What is a “spectator mode” in this analysis of molecular vibration modes I'm reading a paper in which I found this line:

  The other modes have relatively small contributions for the displacements of the heavy atoms and can be considered as spectator modes for the internal conversion

I cannot find a definition or explanation of term “spectator mode”. What is it exactly?

The paper in question is Ab Initio Trajectory Surface-Hopping Study on Ultrafast Deactivation Process of Thiophene. From a quick Google Scholar search that returns 187 hits, it is clear that the expression “spectator mode” for a harmonic vibration mode is not a standard terminology, but is nonetheless used by plenty of authors.

As to the meaning, I think you got it from the everyday meaning of “spectator”: they are vibration modes that do not participate in the phenomenon of interest, i.e. that are not coupled to the chemical or physical process studied (which can be a chemical reaction, an adsorption process or a photophysical process).


Why is the Gaussian Window Function used to enhance resolution in 1D NMR spectra? In a recent question it was explained that Gaussian window functions are not used in 2D NMR spectra as they can introduce artifacts into the spectrum baseline by truncating the FID. Why then are they used with 1D spectra instead of sin or cosine functions?

Under the conditions you measure a plain 1D proton spectrum, you usually never have any roblems with truncation of the FID. You have a relatively long relaxation delay, you don't gain anything by making your acquisition time much shorter than it can be. So a default setup might measure 64k or even 128k points, which is far, far larger than the values you use in 2D NMR.

In two-dimensional spectra there are several aspects that severely limit the number of points you measure:

Measuring more points in the indirect dimension takes time, if you want twice the points you have to measure twice as long. For that reason you only measure as much points in the indirect dimension as you have to, no more.
Processing can become a bit slow and take quite a bit of space when you increase the number of points in the direct dimension in a 2D spectrum. It's not a big problem with current hardware, but with a 3D (or higher) spectrum this gets much, much worse.
If you need to decouple during acquisition time, the acquisition time is severely limited by the probe hardware and heating effects of the decoupling.

So a typical 1D spectrum will have a lot more points measured, beyond the point where the FID has decayed below the noise. In such a case you won't get any truncation artifacts, as the FID was never truncated to begin with. You only need to worry about truncation artifacts if you actually truncate the FID.

In the following image you can see

lila: QSINE (SSB 4) 10k points in FID
green: GM (LB -1, GB 0.5) 10k points in FID
red:  QSINE (SSB 4) 2k points in FID
blue: GM (LB -1, GB 0.5) 2k points in FID

You can clearly see the truncation artifacts in the gaussian-processed spectra with 2k points, but already at 10k points they are not visible anymore. The quadratic cosine of the truncated FID doesn't look all that nice, but it suppresses the truncation artifacts much better than the gaussian window function.

The SSB 4 I used for this example is also a far more aggressive setting than I'd usually use. I chose it because it gives a relatively similar resolution as the gaussian I compared it to. Usually I'd use SSB 2 for most 2D spectra, which emphasizes signal/noise over resolution.


Controlling the shape of metal Is there a way to control the shape of metal? Suppose I have a fast melting metal, is there a way to make it take a specific shape? Of course I don't mean to like put it in a template to take the shape of a template, but rather to control the shape using a computer or something. Is something like this possible or done? If not possible, then why?

Molten or solid at the end?  It can be shaped in an induction field (liquid)  or dimagnetically levitated (liquid to solid).  Magnetoforming, hydroforming.  Spin molten into into an oblate ellipsoid (solid).

http://www.capturedlightning.com/frames/shrinkergallery.html
Z-pinch and such.

Munroe effect for jets and planes (angle charges), and self-forging munitions.


How to know which kind of distortion takes place due to Jahn-Teller effect I have gone through a few videos and a few articles (including Wikipedia's) about Jahn-Teller distortion effect and I'm not able to follow if there is a rule which governs which kind of distortion will take place in which compound. If I want to remember which compound has which distortion, should I remember a rule or memorise the whole list?

If you read and watched videos, you probably already know that the Jahn-Teller effect causes stabilisation in the case of unevenly populated orbitals. Reducing the symmetry of the system renders the orbitals no longer symmetrically equivalent thereby creating an energy difference used to the compound’s advantage.

Let’s separate two cases: 1) octahedral complexes with $\mathrm{d^4}$ and $\mathrm{d^9}$ configuration, in which the $\mathrm{e_g}$ orbitals are unevenly populated; and 2) all the other cases.

Uneven $\mathbf{e_g}$ population

In an octahedral complex with four or nine d-electrons, there will be either one or three within the $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$). It is unfavourable for orbitals of the same energy (as symmetrically equivalent orbitals have) to have different populations so the system distorts itself. The distortion will always be along a single axis and per defintion this axis is always the $z$ axis. Thereby, ligands on the $z$-axis (negative charges in the most simple crystal field model) move away from the metal centre stabilising any orbital with a contribution in $z$ direction and most significantly the $\mathrm{d}_{z^2}$ one.

The effect is greatest in this case because the $\mathrm{e_g}$ orbitals actually point towards the ligands thus rendering them most susceptible to distance effects.

All other cases

In all other cases, the effect is much weaker. This is mainly because no other orbitals are known to point directly at the ligands; even for tetrahedral complexes the $\mathrm{t_2}$ orbitals point in-between them. Therefore, moving a ligand away slightly has much less effect on the electronics of the system.

What exactly happens really depends on the system itself and how it can be stabilised. If you are speaking of an octahedral complex, it will usually distort itself to $D_{\mathrm{4h}}$, i.e. equivalently to the former case because that is the distortion that keeps the highest possible symmetry. In that case again, one would say that the ligands in $z$ direction are moved away. Otherwise you need to figure it out by drawing, thinking and guessing or by studying crystal structures.

Tl;dr

Instead of memorising a list remember the first case and $z$-distortion. Ignore all the other cases for being weak.


Nomenclature of compounds by the IUPAC Are all compounds named according to International Union of Pure and Applied Chemistry (IUPAC) recommendations or is there another system?

Certainly, the most important organization for chemical nomenclature is the International Union of Pure and Applied Chemistry (IUPAC). In most cases, the various IUPAC recommendations represent the best references.

However, not all IUPAC recommendations are universally accepted. For example:

The correct internationally standardized name of element 111 is ‘röntgenium’; IUPAC recommendations spell it ‘roentgenium’, which is explicitly considered wrong by international standards.
According to current IUPAC recommendations, the formula of ions like sulfate is written as $\mathrm{SO_4{}^{2-}}$, which has not yet gained much acceptance. According to ISO and various national standards, the correct internationally standardized notation is $\mathrm{SO_4^{2-}}$.
In the 1970 IUPAC recommendations, a chelating ligand with two coordinating atoms was indicated by the term ‘bidentate’. In the 1990 IUPAC recommendations, the term ‘didentate’ was used rather than ‘bidentate’, for reasons of linguistic consistency. However, the term ‘bidentate’ remained in common usage, which is reflected by the reversion to the previously accepted term ‘bidentate’ in the 2005 IUPAC recommendations.
The @ notation for “fullerenes with metals inside” was proposed in J. Phys. Chem. 1991, 95 (20), 7564–7568 and became generally accepted. The corresponding IUPAC proposal, which was presented in Pure Appl. Chem. 1997, 69 (7), 1411–1434, did not use the @ symbol, but it was not as successful as the first one.

Many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. In addition, various traditional names, which have been coined and used over the years, are retained for use in IUPAC nomenclature. Therefore, current IUPAC recommendations include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including the many retained names. Nevertheless, other names that are generated in accordance with IUPAC recommendations may be used in general nomenclature.

Apart from the large and universally active IUPAC, further smaller and specialized organizations exists, for example the International Union of Biochemistry and Molecular Biology (IUBMB). Such organizations may issue their own recommendations concerning their field of expertise. However, many of these recommendations are actually harmonized with IUPAC recommendations.

Another large and universally active organization that prescribes independent naming rules is the Chemical Abstracts Service (CAS). CAS names may differ from IUPAC names. For example, the preferred IUPAC name of Buckminsterfullerene is (C60-Ih)[5,6]fullerene, whereas the corresponding CAS name is [5,6]fullerene-C60-Ih.


Intuitive Explanation for why more negative charge on ligand does not imply more splitting effect? I have studied coordination chemistry as part of my school curriculum and we covered Crystal Field Theory as our final theory, and the book says that Ligand Field Theory and Molecular Orbital Theory are used to explain the limitations of CFT but are beyond our scope of learning.
One of the limitations of CFT according to the book is that if the ligands are point charges then it should follow that an anionic ligand should exert more splitting effect but that is not the case as we see in the spectro-chemical series.
A good example would be $\ce{S^2-}$ < $\ce{F-}$ < $\mathrm{EDTA^{4-}}$ < $\ce{CO}$
Also we can see that size is not the determining factor as then $\mathrm{EDTA^{4-}}$ wouldn't be so high as compared to $\ce{F-}$
Is there some intuitive explanation or logic for this which I can follow with my current knowledge till CFT without going too deep into MOT or LFT?

Crystal Field Theory assumes that all of the interactions occuring are purely electrostatic and only concern d-orbitals.  This is clearly not true when dealing with some ligands which create a covalent bond.
Most limitations of CFT all stem from the inability to explain non-electrostatic interactions.  For this reason, it works well when the electrostatic aspect is the dominant factor but it completely fails for some cases such as neutral CO ligands.  While you could easily argue electrostatic forces still exist in this situation, clearly they are not dominant and do not explain any observed behavior while still having an effect.


Is there a function to approximate atomic mass from the atomic number? The product of $2Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively few protons. Whereas, the product of $2.5Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively many protons. The product of $((2+2.5)/2)Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise a moderate number of protons.

What function from the $Z$ values produces a value that maximally approximates the $m_\mathrm{a}$ that corresponds to $Z$?

The short answer is that you can find a power-law fit ($1.61Z^{1.1}$) with low average error.

I'd never really thought about it much, but after downloading the IUPAC Atomic Weights, I decided to do some curve fitting.

Here's a linear fit between atomic number and atomic mass:

As you say, the fit isn't very good for small $Z$, but the overall fit isn't bad - the mean absolute error (MAE) is $2.821 \:\mathrm{u}$, and taken as a whole, the data is surprisingly linear. (Well, surprising to me.)

So I thought of a quadratic fit, requiring the intercept to be 0,0 to ensure the best fit for small $Z$:

Looks better, right? Certainly the fit is much better for first and second row elements, but the MAE only reduces to $2.749 \:\mathrm{u}$.

So I went up to a cubic fit, again requiring 0,0 for the intercepts:

Aha, now we're talking! We get the subtle nonlinearities, and the MAE is down to $1.36 \:\mathrm{u}$.

Thanks to the comment by Nicolau below, I performed a power-law fit.

So that gives a power-law fit with MAE of $0.01 \:\mathrm{u}$ and a fairly easy-to remember function:

$m_a \approx 1.61Z^{1.1}$


Cathode gained mass in zinc electrolysis I performed an electrolysis of zinc, using two zinc electrodes and a zinc nitrate solution. However, after leaving the electrolysis to run for 30 min, at about 0.4 A and 6 V, I discovered that the cathode gained more mass than the electrode lost. I know that ideally the anode oxidises zinc into zinc ions, and the cathode reduces zinc ions into zinc metal, suggesting that total mass should be conserved.

