Can a NOESY NMR experiment be halted partway through and give meaningful results? Some NMR experiments can be halted partway through the acquisition and still give perfectly valid results: The traditional 1H experiment for example. Others, like a Saturation Transfer Difference Spectrum will have errors in the magnitude of the peaks in it if it is halted, due to the fact it uses groups of pulses and takes the difference of between them.

Can the standard proton-proton NOESY pulse sequence be halted halfway through and give a usable spectrum, or will this introduce artifacts?

There are two big differences between the STD experiment and a NOESY that matter when aborting the experiment early. The STD is usually run as a 1D experiment, and it's using the phase cycle to substract the on- and off-resonance spectra from each other.

When you acquire a 1D spectrum with multiple scans, the scans are not saved individually but directly added up to one FID, at least on Bruker spectrometers. So if you abort a 1D experiment (you need to execute tr to do that in most cases, as the FID is only written to disk after all scans are finished in most pulse sequences), you can't just throw away the last incomplete phasy cycle. And with experiments like the STD, your actual signal is the difference of two larger signals, so one incomplete substraction can ruin your spectrum.

In a 2D spectrum the increments for the indirect dimension are measured and written to disk sequentially, so you can throw away the last incomplete increment if you would need to. In most cases the incomplete increment will not be saved at all to disk, so you'll end up with one less data point in the indirect dimension, but no incomplete ones that were aborted in the middle of the phase cycle. This depends on how exactly the pulse program is written, but I suspect that most standard sequences handle it this way.

I've stopped HSQC and NOESY spectra before they were finished pretty often, and never observed any artifacts that could be traced to the abortion of the experiment.


Representation of bromelain molecule in two dimensional diagram If I'm not wrong, I learned in 2011 introduction to neuropsychology course that all enzymes are proteins.

As far as I know, any protein, not even the smallest of the TAL group (which I understand to be the smallest currently known protein group), can be represented on a common-size two dimensional diagram (say, A4 page size) simply because it's too big in comparison to other molecules (such as methylcobalamin).

If bromelain is an enzyme which is larger than, for example, TAL proteins, how come it is represented on a common-size two dimensional diagram in PubChem?

I think I got it. The diagram in PubChem is not bromelain molecule, but a tiny fragment of it. This thing is indeed too small to be an enzyme. Also, it has nothing to do with proteins. Look at the formula, there is not a single amino acid residue here. Indeed, scroll down to 3.4.2 and read: "carbohydrate moiety of bromelain". That's what it is, and not the entire enzyme.

If you want to know what the rest of it is made of, then google "bromelain amino acid sequence". It seems to be known for quite a while. No nice pics, though.

On a side note, there is no such thing as "the smallest protein" or even "the smallest known protein", for the border between proteins and polypeptides is rather arbitrary. You are right, though, in that most enzymes are huge proteins, too big to fit on a 2D diagram. So is bromelain.

So it goes.


Can air inside a high temperature (1300C) Kiln cause an explosion? I saw many kiln designs that lack an opening for hot\pressurized air to come out , or any pressure valve.

But when air is heated to such a high temperature (1300C) inside a closed chamber (the kiln), won't the pressure inside increase enough to explode the kiln ?
(These kilns usually operate for several hours under this high temperature)

I think in general the physics would make it unlikely that the air pressure would build up faster than it leaks out, attempting to seal a furnace is unlikely to happen by accident.

A bomb calorimeter or pressure vessel made to specifically seal the inside from outside does not have a loose fitting lid over a cast refractory edge, it is made to much finer tolerances.

Chemical reactions are not likely unless you have a significant mount of vaporisable fuel in the charge and this is contained until at some point it lifts the lid and lets vapour out and air in with glowing elements to ignite the mixture, a flame may result, explosion still unlikely as volume small and not contained any longer.

In practice most kilns have a hole for temperature sensor that is not airtight and would let pressure equalise as they are heated.

Unless a specific atmosphere (inert, reducing, oxidising, carburising) is required in a kiln there is no real need to make them air tight and they rarely are, even then it is often achieved within a protective wrapping of the load and supporting chemistry inside the wrapping like charcoal, cyanides etc).


Can some one please explain alkyl shifts for me Provide a mechanism for the following reaction. Include all intermediates, resonance structures, charges, and electron pushing arrows to obtain full credit.

I’m stuck with this problem. I got as far as step 2 of the mechanism. I know the double bond acts as a nucleophile and attacks one of the hydrogen on sulfuric acid thus creating its conjugated base $\ce{HSO4-}$ leaving a carbocation on carbon 1.
In previous problems I have done the carbocation will rearrange to form a more stable cation. I’m told it’s done by either a hydride shift (which I understand) or an alkyl shift. I’m having a hard time grasping the concept of a six-membered ring going into a five-membered ring. And essentially understanding why there is a shift.

The reaction of the ketal 1 with aqueous H2SO4 is more likely to follow the pathway 1 --> 2 --> 3 to form α,β-unsaturated ketone 3. This is a typical ketal hydrolysis in acid medium. For the proposed reaction pathway 1 --> 4 --> 5 --> 6, species 6 is unstable. Indeed, if it were formed as an intermediate, it would afford ester 7a and ultimately the carboxylic acid 7b. Is there a possibility that the actual reactant doesn't bear methoxy groups but methyl groups? If this is the case, then the reaction 8 --> 9 is more reasonable.


Nomenclature of the following aromatic compound What is the name of the compound $\ce{(C_6H_5)-CH=N-CH_3}$  ?

I doubt,  it is  n-methyl benzalamine . Is it correct ? Besides , what could be other names of it?

Benzalamine seems to be an ocassionally and formerly used general name for compounds of the formula

$$\ce{C6H5-CH=N(R')R}$$

(where C6H5- is phenyl; R′ may be H). This name probably came from the fact that they can be prepared by a reaction of benzaldehyde with an amine:

$$\ce{C6H5-CH=O + R-NH-R' -> C6H5-CH=N(R')R}$$

$\ce{C6H5-CH=NH-CH3}$ could be, in this manner, named methylbenzalamine or N-methylbenzalamine. You can also find name α-methylbenzalamine, which would be a different compound, $\ce{C6H5-CH(CH3)=NH2}$

However, according to the IUPAC nomenclature, the compound class is and imine, or aldimine.

The correct name of $\ce{C6H5-CH=NH-CH3}$ would be

N-methyl-1-phenylmethanimine (preferred IUPAC name, 2013)

Also note that there is cis/trans (Z/E) isomerism possible.


Acidity of phenol and carboxylic acid Carboxylic acid is much more acidic than phenol ($\mathrm{pK}_\mathrm{a}$  difference of ~6). I wonder why? In terms of resonance, there are 5 resonance structures for phenol, and only 2 in carboxyl group.
My thought is 4 out of 5 structures in phenol are minor (lone pair electrons at carbon, less stable), compared to the equivalent resonance in carboxyl group. Is that really the sole answer to the large $\mathrm{pK}_\mathrm{a}$ difference? I am not so convinced.
Although these resonance (4/5) structures are minor in phenol, I think the combined delocalization effect should make it more acidic than a carboxylic acid.

Since you haven't mentioned the carboxylic acid to be measured, I assume the carboxylic acid to be $\ce{HCOOH}$. It has a $\mathrm pK_\mathrm a$ of $3.75$
Now, as you mentioned, phenol (or carbolic acid) has a $\mathrm pK_\mathrm a$ of $10.0$. There is a difference of about $6$ in the two values. This is because of the effective delocalization that takes place in formic acid compared to that of phenol. The best way to check for this is by computing the charges on the atoms in their respective conjugate bases and then check the stabilities of the two.
So first, we check the delocalization on the phenoxide ion.
The Mulliken partial charges on the atoms ($\ce{C-1}$ as carbon connecting to oxygen and we move counter-clockwise. $\ce{H-2}$ signifies the hydrogen bonded to $\ce{C-2}$)
$$
\begin{array} {c}
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{C-1} & 0.307 \\
\ce{C-2} & -0.246 \\
\ce{C-3} & -0.204 \\
\ce{C-4} & -0.243 \\
\ce{C-5} & -0.204\\
\ce{C-6} & -0.246 \\
\end{array}
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{H-1} & 0.103 \\
\ce{H-2} & 0.100 \\
\ce{H-3} & 0.089 \\
\ce{H-4} & 0.100 \\
\ce{H-5} & 0.103\\
\ce{O} & -0.659\\
\end{array}
\end{array}
$$
Now, we check the delocalization on the formate ion,
$$
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{C} & 0.363 \\
\ce{H} & -0.083 \\
\ce{O-1} & -0.639 \\
\ce{O-2} & -0.639 \\
\end{array}
$$
Comparing the two, we can see that the more effective delocalization has taken place in formic acid since the net charge on an individual oxygen atom is less ($-0.659$ in phenoxide ion and $-0.639$ in the formate ion)
Due to this, we can say that the net effective charge is much better delocalized in formic acid which means that the conjugate base is better stabilized. Even though there were five possible resonance structures for the phenoxide ion, there do not contribute to the final resonance hybrid as well as the two equivalent resonance structures for the formate ion.
Since order of conjugate base stability is directly proportional to acidity. Hence formic acid would be more acidic than phenol.
Note: These values were calculated on WebMO with geometry optimization using Gamess DFT B3LYP/3-21G


How can one explain niobium’s weird electronic configuration? As cited in an answer to this question, the ground state electronic configuration of niobium is:

  $\ce{Nb: [Kr] 5s^1 4d^4}$

Why is that so? What factors stabilize this configuration, compared to the obvious $\ce{5s^2 4d^3}$ (Aufbau principle), or the otherwise possible $\ce{5s^0 4d^5}$ (half-filled shell)?

There is an explanation to this that can be generalized, which dips a little into quantum chemistry, which is known as the idea of pairing energy. I'm sure you can look up the specifics, but basically in comparing the possible configurations of $\ce{Nb}$, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, as seen below:

d:   ↿ ↿ ↿ _ _        ↿ ↿ ↿ ↿ _        ↿ ↿ ↿ ↿ ↿     ^
                 OR               OR               |
s:   ⥮               ↿                _            Energy gap (E)

The top row is for the d-orbitals, which are higher in energy, and the bottom row is for the s-orbital, which is lower in energy. There is a quantifiable energy gap between the two as denoted on the side (unique for every element). As you may know, electrons like to get in the configuration that is lowest in energy. At first glance, that might suggest putting as many electrons in the s-orbital (lower energy) as possible, and then filling the rest in the d-orbital. This is known as the Aufbau principle and is widely taught in chemistry classes. It's not wrong, and works most of the time, but the story doesn't end there. There is a cost to pairing the electrons in the lower orbital, two costs actually, which I will define now:

Repulsion energy: Pretty simple, the idea that e- repel, and having two of them in the same orbital will cost some energy. Normally counted as 1 C for every pair of electrons.

