase with respect to the preceding halogen. You'll find more in the wikipedia article as mentioned by ron.


Which one is nitric Acid, HNO3 by itself, or when it is included in H2O? This is a quote from my Chemistry textbook:

  Nitric acid ionises according to the following equation:
  $$\ce{HNO3(l) + H2O(l) -> H3O+(aq) + NO3- (aq)}$$

I know that the formula for nitric acid is $\ce{HNO3}$. But I cannot understand what is the product of this equation. What is $\ce{H3O+(aq) + NO3- (aq)}$? Is it a diluted acid with less concentration?

This is another quote (from my textbook) that confuses me:

  Nitric acid and sulfuric acid are not quite as strong as hydrochloric acid. Chemical and spectroscopic studies have shown that concentrated solutions of both these acids do contain some unionised molecules. For our purpose, however, their ionisation is effectively complete. This is particularly true in dilute solutions.

I cannot understand the part that is in the bold. What are these unionised molecules? I know that these acids are considered strong acids which completely ionise in water solution. If that is the case, then why does the text indicate that there are some unionised molecules?

In your first reaction, $\ce{H2O}$ acts as a base to abstract an $\ce{H+}$ from the nitric acid. The resulting $\ce{H3O+}$, called hydronium, is the conjugate acid, while the $\ce{NO3-}$, called nitrate, is the conjugate base (this is the molecule of nitric acid, but stripped of its $\ce{H+}$). You could view this as a dilution, in the sense that the resulting solution is less acidic than highly concentrated nitric acid would be, but the process of dilution of acid by mixing with water inevitably involves an acid-base reaction.

As to your second question, the unionized molecules are the neutral water and nitric acid on the reactant side, while the ions are the products resulting from the acid-base reaction. Nitric acid is relatively strong, meaning it tends to ionize almost completely in aqueous solution. Additionally, the less acidic the solution already is prior to addition of the acid, the more extensively the added acid will dissociate into ions (to understand this, please read up on acid-base equilibria and Le Chatelier's principle, as the topics are too lengthy and complex to cover adequately here). This is why the author emphasizes that the dissociation is more complete in dilute solutions. Statistically, however, some proportion of the nitric acid molecules will inevitably not dissociate into ions. The stronger an acid, the greater the extent of dissociation, but there is always some statistical chance that a proportion of the molecules will remain unreacted. For strong acids, however, the overwhelming preponderance of the molecules will dissociate, while only a tiny fraction remain unreacted. Hence, for many (if not most) purposes, it's reasonable when dealing with strong acids to proceed as if the dissociation is truly complete, since the unreacted portion is typically negligible in quantity.


Flory Huggins Theory For an ideal chain, Flory Huggins theory gives us a distribution which is like a Gaussian.

$$P(l, r) \sim \left(\frac{3}{2 \pi l^2}\right)^{3/2} e^{\frac{-3 r^2}{2 l^2}}$$

where, $l$ is the monomer to monomer distance(also known as Kuhn length) and $R$ is the size of the monomer

This is great but what are the approximation away from an ideal chain polymer? I want to understand how one can go beyond Flory Huggins Theory.

The assumptions made in the Flory-Huggins theory are

Quasi-solid lattice in the liquid
Inter-changeability of segments (not necessarily the same as the polymer structure
units) of polymer and solvent molecules in the lattice
Independence of lattice constants on composition (artificial)
Polymer molecules are of same size
Average concentration of polymer segments in cells adjacent to cells unoccupied
by the polymeric solute is taken to be equal to the overall average concentration
The expected number of available positions for each successive segment is
overestimated in the formulation as the formula includes double counting of
segments, separated by 2 or more segments in the same chain, that will fall on the
same position twice

In particular the assumption that polymer molecules are of the same size is a very unrealistic one, which was later removed in the Flory-Krigbaum theory


Major product in Reimer-Tiemann Carbonylation Attempt:

This is Reimer-Tiemann Carbonylation reaction in which the ortho product is major when phenol is the substrate. Hence, here I expected the ortho product to be major as well as it would also make hydrogen bonding possible, which stabilizes the compound. Besides, $\ce{-OH}$ is a better activator than $\ce{-OCH3}$.

But the answer given in the module is $C$ and not $A$. Where is my thought process wrong?

The product of the Reimer-Tiemann reaction actually also depends on the counter cation of the base we use. It is found that if we use $\ce{NaOH}$ as the base used for formation of carbene electrophile, ortho product is major. But surprisingly, if we use the base $\ce{KOH}$, the para product is produced in major amount.  The reason behind this is the stability of the transition state via formation of an ordered cyclic arrangement.Due to the smaller size of $\ce{Na+}$, this can get fit between the Oxygen and Chlorine atom in ortho positioned electrophile via electrostatic attraction and give rise to a very ordered transition state which governs the major formation ortho product. See for details, https://chemistry.stackexchange.com/a/98197/58497
On the other hand, $\ce{K+}$ is huge in size related to $\ce{Na+}$, which obviously can't increase the stability of the ortho product by forming a cyclic orderd Transition state. That's why, the para product generally gets preference in the case of $\ce{KOH}$.