A key observation I made was that a white substance was observed on the anode, which probably led to the extra weight. Also, bubbles were observed on one of the electrodes. I am not sure which one though.

Edit: My chemistry teacher suggested this could be related to zinc oxide or zinc hydroxide, but I can’t find much information on this.

If the differences between the two electrode masses is small its probably due to excess electrolyte left on it or a measuring error. The white substance is  dried up zinc nitrate.

Bubbles should have been observed at both electrodes since that would have been the hydrogen, oxygen and nitrous oxide dissociating form the diluted solution of zinc nitrate.

ie: $$\ce{3e¯ + 4H+ + NO3- -> NO + 2H2O}$$ $$\ce{O2(g) + 4H+(aq) + 4e ->  2 H2O}$$

So I'm pretty sure the bubbles didn't have anything to do with the additional mass gain. $%edit$


Can you create a buffer with a strong acid? So I was wading through several textbooks yesterday reviewing the concepts of buffers. I found them lot more challenging than I used to. One thing in particular disturbed me: The books and internet all said "Yeah, buffers work pretty great in as long as the  $\rm pH \approx pK_a\pm 1$." But I am not one to leave these things alone. I finally found the Quantitative explanation I was looking for over at buffer capacity on Wikipedia.

However, the graph inside the link says the buffering capacity becomes great when you solution becomes very acidic or very basic, as well. And, of course, looking at titration curves there seem to be a few "buffer" zones e.g. where the solution is highly acidic or highly basic, the $\rm pH$ doesn't change as much after a certain point:

So, obviously there is something wrong with my reasoning, since nobody calls those areas buffer zones. Is it because it is not useful to have such a highly acid/basic solution? Or is it just because technically, there is no buffering, just small $\rm pH$ changes? Or maybe something else entirely...

I think the issue with buffer systems is that they are resistant to changes from strong acids/bases in their normal state.  For example, biological systems often are buffered so that they have some resistance to changes in pH, which can wreck havoc on biochemistry.

Recall the basic buffer equation:

$\ce{HA <-> H+ + A-}$

where $\ce{HA}$ is a weak acid, and there is an excess of $\ce{A-}$, the conjugate base. Because $\ce{HA}$ is weak, it is a relative poor proton donor, but $\ce{A-}$ is a relatively good proton acceptor.  So if you now add a strong acid, the $\ce{H+}$ it donates reacts with the $\ce{A-}$ to produce $\ce{HA}$, shifting the equilibrium to the left, rather than to the right as one might expect.  Or perhaps, preventing as much shifting to the right as might be expected. This gets to your issue of buffering capacity.

This line of reasoning does not work with strong acids/bases because they donate/accept protons with gusto, and thus are not resistant to shifting the equilibrium.  You would need to be way at the extremes of pH for there to be a small change to pH, as your graph shows.  In contrast, buffers are resistant to pH change in their normal state.


Why does butanoic acid have a higher boiling point than isobutyric acid? Why does butanoic acid have a higher boiling point then isobutyric acid, when they have similar molar mass and molecular structure?

Butanoic acid ($\ce{C4H8O2}$) has a molar mass of 88.11 g/mol, the boiling point is 163 °C, and intermolecular force of H-bond.

Isobutyric acid ($\ce{C4H8O2}$) has a molar mass of 88.11 g/mol, the boiling point is 155 °C, and intermolecular force of H-bond.

In general, linear chains are more easily structured and packed than branched ones.  You need more energy to disturb the more relatively packed structure.This could explain why butanoic acid has a higher boiling point than isobutyric acid, and yet they are isomers.


Which has a lower boiling point, Ozone or CO2? Why? Ozone is a bent molecule, and therefore, Ozone is a polar molecule.
Polar molecules have dipole-dipole intermolecular forces.
CO2 is a linear molecule, and therefore, CO2 is a nonpolar molecule.
Nonpolar molecules have london dispersion intermolecular forces.

Having dipole-dipole intermolecular forces means you have greater intermolecular forces than London dispersion forces, and therefore means that you have a higher boiling point. Therefore, Ozone should have a higher boiling point than CO2.

However, after looking at multiple sources across the internet, Ozone has a lower boiling point than CO2.
Could someone provide an explanation as to why this is true?

The primary thing to realize is that, although the ozone molecule has a bent structure, it is a homonuclear molecule.
This means that between the individual oxygens, the bond moments being formed are quite less, as inherently, both the oxygens have the same Pauling electronegativity values. Only a slight difference in the same will occur, due to the asymmetrical bonding scheme in ozone.

It's worth noting that bond moments in carbon dioxide are much more pronounced than those in ozone, even though they end up being canceled vectorially (like you said).

For comparing the boiling points, we need to see the extent of intermolecular attraction as a measure for the same. Due to the difference in electronegativities between $\ce{C}$ and $\ce{O}$, $\ce{CO2}$ ends up having more partial charges on its constituent atoms as compared to ozone. Hence, the boiling point of carbon dioxide is more than ozone


What is the Chemical Compound that has a NFPA 704 (Fire Diamond) of 2-2-1-ox? I recently saw this fire diamond configuration of:

2 - Flammability;
2 - Health;
1 - Reactivity;
OX - Extra Information

I have been attempting to find a database of terms, but after turning up with nothing, and seeing a similar answer about fire diamonds answered here, I posted this question.

I would be thankful if someone happened to know what compound that this fire diamond is labeled to.

From what I found NIOSH lists 6 compounds with 2-2-1 pattern, but none of them are labeled with oxidizing properties:

ethylhexaldehyde: NFPA 704 2-2-1;
benzyl chloride: NFPA 704 2-2-1;
2-methyl-1-butanol: NFPA 704 2-2-1;
n-octyl mercaptan: NFPA 704 2-2-1;
vinyl toluene: NFPA 704 2-2-1;
methyl chloroacetate: NFPA 704 2-2-1.

On the other hand, there are oxidizers with slightly different NFPA diamond which much as well be the candidates:

ammonium dichromate NFPA 704 2-1-1-OX;
amyl nitrate NFPA 704 2-2-0-OX.

The values might deviate a little in different sources and they also depend on conditioning.


Identifying the oxygen atoms that undergo reduction Identify the oxygen atoms that undergo reduction in the reaction below:
$$\ce{O2 + Cr^3+ -> H2O2 + Cr2O7^2-}$$

I know that the oxygen atoms in $\ce{O2}$ have an oxidation state of $0$.
In the product $\ce{H2O2}$, the oxygen atoms have an oxidation state of $-1$; and in $\ce{Cr2O7^2-}$, oxygen atoms have an oxidation state of $-2$.
Based on how the problem was solved on the answer key it was assumed that all the oxygen atoms from $\ce{O2}$ are reduced to peroxide atoms.
How do we know that there are no oxygen atoms that are oxidized to become chromate oxygens with a $-2$ oxidation state?
I understand that there are other $\ce{H2O}$ oxygen atoms available to provide the chromate oxygens that would change the oxidation state.

The oxygens in the dichromate ion on the right have -2 oxidation number, the chromium being +6.
http://ch302.cm.utexas.edu/worksheets/basic-redox-wkst-KEY.pdf
https://socratic.org/questions/what-is-the-oxidation-state-of-chromium-in-the-dichromate-ion-cr2o7-2

As you seem to have deduced, the peroxide ions are -1 so all oxygens, as shown, have been reduced.


What is the use to know that a compound optically active or not? What is the use to know that a compound optically active or not? Is it important to know that?

Yes, it is important.

For reasonably enantiopure chiral molecules, it is expected to record and publish a specific rotation when it is first described. For the person recording these, they are a little bit annoying, especially if you have a polarimeter as old as the one I had to work with during my PhD. Therefore, you want to cut time as much as possible. By immediately determining that a compound is achiral, you eliminate the need to record a specific rotation because theory predicts it is zero.

Conversely, if you were to simply record all specific rotations of your compounds and then just remove the ones that end up as zero you would likely miss the chiral molecules you synthesised as a racemate. If, however, you take a look at the molecule on paper, are able to determine it should be chiral—given the methods and chemicals employed—but still find a specific rotation of zero you have something that must be discussed in the final paper or thesis.


More acidic - protonated carbonyl or protonated hydroxyl? Consider two acetic acid derivatives, one with a protonated carbonyl oxygen and one with a protonated hydroxyl oxygen.

An argument for the one with the protonated hydroxyl oxygen being more acidic is as follows:

This structure has a lone pair on its protonated hydroxyl oxygen and the oxygen is adjacent to an $sp^2$ carbon. Therefore we can draw an extreme resonance structure in which the hydroxyl oxygen has a +2 formal charge. No such "extreme" structure exists for the one with the protonated carbonyl oxygen.

Now, we can definitely draw this resonance structure, but its contribution is likely very small and bordering on insignificant. I can't see how delocalization of electrons away from a highly electronegative, positively charged oxygen can help stabilize it.

Therefore: Which is more acidic? The argument that the protonated hydroxyl one should be more acidic seems to be flawed at best.

You have outlined the correct reasoning but drawn the wrong conclusion. The protonated carbonyl structure (top left) is stabilized by resonance (bottom left), giving two equivalent resonance structures. Although the protonated hydroxyl (top right) has a resonance structure that can be drawn (bottom right), it should be considered a negligible contributor since there is a double positive charge on a very electronegative atom that would be sp2 hybridized (also toss in the charge separation). Since the protonated hydroxyl is not stabilized by resonance, it is more acidic, i.e., it is more willing to give away the proton (and keep the electrons).


Why is the pressure correction constant 'a' independent of temperature? In the van der Waals equation of state, 'a' which is the correction value is based on inter-molecular forces of attraction? Then should inter-molecular forces decrease when temperature increases, and thereby cause a change in value of 'a'?

Then why is 'a' independent of temperature?

The parameter $a$ is independent of temperature because we defined it this way. There is no other reason. Indeed, van der Waals equation is very much not a law of nature. It is just an empirical approximation. There are other approximations, maybe a dozen or more. You may create your own, where $a$ would decrease with temperature. Wait, it's been done already - look at the Berthelot equation!


Does anyone know where I can find the reaction rate coefficient and design equation for the electrolysis of brine and sodium hydroxide solution? equation 1) $\ce{2 Cl^{–} -> Cl_2 + 2e^{-}}$

equation 2) $\ce{2 H2O + 2 e^{–} -> H2 + 2OH^{–}}$

These equations are the key reactions in the chlor-alkali process. I want to model a reactor for this process based on a CSTR (continuous stirred tank reactor). However I really need reaction rate coefficients (k values) and design equations. Any help would be much appreciated!

You should look in some kind of reaction rate database, e.g. NIST. An example from the database:


Van der Waals nonspecific interaction definition Every definition I see of Van der Waals interactions claim that they are short-range, nonspecific interactions between two chemical species. What does "nonspecific" mean in this context?

