Theoretical versus empirical equations I'm familiar with the fact that theoretical formulae are based on already known theory and are very frequently simplified (some equations don't include interactions between molecules for example) while the purpose of empirical equations (like polynomial equations) is just to follow experimental measurements even if it doesn't make any chemical or physical sense.

Which of these yield more accurate results (compared to reality)? What are the upsides and downsides of each group? Which are more universal? Which would require less compound-specific data (constants) to use it?

Edit: Removed van der Waals equation as an example of theoretical equations since, as pointed out by others, it's actually an empirical equation.

As pointed out already, the van der Waals (vdW) equation is also an empirical equation.  However, I think what you are saying is that the vdW equation is a physically motivated empirical equations where the empirical parameters in principle correspond to something "real" (e.g. molecular size and interaction strength) in contrast to a fit to an arbitrarily chosen fitting function such as a polynomial.

If you have lots of data points so that you can fit many parameters then a simple fit will be more accurate since it tends to have more fitting parameters than the physically motivated equation.  However, the simple fit will likely only be accurate for interpolation, and is still likely to be comparatively less accurate for extrapolation.


Kinetics energy expression of photon I need to calculate the kinetics energy of a photon and I'm trying to deduce a correct expression for calculating it.

So we have that

$K= \frac{1}{2} m_fv_f^2$

So $v_f= c$

For De Broglie equation we have that $m_f = \frac{h}{\lambda c}$

And then

$K = \frac{hc}{2 \lambda}$

But apparently the correct expression is:

$K = \frac{hc}{\lambda}$

Can someone help me figure out what I'm doing wrong?
Thanks!

A photon has no mass and moves at relativistic speeds, therefore you can't use the usual expression for the kinetic energy.

The energy of a photon is given by (as you already stated)
$$
E = \frac{hc}{\lambda}
$$
where $h$ is Planck's constant, $c$ the speed of light and $\lambda$ the wavelength of the photon.

You can derive it from the relativistic energy
$$
E^2 = p^2c^2 + m^2c^4
$$
with $m =0$ and using the fact that for a photon
$$
p = \hbar k = \frac{h\nu}{c} = \frac{h}{\lambda}.
$$


If a sample to be carbon-dated is wrapped in a piece of paper, will you still be able to effectively measure the radioactivity of the sample? If a sample to be carbon-dated is wrapped in a piece of paper, will you still be able to effectively measure the radioactivity of the sample?

Hardly, with or without paper wrap.

Most of the radiation will be absorbed by the sample itself, you'll have a hard time modelling the absorption to derive the actual remaining activity of the carbon. Artifacts are usually not a piece of completely homogeneous material. Also you'll have a hard time finding anyone with a suitable Geiger counter, if you want to do more than guesswork.)

Modern radiocarbon methods are destructive, you take a number of tiny (<1 mg), representative samples out of your artifact, and directly do mass spectrometry on them, without waiting for radioactive decay.


Why is enthalpy change = heat change only for constant pressure? Please help me, a mathematician doing his elective course on Physical Chemistry, out with this very simple question: why does the relation between enthalpy and heat $$\Delta H = \Delta U + p\,\Delta V = \Delta Q$$ hold only under constant pressure? With the integral definition of work, couldn't one just as well write $$\Delta H = \Delta U + \int_{V_i}^{V_f} p(V)\,\mathrm dV = \Delta Q $$ in the case where the pressure is not constant during expansion?
I'm still having some trouble with chemical notation of derivatives/infinitesimals as well as with the physical realities behind the formulae, so there might any number of basic things I'm missing here.
Thanks in advance. Also as this is my first question here, apologies if I did anything wrong or if I missed a similar question.

Firstly, $\int_{V_i}^{V_f} p\, dV\neq p \Delta V$ unless $p$ is constant. Hence the formula holds in that case only. Also, the definition of $H=U+pV$, hence if we wanted $\Delta H$, we should take:$$\Delta H=\Delta U +\int_{V_i}^{V_f}\,d(pV)$$
Here, of course, $d(pV)=pdV+Vdp$ using the product rule.


Change in standard Gibbs energy is always zero? So I have this statement from our professor :

  We have $\Delta G^\circ = -RT\ln K$ and under standard conditions the activities of all substances in the reaction is $1$ ($a=1$), therefore $\Delta G^\circ > 0$.

I dont understand it.If every $a=1$, then $K = 1$ (as the division of activities) and thus $\ln K =0$ which would make $\Delta G^\circ = 0$ as well instead of $>0$,right?

No. $K$ is not a ratio of instantaneous activities! Let's say you hypothetically prepare a system where all components have unit activity. In that case you would have the reaction quotient

$$Q = \prod_i a_i^{\nu_i} = 1$$

but $K$ is the ratio of activities at equilibrium. If that hypothetical system is not at equilibrium, then it will go towards equilibrium. When it reaches equilibrium, the activities are no longer going to be equal to unity, and $K$ will not be equal to 1.


Can we use peaks in any chromatogram to identify a compound Does a compound give a specific chromatogram peak, using which we can identify it in a mixture?

Compounds have specific retention times that are used to identify their respective peaks in chromatograms. Usually these times are dependent on many factors (temperature, column type, mobile phase...), and as such some kind of pre-calibration of the machine with respect to compounds in the mixture has to be done. The area of the peak is then indicative of the amount of substance, also subject to calibration (or comparison with reference substances with known concentrations).

For structure elucidation of previously unknown compounds in mixtures other methods such as GC-MS are much better suited.


Why does a liquids height in a Barometer depend linearly on pressure? Here is a derivation I found relating the pressure P, gravity g, height h, and density $\rho$, of a liquid in a barometer

$$
P=\frac{mg}{A}\\
V=Ah \\
P = \frac{mgh}{V}\\
\rho=\frac{m}{V}\\
P = \rho g h
$$

However, this derivation seems misguided, since $P = \frac{F}{A}$, and P is the pressure of the gas, not of the liquid, as the derivation later relates it to.

Furthermore, it doesn't make sense to me that the height of the liquid and the pressure are necessarily linearly related. Wouldn't this depend on the compressibility of the liquid? Or rather, it seems like depending upon the properties of the liquid, it may be very difficult to change the height past a certain point, for example. In this sense the added energy from increasing the pressure goes into bonds or is stored in electrical interactions.

In a barometer at equilibrium, the pressure of the mercury equals the pressure of the air (at the air/mercury interface).

The compressibility of mercury in principle does make it non-linear.  The density would be 3.8 parts per million higher at the air interface than at the vacuum interface.

Certainly there are other greater sources of error: the vacuum isn't really a vacuum but contains mercury vapor, density varies with temperature, vapor pressure varies with temperature, the size of the tube varies with temperature, there are surface tension effects.

Just like any experimental technique, you need to evaluate sources of error and report a measurement with appropriate uncertainty.


Confusion over boiling point of gallium I am completing a project on gallium, and I need to include its boiling point. I thought that this would be fairly simple to look up, however, it appears that different sources quote different temperatures as the boiling point, normally either ~2200 °C or ~2400 °C.

Does anyone know why two different values are presented, or which value is correct?

Below are some examples of websites quoting each figure.

2200 °C:

https://education.jlab.org/itselemental/ele031.html

http://www.rsc.org/periodic-table/element/31/gallium

https://www.livescience.com/29476-gallium.html

2400 °C:

http://www.chemistryexplained.com/elements/C-K/Gallium.html

https://www.britannica.com/science/gallium

http://www.chemicalelements.com/elements/ga.html

This question has been recently raised in Chemistry & Engineering Letters. The CRC Handbook (2017) in section "MELTING, BOILING, TRIPLE, AND CRITICAL POINTS OF THE ELEMENTS" [1, p. 4-117] lists the following values for gallium, quoting original publication [2]:

\begin{align}
t_\mathrm{tp} (\ce{Ga}) &= \pu{29.7666 ^\circ C} \\
t_\mathrm{m} (\ce{Ga}) &= \pu{29.7646 ^\circ C} \\
t_\mathrm{b} (\ce{Ga}) &= \pu{2229 ^\circ C} \\
\end{align}

  [...] values are given for the solid-liquid-gas triple-point temperature $t_\mathrm{tp}$ and pressure $p_\mathrm{tp}$; normal melting point at $\pu{101.325 kPa}$ pressure $t_\mathrm{m}$; normal boiling point $t_\mathrm{b}$; and critical temperature $t_\mathrm{c}$ and pressure $p_\mathrm{c}$. All temperatures are on the ITS-90 scale.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.
Geiger, F.; Busse, C. A.; Loehrke, R. I. Int J Thermophys 1987, 8 (4), 425–436 DOI: 10.1007/BF00567103.


What are the non-aqueous solvents for sodium polyacrylate? I am attempting to use sodium polyacrylate as a sodium transport mechanism in a sulfur-bromine cell. I can upset the sodium balance with sodium salts, but the contamination introduced by the non-metallic ion from that salt is unacceptable. Water is a solvent that can be ultimately coerced into a solution compatible with the acrylic polymer I wish to matrix the polyacrylate within.

I have tried (with no success) alcohols, acetone, and ketones. Are there non-aqueous solvents I should consider that suit my needs that I have not yet tried?

Sodium polyacrylate can locally organize solvent molecules to become soluble. Since the backbone is huge compared to water, the water aligns with its more positively charged end towards the carboxylate site while the sodium ions end up on the negative end of the water. I can't find the good papers right now, but if you search google scholar for using polyacrylic acid as a draw solute for forward osmosis you can dig way deeper into the mechanisms.

What this means practically is that you need really polar solvents that can line up with the ions. Wikipedia has a good page explaining solvent classification in more detail with a table of solvents including their dipole moment (quantifying the polarity). As suggested by Mithoron, DMSO seems to be a good option. Unfortunately, after that you start getting into much nastier solvents like acetonitrile and DMF.


Enzyme-Kinetics. Michaelis-Menten Curve A question regarding the Michaelis-Menten curve. For those of you who are well-versed in chemical kinetics(or even better, enzyme-kinetics) you most likely have come across this curve before. It is a hyperbolic curve, eventually reaching a plateau, where the Initial velocity(reaction rate) of the reaction is given as a function of the substrate concentration. Befofe I go on to say what has been bothering me I just want to remind that once the curve has reached a plateau this is known as the maximal initial velocity and it is caused by enzyme saturation by the substrate.

Now what has been truly bewildering me is the fact that when these experiments are done the enzyme concentration, which is always kept constant, is always much much lower than the substrate concentration(even when the substrate is at it's lowest concentrations). I'm talking about Enzymes having nanomolar concentrations while the substrate is five or six orders of magnitude higher. Surely the amount of substrate is already more than enough to saturate the enzyme from the get-go with the smallest concentrations. How is this not so?? When I saw the ratio of enzyme molecules to substrates I figured it is way more than enough to saturate the enzymes thus as a result the Michaelis-Menten curve wouldn't make sense and that the curve should start of as a plateau at the maximal initial velocity and just stay that way forever. I hope this wasn't too confusing.

Thanks for any attempts at an explanation

Part of your confusion lies in the all-too common belief that it's the ratio of ligand to binding partner (e.g. substrate and enzyme) which determines the amount of binding. That's not the case. Instead, it's the absolute concentration of free ligand which determines how much is bound.