How will the periodic table change due to redefinition of the kilogram (May 2019)? The definition of a kilogram has recently been changed. What effect, if any, will this change have on data on the periodic table?

Molar masses

The molar mass of the carbon-12 isotope used to be 12 g/mol exactly (by definition). Now, it is experimentally determined, with the current value $\pu{11.9999999958(36) g mol-1}$ (CODATA 2018). So the molar mass of carbon-12 changed. The standard atomic weights (relative atomic masses) shown in the periodic table are not affected by this. However, atomic weights (dimension one) are no longer exactly equal to molar mass divided by 1 g/mol. Instead, they differ by a tiny bit (you would need 9 or 10 significant digits to see the difference).

No visible changes

Here is an animated gif of a portion of the periodic table at ptable.com over the past four years (Jan 23rd 2016, Jan 10th 2017,Jan 6th 2018, May 2nd 2019). Nothing much has changed near carbon:

The standard atomic weights change a bit each year, as better estimates of the isotope compositions of elements with multiple stable isotopes are made. Also, the publishers of this specific periodic table decided to truncate atomic weights to 5 significant figures in 2018. The biggest change occurred when elements in period 7 were discovered and named.

The last image is from May 2nd 2019, before the change of the kilogram (and mole) definition. On May 29th 2019, the data shown at ptable.com is identical.


Do both electrons in a dative bond come from the same atom? The textbook definition of a dative bond is that it is a bond "in which the two electrons derive from the same atom" (credits).

However, I found that the two electrons of a dative bond may come from different atoms, and the two electrons of a "normal" bond may come from the same atom.

For example, consider the following reaction:

$$\ce{CH3CHBrCH3 + NaOH -> CH3CHOHCH3 + NaBr}$$

The $\ce{C-O}$ bond in $\ce{CH3CHOHCH3}$ would be an example where both electrons actually come from the $\ce{O}$ atom.

Actually, one of the electron may not have come from $\ce{O}$, but the $\ce{Na}$ atom!

What further complicates things is that all electrons are equal, so how can we tell their origins?

Your textbook’s definition is inconsistent for exactly the reasons you point out. I am going to use slightly different examples than you did but they work equally well.

A prime example for dative bonding is $\ce{H3N\bond{->}BH3}$, amminborane or ‘inorganic ethane’. The latter name because it is isoelectronic to $\ce{H3C-CH3}$, (organic) ethane. Typically, amminborane is generated by ammonia reacting with $\ce{B2H6}$ — or at least it seems very plausible to create it in that way. Ethane can be created in two different ways:

By the recombination of two methyl radicals:
$$\ce{H3C. + .CH3 -> H3C-CH3}\tag{1}$$
By the attack of a formal methide anion with a formal methanium cation:
$$\ce{Li-CH3 + H3C-OTf -> H3C-CH3 + LiOTf}\tag{2}$$

(The second version is absolutely equivalent to your $\mathrm{S_N2}$ reaction to give isopropanol.)

Going by your textbook’s rules, the first reaction might generate a normal covalent bond, but the second should give a dative bond, since both electrons of the $\ce{C-C}$ bond came from methyllithium and none from methyl triflate. And, of course, we cannot distinguish between these electrons and they are delocalised all across the molecule — that goes for both ethene and amminborane.

A much better definition of dative versus normal covalent bonds does not consider their generation but rather what happens when the bond dissociation energy is added, i.e. their dissociation. In amminborane’s case, the dissociation happens at a rather low energy and will yield ammonia and borane again, so the electrons ‘end up where they were before.’ In more technical terms, the compound dissociates heterolyticly. (Equation $(3)$)

$$\ce{H3N\bond{->}BH3 -> H3N + BH3}\tag{3}$$

Ethane, on the other hand will dissociate homolyticly upon being fed with the (much higher) $\ce{C-C}$ bond dissociation energy. It will reform the two methyl radicals used to generate it in equation $(1)$. (Equation $(4)$)

$$\ce{H3C-CH3 -> H3C. + .CH3}\tag{4}$$


How to eliminate electrostatic charges from titanium dioxide powder I have synthesized $\ce{TiO2}$ nanoparticles via sol-gel method and I need to calcine it at elevated temperatures for the crystal phases formation. The powder is very highly charged and not only difficult to work with but I also lose significant amounts as it gets repelled by quartz glass and other tools and fly all over the places.

Is there a way that I can alleviate the repulsive forces between the powder and the tools that I use?