The definition of the van der Waals force that I like (and apparently so too do the folks who have contributed to the Wikipedia article on the topic) is much more *specific about what is and is not a van der Waals force. This definition originates in the International Union of Pure & Applied Chemistry Compendium of Chemical Terminology

The van der Waals force is the sum of the attractive or repulsive forces between molecules (or between parts of the same molecule) other than those due to covalent bonds, the hydrogen bonds, or the electrostatic interaction of ions with one another or with neutral molecules or charged molecules.

Thus, a van der Waals force is a catch-all term to include all other very weak interactions between atoms and molecules:

Dipole-dipole interactions between permanent dipoles
Dipole-induced dipole interactions
Induced dipole - induced dipole interactions (London dispersion)
$\pi-\pi$ stacking
Halogen bonding
Repulsive forces derived from atoms trying to be inside another atom's van der Waals radius (Steric strain and torsional strain)


Polymer Chemistry NMR question I am extremely new to both polymer chemistry and NMR.
I am struggling with the following question:
A proton NMR is used to attempt to quantify the molecular weight of a poly(ethylene oxide) molecule with methyoxy end groups at each terminus. If the integration of the methyl protons relative to the methylene protons gave a ratio of 1:20, what can  you say about the molecular weight?

Thank you. All help is appreciate including resources for me to be able to read up on my lacking knowledge.

I hope this is not a homework question. Suppose your polymer has $n$ reapeating units and capped with methyl groups at the end as you described. Thus, it should looks like following figure:

Thus it has $n$ $\ce{(-CH2CH2-)}$ units and $2$ $\ce{(-CH3)}$ units. Thus your $\ce{(-O-CH2-)}$ signal is accounted for $4n$ protons while $\ce{(-O-CH3)}$ signal accounted for only $6$ protons. Thus,
$$\frac{4n}{6} = \frac{20}{1}$$
You may find the value of $n$ from this equation and you can calculate the molecular weight of the polymer in hand.


What does this scale mean? $$\log_{10}~\mathrm{pg}/\mathrm{ml}$$

I do not understand the scale.  How do I convert this to a standard concentration (without the logs)?

To expand upon the comments, the y axis in this graph is simply the log of the concentration in pg/ml.  To convert you use the equation $$C = 10^{y}$$ where y is the y value and C is the concentration.  Note that in older literature, log was oftentimes assumed to be base e whereas nowadays we assume base 10.  The figure (thankfully) clarifies this potential discrepancy.

A final note: log values are unitless so the y axis has no units.  The units are reported here so that you know what the appropriate units of the converted value should be.


Intutive understanding of internal energy Our high school teacher told us that once we calculate the formula of internal energy, it is valid everywhere. I cannot understand how? I asked him, he simply answered because it is state function. It made me even more confused.

From first law, $\mathrm dU=\mathrm dq+\mathrm dW$

In constant volume process, volume differntial $\mathrm dV=0$, hence $\mathrm dW=0$, so

$(\mathrm dU)_V=(\mathrm dq)_V$ and $(\mathrm dq)_V=C_V\mathrm dT$

  So, $(\mathrm dU)_V=C_V\mathrm dT$

Why we use this formula everywhere even when volume is changing (since we derived this relation for a constant volume process), at least for ideal gas?

For an arbitrary fluid (not necessarily an ideal gas), a combination of the first- and second laws of thermodynamics tells us that, $$\left(\frac{\partial u}{\partial v}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_v\right]$$where U is the molar internal energy and v is the molar volume.  For a fluid which obeys the ideal gas law Pv=RT, the right hand side of this equation is zero.  So the internal energy of a substance which obeys the ideal gas law is independent of volume.  An ideal gas is a substance which obeys the ideal gas law because the interactive forces between its molecules are negligible.  So, yes, it is because ideal gas molecules do not have attractive forces.


Which mechanism is not seen during hydrolysis of esters? Which mechanism is not seen during the hydrolysis of esters?

AAC2
BAC2
AAL2
BAL2
AAC1

I know that hydrolysis of ester is pseudo-first order reaction but I found this question where there are completely unknown reaction mechanisms.

The reaction mechanism said actually corresponds to the following mechanisms,

$A_{AC^2} $ which means acid catalysed acyl carbon oxygen bond breaking bimolecular mechanism.
$A_{AL^2} $ which means acid catalysed alkyl carbon oxygen bond breaking bimolecular mechanism.
$B_{AC^2} $ which means base catalysed acyl carbon oxygen bond breaking bimolecular mechanism.
$B_{ALC^2} $ which means base catalysed alkyl carbon oxygen bond breaking bimolecular mechanism.

In the last one it is unimolecular, and acid catalysed and acyl carbon oxygen bond breaking mechanism.

In esters, Hydrolysis can occur via breaking of two possible bonds. If the ester is $\ce{R'-CO-O-R}$, then the $\ce{CO-O}$ bond is the acyl carbon oxygen bond, whereas the $\ce{O-R}$ bond is alkyl carbon oxygen bond. Either of these two breaks in the process of hydrolysis.

If you take usual hydrolysis of esters, they can occur in both acid and base catalysed reaction medium condition, and in that case the $\ce{R'-CO}$ part of the ester goes to the acid, and the $\ce{-O-R}$ goes to alcohol. So, this is the acyl carbon oxygen bond breaking and bimolecular as the $\ce{OH-}$(in base catalysed) and $\ce{H2O}$ (in acid actalysed) attacks at the rate determining step. So, $A_{AC^2}$ and $B_{AC^2}$ is easily seen.

In an ester, where the attack of $\ce{OH-}$ or $\ce{H2O}$ at the carbonyl carbon is highly sterically hindered (e.g. in mesitoic ester) the reaction passes through alkyl carbon oxygen bond breaking, and thus the $\ce{R'-CO-O}$ part of ester goes to acid, whereas only the R part goes to alcohol, and they are also bimolecular generally, as in the slow step, both the nucleophile attacking at the alkyl carbon remains present along with the molecule. That's why $A_{AL^2}$ and $B_{AL^2}$ are also seen but it is slightly rare.

But for showing $A_{AC^1}$, the molecule doesn't have any other chance rather than forming an acyl carbocation(as, the carbocation intermidiate follows unimolecular reactions). But if it forms acyl acrbocation in acid medium, the lone pair on the adjacent oxygen is already donated to $\ce{H+}$, so, there is no extra stabilising factor rather than any highly electron donating group, like tert butyl or phenyl attached to carbonyl carbon. But if you have those, the molecule generally prefers the alkyl bond breaking to avoid steric hindrance.So, $A_{AC^1}$ is very very rare to happen.

But $A_{AL^1}$ can be seen where R is tert butyl group. because, it can get stabilisation by forming a carbocation of its own, and the remaining portion of molecule can get stabilisation through resonance.


what happens to the intramolecular bonds within water molecule during phase changes? Does intramolecular bonds within water molecule change during phase changes?

The only intramolecular bonds in water are the O-H bonds. They are not modified during phase changes. The hydrogen bond is not an intramolecular bond. It is an intermolecular bond. These hydrogen bonds are modified during fusion or ebullition.


Sunscreen or sunblock When I went to the supermarket, I noticed that some of the product say sunscreen and some of them say sunblock.

My question split into two part.

1) What are the different chemical substances use in both item? (In general)

2)What is more effective to prevert sunburn? (In general)

Giving advice on preventing sunburns may fall under health advice, which is not permitted on this site. For advice, you should ask a medical professional.
The following information is from The Melanoma Foundation.
What is the difference between sunscreen and sunblock?

Sunscreens can be classified into two major types: chemical and physical. Chemical sunscreens contain special ingredients that act as filters and reduce ultraviolet radiation penetration to the skin. These sunscreens often are colorless and maintain a thin visible film on the skin. These sunscreens usually contain UVB absorbing chemicals and more recently contain UVA absorbers as well.
Physical Sunscreens, most often referred to as sunblocks, are products containing ingredients such a titanium dioxide ($\ce{TiO2}$) and zinc oxide ($\ce{ZnO}$) which physically block ultraviolet radiation (UVR). Sunblocks provide broad protection against both UVB and UVA light. They can be cosmetically unacceptable to many people, because they are often messy, visible and do not easily wash off. However, some new zinc oxide products are available in brightly colored preparations which are popular with young people. The amount of sun protection these sunblocks provide, while potentially high, cannot be quantified in the same manner as sunscreen SPFs. Physical sunscreen is recommended for individuals who have unusual sensitivity to UVR. Most recently on the sun protection scene is sun-protective clothing designed to block UVA and UVB radiation. The effective SPF is greater that 30.

When should you use a sunscreen?

Sunscreens should be used daily if you are going to be in the sun for more than 20 minutes. Most people will receive this amount of sun exposure while performing routine activities. They can be applied under makeup. There are many cosmetic products available today that contain sunscreens for daily use because sun protection is the principal means of preventing premature aging and skin cancer. Sunscreen used on a regular basis actually allows some repair of damaged skin. Because the sun's reflective powers are great - 17 percent on sand and 80 percent on snow - don't reserve the use of these products for only sunny summer days. Even on a cloudy day 80 percent of the sun's ultraviolet rays pass through the clouds. Skiers beware, ultraviolet radiation increases 4 percent for every 1,000-foot increase in altitude.

To add to this last part, sunblock manufacturers usually include instructions with their products on the proper way to use them.


Predict XPS Spectrum of Organic Monolayers on a Semiconductor X-ray photoelectron spectroscopy (XPS) is a particularly useful technique for examining the surfaces of all sorts of materials. Organic monolayers are particularly difficult to characterize completely via this technique due to the fact that many carbon peaks do not differ much by chemical shift. Traditionally quantitative information is obtained via a lot of guess-work and forced peak fitting. It would be better to predict peaks shapes and peak positions beforehand to aid in the interpretation and prevent a terribly mis-guided interpretation. How well can theory help the non-specialist with this? I wouldn't have the time to develop a program to do this either. Preferably one could point me to the right direction in terms of a research paper/software.

Some density functional theory packages, such as Quantum Espresso, can calculate XPS core level shifts using the pseudopotential formalism. See Phys. Rev. B 74, 045430 (2006).


Is there an upper limit on the coordination number of a coordination compound? From the Gold Book, coordination number is:

  In an inorganic coordination entity, the number of σ-bonds between ligands and the central atom. π-bonds are not considered in determining the coordination number.

In high school, we mostly had complexes upto coordination number six (like $\ce{[FeF6]^3-}$). But I wondered if the coordination number could go even higher.

From my search, I was able to reach this research paper (J. Am. Chem. Soc., 1979, 101 (2), pp 334–340 DOI: 10.1021/ja00496a010) where they mention lanthanide complexes of coordination number uptil nine!

So, is nine the upper limit? Or can the coordination number go higher? Also, does the upper limit exist only theoretically, or has the corresponding complex been synthesized?

I don't see why not. It is a matter of having

enough of an interaction between the center and the ligands, usually in the form of donor/acceptor orbital pairs, and
enough space around the coordination center.

These go hand-in-hand, as any atom that will have enough electrons to donate to ligands will also most likely have a larger atomic radius than a first-row transition metal like iron.

