Can ML algorithms be applied to predict the products of a chemical reaction given a set of reactants? How would the training data look like ? Would it be a set of reactions ? What would be a good algorithm to run on the data ?

Wolfram Alpha does have an interface for chemistry, but does not seem really robust. The other databases for chemical reactions seem to work on a static database.

The ultimate goal is to predict the products given a set of reactants. Any suggestions will be helpful !

Adding on top of Emre's answer, product formation depends on certain reactive groups say for e.g., $-OH$ (Alcohol) having more chance of releasing a Hydrogen than $-CHO$ (Aldehyde) or a Methyl $-CH_{3}$. Instead of using reactants as categorical variables you should use reactive groups on that, as variables.

For example if you take a vector of say (Chlorine, Alcoholic_group, Amine) to be coded for Para-Amino Phenol, it will be $(0,1,1)$ as chlorine is absent and for $CCl_{4}$, it will be $(1,0,0)$. But remember the first vector will be same for O-Amino Phenol as they are positional isomers. But both have different properties and reactivities. So categorical is kind of wasted here.

In my more deeper and frank opinion, these values should be continuous. You should be kind of looking for a value to quantify these each dimension of the vector coded. Then you can use classifiers from this vectors to find the probability of reaction happening or for product formation as @Emre mentioned.

So, take a simple, small set of molecules and make a vector having fewer dimensions and test it with categorical and continuous values(Give a value to each dimension based on your chemistry gut opinion). That is the better way to find whether your hypothesis/assumptions are reasonable.

EDIT 1:
My answer basically answers whether reaction is feasible or not but does not answer which products are returned. I am sorry for that. But I hope it kind of points to what you can achieve with ML if you want to take your first step.

Hope this helps.


What is responsible for the smell of incomplete hydrocarbon burning and what sensor can detect it? After I pass the smoke from partially burned fuel oil through a filter, I can still very strongly detect the smell. It is the same thing you smell from the exhaust when an old car or truck goes by.

Even though I can easily smell it with my nose, none of my sensors can detect it. I've tried $\ce{VOC}$, $\ce{O3}$, $\ce{PM_{2.5}}$, $\ce{CO}$, $\ce{NO2}$, & combustible gas detectors and all come up clear on the filtered exhaust.

So what is left in this gas that is responsible for this strong smell (and the resulting headache) and what sensors can I use to detect it?

In analytical chemistry, sensors refer to analyte specific devices, e.g., glucose sensors, pH sensor etc. Smell on the other, generated by a burning complex mixtures such as fuels can rarely be associated with one compound. People have spent their life (e.g. Alan Marshall) studying the constituents of fuels. There are literally thousands and thousands of compounds.

However, if you were in a research lab investigating odors, you would be using something called as a olfactory gas chromatography technique. There is a chromatography column which separates all the gaseous/volatile components and a scientist puts a nose at the end of the column and records the "response". Later you can identify that particular odor by a mass spectrometer.

In short, there is no sensor per se. You need a instrumentation which would fit in a kitchen sized area to determine what causes that particular odor after burning an extremely complicated mixture such as fuel oil.

Here is how it looks like:


Bonding in alkali metal complexes How exactly is the bonding described in complexes involving alkali metals? In particular, how much of a covalent component is there?
Take three examples:

$\ce{[Na(H2O)6]+}$: This previous answer by @tschoppi suggests that the bonding in this complex is purely electrostatic in nature due to the lack of available d-orbitals for covalent bonding. This seems like it might be true because it would explain why the water complexes are not strongly bound.

$\ce{Na2(edta)}$: This previous question suggests that the sodium here is not complexed with the EDTA but rather that it again forms an essentially ionic (i.e electrostatic) bond with little covalent character.

$\mathrm{[Na(\text{15-crown-5})]^+}$: This is very stable, suggesting a strong bond of some sort, probably with some covalent character. However, it doesn't seem to be radically different from the previous ligands, particularly EDTA, which has similar oxygens available for bonding.

The bonding of alkali cations to water, to EDTA, and to crown ethers is strictly electrostatic in all three cases, at least based upon QTAIM analysis$^{[1]}$ of representative systems.

For this analysis, I ran gas-phase quantum chemical optimizations on $[\ce{Li}(\text{12-crown-4})]^+$, $[\ce{Na}(\text{EDTA})]^{3-}$, and $[\ce{Na}(\ce{H2O})_6]^+$ using ORCA v3.0.3. (I can post the computation inputs/outputs if anyone is interested.) I also optimized $\ce{NaCl}$ as an additional system with uncontroversial ionic bonding. While the gas-phase results will poorly represent a variety of aspects of the dissolved species, I think they should capture well enough the characteristics of the wavefunction necessary to study the types of bonding involved. Other gas-phase calculations I've run on, e.g., $[\ce{Cu}(\ce{H2O})_6]^{2+}$ have optimized smoothly to the expected octahedral ligand arrangement, lending credence to this assumption.

Some notes on the optimizations:

Geometry optimization of $[\ce{Li}(\text{12-crown-4})]^+$ converged readily.
For $[\ce{Na}(\text{EDTA})]^{3-}$, despite starting in a prototypical octahedral chelation geometry, the EDTA molecule immediately began to 'loosen its grip' on the $\ce{Na+}$, clearly heading toward a less compact configuration. In fact, one of the oxygen atoms actually discarded its bonding interaction with the $\ce{Na+}$ in favor of an intramolecular hydrogen bond.  I performed the QTAIM analysis on the last geometry obtained before ORCA reached its iteration limit in the geometry optimzation procedure.
Similarly, for $[\ce{Na}(\ce{H2O})_6]^+$, despite starting the optimization in a prototypical octahedral coordination geometry, the water ligands 'wandered' throughout the optimization process, in a fashion loosely reminiscent of a molecular dynamics run. As with the EDTA complex, I just chose the final geometry after the optimizer halted as the one on which I ran the QTAIM analysis.

For illustration, I generated GIF animations of the $[\ce{Na}(\text{EDTA})]^{3-}$ and $[\ce{Na}(\ce{H2O})_6]^+$ optimizations. The individual frames were produced with the 'Movie Maker' module of VMD 1.9.1 and the final GIFs were generated using EZgif.com.

Several metrics generated by QTAIM analysis argue for strictly ionic bonding of the alkali metal atoms in these systems:

Low electron density $(\rho \substack{<\\\sim}0.1)$ and positive density Laplacian $(\nabla^2\rho>0)$ at the relevant line critical points between the alkali cation and the coordinating oxygen (or nitrogen) atoms
High localization index $(\mathrm{LI}\substack{>\\\sim}0.9)$ in the alkali atom basin
Low delocalization index $(\mathrm{DI}\substack{<\\\sim}0.75)$ between the alkali atom basin and those of the coordinating atoms

Strictly speaking, in addition to $\#1$ above, the terms of the Laplacian perpendicular to the bond path must also be small in magnitude in order to diagnose ionic bond character$^{[2]}$, but I have omitted those values here.

The table below presents these metrics for the above representative chemical systems, as well as some reference molecules possessing uncontroversial bonding types.  Due to the fairly regular geometry of $[\ce{Li}(\text{12-crown-4})]^+$, the metrics didn't vary much throughout the system and I averaged them for each bond and atom type.  Despite the lack of a deep energy well, the metrics were also similar for the six oxygens in $[\ce{Na}(\ce{H2O})_6]^+$, so I averaged the values here as well.  For the irregular geometry of $[\ce{Na}(\text{EDTA})]^{3-}$, the values differed enough that I chose to report them indivdually; the numbering of the coordinating atoms is arbitrary.  All non-literature data were generated by MultiWFN v3.3.7 on Windows 7 using the default settings and the "Medium" grid for generation of the atomic basins for the $\mathrm{DI}$ and $\mathrm{LI}$ calculations. All values are in atomic units: $\rho\equiv{e^-\over\mathrm{Bohr}^3}$ and $\nabla^2\rho\equiv{e^-\over\mathrm{Bohr}^5}$ ($\mathrm{DI}$ and $\mathrm{LI}$ are dimensionless).

$$
~ \\
\textbf{QTAIM Results} \\
\begin{array}{ccccccccc}
\hline
\mathbf{Species} & ~ & \mathbf{Bond} & \rho_\mathbf{LCP} &
    \mathbf{\left(\nabla^2\rho\right)_\mathbf{LCP}} & \mathbf{DI} & ~ &
    \mathbf{Atom} & \mathbf{LI} \\
\hline
 & &  \ce{Li-O} & 0.036 & +0.26 & 0.077 & & \mathrm{Li} & 0.922 \\
[\ce{Li}(\text{12-crown-4})]^+ & & \ce{O-C} & 0.25 & -0.50 & 0.856 & &
     \ce O & 0.870 \\
 & & \ce{C-C} & 0.26 & -0.70 & 0.949 & & \ce C & 0.656 \\
 & & \ce{C-H} & 0.28 & -1.01 & 0.901 & & \ce H & 0.420\\
\hline
& & \ce{Na-O}1 & 0.015 & +0.086 & 0.064 & & \ce{Na} & 0.974 \\
& & \ce{Na-O}2 & 0.021 & +0.131 & 0.096 & & \ce{O}1 & 0.899 \\
[\ce{Na}(\text{EDTA})]^{3-} & & \ce{Na-O}3 & 0.020 & +0.121 & 0.089 & &
     \ce{O}2 & 0.900 \\
& & \ce{Na-N}1 & 0.014 & +0.074 & 0.056 & & \ce{O}3 & 0.900 \\
& & \ce{Na-N}2 & 0.018 & +0.094 & 0.065 & & \ce{N}1 & 0.766 \\
& &  & & & & & \ce{N}2 & 0.769 \\
\hline
[\ce{Na}(\ce{H2O})_6]^+ & & \ce{Na-O} & 0.014 & +0.086 & 0.057 & & \ce{Na} & 0.975 \\
& & & & & & & \ce{O} & 0.916 \\
\hline
[\ce{NaCl}]^0 & & \ce{Na-Cl} & 0.035 & +0.197 & 0.321 & & \ce{Na} & 0.978 \\
& & & & & & & \ce{Cl} & 0.975 \\
\hline
[\ce{LiF}]^{0\,\ddagger} & & \ce{Li-F} & 0.079 & +0.749 & 0.179 & & \ce{Li} & 0.957 \\
& & & & & & & \ce{F} & 0.991 \\
\hline
[\ce{CH4}]^{0\,\ddagger} & & \ce{C-H} & 0.291 & -1.168 & 0.980 & & \ce{C} & 0.661 \\
& & & & & & & \ce{H} & 0.469 \\
\hline
^\ddagger~_{\text{Values from Ref. [2]}}
\end{array}
$$