Let's take a microscopic view, and look at a single enzyme molecule. The enzyme doesn't know or care about what the other enzymes are doing, it only knows about its own local environment. To bind a substrate molecule, it needs to encounter (collide with) a substrate molecule in the correct orientation. How frequently it does that is directly proportional to the amount of free, bind-able substrate molecules there are in solution. For bound enzymes, (ignoring catalytic turn over for the moment), transitioning to the unbound state is a first-order process, independent of the concentration of anything else in the solution. (The total macroscopic rate of unbinding events depends on the number of bound complexes in solution, but for an individual bound enzyme it's concentration independent.) So for a single enzyme molecule, the fraction of the time it spends in the bound state is only dependent on the concentration of free ligand, so the time average of a large number of molecules is going to reflect that same distribution, by the ergodic hypothesis.

If binding is difficult but release is easy, then you're going to need a large amount of substrate in order to push the rate of binding to be comparable to the rate of release. If binding is easy but release takes a long time, you won't need as much substrate. That's what the KD of a binder (and roughly KM of an enzyme) measures. It's the concentration of the free ligand which makes the effective on-rate an the off-rate identical, and thus results in enzymes binding about half the time (or equivalently, at any one time half the enzymes are in the bound state).

So it doesn't matter if you have 100 molecules of substrate for every molecule of enzyme - if you're dilute enough, the concentration of substrate will be too low, the rate of enzyme/substrate encounters will fall off, and any enzyme-substrate complexes which form will fall apart long before some other one can form.

Enzyme/substrate (and ligand/binder) ratios are important, but mainly because it's the concentration of free substrate, rather than the concentration of total substrate, that's involved in the calculation. As your enzyme binds substrate, it depletes the pool of free substrate. If your enzyme concentration is large in comparison to your concentration of substrate, substrate going to the enzyme-bound state could cause a significant drop in free substrate concentrations, leading the nominal substrate concentration to be different from the actual free substrate concentration. (This is also part of why people primarily work with initial rates - the amount of free substrate drops over time due to catalysis.)

Caveat: KM is not quite the binding affinity of an enzyme for the substrate. In addition to the standard off rate of dissociation, there's also the "off rate" due to catalysis. There's also the lifetime of catalytic intermediates and the product-bound state which comes into play. With typical Michaelis-Menten assumptions - single step catalysis which is very slow compared to binding, with rapid product dissociation - the difference is minimal, but for many enzyme systems those assumptions don't hold.


Ultrasound Vegetable Glycerin homogenization We are trying to homogenize VG (USP) with propylene-glycol using ultrasonic homogenizer with titanium probe, but we are experiencing very strange - bad odor from the emulsion after sonication.
We tried different amplitudes, different time settings, different power settings. We can control amplitude for 2 modes only, I believe they are 50% and 100%.
Also if I am sonicating pure vegetable oil for just 5-10 sec I can smell that strange odor, it is very similar to that, I am getting with VG.
I just want to know, what this smell can come from? From our probe? No it is clean, we can sonicate water well.

Depending on intensity, ultrasound can produce intense heat at some locations. In fact, cavitation can induce temperatures far exceeding 5,000 K. Even a small ultrasonic room humidifier can burn through a plastic comb held in the upwelling stream. The sound intensity is decomposing the oil and/or propylene glycol.

Try lower frequencies and lower sound intensity. You might experiment with different shape containers, as well, to avoid focusing the sound into a high-intensity spot.


Is ruthenium really invulnerable to aqua regia? Recently, I answered a question "What metals aren't dissolved in/attacked by aqua regia?" in which I finally concluded "ruthenium is resistant to aqua regia in any form at any concentration of acid".Fellow answerer  @Linear Christmas also supported my statement

However, in my answer, I pointed out that something is not right. In the blockquote I wrote:

  Ruthenium can be dissolved in aqua regia only when in the presence of oxygen (Why? will be asked in a separate question)

It is written that ruthenium can only be attacked in aqua regia in the presence of oxygen. And for this statement, this question is asked. This statement can also be found in the wikipedia article of noble metal.

In the wikipedia article of ruthenium it is clearly written that:

  Ruthenium (...) is not attacked by acids (even aqua regia)

Even in the handbook, it is written that (statement can be found in my answer):

  Ruthenium is by far the most chemically resistant of the noble metal,
  unattacked either in hot or cold aqua regia.

Also this video by Nurdrage shows that ruthenium is invulnerable to aqua regia or any kind of acid.

However in this google book, it is written:

  Pure ruthenium powder or sponge is not appreciable attacked by hot, single or mixed mineral acid(aqua regia).

Why it is written "not appreciable attacked"? Maybe it should have wrote invulnerable or resistant. I am assuming that the corrosion rate is not appreciable/very slow i.e. corrosion is too slow to get noticed by naked eye.

Is my assumption correct? Can ruthenium be attacked in aqua regia only in presence of oxygen? Why wikipedia wrote that statement regardless of the fact that ruthenium is resistant to aqua regia in any form at any concentration of acid? I have not found any evidence that shows ruthenium is vulnerable to aqua regia.

Page 807 of A Textbook of Inorganic Chemistry says that iridium, rhodium. and ruthenium are "almost unaffected" by aqua regia.  So "completely invulnerable" = no, but "resistant" = definitely yes.


What conditions would make SN1 products the major products over E1 products, and vice versa? It seems that if a carbocation intermediate is formed through solvolysis, then the product(s) are going to always be a pretty even mix of SN1 and E1 products.

Are there any conditions that greatly favor one mechanism? I thought that a very bulky solvent - like t-$\ce{BuOH}$ - would favor E1. However, the carbocation is $\ce{sp^2}$ hybridized, so there still wouldn't be a lot of steric hindrance to prevent SN1, right?

In a practical way, you would have to consider each particular case, there are tons of reasons that could make a reaction to yield mostly E1 or SN1 product (it will be always a mixture, that can vary from 50:50 to 0.01:99,9, I hope I'm making myself clear, they are always competing reactions).

But there are several factors that contribute to make one of the two more favorable than the other, for example:

The kind of the reactant employed:

If you use hydrogen halides, ($\ce{HCl, HBr, HI}$) you will mostly obtain SN1 products.

If you use acids like $\ce{H2SO4, H3PO4~or~TsOH}$ you will mostly obtain E1.

Why is this? The reagents/acids of the first case, are very good nucleophiles. The reagents lisetd on the second case, are awful nucleophiles.
A good nucleophile is key for a substitution. But for elimination you only need something that can abstract a product in the alpha position to the leaving group.

On the other hand, very strong and small bases, can favour an elimination through E2 mechanism, with almost no substitution taking place. E1 mechanisms is usually observed only under mild basic conditions.

Solvents, like you say, also can favour one type or another:

In some cases, solvent acts a base for E1 reactions. (i.e. $\ce{EtOH -> EtOH2+}$). Bulkiness of the solvent can cause both E1 and SN1 (in case that the solvent itself is the nucleophile, which usually happens) to be slower: it has to approach the compound that undergoes the reaction anyway. Actually is easier to approach a hydrogen than a carbon, so probably will favour SN1 over E1, BUT, do NOT take this as a determining factor at all. The alpha carbon from which you need to abstract the proton might be more or less hindered.

Solvents with low nucleophility also decreases the % of substitution reaction.

Both E1 and SN1 are favoured in protic solvents, in general.

To sum up, you will always have both products, but yes, there are factors that can be modified in order to maximize one of the two reactions. There are some general considerations, like the ones I posted, but in practice, each reaction is a world.


I need to report MoS2 values from Mo and Sulfur numbers I'm having a hard time figuring out how to do this and yes, I have the molecular weights, just don't know how to calculate. Been 20 years since college.

EDIT: I'm working with Copper ores. One customer reported Mo% as 5.2, S% as 5.3. The other customer reports $\ce{MoS2}$ values (this particular one is 12.34%) I need to compare these values.

EDIT #2::  I need to report 5.2% Mo and 5.3% Sulfur as ONE value - As "MoS2" if that is possible.

As the molecular mass of molybdenum is $95.95~\mathrm{g/mol} \approx 96~\mathrm{g/mol}$ and the molecular mass of sulfur is $32.06~\mathrm{g/mol} \approx 32~\mathrm{g/mol}$ one can construct a ratio between both molecular masses that in turn can be solved easily.
$$\frac{96~a}{32~b}=\frac{5.2}{5.3}=0.98\approx1$$
Which gives us after the multiplication with 32/96 on both sides
$$\frac{a}{b}=1\cdot\frac{32}{96}=\frac{1}{3}$$
So customer 1's determined "compound" is not $\ce{MoS2}$ but more $\ce{MoS3}$. Now I would guess that they don't analyze the copper ore for the compound $\ce{MoS2}$ but for both elements $\ce{Mo}$ and $\ce{S}$ separately which is why they came up with $5.2~\%~\ce{Mo}$ and $5.3~\%~\ce{S}$.

In my opinion, you should ask customer 2 to split up it's $12.34~\%~\ce{MoS2}$ into the percentages of each element and compare them with $5.2~\%$ and $5.3~\%$.

Using the ratio for $\ce{MoS2}$ which is $60:40$ one would estimate those $12.34~\%$ to split up into $7.40~\%~\ce{Mo}$ and $4.94~\%~\ce{S}$ or into $6.16~\%~\ce{Mo}$ and $6.18\%~\ce{S}$ if it is more like $\ce{MoS3}$.

Or to answer your question ... probably you can simply add both values to give you $10.5~\%$ which nonetheless will not be equal to $\ce{MoS2}$ based upon the data you are given!


Role of phosphoric acid in redox titrations For redox titrations of $\ce{Fe^2+}$ versus $\ce{KMnO_4}$ or versus $\ce{K_2Cr_2O_7}$  , role of phosphoric acid is quite unclear to me. In dichromometric titration, $\ce{H_3PO_4}$ complexes $\ce{Fe^3+}$ and thus lowers the  potential of the $\ce{Fe^3+/Fe^2+}$ system, but if it is a permanganometric titration, $\ce{H_3PO_4}$ strictly complexes to mask the yellow colour of $\ce{FeCl_3}$ but plays no part in lowering the potential of the $\ce{Fe^3+/Fe^2+}$ system. (role of ZR solution if $\ce{Cl-}$is present in the system )

My professor strictly mentions this as a key point of difference between roles of phosphoric acid in these two types of titrations and we cannot write the lowering potential point for the latter but that makes no sense to me. I do not understand why $\ce{H_3PO_4}$ in titrations versus $\ce{KMnO_4}$ will not reduce the potential of $\ce{Fe^3+/Fe^2+}$ system.

I don't know why your professor is unnecessarily creating an example which has no real value! No one would do potentiometric titrations of iron in the presence of phosphate.

(i) Potentiometric titrations have nothing to do with color or need masking.
(ii) One would need to do a preliminary separation of iron via ion exchange resin.