The best way I've found is to use an anti-static gun (sold by chemical companies such as Sigma-Aldrich).

If this isn't available, taking gloves off helps enormously, as the gloves can easily become 'static' in the same way as rubbing a balloon against some fabric. Gently tapping the flask against a metal object also helps, as the metal 'takes' the static away.

I have also known chemists to wear the same anti-static/grounding bracelets used by people who work with computers so as not to damage the electrics internally, however for a bench-chemist only occasionally working with fine powders, this would probably be over kill.


Term for compound of carbon, hydrogen, and oxygen Is there a term for a compound of carbon, hydrogen, and oxygen?

There is hydrocarbon for a compound of carbon and hydrogen.  There is also carbohydrate but that requires the ratio of hydrogen to oxygen to be 2:1.

As well as the carbohydrates, there are many other compounds of these three elements: alcohols, aldehydes, ketones, esters, etc.  Yet, I cannot find a term for the whole group.  Have I missed something or just no need is felt for such a term?

There is no term for this group as it is too broad and diverse.  Even the narrower term hydrocarbon is usually qualified e.g. aliphatic, olefinic, aromatic.

With thanks to Karl.


Deriving a reduction potential from two other reduction potentials Given:

  \begin{align}
\ce{Co^{3+}(aq) + e-  &-> Co^{2+}(aq)} &   E° &= \pu{+1.82 V} \\
\ce{Co^{2+}(aq) + 2e- &->  Co(s)}      &   E° &= \pu{-0.28 V}
\end{align}

  what is the standard reduction potential for $\ce{Co^{3+}(aq) +3e-  -> Co(s)}$?

If I add up these two half-equations to get the desired half-equation, I get $E^\circ = \pu{+1.54 V}$. But the correct answer is $\pu{+ 0.42 V}$!

Am I missing something?

That is because you cannot simply add the electrode potentials algebraically. In both cases, the electrode potentials have to be multiplied by $n$. What you can do instead is use Gibbs free energy change for each reaction which then can be added algebraically.

$$\Delta G^\circ = -nFE^\circ$$

Using this you can get the $\Delta G^\circ$ for each reaction which can be then added algebraically in the same way you add the reactions. The resultant $\Delta G^\circ_\mathrm{net}$, on equating with $-nFE^\circ_\mathrm{net}$, we can get the value of $E^\circ_\mathrm{net}$, which should be your answer.

In this approach, since the value of $n$ for both the reactions are different, the relative weightage of the electrode potentials of each reaction in determining the resultant $E^\circ$ will differ and therefore the final $E^\circ$ will be different than what you get by just adding up the electrode potentials.

For your case:

$$\begin{align}
\ce{Co^3+ + e- &-> Co^2+} & E^\circ_1 &= \pu{+1.82 V} & \Delta G_1^\circ &= \pu{-175.6 kJ mol-1} \tag{1} \\
\ce{Co^2+ + 2e- &-> Co} & E^\circ_2 &= \pu{-0.28 V} & \Delta G_2^\circ &= \pu{+54.0 kJ mol-1} \tag{2} \\
\end{align}$$

$$\begin{align}
\Delta G^\circ_\mathrm{net} &= \Delta G_1^\circ + \Delta G_2^\circ \\
&= \pu{-121.6 kJ mol-1} \\[3pt]
E^\circ_\mathrm{net} &= -\frac{\pu{-121.6 kJ mol-1}}{3F} \\
&= \pu{+0.42 V}
\end{align}$$


Curly arrow mechanism for reaction between sodium borohydride and water? I was asked to compare the reaction between sodium hydride and water and sodium borohydride and water. I know that the former acts more violently than the latter, but apparently, I must use a curly arrow mechanism to show this.

I think that the HOMO for the latter reaction is the oxygen lone pairs, and the LUMO is the B-H bond, but I'm having difficulty moving on from there.

What attacks what first?

Their mechanisms are easily searchable online.
This is from 2009 from art-xy.com:

Notice that the reaction would be different in the presence of metal catalysts:
$$\ce{NaBH4 + 2H2O ->[metal catalyst] NaBO2 + 4H2}$$
This is from 2007 from chembook.co.uk:

Which reaction is more violent is already explained by Jan:

In one case [$\ce{NaH}$], you have free hydrides, which go ‘Oh my god, I see a proton, I’m going to jump at it!’ and in the other case [$\ce{NaBH4}$] you have hydrides thinking the same thing, but being pulled back by a boron saying ‘No, stay here, I need your electrons! I’m poor!’

Therefore the former is more violent.


Is every chemical reaction in equilibrium? I read that every chemical reaction is theoretically in equilibrium in an old textbook. If this is true how can a reaction be one way?