Exchange energy: This is a little tricky, and probably the main reason this isn't taught until later in your chemistry education. Basically (due to quantum chemistry which I won't bore you with), there is a beneficial energy associated with having pairs of like energy, like spin electrons. Basically, for every pair of electrons at the same energy level (or same orbital shell in this case) and same spin (so, if you had 2 e- in the same orbital, no dice, since they have to be opposite spin), you accrue 1 K exchange energy, which is a stabilizing energy. (This is very simplified, but really "stabilizing energy" is nothing more than negative energy. I hope your thermodynamics is in good shape!) The thing with exchange (or K) energy is that you get one for every pair, so in the case:

↿ ↿ ↿

from say a p-subshell, you would get 3 K, for each pair, while from this example:

⥮ ↿ ↿ ↿ ↿

from a $\ce{d^6}$, you would get 10 K (for each unique pair, and none for the opposite spin e-)

This K is quantifiable as well (and like the repulsion energy is unique for each atom).

Thus, the combination of these two energies when compared to the band gap determines the state of the electron configuration. Using the example we started with:

d:   ↿ ↿ ↿ _ _         ↿ ↿ ↿ ↿ _         ↿ ↿ ↿ ↿ ↿    ^
s:   ⥮           OR   ↿           OR   _            |
PE:  3K + 1C          6K + 0C          10K + 0C     Energy gap (E)

You can see from the example that shoving 1 e- up from the s to the d-subshell results in a loss of 1C (losing positive or "destabilizing" repulsive energy) and gaining 3K (gaining negative or "stabilizing" exchange energy). Therefore, if the sum of these two is greater than the energy gap (i.e. 3K - 1C > E) then the electron will indeed be found in the d shell in $\ce{Nb}$'s ground state. Which is indeed the case for $\ce{Nb}$.

Next, lets look at perhaps exciting the second s e- up to the d-subshell. We gain 4 additional K but don't lose any C, and we must again overcome the energy gap for this electron to be found in the d-subshell.

It turns out that for $\ce{Nb}$: 4K + 0C < E  (remember that C is considered a negative value, which we're not losing any of), so $\ce{Nb}$ is ultimately found in the $\ce{5s^1 4d^4}$ configuration.


Calculating time needed to coat an object using the process of electrolysis Suppose we have an object with a surface area of $S=\pu{1.5 dm2}$, and we need to coat it with hard chrome. The thickness of the coat should be $d=\pu{80 \mu m}$. The mixture used for coating is of the following mixture: $\ce{CrO3}$  $\pu{290 g/L}$, $\ce{H2SO4}$ $\pu{2.5 g/L}$. The current density is $J = \pu{50 A dm-2}$. The current utilization factor is $\mu = 15\%$. Using Faraday's law of electrolysis:
$$\frac{m}{M}=\frac{q}{F}$$
and the fact that
$$J_{\mathrm{actual}}=0.15\cdot\frac{q}{S\Delta t}$$ where $S$ represents the surface through which the constant current passes through, I got $\Delta t = \pu{1425 s}$, however the answer is $\Delta t = \pu{2 h 22 min}$.

The trick you're missing here is that you don't account for how many electrons are needed to reduce $\ce{CrO3}$ to $\ce{Cr}$.
Cr has an oxidation state of 6 in $\ce{CrO3}$, so it takes 6 electrons to reduce every chromium atom.
The full form of Faraday's law is $$\frac{m}{M} = \frac{q}{zF}$$  where $z$ is the number of electrons transferred.
Therefore, you need 6 times as much charge as you initially calculated, which works out to 8550 seconds or 2:22:30.


Regioselectivity in Claisen condensation and aldol reaction I'm having a hard time determining regio-selectivity in those two reactions.

At the image above I've drawn one example for each on of them. My question is how do we choose where the deprotonation will occur. My approach was that the base will deprotonate the substrate in order to produce the most stable enolate. In the Claisen condensation, A should be the main product since the -CH2- group is present thus corresponding the mechanism. For B there will be no acidic hydrogens so it is not favoured, or at least that's what I thought.

Moving to the aldol condensation I can't figure out which product will be the main. Theoretically in C, the enolate is more stable than that of D, but the final unsaturated ketone isn't more stable in D?

Either way, I can't comprehend how can we determine what enolate will be produced each time.

Claisen and aldol condensations are thermodynamically controlled,* so it is not a question of which enolate forms, it is a question of which product forms.

In your first example, product A can be deprotonated by ethoxide as there is still an acidic hydrogen between the two carbonyls. Product B, on the other hand, has a quaternary carbon between the two carbonyls and cannot be deprotonated.

Because the pKa of a 1,3-dicarbonyl (~10) is significantly smaller than the pKa of ethanol (~15),** this deprotonation of A is almost complete and it therefore pulls the equilibrium position over to one which favours product A, simply by le Chatelier's principle. In fact if you somehow make product B via a different route and then treat it with ethoxide, you should exclusively get product A out of your flask, via retro-Claisen and Claisen condensation.

For the second example I would expect product D (tetrasubstituted alkene) to be preferentially formed over C (trisubstituted).

* Unless you use kinetic deprotonation conditions such as LDA. In this case you will then have to look at the stability of the transition states leading to enolate formation, but that's a story for another day.

** Very rough values.


Does chelation take ions out of solution Does EDTA take $\ce{Mg^{2+}}$ and $\ce{Mn^{2+}}$ out of solution, so that it can be removed by a filter? Or does it just take it out of any chemical reactions going on in the solution?

No- still too small for normal filters. Those ions are in water complexes anyway. It doesn't stop (prevent) chemical reactions rather alter the thermodynamics. EDTA and other organic complexes are the common way to allow animals to absorb ions that would otherwise tend to form oxides when swallowed so its biologically (chemically) available.


Why does Chem3D minimize nitrogen dioxide (radical) so badly? I build from text O-N+O (the SMILES of this is O=[N+][O-]), I delete the hydrogen atom added automatically on N, then I start MM2 Minimization. Both the bond length and the angle differ from what I see in the top right image here: https://en.wikipedia.org/wiki/Nitrogen_dioxide

By the looks of it, Chem3D uses generalised classical force fields to describe the molecular interactions.

Such force fields are optimised to provide a reasonable fit to multiple properties for lots of different molecules. For example, they reproduce reasonable bond lengths not only for $\ce{NO2}$ but also for the various bonds in organic molecules.

They are also constructed to reasonably reproduce solvation properties which depend on the molecule-specific partial charges, which in turn alter bond lengths and angles.

So it is unlikely that you have done anything wrong.

If you require a more accurate model, either apply more accurate force fields (e.g. those fitted specifically to $\ce{}$) or switch to quantum chemistry methods such as DFT.


What is involved in this firey demonstration? In this rather shaky Facebook video (sorry I can't provide a better link), a science teacher takes a burning beaker of something and spills it out onto the floor.  It then proceeds to shoot fire out in all directions, and we can see embers burning themselves out, leaving black ash of some sort.

I've seen teachers do similar stunts with alcohol fires, but those burn blue.  This one is much more explosive and stays yellow.

Is there any way to tell what the teacher is using here?

This is liquid methane, which boils at a temperature of −161 °C. (Methane can be condensed in a cold trap that is cooled with liquid nitrogen.)


Difference between relaxation and resonance leading to an absorption spectral feature? I need help understanding the physics behind this insightful comment below the question Does water really have strong EM absorption at 3 kHz in solid and 2 GHz in liquid? Why the huge shift?:

  Please note: This is a relaxation spectrum. The transitions do not come from a resonance (absorption, like in NMR, IR, etc.), but an applied electric field has a lossy (at low frequencies) or elastic (at higher f.) effect on some mode of molecular orientation of electrical dipoles in your sample.

The following comment links to https://en.wikipedia.org/wiki/Dielectric_spectroscopy which confirms this.

Both mechanisms can to a peak in the imaginary part of the dielectric permittivity at a certain frequency or energy, and at this point the slope of real part is steepest.

But is there a simple yet intuitive way to understand the physical difference between a spectral absorption feature due to a resonance process versus a relaxation process?

The difference can be viewed in two ways, I guess.

Firstly, for relaxation spectroscopy, you apply a forced oscillation on your sample. That can be mechanical (dynamic rheology, mHz to .1kHz, several kHz with very special, not commerically available equipment) or electrical (dielectric spectroscopy, low mHz to GHz range). The mathematical concepts behind both methods are very similar. For vibrational, MR, UV, etc. spectroscopy, the sample does just not interact with frequencies outside of the linewidth of some QM transition. The width of a resonant peak is given by the energy needed for the transition, and the lifetime of the excited state. The width of a feature in a relaxation spectrum is typically at least half an order of magnitude, from "sample can follow the external stimulus and stay close to a (dynamic) equillibrium" to "external stimulus is much too fast for the sample to follow at all".

The other point is that in relaxation spectroscopy, energy is transferred and dissipated into the sample below a specific frequency (invariably as heat), and reflected (elastically) at higher frequencies. The energy is not quantised, because the induced change in the sample is translatory. The peak frequency itself has no real significance, it's inverse is called a "relaxation time" of the sample. A statistical quantity, mostly. For QM transitions, energy is only transmitted at the given frequency/energy, and nothing happens otherwise, except Rayleigh scattering.

Basically, resonant spectroscopy deals with individual quantum mechanical species, and relaxation spectroscopy deals with the classical, unquantised newtonian properties of the whole sample. NMR is a bit of a hybrid: The induced transitions are purest QM, but you can only observe the classic induction of the ensemble, never individual photons.

Technically, relaxation spectra are usually taken by applying the stimulus at discrete frequencies, stepwise, and you want to have a phase-sensitive detector (voltage - current, deformation - torque).


Siiloxane polymer experiment I believe the first reaction would be a substitution of $\ce{SiMe3}$ for $\ce{I}$ and creating $\ce{SiMe3Cl}$.
The second process I think would involve the $\ce{Mg}$ removing the two $\ce{Cl}$ and forming diphenylsilane, which would likely be a repeating unit in a silicone polymer.
Could any explain the mechanism for these reactions?

A)SiMe3 Increases the inductivity and reduces Aromaticity,
so by free radical (uv termination) its likely to have p-iodideSi(Me)3Bencene.

B)Dimer crystal solution in Cl- ions.  (Ph)2Si(Cl)Mg+---ClSi(Ph)2
Or Solid crystals of (Ph)2Si(Cl)MgCl.


Why is pH used as a measure of acidity if not only hydronium ions can act as a acid when in solution? The Brønsted-Lowry theory tells us that the when an acid such as $\ce{HCl}$ dissolves, the following reaction takes place:
$$\ce{HCl + H2O <--> Cl- + H3O+}$$
It is said that the hydroxonium ion is what functions as an acid when a base is added to such a solution. However, there will indeed remain some $\ce{HCl}$ molecules in solution. So wouldn't they also act as an acid in some instances?

This problem becomes more apparent when we look at weak acids such as ethanoic acid. There are many many ethanoic acid molecules in comparison with the hydroxonium ions. Why is the acid 'weak' if the ethanoic acid molecules can just react with the base anyways?