As can clearly be seen, the interactions between the alkali metals and the ligating atoms are uniformly of ionic character: low electron density and positive Laplacian at the LCP; low $\mathrm{DI}$ between the paired basins, and quite high $\mathrm{LI}$ for the alkali metal basins. This is consistent with the ionic species $\ce{LiF}$ and $\ce{NaCl}$ in the table, and is in marked contrast to the covalent systems included.

$^{[1]}$ More information can be found on Wikipedia and at RFW Bader's page at McMaster University.  A more readable (but unfortunately paywalled) description is provided in Ref. [2].
$^{[2]}$Bader & Matta, Found Chem 15: 253 (2013). doi:10.1007/s10698-012-9153-1.


Differentiating Al2O3 from SiO2 using chemical methods I have granules of $\ce{Al2O3}$ and granules of $\ce{SiO2}$. They look almost identical and I have mixed them up. Is there a simple chemical test that would allow me to differentiate between these granules?

Although not a chemical method, using flotation in bromine to separate your mixture will work at room temperature. Unlike the chemical methods given in other answers, this physical method of separation does not destroy or otherwise degrade the components of your mixture, so you can recover them.

The density of bromine at 20 $^\circ$C is 3.1 g$\cdot$cm$^{-3}$; the densities of $\ce{Al2O3}$ and $\ce{SiO2}$ at the same temperature are 3.95 g$\cdot$cm$^{-3}$ and 2.65 g$\cdot$cm$^{-3}$, respectively. The $\ce{SiO2}$ will float and the $\ce{Al2O3}$ will sink.

As noted in the comment thread above, this method will work unless the particle size is too small.

And as noted in the comments below, sodium polytungstate solution (with a density of 3.1 g$\cdot$cm$^{-3}$) is an alternative to bromine.

Because there are likely a large number of substances with a density in the range required to separate your mixture, the suggestions here represent a small number of those choices.


Altitudes effect on taste Over dinner, some vegan friends of mine discussing plans to open a 5 star organic restaurant inside the cabin of an airbus A380.

Exactly what I thought too! But they raised an interesting point: apparently the taste of certain foods changes depending on the altitude you consume it at.

I looked into it a little and discovered lots of interesting facts about Heston Blumenthal cooking salmon and whatnot for astronauts, but not must hard-chemistry behind why our sense of taste depletes the higher in the atmosphere we go.

So what causes such a taste discrepancy? Is it compounds within the food that "lose" their flavour, or it just our perception of taste that changes?

The difference in perception of taste is altered both by the lower air humidity as well by the lowered cabin pressure while on flight, the later typically kept to what corresponds to an altitude of about 2'400 m above sea level (source).

Regarding the actual changes in taste, BBC Future once nicely shed light on context, causes and outcome; in probably more accessible words than ongoing research articles like this about it.


Why alkali metals have bcc structure? It is well known that metals are typically, well, metallic and usually favor the most dense packing, like hcp or fcc cells. This may be easily rationalized as dense packing of metal ions in a sea of electrons, since metals typically have low electronegativity. The exceptions, however, do exist.

However, the simplest metals, where one would except the metallicity to define everything, have bcc cell, and not the densest packing. I'm speaking about alkali metals. Interestingly, alkali earths favor dense packings (hcp for Be, Mg and fcc for $\ce{Ca},\ce{Sr}$) except for $\ce{Ba}$ and $\ce{Ra}$ (again, favor bcc). So, the question is: why alkali metals have bcc cell and what guards relative stability of bcc/hcp/fcc cell stability in $\ce{Li/Ba}$ square? Bonus points for the fitting description for the p-block metals. I'm pretty sure there is no simple answer for d-block, as valent interactions involving d-electrons become the deciding factor there.

For the moment I will focus on Na, although I may mention other elements in comparison.

First, it is true that the STP stable phase of Na is bcc. However, the energy difference between the bcc, fcc, and hcp phases is quite small. As one example, the SGTE database (A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991)) gives the Gibbs free energies (relative to BCC) in Joules/mole as FCC = -50 + 1.3T, and HCP = -104 + 2T. These are really small; normally one would expect kJ/mole or 10's of kJ/mole type differences. Some experiments indicate that Na transforms to hcp below 51K. It is also known that at lower temperatures (35K), Na undergoes a martensitic phase transition. So, bcc is stable at STP, but not very from a big-picture thermodynamics view.

For a more detailed view, lets start with Ashcroft & Mermin's Solid State Physics book. They start their description of the alkali metals (Chapter 15) with

  The alkali metals have singly charged ions (whose core electrons from the tightly bound rare gas configuration and therefore give rise to very low-lying, very narrow, filled, tight-binding bands) outside of which a single conduction electron moves. If we treated the conduction electrons in the metal as completely free, than the Fermi surface would be a sphere of radius $k_{F}$...

Then we get

  De Haas-van Alphen measurements of the Fermi surface confirm this free electron picture to a remarkable degree of precision, especially in Na an K, where deviations in $k_{F}$ from the free electron value are at most a few parts per thousand... Thus, the alkalis furnish a spectacular example of the accuracy of the Sommerfeld free electron model. It would be wrong to conclude from this, however, that the effective crystalline potential is minute in the alkali metals. What it does suggest is that the weak pseudopotential method (Chapter 11) is well suited to describing the conduction electrons in the alkalis. Furthermore, even the pseudopotential need not be minute, for except near the Bragg planes the deviation from free electron behavior occurs only to second order in the perturbing potential...

In other words, they (including Na) really are 'simple' metals, but even simple metals might not be as simple as one might expect.  Note that nowhere is the question of fcc vs hcp vs bcc brought up, but this is a textbook. However, being bcc does not prevent the electrons from being close to free.

Moving on to more recent work on the crystal and electronic structure, I'll turn to On the Constitution of Sodium at Higher Densities (which has Ashcroft as a co-author, likely resulting in the excellent writing as well as the references to Wigner and Seitz' work on sodium from the 1930's). Some quotes relevant to the problem at hand include:

  Under atmospheric conditions, the ground state electronic
  structure of sodium therefore conforms to the classic
  nearly free electron system of Wigner and Seitz ...,
  having a single valence electron and an equilibrating density
  where the core volume is certainly an exceedingly
  small fraction of the cell volume. The pseudopotential
  for sodium at this density is quite well approximated as
  local, and under moderate compression a purely local picture
  also continues to suffice: The primary Fourier component
  of the pseudopotential $V_{G}$ diminishes relative to
  the growing bandwidth. But beyond an approximately
  3-fold compression in volume this simple interpretation
  no longer holds. In Fig. 1 we plot the band structure of
  fcc sodium at $r_{s} = 2.3$ and compare it with that of lithium
  at the same density. The bands near points K, W, and L
  of Na are quite different from those of Li, and, interestingly,
  the Fermi surface does remain nearly spherical in
  Na at these elevated densities.

and

  We have seen here that at the relevant length scales the electron-ion interaction in sodium is no longer weak under compression ..., and its putative "simple" metallic behavior at low densities appears to be an accident of the relative core and unit cell volumes present at one atmosphere.

Stepping deeper in to the simulations, one might look at First-principles study of the structural properties of alkali metals for pseudopotential calculations of bcc, fcc, and hcp in Li, Na, and K. Another interesting paper might be Structural phase stability in third-period simple metals, where calculations on fcc, hcp, and bcc for Na, Mg, Al, and Si are shown. One point of interest in this paper would be Figure 9, where they show the calculated interatomic pair potentials extracted from their calculations, including showing where the nearest neighbors fall. For Na, while the fcc nearest neighbors are closest to the minimum of the pair potential, the first and second bcc neighbors both are in that first minimum. This gives 14 nearest neighbors for bcc in the big minimum vs only 12 for fcc, yielding the possibility (and reality) that the bcc phase can be slightly more stable than fcc.

So, to wrap up:

Being bcc does not limit the simple nearly-free electron behavior for metals.
For sodium, the energy differences between bcc, fcc, and hcp are really small. The observed stability of bcc at STP is the result of slight differences in the effective pair potential (as well as dynamic effects such as the electron-phonon coupling).
Lower temperature or higher pressure push the balance towards other phases.

I would be happy to expand discussion on specific points if so desired.


Is greater relative AA battery capacity at high currents indicative of greater capacity at low currents? I originally posted this on EE.SE but might be better here:

Battery showdown compared the mAh and mWh capacity of different AA batteries discharged at 200mA.