For example, here is a thorium complex that has CN = 15 in its crystal structure and CN = 16 in an isolated gas-phase density functional calculation:

Daly, Scott R.; Piccoli, Paula M. B.; Schultz, Arthur J.; Todorova, Tanya K.; Gagliardi, Laura; Girolami, Gregory S. Synthesis and Properties of a Fifteen-Coordinate Complex: The Thorium Aminodiboronate $\ce{[Th(H3BNMe2BH3)4]}$. Angew. Chem. Int. Ed. 2010, 49, 3379-3381, DOI: 10.1002/anie.200905797


How can I predict whether a reaction will occur? I'd like to know how can I predict if two substances will react. I already know that there's a reaction if one of the products is a weak electrolyte or a precipitate but that leads me to the question:

  If we have two substances such as salts and they react with one another forming two other salts, when is the reaction double replacement and when there is no reaction?

I'll try giving two examples from the problems I came across:

Which one of the following substances will react with one another in a solution –

$\ce{(NH4)2CO3, CuCl2, K2SO4, Ba(NO3)2, HCl}$
$\ce{Na2CO3, Ba(NO3)2, Na2SO4, AgNO3, CaCl2}$

For the first example, I definitely wouldn't want to react $\ce{CuCl2}$ with $\ce{HCl}$, because then I would end up getting the same products again. I'm primarily driven by the motive to create something new, albeit the reaction must be a feasible one. Many factors influence feasibility, and you've mentioned the formation of a precipitate is one of them.
Reactions involving neutralization:
$\ce{HCl}$ would react with $\ce{(NH4)2CO3}$, $\ce{K2SO4}$ and $\ce{Ba(NO3)2}$, as the reactions give out $\ce{H2CO3}$, $\ce{H2SO4}$ and $\ce{HNO3}$ respectively which are weaker acids.
Reactions involving precipitation:
I see a very tempting reaction here:
$$\ce{K2SO4 + Ba(NO3)2 -> 2KNO3 + BaSO4\downarrow}$$
$\ce{BaSO4}$ is a very good precipitate, and I highly emphasize this fact. Just as a fun fact: It can be partially soluble if you try to dissolve it in boiling concentrated $\ce{HCl}$ (or boiling concentrated $\ce{H2SO4}$).
On similar grounds, $\ce{Ba(NO3)2}$ would react with $\ce{(NH4)2CO3}$ to give out the precipitate of $\ce{BaCO3}$. How do I know this? Partly because of the fact that while testing for the presence of $\ce{CO3^2-}$ ions in a laboratory we use the lime water test. Instead of using $\ce{Ca(OH)2}$, baryta water or $\ce{Ba(OH)2}$ is also used at times yielding strikingly similar results.
Also, I know this partly because $\ce{CO3^2-}$ ion is notorious for forming precipitates. It forms a precipitate with a large number of ions, some exceptions (they're the only exceptions I know) being $\ce{Na+, K+, Rb+, Cs+}$ and $\ce{NH4+}$.

However, what happens to the reactions like these?

$$\ce{(NH4)2CO3 + K2SO4 <=> K2CO3 + (NH4)2SO4}$$

Neither of the product is a precipitate. Hey that's easy? $\ce{K2SO4}$ is made up of $\ce{KOH}$ and $\ce{H2SO4}$. And look the base is the same for $\ce{K2CO3}$, that means I've made carbonic acid from sulfuric acid and thus the reaction takes place. Right? Wrong.
In your haste you forgot to take note that I could've used the same logic if I were to deal with $\ce{(NH4)2CO3}$ and $\ce{(NH4)2SO4}$. Wait! This reverses my previous reasoning! Yes it does. Or maybe we get a little confused.
Perhaps these cations are spectators, shall I strip them off? Yeah, let's give it a try too. We're now (sadly) left with:
$$\ce{CO3^2- + SO4^2- <=> CO3^2- + SO4^2-}$$
That's bizarre. What's happening? Nothing. And that's what it is in its true form, the ions basically are just combining with each other, there are four ions, two cations and two anions, four possible combinations. Does that make it a reaction? Perhaps not. It's just an equilibrium, there's no going forward or backward. When we say there's a reaction taking place, we are implying that the equilibrium sways to the the product side, which apparently isn't the case here.


Why any non polar compounds dissolve in nonpolar substance like Benzene what is the reason behind it? In case of ionic solvent the ionic solute dissolve in it because due to polar nature of water it break the bond between the ions through hydration energy and the ionic solvent dissolve in it.
So I want to ask that for what reason the non polar solute dissolve in non polar solvent ?
Means a basic framework like due to what reason it happens?
I searched on Google but I don't get slim and articles or content

The simple answer is entropy. The explanation not so much.  If you consider solvation (a specific type of dissolving) to really be a melting process you can consider the resulting solution a eutectic/peritectic mixture.

To explain Figure 1 is the phase diagram of tin-lead compounds.  You could consider tin to be soluble in lead since at $300^\circ C$ lead is liquid and tin is solid but when tin is added to lead, it dissolves.  Or you could consider the mixture of the two to be a eutectic that lowers the finally melting temperature of both substances to $183^\circ C$ (at 61.9 wt% lead) which is commonly the case.  In fact if you were to make a phase diagram for any solute and solvent you would see similar behaviour.

Now sometimes the solvent forms a solid complex such as $\ce{CaCl2 *6H2O}$ which casuses a discontinuity. in Figure 2, point A is a eutectic point but points B and C are peritectic which is a result of the solid phase having a different equillibrium complex ($\ce{CaCl2*6H2O}$ vs. $\ce{CaCl2*4H2O}$.

That was some concept now for some math.  The entropy of mixing can be determined for a solution by: $\Delta S_{mix} = -R[x_a ln(x_a) + x_b ln(x_b)] $ where $x_a$ and $x_b$ are the molar ratios of the components  ($x_a = 1 - x_b$). Since $\Delta S_{mix}$ is always positive, $\Delta G_{mix}$ will always be negative as long as the heat of mixing ($\Delta H_{mix}$) is not too positive.

Figure 1: Eutectic Mixture Example:

Figure 2: Peritectic Example:


Mechanism of a dephenylation reaction I recently came across this reaction in my undergraduate labs. The experiment is taken from this article: J. Chem. Educ., 2008, 85, 413.

The condensation of 2-methoxynaphthalene with cinnamoyl chloride leads to the formation of 9-hydroxyphenalenone:

I think it's fairly clear that the first step is a Friedel-Crafts acylation. Subsequently to form the new C-C bond, there must be a Michael addition of some sort; however this would imply that the phenyl group leaves as "$\ce{Ph-}$". There are two things I'm not very sure about:

How does the phenyl group even leave?
How does the oxygen get demethylated? I am guessing that the mechanism should be similar to that with $\ce{BBr3}$, i.e. coordination of oxygen to the Lewis acid followed by nucleophilic attack of $\ce{Cl-}$ on the methyl carbon. Would that be right?

I looked up the original synthesis, which was reported quite a long time ago, in J. Org. Chem., 1941, 6, 558. The authors write that the reaction proceeds via the intermediacy of this compound:

which suggests that the initial nucleophile in the Michael addition is not the aromatic ring but rather the -OMe (or -OH) group. Unfortunately, that just makes things more confusing.

First, the methoxy- group can be demethylated by $\ce{AlCl3}$, beginning with $\ce{AlCl3}$ coordination to the ether oxygen, followed by an $\mathrm{S_N2}$ reaction:

$\hspace{3.6cm}$

$\hspace{0cm}$

The oxygen activates naphthalene's position 1, and the acid chloride will undergo a Friedel-Crafts acylation (as you noted):

$\hspace{.5cm}$

Being a strong Lewis acid, $\ce{AlCl3}$ can coordinate with the carbonyl oxygen. This activates position 4 of the $\alpha$-$\beta$ unsaturated carbonyl towards electrophilic addition:

$\hspace{.75cm}$

Electrophilic addition occurs at naphthalene's spatially adjacent position 8, the carbocation stabilized through the extensive conjugation of the molecule (as well as by the oxygen, as you noted):

$\hspace{1.75cm}$

The spatially adjacent phenyl group accepts the proton and is ejected as benzene:

$\hspace{1cm}$

Addition of water evolves $\ce{HCl}$ and precipitates $\ce{Al(OH)3}$, yielding the final product:

$\hspace{1.8cm}$


The equivalent weight of HNO3 in the following reaction The equivalent weight of HNO3 in the following reaction is ‘x’, then the value of x/21 is

3Cu+8HNO3⟶3Cu(NO3)2+2NO+4H2O

in this I atempted by using redox formulas but my answer is comming 1 but correct answer is 4

The term equivalent weight is used in conjuction with normality (which is quite obsolete, except in the industry). Anyway, whenever you determine equivalent weight, you have to know beforehand whether you are considering a given "molecule" as an acid or as an reducing/oxidizing agent. This is where the confusion sets in.

Nitric acid as an acid is a monoprotic acid: It's equivalent weight is the same as its formula weight

Calculating the equivalent weight of nitric acid as an oxidizing agent is another story, it depends on how the redox equation is written.

From your equation nitric acid is acting as an oxidizing agent because there is no nitrate ion on the right hand side and the oxidation state of N is different on both sides:

$\ce{HNO3}$ is converting to NO. In the answer above, you already have the equation

$\ce{ NO3−}$ +3e => NO

In this case, the equivalent weight of nitric acid is 1/3 (formula weight).

As you can see equivalent weight is very much dependent on the way you write the equation. This will come with practice.


Which one is more basic: acetamide or aniline? Which one is more basic: acetamide or aniline?

I had approached the question by protonating both of the compounds. Then I judged them on the basis of the stability of the corresponding conjugate acid.

In $\ce{CH3CONH3+}$, due to resonance, a positive charge is created on the oxygen atom. So, resonance is not very effective here. Hence $\ce{ CH3CONH2}$ is not a good base.
In $\ce{C6H5NH3+}$, there is far better resonance which takes place.

From the above observations, I concluded that aniline is more basic.

I had another view that $\ce{C6H5NH2}$ is a basic type of compound rather being salt-like as is $\ce{CH3CONH2}$.

Am I going in the right direction to solve the problem?

You can consider this in the following way.

If amide undergoes protonation, it forms a positively charged nitrogen atom next to the strong electron-withdrawing group. This destabilises the molecule, so most probably it will not act as a base. Instead, it will act as a moderate acid with a strong base, since conjugate anion will be stabilized by the adjacent carbonyl group.

In the case of aniline, the phenyl group is not a strong withdrawing group. Still, the phenyl group withdraws electron density from the nitrogen atom through resonance. However, anilines can be protonated by strong acids. You can observe that during the formation of a diazonium salt, where aniline is protonated by strong acids like $\ce{HCl}$.


Heat exchange in adiabatic process I know my teacher told me that

  In adiabatic process there is no exchange for heat between system and surrounding.

Then he came to point. From first law of thermodynamics

$$∆U = q + W \tag{1}$$

Since in adiabatic process there is no heat exchange then,

$$∆U = W \tag{2}$$

And then he said as we know

$$∆U = nC_v∆T \tag{3}$$

So,

$$W = nC_v∆T \tag{4}$$

I don't get it. Above equation tells us that the work done when a system is given heat at constant volume equals to $nC_v∆T.$
But how can this be applied in adiabatic process, since there is no exchange of heat?