Bond length comparison between sulfur tetrafluoride and thionyl tetrafluoride I'm comparing sulfur tetrafluoride and thionyl tetrafluoride. The lone pair has more s character compared to a double bond. Therefore, sulfur in $\ce{SF4}$ should have a greater s character compared to sulfur in $\ce{SOF4}$.
Since $\ce{SF4}$ has a greater s character it's bond length would be comparativly smaller than that of $\ce{SOF4}$.
Therefore, $\ce{X}>\ce{Y}$. But my book says otherwise.
Did I go wrong anywhere?

The simplest way to compare these two compounds is to compare sulfur’s oxidation state. In $\ce{SF4}$, sulfur is in the $(\mathrm{+IV})$ state. In $\ce{SOF4}$, it is $(\mathrm{+VI})$. The higher a compound’s oxidation state, the higher its formal positive charge and thus the more stable (i.e. lower in energy) its orbitals are. Orbitals reduced in energy means contracted orbitals which in turn leads to shorter bonds.

Other than that, the two compounds are pretty identical. Both feature two two-electron-two-centre $\ce{S-F}$ single bonds and one four-electron-three-centre $\ce{F\bond{...}S\bond{...}F}$ bond; the latter can be understood with the following resonance depiction:

$$\ce{F^-\ \ S^+-F <-> F-S^+\ \ F-}$$

Even the structures are comparable although the angles are slightly different.


How to prepare 2-bromobutane from 1-butanol? So my answer (g then b) was incorrect. I thought that the bulky strong base of b would do E2 on 2-bromobutane, making it but-1-ene, then just add $\ce{HBr}$. What is wrong with my way of thinking?

The hydroxyl fragment isn't a good leaving group, unless the medium is acidic. Using just HBr, you can convert the hydroxyl group into an oxonium ion (the H2O fragment would be a good leaving group). Upon heterolysis of the C-O bond, a hydride shift occurs simultaneously to create a secondary carbocation. The exogenous bromide can act as a nucleophile towards the secondary carbocation to form 2-bromobutane.


Which d orbitals can form sigma, pi, delta bonds? If two $\mathrm d_{xy}$ orbitals approach each other on $x=y,$ $z=0$, would a σ bond be formed? I would think so.

Can $\mathrm d_{z^2}$ form π bond with another $\mathrm d_{z^2}?$ (As all others can on $x$ or $y$ or $z$ as internuclear axis?) I am not sure but overlapping similar to p could be observed.

I would assume all except $\mathrm d_{z^2} - \mathrm d_{z^2}$ form δ bond. (both of same type like $\mathrm d_{xy}$ with $\mathrm d_{xy}.)$

What are the faults in my logic?

Sigma, pi and delta denote how many planar nodes are in the bond. Sigma bonds have no node, pi bonds have one and delta bonds have two. You can tell what kind of bond forms by how the orbitals overlap. Two single lobes form a sigma bond, two pairs of lobes form a pi bond and two quartets form a delta bond.


Preparation of alkyl halide from alcohols $\ce{R-OH + HBr -> R-Br + H2O}$

An $\ce{R+}$ carbocation is formed during this reaction.

If $\ce{R}$ is any alkyl group instead of methyl or ethyl, can the beta hydrogen to the hydroxy group be removed to form a double bond, similar to a dehydration??

It can happen.

It practice, you avoid it happening by keeping the temperature low. $\ce{Br-}$ is a better nucleophile than $\ce{H2O}$ is a base, so at lower temperatures, the substitution is faster.

If the elimination happens, alkenes react with $\ce{HBr}$ to form alkyl bromides. As long as acid is present, the carbocation can reform. For example, with isopropyl alcohol

Direct SN1 Substitution
$$\ce{(CH3)2CH-OH + HBr -> (CH3)2CH-OH2+ + Br-}$$
$$\ce{(CH3)2CH-OH2+ -> (CH3)2CH+ + H2O}$$
$$\ce{(CH3)2CH+ + Br- -> (CH3)2CH-Br}$$

Substitution by elimination-addition
$$\ce{(CH3)2CH-OH + HBr -> (CH3)2CH-OH2+ + Br-}$$
$$\ce{(CH3)2CH-OH2+ -> (CH3)2CH+ + H2O}$$
$$\ce{(CH3)2CH+ + H2O <=> CH3-CH=CH2 + H3O+}$$
$$\ce{(CH3)2CH+ + Br- -> (CH3)2CH-Br}$$

In some cases, two processes may appear to generate different alkyl halides, but this would be no different than if a carbocation rearrangement occurred.


Would sodium salicylate dissolve in HCl? I've been searching around the web and the consensus seems to be sodium salicylate dissolves poorly in HCl because it would form salicylic acid. However, I don't understand why forming salicylic acid product (which is poorly soluble in water) would make sodium salicylate (which is the reactant) dissolve poorly in HCl. If HCl doesn't contribute to making it more soluble, wouldn't the polarity of the water molecule dissolve it anyway?

HCl is in almost all cases already aqueous for common chemistry, mostly dissociated into $\ce{H^+} + \ce{Cl^-}$ because it is a strong acid, which means water is the primary solvent, not HCl. Pure HCl under ambient conditions is actually a gas (thus certainly not a pure liquid solvent).

That leaves the compound in question insoluble in the water after being protonated by $\ce{H^+}$ to form the insoluble acid molecule.

The process I mentioned in the comment is illustrated here:


Will sodium sulfite effect the oxidization of toluene to benzoic acid with potassium permanganate? I am trying to oxidize toluene to a mixed salt $\ce{K/Na}$ benzoate/ benzoic acid using the potassium permanganate method, but I am using OTC sourced $\ce{KMnO4}$ which has a significant amount of sodium sulfite contamination.

Will that hinder the reaction?  And is there a practical method for filtering said contaminants?

How do you know your permanganate contains sodium sulphite?

It seems highly unlikely because $\ce{KMnO4}$ is a very strong oxidiser and sulphite is very easily oxidised to sulphate:

$$\ce{SO3^{2-}\to SO4^{2-} +2e-}$$

Also, it seems improbable to find a sodium salt in a potassium salt.

Even if the $\ce{KMnO4}$ contained some "surviving" $\ce{Na2SO3}$, during the oxidation of the toluene, it would quickly be oxidised to sulphate. At worst your benzoic acid yield would suffer slightly.


Bravais Crystal Lattices and Voids What is the difference between Hexagonal Close Packing (HCP) structures and the hexagonal crystal Bravais lattice? I've looked up some sites but can't seem to understand them. It would help if you backed up the answer with your specific explanation of Bravais lattices and unit cells.

Edit:
I thought the difference was that the lattice consisted of many unit cells and the unit cell packing could be like hexagonal, cubic or simple. But then I looked at the stoichiometric defects which talked about molecules being in interstitial spaces. But in the unit cell, the spaces are already filled. So how is that possible.

2 brief observations about the answer by Mitchell.

a hexagonal lattice does not exist in 2-dimensions.

By definition,A lattice is an infinite array of points in space, in which each point has identical surroundings to all others. For example consider a 2d cubic lattice; now, for example, choose one of the lattice point (we call it point A) and connect it with an arrow with the point just above and with another arrow with the point on its right. Now move these two arrows as a solid object to another point of the lattice (we call it point B); you will see that the arrow will again connect this other lattice point with other two lattice points, one just above, the other at its right. You find the same thing for both point A and point B, whatever choice you make.
You can do the same job with a hexagonal arrangment of points. Choose again an arbitrary point A in the hexagonal arrangement and plot the two arrows connecting point A to two first neighbouring point. Then again move the two arrows as a solid object from point A to one of the 3 first neighbouring three points. You will see that now arrows will end where no point is present in the space. This is because the 2D hexagonal arrangment of points is actually constituted by two different sublattices, whose symmetries are trigonal.

the different crystal systems are defined by the symmetry operations that are present, not by their metric.

The group of the symmetry operations define the symmetry along the different main crystallographic directions. For example, in the cubic system we have a=b=c and alfa=beta=gamma=90 metric because the symmetry operations that are present impose this.
The tetragonal symmetry simply dictates a=b and alfa=beta=gamma=90; there's no METRIC restriction on the c-axis. The SYMMETRY along the c-axis is different from that along a and b, but in principle it is possible to find a crystal structure that has a tetragonal structure (= symmetry), but a cubic metric, that is a=b=c (and it is still tetragonal!). there are several compounds where this is observed. In conclusion, inequalities strictly hold on symmetry, not on the metric of the crystal structure.


Cyclisation of acetone on reaction with conc. H2SO4 Mechanism for reaction of acetone with concentrated $\ce{H2SO4}$:
The product is supposed to be mesitylene, and on various sites I found the reason behind it to be aldol condensation And some sites provided me with unusual products( as for a high school student)

Does this reaction have a reasonable reaction mechanism or do we just have to mug this up?

What are these cyclisation kinda mechanisms called?

Do they have a category that is to be understood?

You may have noticed that your question received a downvote. This was probably because you did not show any effort toward a solution of the question. I posted in a Comment some clues to assist you in a solution. In the interim an Answer was posted that showed the compounds formed in the reaction of acetone with H2SO4 but it did not illustrate a mechanism for the formation of mesitylene, which was your question. Here is my contribution to your understanding.
The reaction between the keto form of acetone 1a and its enol 1b forms aldol 2. Acid-catalyzed dehydration of 2 via the enol 3 leads to mesityl oxide 4. The loss of water from 3 may be stepwise but, to save space, I have presented the loss of water in a single operation. The third unit of acetone is incorporated via the vinylogous enol 4b to form the dienone 5a. Enolization of 5a forms 5b' that allows rotation about the red single bond in 5a and 5b' to form conformation 5b". This species may cyclize by an allowed thermal electrocyclization to form cyclohexadiene 8 which is readily dehydrated by acid to mesitylene 9. Alternatively, enol 5b' may reketonize to 6 and undergo cyclization to cation 7 and thence on to mesitylene 9. continued...

Addendum: An old name for acetone was mesite, the source of mesitylene. Read about it here and here.


What happens if you add NaCl during the neutralization reaction between HCl and NaOH? In class, I was taught that $\ce{NaOH}$ and $\ce{HCl}$ react completely (no equilibrium). I get the equation: $\ce{NaOH + HCl -> H2O + NaCl}$.

However, if I were to add a lot of $\ce{NaCl}$ to the solution, would some $\ce{NaOH}$ and $\ce{HCl}$ be formed according to Le Chatelier's principle?

You yourself said that there is "no equilibrium", but then forgot that Le Chatelier's principle only applies to any system at equilibrium!  Recall its definition:

  Le Chatelier's principle: a principle stating that if a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

In your case, as also MaxW pointed out above, $\ce{NaOH}$, $\ce{HCl}$ and $\ce{NaCl}$ are all completely ionized in water (and their ions are hydrated/solvated by water molecules). There is no such thing as an "equilibrium" in this case, as the reaction, being an ionic reaction, completely proceeds in the forward direction.