Would the relative capacities of each battery type (within the same chemistry type) be the same at much lower current draw, e.g. 1mA? If not, why not? Are there things battery manufacturers do to 'tune' batteries to a certain discharge range?

Furthermore, I've noted that as current goes down alkaline batteries approach lithium batteries in capacity. Is there a current below which alkaline batteries have more than lithium?

Image from battery showdown:

This is a great question and I'm sorry it wasn't answered sooner. I'll give the caveat that I am not a card-carrying electrochemist, and I'm definitely not a battery researcher, even though I have worked with them in the past. I'll give my best effort.

As you note, in principal, the capacity of a battery is tracked with units of mA$\cdot$hr. In theory, this should be constant no matter the current draw.

That doesn't mean the capacity is uniform across a particular size/shape designation. The "AA" type specifies a particular shape/volume dimension, but different chemistry will give different electrochemical capacities. For example, I have plastic shells that you put a AA battery to fit "C" or "D" types - the downside is less electrochemical capacity (~2500 mA$\cdot$hr for a typical AA vs. ~8000 mA$\cdot$hr for a C cell).

One part of the electrochemical capacity is how much of the volume is filled with electro-active materials, rather than electrolyte, fillers designed to dissipate heat, or for other safety margins.

In practice, as seen in the showdown, nothing ever matches the theoretical capacity, and different batteries and different chemistries yield different practical capacities depending on the current draw.

Results @ 200 mA
Results @ 1000 mA

We see that the lithium cells get ~3000 mA$\cdot$hr under either current. The alkaline cells get ~2500 mA$\cdot$hr at low current, and much less under high current ~1500 mA$\cdot$hr.

So what's the difference?

First, as we mentioned, not every battery has the same capacity, even of the same battery type (size/shape/volume). Assume for the sake of argument that we have two 3000 mA$\cdot$hr batteries: one lithium (of some sort) and one alkaline.

Above a certain current threshold, lower currents will let batteries get closer to their theoretical capacities.

The problem is actually simple. Besides electrons, in every battery ions move through an electrolyte.

Alkaline cells use mobile $\ce{OH-}$ ions.
Lithium cells use mobile $\ce{Li+}$ ions.

At the end of the day, discharge requires ionic conductivity to match the electrical current. So which is likely to have a higher ionic conductivity? Well, $\ce{Li+}$ is smaller and usually has higher ionic conductivity.

In principal, at low current, an alkaline cell with the same capacity rating as a lithium cell, say 3000 mA$\cdot$hr should have about the same capacity. I'm not sure I've ever seen an alkaline AA battery with such high capacity -- they usually come around 2500-2700 mA$\cdot$hr theoretical capacity.

Finally, you ask if manufacturers "tune" the batteries. Put simply, yes. Besides safety considerations, different cells add compounds to minimize self-discharge, corrosion, shelf-life, etc. These can affect the amount of electrochemical capacity.

UPDATE

I realized I should clarify the very low current comment. If the current is low enough, then the rate of ionic mobility will be high enough for the chemical part of a battery to keep up with the electrical draw. So decreasing current draw will improve the capacity, up to this point - and then the real capacity should remain the same no matter how low the current draw.


Why do ethanethiol molecules have stronger London dispersion forces than those of ethanol? I know that, molecules with higher molecular weight tend to form London forces. Is there any other reason which makes Ethanethiol to have strong London dispersion forces as compared to Ethanol?

Oooh, interesting question. Let me have a go at this and we'll see if there's someone else who can put it in better terms.

So, the way I tried to reason this out is to remind myself that every single bond, in each of these two molecules, is the same except the ones where C is bonded to O or S and the ones where H is bonded to O or S. We then pay attention to these bonds, particularly.

Now, at this point, we also have to remind ourselves that O is more electronegative than S and therefore, pulls electrons towards itself much more than S does and both of these atoms are a tad bit more electronegative than C or H. Therefore, the region of higher electron density in ethanol is O and the region of higher electron density in ethanethiol is S. We also have to remember that O has a more negative partial charge than S does. We also need to remind ourselves that S is larger than O which, by the way, is also a contributing factor as to why O is more electronegative than S.

I then uses these facts and try to make sure my reasoning is sound; If we look at the way london dispersion forces work, we realize that it is easier for electrons to fluctuate between molecules when they are within the bonds than when they are within the atoms constituting the molecules. In other words, when the electron density is concentrated in the bond, as is the case when there is 0 electronegativity difference, then london dispersion forces are stronger. (Electron fluctuation, by the way, is the basis for dispersion forces)

Using the reasoning above, it becomes natural to think that the dispersion forces in ethanthiol are stronger, since the electron density is more concentrated in the bonds than it is when you consider ethanol. Therefore, it is easier for the electrons to fluctuate between two ethanethiol molecules than between two ethanol molecules, which is why ethanethiol molecules have stronger dispersion forces between themselves as opposed to the ethanol molecules.

Note: However, ethanol molecules have a much more powerful intermolecular force available to them, which is hydrogen bonding. So, clearly, ethanol molecules have a higher boiling point or whatnot than ethanethiol. Just to make sure that this is clear enough.


Unique Solubility Expression for Sulfide Salts While looking at aqeuous $\ce{K_{sp}}$ values in the back of my textbook, I noticed that there is a asterisk by all of the sulfide compounds (such as $\ce{HgS}$), and the asterisk says, "For a solubility equilibrium of the type $\ce{MS_{(s)} + H_2O_{(l)} <=> M^{2+}_{(aq)} + HS^{-}_{(aq)} + OH^{-}_{(aq)}}$." Now, why is it this expression, instead of the expected $\ce{MS_{(s)} <=> M^{2+}_{(aq)} + S^{2-}_{(aq)}}$?

Upon examination, I see that the given expression is just the expected expression and then tacking on the base equilibrium of the $\ce{S^{2-}}$ ion, which is $\ce{S^{2-}_{(aq)} + H_2O_{(l)} <=>HS^{-}_{(aq)} + OH^{-}_{(aq)}}$. But why do we involve the base equilibrium of $\ce{S^{2-}}$ when we don't involve other quite basic anions, such as $\ce{F^-}$ or $\ce{CO_3^{2-}}$?

To summarize the comments, the sulfide anion is a strong base. This site (sourced from the CRC Handbook of Chemistry and Physics) lists the second $K_a$ of $\ce{H2S}$ as $1\times 10^{-19}$, which is less acidic than water ($1\times 10^{-15.7}$).

Thus the $K_b$ of $\ce{S^2-}$, which is the equilibrium constant for the reaction with water, is $1\times 10^5$.

$$\ce{S^2- + H2O -> HS- + OH-}\ K_b= 1\times 10^5$$

After wrangling the algebra, if you start with a $1\ \mathrm{M}$ solution of $\ce{S^2-}$, at equilibrium $[\ce{S^2-}]=3.16\times 10^{-10}\ \mathrm{M}$ - almost all of it is converted to $\ce{HS-}$

Compare for some other anions (assume starting with total concentration $[\ce{A-}]+[\ce{HA}]=1\ \mathrm{M}$

$$\begin{array}{c|c|c|c|c}
\ce{A-} & K_b & [\ce{A-}]_{eq} & [\ce{HA}]_{eq} & \text{% protonated}\\ \hline
\ce{S^2-} & 1\times 10^5 & 3.16 \times 10^{-10}\ \mathrm{M} & \sim1\ \mathrm{M} & \sim 100\% \\
\ce{CO3^2-} & 2.1\times 10^{-4} & 0.99\ \mathrm{M} & 1.45\times 10^{-2}\ \mathrm{M} & \sim 1\% \\
\ce{F-} & 3.13\times 10^{-11} & \sim1\ \mathrm{M} & 5.59\times 10^{-6}\ \mathrm{M} & \sim 0\%
\end{array}$$


Is identifying a chemical equivalent to be able to produce it? This question is a follow up of another thread, where we are discussing Theresa May's accusation of the Russian government which is supposedly behind the recent chemical attack in the UK.

Soon after the attack the UK implemented a chemical analysis which again supposedly shows that this compound can be produced by only one government in the world: the Russian government.

So, my question is, if the UK Chemical Weapons Institute can carry out such a chemical analysis and figure out what chemical weapon was used, does it mean that they themselves possess this technology?

On 14 March 2018, at UN Security Council Russia's representative Vasily Nebenzya claimed:

  For the British specialists to be perfectly confident that this was a Novichok agent and not any other kind, they would need a control standard for proof. It [the substance used in the attack] must be compared to a control substance... They have a collection and they have the formula. In other words, if the UK is so firmly convinced this is Novichok, they have samples and formula and are capable of formulating it themselves.

Nebenzya in his speech refers to some professional chemist and I wonder what other professional chemists think of this.

Further reading:

A BBC article explaining the Novichok agent; the Wikipedia page about the same as well
Another BBC article detailing the timeline of political events.

No, the capability to identify a compound does not imply capability to synthesize that compound. Indeed, "analysis" and "synthesis" represent two branches of chemistry, and many chemists spend years trying to synthesize compounds of known structure.

Modern structure determination relies on roughly two steps. First is separation of sample mixtures into constituent parts, using various forms of chromatography. Next is analysis of chemical composition and bonding patterns using techniques such as IR spectroscopy and nuclear magnetic resonance spectroscopy, as well as mass spectrometry (MS), which weighs molecules (and their destructively ionized fragments) with extraordinary precision and dynamic range.

Advanced MS instruments can analyze very small amounts of collected samples, without special preparation, and at atmospheric pressure. Pure samples are not required for the instruments typically used in forensics and product testing (DART). When a pure sample is desired, one or more separation devices can feed the MS instrument in-line (LC-MS, GC-MS, MS-MS).