The problem is that we were taught incorrectly in freshman physics.  They told us that $Q=nC\Delta T$, where C is called the heat capacity.  However, they neglected to tell us that this only applies if no work is done (i.e., the volume is essentially constant, as for a solid or liquid).  Otherwise, it gives the wrong answer for the heat Q.

"Heat capacity" is a bit of a misnomer for C.  From what we then learn in thermodynamics, a better term would be the "internal energy capacity," because that is how Cv is more properly defined.  For an ideal gas,  $$nC_v=\frac{\Delta U}{\Delta T}$$ or $$\Delta U=nC_v\Delta T$$This has nothing to do with the heat added unless the volume is constant and no work is done.  In that case, $$Q=\Delta U=nC_v\Delta T$$So the freshman form of the equation is recovered only if no work is done.


dsp3 and sp3d hybridization — what is the difference and which applies to square pyramidal? What is the difference between $\mathrm{dsp^3}$ and $\mathrm{sp^3d}$ hybridization?
Are they one in the same?

Also, I have a book that says that in compounds where the central atom is $\mathrm{dsp^3}$ hybridized, it's shape is square pyramidal, like that of $\ce{BrF5},$ but the hybridization of $\ce{Br}$ in $\ce{BrF5}$ is $\mathrm{sp^3d^2}.$ Why is it so? Is my book wrong or am I missing something?

We can split this into two cases: main group elements and transition elements.

Main group elements do not use their d-orbitals for bonds and thus cannot be hybridised in any d-containing way. Bromine in $\ce{BrF5}$, to cite your example, is unhybridised. The $\mathrm{p}_z$ orbital forms a traditional two-electron-two-centre bond with one of the fluorines. The $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals each form a four-electron-three-centre bond with two additional fluorines for a total count of five fluorines. The bond order of these four-electron-three-centre bonds is $\frac12$.

Bromine’s 3d orbitals are completely populated and effectively belong to the core orbitals. They do not participate in bonding. Bromine’s 4d orbitals are energetically even further removed than the 5s orbital is — remember the aufbau principle. Thus, neither the lower nor the higher d-orbitals can participate in bonding to any notable extent.

This extends well to all main group elements as hinted above. Practically no main group element compounds will include d-orbitals.

For transition metals the case is slightly different since we are in the midst of the d-block where d-orbitals can play a role. However, invoking hybridisation for transition metals puts a big blanket over the bonding situation declaring things equal which really aren’t. A much better description for transition metals would be to use a molecular orbital scheme as I have done in countless answers of mine. Most importantly, the bonding situation can, in my opinion, not be understood without this depiction.

Additionally, many transition metal geometries provide unintuitive ‘hybridisation’ solutions — octahedral being the notable exception which actually hybridises six orbitals (three d-orbitals are nonbonding). For example, a tetrahedral metal complex should be understood as using all orbitals for bonding. Likewise, the square-planar ligand arrangement only has one nonbonding d-orbital.

My suggestion is never to use hybridisation approaches for transition metal complexes.

Formally, one may say that putting the d before s and p implies a lower-shell d-orbital. That would be the transition metal case above. And placing the d behind s and p would use d-orbitals from the same shell. That would be the main group case. Thus, the difference between the two (tl;dr) is:

The first is very strongly discouraged, the second is wrong.


How did Millikan measure the mass of these oil droplets? I was reading the R.A. Millikan's oil drop experiment and the following part confused me:

By measuring the rate of fall, Millikan was able to measure the fall of the oil droplets (which were dropped through a tiny hole in the upper electrical condenser).

In Newtonian physics, the rate of fall (or velocity) of a freely falling body does not depend on the mass of the object. So, how did Millikan measure the mass of these oil droplets? Does anyone know the equations he used?

By measuring the rate of fall, Millikan was able to measure the fall of the oil droplets (which were dropped through a tiny hole in the upper electrical condenser)

The rate of fall here means terminal velocity of oil, $v_\mathrm{t}.$ Oil was made to drop at constant rate so no acceleration. Drag force on oil drop is:
$$F_\mathrm{d} = 6\pi r\eta v_\mathrm{t},\tag{1}$$
where $r$ is the radius of oil drop and $\eta$ is the viscosity of air.
Assume oil droplet is spherical. Apparent weight of oil droplet in air is the true weight minus the upthrust (which equals the weight of air displaced by the oil drop). So, it can be written as:
$$w = \frac{4\pi r^3(\rho - \rho_\mathrm{air})g}{3}.\tag{2}$$
No acceleration is taking place so the total force acting on it must be zero and thus the two forces must cancel each other (i.e $F_\mathrm{d} = w).$ Solving for $r$ gives
$$r = \sqrt{\frac{9\eta v_\mathrm{t}}{2g(\rho - \rho_\mathrm{air})}}.\tag{3}$$
The density of oil is known so by known so by putting $r,$ you can calculate the mass of oil droplet as
$$m = \rho V = \rho\frac{4\pi r^3}{3}.\tag{4}$$


Wittig reaction with benzaldehyde Somehow I am told that my product is incorrect. Why is my product incorrect?

As far as I can see this is a Wittig reaction resulting in a two carbon increase with removal of the oxygen.

The only thing that strikes me as odd is that the Wittig is conjugated to a bromide salt. The only way this can happen is if the phosphorous part has an overall positive formal charge. Does this mean that the n-butyllithium will abstract a proton and leave the carbon backbone of the Wittig as $\ce{CH3CH}$?

Even so this doesn't seem to affect my final product ...

Below is the preparation of the Wittig reagent you are given. Yes, the intermediate is a phosphonium salt, with a formal positive charge on phosphorus. Treatment with base (BuLi) deprotonates at the adjacent carbon giving a phosphorus ylide.

As for the Wittig reaction itself, one postulated mechanism (depicted in Organic Chemistry by Maitland Jones) is shown below. Nucleophilic attack of the ylide onto the aldehyde gives an open betaine intermediate, which closes to form an oxaphosphatane. The strained 4-membered ring decomposes readily to give an alkene and triphenylphosphine oxide.

The stereochemistry of the Wittig is related to the nature of the ylide used. Unstabilized ylides, such as the one here, predominantly give the Z-configured alkene, while stabilized ylides (usually through conjugation to a carbonyl) give an E-configured alkene as the major product. With substrates similar to the ones here, I obtained a 4:1 ratio of Z:E products, so the selectivity is not perfect, as K_P points out.

The stereochemical outcome is intimately related to the mechanism, and according to my copy of Jerry March's Advanced Organic Chemistry (4th ed), there is no evidence for the betaine intermediate. Instead, the oxaphosphatane is formed directly through a [2+2] cycloaddition mechanism. This mechanism is attractive because the stereochemical demands of the cycloaddition account for the preference of the (less stable) Z-product. However, this does not explain why different ylides give different selectivities.


How is it thermodynamically possible to obtain energy from a piece of coal? If we put a piece of coal in open air we don't get any energy of it. But if we put fire to it in open air we get energy. Explain it with help of Gibbs free energy.

Coal in open air can and does sponstaneously combust.

See for example Spontaneous Heating and Fire in Coal Mines and Coal’s Spontaneous Combustion Problem.

At low temperature the change in Gibbs free energy of coal oxidation is still strongly negative, just like at high temperature.

But at low temperature, the reaction rate is low.

Still, if the rate at which heat is dissipated is less that the rate at which heat is produced, temperature increases and the reaction rate increases until there is appreciable combusiton.


How can you extract palladium from palladium chloride Hey guys I am working on a school project, focused on how precious metals can be recycled from industrial products. I am very much not an expert on the subject of chemistry. But I am just trying to get basic understanding on how to extract palladium from a catalyst like palladium chloride.
Thanks in advance,
Ilias

Yes, there are several methods of reducing palladium chloride to elemental chloride.

By converting Pd(II) salt to palladium acetate ($\ce{[Pd(CH3CO2)2]3}$) and decomposing it at 200-300 $\pu{^\circ C}$. (link)

By using reducing agents

carbon monoxide
$$\ce{CO + PdCl2 + H2O → Pd + CO2 + 2HCl}$$
This is a qualitative reaction to determine the presence of carbon monoxide in the
air. A paper moistened with a solution of palladium chloride turns black in the presence of carbon monoxide due to the formation of finely dispersed metallic palladium. The reaction proceeds at r.t. The yield is high (84%).

Hydrogen
$$\ce{PdCl2 + H2 -> Pd + 2HCl}$$
Pd(II) salts can be reduced to finely divided elemental palladium by using
hydrogen. The reaction proceeds at r.t. but the reaction rate can be increased by
increasing the temperature. See this paper for more details.

Using $\ce{SnCl2}$

The reaction proceeds by formation of a complex. Yield is also high (90%). See here for more details.

Other reducing agents ($\ce{TiCl3, Fe, Cu, Zn}$)

Decomposition of palladium chloride. But the reaction is unreliable. The reaction proceeds at high temperature (590-740 $\pu{^\circ C}$). May release fumes of chlorine and hydrogen chloride which might be dangerous. Palladium obtained may be impure because of formation of various types of oxides. (link)

Decomposition of potassium tetra/hexachloropallade

Calcination in air. Dissolve the salt in a suitable solvent (usually water), "paint" the solution on a substrate, allow the solvent to evaporate and then start calcination. Pd(II) would become Pd(0), leaving a thin metal film on the substrate.

Binding of Pd-salt into an organic matrix (e.g., a zeolitic inorganic-organic polymer electrolyte), and then calcinate the product. However, using this approach, calcination should be done under vacuum or under an inert atmosphere (otherwise, oxygen might yield Pd oxide systems). In this case, Pd ends up as metal nanoparticles into a carbonaceous matrix.

Solution phase reduction. Impregnate silica with your metal salt solution, add a reducing agent like formic acid, ethylene glycol, $\ce{NaBH4}$ or hydrazine, heat at $\ce{150 ^\circ C}$ for ethylene glycol and at $\ce{80 ^\circ C}$  for the rest of the reducing agents, you will get Pd decorated $\ce{SiO2}$ particles. Nano particles are prone to oxidation at higher temperatures, so performing the reaction in inert atmosphere like $\ce{N2}$ or Argon will ensure the Pd(0) particles.

by Emeraldine base to form palladium nanoparticles. See here

Details about the points can be found in this researchgate article. But since this is a school project, points 3-7 are out of scope. The best method would be the "reduction through carbon monoxide".


Equilibrium constant and reaction Quotient does reaction quotient After equilibrium has been achieved make any sense?
At equilibrium point rates of forward and backward reaction are same and the system is in dynamic equilibrium.The graph of concentration vs time shows a straight line parallel to x axis after equilibrium point.
Reaction quotient refers to ratio of concentration at any stage.This means that after equilibrium point reaction Quotient value will same as equilibrium constant as no more products are formed.So how can it have a value more than that of K ?!
I have been thinking about this for hours but I get more confused with time.