To further explain where my problem is, consider the scale which we use to measure the 'acidity' of a solution, the pH scale. Consider a scenario where I have 2 different solutions of ethanoic acid and $\ce{HCl}$, both of the same concentration. The $\ce{HCl}$ solution will definitely have a lower pH and will be considered 'more acidic'. But would it really be correct to call the $\ce{HCl}$ solution 'more acidic' just because the $\ce{HCl}$ has dissociated to produce a much higher hydronium ion concentration? The ethanoic acid solution will have a lower hydrogen ion concentration, but the ethanoic acid molecules can still act as an acid … The main problem here is with using pH as a measure of acidity if molecules which haven't dissociated may also act as acids …

HCl being a strong acid, it is quantitatively dissociated in water, meaning that it fully dissociates into hydronium and chlorides. The hydronium are therefore the main contributors to the acidity of the solution. In a solution of ethanoic acid, the majority of acid molecules are undissociated. This is what "weak acid" means.

The amount of base necessary to neutralise a given acidic substance can be used as a scale of acid "strength" in chemical risk assessment.


Why do the two cycloheptadiene isomers both show four distinct shifts in carbon-13 NMR? In the following question I have filled in the correct answer for i:

For ii I thought that should be three for Q and two for P,

For Q there are three distinct enviroments for the C atoms:

$\ce{-\mathbf{C}H2-{}}$
$\ce{-\mathbf{C}H2-C=C-{}}$
$\ce{-\mathbf{C}=C<}$

and for P there are two:

$\ce{-\mathbf{C}H2-C=C-{}}$
$\ce{-\mathbf{C}=C<}$

Howevere the correct answer is 4 for both P and Q.

So then I thought maybe they want me to include the splitting of the peaks but in both P and Q there is too much splitting for this to be the case...

Why is the correct answer 4?

There are indeed 4 carbon environments in both of these molecules, shown below;

A plane of symmetry can be drawn through each, with one environment having only one carbon nucleus. All other sites have 2 carbons.

I don't know that any words will make it any clearer than the picture, but if you aren't sure after looking at this, I can elaborate.


Difference between water solubilities of phenol and cyclohexanol? The solubility of phenol is 83 g/L, while that of cyclohexanol is 36 g/L. Both of them have a hydroxy group, and can form hydrogen bonds with water. Also, the hydrophobic part of phenol, as well as cyclohexanol is almost the same. Then what is the reason for phenol's better water solubility in water compared to cyclohexanol?

In addition to the acidity difference described in the comments, there is a more subtle effect involving conjugation in the undissociated phenol.  When the phenyl group withdraws some electron density from the π-donating hydroxyl group, it gains some negative charge and so is better able to attract the hydrogen ends of the water dipoles.  Ultimately this is a manifestation of the polarizability of the delocalized π electrons.


Which singlet dihalocarbene has the strongest carbon-halogen bond? There are three singlet carbenes - difluorocarbene, dichlorocarbene and dibromocarbene. Which has the strongest carbon halogen bond?

Obviously backbonding is a point to be considered, as carbene is an electron deficient species. As fluorine is the most easiest to overlap with, the $\ce{C-F}$ bond will be strongest.

But I am not able to get that whether it is possible for the singlet carbene to overlap or share electron density with empty d-orbitals on chlorine and bromine atoms?

And which is more probable, backbonding from halogens to carbene empty orbital, or the other way round (from carbene filled orbital to vacant d-orbitals)?

Singlet carbene is zwitteronic, meaning the $\mathrm{sp^2}$ orbital is doubly occupied resulting in a negative $\mathrm{sp^2}$ orbital and a slightly positive empty $\mathrm{p}$ orbital. Therefore it attracts electron on one orbital and repulses them around another orbital. The huge reactivity of the singlet carbene comes from this zwitterionic character.

Below is the picture electrostatic potential of on the left a singlet carbene with doubly occupied sp2 orbital and the right triplet carbene.

It can be argued in many ways how a halogen stabilizes the singlet state of carbene.

Two explanation usually used are:

Electronegative species increase the $\mathrm{s}$ character of σ orbital (Bent's Law).
Electron donation from the electron pairs into the empty carbene $\mathrm{p}$ orbital.

Though the second one is usually used, I suggest you follow the first (more physical). Adding electron negative substituents increases the s character of the doubly occupied sigma in-plane orbital of carbene stabilizing it, significantly decreasing the energy of the system. Following this concept you can see that flourine should provide the maximum stabilization and so on.

$$\ce{F} < \ce{Cl} < \ce{Br} \ldots$$


Removing PTFE marks on glass bottles after using magnetic stir bars I am using magnetic stir bars to agitate solutions for 4 weeks at a time.  I have noticed at the bottom of the glass bottles that white marks have accumulated where the stir bar spins.  Even with soaking, thorough washing and scrubbing, acid washing, and sonicating, the marks remain.  I suspect these marks are white PTFE that has eroded from the stir bars then stuck to the glass.

I have 2 questions:

Are the marks removable?  I know PTFE is very resistant to most solvents, which makes removal seem difficult.
Is it necessary to remove these marks?  Perhaps because PTFE is so resistant to solvents, it will not affect my analysis.

I am extracting ammonium from soil extracts.  The soil extract is an acidified water solution, which I immediately add MgO to, in order to increase the pH to above 9.  This pushes the NH4+ into the headspace of the bottle.  Floating on the solution is a glass filter in a sealed Teflon packet.  The ammonium ions diffuse through the Teflon, and react with citric acid on the filter to make an NH4Cl salt.  This is then analyzed on an EA-MS for d15N.

Stir bar has a very rough grinding action. It is quite well known among silica manufacturers that if you stir micron sized silica with a stir bar for a long time in a reaction mixture, the spherical particles are damaged to some extent. In such cases, overhead stirring is preferred. This is not applicable to your case , I guess. If you had soil as well, the scenario is no different. It is a grinding action of the stir bar on glass. The "frosting" of glass is like rubbing a sand paper on glass for a long time. Nothing can heal this, except a glass-blower's flame treatment. Of course this is not worth it, as it is not going to affect the analysis of ammonia. I don't think that this is PTFE on glass. It will not stick to glass and secondly it is so hydrophobic that it would be floating on your solution rather than firmly sticking to glass.


How to reduce surface tension of oil? I would like to create tiny colored drops slowly falling down in a transparent liquid.

The diameter of the droplets must be around 2mm, and their speed must be approximately 0.5 mm / sec.

For that, I need the colored liquid to be slightly more dense than the static liquid.

The easiest setup is to drop water into oil with a syringe.

However, when I drop water from air, the oil surface tension prevents water drops to fall slowly into the oil.

And when I put the needle in the oil, the water accumulates around it and creates huge drops.

I would also like to pump back the water at the bottom to re-inject it later on; so the two liquids must be immiscible.

Is there a simple way to solve this problem?

A better setup would be to increase the oil density so that I can drop it in water (which is cheaper, so much better if I need to scale up my system). I asked another question for that purpose but I could not find any cheap heavy compound miscible with oil.

You may ease the droplets of water to "fall further down" across the oil by adding some of dishwasher into the water.

As just tested with a glass filled with some of sunflower oil, the water-dishwater droplets reaching the surface air./.oil will deform and flatten when they fall from about 30 cm above on said surface. Eventually however, cohesion and surface tension of water contracts these and they pass across the oil in droplike shape to reach the bottom of the glass where multiple water-dishwater droplets may coalesce.

Limitation:  Since dishwasher is a surfactant, however, it renders the two phases miscible in each other.  Stirring the content of this glas will distroy the bounderies of the two condensed phases.  (Which might be not your concern, though, reading "I could not find any cheap heavy compound miscible with oil [fit for this purpose]".)  It won't create you a thing similar like a "sand clock" which may be used again and again.

Addition since the original post was edited:  James Guides indeed added a good thought to the table.  Perhaps you do not need a liquid A imiscible to an other liquid B to study the fall of droplets, and you may use spheres instead.  Depending on perspective, this then is a problem of fluid mechanics (mécanique des fluides) as in applied biochemistry (e.g., here) and rheology (e.g., sphere viscosimeter).  In a steady state, the settling velocity $V_s$ may be described as

\begin{equation}
V_s = \frac{2}{9} \frac{r^2 g (\varrho_p - \varrho_l)}{\mu}
\end{equation}

with
$r$ the Stokes radius of the particle
$g$ earth's gravitational accleration
$\varrho$ the densities of of the particle ($\varrho_p$), or of the liquid ($\varrho_l$), and
$\mu$, the dynamic fluid viscosity (which depends even more on temperature, than $\varrho$).

Other than the two fields mentioned above, this relation is of importance havesting small crystals for X-ray diffraction analysis.  From there I know about documenting examples as in the screen photo below

taken for a large, a medium, or a small sphere of Teflon falling in a column of glycerol.  It equally allows to understand why you can't harvest small objects in a viscous medium from a bottom of a container with a spoon and have to resort to a fork.  (And with 2 mm of diameter, Brownian motion wont inhibit yet their settlment.)


Conformers of a meso-compound In a particular question in one of my reference book, I was asked if the given compound (meso-butane-2,3-diol) was a meso compound or not.  In it’s solution part, to show that the compound was actually meso, they did the following.

As I was not convinced about rotation of C2, I searched the web where I found this which says

  While rotating the bonds about any atom
  (chiral) the stereochemistry of the molecule
  does not changes.

This is what I am confused. I mean if we rotate meso-butane-2,3-diol in the given way such that we have a conformer other than eclipsed or staggered, wouldn’t it be optically active, without any plane or centre of symmetry, which means rotation does change stereochemistry. Where am I going wrong?

The method shown in OP's question to find a meso-compound is the way most people used. However, I usually used the Cahn-Ingold-Prelog (CIP) assignments of chiral centers for this purpose. It is a fact that the mirror image of (R)-chiral center is corresponding (S)-chiral center. For example, bromochlorofluoromethane is a chiral compound and the morror image of its (R)-stereoisomer is its (S)-stereoisomer as depicted in the diagram below:

When (S)-isomer is rotated $180^\circ$ vertically, the resultant structure would not be superimposed on (R)-isomer, as shown in the diagram. This is to justified the above mentioned fact.

Now, let's assign the compound in hands with CIP system. I'd call it (2 R, 3 S)-butane-2,3-diol (it could also be (2 S, 3 R)-butane-2,3-diol, based on the numbering). Either way, carbon #2 has three different groups attached to it, which are identical to those on carbon #2. Based on the given assignments on carbon #2  and #3, if you'd put a mirror between these carbons, you must see the mirror image of one part (carbon #2) on other side (carbon #3) as depicted in the diagram. Therefore, the compound has plane of symmetry (the mirror) and hence, it is the meso-isomer.

To conclude this fact, I included two enanthiomers of this compound ((2 R, 3 R)-butane-2,3-diol and (2 S, 3 S)-butane-2,3-diol), none of which gives mirror images when a mirror placed in between two chiral carbons (see the diagram).