MS can measure molecular weight so precisely that the chemical formula can be deduced directly based on the fractional weights of the elements due to varying isotopic abundance. Often, a chemical formula is almost sufficient for unique identification. For example, the NIST Chemical WebBook contains only 4 entries with the same formula as the VX agent, all for VX or nearly identical compounds. The PubChem database contains only 1 (VX itself).

It is also possible to identify compounds using known chemical or physical interactions linked to a readout such as a color change. This principle is used in devices such as home pregnancy tests and lead test kits. The US military even has custom field test kits for some chemical weapons.

It should be noted that compounds introduced into an organism may be identified indirectly via metabolites found in blood or urine, even if the original compound is long gone. Again, VX agent is a suitable example.


does Hydrochloric Acid react with PET? From someone coming from electronic engineering not skilled in chemistry,

There are transparent electric conductors in form of ITO ( Indium tin oxide ) coated glass or transparent PET films ( Polyethylene terephthalate ). By etching the ITO coating we can create transparent circuit boards and with the case of ITO coated PET also flexible transparent circuit boards.

There seems to be very little info on what can be used for etching ITO PET sheets.
For ITO coated glass, people use HCL with no issue, but will HCL also work to etch ITO but leave the transparent PET film intact, or
do PET and HCL react?

You can etch ITO on PTE by mean of conc. HCl solution without problems. The treatment duration depends on the ITO thickness but is limited to minutes.

Just make few samples (residuals from cutting etc will suffice to make this test) to find out how long it takes to fully remove ITO.

No reference need it as I do etching of ITO coated PTE electrodes since years.

For sake of completeness let me add that I never investigated PTE itself prior and after the etching so I do not know if there are changes which, for instance, may interfere with long (very long) stability etc.

In my case it suffices that PTE maintains its transparency and still behave as a robust but flexible substrate. It seems it could be your case, too.

Just to be pedantic. Obviously after etching the foil must be washed thoroughly.


Brown color at end point during titration with phenolphthalein as indicator We did a titration experiment at school and I did not understand one part of it.

The scenario:

Titrate : White wine (ethanoic acid)
Titrant: $\ce{NaOH}$
Indicator:  Phenolphthalein

What happened: We carried out the titration and the end point was a brown colour!

How does that happen? Why is the color brown and not pink?

Assuming there are no major color-influencing side reactions, the subtractive color mixing as follows:

#FFA8D9 (Pink, phenolphtaleine) + #DAA520 (Goldenrod, white wine) = #D2691E (Chocolate, result)

actually already gives brown(ish) color. It's hard to pick the right color codes, especially for the white whine, but OP can probably try to perform the same operation with the color he remembered/picked from the photo.


Acidity comparison between cyclopropene and cyclopropane There many sources on the internet saying that cyclopropane is more acidic than cyclopropene because the conjugate base in latter would have antiaromaticity which would destabilize it.

But in this explanation we are assuming that the central $\ce{C-H}$ bond in cyclopropene ionizes. I argue that the $\ce{C(sp^2)-H(s)}$ bond should ionize because of two factors:

The $\ce{sp^2}$ carbon would better accommodate the negative charge than the central $\ce{sp^3}$ carbon.
Consequent conjugate base would not contain anti aromaticity.

My logic can also be rationalized by considering acidity order between straight chain alkanes and alkenes.

This would mean inversion of the order which is incorrect according to experimental data. What is the flaw in my reasoning?

This reference does in fact indicate that cyclopropene ($pK_a=29$) has a much higher a acidity than cyclopropane ($pK_a=44$).  Deprotonation at the vinylic hydrogen would account for this.  That cyclopropene is deprotonated at a vinylic carbon is briefly summarized here with a primary reference given.


Would a loaded cartridge ignite/explode when hit from the side or front by a bullet? This question https://physics.stackexchange.com/a/193472/93888 explains what would happen after a cartridge is ignited: the bullet gets propelled by the gases in one direction and the cartridge (with the person, in the case of the question) in the opposite direction.

But would a cartridge explode/ignite if hit from the side (the rear part where the gunpowder is, not the side of the bullet) or the front by a bullet? Or is the rear the only side able to start the reaction?

Maybe

Unfortunately, the correct answer is “it depends.” Small arms cartridges struck by small arms projectiles generally do not ignite – but they do sometimes.

The issue is quite important in the armed forces. If you come under enemy fire, the last thing you want is for a near miss to cause your own ammunition to explode. There has been formal research on the subject for decades. However, it can be quite difficult to prevent.

There are formal test protocols to assess the safety of new ammunition types, and they can involve munitions struck by a range of different “stimuli” from various distances, points of impact and angles. Depending on the exact set-up, the response can vary from 100% explosion to barely scratched.


Why is H2SO4 a good dehydrating agent while other strong acids are not? I understand that $\ce{H2SO4}$ is a very good dehydrating agent because it is a strong acid, so that it quickly donates $\ce{H+}$ to hydroxide ions in water to form $\ce{HSO4-}$ and $\ce{H3O+}$ .But we know that there are a lot of other acids stronger than $\ce{H2SO4}$ like $\ce{HClO4}$ which are not used as dehydrating agents. They can also readily give $\ce{H+}$ to hydroxide ions in water and act as dehydrating agents but we don't use them for this purpose. Why is that?

As pointed out by Maurice and myself, the argument provided in the answer by iad22agp is rather incorrect. Sulfuric acid is a dehydrating agent just because it is available in concentrated form whereas rest of the common acids like HCl(aq) are not. No, it is not a concentration effect. First of all, it is not possible to have HCl(aq) more concentrated than 38% wt/wt under normal conditions. This is already a saturated solution of the gas.
Secondly, liquified HCl cannot be a good dehydrating because as everyone knows HCl g fumes heavily in the presence of moisture. It cannot fulfill the criterion of a dehydrating or dessicant. The same arguments apply to conc. nitric acid as well.
The two main reasons which make conc. sulfuric acid a good dehydrating agent are (a) it has a very low vapor pressure and (b) secondly the hydration of sulfuric acid is quite exothermic as compared to other common acids. Maybe a quantum mechanical study reveal "why" this is so. Acid strength alone cannot be used to predict the dehydrating capability. One has to think about the vapor pressure as well.
General chemistry book also preach that sugars are dehydrated by conc. sulfuric acid. It is actually a wrong notion to say that sulfuric acid extracts water out of sugar. There is no water molecule in sugar. This is a chemical reaction between sugar and conc. sulfuric which generates water and carbon. The true demo for sulfuric acid dehydration is the dehydration of hydrated copper sulfate. Adding a few drops on the blue crystals makes them white (anhydrous) copper sulfate almost immediately.


Finding the mass of NaNO3 in original mixture from preciptate A 1.61-g sample of a mixture of $\ce{AgNO3}$ and $\ce{NaNO3}$ is treated with excess $\ce{Na2S}\:(\mathrm{aq})$. The precipitate is filtered off, dried and weighed. The dried precipitate weighs 0.44 g. What is the percentage by mass of $\ce{NaNO3}$ in the original mixture?

I attempted this problem by using two equations: one with $\ce{AgNO3}$ and $\ce{Na2S}$, and the other with the other reactant and $\ce{Na2S}$. Then I wrote an equation in moles, using m and 1.61-m as the mass of the reactants. My answer was 2.34 g, which is greater than the original so I know I did something wrong.

The dried precipitate is $\ce{Ag2S}$.
The number of moles of $\ce{Ag2S}$ is:
$$n=\frac{m_{\ce{Ag2S}}}{M_{\ce{Ag2S}}}=\frac{0.44}{32+108\times 2}=\frac{0.44}{248}=0.00177\,\mathrm{mol}$$
The number of moles of ion $\ce{Ag(I)}$ is$$n'=2n=0.00355\,\mathrm{mol}$$

On the other hand, the percentage by mass of $\ce{AgNO3}$ in the original mixture is: $$\frac{m_{\ce{AgNO3}}}{m_{\ce{(AgNO3 + NaNO3)}}}\times 100=\frac{n'\times M_{\ce{AgNO3}}}{m_{\ce{(AgNO3 +NaNO3)}}}\times 100=\frac{0.00355 \times (108+14+16 \times 3)}{1.61}\times 100=37.48\%$$

So, the percentage by mass of $\ce{NaNO3}$ is: $100-37.48=62.52\%$


How to determine if a reaction proceeds via SN1, SN2, E1 or E2 mechanism? Give the main product and reaction type: SN1, SN2, E1 ,  E2.

As the 1-bromohexane is primary and the nucleophile a strong unhindered base, the reaction should be a SN2 reaction. However, the solvent $\ce{EtOH}$ is a polar protic solvent which favours SN1/E1 reactions. In addition, the temperature of the reaction is quite high and high temperatures favour the formation of elimination reactions, thus the reaction should be E1.
But an E1 doesn't require a strong base, or else an E2 reaction would take place at the beginning. So does that make this reaction an E2 reaction? Am I thinking correctly on how to consider all the different parts of the reaction?

$\ce{EtO-}$ is a strong base, as well as a strong nucleophile, and the temperature is not too high. Therefore a mixture of products is most likely to be formed, where substitution will lead to the major and elimination will lead to the minor product.
Also, the reactant is a primary halide. So, in substitution, SN2 will be favoured, whereas in elimination E2 will be favoured.
So, final product will be,

Ethoxy hexane (via SN2) (major)
Hex-1-ene (via E2) (minor). (Maybe, a very little amount of hex-2-ene may also form by very very small fraction of E1).