$K$ is just the specific value of $Q$ for which $\Delta_\mathrm r G = 0$.
In that case:
$$\begin{align}
\Delta_\mathrm r G &= \Delta_\mathrm r G^\circ + RT \ln Q \\
0 &= \Delta_\mathrm r G^\circ + RT \ln K \\
\Delta_\mathrm r G^\circ &= -RT \ln K
\end{align}$$

For any system not at the equilibrium it is possible that $Q < K$ or $Q > K$. What is the "forward direction" is just a convention, it depends on how you decide to write the equilibrium. There is nothing special in one direction or the other. The system will "move" toward the direction that leads to equilibrium in both cases. In the first case the reaction will consume reactants to generate products, while in the second case the opposite will happen.
You can express this idea also in terms of the ratio $\frac Q K$. If it is > 1 then the reaction will consume reactants. This is also often referred to as the "Le Chatelier principle".


How to remove white coating on the seashells after lemon acid and soap? What is the way to dissolve oxalates? I have tried to clean the seashells with lemon (or citric, as I was corrected) acid and then I washed them with soap (with pH above 10). There is a white coating on the shells now, which disappears when is wet and appears back when is dry. I suspect I accidentially covered the shells with some oxalate. Please help.

Seashells' outer layer (periostracum) mainly consists of conchiolin and polymorphic modifications of $\ce{CaCO3}$ (aragonite, calcite). To sum it up, I see two possible sources of the white coating after treating the seashells with citric acid:

Formation of calcium citrate, which has limited water solubility ($\pu{0.9 g/L}$ at $\pu{20^\circ C}$):
$$\ce{3CaCO3 + 2C6H8O7 -> Ca3(C6H5O7)2 v + 3CO2 ^ + 3H2O}$$
You can try to use acetic acid instead of citric acid in order to produce calcium acetate, which is very soluble in water ($\pu{347 g/L}$ at $\pu{20^\circ C}$). I don't think you can easily dissolve calcium citrate with acetic acid because calcium is complexated by citrate, and this is a very stable formation. The problem here is that the prolonged exposure to acidic medium can further damage the outer layer as more calcite will be washed away.
Conchiolin porous network remained after $\ce{CaCO3}$ has been eliminated with an acid might also appear white when dry and then, when the pores are filled with water, return to it's original appearance. This is rather a speculative theory, but if this is the case, you will need to perform mechanical surface polishing in order to get rid of the porous layer. This can potentially damage the seashells and should be done cautiously.


Deriving the equation for ionic current High-schooler here reading up on electrochemistry.....

I came across this equation (and its derivation) in the book Physical Chemistry, by Wallwork and Grant:

Now the thing is, the equation is valid for 'uni-univalent' electrolytes (like KCl). But I want a more....'versatile' formula.

So I attempted to modify that equation to suit electrolytes of the type AB (i.e- magnitude of charge on a formula unit cation is equal to that of the charge on a formula unit anion).

I wrote down the whole thing (my formatting skills are terrible, so to type it out would be a HUGE pain in the neck), pardon the handwriting, I hope it's legible enough....

As for that last question.....I was wondering if the equation could be made even more 'versatile' 3:)

( @getafix don't think of me as clingy but...I'm counting on you for this one too! )

I will be using the following equation from my previous answer

  $$I = (z \mu \nu c F E)A$$

You are interested in an electrolyte of the type $\ce{A^{2+}}$ $\ce{B^{2-}}$.

So, the "cation" current is

$$I_+ = (z_+ \mu_+ \nu_+ c F E)A$$

and similarly the "anion" current is

$$I_- = (z_- \mu_- \nu_- c F E)A$$

Here, $z_+ = z_- = z$  and $\nu_+ = \nu_- = \nu$. The mobilities, however, are different, so $\mu_+ \neq \mu_-$

So combining the contribution from the cations and anions in the solution, we get a total current

$$\mathrm{I}_{\text{tot}} = (z \nu c F EA)(\mu_+ + \mu_-)$$

After substituting  $\ E = \frac{V}{L}$,  one gets the desired result

$$\mathrm{I}_{\text{tot}} = \left (z \nu c FA \frac{V}{L}\right ) (\mu_+ + \mu_-)$$

In general, if $z_+ \neq z_- $  and $\nu_+ \neq \nu_-$ then simply write down the individual expressions for cationic and anionic contributions to current and add them.


According to Markovnikov rule and stability of carbocation which is most likely to form if HI "excess" is not given? Problem

Answer                          >
I thought "I" can stabilizes carbocation formed intially by resonance so there can be a shift of (+) to the carbon close to "I" and hence forming the product 1,1-diiodopropane but given is 1,2 product , whats wrong in this ? Or is that H+ and +I combined effect of CH3 is slightly greater than compared to resonance of "I"?

There are three reasons as far as I can see as to why the 1,2 product is favored. There may be more.

Steric hindrance
The iodine atom is large in size compared to the carbon atom. Having two iodine atoms on successive carbon atoms in a hydrocarbon is quite unstable. In fact, should two successive carbon atoms possess iodine substituents, they usually undergo elimination reactions. And here the 1,1 product would have two iodine groups on the same carbon atom!

Inductive effect of halogens
While it is true that halogen atoms have a positive mesomeric effect, this effect is almost always dominated by their negative inductive effect. If you see a halogen atom, you can usually neglect it's +M effects in favor of it's −I effects.

Carbocation stability
The carbocation formed in the process of formation of the 1,2 product is more stable (due to hyperconjugation) than the one formed in the process of formation of the 1,1 product (again, due to the −I effect of iodine).

Either way,  1,2-diiodopropane is more stable than  1,1-diiodopropane.


Extraction of glucose and fructose from golden syrup Not so long ago, I asked how to make golden syrup.

Now I am wondering how to extract glucose and fructose from it. I would then like to separate the fructose from the glucose (while keeping both of the materials.)

When I made the syrup I used citric acid as a catalyst.

How would you start the extraction if I only need glucose and fructose (separated)? I have no idea.

Extraction in terms of selectively extracting either glucose or fructose from the syrup is going to be difficult to impossible, at least as far as achieving any reasonable degree of purity is concerned.  The glucose-fructose separation method preferred by Dow Chemical is called ligand exchange chromatography.  According to this publication:

  Unlike the macroporous structure of DOWEX™ resins used in deashing,
  DOWEX MONOSPHERE™ 99 separation resins are gel beads which have a
  smooth, uniform surface.  While macroporous resins are opaque, gel
  resins are translucent.  Chromatographic separation resins are
  functionalized and the beads contain a significant amount of water.
  While going through the column sections, the sugars to be separated
  dissolve in the water contained within the beads.  Inside the  bead,
  the dissolved sugars interact with the calcium ions held by the resin.
  Fructose, glucose and water form weak ligand complexes with the
  calcium ion.  A stronger interaction in the fructose/calcium ion
  complex than in the glucose/calcium ion complex is the basis of the
  mechanism of separation of fructose from glucose.

As this is a readily available commercial product designed for use on massive scales, constructing a simple separation column based on these separation resins as described in the publication above should be a very doable, cost-effective approach to your problem.


How can different no. of unpaired electrons lead to different colours of co-ordination compounds? To make the question clear I would like to take an example. $\ce{K_3[Fe(CN)6]}$ and $\ce{K3[FeF6]}$ shows different colors in solutions. It is given in my textbook that it is because of different number of unpaired electrons.
But, how can the no. of unpaired electrons affect the frequency of light emitted?

The colors of co-ordination compounds are due to d-d transition of electrons. $\ce{CN^-}$ is a strong field ligand while $\ce{F^-}$ is a weak field ligand. Thus, there will be difference in the crystal field splitting energy in both of the given compounds. This, results in a different energy gap between the eg and t2g energy levels in them. Thus, when electrons undergo transition different frequencies corresponding to different energy gap is emitted.

I was hoping this as the answer. But, in my textbook the reason is attributed to only the difference in no. of unpaired electrons. Why is it so?

The crystal field splitting does definitely contribute.But it is not the only thing to contribute to the colour of the coordinate complex solution. The colour is equivalent to the spectra (emission or the absorption) of the solution, and hence by enumerating the factors contributing to the emission spectra of a solution, we can compare the colours of two different solutions.

It usually is an unpaired electron which shifts between the various available orbitals and paired electrons rarely do so, since the energy required to unpair acts as a barrier for its' transitions. Moreover, there might also be spin inversions (see Hyperfine structure) which might give additional lines in the spectra. Presence of unpaired electrons helps in producing additional energy levels in the electric and magnetic fields of the ligands (zeeman and stark effect), providing additional lines in the spectra. All these establish the uniqueness of the spectral distribution which the solution absorbs, and hence the colour.

The number of unpaired electrons also determines the resultant magnetic field of other species in the solution, which alters the energy levels of the solvating molecules.


Finding the equilibrium temperature between two substances in a constant pressure calorimeter I need to find the temperature two things equilibrate to in a constant pressure calorimeter given only their specific heats, masses, and starting temps.

The only thing I know for specific heat is that. $c = \frac{J}{G\times \Delta T}$, and how to work that around to find if one is missing, and well as $\frac{q}{\Delta T}$ for doing things with calorimeters, though a different kind. That's the only equation I know for this stuff. In this question I have neither the energy nor the change in temp. I know how to solver it if the two substances have equal specific heats, then you just find the mid point between temperatures. But two different substances, I have no idea. I tried to do it the hard way of just slowly taking of/adding one *C of each and matching up the energies required but that got me nowhere.

Assuming the heat absorption of the calorimeter is negligible (so if it's made of an insulating material, the only changes in heat will be of the contents themselves).

Substance 1 (the one at the higher initial temperature) has a heat = $q_1$

Substance 2 has a heat of $q_2$

Since substance 1 will lose heat until it comes to equilibrium with substance 2, the heat lost by 1 will be absorbed by 2, so $q_1 = -q_2$

If $q_1 = m_1c_{p1}(T_{{1}_{initial}} - T_{{1}_{final}})$ and $q_2 = m_2c_{p2}(T_{{2}_{initial}} - T_{{2}_{final}})$ and $T_{1_{final}} = T_{2_{final}} = T_f$ you can solve for $T_f$.

If the heat absorption of the calorimeter was not negligible, then $q_1 + q_2 + q_{cal} = 0$


A non-life process making glucose from carbon dioxide and water using electricity or light? update (Sept. 2018): Here is a bit of news that someone might be inclined to cite in a "No, not yet, but..." answer.

NASA has announced a CO2 conversion challenge!

Get out of your element! Convert CO2 into molecules to power bio-manufacturing in space.
Convert CO2 to Glucose:
Help us discover ways to develop novel synthesis technologies that use carbon dioxide (CO2) as the sole carbon source to generate molecules that can be used to manufacture a variety of products, including "substrates" for use in microbial bioreactors.
Because CO2 is readily abundant within the Martian atmosphere, such technologies will translate into in-situ manufacturing of products to enable humans to live and thrive on the planet, and also be implemented on Earth by using both waste and atmospheric CO2 as a resource.