Density range of table salt? Does anyone have a reference for the density range of common table salt $(\ce{NaCl})?$
A pure $\ce{NaCl}$ crystal has a density of $\pu{2.16 g/cm^3}.$  However, the salt granules in table salt don't pack perfectly — there's a lot of air mixed in.
I carefully measured some Morton's iodized table salt at home $(> 99\%$ $\ce{NaCl};$ remainder is calcium silicate, dextrose, and $\ce{KI}),$ and got a density of $\pu{1.40 g/cm^3}*,$ which gives a packing fraction of $1.40/2.16 × 100\% = 64.8\%.$
Interestingly, this is (within my measurement error) essentially the same as the $64\%$ random close packing limit for monodisperse frictionless hard spheres.
But I don't know how much variation there is in the density of table salt, and have been unable to find a reference online.
*Here is how I measured the density:
I started with a metal 1 tbsp measuring spoon.  I didn't trust that its volume was actually 1 tbsp, so I filled it with water and measured the weight of the water (14.25 g) with a calibrated centigram scale, and its temperature (76 F) with a thermometer.  Since water @ 76 F has a density of $\pu{0.997189 g/cm^3}$, the volume of the measuring spoon was:
$$V_{spoon} = \frac{\pu{14.25 g}}{\pu{0.997189 g/cm^3}} = \pu{14.2902 cm^3},$$
as compared with the actual volume of a tablespoon, which is $\pu{14.7686 cm^3.}$
I then weighed a level tablespoon of salt (20.00 g) and, from this, determined that
$$\rho_{table salt} = \frac{\pu{20.00 g}}{\pu{14.2902 cm^3}}= \frac{\pu{1.40 g}}{\pu{cm^3}}$$

The density of $\pu{2.17 g/cm³}$ refers to the bulk density, i.e. within a crystal of NaCl.  In chemical engineering, the terms of powder density, tapped powder density and settled apparent density take into account for the air between the grains of a solid. Especially the later recognizes that there may be a difference between the solid simply poured into a container, and after light compression (still with air gaps between the grains) after applying a little pressure e.g. if you shake and knock the tin filling with freshly ground coffee powder.
References like this, this, this, or this .pdf state powder densities of $\pu{1.378 g/cm³}$ (fine table salt), $\pu{1.282 g/cm^3}$ (granulated salt, again from here), and $\pu{1.089 g/cm^3}$ (rock salt). From these values, your estimate of $\pu{1.40 g/cm^3}$ seems plausible.
However, these data lack to state the typical size of the grains (think about the diameter), as well as the dispersion of the grain sizes (presence of larger and smaller grains, equally known as particle-size distribution) of the samples characterized. Both influence the packing of the grains and thus the recorded density.  In this perspective, the softer / more airy harvest of fleur de sel possibly packs much less dense.


Why is CCSD(T)/aug-cc-pVTZ not a suitable method for the geometry optimisation of omeprazole? I know that CCSD(T) with the aug-cc-pVTZ basis set for geometry optimization of omeprazole (5-methoxy-2-[(4-methoxy-3,5-dimethylpyridin-2-yl)methanesulfinyl]-1H-benzimidazole, the active substance in Losec) is not suitable using normal computers. The substance is:

(omeprazole)

Can someone explain to me what exactly makes these methods too expensive for this substance?

To transform the question into something simpler, let's assume CISD instead CCSD(T), which is cheaper and share the common expensive step.

The molecule has 182 electrons and 1545 basis functions in the basis you specified, that means approx. 1400 virtual orbitals.

Therefore, the number of double excitation $ij \rightarrow ab$ is approximately $180 \cdot 180 \cdot 1400 \cdot 1400 = 32400 \cdot 1960000 = 63504\cdot10^6$

Unfortunately, in each of this excitation you have to calculate integral of type $(ab|cd)$, and there are $1400^4$ of them.

Such a big problem is out of reach for traditional CCSD, but tractable with more approximate methods.


Synthesis of glycerol from methane? Any ideas on how to get $\ce{C3H8O3}$ (Glycerol) from $\ce{CH4}$?

My theory is that by reacting $\ce{CH4}$ with $\ce{Cl2}$ (or $\ce{Br2}$), getting $\ce{CH3Cl}$, then Wurtz's method to $\ce{CH3CH3}$, adding $\ce{Cl2}$ (or $\ce{Br2}$) again, getting $\ce{CH3CH2Cl}$, then reacting with $\ce{NaOH}$ to get $\ce{CH3CH2OH}$, and finally $\ce{CH3CH2OH + CO2 + H2O}$ and getting glycerol ($\ce{C3H8O3}$), but my way seems rather off to me so I need some help.

Your synthesis is unlikely to become a huge commercial success ;)

Methylchloride is a technical product, but usually made by reaction of methanol with hydrogen chloride:

$$\ce{CH3OH + HCl -> CH3Cl + H2O}$$

This works nicely without side products, because

$\ce{CH3Cl}$ has a low solubility in water (around $\mathrm{5\,g\cdot L^{-1}}$)
under standard conditions, $\ce{CH3Cl}$ is a gas (bp around $\mathrm{-28\,°C}$)

The radical halogenation of methane will usually lead to various halogenation products and thus is not a good idea. The same is true for the similar reaction of ethane.

In addition, I fail to see how the reaction of ethanol and carbon dioxide will lead to glycerol.

You should further take into account that there's plenty of glycerol available as a side product in the production of biodiesel.

From a commercial and ecological point of view, methane therefore is better converted to cabon monoxide and hydrogen by steam reformation. Further hydrogen may be obtained from a subsequent shift reaction and the resulting carbon dioxide may be reacted with gycerol to yield 1,2-glycerol carbonate.


Why did Dalton assume that atoms are spherical? One of the primary components of Dalton's theory is to "express" an atom as an unbreakable spherical shape filled with whatever. The question is, why didn't he suggest any other shape, like for example, cubes?

I don't know what Dalton was thinking exactly, but I do know that for a very long time philosophers and scientists have assumed that things are spherical in the absence of any indication that they are not.

The reason for this was originally related to religious beliefs, and the idea that God or (The Gods) would only create perfect objects, and the sphere (being perfect) was the obvious choice for heavenly objects and subatomic particles.

The assumption that things are spherical is still used very often, although it is not usually directly related to religious beliefs anymore. These days, the idea is that you should always use the simplest possible explanation that works. The simplest geometric shape is a sphere - you only need one dimension to describe it, and it has some other interesting properties like minimized surface area-to-volume ratio. Mathematics can also be greatly simplified by using spherical coordinates when things are spherical, which is another big advantage.

Therefore, when we imagine particles that we can't see, a sphere is a natural choice, and I would guess that Dalton was using one or both of these lines of reasoning when he came up with the theory.


Are all stable Xenes (graphene-like 2D honeycomb sheets) buckled? background: Per this abstract

  Silicene, germanene and stanene are part of a monoelemental class of two-dimensional (2D) crystals termed 2D-Xenes (X = Si, Ge, Sn and so on) which, together with their ligand-functionalized derivatives referred to as Xanes, are comprised of group IVA atoms arranged in a honeycomb lattice — similar to graphene but with varying degrees of buckling.

question: This comment notes

  Graphene is not flat, either. Cannot be, because only 3D crystallites have perfect long-range order. Landau-Peierls instability this is called.

How general is this? Can we expect all stable 2D crystals to be buckled, and any that would be flat and not buckled to "self-destruct" due to instabilities related to long-range order?

According to the authors of Ref. 1 this is a general property. They provide an explanation for the stability of such sheets based on formation of particular buckled geometries:

  The discovery of a flat two-dimensional crystal known as graphene has contradicted Landau−Peierls−Mermin−Wagner arguments that there is no stable flat form of such crystals. Here, we show that the “flat” shape of graphene arises due to a microscopic buckling at the smallest possible interatomic scale. We show that the graphene, silicene, and other two-dimensional crystals are stable due to transverse short-range displacements of appropriate atoms.

This explanation is corroborated by experimental observations:

  On the other hand, the generality of our prediction of the buckled universal shape has been revealed in recent experiments on silicene [...]

The authors note in the introduction that their explanation is only one among three that at that time were being seriously considered. The two other arguments are that (1) the sheets are trapped in a quenched metastable state that is not subject to large thermal perturbations compared to the lattice strength and (2) the sheets are slightly warped in the third dimension, the deformation suppressing thermal excursions.

The reference cited above is from 2012. Low-buckled 2D lattices have been reported more recently, for instance for germanene in Ref. 2.

Reference

A. O’Hare, F. V. Kusmartsev, and K. I. Kugel .  A Stable “Flat″ Form of Two-Dimensional Crystals: Could Graphene, Silicene, Germanene Be Minigap Semiconductors?
Nano Letters 2012 12 (2), 1045-1052. DOI: 10.1021/nl204283q
Jincheng Zhuang, Nan Gao, Zhi Li, Xun Xu, Jiaou Wang, Jijun Zhao, Shi Xue Dou, and Yi Du . Cooperative Electron–Phonon Coupling and Buckled Structure in Germanene on Au(111). ACS Nano 2017, 11 (4) , 3553-3559. DOI: 10.1021/acsnano.7b00687.


Electrolysis of salt water into sodium hydroxide with YSP I want to electrolize salt ($\ce{NaCl}$) water into a solution of sodium hydroxide for soap making. The salt I have has yellow prussiate of soda (sodium ferrocyanide) added as the anticaking agent. What will happen to the ferrocyanide ($\ce{Fe(CN)_6}$) molecules during the electrolysis process?

I will be using an old battery charger to perform the electrolysis, $20\, \mathrm{V}$, $0.9\, \mathrm{A}$ or $1.8\, \mathrm{A}$ settings.

If you want sodium hydroxide for soap making you should just buy some: it is a cheap and common chemical. This is because there are effective industrial processes to make it in vast quantities.

Trying to make it yourself is not easy and will cost you far more effort than just buying it. The least of your problems will be what happens to contaminants like the hexacyanoferrate (which is present in such low concentrations it isn't really significant). Have a look at the way industry makes bulk hydroxide and chlorine (e.g. here); it isn't as simple as just electrolysing brine. All the processes involve sophisticated ways to separate the products produced at the different electrodes. You don't get decent quantities of either product if you don't maintain this separation (one method uses flowing mercury; another expensive ion-exchange membranes). In addition to this, uncontrolled electrolysis is dangerous as it produces toxic and explosive gases. You really don't want to try it at home.

Since the most basic components for an effective reaction will cost you far more than the couple of dollars a large bottle of pure sodium hydroxide will cost, buy the chemical and ignore the electrolysis as a route to what you want.


What is the reaction of ammonium nitrate with hypochlorous acid What would the product of ammonium nitrate and hypochlorous acid? Would there be any reaction? Would the HOCl get broken down? Trying to determine if the free chlorine content would be diminished upon mixing in an aqueous solution.

Do not for goodness sake mix chlorine bleach with ammonium compounds or ammonia, it has a good chance of ending badly.

I am not sure of the curved arrow mechanism, but what I suspect will happen is that the ammonia will attack the HO-Cl using the OH- as a leaving group. So by making the system less basic it will favour the reaction.

This will convert ammonia into NH2Cl, which will then be converted into NHCl2 and finally into nitrogen trichloride (NCl3). When I was a PhD student one of the other people has to make NCl3, he did it with great care. He did the reaction in such a way that the NCl3 was diluted in an inert solvent.

The scary thing about nitrogen trichloride is that it is a very unstable and shock sensitive explosive. One of my uncles is a chemist like me, he once witnessed an accident in which a man lost several fingers playing around with nitrogen iodide.