Resources to learn second quantization formalism and how to implement it in code for quantum/computational chemistry research I want to know what are some useful and practical resources to learn the formalism of second quantization as it pertains to fermions to be used in solving Quantum Chemistry problems. I am particularly asking for resources that go beyond the symbolic formalism and have guides/examples on how to implement the second quantization ideas in code (regardless of comp. language).
The pink book: Molecular Electronic-Structure Theory from Helgaker is a good start on the topic with all the formalism but I am looking for other resources with coding examples or practical implemmentations?

You could check out any code set that implements configuration interaction wave function methods.  The Journal of Chemical Physics published an open-access article reviewing the state of ab-initio software, and it lists code sets that use CI.
 https://aip.scitation.org/doi/pdf/10.1063/5.0023185
Notably, NWChem uses the second quantization approach and is described in this open resource paper:  https://arxiv.org/pdf/1904.01131.pdf
Finding algorithms to rapidly create valid CI expansions was a big research area in the late 1970's, and the Graphical Unitary Group Approach was created.  There are several open source packages that use the GUGA, including GAMESS. https://www.msg.chem.iastate.edu/gamess/download.html   You can get GAMESS by requesting access and registering.  You could then review the implementation assuming a basic knowledge of FORTRAN.


Alkylation of 1-naphthol in trifluoroethanol versus in DMSO I am supposed to work out which product is formed in which solvent when naphthalen-1-olate is treated with benzyl iodide.

DMSO is a polar aprotic solvent, as it does not possess an acidic hydrogen. So, the negative charge on oxygen will be concentrated as the anion will be poorly solvated. Hence, it should act as a hard nucleophile. Since alkyl halides such as BnI are soft electrophiles, I think that the substituent should preferentially add to the ring.

In 2,2,2-trifluoroethan-1-ol (TFE), I think the reaction would proceed via an SN1 mechanism. Now, the carbocation is a hard electrophile, and thus O-alkylation might be observed.

But considering the better solvation of the negatively charged oxygen in TFE, it would also be more hindered, which suggests that C-alkylation might also be seen.

Which products are actually formed, and am I on the right track?

The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:

Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.

In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.

EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."


Is cyclopentadiene or cycloheptatriene more acidic? Which of the following cyclic compounds are more acidic?

The most acidic proton is positioned on the carbon that is at the top of the above drawings (the methylene hydrogens) on each of the two species, as deprotonation allows resonance.
I am aware of Hückel's rule, which states that an aromatic species has $4n+2$ π-electrons. Only the five membered ring would fulfil this requirement. However, I can draw resonance structures where the seven membered conjugate base has the double bond at each position on the ring. I ask why is it not aromatic? Which conjugate base is more stable? Essentially it's a case of aromaticity vs number of resonance structures.

Essentially it's a case of aromaticity vs number of resonance
  structures.

Well put!

Aromaticity is a very strong driving force so aromaticity wins out; Huckel's rule is more important than the number of resonance structures.  Therefore cyclopentadiene is more acidic than cycloheptatriene.  Even though we can draw 7 resonance structures (one with a negative charge on each carbon atom) for the tropylium anion and only 5 for the cyclopentadienyl anion, Huckel's rule tells us that the cyclopentadiene anion ($4n+2$ π electrons, $n=1$) is aromatic while the tropylium anion with 8 π electrons (doesn't fit $4n+2$) is not.

Another example is cyclooctatetraene.

Like benzene, we could draw resonance structures by shifting the double bonds in this molecule too.  But this molecule is not aromatic, it does not fit the $4n+2$ rule.  In fact, Huckel says with 8 π electrons it is antiaromatic.  To avoid this destabilization cyclooctatetraene adopts a tub-like conformation.

(image source)

Again aromaticity trumps resonance structures.

Edit: Huckel's Rule: Aromaticity - Antiaromaticity

If we look at the energetic positioning of the molecular orbitals (MO's) in a cyclic, conjugated polyene, we can quickly understand the basis for Huckel's rule.

We can use Frost diagrams to determine the energetic positioning of these MO's.  Here is the diagram for benzene, we see that all of the pi electrons are in bonding MO's, the molecule is a stable aromatic that fits the $4n+2$ rule ($n=1$).

Here is the diagram for cyclooctatetraene, and we see that not all of the electrons are in bonding MO's, two π electrons are in non-bonding MO's.  They don't contribute to bonding or stabilization.

This is a $4n$ π electron compound ($n=2$) and accordingly would be classified as antiaromatic by Huckel's rule.  As mentioned above, the compound adopts a non-planar conformation to avoid this destabilization.


Nomenclature in case when same fucnctional group is present more than once here since Only CN group is present, it is the major functional group. In this molec. we cannot have a chain which has carbons of all CN group. hence in such compounds we do not take any carbon of cn in the parent chain.so answer should be 3.
but few books and many reference texts are showing
name as option d.

The IUPAC rules do not seem to have an explicit rule for general cases when more than two cyano groups are attached to the (linear) alkane chain, like it does for aldehydes (see Is propane-1,2,3-tricarbaldehyde the correct IUPAC name of CHO-CH2-CH(CHO)-CH2-CHO?), but do have a related one:

  P-66.5.1.1.2 If an unbranched alkane is linked to more than two terminal cyano groups, all cyano groups are named from the parent hydride by the substitutive suffix ‘carbonitrile’, preceded by appropriate multiplying prefix and locants.
  Example:

  $\ce{\overset{4}{C}H3-[CH2]2-\overset{1}{C}(CN)3}$
  butane-1,1,1-tricarbonitrile (PIN)

I think that similar usage in general is implicitly allowed in the rule

  P-66.5.1.1.4 When a group is present that has priority for citation as the principal characteristic group or When all $\ce{-CN}$ groups cannot be expressed as the principal characteristic group, the $\ce{-CN}$ group is designated by the preferred prefix ‘cyano’.

  (…)

(Three examples follow, two of them have another higher-priority group $\ce{-COOH}$; the other one has three cyano groups, not on linear chain but one of them is attached via a chain branch, $\ce{NC[CH2]2CH(CH2CN)[CH2]2CN}$, 4-cyanomethyl(heptanedinitrile).)

I our case, we can express all the groups as principal characteristic group, so the correct name should be propane-1,2,3-tricarbonitrile.


"Asymmetric molecules are necessarily polar" Is a molecule with no symmetry necessarily polar?

Can a polar molecule still have some elements of symmetry (i.e. mirror image is the same as itself)? I think so ... because we can have achiral tetrahedral carbon molecules that are polar - i.e. methylene chloride.

So, is it proper to say that a molecule that lacks symmetry is necessarily polar?

Or can a more nuanced statement based on symmetry be made? Is there a "minimum" level of symmetry needed to achieve non-polarity?

Let's start by defining a polar molecule as a molecule with a dipole moment.

  Is a molecule with no symmetry necessarily polar?

Yes, a molecule with no symmetry belongs to point group $\ce{C_1}$ (e.g. the only symmetry element it contains is a $\ce{C_1}$ axis, rotation of the molecule 360/1 degrees converts the molecule back into itself).  The molecule is asymmetric and polar.

  Can a polar molecule still have some elements of symmetry (i.e. mirror
  image is the same as itself)? I think so ... because we can have
  achiral tetrahedral carbon molecules that are polar - i.e. methylene
  chloride.

Yes, methylene chloride, vinyl chloride, methanol, chloroform, etc. all contain planes of symmetry, but are still polar.

  So, is it proper to say that a molecule that lacks symmetry is
  necessarily polar?

If I understand correctly, this is a repeat of your first question.

  Or can a more nuanced statement based on symmetry be made? Is there a
  "minimum" level of symmetry needed to achieve non-polarity?

Yes

All asymmetric molecules are polar.  Such molecules belong to point group $\ce{C_1}$
Molecules belonging to point group $\ce{C_{n}}$ with n>1 are polar. 1,3-dimethylallene is an example (it has $\ce{C_2}$ symmetry, rotate it by 360/2 degrees about its $\ce{C_{2}}$ axis and it converts back to its original representation)
Molecules belonging to point group $\ce{C_{s}}$ are polar.  A sigma plane either contains or bisects molecules in this point group, and this plane is the only symmetry element.  Methanol, bromocyclopropane, vinyl chloride, etc. belong to this point group and are polar.
Molecules belonging to point group ${C_{nv}}$ are polar.  Formaldehyde and methylene chloride (both ${C_{2v}}$), chloroform (${C_{3v}}$) and hydrogen chloride (${C_{\infty v}}$) are examples.  This point group contains a ${C_{n}}$ axis and n sigma planes containing the axis.

If the molecule is not in one of these point groups, then it is not polar and does not have a dipole moment (or said differently, molecules in all other point groups [${D_n, S_n, C_{nh}, D_{nd}, D_{nh}, T_d, O_h}$ and higher] are not polar and do not have dipole moments).


The ionization energy of H₂⁺ molecule ion I am trying to compute the ionization energy of $\ce{H2+}$ molecule ion from the electronic energy spectrum. The question is whether one should use the purely electronic Hamiltonian or the Hamiltonian with the internuclear repulsion term to do that? I illustrated two ways to compute the ionization energy below. Which one is correct?

Notice that, nuclei are not at the equilibrium distance since in my problem I can manage this distance by external forces.

I would like to clarify the matter since from OP's own answer I have a feeling he/she does not quite understand it.

So, first of all, in general the ionisation energy has to be calculated as the difference in total energy and not just the electronic one.