Yes, every chemical reaction can theoretically be in equilibrium.  Every reaction is reversible.  See my answer to chem.SE question 43258 for more details.
This includes even precipitation reactions and reactions that release gases.  Equilibrium isn't just for liquids!  Multiphase equilibria exist.
The only thing that stops chemical reactions from being "in equilibrium" is the lack of the proper number of molecules.  For a reaction to be in equilibrium, the concentrations of reactants and products must be related by the equilibrium constant.
$$ \ce{ A <=> B} $$
$$ K = \frac{[B]}{[A]} $$

When equilibrium constants are extremely large or small, then extremely large numbers of molecules are required to satisfy this equation.  If $K = 10^{30}$, then at equilibrium there will be $10^{30}$ molecules of B for every molecule of A.  Another way to look at this is that for equilibrium to happen, there need to be at least $10^{30}$ molecules of B, i.e. more than one million moles of B, in the system for there to be "enough" B to guarantee an equilibrium, i.e. to guarantee that there will be a well-defined "equilibrium" concentration of A.

When this many molecules are not present, then there is no meaningful equilibrium.  For very large (or very small) equilibrium constants, it will be very difficult to obtain an equilibrium.  In addition to needing a megamole-sized system (or bigger), the system will have to be well-mixed, isothermal, and isobaric.  That's not easy to achieve on such large scales!

Update Commenters suggest that "irreversible" reactions do not have an equilibrium.  This is true, but tautological.  In the real world, all reactions are reversible, at least to a (perhaps vanishingly small) degree.  To say otherwise would violate microscopic reversibility.  A reaction that was 100% irreverible would have an equilibrium constant of infinity.  But if $K= \infty$, then $\Delta G^{\circ} = -RT \ln{K}$ would turn into $\Delta G^{\circ} = -\infty$. So to get infinite energy we would just have to use 100% irreversible reactions!  Hopefully the problems with the idea of "irreversible" reactions are becoming apparent.


Do coefficients matter in rate law? For reaction $\ce{2A→B}$ (elementary step), according to the rate law, rate $= k [A]^2$.

In some calculations, we use $k[A]^2$ as the production rate of B. Why isn't it ${1 \over 2} k[A]^2$?

In this case, is the consumption rate of A also $k[A]^2$? If so, why are they the same when in fact two molecules of A produce one molecule of B?

I'm really confused about this part. Can anyone explain why coefficient doesn't play a role in this calculation? Thanks!

This confusion pops up often around here.

The problem usually comes down to insufficient consideration of units.

When you say "rate", what do you mean?  It has units of mol per volume per second, e.g. $\mathrm{\frac{mol}{L\cdot s}}$.  But that isn't enough information.  You need to know moles of what.

First example: if the rate is defined as loss of A

For example, if you say that the "rate" means the disappearance of A, then the units of the rate are really $\mathrm{\frac{mol~of~A}{L\cdot s}}$.  If so, then as long as:

The reaction is $\ce{2 A->B}$; and
The rate as you have defined it can be expressed by $r_A = k_1~[\ce{A}]^2$ ...

Then the rate of appearance of B will be $r_B = \frac{1}{2} k_1~[\ce{A}]^2$

Second example: if the "rate" is defined as appearance of B

If conversely you suppose that the "rate" refers to the appearance of B, then the units of rate are really $\mathrm{\frac{mol~of~B}{L\cdot s}}$, and if so, then as long as

The reaction is $\ce{2 A->B}$; and
The rate as you have defined it can be expressed by $r_B = k_2~[\ce{A}]^2$ ...

Then the rate of disappearance of A will be $r_A = 2 k_2~[\ce{A}]^2$

Third example: a totally separate definition of rate

You say its an elementary reaction.  Let's say that we didn't know that and wrote the reaction as $\ce{200 A -> 100 B}$.  The rate, according to this new example definition, is the rate of the reaction.  The units of the rate are $\mathrm{\frac{mol~of~"reaction"}{L\cdot s}}$. If the reaction is still 2nd-order then the rate of the reaction will be $$r_r = k_3~[\ce{A}]^2$$

According to our reaction, 200 moles of A disappear for each mole of reaction, so $$r_A = 200 k_3~[\ce{A}]^2$$  and 100 moles of B appear, so $$r_B = 100 k_3~[\ce{A}]^2$$

Final thoughts

You will notice that I used different subscripts to distinguish between the $k$ values for the different assumptions.  That's because they aren't equal to each other.  The value of the "rate constant" $k$ depends on how you define the rate.

No matter how we define "the rate" and what we define as the stoichiometry (e.g. $\ce{2A -> B}$ or $\ce{A -> 1/2B}$ or $\ce{200 A -> 100 B}$, the rates must be equal.  Therefore:

$$r_A = 200 k_3~[\ce{A}]^2 = 2 k_2~[\ce{A}]^2 = k_1~[\ce{A}]^2$$

That is,

$$ 200 k_3 = 2 k_2 = k_1$$

But no matter which way we decide to write it, it is still a second-order reaction because "the rate", no matter how we define it, is proportional to $[\ce{A}]^2$.