But back to nitrogen trichloride, as well as being explosive it is also a harmful and irritating substance. I would suggest you have nothing to do with it !


Cauchy-Schwarz inequality for integral derivatives The Cauchy-Schwarz inequality for regular two electron integrals is given as:

$$\left(\left.ab\right|cd\right)\leq\sqrt{\left(\left.ab\right|ab\right)}\sqrt{\left(\left.cd\right|cd\right)}$$

Now this can be differentiated with respect to $x$:

$$\frac{\partial}{\partial x}\left(\left.ab\right|cd\right)\leq\frac{\partial}{\partial x}\sqrt{\left(\left.ab\right|ab\right)}\sqrt{\left(\left.cd\right|cd\right)}$$

By using the chainrule for $\frac{d\sqrt{f(x)}}{dx}=\frac{df(x)}{dx}\frac{1}{2\sqrt{f(x)}}$, now giving:

$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)+\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)+\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)+\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{2\sqrt{\left(\left.ab\right|ab\right)}}\left[\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)+\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)+\left(ab\left|\frac{\partial a}{\partial x}b\right.\right)+\left(ab\left|a\frac{\partial b}{\partial x}\right.\right)\right]+\frac{\sqrt{\left(\left.ab\right|ab\right)}}{2\sqrt{\left(\left.cd\right|cd\right)}}\left[\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)+\left(\left.c\frac{\partial d}{\partial x}\right|cd\right)+\left(cd\left|\frac{\partial c}{\partial x}d\right.\right)+\left(cd\left|c\frac{\partial d}{\partial x}\right.\right)\right]$$

By using the symmetry of the two electron integrals $(ab|cd)=(cd|ab)$, it can simplified to:

$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)+\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)+\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)+\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left[\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)+\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)\right]+\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left[\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)+\left(\left.c\frac{\partial d}{\partial x}\right|cd\right)\right] \tag{1}$$

Now by comparing terms it can be seen that:

$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)$$

$$\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)$$

$$\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)\leq\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)$$

$$\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left(cd\left|c\frac{\partial d}{\partial x}\right.\right)$$

For geometrical detrivatices $x$ would be a cartesian direction of a nuclie. Now my question is, for geometrical derivatives when would be the right point to use the Cauchy-Schwarz inequlity for screening the integrals?

I think I could either $(1)$ and screen the total derivative $\frac{\partial}{\partial x}(ab|cd)$, or use the last four equations and screen the four different (up to three for geometrical derivatives) contributions individually.

Also what would be a reasonable threshold for derivatives?

For regular integrals $10^{-10}$ or lower is often used.

This is indeed a way that pre-screening of two-electron integrals is done, and if you thought of this yourself, then I'm very impressed! Though I doubt that is too much of a compliment.

Unfortunately, Tyberius is correct that you cannot be sure this inequality will hold after taking the derivative. Fortunately, the Cauchy-Schwartz inequality holds for any arbitrary vectors which have an inner-product. This means that you can simply switch the order you are doing things and it will work.

For what comes below, I will basically replicate some work from a paper by Reinhart Alrichs et al., and will use their notation as well [1].

In what follows, a superscipt $i$ represents a derivative in any of the cartesian $x$, $y$, or $z$ directions.

The Cauchy-Schwarz inequality states that,
$$\tag{1}\label{eq:1}
|(ab|cd)|\le Q_{ab}Q_{cd}
$$
where,
$$\tag{2}\label{eq:2}
Q_{ab}=(ab|ab)^{\frac{1}{2}}
$$
and the same for $Q_{cd}$.

If we take the derivative in the $i$ direction of $(\ref{eq:1})$, we have,
$$\tag{3}\label{eq:3}
(ab|cd)^i=(a^ib|cd)+(ab^i|cd)+(ab|c^id)+(ab|cd^i)
$$
which is contained in the work you have done, but with some extra stuff from the original inequality.

Each of these terms, however, are still a part of the original space and hence still have a well-defined inner-product. This means we can apply the Cauchy-Schwarz inequality directly to these derivatives as a means of screening the integrals.

That is,
$$\tag{4}\label{eq:4}
|(a^ib|cd)|\le Q_{a^ib}Q_{cd}
$$
and so one for the other derivatives.

The advantage of this is that one can give an upper bound for the size of these integrals based on calculations which have to be done for the other integrals anyways, and some simple to compute auxiliary quantities. See [1] for details on these auxiliary quantities (eqs. 10-19). I do not fully understand this part of the procedure, so I'm not restating it here.

Next, you asked where one should do the screening. The authors of [1] advocate what they call a "global screening" methodology. This means that all of the quantities required for calculating the upper-bound are pre-computed (before any integral loops but after any checks for symmetry), and compared to some pre-specified threshold, say $10^{-10}$. Although we do have to compute some new things, these are much cheaper than the integrals themselves and thus save quite a bit of time.

Ref. [1] also goes into detail about some test cases as well as doing the same screening for second-derivatives and hence the application of this pre-screening to Hessians.

A quite different approach which is very interesting to read is given in ref. [2], which came after and improves on [1]. This method, quite naturally I think, takes advantage of a generalization of Cauchy-Schwarz known as Holder's Inequality. For our purposes, the relevant form of Holder's inequality is (see the section titled Lebesgue Measure or ref. [2]),
$$\tag{5}
\int|f(x)g(x)|dx\le\left(\int|f(x)|^mdx\right)^{\frac{1}{m}} \left(\int|g(x)|^n dx\right)^{\frac{1}{n}}
$$
for $\frac{1}{m}+\frac{1}{n}=1$.

Notice, however, that while this strengthens the inequality by giving a class of possibilities for different $m$ and $n$, it weakens the inequality in that we may only use one variable $x$. Thus, to apply this to 2-electron integrals, the authors of ref. [2] first integrate out one of the variables, which gives us the same inequality for our physical situation as,
$$\tag{6}
\int|(ab|cd)|\le\left(\int|V_{ab}|^md\textbf{r}\right)^{\frac{1}{m}} \left(\int|(cd|cd)|^n d\textbf{r}\right)^{\frac{1}{n}}
$$
where $V_{ab}$ is the potential due to electron 1. Thus, in this case the screening must always take place after the first coordinate is integrated-out. This method, however, is very flexible in that it apparently gives tighter bounds (a smaller maximum), and also has many possible solutions. Ref. [2] discusses four of these possible solutions, and compares to the Schwarz pre-screening method, which is a special case of what they derive. Again, some special quantities must be pre-computed, but this is not such a big deal for the time it saves.

Ref. [2] does not discuss derivatives, though I assume the same method ought to possible as in the first part, only things might be complicated by the fact we need we need to integrate out a coordinate. Maybe not in the first derivative case, but in the second derivative case, this would present a problem as far as I can see.

[1] Horn, H., Weiß, H., Haser, M., Ehrig, M., & Ahlrichs, R. (1991). Prescreening of two‐electron integral derivatives in SCF gradient and Hessian calculations. Journal of computational chemistry, 12(9), 1058-1064.

[2] Gill, P. M., Johnson, B. G., & Pople, J. A. (1994). A simple yet powerful upper bound for Coulomb integrals. Chemical physics letters, 217(1-2), 65-68.


Tunneling in chemical reactions We know that quantum tunneling is the reason behind several natural phenomenon like alpha decay and thermonuclear fusion inside the stars. How can it influence chemical reactions by tunnelling a species through the activation energy? If so how does it influence the kinetics and fraction of molecules taking part in the reaction.

The probability of tunnelling at an energy $E$ is given by $p(E)\approx e^{-bA\sqrt{m}}$ where $A$ is proportional to the area of the potential energy barrier above  energy $E$, i.e. the top part of the potential barrier, $m$ the mass and $b$ some constants, $\pi, \hbar$. etc. Thus for a given mass and energy if the barrier is narrow, so $A$ is small, tunnelling is more likely than if the barrier is wide. At a given energy for the same barrier if the mass is large tunnelling is small. Thus we tend to see tunnelling only with H and D  and not Cl atoms for example. Tunnelling is also important in electron transfer reactions.

As there is not just a single energy in a reaction but a distribution of energies, according to the Boltzmann distribution, it is necessary to modify the expression above to average over the energy but the basic result is the same, which is that the reaction rate constant is reduced by the factor $p(E)$.

[If the potential energy barrier is $V(x)$ then $\displaystyle p(E)=\exp\left(-\frac{2\sqrt{m}}{\hbar}\int_{x_1}^{x_2}\sqrt{V(x)-E}\right)$ where $x_{1,2}$ are the points either side of  the barrier with energy $E$.]


Why does potassium react more violently with water than lithium? Recently, I was telling my friends about the violent reaction that takes place when you throw potassium into water. Soon after, a friend of mine claimed that lithium would react more violently than potassium.

I disagreed with him, because potassium is more electropositive than lithium and thus more reactive.

My friend claimed lithium to be more reactive than potassium due to its position in the reactivity series of metals:

$$\ce{Li > Cs > Rb > K > Ba > Sr > Ca > Na > Mg}$$

Then we found out that potassium reacts indeed more violently in water.
But what about his argument? Why isn't he right?

For the reaction,
$$\ce{M -> M+ + e-}$$

the heat liberated is highest for lithium owing to its high negative $E^\circ$ value so one would think that the reaction must be most vigorous.

The reason behind the more violent reactivity of potassium rather than lithium lies in kinetics and not in thermodynamics.

No doubt, maximum energy is evolved with lithium but the vaporization and ionization will also consume maximum energy (the melting point and ionization energy of lithium are the highest) and so the reaction proceeds gently. On the other hand, potassium has a lower melting point and ionization enthalpy. The heat of reaction is sufficient to melt it. The molten metal spreads over the water and exposes a larger surface to water. Also, the hydrated radius of lithium is the greatest out of all alkali metals. This reduces the ionic mobility which in turn reduces the speed of the molten metal.

That's why potassium gives a more violent reaction with water.

Reference:

Kumar, Prabhat Conceptual Inorganic Chemistry; Shri Balaji Publications: Muzaffarnagar, U.P., 2014.


Physical state of the carbon denoted with "gr" I would like to know what does a gr mean when used to represent the physical state of the carbon, for example, in the synthesis of the glucose

$$\ce{6C_{(gr)} + 6H2_{(g)} + 3O2_{(g)}-> C6H12O6_{(\ell)}} \quad,$$

what does the $\mathrm{gr}$ mean?

The letters in brackets indicate state of aggregation (includes state of matter, crystal structure (solids) and atomicity (gases)), therefore $\ce{C_{(gr)}}$ most likely refers to graphite. Often one can also see $\ce{C(s, gr)}$ or $\ce{C(s, graphite)}$ There is a nice demonstration in [1, p. 248]:

  $$\ce{C(gr) + O2(g) <=> CO2(g)}$$
  Here $\ce{(g)}$ indicates that the compound is gas, [...], and $\ce{(gr)}$ that carbon is graphite (not many people will burn diamonds, even if they are chemists).

P. S. According to IUPAC recommendations, subscript notation is deprecated.

Metiu, H. Physical Chemistry: Thermodynamics; Garland Science, 2006. ISBN 978-1-315-31491-4.