Secondly, for molecules there are two types ionization energies: adiabatic and vertical:

  The adiabatic ionisation energy refers to the formation of the
  molecular ion in its ground vibrational state and the vertical
  ionization energy applies to the transition to the molecular ion
  without change in geometry. (From IUPAC Gold Book)

Note that in accordance with the Franck–Condon principle the most probable transition is the vertical one. Consequently, usually, when we say "ionisation energy" we actually mean the "vertical ionisation energy".

Finally, if the total energy is understood in the clamped nuclei approximation, i.e. as the electronic energy $E_{\mathrm{e}}$ together with the nuclear repulsion energy $V_{\mathrm{nn}}$, then for the vertical ionisation energy one could indeed use the difference in electronic energy rather than in total one simply because the nuclear repulsion energy stays the same. Mathematically, $V'_{\mathrm{nn}} = V_{\mathrm{nn}}$ since geometry is not changing, and thus,
$$
  E_{\mathrm{i}}
  =
  U' - U
  =
  (E'_{\mathrm{e}} + V'_{\mathrm{nn}})
  -
  (E_{\mathrm{e}} + V_{\mathrm{nn}})
  =
  E'_{\mathrm{e}} - E_{\mathrm{e}} \, ,
$$
where primed quantities refers to the resulting molecular specie and non-primed are that of the original molecular specie.

So, yes, for the vertical ionisation energy in the clamped nuclei approximation one can use the difference in electronic energy. And one can do so regardless of what are the molecular species in question. For the specific case of the vertical ionisation of $\ce{H2+}$ the electronic energy of the resulting $\ce{H2^2+}$ specie is obviously zero, so we specifically get what OP suggested
$$
  E_{\mathrm{i}}
  =
  E'_{\mathrm{e}} - E_{\mathrm{e}}
  =
  0 - E_{\mathrm{e}}
  =
  - E_{\mathrm{e}} \, .
$$
However, this is just a trivial simplification of a general formula for the vertical ionisation energy for the case of a zero-electron resulting specie.


How different concentration of HCl on magnesium affect the enthalpy change I had conducted an experiment where I reacted a $0.3\ \mathrm{g}$ strip of magnesium with $100\ \mathrm{mL}$ of varying concentrations of $\ce{HCl}$: $1\ \mathrm{mol/L}$, $2\ \mathrm{mol/L}$ and $3\ \mathrm{mol/L}$. From the data I collected, I noticed that the reaction with $1\ \mathrm{mol/L}$ $\ce{HCl}$ produced the most energy and the reaction with $3\ \mathrm{mol/L}$ $\ce{HCl}$ the least. I just wanted to know if my data is correct or not because I hypothesized that the $3\ \mathrm{mol/L}$ reaction would produce the most heat energy.

In fact all three reactions produced the same amount of heat energy and hence had the same enthalpy change. By changing the concentration of the HCl only effects the reaction rate and how much energy is produced in a certain time. However eventually, each reaction will produce the same amount of energy provided that all the Mg reacts which does happen in your experiment as HCl is in excess. The reason for this can be explained quick easily using the formula: $$\Delta H=\Delta U+pV$$ Then using the ideal gas equation, we can substitute $pV$ with $\Delta n(g)RT$. Note that the only variable that actually changes is n(g) which is the change in the number of moles of gas molecules. So the equation becomes: $$\Delta H= \Delta U + \Delta  n(g)RT$$ So the only thing that effects the enthalpy change of the reaction in the change in internal energy (U) and the number of moles of gas molecules formed which are both constant for all three reactions.


What chemical properties that allow for colour exist in the dark? Of course if there's no light around, there's no colour that you can see. On the other hand, the wall must have some property that makes it be blue. That property is still there in the dark.
— Mike W.

Is Mike referring to the property of excited states, or the property that the states are excitable (via split d-orbitals or conjugated orbitals)?
I understand that anything that we see as coloured but not glowing is transmitting unabsorbed light — the complementary colour of the absorbed light — which is only possible when electrons jumping from ground to excited states.
What I'd like some clarity on is whether those excited electrons fall back to their ground state in the absence of light (another potential reason why colour does not exist in the absence of light). If the electrons do, then I guess the property Mike's referring to is the excitability of the electrons — which makes sense.

The chemical property that creates colour is the ability to absorb light of a specific visible wavelength. There is more than one way to do this.

Mostly colour is caused by the existence of electronic transitions in substances that match the energy of some wavelength of light so when light hits the substance, some is absorbed by exciting electrons from a lower to a higher energy level. The higher energy levels will not be populated unless there is some energy input to excite the electrons. But the structure that enables this absorption to exist is present whether or not there is any light.

There are other ways a substance can create colour. For example, some compounds do it by interference: the physical structure contains features that have the same scale as wavelengths of some colour of light. When light hits the substance the reflected light creates interference patterns that can selectively eliminate some wavelengths. This is why oil films on water show rainbow colours and why the mineral opal shows beautiful iridescence. again, though, the cause of colour is present whether there is light or not.

In short, the cause of colour is sometimes a chemical thing (involving electrons being excited) and sometimes a physical thing (involving interference) but, either way, the property that creates it exists even when there is no light.


Bonding of Lithium and Argon I saw a meme that was joking around about Lithium and Argon bonding (see pic below). It got me wondering: Can Lithium and Argon bond in any circumstance?

Argon is a noble gas. Meaning, it does not simply "bond" with anything. These special gasses have a VERY strong resistance to joining with other atoms. So, to answer your question, no, you can not bond Argon and Lithium under any circumstance. Anyone who says they have done so is a LiAr ;)

Also, since Lithium is highly reactive and flammable and Argon is used in neon, I would not suggest trying to combine these two without a proper location and gear.


Writing equilibrium expressions with aqueous species If we have the reaction: $$\ce{A +BC_{(aq)} <=> D}$$ where $\ce{A}$ and $\ce{D}$ are generic compounds (neither is pure solid nor pure liquid) and $\ce{BC}$ is an ionic compound dissociated in water (consisting of $\ce{B+}$ and $\ce{C-}$ ions), would the $K_{\mathrm{c}}$ expression be written as

$$K_{\mathrm{c}}=\frac{[\ce{D}]}{[\ce{A}][\ce{BC}]}$$

or

$$K_{\mathrm{c}}=\frac{[\ce{D}]}{[\ce{A}][\ce{B+}][\ce{C-}]}?$$

I will say the latter form is more correct, and in fact an important way to consider equilibrium with solubility and the common ion effect.  You should not count any solids in equilibrium expressions though as they do not determine the final equilibrium.


Can other substances be dissolved in a saturated solution? If I have a liter of water fully saturated with sucrose would it be possible to dissolve something like salt or any other substance in the water? Or when the solution is saturated, is it impossible to dissolve another solute in it?

Saturating a liquid with one solute does not mean that the liquid will no longer dissolve another solute. However, you can expect the solubility of the second solute to be different, generally lower, than in the neat solvent.

One relevant concept here (though not specifically applicable to sucrose), in the case of ionic solutes, is the common-ion effect. According to this Wikipedia article:

  The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions.

Regardless of whether the solute is ionic or not, when you are given a solubility value, for example, the solubility of some compound in water as $\pu{g solute/100mL water}$, this value is only relevant to the solubility in pure water. Once you have saturated water with sucrose, you then have a very different solvent system. There are no simple means of determining what the solubility of a particular solute will be in a solution saturated with another solute. However, there will, in general, be some solubility of additional solvents except as in the case described for the common ion effect.

Please don't hesitate to ask for clarifications in the comments below if I have misunderstood your question or left anything out.


Acid-base related chemistry What is the function of silica, $\ce{SiO2}$ in the equation
$$\ce{CaO + SiO2 -> CaSiO3}$$
  a) A basic oxide
  b) A acidic oxide
  c) A reducing agent
  d) An oxidising agent

Please explain.

Since this sounds like a homework question, I will not provide a full solution!
Is $\ce{SiO2}$ an oxidizing or reducing agent in the reaction?
Figure out the oxidation state for the $\ce{Ca}$ and $\ce{Si}$ in the starting materials and the product. As a rule of thumb, consider the oxidation state of (-2) for $\ce{O}$. Unless peroxides are concerned, this works well.

 $$\ce{Ca^{(+2)}O + Si^{(+4)}O2 -> Ca^{(+2)}Si^{(+4)}O3}$$
 Apparently, there is no change - this is not a redox reaction!

Is $\ce{SiO2}$ a basic or an acidic oxide?

How are basic and acidic oxides defined?
Are there any rules (of thumb) for the oxides of metals and non-metals?

 Metal oxides are usually basic oxides. These can be considered anhydrides of the corresponding metal hydroxides, $\ce{CaO}$ is a typical example: $\ce{CaO + H2O -> Ca(OH)2}$
 What about $\ce{SiO2}$?


What has old Potassium metabisulfite turned into? I found a glass jar labeled potassium metabisulfite in my abandoned photo lab.
The jar has been standing with a loosely fitting lid for thirty years in a cool and dry place.

The jar contains a dry substance now like matte flakes. The substance does not react visibly to hydrochloric acid and it does not smell of sulfur dioxide.

What substance has the potassium metabisulfite turned into? Just curious.

EDIT. The substance does not dissolve visibly in cold or boiling water. A few flakes poured in water remain on the bottom seemingly unaltered.

MORE EDIT. When heated to red heat in a test tube, it does not melt, the coarse powder is not sintered. No water is given off. Some sulfur is sublimated, a yellow deposit in the cold part of the test tube and a smell of sulfur. Not much sulfur is given off, the volume of the original substance is not visibly altered.

According to my research, I found the reaction of oxidation of potassium metabisulfite but the products of this reaction does not seem to match with your experimental observations.