How am I supposed to answer this? First the question gives no units or that these are covalent radii or what, so I assume they are in nanometers and assume an atomic radius. I check my periodic table app and it says a 92pm atomic radius, 75pm covalent and 155pm van der waal for the N atom, which happens to be the first option I crossed out. Now what? BTW, this question is from a previous national aptitude test.

Effectively, the question is giving the radii in some sort of reduced units, where it's expressing the size of each species relative to the size of something else.

Thus one can interpret the choices in this question as providing relative sizes.  So, for instance, choice "A" says the ion is 2.60/.77 = 3.38 times the size of the neutral atom, and so on.

Given your knowledge of the actual sizes of the ion and the neutral atom, you should now be able to choose the best answer.


Mixing polar and non polar substances According to the theory of London dispersion forces, intermolecular forces are created in non polar substances by induction of instantaneous dipole to neighbouring molecules. If that is the case then when we mix water and oil, water being polar should induce polarity in oil and they should mix. But they don't. Doesn't water induce polarity to oil? Do I have a false concept or something.

Water is already very 'happy' with his own guys due to strong hydrogen bonds . While van der waals force is so weak that its presence is so little for it to be seen while there is hydrogen bond. So even though there is some dipole induced it isn't enough.

  the total energy is lower than when they are mixed. The system is more stable if unmixed. -Chin Yeh


Why did they ‘split’ this H2+ ion in the ionic equation? I am asked this chemistry question in my textbook:

  Write an ionic equation for this reaction:
  $$\ce{Mg(s) + H2SO4(aq) -> MgSO4(aq) + H2(g)}$$

This is my thought process and answer:
$\require{cancel}$
\begin{align}
  \ce{Mg(s) + H2SO4(aq) &-> MgSO4(aq) + H2(g)}\\
  \ce{Mg(s) + H2+(aq) + SO4^{2-}(aq)
    &-> Mg^{2+}(aq) + SO4 ^{2-}(aq) + H2(g)}\\
  \ce{Mg(s) + H2+(aq) + \cancel{\ce{SO4^{2-}(aq)}}
    &-> Mg^{2+}(aq) + \cancel{\ce{SO4^2-(aq)}} + H2(g)}\\\hline
  \ce{Mg(s) + H2+(aq) &-> Mg^{2+}(aq) + H2(g)}
\end{align}

This was the answer:

  $$\ce{Mg(s) + 2H+(aq) -> Mg^{2+}(aq) + H2(g)}$$

Why did they ‘split’ the $\ce{H2+(aq)}$ ion into $\ce{2H+(aq)}$?

$\ce{H2+ (aq)}$ ions don’t exist. Furthermore, even if they did, they would not make a neutral compound $\ce{H2SO4}$ with the sulfate ion, since the sulfate ion has a $2-$ charge.

So you must split $\ce{H2SO4(aq)}$ into $\ce{2 H+}$ and $\ce{SO4^2-}$ ions.


Why does zinc sulfide glow? Zinc sulfide is used in road signs because when light from cars hit the road signs, it glows. What actually is the reason for the glowing of road signs (actually glowing of $\ce{ZnS}$)?

In the Wikipedia article Phosphorescent paint,

  Phosphorescent paint is commonly called "glow-in-the-dark" paint. It is made from phosphors such as silver-activated zinc sulfide or doped strontium aluminate, and typically glows a pale green to greenish-blue color. The mechanism for producing light is similar to that of fluorescent paint

The process is summarised in the article Fluorescence vs. Phosphorescence,
The explanation from the article:

  Phosphorescent materials produce light in a similar way as does fluorescence materials. A visible difference between these two types of luminescence, the ability of phosphorescence materials to glow after the excitation energy source is removed.

The website has a diagram exhibiting the process (in a bit of a simplistic, but understandable way - just replace the 'UV' with 'car headlights etc'):

Essentially, photons are emitted from every step back from the excited state to the ground state, unlike fluorescence, where the photon is emitted when the electron goes straight to the ground state.


How might one reduce the concentration of bubbles in their ice, from ice cube trays? So I have been freezing ice a lot recently, and I will come out and say it, as it turns out it is, albeit unlikely, it is sometimes possible for ice to "explode." I was not so aware of this before, but from now on I consume my ice beverages with deliberate care.

The first action I have taken is to make sure no bubbles are visibly present before freezing.

I was thinking about how one might reduce these frozen bubbles when making ice with ice trays and a freezer (the bubbles are sometimes a few millimeters across). I was thinking perhaps the bubbles are more likely to form if there are more dissolved minerals and other compounds commonly found in tap water. I think it is time to review osmosis to see if anything will shed light on this idea. My first instinct was to distill water and freeze it to see if there would be less bubbles, but since there will always be dissolved gases even in a freezer it does not follow that this will greatly affect anything.

Is it the case that at lower temperatures water will not be able to hold as many dissolved gases? If that is the case, then it might be worthwhile to lower the temperature of ice in the fridge or to a temperature just above freezing and allow the solution to stabilize, and then put it in the freezer.

Your insight is appreciated, thank you.

IIRC, the solubility of virtually all gases in water decreases with temperature. This is the reason why heating water will result in bubble nucleation, and this can be exploited to produce very clear ice by boiling water, perhaps several times, before putting it into the trays.


Solubility in homogenous mixtures of solvents Alcohol and water are miscible.  I am coming across some chemicals that are soluble in alcohol and insoluble in water.  If I know solubility in two solvents separately, is solubility in a mixture of the two always straightforward, or are there confounding variables?

E.g., if some chemical will dissolve to 100g/L alcohol at 25°C, then will its solubility in a 50/50 mixture by volume of water and alcohol at 25°C always be 50g/L?

Does this generalize to all miscible solvents?

The complication comes from solubility of the solvents with each other whilst simultaneously having some solvation power of a solute. I have made a compound before that was so water soluble that it caused sodium chloride to precipitate out. A real example was of an organic compound that had 1%w/v solubility in ethanol but around 6.5%w/v in toluene. In a 50:50 mixture,  the compound had solubility of 13%w/v. A completely non-linear result! The solubility was even higher in 90:10 toluene-ethanol. So, the complex nature of three way interactions mean it would be very difficult to predict these solubility graphs just using two data points of the two pure solvents.
If we asked the same question of boiling points, we would point straightaway to azeotropes. So, analogously, other properties exhibit this behaviour. However,  in this instance the nature of the solute as a third component makes it too complicated to predict without having the wealth of property data available for all components. If this is known and you have the software, you may have a chance.


How to calculate the enthalpy of formation of hydrazine from tabulated values? I've been given a problem asking:

  Estimate the standard enthalpy of formation of hydrazine, $\ce{N2H4(g)}$, from the following data.
  \begin{align}
\Delta{}H_\mathrm{B}(\ce{H2}) &= 436~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N2}) &= 944~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N–N}) &= 163~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N–H}) &= 388~\mathrm{kJ/mol}\\
\end{align}

While I know that $\Delta{}H_\mathrm{B}(\ce{N–N})$ refers to the energy released from breaking that bond, I'm not sure what $\Delta{}H_\mathrm{B}(\ce{H2})$ or $\Delta{}H_\mathrm{B}(\ce{N2})$ refers to. Could someone please help me understand?

In order to "make" hydrazine, $\ce{H2N-NH2}$ from hydrogen and nitrogen atoms, you have to

break $\ce{N2}$ once, giving 2 nitrogen atoms
break $\ce{H2}$ twice, giving 4 hydrogen atoms
form one $\ce{N-N}$ bond
form four $\ce{N-H}$ bonds

You have all the enthalpies that you need. Just pay attention to the coefficients and the signs.


Do chemists refer to water as "dihydrogen monoxide"? Is the name "Dihydrogen monoxide" actually what chemists would use to refer to $\ce{H2O}$ (assuming there was no common name, "water")?

Of course, this is all over the internet. I'm a little skeptical though because the similar chemical $\ce{H2S}$ is called "hydrogen sulfide", not "dihydrogen monosulfide".

No, it's not. The "dihydrogen monoxide" name is used as part of a hoax. In the scientific community, there are chemical names for water, and which one is used in the literature generally depends on how it interacts with something else (hydroxic acid and hydrogen hydroxide were two I heard most often in acid-base reactions).
IUPAC, the standards committee that sets standard names for chemical structures, suggests "oxidane" as a starting point for the construction of other names for chemicals that are derived from water. However, they do not suggest "oxidane" itself be used to refer to plain water.
Most chemists would use "water", even when writing scientific papers.


How does one experimentally determine virial coefficients? The Virial expansion is a systematic correction to the ideal gas law which takes the form of an expansion around the number density, $\rho$, of the gas in question. It looks like this: $$\frac{p}{k_bT}=\rho+B_2(T)\rho^2+B_3(T)\rho^3+\cdots$$
The form of this equation is not surprising. An ideal gas is simply one for which the coefficients $B_2$, $B_3$, etc. are equal to zero. These are called the second and third Virial coefficients.

According to Wikipedia (and I have heard this elsewhere), the Virial coefficients systematically correct the ideal gas law by considering specific interactions. To quote the Wikipedia article,

  The second virial coefficient $B_{2}$ depends only on the pair interaction between the particles, the third ($B_{3}$) depends on 2- and non-additive 3-body interactions, and so on.

This quote gives me quite a good idea of how I would determine these coefficients theoretically using $\textit{ab initio}$ calculations, but I can't imagine how I could calculate only the 2-body interactions experimentally...

Indeed, this paper seems to present one way to determine them by the so-called Burnett method, but I don't have a way of accessing this paper and have never heard of this method.

So, if anyone could give me a summary of how to determine these coefficients, I would appreciate that and am quite interested to hear the answers.

These coefficients can be calculated based on 2 ways:

1) Based on experimental data

In this method, from pressure-volume-temperature (PVT) data these coefficients will be calculated. Consider the following equation:
$$\vartheta\left(\frac{P\vartheta}{RT} - 1\right) = B(T) + \frac{C(T)}{\vartheta} +\ldots$$
$$\vartheta = \frac{1}{\rho} =\mathrm{Molar\ volume}$$

This equation is a re arange form of Virial equation. Now accroding to this equation, if you plot the left hand side of aforementioned equation versus molar density, the intercept of that line represents second virial coefficient and the slope of that line is the third virial coefficient. If the curve was not a straight line, we should consider more virial coefficients for them.

2) From potential functions.

In this method, the virial coefficients are calculated by potential functions such as L-J model, Kihara model, Exp-6 model and etc. The basis of This method is statistical thermodynamics.


Why do molecules having a higher relative molecular mass have stronger inter-molecular forces? Why do molecules like alkanes with higher relative molecular
mass ($M_\mathrm r$) have stronger intermolecular forces? For example, methane ($\ce{CH4}$) has a weaker intermolecular force than pentane ($\ce{C5H12}$). Pentane has a higher $M_\mathrm r$ than methane and therefore pentane possesses stronger intermolecular force than methane. These are members of the same family in the homologous series, known as alkanes.