It looks like this will be a multi-year project.

original question (Mar. 2017):
A large mass fraction of foodstuff a person would need on a long trip (for example, in space or otherwise not on Earth) would come from its caloric (energy) value. Scenarios have been explored for decades to use solar photovoltaic (or other electric) power to generate light to grow plants locally for food. Light provides energy for photosynthesis which combines $\ce{CO2}$ and $\ce{H2O}$ and makes glucose, (later converted to and stored as starch) which then provides caloric value to food.
$$\ce{6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)}$$
I'd like to know if a purely non-life based self-contained chemical process has been developed or is at least close to continuously converting $\ce{CO2}$ and $\ce{H2O}$ to glucose, using electricity directly, or electrically generated light.
By "non-life based" I mean it's not a bioreactor based on living cells. It could certainly use processes that are similar to those used in organisms.
The process would have to be potentially able to run stand-alone with minimum consumables, in the same sense of the term "process" as is mean for the Haber Process.
I am not asking if you think it could be done, but if there is a way that it could be done, based on well-documented research towards implementing complete Artificial Photosynthesis using light or electricity, not just a demonstration where a laboratory technician regularly adds separately prepared reagents and precursors for each step.

Popular research these days for CO2 fixation involves the following enzymes with associated conversion rate:

RuBisCO - 10 molecules/sec
ECR(Enoxyl-CoA carboxylase/reductase) - 80
molecules/sec

The major noncarbohydrate precursors are lactate, amino acids, and glycerol.
Here are some links to current research:

TiO2-catalyzed synthesis of sugars from formaldehyde
CO2 reduction -> Hydrocarbon eg. Formaldehydes/acids (CH2O).
Then Formose/Butlerov reaction (catalyst:base+Ca/Mg) -> sugar
Volume 1 of Comprehensive Glycoscience
CO2 to Glucose Whitepaper

Some topics for research:

Gluconeogenesis
Formose reaction
Calvin cycle (absence of light)


Need something that is dishwasher safe that breaks down paraffin I work for a company that has a lab where there work leaves a lot of dishes coated in paraffin. Today I learned that they wash these in a non commercial dishwasher to remove the stuff.... one issue they paraffin solidifies and gunks up the washer. They have been replacing these every few months or so. After talking with the lab staff, they expressed concern and are open to using an extra step to remove the paraffin before washing. What most of the internet has to say is obviously do not wash paraffin in a dishwasher. Feeling a little s.o.l here, does anyone know anything that could potentially help, so we can end the vicious cycle of dishwasher destruction.

This may be doable, per a source, to quote:

Paraffin wax is mostly found as a white, odorless, tasteless, waxy solid, with a typical melting point between about 46 and 68 °C (115 and 154 °F).

So as a pre-treatment, try applying a jet of steam to facilitate the removal of the paraffin wax.


Why doesn't a fractured salt crystal stick back together? If you crack a large crystal of a salt like NaCl, why don't the pieces spontaneously bond back together on contact? The bonding is an exothermic process and the charged ions are presumably stuck approximately right where they were before the facture.

You would have to get the two surfaces (or rather their crystallographic axes) together at the atomically perfect original angle to each other. Otherwise no overall (ionic) binding energy will result: Even if the surface is perfectly even, half the ions will be at places where they feel repulsion rather than attraction by the other side.

With polished metal surfaces, this works much better, also because metal atoms are (in "average") more mobile on a surface than ions.

Another point is humidity. Many surfaces, and certainly ionic substances, adsorb a layer of water from the air, practically instantly. It'll take quite a while before that has diffused out again when you press the surfaces against each other.


How to check purity of mercury metal Mercury is a very good solvent for many metals. We have mercury metal available, how do we check for its purity? Is there some kind of test available to verify its purity? And, more helpfully, which kind of metals are dissolved in it?

In case your mercury didn't come with an assay of the supplier, or has been used before, you might want to remove oxidizable metals by repeatedly dripping the mercury though diluted nitric acid and distilled water.

Subsequent distillation of mercury is certainly possible (at 25 mm Hg or lower), and has been reported by E. H. Riesenfeld and W. Haase in Ber. Dtsch. Chem. Ges., 1925, 58, 2828-2834 (
DOI).

In their article Über die Destillation von gold-haltigem Quecksilber ( = On the distillation of gold-containing mercury) the authors examined the distillation of gold-saturated mercury and stated that after two distillations, the gold content in the purified mercury was lower than 1.5 ng $\ce{Au}$ in 1 g $\ce{Hg}$.

UPDATE

In Joel H. Hildebrand, PURIFICATION OF MERCURY, J. Am. Chem. Soc., 1909, 31, 933-935, (DOI), the author seems to discuss further different methods,

THE PURIFICATION OF MERCURY BY AN ELECTROLYTIC METHOD
Science, 1933,
78(2027), 414-415, (DOI) might be woth a look too.

However, please notice that these are rather old references. Please make sure, that the methods proposed are still in agreement with currents standards of lab safety and waste management!


The decomposition of NaCN in water One method for making NaCN is the reaction of $\ce {2K4FeCN6 + 7H2SO4 = 12HCN + 2FeSO4 +2KHSO4 +K2SO4}$. The $\ce {HCN}$ is distilled into 50% $\ce {NaOH}$...after awhile the solution turns dark brown and you can smell ammonia.The reaction is likely $\ce {8NaCN +3H2CO3}$ (from water and $\ce {CO2}$)$\ce {=2NH3 + 6NaOCN + Na2CO3 + 2C}$.
What can be done to reverse this reaction? Is there any way to get rid of the dark brown color? The Merck Index says water solutions of $\ce {NaCN}$ rapidly decompose...if that is true then how are any $\ce {S_N2}$ reactions done with $\ce {NaCN}$ in water--ethanol solution? The internet has next to nothing on this problem.

Sodium cyanide is slowly decomposed by water, but the implication is that it can still be used over short time periods provided that acids are kept away and proper safety practices are followed.  See here.


Is Jet Fuel/Kerosene still toxic after drying out? A friend of mine is currently living in India. The people she is living with just treated the wooden door with either kerosene or jet fuel (its not clear which of the both, as the language barrier makes communication rather difficult). We do now wonder if touching or even being close to this door will be still dangerous after the kerosene/jet fuel dried out.

Kerosene is not a pure compound. To quote Wikipedia:

Regardless of crude oil source or processing history, kerosene's major components are branched and straight chain alkanes and naphthenes (cycloalkanes), which normally account for at least $70~\%$ by volume. Aromatic hydrocarbons in this boiling range, such as alkylbenzenes (single ring) and alkylnaphthalenes (double ring), do not normally exceed $25~\%$ by volume of kerosene streams. Olefins are usually not present at more than $5~\%$ by volume.[10]

[10]: American Institute of Petroleum (September 2010). "Kerosene/Jet Fuel Assessment Document" (PDF). EPA. p. 8. Retrieved October 28, 2016.

Because kerosene is a complex mixture, it is hard to say something specifically. However, the compounds noted are basically long chain alkanes and substituted cycloalkanes, a smaller fraction being alkylated aromatic rings. Benzene and naphthalene do not seem to be significant contributors.
None of these are classified as toxic. Most importantly, while alkylated benzenes and alkylated naphthalenes may still intercalate into DNA making them potential carcinogens, they are metabolised by a completely different pathway from benzene proper, making their metabolites much less dangerous. (Rather than the very reactive 1,2-epoxybenzene, various benzoic acids are the primary oxidation products.) That doesn’t mean they’re harmless, but toxic is the wrong designation and should be reserved for more dangerous compounds.

A general rule for all compounds is:

As long as you smell them, they can affect you.

The desired effect of painting a door with kerosene is, however, that the kerosene penetrates the door and stays there. The least-volatile constituents will remain on the door itself, protecting it from damage. If all the kerosene diffused away, the method would not be used because it would be worthless. Therefore, after the door has vented properly, letting the volatile components disappear, there is practically nothing left that will be harmful to any noteworthy extent.
In fact, I am include to proclaim that your average coloured paint may well contain more harmful chemicals that the kerosene does.
Yes, you can touch it, no you do not need space suits to be safe.


How to dissolve PLA (polylactic acid)? I'm a maker and I've been interested in 3d printing for years.

I have found a technique for dissolve the PLA supports in a model printed in 3D in HIPS

I have printed the model in HIPS and the support in PLA
I have put the 3D model in a ultrasonic cleaner in the solution with water and sodium hydroxide

(1 liter H2O + 200 grams NaOH)

The PLA has dissolved after 3 hours with temperature 65° and 1 hour of ultrasonic sound.

I attach a youtube link with a similar experiment

https://www.youtube.com/watch?v=uWYn5wJ72zY

Is it possibile use another substance / solution less dangerous and less toxic?

In theory, the PLA is biodegradable. Is it correct?

Ps:
I want use this combination of materials HIPS for model and PLA for Support.

Thank you in advance for the answers.

Ethyl acetate disolves PLA - which is typically the ingredient in non-acetone nail polish removers.

https://en.wikipedia.org/wiki/Ethyl_acetate

According to Wikipedia:

  "If the filament is soaked in a small amount of ethyl acetate, it will dissolve and can be used to clean 3D printing extruder heads or remove PLA supports. The boiling point of ethyl acetate is low enough to also smooth PLA in a vapor chamber, similar to ABS and acetone."


How should we express concentrations and partial pressures in ΔG=-RTlnK? I understand that $K$ will be dimensionless. Let's say an equilibrium involving $K_p$ and the reaction is $\ce{A -> B}$, $K_p$ will be dimensionless. If the reaction $\ce{A + B -> C}$, then based on the formula of $K_p$ will be:

$$ \frac{(\text{partial pressure of C})}{[(\text{partial pressure B})\times(\text{partial pressure of A})]}$$

Since partial pressure can be in $\mathrm{Pa}$, or $\mathrm{bar/atm}$, with the difference between them will be roughly 100000 times. So the $K_p$s will have 100000 difference. Putting the $K_p$ into $\Delta G=-RT\ln K$, there will be a huge difference of $\Delta G$. So what units of $K_p$ should we use in this case?

What units of K should we use in ΔG = -RT lnK?

The full formula is:

$$\Delta_r G^\circ = - R T \ln K$$

The circle after the G refers to a standard state, where reactants and products are present at standard concentrations/pressures. The equilibrium constant has to be given with reference to the same standard state. The subscript r stands for reaction. The balanced chemical reaction has to match between the Gibbs energy of reaction and the equilibrium constant expression. Let's say our balanced reaction is:

$$\ce{ \nu_A A + \nu_B B + \nu_C C + ... <=> \nu_Y Y + \nu_Z Z + ...}$$

Here, the different $\nu$ are the stoichiometric coefficients.

The correct equilibrium expression for this reaction uses activities of species raised to their respective stoichiometric coefficient. Activities are dimensionless, and are equal to one at standard state. For dilute solutions, these can be estimated as the ratio of a concentration (or partial pressure) divided by the standard concentration (or partial pressure):

$$ a_\text{species} \approx \frac{c_\text{species}}{c^\circ_\text{species}} = \frac{p_\text{species}}{p^\circ_\text{species}}$$

As a consequence, the equilibrium constant expression will always be dimensionless, and will not depend on the concentration measure used, as long as the standard state concentration is expressed using the same measure.