  The jar has been standing with a a loose fitting lid for thirty years in a cool and dry place.

This means that your old  potassium metabisulfite has been subjected to atmospheric oxidation in a cool and dry place and thus the temperature of the surrounding did not exceed the room temperature.

$$\ce{2K2S2O5 + O2 ->[298K] 2K2SO4 + 2SO2}$$

(source)

  The jar contains a dry substance now like matte flakes. The substance does not react visibly to hydrochloric acid and it does not smell of sulfur dioxide.

Yes, potassium sulfate has a flaky, crystalline appearance. Potassium sulfate does not react with hydrochloric acid. But one of the product is sulfur dioxide and the flakes smell of sulfur dioxide.

  The substance does not dissolve visibly in cold or boiling water. A
  few flakes poured in water remain on the bottom seemingly unaltered.

Potassium sulfate is soluble in water: 120 g/L (25 °C).

  When heated to red heat in a test tube, it does not melt, the coarse
  powder is not sintered. No water is given off. Some sulfur is
  sublimated, a yellow deposit in the cold part of the test tube and a
  smell of sulfur. Not much sulfur is given off, the volume of the
  original substance is not visibly altered.

Potassium sulfate does not decompose even at high temperatures. But if the surrounding has carbon monoxide, it gets reduced to potassium sulfide which has a yellow-brown appearance when impure.

Summarizing as a table,


Why is potassium cation harder than copper? According to HSAB theory, the smaller radius and the less polarizable an acid is, the harder it is. According to Wikipedia, the ionic radii of $\ce{ K+}$ is much larger than $\ce{Cu+}$, which makes sense because d-electrons are poor shielders. Also, $\ce{Cu+}$ has larger electron density but it has a higher effective nuclear charge in comparison to potassium ion because of the poor shielding ability of d-electrons. Yet why is $\ce{K+}$  ion considered hard while $\ce{Cu+}$ ions are considered soft?

$\ce{Cu+}$ has a kernel of $\ce{K+}$, with an extra 10 protons in the nucleus and a $\mathrm{3d^{10}}$ electron shell around it. $\ce{Cu+}$ is smaller because it has more positive charge pulling those electrons in. But size isn't everything!

The first ionization potential of potassium ($\pu{4.34 V}$) is much lower than the first ionization potential of copper ($\pu{7.73 V}$). But from there on, the situation changes: Second ionization potential: $\pu{31.63 V}$ for $\ce{K}$ vs $\pu{20.29 V}$ for $\ce{Cu}$; third ionization potential: $\pu{45.72 V}$ for $\ce{K}$ vs $\pu{36.83 V}$ for $\ce{Cu}$; fourth ionization potential: $\pu{60.91 V}$ for $\ce{K}$ vs $\pu{55.2 V}$ for $\ce{Cu}$; and fifth ionization potential: $\pu{82.66 V}$ for $\ce{K}$ vs $\pu{79.9 V}$ for $\ce{Cu}$. It is not till the sixth ionization potential that potassium is again easier to ionize than copper: $\pu{100.0 V}$ for $\ce{K}$ vs $\pu{103 V}$ for $\ce{Cu}$.

So, although the smaller size of the $\ce{Cu+}$ ion suggests that it would be harder (after all, it's stuffed with more electrons!), it turns out that the $\ce{Cu+}$ ion has a softer (i.e. more easily ionized, therefore more easily polarized) outer skin.

Perhaps this is not predictable unless you take into account more than just ionic radii, but at least it is explainable.


Will an aqueous sodium hydroxide solution degrade over time? I have a chiller bath, which is attached to a 150 ton cooling unit through 4" PVC pipe. The bath itself holds around 5000 gallons of water, and is made of 304 stainless steel. It constantly circulates through the PVC into the cooling units outside via a 20hp pump. The outside cooling units are made of steel, although I do not know the exact grade of the steel. It was however built for water circulation. I use a sodium hydroxide solution to raise the pH of the water, therefore preventing rust and keeping my water clean. The problem is that I constantly have to add the solution to keep the pH at a level that is acceptable (9-11). I have watched it enough to say that on average the pH drops by about 0.3 a day. I am wondering where all my sodium hydroxide solution is going. Thank you for the responses so far, I apologize for not being more clear on the problem.

Usually, the biggest problems with preventing a sodium hydroxide solution from changing in $\ce{NaOH}$ concentration, or pH, over time are adsorption of $\ce{CO2}$ from the air, and evaporation of water via permeation through the container holding the solution.

If we can really say that it is inside a closed system as you've specified, then any degradation over reasonable time-scales (i.e. months to years) should be negligible.

However, you need to be sure about your "closed system". Is it glass, which could slowly react with the $\ce{NaOH}$? Is it a plastic that could be slightly permeable to water vapor, air (and thus $\ce{CO2}$), etc.? Is it, heaven forbid, a metal that would react and degrade your $\ce{NaOH}$? You need to think about these things at every contact point (vapor or liquid) throughout your system.


Determining change in entropy and enthalpy during mixing of gases Question taken from KVPY 2011. Answer key is A.

Logically it makes sense that $\Delta S_{\text{system}}>0$, because the process is spontaneous and as per second law, this means $\Delta S_{\text{universe}}>0)$. System is isolated ($\Delta S_{\text{surrounding}}=0$), hence system's entropy must be increasing.

However, it is also given that $$\Delta S_{\text{system}}=\frac{\Delta q}T$$ So does this mean that heat is transferred in some way? How is that possible for an isolated system?

Also, if there is no heat transfer, we can conclude that $\Delta H = 0$. But is that true?

Your equation for $\Delta S_{system}$ is incorrect.  The correct equation is $\Delta S_{system}=\int{\frac{dq_{rev}}{T}}$.  I hope you understand the difference, and the fact that the subscript "rev" refers exclusively to a reversible path between the initial and final states of the system.  To get that, you need to forget entirely about the actual irreversible path, and instead focus on the two end states.  You then have to devise a reversible path between these same two states so that you can determine the value of the integral.  For your problem, this recipe will yield a positive value for $\Delta S_{system}.$

Regarding the enthalpy change, you are aware that the enthalpy of mixing of ideal gases is zero, correct?


How to redefine internal pressure in terms of κ and α? The formula for internal pressure of a system is:
$$\pi_T=T\left(\frac{\partial p}{\partial T}\right)_{\!V}-p$$

My professor said we can redefine this equation in terms of either the

thermal compressibility, $\kappa$
expansion coefficient, $\alpha$

How do we do this?

I was able to derive for a van der Waals gas:

$\Large\kappa = \frac{V-nb}{pV-\frac{an^2}{v}+\frac{2abn^3}{v^2}}$
$\Large \alpha= \frac{nR}{PV-\frac{an^2}{v}+\frac{2abn^3}{v^2}}$

$$dV=\left(\frac{\partial V}{\partial T}\right)_pdT+\left(\frac{\partial V}{\partial p}\right)_Tdp$$So at constant volume dV = 0 and $$\left(\frac{\partial p}{\partial T}\right)_V=-\frac{\left(\frac{\partial V}{\partial T}\right)_p}{\left(\frac{\partial V}{\partial p}\right)_T}$$

The rest is straightforward.


Why is NaHCO3 a basic salt? When NaHCO3 is added to water should give an H$^+$ ion in water and this should make it acidic. Why is it a basic salt? Does it have something to do with H2CO3 being a weak acid?

The thing is that $\ce{HCO3-}$ can act as an acid and a base. These substances are  called amphiprotic (other examples include $\ce{HSO4-}$ and $\ce{H2PO4-}$). So it can donate a proton to water to form $\ce{H3O+}$ but it can also accept a proton from water to form $\ce{OH-}$. So really, the  question is which process occurs to a greater extent?.

You can't answer that question by simply looking at the molecule. Instead we need to know the $\mathrm{K_a}$ values for $\ce{H2CO3-}$ and $\ce{HCO3-}$. By comparing these values and doing a few calculations we can determine whether $\ce{HCO3-}$ is basic or acidic. It turns out $\ce{HCO3-}$ is actually basic.

However I am assume that equilibrium constants are probably above the level that you are learning at. So this is just one of these things which you should just know. For example you should know that despite $\ce{HCO3-}$ being basic, $\ce{HSO4-}$ and $\ce{H2PO4-}$ are actually acidic.


Why H3O+ exists and H3O- doesn't? I really don't know if it is even logical to ask that $\ce{H3O-}$ doesn't exist but someone asked me and I wasn't clear. Hope that some one will make me clear.

I'm not a big fan of MOs when this can be explained using just AOs. For sure, MO theory gives a much better and profound explanation, and I'm surely waiting for someone to do that, but that's what I would answer if the kids asked me this.

Oxygen has 6 electrons on its outer shell, which has $\mathrm{2s}$ and $\mathrm{2p}$ orbitals. The arrangement of these electrons is such that $\mathrm{2s}$ sublevel is completed, and $\mathrm{2p}$ sublevel contains 4 electrons. Following Hund's rule,  two electrons occupy one of three $\mathrm{2p}$ orbitals and other two come by one to each of $\mathrm{2p}$ orbitals left.

When water forms, oxygen shares two non-bonded electrons with hydrogen atoms, forming a covalent bond with them.

$\ce{H3O+}$ forms via donor-acceptor bonding, when oxygen supplies its lone pair, which it has two, to form a bond with $\ce{H+}$ ion.

To form $\ce{H3O-}$, you'll need an interaction between water molecule and $\ce{H-}$ ion. Even if we ignore the fact that hydride is a powerful base ($\mathrm pK_\mathrm a$ is around $42$) and will react violently with water, it would be impossible. In this theoretical case neither of the species has an orbital suitable to accept a lone pair from another species. Each has a complete outer shell.