My issue is why does pentane have stronger intermolecular force than methane? Why is there a relationship between relative molecular mass ($M_\mathrm r$) and inter-molecular forces. The stronger the intermolecular force the higher the boiling point and the melting point?

There are a lot of reasons for why some molecules have stronger IMF's than others, but the trend of increasing IMF for increasing relative molecular mass ($M_\mathrm r$) is due to an increase in London Dispersion Forces, part of the greater set of van der Waals forces. As you may or may not know, these alkanes are non-polar molecules (that is, they don't have a real internal dipole--all $\ce{C-C}$ bonds are non-polar and the electronegativity of the $\ce{C-H}$ bond is close enough that those bonds are also considered non-polar). This matters because polar interactions are a relatively strong IMF.

Let's take a detour into e- distributions. Typically, electrons have the highest probability of being evenly distributed about the atoms in a molecule, proportional to the electronegativity of those atoms. That being said, that is only the most probable state, and so, of course, there are times when the e- are NOT evenly distributed. Remember how alkanes normally aren't polar? Well, when the electrons get unbalanced in the molecule, surprise! An Instantaneous Dipole is temporarily introduced.

What does this mean? Well, when an alkane is next to another alkane that is subject to an instantaneous dipole, the electrostatic forces from the quasi-polarized alkane cause the other alkane to also attain a lopsided distribution of electrons, causing it to gain a temporary polarity, and this effect both attracts these molecules while also propagating throughout the surrounding molecules.

So now that you have a little understanding of what LDF's or instantaneous dipoles are, why the trend with molecular weight? The short answer is that larger molecules have more e-, and are affected by LDF's more than smaller molecules with less e-. With more electrons and more atoms for those electrons to be distributed over, there is a greater chance for frequent or large instantaneous dipoles, increasing the overall Van der Waals forces.


Are allylic three degree alcohols oxidized by the Jones' reagent via a classical carbocation intermediate? I know the mechanism for Jones' oxidation is (picture from Organic Chemistry by Clayden):

and it proceeds through a chromate ester. And in general we understand that the Jones' reagent $\ce{H2CrO4}$ cannot oxidize three degree alcohols.

However, in my book, a question asked to oxidize the following three degree, allylic alcohol:

The solution proceeded via a carbocation intermediate to eventually form 3-methylcyclohex-2-en-1-one.

I wish to ask if this a correct solution. Does this reaction really occur? That too in the way it is described? If so, why isn't it described in any other book, even as an exceptional case? (Most other books simply accept that three degree alcohols cannot be oxidised) Or is this a by-product or a low-yield reaction?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q19; Alcohols, Ethers, Epoxides

It makes sense. Jones's oxidation occurs in presence of a strong acid. Hence, the oxygen can get protonated and leave as a water molecule, giving rise to a tertiary carbocation. The charge can then delocalize, the secondary alcohol get formed, and this alcohol then undergo oxidation. The oxidation will probably drive the equilibrium towards the product.


What factors might affect the reaction between potassium permanganate and ethylene glycol? Our chemistry club has, for the last few years, been showing off some reactions to prospective students. One of the most popular tricks we have is a bit of an explosive one: we mix some potassium permanganate with ethylene glycol in a small bottle. The permanganate oxidizes the glycol, and we usually end up with an intense, three foot flame that shoots out of the bottle, followed by some popping, and finally, the whole thing burning to the ground.

Usually.

Recently, we've been using the exact same mixtures with the same ratios and getting wildly differing results. This entire week, all we've gotten is the glycol bubbling a little and maybe spitting out some of the permanganate, and not much more. The ratios we use are almost the same, so I'm not sure what could be causing this.

Is this reaction sensitive to age of the reactants? Ambient temperature? Humidity? What could cause the reaction to not work normally?

P.S. In spite of the home-experiment tag, the chemicals used here are lab-grade (from Sigma-Aldrich), though I'd be interested in how the purity might affect the reaction as well.

Capsules of potassium permanganate injected with ethylene glycol are used as aerial ignition devices in bush and forest firefighting. They are considered a reasonably safe option because of the delayed reponse prior to ignition. They are sometimes known as dragon eggs or fireballs. The rate of chemical reaction is very much dependent upon

the particle size of the potassium permanganate. Powdered permanganate (greater surface area) works much better than larger crystals.
the concentration of the ethylene glycol. Dilute glycol may not lead to ignition at all. Perhaps your glycol solution is very wet.
ambient temperature. Unless you are trying this in the middle of winter outside in the snow, this is unlikely to be the cause of disappointment.


Calculate accelerating voltage required to separate two particles in a magnetic sector analyser So, in mass spectrometry we have the following equation to describe a magnetic sector analyser,
$$\frac{m}{z}=\frac{B^2r^2e}{2V}.$$

Lets say that I have two particles with different mass/charge ratios,
$$\frac{m_1}{z_1} = 120.9,\quad \frac{m_2}{z_2} = 121.0,$$
and I have $B=2\text{T}$, $r=25\text{cm}$. For example, what accelerating voltage $V$ do I need to apply to separate these two particles? What criteria do I have to determine whether two particles are separated?

** ** EDIT ** **
I don't know where to start with this. So even though you don't generally answer homework-like questions on this forum, some pointers in the right direction would be greatly appreciated!

Baseline separating C-12 from C-13 from C-14 requires accelerator mass spectrometry,

http://en.wikipedia.org/wiki/Accelerator_mass_spectrometry

Your example requires 100 times the finesse.  I'm not optimistic for reduction to practice vs. a paper solution.  Your target could be a charge-coupled device or a channel plate amplifier.  Observe the spatial separation of signals above noise.  A Faraday cup works differently.  Time-of-flight is not optimistic.

You might diddle FT ion cyclotron resonance in a high-Q cavity.


How to name long hydrocarbon chains? It is not that I didn't google it before asking it here, I could it find it anywhere. My question must be simple but I'm really stuck upon it. How do you name long hydrocarbon chains? Like $\ce{C21H44}$, $\ce{C30H62}$, $\ce{C100H202}$ etc.

I found this when I googled but I didn't quite understand it.

Use the table mentioned in the linked article, §NT-1.1:

1   mono- or hen-*  10  deca-       100 hecta-      1000    kilia-
2   di- or do-*     20  icosa-**    200 dicta-      2000    dilia-
3   tri-            30  triaconta-  300 tricta-     3000    trilia-
4   tetra-          40  tetraconta- 400 tetracta-   4000    tetralia-
5   penta-          50  pentaconta- 500 pentacta-   5000    pentalia-
6   hexa-           60  hexaconta-  600 hexacta-    6000    hexalia-
7   hepta-          70  heptaconta- 700 heptacta-   7000    heptalia-
8   octa-           80  octaconta-  800 octacta-    8000    octalia-
9   nona-           90  nonaconta-  900 nonacta-    9000    nonalia-

just enumerate the digits of the carbon and add "-ne".

So $\ce{C21H44}$ becomes "hen-" "icosa-" "ne" = henicosane / heneicosane (the latter is just a spelling variation).

There are exceptions like propane ($\ce{C3H8}$) but these only occur for small molecules, up to $\ce{C4H10}$.


Completely chemically inert metal alloy? Background:
I was a chemistry major and have recently started watching this series of videos by NurdRage.
I learned that Platinum dissolves in Aqua Regia but Ruthenium is only attacked by high temperature halogens and some alkalis.
Question:
Would a Ruthenium and Platinum alloy be resistant to both acid and alkali?
But more importantly, is there any alloy or pure metal that is completely chemically inert to every strong oxidizing reagent? Is there any research about the most chemically inert metal alloys?

Due to electronegativities no alloy would be thermodynamically stable in the presence of fluorine (F$_2$). Even in an alloy the constituent metal atoms are either "sharing" electrons or receiving/donating electrons. Fluorine is the element with the highest known electronegativity and would "pry" the electrons from the electron-accepting metal atoms.


What happens if don't use MgCl₂ in buffer solution in Ca²⁺/Mg²⁺ titration? What happens to a complexometric titration of calcium carbonate (analyte) with $\ce{EDTA}$ (titrant) when no magnesium(II) chloride crystals are added to the buffer ($\ce{NH3/NH4^+ + EDTA}$) solution? In a water analysis (where water sample is the analyte and $\ce{EDTA}$ is the titrant)?

Summary The Magnesium is added because the color change is easier to spot with it, even though calcium also changes the color of the solution in presence of TBT.

Long explanation and experiment

Even though your question might looks silly at first, I know it makes sense because I work with Soil Analysis, and the titrimetric determination of $\ce{Ca^2+/Mg^2+}$ is performed using the same pH 10 $\ce{NH3/NH4^+}$ buffer, with the addition of a small amount $\ce{MgSO4}$ and $\ce{EDTA}$ to the buffer.

As far as I know, and according to a fast online search, the addition of magnesium is to ensure the visibility of the end point when using Eriochrome Black T (EBT) as indicator. Supposedly, the color change wouldn't be easily spotted without the magnesium addition. This is how it's supposed to happen:

Considering there will be no change of color in EBT only due to Calcium, a small amount of magnesium is necessary to interact with the indicator and dye the solution. The equivalent number of moles of EDTA is added to the buffer, otherwise the results would be overestimated. Upon completion of the titration, the remaming $\ce{Mg^2+}$ complexed with TBT would be chelated by EDTA and the endpoint detected.

This is important in basically two scenarios:

There is no magnesium in your sample.
The buffer used is not pH 10, but pH 12, exactly to remove magnesium and detect only calcium.

In both cases, you need some (known) amount of magnesium to see the color change.

The problem with this is that Calcium is supposed to color TBT. It's even the example given on TBT's Wikipedia page, and there's also a full article about the interactions of magnesium and TBT. So why would you add magnesium?

To answer your queston - which also became my own - I did a qualitative experiment, because I'm in the lab right now. And the result was interesting.

I added small amounts of Calcium Chloride and Magnesium Sulphate to two different beakers containing purified water, and then added a few drops of TBT. I also added TBT to the purified water as a blank. This is what I got:

As you can see, the $\ce{Ca^2+}$ solution did change the color of the indicator, but not nearly as much as $\ce{Mg^2+}$ did. If you think about real-sample analysis, where usually the water (or soil extract) contains other coloured materials, making sure the endpoint is easily visible is very important. As a colorblind, I already find this change in color a bit hard, and for $\ce{Ca^2+}$ alone I suppose it would be even harder without the magnesium.

So, finally: The Magnesium is added because the color change is easier to spot, even though calcium changes the color of the solution in presence of TBT.


Is borazine aromatic? I came across this structure which has got pi bonds only in its canonical forms (which are also unstable compared to the Ist structure), is it aromatic?

Also, is it compulsory for an aromatic compound to contain a carbon ring? Can't a conjugate system without carbon be aromatic (as in this case)?

The modern definition of aromaticity from deep theoreticians is that the π-system needs to support aromatic ring currents. Borazine can support it, so it is technically aromatic.