Is there anything in the literature that supports revolutionary claims for this "new" sulfuric-acid battery? I've got the following from a supposed new energy-storage technologist. It's supposed to be a fancy new cheap, high-powered, highly scalable battery. I want to know if this is known chemistry, and if there's anything genuinely revolutionary about it:

  "Basic materials in MEG's E-SOURCE™ power cells are lead metal, absorptive glass mat paper, and sulfuric acid solution. Materials unique to the MEG advanced power cell are S2 glass fiber yarn; sodium sulfate; Teflon emulsion, glass-filled polypropylene; bituminous, petroleum-based wax; and metal-free nano-oxides. "

The lead is apparently in the form of lead oxide.

The claim is that this is a radical new cheap energy-storage system - but someone makes that claim for yet another technology every week or two. Is there anything in the recent literature that would give credibility to this one?

There is nothing in the chemistry of what they describe that is different in any way to a conventional Lead-Acid battery. What they really seem to claim is improvement in materials in the battery not to change the electrochemistry but to improve other aspects of battery operation. For example:

MEG's E-SOURCE™ power cell technology is based on a lead-fiberglass composite material that is one-third the weight and 50 times the strength of metal alloys used in conventional batteries. To date, more than $100 million has been invested in research, development, manufacturing, and market testing of this technology in the United States and several international markets.

This suggests they have built a better lead-acid battery rather than something more radical. This is not ridiculous. They claim the following advantages:

Extraordinary manufacturing and user cost savings
Significant weight reduction compared to conventional lead-acid    batteries
Exceptionally long service life
Dramatically fast recharge speeds at all operating temperatures
Safe, simple, and low-cost maintenance

It is not impossible to imagine clever overall design and improved microstructural components and electrode materials could deliver these sorts of improvements and it is certainly far easier than inventing a whole new battery electrochemistry. For example, charging speed is often dominated by the physical material properties of the electrodes and not the underlying electrode chemistry, so clever design can mitigate the physical limits on charging speed such as by increasing the effective surface area.


Is 2,3-dimethylprop-1-ene a possible organic structure? Is 2,3-dimethylprop-1-ene a possible organic structure?

I am not sure how to approach this, what would the correct name of this molecule be?

2-methylbut-1-ene.

For future reference: an alkyl substituent at the 3 position of a propyl group probably indicates you did not select the longest carbon chain to name the parent molecule.

So your answer was not 'formally' correct.  That being said there was only possible structure to be drawn from the name you used, so correct or otherwise, it is effective communication.


Test presence of oxygen > 1% I want to qualitatively test for the presence of oxygen within argon gas. I would like the test to be positive if approx $1\%$ or more is oxygen, but definitely negative if less than $0.1\%$. Anywhere in between is fine—A quantitative test is a bonus. I was thinking of heating iron filings in the gas to deduce it. I'm not sure what apparatus and setup  I would need since heating gas expands, I would like to use ultra low tech equipment. I'm thinking of using water or cooking oil displacement to hold the extra gas, but I don't know if argon would dissolve, or how much oxygen would dissolve. Preferably a nonflammable, certainly nontoxic liquids (or tests).

This should do it

http://www.sciencebuddies.org/science-fair-projects/project_ideas/Weather_p004.shtml


If you run an electrolytic cell long enough, does the conductivity decrease? I've never seen this discussed so I wanted to ask here.  Consider the electrolysis of water with inert electrodes and electrolytes that won't be oxidized or reduced, such as sodium sulfate.  Since the migration of electrolytes in an electrolytic cell is responsible for the electrical current, does the gradual sequestering of electrolytes at the electrodes decrease the electrolyte concentration in the middle (between the electrodes), and cause a decrease in conductivity?

A couple thoughts occur to me:

the $\ce{H+}$ and $\ce{OH-}$ will diffuse from the electrodes into the cell water.  As the $\ce{H+}$ and $\ce{OH-}$ diffuse back into bulk they could carry the electrolytes with them.
the battery could be disconnected momentarily to mix the electrolyte solution.

...does the gradual sequestering of electrolytes at the electrodes decrease the electrolyte concentration in the middle (between the electrodes), and cause a decrease in conductivity?

In a word, no.  But, it is actually very complicated.  The electrolyte in most simple lab electrolytic cells for water is not very viscous.  This means that convective transport dominates over diffusion, except very very close to the electrode surfaces, i.e. at the boundary layer, which is hundreds of nanometers to maybe a few hundred micrometers thick, depending on conditions.  Outside the boundary layer, i.e. in the vast majority of the space "between the electrodes", no, the concentration of the electrolyte will not decrease.  In fact it may build up to higher concentrations, since water electrolysis involves loss of water, leaving more electrolyte per unit of solvent.

Inside the boundary layer, a phenomenon that does distort the bulk concentration of electrolyte can happen.  This phenomenon is called concentration polarization.  For water electrolysis, this means that the cathode (which generates hydrogen gas and consumes aqueous protons) would be depleted in protons, i.e. would be more alkaline than the bulk solution.  The anode's boundary layer, which generates protons from water while evolving oxygen) would have an excess of protons. The cathode boundary layer would have excess sodium cation and the anode excess sulfate anion to maintain electrical neutrality.

The formation of the boundary layer is very fast, taking less than a second, not gradual.  But the boundary layers are very small.

So in the end, you had it right with your first idea:

  the $\ce{H+}$ and $\ce{OH-}$ will diffuse from the electrodes into the cell water. As the $\ce{H+}$ and $\ce{OH-}$ diffuse back into bulk they could carry the electrolytes with them.


Raman activity in sulfur I was wondering why sulfur is Raman active but other elements (e.g. carbon) are not.

In a review (PDF, CalTech) of the properties of elemental sulfur, the author notes that the most stable form of sulfur at STP is in fact cycloocta-sulfur, $\ce{S8}$, with other polyatomic species also found. Elemental sulfur is commonly formed as a molecular solid, with arrangements of $\ce{S_n}$ rings or chains, and the S-S bonds can be Raman active.

The paper offers depictions of some of those structures, and also tabulates the principal Raman bands for selected allotropes of elemental sulfur.


Aspirin Hydrolysis at pH 2 Conducting an experiment, I found that very little salicylic acid was evolved performing hydrolysis at $\mathrm{pH \ 2}$ and at $\pu{45 ^\circ C}$. In fact, far more salicylic acid was evolved at $\mathrm{pH \ 7}$. This comes as a surprise to me because I expected an acid-catalyzed ester hydrolysis to actually quicken the rate of reaction and thereby evolve MORE salicylic acid.
Is aspirin generally more stable at lower $\mathrm{pH}$s then?
UPDATE 1: This graph shows that aspirin hydrolysis doesn’t lend itself well in low $\mathrm{pH}$ conditions, specifically at $\mathrm{pH \ 2}$.
UPDATE 2: No answer has been found — a reword to this question can perhaps be “Why is aspirin less susceptible to hydrolysis at $\mathrm{pH \ 2}$ when theoretically, lower $\mathrm{pH}$s are advantageous for ester hydrolysis?”

Source: https://compoundingtoday.com/Newsletter/Science_and_Tech_1112.cfm
What seems to be the explanation to this?

To understand the pH rate profile of the Aspirin hydrolysis, we have to look at all the possible mechanisms through which the hydrolysis may take place.
A good rule of thumb for interpreting a pH rate profile is: going from left to right, when the line curves upward (i.e. the rate increases suddenly), a new mechanism is dominating the reaction. When the line curves downward (i.e. the rate drops suddenly), that generally indicates that there is a change in ionisation/dissociation or that the rate-limiting step has changed for the same mechanism.
If you look at the graph for Aspirin, you will see that the graph bends up two times, around pH 2.5 and pH 8.5. So, there must be 3 different mechanisms throughout the whole pH range. (There is also a downward bend at pH ~5, that's not important at the moment).
So you can divide the graph into 3 segments (separated by blue line):

We need 3 mechanisms—(A), (B) and (C). Let's deal with them one at a time.

(A)

At pH < ~2.5 i.e. strongly acidic enviornment, the main components in the solution are the protonated form of aspirin (pKa of aspirin ~ 3.5), and $\ce{H3O+}$. A mechanism can be drawn:

This is the usual acidic ester hydrolysis mechanism. I haven't drawn the last few steps for brevity.
The rate of the reaction is: $$\ce{r=k_1[aspirin][H3O+]}$$ $$\mathrm{usually\,written\, as,}\;\; \ce{r=k_{obs}[aspirin]}$$
Which means, $\ce{k_{obs}=k_1[H3O+]} \Rightarrow \ce{log(k_{obs})=log(k_1) +log([H3O+])} \Rightarrow \ce{log(k_{obs})=log(k_1) - pH}$. You can immediately see why the graph decreases linearly as pH increases.

(B)

When pH > ~2.5, the solution contains both the protonated and the deprotonated form of aspirin, along with a lower concentration of $\ce{H3O+}$ as the solution is not strongly acidic anymore. As the rate of the reaction has increases suddenly, that must mean there is another mechanism in effect now, which overshadows the previous acid catalysed mechanism.
This part is difficult, because multiple mechanisms can be drawn, and it is difficult to distinguish which one is actually happening. Here the graph bends down once, and that happens around pH 4, close to the pKa of aspirin(3.5) which should clue you in to the fact that the mechanism here is somehow affected by the deprotonation of aspirin.
The rate rises at first, until pH 4 and then it stays almost constant until pH 8.5 (where (C) starts). This indicates that the mechanism is not catalysed by acid or base (because then the graph can't stay flat against changing pH), and that the mechanism involves the deprotonated form of aspirin (because once all the aspirin is deprotonated, the graph becomes flat).
One possible mechanism:

Note that we can't draw a mechanism where water attacks the ester directly, because water is not a good enough nucleophile, and also because the anion has to participate in the reaction somehow, as I said earlier.
Another possible(*) mechanism would be direct attack of the carboxyl anion on the ester:

Here an anhydride intermediate is formed, which is then hydrolysed as usual, and I have not drawn that part of the mechanism.
(*) I remember hearing in a lecture, that this mechanism via anhydride has been proven to be wrong, based on isotope substitution studies. I can't find a reference right now, so I will leave the mechanism in the answer, but I am pretty sure the first mechanism is the correct one.
Here, $\ce{[aspirin-]}$ and $\ce{[aspirin-H]}$ represents the deprotonated and protonated form of aspirin, respectively. The rate of the reaction is: $$\ce{r=k2[aspirin-]}$$ But the way the experimental measurements are taken means that the rate is expressed in terms of the total aspirin concentration:
$$\ce{r=k_{obs}[aspirin]}$$
From the acid dissociation equilibrium, we can write:$$\ce{[aspirin]=[aspirin-H] +[aspirin-]=\frac{[aspirin-][H3O+]}{K_a} +[aspirin-]=[aspirin-]\left( 1 +\frac{[H3O+]}{K_a}\right)}$$
We can substitute into the first equation: $$\ce{r=k2[aspirin-]= \frac{k2}{\left( 1 +\frac{[H3O+]}{K_a}\right)}[aspirin]}$$
Which gives:
$$\ce{k_{obs}=\frac{k2}{\left( 1 +\frac{[H3O+]}{K_a}\right)}}$$
In the region pH ~2.5 to 4, $\mathrm{k_{obs}}$ increases as $\ce{[H3O+]}$ decreases. After pH ~4, $\ce{[H3O+]<<K_a}$ so, $\mathrm{k_{obs}\approx k_2}$ and the graph becomes almost flat.