Electrochemical series - when not at standard conditions Is there a reason why this question's answer is treating it as if it is at standard conditions? Isn't it using 0.2M aqueous solutions, so the electrochemical series won't be valid for this question? The answer seems to ignore this.

Thanks

While the electrochemical series does assume standard-state conditions, we will show that in certain cases it will also work for non-standard-state conditions. One recalls that the Nernst equation is given by $$E = E^\circ - \frac{RT}{nF}\,\ln Q,$$ and can be used to determine electrochemical cell potentials when the cells are not at standard state. This works directly for the second reaction you list, as $Q = \ce{[Sn^2+]}/\ce{[Cu^2+]} = 1$ and $\ln Q = 0$, so the second term in the Nernst equation vanishes, but will not work for the first reaction, for which $Q = \ce{[Sn^2+]}/\ce{[Ag+]^2} \neq 1$.

In this case, it is reasonable to assume that one knows both (i) that the correction due to the second term is relatively small, and (ii) that $\ce{Ag}$ and $\ce{Cu}$ are relatively inert metals, and strongly favor being in their elemental form relative to other metals. The quoted result then follows.


Chemical shifts of OH protons The chemical shift of a proton is an indication of the amount of shielding it receives from the electron density on it or other magnetic anisotropy effects present in the molecule. For the protons of the carboxylic acid functionality, the chemical shift typically exceeds $\pu{10 ppm}$. Baseline estimates given for $\ce {CH, CH2, CH3}$ protons are $ 1.7, 1.3$ and $\pu{0.9 ppm}$ respectively. However, we are often able to observe the chemical shifts of alcohol protons in the range of $\pu{1-2 ppm}$. This is rather counterintuitive since the electronegativity of oxygen is markedly higher than that of carbon and hence we would observe a mucher greater deshielding of the $\ce{OH}$ proton. For alcoholic protons, there is likely to be the presence of hydrogen bonding effects. Is the presence of these $\ce {H}$ bonding effects responsible for the less than expected deshielding?

Hydrogen bonding leads to deshielding in alcohols. Isolated ethanol in the gas phase has a shift of $\pu{0.55 ppm}$, while the shift is $\pu{0.8 ppm}$ in neat ethanol, so there is a shielding effect due to H-bonding. However, the OH shift is extremely sensitive to self association and/or solvent, resulting typically in greater deshielding. See for instance the table of shifts in reference [1].

According to Hore (in his Oxford NMR book), the charge distribution around the proton is polarized (mainly because the proton is attracted by the H-bond acceptor), effectively deshielding it. In any event, the hydrogen bond acceptor $\ce{O}$ transfers charge to the donor $\ce{O}$. The intervening H apparently gets the hard end of the bargain.

[1]: Fulmer et al, Organometallics 2010, 29, 2176–2179


Why does molten salt turn black? Lately I have been seeing quite a few videos on YouTube about molten salt, specifically table salt (NaCl). One of the things that interests me about it is that regardless of what melting technique or type of crucible, the salt always ends up black afterwards.

Is the issue that the salt is highly reactive and is pulling chemicals out of the crucible regardless of what the crucible is made of? Wouldn't that mean that different crucibles would impart different colorations on the end result? That is the speculation I've seen on the comments of many videos (for example), but nothing definitive.

Or is it perhaps an optical change due to the change of the crystal structure? Or maybe something in the salt itself oxidizing or reacting with something in the air?

Part of the color change might be due to creation of F-centers, crystal defects where there are electrons filling the place of an atom. A corresponding H-center (for halogen interstitial) may also be created for each F-center.

That said, such color changes tend to be rather pale, and as you state, $\ce{NaCl}$ heated to melting in an iron crucible may appear dark or even black. This is likely due to dissolution and oxidation of the crucible. The molten salt attacks the metal at grain boundaries and also dissolves protective oxide coatings, allowing oxygen from the air to attack the underlying metal. Even in a closed system, such as a molten (mixed) salt thermal energy storage system, corrosion is a problem.

The color change could also be due to trace impurities in the "table salt". For example, sodium or potassium ferrocyanide, $\ce{Na4[Fe(CN)6]·10H2O}$, is intentionally added to keep the salt free-flowing despite humidity changes. This might react at high temperatures to form a deeply-colored compound.

To find out what is causing the discoloration, you might try heating a small sample of high purity $\ce{NaCl}$ in a ceramic or platinum (a bit pricey) crucible, and you could analyze heated table salt for dissolved iron compounds.