Aromatic systems that do not contain carbon are not really all that common, but they are known. Pentazole was detected, for example. The problem is, that for aromaticity to be a thing we need a π-system, and π-systems effectively restrict us to BCNOS elements, since other elements are not eager to form π-bonds. Aromatic systems are usually 5- or 6-membered rings. Add in the fact that rings with a lot of N/O atoms often happily go kaboom after a funny look, and it's not surprising we have very few actual examples here. Still, some people insist on doing unreasonable things with results from as far back as the 50s.


Can you find the mass of solvent with mass of solute, volume of solution, and solution density? If you have, for example, $\pu{10 g}$ of a solute dissolved into $\pu{100 ml}$ of a solution, and the solution density is $\pu{1.5 g ml^-1}$, can you just multiply $100\ \mathrm{ml} \times 1.5\ \mathrm{\frac{g}{ml}}$, then subtract the $\pu{10 g}$ of solute from the result to get the mass of the solvent in grams?

Does that work or am I trying to do do this wrong?

That works!  In most chemistry (but not nuclear reactions!),  mass is always conserved, so

$$m_\text{solution} = m_\text{solvent} + m_\text{solutes}$$

Mass is also equal to density times volume, so

$$m_\text{solution} = \rho_\text{solution} V_\text{solution} = m_\text{solvent} + m_\text{solutes}$$

Solving for $m_\text{solvent}$:

$$m_\text{solvent} = \rho_\text{solution} V_\text{solution} - m_\text{solutes}$$

Which is exactly the calculation you did.


What is the main contributor of the color of honey? I was told that the color of honey contains so much chemistry and that was interesting to me and after doing a couple of research I found that the color of honey depends on the presence of plant pigments including carotene, xanthophylls, anthocyanins, flavonoids, polyphenols, as well as amino acids and mineral salts. So as we can see there are many factors to the color of honey but my question is which out of all of the listed above is the main contributor and is the most affected by the increase in temperature (for example for acidity the main contributor is gluconic acid).

I found that the color of honey depends on the presence of plant
pigments including carotene, xanthophylls, anthocyanins, flavonoids,
polyphenols, as well as amino acids and mineral salts

Yes, you are right about that. The color differences is due to the difference in levels of minerals, anti-oxidants, and other trace elements that are present. In particular, levels of polyphenols, flavonoids, and carotenoid have been observed to have high correlation with color. Polyphenols are found in high concentration in darker honeys, whereas flavonoids are found in higher concentrations in lighter honeys.
References

Antioxidant and antimicrobial capacity of several monofloral Cuban honeys and their correlation with color, polyphenol content and other chemical compounds by Jose M.Alvarez-Suarez Sara Tulipani, Daimy Díaz, Yadiley Estevez, Stefania Romandini, Francesca Giampieri, Elisabetta Damiani, Paola Astolfi, Stefano Bompadre, Maurizio Battino, Volume 48, Issues 8–9, August–September 2010, Pages 2490-2499, DOI: https://doi.org/10.1016/j.fct.2010.06.021
Total Phenol, Antioxidant Activity, Flavonoids, Anthocyanins and Color of Honey as Affected by Floral Origin Found in the Arid and Semiarid Mediterranean Areas by
Taha M. Rababah  Mohamad Al‐Omoush  Susan Brewer  Mohammad Alhamad  Wade Yang  Mohammad Alrababah  Abd Al‐Majeed Al‐Ghzawi  Muhammad Al‐ u'datt  Khalil Ereifej  Fawzi Alsheyab  Ranya Esoh  Ali Almajwal, Volume38, Issue3, June 2014, Pages 1119-1128, DOI: 10.1111/jfpp.12071


How to write the redox reaction of chlorate to chlorine and oxygen using half rections? I know how to solve redox reactions, but this one is really confusing me since I can't really write half reactions for it.

$$\ce{ClO3- -> Cl2 + O2}$$

The answer is apparently:

$$\ce{2H+ + 2ClO3- -> 5/2 O2 + Cl2 + H2O}$$

This makes sense. I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine(V)'s so you have to use a water to make the charge balance work out. I just don't see how you'd come to this conclusion on your own.

This is tricky and one way to think about it that I can offer is to treat $\ce{ClO3-}$ as a combination of $\ce{Cl^5+}$ and $\ce{O^2-}$. Since the problem is that the starting material is undergoing disproportionation with different elements being reduced/oxidised, if you could split the starting material into the two different elements, then you could circumvent the issue. Obviously the bonding in the chlorate ion is not ionic but this procedure allows you to construct "half-equations" and then an overall equation.

I would not suggest that you write this in an exam.

The two half-reactions are then:

$$\begin{align}
\ce{Cl^5+ + 5e-} &\longrightarrow \ce{1/2Cl2} \tag{1} \\
\ce{O^2-} &\longrightarrow \ce{1/2 O2 + 2e-} \tag{2}
\end{align}$$

$[(1) \times 2] + [(2) \times 5]$ gives

$$\ce{2 Cl^5+ + 5O^2- -> Cl2 + 5/2 O2}$$

In order to "re-form" your chlorate, you need three oxide ions per Cl ion, so you need to add one more oxide to both sides:

$$\ce{2 Cl^5+ + 6O^2- -> Cl2 + 5/2 O2 + O^2-}$$

Now you can get back your chlorate:

$$\ce{2 ClO3- -> Cl2 + 5/2 O2 + O^2-}$$

Now it's highly unlikely that there are going to be free oxide ions hanging around, so we just protonate them by adding $\ce{2 H+}$ on both sides. The oxide ion simply becomes water:

$$\ce{2 ClO3- + 2H+ -> Cl2 + 5/2 O2 + H2O}$$

  I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine (V)'s so you have to use a water to make the charge balance work out.

Yeah, if all six oxygens became $\ce{O2}$ then they would lose 12 electrons. Therefore, one of the oxygens does not become $\ce{O2}$; it becomes $\ce{H2O}$ and remains as oxygen(-2). It is not about charge balance.


Can rocket candy explode during the preparation? I would like to follow the steps in this video: https://www.youtube.com/watch?v=hyPysthIRss

After I mixed the Potassium nitrate KNO3 and the sugar, I start to cook it in a pot, and stir it very often. After 25-30 min, the KNO3 and the sugar will melt together, and become a brown mass.

My question is: Can this mixture explode during the heating somehow? Or it can just burn, not explode? How to prevent it, and do it the most secure way?

See James Yawn's website ${\it Recrystalized\; Rocketry}$, which is all about solid propellant rocketry of the type you're interested in. It's really an exceptionally comprehensive site.

Can this mixture explode during the heating somehow?

It appears that it's more likely to catch fire, and do so rather vigorously, as in cases where it's exposed to a spark or an open flame. Because the system (pan and mixture of $\ce{KNO3}$ and sugar) would be open to the atmosphere (and not confined to a fixed volume), there's more of a risk of fire than of an explosion. As reported by James Yawn in the subsection on "Skillet Rcandy":

This stuff looks innocent enough in the pan.  But if you have ever seen a quantity this size burn you will have more respect for it.  This batch could send a tongue of flame 20 feet in the air and engulf the person stirring it, if only briefly.  Burns to the body would likely be second degree, and I am willing to tolerate that level of risk.  But eyeballs don't recover from this kind of trauma easily.

See this safety page from James Yawn's site.

Or it can just burn, not explode?

Yes, it can just burn: that's essentially what's going on in the rocket engine. The exhaust from the burn is channeled in one direction (${\it via}$ the hollow core in the engine itself), resulting in thrust in the other direction.

How to prevent it, and do it the most secure way?

From James Yawn's site:

Basic Safety Precautions
At present, I do NOT recommend making this propellant in a household kitchen.
Make small batches only!  No more than one standard recipe, which yields about 160 grams of propellant.
Wear protective gear. You should have a full-face shield, long sleeved heavy cotton shirt, heavy apron and heat-resistant gloves.  These must be worn at appropriate points in the process.
Do not allow children or rambunctious pets in the room where you are working.
Do not allow distracting, inattentive, inebriated or otherwise incompetent adults in the room, especially if they smoke.
Be vigilant of possible sources of sparks, flame, or high temperature that could ignite propellant on contact.
If you smoke, place your cigarettes and lighter in another room so you will not light up unconsciously.
Clear your work-area of any flammable materials.
Have a good fire extinguisher available. Also a 5-gallon bucket of water nearby and a garden hose outside.
Make sure there is good ventilation, and a quick escape route to fresh air.
Avoid stupid rocket tricks.
Avoid excessive exposure to Potassium Nitrate.

All of the above points are elaborated upon at the linked page.


Photodissociated iodine laser and population inversion Iodine molecules ($\ce{I2}$) can absorb in the visible region and dissociate into $\ce{2 I^.}$ radicals. One of the I atoms is in ground electronic state $\mathrm{^2P_{3/2}}$ and the other I atom is in the excited $\mathrm{^2P_{1/2}}$. So, photodissociation of $\ce{I2}$ produces 50% of the species in the excited state and 50% in ground state.
The $\mathrm{^2P_{1/2}\rightarrow^2P_{3/2}}$ of iodine has been extensively used in laser production, in chemical iodine laser or methyl iodide photodissociation laser for example. But in those cases, most of the $\ce{I}$ atoms are produced in the excited state. The criteria for laser production is population inversion, where >50% of the species are in the excited state.
Here I am not sure if the photodissociation of iodine molecule satisfies the criteria of population inversion, as exactly 50% of the species is in excited state. So my question is—can we consider 50% excited species as population inversion and if so can photodissociated $\ce{I_2}$ generate laser?

You are making a chemical laser. A laser depend on keeping a positive difference in population between the two levels involved (upper >lower), and of course feedback for a laser oscillator. It does not matter how many I* you make as long as the lasing transition does not go the the lowest ground state level. You must make the transition end at a higher level in the ground state I atoms that has no (i.e. insignificant) thermal population, thus a make a 3 level laser meaning that the laser radiation cannot pump  this lower level to the higher level from which lasing occurs.  You must of course keep producing I* atoms.


Acidic strength of oxides of nitrogen Why do we compare acidic strength of oxides of nitrogen even though they don't have hydrogen?

Compounds containing hydrogen are not only acidic but we have other concepts also. Like in your case it can be explained through Lewis acid-base concept.

Lewis concept: A base is defined as a substance which can furnish a pair of electrons to form a coordinate bond whereas an acid is a substance which can accept a pair of electron.

For example, $\ce{NH3}$ is a Lewis base since on nitrogen we have one electron pair which easily be donated to form a coordinate bond. As for acid, Take any two oxides of nitrogen for example $\ce{NO2}$ and $\ce{N2O5}$ . First check the oxidation state of nitrogen in two cases, here it is +4 and +5 . Since a substance which can accept a pair of electrons is an acid, so the nitrogen with more positive oxidation state will relatively accept pair of electrons with more easily, therefore $\ce{N2O5}$ is a better acid then $\ce{NO2}$.