(C)

After pH 8.5, the main components of the solution are the deprotonated form of aspirin and $\ce{OH-}$. So, the base-catalysed mechanism becomes dominant.

As all the aspirin is in deprotonated form, we can write
$$\ce{r=k_3[aspirin][OH-]}$$
Which means, $\ce{k_{obs}=k_3[OH-]} \Rightarrow \ce{log(k_{obs})=log(k_3) - 14 + pH}$
So the graph rises linearly with increasing pH.
[P.S. The solvent does not appear in the rate equations.]


Stereogenic C and enantiomers Why is it said that the presence of a carbon with 4 different groups is necessary for it to exist as an enantiomeric form?

Lets take 2-fluoropropane for example.

When I draw the structure of it in such a way so that the 2 CH3 groups are adjacent to each other (on paper ) and take its mirror image I see that I got a mirror image which is not superimposable on the original structure. My teacher told me I am only allowed to rotate it in the plane of the paper and only by 180° without lifting it out.

This seems to tell me that they are a pair of enantiomers( non superimposable mirror images).

On the other hand when I draw the structure such that the 2 CH3 groups are on opposite sides of the central carbon and then take a mirror image of it , rotate it by 180° in the plane of the paper I get the original structure.
This tells me that they are not stereoisomers.

Also where is the mirror supposed to be placed - up down left right? Any convention (here I have placed the mirror on the right)?

When you say:

 When I draw the structure of it in such a way so that the 2 CH3 groups are
 adjacent to each other (on paper ) and take its mirror image I see that I got a
 mirror image which is not superimposable on the original structure.

and then:

On the other hand when I draw the structure such that the 2 CH3 groups are on
opposite sides of the central carbon and then take a mirror image of it ,
rotate it by 180° in the plane of the paper I get the original structure.

This tells me you are drawing structures where the bonds are at right angles to each other, like this:

     F                              CH3
     |                              |
 CH3-C-CH3        and           CH3-C-H
     |                              |
     H                              F

This will make the seeing of the structures much harder, you can't quite see the 'tetrahedral'.  In fact both of the drawings you have indicated are the same, only the connectivity matters -- that is the central carbon is bonded to a fluorine, a hydrogen and two methyl groups.

This is a good place to use models.  And if you don't want to buy tinker toys, tooth picks and colored marshmallows (or cheeses) will work just as well.

To resolve your conundrum, expand requirements such that if there exists any arrangement of the groups in your drawing that can form a superimposable mirror image, the molecule is not chiral.

Also consider using dashes and wedges, such as:


How Can I Induce Corrosion Electrochemically? I'm very interested in measuring corrosion in metal alloys, particularly in salt solutions. I understand metals corrode over time, however, I was wondering if it's possible to electrochemically "speed up" or induce further corrosion in saline solutions? How would I go about this electrochemically? I've considered applying a specific current / potential and utilising it to induce oxidation, however, I have not found much in the way of literature actively seeking to induce / speed up corrosion, most are content to investigate over longer time periods.

Many thanks.

Corrosion is a more or less involuntary/undesired reaction. If you apply a current, the corrosion is defined by the current you apply, so you are controlling the system, not just observing it. In general, electrochemical forcing puts your system into a highly unnatural regime, but can certainly provide interesting information.

Corrosion can be affected (in ways that you can determine and document) by temperature, by contact with other metals, by the corrosivity of the solution (concentration and specific ions), and by inhibitors.

Some time ago, we were interested in stress-corrosion cracking (failure) of stainless steels. Hydrogen, (slowly generated in a natural environment) would embrittle the steel. By increasing the corrosivity of the solution (chlorides) and adding an activator (sulfides), we were able to speed up the hydrogen evolution/absorption so that steel specimens under tensile stress would fail within days or weeks, rather than months or years in nature. This enabled choosing alloy compositions that were most resistant to stress-corrosion cracking.


Molarity is dependent on temperature? Okay I know that as the temperature rises the number of particles in a given volume decreases. The molar volume likewise increases. So the number of molecules per liter would decrease as the temperature rises meaning that a 1M solution is only 1M at a specific temperature.

But why is molarity dependent on temperature? I mean sure the density does decrease as the temperature rises but still there is a small chance that the molarity will stay the same.

First, there is very very little chance it will remain same. Density will surely change with temperature.

Secondly, even if there is a small chance that it will remain same, then also there is an even greater chance that it will change.

So we will have to look at the general case and conclude that molarity depends on temperature.


What makes E1 more favourable than SN1? Each goes through a carbocation intermediate which is enormously reactive so what would promote attack towards a proton (E1) rather than towards the carbocation for substitution. Besides an increase in temperature I can't think of much.

The most important factor is the nucleophilicity versus basicity of the attacking species.

Taking the trimethylcarbenium cation as a classic example:

Sterically hindered bases such as lithium diisopropyl amide (LDA) will favour elimination because it is physically much easier for them to remove a small proton from the outside of the cation than to reach the carbon atom at the centre surrounded by three methyl groups and form a new bond there. On the other hand, small species such as cyanide and water will favour substitution because because it is physically easier for them to reach the central carbon.

Additionally the strength of the base influences which reaction takes place. LDA is a strong base and the methyl hydrogens are quite acidic means that the elimination reaction is much more likely. However, weak bases such as iodide and water will favour substitution because they are less basic and so they are less likely to abstract an acidic proton from one of the methyl groups.

It should also be noted that the nature of the solvent has a strong effect both on the stability of the cation but also on the nucleophilicity of the attacking species. This page has some discussion on the effects of solvents on nucleophilicity and the website also has good discussions of elimination and substitution mechanisms.

http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alhalrx1.htm#hal1


What is the charge on the copper ion in antlerite? Antlerite is $\ce{Cu3(SO4)(OH)4}$.

I'm trying to get the charge on the copper ion in this substance.  So far I have the mass of $1$ mole of antlerite, its percent oxygen content, and the number of individual hydroxide ions in a $\pu{955 mg}$ bead of the substance, if that information helps at all.

I need help to solve this question.

Ok, here is a short answer. Since you already know the chemical formula it's actually quite easy provided you are familiar with the process of determining the oxidation numbers in chemical compound. From the Wikipedia article (and the way the chemical formula is written indicates that too) you can get the information that the $\ce{SO4}$ part of the formula is sulfate and sulfate ions carry a charge of $-2$, i.e. $\ce{SO4^{2-}}$. Furthermore, in most cases $\ce{H}$ gets an oxidation number of $+1$ and $\ce{O}$ gets an oxidation number of $-2$ so that the 4 $\ce{OH}$ units in the compound each have a charge of $1 - 2 = -1$, so you have $\ce{OH^{-}}$, i.e. hydroxyl groups - this is supported by the way those $\ce{O}$'s and $\ce{H}$'s are grouped, if you see "$\ce{OH}$" in a compound it usually means "hydroxyl group". So, you have one sulfate group with a charge of $-2$ and four hydroxyl groups with a charge of $-1$ each, giving you a charge of $-6$ in total. But your compound $\ce{Cu3(SO4)(OH)4}$ is neutral, i.e. has a charge of $0$, so the three $\ce{Cu}$ have to counteract the negative charge exerted by the anions, in other words you have to distribute a charge of $+6$ between the three copper atoms. If you assume that the charge is evenly distributed, i.e. that all the $\ce{Cu}$ are the same (this need not necessarily be the case, you can also have mixed oxides like $\ce{Fe3O4}$ where you have $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ within the same mineral), you get to a charge of $+2$ for each copper ion, i.e. $\ce{Cu^{2+}}$.


MO Diagram "greatest stabilisation when orbitals are close in energy" I was studying MO diagrams in Chemistry and we were given a diagram explaining this:

I understood the diagram furthest to the left describing the situation where the Electrophile (E) and Nucleophile (Nu) result in the largest energy gain when both are equal in energy.

However, the lecturer later said that when there is a large difference in energy between E and Nu (far right), the energy gain is negligible. He didn't go further to explaining why this was true.

I tried looking on the internet for an answer but was led to s-p mixing and content I am not familiar with nor find relevant to the question.

I'm confused here. What explains this phenomena? Or, is it just an effect observed without explanation - like a Postulate?

Fundamentally it's mathematics. The energies of the MOs are the eigenvalues of the matrix $$\begin{pmatrix}\alpha_1 & \beta \\ \beta & \alpha_2 \end{pmatrix},$$ where $(\alpha_1,\alpha_2)$ are the energies of the two constituent orbitals and $\beta$ is (loosely speaking) the overlap between them.

For the same size of $\beta$, if $\alpha_1 \approx \alpha_2$, then the eigenvalues can differ quite a lot from $(\alpha_1,\alpha_2)$. On the other hand, if $\alpha_1$ is very different from $\alpha_2$, then the eigenvalues will be closer to $(\alpha_1,\alpha_2)$.

As an illustration consider the matrix

$$\begin{pmatrix}1.1 & 0.5 \\ 0.5 & 1 \end{pmatrix}$$

This has eigenvalues of $1.55$ and $0.55$, which are relatively distant from the "original energies" of $1.1$ and $1$. On the other hand, the matrix

$$\begin{pmatrix}3 & 0.5 \\ 0.5 & 1 \end{pmatrix}$$

has eigenvalues $3.12$ and $0.88$, which are closer to $3$ and $1$.

Taking a more abstract perspective, the eigenvalues of $$\begin{pmatrix}\alpha_1 & \beta \\ \beta & \alpha_2 \end{pmatrix}$$ are

$$\lambda_\pm = \frac{\alpha_1 + \alpha_2}{2} \pm \frac{\sqrt{(\alpha_1 - \alpha_2)^2 + 4\beta^2}}{2}$$

and you can play around with this expression to gain some insight. For example, in the limit where  $(\alpha_1 - \alpha_2)^2 \gg 4\beta^2$ (corresponding to a large energy difference between the original interacting orbitals), the eigenvalues reduce to

$$\lambda_\pm \to \frac{\alpha_1 + \alpha_2}{2} \pm \frac{\alpha_1 - \alpha_2}{2}$$

which are simply $\alpha_1$ and $\alpha_2$.


stone which turns to black on exposing to atmosphere? My mom is using a stone like substance as a dye for hair recommended by a herbalist. That stone becomes black when exposed to air. Do you know which compound is this?

There is a compound called cerussite or 'natural white lead' which turns to black 'lead sulphide' when it reacts with hydrogen sulphide in the atmosphere. It was commonly used as make up by women in medieval times and is sometimes still used as a white pigment in 'lead paint'.

Of course, we all now know lead is toxic to many organs and tissues including the heart, bones, intestines, kidneys, and reproductive and nervous systems, as well as causing potentially permanent learning and behavior disorders, particularly in children.