BTW, F-centers and other crystal defects can be created intentionally with optical lasers and x-rays. If memory serves, there was an article in Scientific American's Amateur Scientist about creating F-centers using an electrostatic generator.


Why does hydrogen bonding in salicylic acid make it more acidic? When salicylic acid deprotonates (losing the proton from the carboxyl group) it forms a hydrogen bond with between the oxygen in the carboxylate anion and the hydrogen in the alcohol group. However, surely there will be a steric clash between these groups which would surely be make more significant by the long hydrogen bond? Could the hydrogen bond still form with the carboxyl group twited out of the plane? Also, why doesn't it hydrogen bond intramolecularly before deprotonation which would disfavor the loss of a proton?

If the COO(-) group on salicylic acid's conjugate base didn't hydrogen bond, then wouldn't the COO(-) therefore be more inclined to pick up a proton and return the entire molecule to salicylic acid? In other words H-bonding is a stabilizing interaction. You can view H-bonding as diluting the negative charge density in the COO(-) group. This in turn makes proton abstraction less of a favorable move.

Structure of salicyclic acid's conjugate base:


How to set the memory for the Gaussian 09 cubegen utility? I am connected to a supercomputer via SSH and my cluster uses Linux Scientific, I am using the Bash (Unix Shell), I load the modules of Gaussian 09 and Gaussview. I am trying to print orbital densities for a very large tetramer system, but when I try to generate the orbitals on Gaussview, the system crashes because of insufficient memory. On the internet, I found that using the "cubegen" keyword I can generate the cube files of the orbitals. The keyword gives me the power to modify the amount of memory but I do not know how. I go to interactive mode so I can use computational resources directly and on the command shell I do the following:
 (I try changing Gaussian's memory's environmental variable directly at the folder I am at)

GAUSS_MEMDEF=16gb

then,

cubegen 16000000 MO=HOMO xxx.fchk xxx.cub 0 h

But it still crashes with the following message,

Out-of-memory error in routine DecCub-RdMO (IEnd=      34279201 MxCore=      33546152)
Use %mem=33MW to provide the minimum amount of memory required to complete this step.
Error termination via Lnk1e at Wed Sep 28 16:41:41 2016.
Segmentation fault (core dumped)

How can I solve this problem? How can I allocate memory to Gaussian from the command window (I do not know where Gaussian folders are located since I just load it as a module, given that the software is already installed on the system)

By reading the information from Gaussian about cubegen, you would find out, that use of cubegen is like the following:

  cubegen nprocs kind fchkfile cubefile npts format cubefile2

They further elaborate about nprocs:

  nprocs

  Number of shared memory processors used for electrostatic potential calculations. A value of 0 is equivalent to 1 (it is the default). Note that this parameter must be included if other parameters are specified. Previously this parameter was used to specify the amount of memory to allocate. The GAUSS_MEMDEF environment variable should be used instead.

You can read about setting GAUSS_MEMDEF on the site about the default.route file.

  Environment Variables

  All of these directives can also be set via environment variables or UNIX command line arguments. The environment variable GAUSS_XDEF provides a line equivalent to -X- in the Default.Route file. Similarly, the command line argument below specifies the same setting:

  $ g09 -x="value"

  Similarly, all of the following have the equivalent effect:

  Default.Route file: -M- 4GB
  bash shell: export GAUSS_MDEF=4GB
  tcsh shell: setenv GAUSS_MDEF 4GB
  command line: g09 -m="4GB" …

Based on that, you could try the way by exporting GAUSS_MDEF to the amount that you want, or by using the commandline. But I have no idea if the latter works with cubegen instead of g09. Let me know. :)


When a carboxylic acid acts as a nucleoplile, why does it attack through the carbonyl bond? Here's a couple of examples of what I mean. Firstly, in the reaction with thionyl chloride to make a carboxylic acid:

Secondly, in this specific example of a hydroboration reaction:

I don't really understand why you cant attack from the hydroxide group and then proton transfer. Any suggestions?

For both the examples that you mentioned, the carbonyl oxygen has its lone pair localized while the other oxygen's lone pair is delocalized with the double bond. The better nucleophile is the one which has greater electron density.


Is the haloform reaction possible with fluorine and pseudohalogens also? The haloform reaction is:
$$\ce{RCOCH3 +3X2 +4OH^- -> RCOO^- +CHX3 + 3X- + 3H2O}$$
where, $\ce{X}$ is $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$. However, I was wondering if this reaction works with fluorine and pseudohalogens? For instance would the following reactions occur?
$$\ce{RCOCH3 +3N2O4 +4OH^- -> RCOO^- +CH(NO2)3 + 3NO2- + 3H2O}$$
$$\ce{RCOCH3 +3(CN)2 +4OH^- -> RCOO^- +CH(CN)3 + 3CN- + 3H2O}$$

TL;Dr: Highly unlikely to be made using the general haloform reaction.
Long answer:
The mechanism of haloform reaction involves hypohalite ions resulting from the reaction between the base and the halogen.
$$\ce{X2 + NaOH -> X- + XO- + H2O}$$
If we use fluorine as halogen, the resulting reaction will produce hypofluorite ion but this ion is extremely unstable and as such fluoroform can't be made using this reaction.  Trifluouromethane is known and used industrially in the manufacture of polytetrafluoroethylene ("Teflon"), but it is commonly made by anion exchange with another haloform, typically chloroform plus $\ce{HF}$.