Create my own solar cells? How would I grow my own and build functional solar cells? I've read and researched that they're mostly a crystal structure of something like silicon. I would assume I'd be able to grow my own crystals and sandwich this in a process of some sort with a p-type and n-type, etc.

You have to study existed architectures. First, it is necessary to distinguish between photo-conductive and photovoltaic elements. In photovoltaic elements one needs a mechanism of separation of positive and negative charges formed under an irradiation of the semiconductor. Even a simple p-n junction can do this. But, this is all about efficiency. The solar cell has to absorb as much light as possible without reflections, and, at the same time, the charge transport inside should be without losses that is impossible in thick structures. Also we have to care about the absorption spectrum. To absorb in all spectrum of the solar radiation, people make cascade structures with different materials absorbing in different ranges of spectrum.


Chemical methods to extract Oxygen from Rust Let's say I have a large quantity of rust and want to extract the oxygen from it in a way that allows me to collect the oxygen in a container.

What chemical process can I use to do this and what equipment does it require?

It's preferable if the process is doable at home so I can demonstrate the process but it's not mandatory.

Good question; unfortunately, no good answers. The problem is that oxygen, pure oxygen, doesn't like to exist as such; it will form a diatom like many gases, but it's very electronegative, and so it will prefer to get the extra electrons from something more willing to donate them, like the iron. This is why "oxidation" has that name; until fluorine was isolated in the mid-1800s, oxygen was the strongest known electron acceptor ("oxidizer"). It's easier to run a reaction that gives the oxygen something else to oxidize, producing the raw metal, than to get the iron to complex with something else and release the oxygen. In at least one circumstance, this is very good for humans and other aerobic organisms; oxygen complexes with the iron in your hemoglobin for transport to your cells. In most other circumstances, we could do without this behavior.

Rust is insoluble in water but will dissolve in strong acids. It will react with hydrochloric acid to form iron chloride and water (sometimes producing hydrogen gas if there isn't enough oxygen to go around). The easiest "household" method to obtain oxygen from iron oxide, therefore, is to dissolve the rust in muriatic acid forming iron chloride and water, neutralize any remaining HCl by adding lye to form sodium (and/or potassium) chloride and water, distill the water from the chloride salts by boiling and then cooling the vapor, then electrolyze the water with a battery to produce oxygen and hydrogen. That's cheating, though. First, water's water, and you're adding far more to the party with the acid and base solutions than you would ever get from the rust. Second, muriatic acid and lye, while available in relatively pure form for pool maintenance and for soapmaking respectively, are not exactly everyday household chemicals in those forms (both are components of other household chemicals, such as drain cleaner and lime scale remover, but are somewhat difficult to isolate from them).

Theoretically, if you heat the rust enough, the components will dissociate and you'll be left with pure iron and oxygen. Unfortunately, even with carbon in the mix (producing elemental iron metal and carbon dioxide, which is the reaction behind iron ore refinement and recycling), the minimum temperature required for this reaction to become favorable is 842K = 569*C ~= 1056.2*F. I can't find data on iron-oxygen bond enthalpy or entropy, which is needed to calculate the spontaneous decomposition temperature of the compound, but logic states it's much higher than with added carbon, because you're not getting any energy release by forming the carbon-oxygen bond, so every Joule of energy needed to break the iron-oxygen bonds has to come from outside the material. Theoretically it would cost you about 2.5kWh/kg of heat added to the iron; in practice the amount of heat you'd have to generate (and lose to various inefficiencies) to get the metal that hot is beyond any household device's capabilities, unless your hobby happens to be glassblowing.

The "thermite reaction" between iron oxides and aluminum under heat is another well-known method of separating the iron and oxygen. The rust and aluminum powder, given some initiating heat, will exchange the oxygen to form alumina (Al2O3) and pure iron, and enough heat to keep the reaction going (with plenty to spare; a thermite reaction is too bright to watch without a welder's mask and in fact is used in some situations to weld iron). Two problems with this; first, it's illegal in the US to make thermite without BATFE approval, and second, the reason it works at all is because aluminum oxide is ridiculously stable, more so than iron oxide, and so it takes even more energy to reduce the oxide back down to the raw metal (about 8.2 kWh/kg for aluminum, theoretical, versus 2.25kWh/kg for iron). This is evidenced by the highly exothermic nature of the thermite reaction itself; if creating aluminum oxide produces far more than enough heat to smelt the iron to get the oxygen, think of how much would be needed to go the other way.

Pretty much the only direct route to releasing oxygen gas from rust without a blast furnace is to introduce an even more powerful oxidizer. There are very few known to modern science, and none of them are "household chemicals" to say the least. The easiest to work with would be elemental fluorine gas, and the experienced chemists in the crowd will know that for the extremely relative statement that it is. Fluorine is the most electronegative element in the known universe, and so pretty much any substance that isn't already highly fluorinated (like some polymers, such as Teflon) will react pretty vigorously with it. That includes the water, proteins, enzymes and other compounds in your own body; a fluorine gas leak is a "drop everything and run" type of emergency. As used on rust, the fluorine will simply barge in and displace the oxygen from the rust to form a white to pale green iron fluoride, liberating the oxygen as a gas. You then have to separate the oxygen from the excess fluorine, and one of the standard ways in industry to catch excess fluorine is to pass the exhaust gas over or through a powdered sodium oxide filter, which will form sodium fluoride and liberate more oxygen in a very similar reaction as you got with the iron. Sodium oxide isn't a pleasant chemical to work with either; add water (such as by breathing in any loose powder) and it forms the strong base sodium hydroxide, which eats flesh about as efficiently as anything else you could think of.


Software for determining symmetry elements and point group of drawn molecules The title represents my overall question. I would like to have a software that able to draw molecules (especially coordination complexes) and determine the symmetry elements and point group of the molecules. I have a PC with Windows 10 system and an iPad Pro.
Thank you.

Coordination compounds / complexes may be a bit tricky in terms of «what do you define as a bond» (covalent / dative bond) and the convention what a stroke between atoms expresses (e.g. in the representation in compounds like ferrocene).  But if you already have a .mol or .pdb file in available, you could try Jmol:

(credit, includes some details how to use the program this way)
While the example is about methane as a simple organic molecule, Jmol is well capable to interact with crystal structure models, and displaying symmetries of the unit cell, too:

(credit, Hanson, J. Appl. Cryst. 43, 2010, 1250-1260 [open access])
including interactive galleries like Symmetry@Otterbein both about molecular and crystal symmetries.


Rate of reaction of silver salts of carboxylic acids having different alkyl groups towards Hunsdiecker Reaction Hunsdiecker reaction follows free radical mechanism but it is given that rate of reaction with different alkyl groups (R) attached to silver salts of carboxylic acid follows the order as follows

Primary > Secondary  > Tertiary  (Rate)

But free radical stability is exactly opposite!

Primary < Secondary < Tertiary  (Stability Free Radicals)

What can be the reason for this kind of trend in this reaction?

I'm attaching the reaction mechanism for reference.

If the free radical is less stable, it will react faster.
So once the system is started, the propagation (step 4) will proceed faster for primary radicals.
That means that it is the limiting step and that step 3 is faster than 4 for any type of R.


Why has the mole been redefined for 2019? The mole is still defined as an SI unit representing the number of constituent particles (whether they be atoms, molecules etc.) in 12 grams of carbon isotope 12 (carbon-12).

Effective from May 20 this year the definition of the mole will change to a unit representing Avogadro constant number of particles.

Since the Avogadro constant was and will be still tied to the mole (number of particles in a mole), I think this also changed the definition of the Avogadro constant, though I'm not sure. Although the old and new definitions define the constant as the number of particles contained in the mole, it seems that the definition of the mole has changed from the earlier carbon-12 definition to simply the Avogadro constant.

First of all I would like to know if anyone knows why this redefinition happened now (it was decided at the end of the last year).

Also, please correct anything wrong I've said. This is rather confusing for me and the definitions seems rather circular. The new definitions are that Avogadro's constant is the amount of particles in a mole, and a mole is defined by the Avogadro constant.

  Previous definition:
  The mole is the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.
  2019 definition:
  The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly 6.02214076 × 1023 elementary entities. This number is the fixed numerical value of the Avogadro constant, NA, when expressed in the unit mol−1 and is called the Avogadro number.
  Avogadro constant - Wikipedia

You have already mentioned the last definition of the mole before the proposed change:

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is “mol”.
(…)

According to this definition, the mole still depends on the definition of the kilogram. The main purpose of the proposed change is to find an independent definition. Another purpose is the possibility of linking the definition to an exact value (similar to the speed of light.)
The formal decision for the change was made during the 26th meeting (13–16 November 2018) of the General Conference on Weights and Measures (CGPM).
The motivation behind the change of the definition of the mole can, for example, be found in the resolutions of the 24th meeting (17–21 October 2011) of the General Conference on Weights and Measures (CGPM); however, the discussions and preparations started already much earlier.

On the possible future revision of the International System of Units, the SI
(…)
considering
(…)
that of the seven base units of the SI, only the kilogram is still defined in terms of a material artefact, namely, the international prototype of the kilogram (1st meeting of the CGPM, 1889, 3rd meeting of the CGPM, 1901), and that the definitions of the ampere, mole and candela depend on the kilogram,
(…)
that it is also possible to redefine the mole so that it is linked to an exact numerical value of the Avogadro constant $N_\mathrm A$, and is thus no longer dependent on the definition of the kilogram even when the kilogram is defined so that it is linked to an exact numerical value of $h$, thereby emphasizing the distinction between amount of substance and mass,
(…)


Gas spectra in the 400 nm - 1000 nm range Disclaimer: My background is computer science.

Is it possible to detect any gas using a visible light and near infrared (400 nm - 1000 nm) hyper spectral imaging camera? This can be related to gas leaks, pollution, gasses released in fires... From what I found when googling, this is not really possible, but I though to check here to be sure.

The answer is that you can detect all types of gases if you have the correct light source and use Raman spectroscopy. Raman is more adaptable than absorption or emission spectroscopy as it is a scattering process and the laser does not have to operate at the wavelengths that the molecules absorb, thus one laser can be used for all samples. A suitable source is going to have to be a narrow band laser (although when Raman discovered this effect he used sunlight). You will also need notch -filters to remove the laser wavelength and prevent it from saturating your detector. Not cheap.


Structure of lead(II,IV) oxide I need help sorting out the molecular structure of tri-lead tetroxide, aka lead(II,IV) oxide $\ce{Pb3O4}$.

This data sheet shows this image:

The Wikipedia's lead (II,IV) oxide's link to a Jmol interactive image looks like this:

Even the Wikipedia article's Talk Page's discussion on the naming and structure makes for interesting reading.

The first drawing shows a $\ce{Pb-O-Pb-O-Pb-O -}$ ring, the second shows a 4-bond $\ce{Pb}$ in the middle with two 2-bond $\ce{Pb}$'s on either side.