  Since partial pressure can be in Pa, or bar/atm, with the difference between them will be roughly 100000 times.

No, because you are dividing by the standard state, using the same concentration measure (and for ease of cancelling units, in the same units). For example, let's say the standard state of a species in the gas phase is defined as one molar. At ambient conditions, this would correspond to a partial pressure of about 0.04 atm. If the equilibrium concentration is - say - 2 M, that would correspond to a partial pressure of about 0.08 atm. Either measure of concentration gives the same result: The species is present at twice the concentration (partial pressure) as the standard state. No matter what concentration measure I choose, I get the same value for the equilibrium constant.

The same goes for switching pressure units. As you are always dividing by the same standard pressure, you always get the same result (either by converting the units of the partial pressure, or the units of the standard partial pressure, and then cancelling).

Switching to a different standard state

In biology and biochemistry, a different standard state is used. This affects both the values of the standard Gibbs energy of reaction, and the values of equilibrium constants. However, these changes are self-consistent, i.e. you still can calculate K from the Gibbs energy and vice versa, and you can convert easily from one convention of the standard state to the other.


If the atomic number is # of protons, why does emission of a beta (electron) particle increase the atomic number? Atomic number: the number of protons in the nucleus of an atom, which is characteristic of a chemical element and determines its place in the periodic table.

Beta emission:

$$\ce{^14_6C -> ^14_7N + ^0_{-1}\beta}$$

Why does emission of an electron increase the atomic number, when the atomic number is the number of protons, not electrons?

During beta particle emission one neutron is converted into proton and electron, to balance charge and electron can't remain inside the nucleus hence emitted from the nucleus.


Which complex is the most stable? Between $\ce{[Rh(Ph2PCH2PPh2)3]^{3+}}$ and $\ce{[Rh(Ph2PCH2CH2PPh2)3]^{3+}}$, which one is the most stable thermodynamically? I think it's about the size of the chelate ring effecting their stability, but i'm not sure.

It is risky to predict just by looking at the structure, of course there is the ring size effect and one could easily conclude that $\ce{[Rh(Ph2PCH2CH2PPh2)3]3+}$ is the most thermodynamically stable compound since you would have a 5-membered ring instead of a 4-membered as in $\ce{Rh(Ph2PCH2PPh2)3]3+}$. However, there is also the contribution of the electronic effect, which is the increase or decrease in electron density of the metal ion caused by the distribution of ligands around the metal center. The combination of electronic and steric effects was reviewed by C.A Tolman in 1977 for monodentate ligands and he proposed a way of measuring these effects experimentally and quantitatively, they are the so-called Tolman's parameters (Tolman, C. A. (1977). "Steric effects of phosphorus ligands in organometallic chemistry and homogeneous catalysis". Chem. Rev. 77 (3): 313–348.)

For bidentate ligands such as DPPE (2 $\ce{-CH2 -}$ groups between phosphorus atoms) or DPPM (1 $\ce{-CH2 -}$  group between phosphorus atoms) another important parameter is the bite angle "β" (the circular sector angle between the metal and the two phosphorus atoms). For example, in square-planar and octahedral complexes the ideal bite angle is 90º , deviations from this value will electronically destabilize the complex (KUMAR, M.  et al. Importance of Long-Range Noncovalent Interactions in the Regioselectivity of Rhodium-Xantphos-Catalyzed Hydroformylation. Organometallics, v. 34, n. 6, p. 1062-1073, Mar 2015.).

Bidentate phosphines with a natural bite angle of 120° preferentially occupy diequatorial sites in a trigonal bipyramidal complex whereas a bidentate phosphine with a natural bite angle of 90° preferentially occupy apical-equatorial positions (Casey, C. P.; Whiteker, G. T.; Melville, M. G.; Petrovich, L. M.; Gavney, J. A.; Powell, D. R. (1992). "Diphosphines with natural bite angles near 120° increase selectivity for n-aldehyde formation in rhodium-catalyzed hydroformylation". J. Am. Chem. Soc. 114 (2): 5535–5543.).

DPPM has a bite angle of 78º while DPPE has a bite angle of 85º (closer to 90º), so considering only the angular aspect, the $\ce{Rh-DPPE}$ complex would be more stable (Ligand Bite Angle Effects in Metal-catalyzed $\ce{C−C}$ Bond Formation
Piet W. N. M. van Leeuwen,, Paul C. J. Kamer,Joost N. H. Reek, and, and Peter Dierkes, Chemical Reviews 2000 100 (8), 2741-2770)*. Electronically, I assume that the 2  $\ce{-CH2 -}$ groups of DPPE could donate more electronic density to the P atoms than DPPM, which would somehow strengthen the $\ce{Rh-P}$ bonds.

However, it is the combination of sterical + electronic effects that can unequivocally decide which of the two complexes are more stable, and this would only be possible using computational chemistry or experimentally, though the above can be used when discussing about it on a statement or paper.


Hypervalency in elements in the second period In my experience, most texts that address hypervalency say that it only occurs from elements in the 3rd period and onwards. This explains the occurrence of $\ce{Cl2O7}$ or chlorine heptoxide. However, some 2nd period nonmetals like $\ce{C}$ and $\ce{O}$ show hypervalency.

Examples:

$\ce{CH5}$ - This is unlikely to occur but it does sometimes happen that carbon bonds to 5 atoms instead of 4.
$\ce{H3O+}$ - Here oxygen is hypervalent.

How is it possible for carbon and oxygen to each have 9 electrons if each orbital only holds 2 electrons? Do they switch between electrons or something?

Normally when we talk about a single covalent bond, we are referring to a 2-centre 2-electron bond, which means that there are two electrons holding two atoms together.

Carbon never forms 5 bonds. The only exception that I know of is the $\ce{CH5+}$ methanium cation, the bonding in which can be explained by a 3-centre-2-electron bond. The same kind of bond appears in diborane ($\ce{B2H6}$). In both cases, the octet rule (or duplet rule in the case of the bridging hydrogens in diborane) is not violated. It is just that those 2 electrons are shared amongst 3 different atoms, so each "bond" is effectively half a bond (in MO theory parlance we say that the bond order is 0.5). You could think of it as three of the C-H bonds being normal 2-electron bonds, and two of the C-H bonds being half-bonds (having one electron each). The total number of electrons around carbon is therefore $3 \times 2 + 1 + 1 = 8$.

The neutral species $\ce{CH5}$ does not exist, because it has one more electron than the $\ce{CH5+}$ cation. That would mean that you either have to put 9 electrons around carbon, or put 3 electrons around hydrogen, both of which are of course not allowed.

The hydronium ion $\ce{H3O+}$ is not actually hypervalent. It is similar to the ammonium ion $\ce{NH4+}$ in that a dative bond is formed from the lone pair on O to a $\ce{H+}$ ion.


How to estimate the concentration of OH that comes from just the autoprolysis of water in a basic solution? In my analytical chemistry course, a question reads:

What is the pH of a 0.10 M solution of THAM?  Is it valid to neglect the autoprotolysis of water?  Justify and explain your answer by estimating the concentration of $\ce{OH-}$ that comes from just water in this solution.

I'm guessing it has something to do with ionization of water ($\mathrm{K}_\mathrm{w}=1 \times 10^{-14}$), any insight would be greatly appreciated.  The calculated pH neglecting autoprotolysis of water is 10.5.

What auto-ionization means is best explained this way: If you put in  0.10000000 M of NaOH in pure water, how much hydroxide is there at equilibrium? You can imagine that "initially", adding that amount to a neutral solution gives you $\ce{OH^-}$ = 0.10000010, and that $\ce{[H^+]}$ is 0.00000010. However, the ion-product at this point is much greater than $K_w$, so some of the hydronium and hydroxide must react to bring the product down. If you solved the equation $(0.10000010-x)(0.00000010-x) = K_w$, and used $0.10000010-x$ to get $[OH^-]$, you would arrive at the concentration taking into account auto-ionization. You can see that $[OH^-]$ would just be a very minute amount greater than 0.100000000  because x is very small. The x is the result of taking into account the auto-ionization. As you can see, 0.1 dwarfes x, so we could have ignored auto-ionization and called it 0.1. Similarly, the hydronnium concentration would be different than if you had just divided $K_w$ by 0.1 rather than set up an equilibrium equation--but by so very little.

Now for your problem,

Let me first try to find the $K_b$ for THAM. If the pH of a 0.10 M solution is 10.5, then $\ce{[H^+]}$ $3.16*10^{-11}$. Using the mass balance for THAM, (I'll denote THAM as $\ce{B}$ and the conjugate acid as $\ce{BH^+}$),

$\ce{[B]}$ + $\ce{[BH^{+}]}$ = 0.10 M

$\ce{OH^-}$ = $3.16*10^{-4}$ (assuming 298 K) because  $\ce{H^+ }$ = $3.16*10^{-11}$

If you neglige the autodissociation of water, in other words, that all of the hydroxide in the solution is due to the THAM, your charge balance equation will be that $\ce{[BH^+]}$ is the same as $[OH^-]$ (by stoichiometry). Then, $\ce{[B]}$ = 0.09968, and $K_b$ is $1.00 * 10^{-6}$.

This should tie everything together:

Before, we had said that $[BH^+] = [OH^-]$. Instead, if you took into account the auto-ionization, you get the equation that $[BH^+] + [H^+ ] = [OH^-]$. Why is it justified to ignore $[H^+]$? Because the solution is so basic, $[H^+]$ is so small that if you tried to take auto-ionization into account, the pH would be roughly the same within quite a few decimal places. The change in $[OH^-]$ that comes about due to auto-ionization is dwarfed by the basicity of the THAM addition, just like the 0.100000 M NaOH example that I used. In other words, if you make the solution intensely basic, you can afford to ignore the effects of auto-ionizaton of water, and similarly for acids. In general, it is OK to ignore the auto-ionization process of water if the solution is more than just slightly acidic/basic.
As an interesting note, try to calculate the pH of a $10^{-10}$ M solution of hydrochloric acid without taking into account auto-ionization. You get a pH of 10, which is definitely not right.


What is the mathematical foundation for the probabilistic nature of chemistry? Is this best answered through quantum theory, Boltzmann statistics, or what?

For example, in a reaction energy diagram, only some of the reactants go over to the side of the products, and then get stuck there. So the progression of the reaction can be thought of in probabilities - what are the chances that something will go to products, and what are the chances the products will revert back. So you'd use the Arrhenius equation here. Could you dig deeper and find a quantum mechanical reason for this?

Actually, a huge amount of the probablistic nature of chemistry is simply the Law of Large Numbers (LLN). A mole of any substance is approximately $6\times10^{23}$ molecules, which is 600 sextillion molecules! The LLN kicks in at quantities much smaller than the sextillion level. Put simply, there are so many interactions occurring in any macro-level reaction that these are inevitably going to converge to the theoretical mean by the LLN.

Now your question adds in the issue of activation energy, but I still think it fits this model, because activation energy tends to produce a sharp difference: too low, and little reaction happens; but high enough, and we mostly go to completion. Once activation energy is high enough, the LLN applies.