For many years, chemists have been unable to isolate cyanoform as a neat, free acid. However, in September 2015, it was successfully isolated by a team of scientists at Ludwig Maximilian University of Munich. It is stable at temperatures below – 40 °C. So, it requires drastic conditions to synthesize cyanoform and is only stable at low temperatures. So, it is highly unlikely to be made at ambient conditions using laboratory grade reagents.

Nitroform has been made in the laboratory by the hydrolysis of tetranitromethane under mild basic conditions. But can it be made using general haloform reaction? Yes, it can technically be made but there are face several problems. Let's start with the starting reaction i.e. reaction between base and nitrogen dioxide which produces nitrite and nitrate ions:
$$\ce{2OH- + 2NO2 -> NO2- + NO3- + H2O (+other nitrogen oxides)}$$

The nitrite that forms is not that great reactive oxidizer.
There might be an undesirable side reaction which is kind of a Meyer reaction to form undesirable nitrolic acid/psudonitrol which in basic conditions gives blood red/blue colored solution.
The reaction involves huge amount of nitrogen oxides in general which is not very economical to deal with.
Multiple SN1 reactions occurs that generates a lot of isomers and other nitro compounds of compounds along with a whole lot of nitrite esters.
Steric hindrance might play a role and the reaction would not proceed properly.
The nitroformate ion ($\ce{C(NO2)3^-}$) might ionize to form different ions: $\ce{CH(NO2)2^-}$, $\ce{CH2NO2^-}$.
Nitroform is itself a strong oxidizer.

There are discussions in forums regarding trinitromethane synthesis using general haloform reaction but this ended up being fruitless. It even discussed using copper salts as catalyst and even employed haloform itself to react with $\ce{Ag/NaNO2}$ in DMF/ether but these syntheses posed so much problems and the yield was generally very, very poor.


How is iron(III) chloride changed from solid to solution? I'm conducting an experiment using iron(III) chloride in aqueous solution, but am also aware that it's usually a solid.

How is iron(III) chloride experimentally converted to a solution?

And can you please explain how the conversion happens and its mechanism.

You can prepare a solution of $\ce{FeCl3}$ by dissolving the appropriate amount in water.

Let's say you want a $0.1 \text{ M}$ solution of $\ce{FeCl3}$. You would do this by dissolving 0.1 mole of $\ce{FeCl3}$ in 1 liter of water.

$$0.1000 \text{ mol} \times \dfrac{162.2\text{ g}}{1\text{ mol}}=16.22 \text{ g}$$

You would need to dissolve 16.22 g of $\ce{FeCl3}$ in enough water to make 1 liter of solution.

This solution is not stable over time. $\ce{FeCl3}$ hydrolyzes in water. See my answer to this question. The hydrolysis can be slowed/reversed by adding $\ce{HCl}$. Since solutions of $\ce{FeCl3}$ are already acidic, there is no harm usually in adding more acid. For this 0.1 molar solution, you might choose to dissolve 16.22 g of $\ce{FeCl3}$ in enough 0.1 M HCl to make one liter of solution.


How can I detect a specific molecule on a surface? Based on the comments and answers, this question wasn't very clear, so I have completely overhauled this question to make it much clearer in another post:
How to create the equivalent of luminol for a compound?

Original Post/Question:

I'd like to be able to detect trace amounts of the chemical 'monomorine I' (5-methyl-3-butyloctahydroindolizine) on a surface. (It is an ant pheromone)

Is there an easy way to determine if the molecule is visible in IR or UV? (before I spend the money on this equipment)

Is there a strategy for causing a visible reaction? or even a reaction visible by IR or UV?

Advice on where to even begin would be a huge help.

(I do not have a chemistry lab with a Mass Spectrometer at my disposal)

Simple UV spectroscopy is not an option.

You won't find anything at all above 254 nm - probably just a a weak band in the 220 nm range. Indolizidines are pretty dead in UV.

Derivatisation towards a fluorescent compound would require clevage of a $\ce{C-N}$ bond, formally a dealkylation of the tertiary bridgehead amine.

I have no idea if a reagent like 2-naphthyl chloroformate, which does form fluorescent carbamates upon heating with tertiary amines will work here and honestly, I wouldn't bother to try.