Is there a "right way" and a "wrong way" to draw a single molecule of $\ce{Pb3O4}$ that represents how it might actually be configured? Are both, either, or neither of the above representations "right"?

What you call a molecular structure is in reality an asymmetric unit. These two terms are not related, and in general are not interchangeable. The same way $\ce{Pb3O4}$ is not a molecular formula, but a formula unit.

All pictures you presented are correct in terms that they do represent an asymmetric unit, but actually an IUCr-recommended asymmetric unit (can be generated with Olex2's command standardise) looks like this:

$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#575961}{\Large\bullet}~\ce{Pb}$.

... and with $Z = 4$ these are packed in a unit cell with $Pbam$ space group accordingly:

The crystal structure itself can be represented as infinite layers of cross-linked chains made of $\ce{[PbO6]}$ octahedra, which are stacked via $\ce{- O-Pb-O -}$ bridges:

Crystallographic data is taken from ICSD #9753 [1].

Gavarri, J. R.; Weigel, D.; Hewat, A. W. Journal of Solid State Chemistry 1978, 23 (3), 327–339. DOI 10.1016/0022-4596(78)90081-6.


Finding the concentration of each ion in saturated solution of lead(II) hydroxide Question

If the solubility of $\ce{Pb(OH)2}$ is $\pu{0.155 g/L},$ then the concentration of each ion in a saturated solution of $\ce{Pb(OH)2}$ is
A. $[\ce{Pb^2+}] = \pu{0.155 g/L}$ and $[\ce{OH-}] = \pu{0.155 g/L}$
B. $[\ce{Pb^2+}] = \pu{0.052 g/L}$ and $[\ce{OH-}] = \pu{0.103 g/L}$
C. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{1.29E-3 mol/L}$
D. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{6.43E-4 mol/L}$

Answer

 C. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{1.29E-3 mol/L}$

My Approach
What I did is that I wrote the dissociation equation:
$$\ce{Pb(OH)2 <=> Pb^2+ + 2 OH-}$$
Then I used to mole ratio to discover that:
$$
\begin{align}
γ(\ce{Pb^2+}) &= \pu{0.155 g/L}\\
γ(\ce{OH-}) &= \pu{0.31 g/L}
\end{align}
$$
I then converted the mass concentration $γ$ into molar concentration $c$ since the $\pu{mol/L}$ is commonly used to describe solubility:
$$\pu{0.155 g/L} × \frac{\pu{1 mol}}{\pu{207.2 g}} = \pu{7.48E-4 mol/L}$$
However, that's not the answer. What did I do wrong?

The mistake you made is that you assumed mass concentrations of the ions from the stoichiometric relation, which is only applicable to the amount of substance, and, as a consequence, amount concentration.

You have to convert mass concentration $γ$ to the molar concentration c first:

$$c(\ce{Pb(OH)2}) = \frac{γ(\ce{Pb(OH)2})}{M(\ce{Pb(OH)2})} = \frac{\pu{0.155 g L-1}}{\pu{241.21 g mol-1}} = \pu{6.43E-4 mol L-1}$$

And, since it's a saturated solution, then a nearly complete dissociation can be assumed:

$$[\ce{Pb^2+}] = c(\ce{Pb(OH)2}) = \pu{6.43E-4 mol L-1}$$

$$[\ce{OH-}] = 2 × c(\ce{Pb(OH)2}) = 2 × \pu{6.43E-4 mol L-1} = \pu{1.29E-3 mol L-1}$$


Trichloride ion (c.f. triiodide ion) While triiodide ion is commonly known, why is that trichloride ion (and its salts) almost unheard of? While it does exist according to literature, why is the stability of trichloride ion in aqueous environment much worse than its bromine and iodine counterparts? Aren't elements down the group bulkier and hence forming weaker covalent bonds with longer bond lengths?

As @nilay points out, we can make "polychlorides" with "nonclassical" stoichiometries. But for more typical laboratory conditions it is true that heavier halogens are more favorable for forming polyatomic ions.
This is because polyhalide ions involve two competing forces. On the one hand, the delocalized covalent bonding gives more stability to a polyatomic ion than we would have with the separate monoatomic ions and diatomic molecules. On the other hand, surrounding counterions or dipoles are more strongly attracted to compact, spherical monoatomic anions than to bulky polyatomic ones. With lighter halogens forming the most compact monoatomic anions, the electrostatic effect wins out and we need severe conditions to overcome it. Heavier halogens, with larger atoms and more polarizable electrons, make the polyatomic ions more competitive, as in $\ce{C_5H_6N+Br_3−}$ (http://www.organic-chemistry.org/chemicals/oxidations/pyridiniumhydrobromideperbromide.shtm).


Distilled water to clean greasy stains: Why does that work? I noticed that distilled water works well to clean away fingerprints on touchscreens or oil stains on a ceramic field after frying an egg.

But water and oil/fat are not supposed to mix, so why does that even work?

There may be two different factors at play here.

First, it isn't necessarily the water that is doing most of the cleaning. I presume the water is used to wet a cloth and the cloth then wipes the surface. So the cloth may then physically remove surface contaminants. The role of the water may be just to improve the cloth's surface contact hence improving its ability to physically remove the contaminants.

Secondly, distilled water might do a better job than tap water as it has fewer dissolved solids which would leave other stains on the surface when the water evaporates.

However, the only way to be sure is to do a comparative experiment using different cleaning fluids. You might try tap water, distilled water and distilled water with a surfactant (e.g. a liquid soap). You might find that the best result uses the surfactant but that pure distilled water does a less good, but good enough job. Science isn't about single isolated experiments: we usually need to compare alternatives to know what is best.


SF3Cl Vsepr Most stable structure I was asked to " Use VSEPR theory to predict the most stable structure for $\ce{SF3Cl}$ and justify your answer. "

I managed to interpret the structure as seesaw and with a lone pair on the equatorial position. However, I had chosen that the most stable structure was that with Chlorine on the equatorial position because I thought that the Chlorine-Sulfur bond length would be greater than that of Chlorine-Fluorine.

However, the answer key stated the Fluorine was more electronegative, and therefore would be more stable on the axial (with chlorine as equatorial).

I was wondering if the answer key was accurate and if so, how would I compare the effects of electronegativity and bond length? I had considered electronegativity but I had thought it would simply make the bond even shorter, increasing repulsion.

Thanks!

The bond lengths have little to do with the stability of the atom positioning around the central $\ce{S}$ atom. Rather, the relative sizes of the electron clouds are what should be considered when positioning these atoms. An atom of $\ce{Cl}$ has a 8 more electrons than does an atom of $\ce{F}$, so naturally its electron cloud will be larger and therefore more repulsive. The $\ce{Cl}$ should therefore be positioned in the equatorial position for the same reason you predicted that the lone pair would be present in the equatorial position--because these positions offer the least repulsion between the electron clouds of the atoms.

In the axial position, the 3 equatorial positions, present at 90 degree angles to the axial position, repel the electron cloud.

In an equatorial position, however, only the 2 axial positions are present at 90 degree angles to repel the electron cloud. The other 2 equatorial positions are positioned away from the equatorial bond in question, and repel its electron cloud only slightly. For this reason unbonded electron pairs and atoms with large electron clouds occupy these spaces first-- they offer the least repulsion and the most stable positioning.


Does storing a hygroscopic reagent in a bottle makes it prone to absorb humidity from the air trapped inside? I've got a full bottle of 99,5% acetone, which I'd really like to keep it as dry as possible, though without using molecular sieves or other dessicants.

Once I open the bottle for the first time, use part of the reagent (say 25% volume) and then close the bottle again, will the remaining 75% acetone begin to absorb water from the 25% trapped air inside the bottle?

Also, does vapor pressure plays a role on this?

Does storing a hygroscopic reagent in a bottle makes it prone to absorb humidity from the air trapped inside?

Yes, there is that tendency. I'll make a couple of points which I'll number for possible further discussion.

(1) 99.5% acetone is probably 0.5% water. So the "extra" water from the air above the acetone in the bottle (from one opening and closing of the cap) would be negligible compared to the amount of water already in the acetone. Obviously when using the acetone, uncap it and recap it ASAP to keep it as dry as possible. The "problem" is opening and closing the bottle many times.

(2) I looked quickly at Wikipedia, and do a quick a google search and didn't find any water-acetone azeotrope. So the distillate would be "pure".

(3) I'm not sure how pure a simple distillation in air of the acetone would be in removing water. I'd guess that the resulting purity would be like your 99.5% starting purity.

(4) Vapor Pressure would play a role in that the acetone would have a partial pressure in the bottle. So not all of the void in the bottle would be air. Acetone doesn't "boil" at room temperature though so the partial pressure won't be enough to keep some air from seeping in and out of the bottle. In other words the void in the bottle is mostly air with just some acetone.

(5) The !@#$%^&*( bottles "breath" as the air pressure changes if they are not absolutely sealed.


MaVa = MbVb - Limitations and Uses We know that for strong acids with a $1:1$ mole ratio between all particles involved are suitable for use with the aforementioned formula for titration problems. I want to know the limits on this formula, that is:

Consider if we had $$\ce{H2SO4(aq) + 2KOH(aq) -> 2H2O(l) + K2SO4(aq)}$$

Since the moles of $\ce{H+}$ = moles of $\ce{~^{-}OH}$ in order to neutralize, how do we apply $M_aV_a=M_bV_b$ to this situation, if at all possible?

Secondly, does this formula work if we had weak acids? I would think not, because with weak acids we don't have the benefit of complete dissociation and thus not a $1:1 = \text{acid}:\ce{H+}$ ion ratio nor a $1:1 = \text{base}:\ce{~^{-}OH}$ ratio.

Regarding the ionization extent of sulfuric acid solutions:

$\text{1)}$ The first ionization is complete at most molarities (until you start reaching extremely high molarities - i.e. $\text{18 M}$). So sulfuric acid is a strong acid.

$\text{2)}$ The second ionization (i.e. the ionization of the hydrogen sulfate ion, $\ce{HOSO3^-}$, can be complete at low starting concentrations of hydrogen sulfate ion. This is because $K_a(\ce{HOSO3^-}) = 1.2 \times 10^{-2}$ is rather sizable!

So we can generalize these two cases of sulfuric acid solutions:

For very dilute sulfuric acid solutions:

$[\ce{H3O^+}]=2M_i - [\ce{HOSO3^-}] = [\ce{HOSO3^-}]+ 2\times[\ce{SO4^{2-}}]$, in which we generally take $[\ce{HOSO_3^-}]$ as negligible (i.e. = $\epsilon \to 0$).

For non-dilute solutions of sulfuric acid, we only have these relationships:

$$[\ce{H3O^+}]=M_i +[\ce{SO4^{2-}}] = [\ce{HOSO3^-}]+2[\ce{SO4^{2-}}]$$

As we can see, the first relationship (one for very dilute solutions) follows from the second because in a very dilute solution, $[\ce{HOSO3^-}]\approx 0$ - i.e. it has fully ionized into $[\ce{SO4^{2-}}]$.