For this reason, it's use has diminished over time, except perhaps amongst New Age crystal 'healers' who claim it can "infuse your aura with light" and "bring amazing change into your life", not only through a lowering of general intelligence, but, it can also help prevent bacterial infections.

Having said all this, you may be surprised to know that another toxic lead-based white crystalline substance called lead acetate is approved by the FDA for use in 'progressive' hair dyes, where it reacts with the sulphur in hair proteins to slowly turn grey hair black, over several applications of course, so 'no-one' will notice.

image of lead(II) acetate crystals, courtesy of Wikipedia

PS: In case you were wondering why lead-based products are still used, given that there are many alternatives out there, many of which are labelled 'non-toxic', 'natural' and 'organic'...

the National Cancer Institute reports that there are over 5000 different chemicals used in hair dyes, including toxic coal-tars, aromatic amines and phenols. Some of these are also carcinogenic (cancer-causing agents) and can be absorbed through the skin or inhaled. Whilst the industry has been working towards eliminating some of these chemicals since the late 1970's, people who regularly began using hair dyes before 1980 may be at increased risk of developing non-Hodgkin lymphoma (a form of blood cancer).

The International Agency for Research on Cancer (IARC) has said that "whilst workplace exposure as a hairdresser or barber is probably carcinogenic to humans", personal hair dye use is “not classifiable as to its carcinogenicity to humans,” based on "a lack of evidence from studies in people".


What is the difference between a chemical state and just a combination? I don't think a combination also can be called a "state" i.e. that the chemical state of vapour or solid would be just a combination. When I read Wikipedia about "chemical state" it refers to just a combination. But AFAIK, an entity that has just combination does not have a state. So how do we know what is a state and what is just a combination?

You seem to be confused with "state of matter"/"phase" and "chemical state".

The chemical state encompasses all the macroscopic parameters necessary to fully define a given system. For example, "pure $\ce{O2}$ at $1\:\mathrm{atm}$ and $100^{\mathrm o}\:\mathrm{C}$" is an example of chemical state. A more complicated example would be something like: "10 moles of liquid $\ce{O2}$ at $-100^{\mathrm o}\:\mathrm{C}$, mixed with 1 mole of gaseous $\ce{O2}$ at $1\:\mathrm{atm}$ and $100^{\mathrm o}\:\mathrm{C}$ and 2 moles of gaseous $\ce{N2}$ at $1\:\mathrm{atm}$ and $200^{\mathrm o}\:\mathrm{C}$" (Yes, this isn't at equilibrium)

On the other hand, the phase or sate of matter only applies to a homogeneous system. A gas is in gas phase, but a system consisting of a gas and a solid doesn't have a phase/state of matter. Note that if a gas/solid is dissolved in liquid, then the phase is liquid since solutions are homogeneous.


Electronegativity trends: Is it more important to prioritize families or periods? I know that an atom that an atom is more electronegative as it moves up a group or right on a period (in terms of the periodic table).

So if you have two atom that is diagonally across from each other, which one is more electronegative? Do you prioritize groups or periods when determining electronegativity?

For example: Which is more electronegative - Sulfur, or Bromine?

Finally, does this priority also work for first ionization and atomic radius trends. (in other words, if you prioritize groups when determining electronegativity, do you also prioritize groups when determining the other two periodic table trends?)

My guess is that they would be equal, but I may be wrong.

P.S. I would be most likely to give best answer to the person who answers all my questions as a numbered list.

There's no hard and fast rule.  In fact, there's actually no rule per se that states that electronegativity does increase along the periods and decreases down the groups.  It's just a general trend, which can be and is broken on occasion.  For instance, Zinc, to the right of Copper, has an electronegativity of 1.66, while copper is 1.75, which "breaks" the rule that electronegativity increases along the periods.

As far as which one takes precedence, it's a case by case scenario.  In some cases, down one and right one would be very close, like you guessed, as with Scandium (1.20) and Zirconium (1.22).  However, if you go over to the non-metals, some of them increase radically in a down-one-right-one scenario (particularly from 3rd period to 4th period, such as Aluminum [1.47] to Germanium [2.02], Phosphorus [2.06] to Selenium [2.48], etc.), while some of them decrease radically (particularly from 2nd to 3rd period and 4th to 5th, such as Boron to Silicon, Oxygen to Chlorine, Arsenic to Tellurium, etc.)

Bottom line:  There's no rule.  You just have to either learn them by heart or check a periodic table such as this one:


Can all substances be supercooled? With optimal conditions can we supercool any compound or is it that just water and a certain set of others can be supercooled? If not, why?

Supercooling is a general phenomenon and it is not limited to any particular substance, although the degree you can supercool something will vary. The idea in thermodynamics is that at the melting point the free energy of the solid and the liquid is equal and a lower temperature will favour the formation of the solid.

However, when there is only liquid present, you will first have to form a small crystal nucleus which is not as stable as the bulk solid i.e. has a higher free energy. This is basically a kinetic barrier which prevents forming the solid.

The formation energy of such a nucleus will depend on the interfacial energy of the crystal-liquid system. In a substance where this interfacial free energy is very little there won't be too much supercooling.


Explaning Cis & trans isomers through VSEPR Total rewrite as was a panic-stricken 2am question.

Unfortunately I wrote on an application that VSEPR theory is a reason of Cis-Trans isomerism (specifically in fatty acids however I doubt that would make any difference). Although in my head it seemed to work at the time, I appreciated about a week after that I was actually thinking of the wrong type of thing.

My question is,

  How can VSEPR and/or the electron configurations in an unsaturated fatty acid explain how cis and trans isomerism arises?

As far as I'm aware, the Pi bond prevents (resists?) rotation around the C=C link, but I feel as if I'm missing something.

Also,

  Does the geometry caused by the repulsion of H's and alkyl electrons on either side of the C=C bond affect the isomers differently depending on whether or not they are Cis or Trans?

One of the below posters also highlighted a good question, does anybody have any insight on this?:
"I think the main question is whether there's a preference for cis or trans isomers in a fatty acid based on steric/VSEPR considerations"
(Thank you for your response)

Thanks for your help

Bonus related question:

  Is there any preference for a specific stereoisomer in the dehydrogenation of a fatty acid's alkyl chain? If so, what are they and why?

Answered my own bonus question (I believe): In case anybody was interested,

Trans seems to be favoured due to the methyl group on either side of a Cis bond repelling eachother (due to to both the methyl groups wanting to retain their tetrahedral structure, hydrogens are repelling one another and hence distorting things), and hence the trans is more energetically favourable.


melting and boiling ...really equilibrium? Why are melting and boiling considered equilibrium processes even though the amount (concentration)  of both phases keep changing i.e from solid to liquid and so on?

I think what you are asking is this:  Equilibria for chemical reactions typically* (see note at end) require specific ratios of products to reactants (as expressed by the equilibrium constant).  By contrast, equilibria for phase transitions don't require specific ratios of products to reactants.   [For instance, at the phase transition between ice and water, ice and water can be in equilibrium with any relative amounts of ice and water.] Why is this?

Consider a gas-phase chemical reaction, say

$$\ce{A(g) + B(g) <=> C(g)}.$$

At equilibrium, the chemical potentials of all species are the same.  Associated with this is an equilibrium constant,

$$K_p(T) = \frac{\displaystyle\frac{p_\ce{C}}{p^\circ}}{\displaystyle\frac{p_\ce{A}}{p^\circ}\cdot\frac{p_\ce{B}}{p^\circ}},$$

which specifies a strict relationship between the partial pressures of the reactants and products.

Alternately, consider a solid-liquid phase transition,

$$\ce{A(s) -> A(l)}.$$

At the melting temperature, the chemical potentials of the two phases are the same.  However, by contrast with the example gas-phase reaction, this is true independent of the relative amounts of $\ce{A(s)}$ and $\ce{A(l)}.$

The question then arises: why the difference between the two?  The answer is that the particular mathematical form of the equilibrium constant—and the attendant constraint it imposes on the relative amounts of reactants and products—arises from the entropy of mixing, and there is no entropy of mixing in a phase transition!  At equilibrium, in the example gas-phase reaction, the entropy of mixing just balances the relative free energies of the reactants and products.  That's why we have an equilibrium.  If it weren't for the entropy of mixing, reactions would always go completely to the side whose pure components had the lower free energy (which is exactly what happens in a phase transition).

By contrast, there is no entropy of mixing term in a phase change, because the difference in phases keeps the two components separated.  And without an entropy of mixing term, the relative amounts of the two phases become irrelevant to their relative chemical potential, and thus irrelevant to the equilibrium state (by which I mean any relative amount of the two phases is allowed at the phase transition temperature).

This also explains why, when we are (say) somewhat above the melting temperature, we don't have a small amount of solid in equilibrium with a large amount of liquid (as would be the case if this behaved like a chemical reaction).  The reason for this is there's no entropy of mixing to favor this mixed state over the liquid alone, i.e., over the pure substance with the lower chemical potential (which, above the melting temperature, is the liquid state).  [Or more accurately, while there is an entropy of mixing, because the two phases aren't mixed at a molecular level—phase is a bulk property (i.e., a collective property of many molecules), not a molecular one—the entropy of mixing term (which is essentially an exchange entropy) is negligible in this case.]

N.B.: Just for clarity, let me emphasize that, as a substance is melting or boiling, it is not at equilibrium.  It's not at equilibrium during a phase transition until there is no macroscopic change in the system, i.e., until it settles into a fixed ratio of one phase to another (and that it can be at equilibrium at any ratio).

*Incidentally, this is also why, when we have a mixed-phase chemical reaction where one phase can't mix with the others, that phase isn't included in the equilibrium expression.  For instance, for

$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)},$$ we find that

$$K_p(T) = \frac{p_\ce{CO2}}{p^\circ}$$


Why aren't atomic radii calculated using Schrödinger's equation? The atomic radii are estimated using a variety of methods: most of these involve dividing their bond length by 2. But that is a very crude way of measuring atomic radii. I mean, atoms overlap each other when bonded, surely the actual atomic radius is much more than the one obtained by this method.

Why don't they simply use the Schrodinger's wave equation to calculate atomic radii. If it can predict atomic orbitals to such accuracy, surely it can predict something as simple as atomic radius?

Erwin with his psi can do
Calculations quite a few.
But one thing has not been seen:
Just what does psi really mean?

(Erich Hückel, english translation by Felix Bloch)

It is not easy to perform the calculation that you called simple. The Hamilton operator $\hat{H}$ for a multi-body problem of $K$ cores and $N$ electrons

\[
\hat{H} = - \frac{1}{2}\sum_a^K\frac{1}{M_a}\Delta_a - \frac{1}{2}\sum_i^N\frac{1}{M_i}\Delta_i - \sum_i^N\sum_a^K \frac{Z_a}{r_{ai}} + \sum_{i < j}^N\frac{1}{r_{ij}}+ \sum_{a < b}^K\frac{Z_aZ_b}{R_{ab}}
\]

is still a beast for $K=1$ with all the core-electron and electron-electron interactions.

So, while the question is legitimate, the calculation simply cannot be done without a lot of approximations, such as the Hartree-Fock method.

By the way, for the determination of the atomic radii of solid metals, there's x-ray diffraction, which I wouldn't call crude.