As pointed out in Martin's comment, GC/MS after washing/extraction probably is the way to go.


How to filter out sulfur dioxide Lets say I have a container that contains a mixture of air and sulfur dioxide. I want to be able to filter out the sulfur dioxide from the air, how can I do this either through the form of a filter or a chemical reaction?

Just add your gas mix in a container with NaOH solution and shake well or bubble it continuously through the solution. SO2 dissolves in water to form H2SO3 (and H2SO4 in the atmosphere when it rains resulting in what we call acid rain) which will be neutalized by NaOH to give dissolved Na2SO3. Your air should be freed of most SO2. This is how it is done in the industry. There are many other ways as well but most rely in neuralizing the SO2. Google Flue-gas desulfurization for more details.


Will this reaction occur? I was studying reductive behaviour of $\ce{H2S}$. So, there is one reaction in my textbook.

$$\ce{3H2S + K2Cr2O7 + 4H2SO4 -> 3S + Cr2(SO4)3 + K2SO4 + 7H2O}$$

This reaction oxidizes $\ce{S^2-}$ to $\ce{S}$. But I thought why not $\ce{SO2}$?

$$\ce{H2S + K2Cr2O7 + 4H2SO4 -> SO2 + Cr2(SO4)3 + K2SO4 + 5H2O}$$

The reaction seems stoichiometrically plausible, but this reaction must need highly concentrated oxidants. I know the 1st reaction happens at low concentration of $\ce{H2SO4}$. Maybe then using high concentrated $\ce{H2SO4}$ would do the work for 2nd one.

I am not sure, though I am confident that the 2nd reaction will happen. What do you think?

But I thought why not $\ce{SO2}$

While the reaction is stechiometrically correct, it is hard to perform.

In excess of $\ce{H2S}$ sulfur will be produced.
In excess of oxidant sulfate will be produced.

And exact balance is impossible to achieve.

That said, $\ce{SO2}$ reacts with  $\ce{H2S}$ finally producing sulfur. This reaction, just like many inorganic reactions, is an unholy mess of many different processes with many products obtained simulationaly and the only way to produce understandable and reproduceable results is to simplify the situation. One of the way to simplify the situation is to use a vast excess of one of the reagents.


How to convert chemical rate of reaction into time? I am very new to chemistry. I am using a simulator for knowing rate of chemical reaction whose unit is $\mathrm{(kmol/m^3)/s}$. I want to convert it into time taken only. I have got concentration of each chemical can I convert it into time.

Well, there isn't usually a specific time that a reaction reaches completion or equilibrium. You need to choose an arbitrary limit, such as "when 90% of the reactant has been converted to product".

You can calculate your time from an integrated rate law. This will require you to know what order your reaction is (perhaps CSE can help you there).

For example, for a first-order reaction, the integrated rate law is:

$C_t = C_0e^{-kt}$

Or, in a form more convenient for you:

$t = \frac{-1}{k}ln(\frac{C_t}{C_0})$

where $k$ is the rate constant of the reaction, $C_0$ is the initial concentration, and $C_t$ is the concentration at the time $t$ you are interested in.

If you choose the arbitrary limit I mentioned earlier (convenient for demonstration, but you'd probably want more product formed that 90%), at your "completion" time there would be 10% reactant left. The math would look like:

$\frac{C_t}{C_0} = \frac{10%}{100%} = \frac{1}{10}$

$t = \frac{-1}{k}ln(\frac{1}{10}) = \frac{1}{k}ln(10) \approx \frac{2.303}{k}$

where you have now found the time it takes for the reaction to proceed to your chosen point. Note that if you choose 100% conversion, the remaining reactant concentration would be 0. This gives you $ln(0)$, which is not defined for real numbers. To get a more meaningful understanding, you can take the limit as the reactant concentration approaches zero (the reaction approaching completion), and you will find:

$t_{completion} = \lim_{C_t \to 0}\frac{-1}{k}ln(\frac{C_t}{C_0})$
$= \frac{-1}{k}\lim_{u \to 0}ln(u) = \frac{-1}{k}(-\infty) = \infty$

And thus by our most useful models, it takes literally forever for every single molecule to react. This is true for any reaction order. This is why we cannot easily say when a reaction is "done", and we must pick an arbitrary point.

Note that all the math examples I presented only apply to first-order reactions.

To find the time your reaction "takes" you will have to: 1) choose an arbitrary point to call completion, 2) know the order of the reaction, and 3) know the rate constant of the reaction.

P.S. If you wanted, you could set $\frac{C_t}{C_0} = \frac{1}{2}$ in the equation solved for $t$, and you will derive the half-life formula $t_{1/2} = \frac{ln2}{k}$ mentioned elsewhere.