Edit: it really exists though. Wow.

Sources:

Probing the transition state via photoelectron and photodetachment spectroscopy of H3O−
Calculation of the photodetachment spectrum for H3O−


Why using smaller portions are better for extractions Why is using one smaller portion more efficient than using one large portion for extraction. No math answer please thank you

Suppose you have 1 liter of water solution of something you want to extract, 1 liter of solvent, and a partition coefficient of 1 (not very practical, but this is just to make the math simpler). You mix them all together, the compound distributes between water and solvent at 1:1 ratio, then you separate the solvent and end up with 50% of your compound. Can we do better than that?

OK, now let's divide the solvent in two equal portions. The first portion of 500 ml takes $1\over3$ of our compound, then the second portion takes $1\over3$ of the rest, or $2\over9$ of the original amount, thus making the total of ${5\over9}\approx55\%$, which is somewhat better than $50\%$.

Believe it or check it yourself, splitting the solvent into more portions will get us even more of our compound. You may call 5% a marginal gain, but with more favorable partition coefficients, the figures may grow quite a while.

And finally, if we split the original solution as well, and apply the portions in the right order, we may get pretty close to the industrial process.


Why is active mass of a pure solid or liquid always taken as unity? Active mass is defined as the molar concentration ie. number of Gram-moles per litre.

My book then wrote

  Active mass of pure solid/liquid is always 1 .

The book reasoned that

  Molar concentration is directly proportional to density. Since density of solid or liquid always remains constant, the active mass is taken as 1 .

Really? Doesn't the density of the solid reactant decrease when the forward reaction is about to reach the equilibrium??

In a word, I am greatly confused. What is the reason for taking the active mass 1 ? Please help.

Does the activity of a solid or liquid change over the course of a reaction?

The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the equilibrium constant and the concentrations of the species involved.

The equilibrium constant is based on activities, which you can think of as the ratio of the concentration (molar density) at standard state to the concentration in the reaction. Since solid/liquid density is constant at a given temperature, activity is too.  It takes quite a large change of temperature or pressure to make a substantial change in the density of a substance, assuming no phase changes occur.

What about gases, why don't they show the same phenomenon and have fixed activities? (Asked in comments.)

For solids or liquids, as mass decreases the volume also decreases by a proportional amount. For example, $\mathrm{100mL}$ $\ce{H2O}$ $\mathrm{= 100g}$ and half as much, $\mathrm{50mL}$ $\ce{H2O}$ $\mathrm{= 50g}$. Both have molar densities of $\mathrm{55.5 mol/L}$.) Gases keep the volume of their container, so if you remove some via a reaction, the mass of the gas decreases, but its volume remains the same, and its concentration decreases as a result.

Why is active mass given a value of 1 then?  Is this simply a convention?  Shouldn't its value depend on the compound? (Asked in comments.)

No, it's not a convention. Activities, active mass or any other activities, are really ratios. The ratio here is a comparison between the activity under experimental conditions and the activity at standard state under standard conditions.  For gases that is 1 bar of pressure at 25°C, for solutions that is 1 molar at 25°C and 1 bar, and for solids and liquids that is the molar density of the substance in its most stable phase and allotrope at 25°C and 1 bar.  The activities of real substances require some additional work and there are particular compounds that do not behave quite as nicely as expected, but the basic idea of activity being the ratio between experimental state and standard state is sound.

With that in mind, consider the activity of solids and liquids.  Their standard state is their density at standard conditions.  Their experimental state is their density at experimental conditions.  It takes significant pressure and temperature changes to cause substantial changes in the density of a solid or liquid, so their experimental state is approximately equal to their standard state, and the ratio of these two states is therefore quite close to 1.


The negative absorbance values in the baseline measurement In the spectrophotometric analysis(uv-vis-NIR) of ITO Nanoparticles in tetrachloroethane using a double beam instrument baseline was measured with 2 identical cuvette with solvent only. This gives me only negative values for absorbance. Does this mean there is significant difference in transmission between the blank and the reference even if they are the same compound?

Negative absorbance in a double beam instrument has no physical meaning because absorbance, by definition, cannot be negative. It ranges from $0$ to $\infty$.
Instrumentally, all that means is that somehow the sample cuvet is transmitting more light than the reference cuvet. In other words, the measured absorbance of the sample is less than than measured absorbance of the reference cell.  Usually this can happen when the cuvets are not optically matched (i.e., they may look identical physically but their optical transmission characteristics are very slightly different).
Try swapping the reference cuvet with the sample cuvet and see if the readings remain negative again. If you are getting large negative absorbance values with the same solvent, try another cuvet set and if the problem persists, the spectrophotometer must be checked. However, if the negative values are quite small such -0.001 or -0.0002, this is tolerable and you can subtract the baseline from the measured signal in order to correct this systematic error.


How do I synthesize 3-phenyloxirane-2-carbaldehyde from chloroacetonitrile? How do I synthesize 3-phenyloxirane-2-carbaldehyde from chloroacetonitrile?

My proposed route involves:

hydrolysis of the nitrile to form a carboxylic acid, and subsequent esterification
then, a Darzens reaction with benzaldehyde
reduction of the ester function with DIBALH and oxidation to an aldehyde with $\ce{H2CrO4}$.

Is there a better method?

The Darzens condensation can be carried out directly from the alpha-chloronitrile. The exact reaction needed here (benzaldehyde + chloroacetonitrile) is listed 15 times in SciFinder using different base catalysts (hydroxide or an alkoxide) and conditions (sometimes with phase-transfer catalysis). The epoxide will be formed as a mixture of stereoisomers, but often chemical yields are high (reportedly upto 100%).

With the epoxide in place, the nitrile can be reduced with DIBAL to give an intermediate imine, which is converted to the aldehyde after hydrolysis. This is pretty much the same strategy you outlined, but with a couple fewer steps.

As for sources of additional synthesis problems, old UC-Berkeley exams are usually very good. Check Chem 3A and 3B at this archive.


What is the difference between nuclear fission and nuclear fusion? I always mix up the definition of nuclear fission and nuclear fusion. So, what is nuclear fission and nuclear fusion, and an example for each?

Nuclear fission involves the splitting of a heavy nucleus into two lighter fragments plus the release of energy.  As an example here is one route by which $\ce{^235U}$, a fissile material, can split apart.

$$\ce{^{235}_{92}U + ^{1}_{0}n -> ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n + 200 MeV}$$

Note the 3 neutrons generated in the fission process, they are part of the chain reaction; they allow the reaction to propagate at an even faster rate as they go on to react with 3 other $\ce{^{235}_{92}U}$ nuclei.

Nuclear fusion involves the fusing together of nuclei to form a heavier element.  Again, energy is released in the process.  Here is an example of a fusion reaction involving the combination of 2 deuterium nuclei.

$$\ce{^{2}_{1}H + ^{2}_{1}H -> ^{4}_{2}He + ^{1}_{0}n + 17$.$6 MeV}$$


Is chromium (III) green or violet in solution? Is chromium (III) green or violet in solution? In terms of oxidation reactions between chromate and $\ce{Cr}$ (III) it forms a green color in solution, but when I searched online in different sources it seems to say that it is actually purple. Which is the right answer?

Even in water this depends on the hydration process, $\ce{[Cr(H2O)6]Cl3}$ is purple. Also the anhydrous $\ce{CrCl3}$ when you buy it is a purple compound. The chloride-ligands exchange for water ligands like $\ce{[Cr(H2O)5Cl]Cl2 * H2O}$ which is blue-green. And in the end the  $\ce{[Cr(H2O)3Cl3]*(H2O)3}$ is green. Some ions like fluoride are said to influence this process but I have not been able to confirm that yet. For Cr(III) there's this interesting phenomenon, and I think I already asked for that, that this exchange seems to be quite slow. I once prepared a Cr(III) citrate solution and it was blue-green, but after about a week it was deep purple. So this exchange may sometimes take some days.


sodium persulfate microwave heating cause crystallized particles I use Sodium persulfate to etch a copper board to produce a PCB. I mix 1 dl sodium persulfate with 1 liter hot water. When the water get cold, I heat it up by microwave. After the microwave, I see such crystallized particles at the bottom of the jar. Please note that there were not any solid particles in the acid before the microwave. I wonder what are these particles? is it sodium persulfate? if it is so, why it is crystallized when it is heated up?
Solubility of water increases as the temperature of the water increases, right? Then why it causes solid particles in the car?

Btw, if I put these crystallized particles into the same solvent, they don't dissolve, however, if I put them into a new hot water cup, they dissolve.
If it is sodium persulfate and if they are crystallized because of the microwave, can't I reuse them again?
If it is copper particles, isn't it supposed to be in such color?

It looks to me like copper sulphate ($\ce{CuSO4·nH2O}$). If I'm right then you can't use it for etching. For this you need oxidant (persulphate). Try to solve this solid in water. You should get light blue solution. Stick into it a nail (i.e some metal iron source). If it cover itself with a brownish powder like in wiki-link you gave, then it's a copper(+2) salt.


Is the reaction involving a proton transfer a redox reaction? $$
\begin{align}
\ce{H2S (aq) + NH3 (aq) &→ NH4+ (aq) + HS- (aq)}\tag{1}\\
\ce{Pb(s) + 2 FeCl3 (aq) &→ 2FeCl2 (aq) + PbCl2 (aq)}\tag{2}
\end{align}
$$

  Which of the following statements correctly describes both of the equations above?

  A. Both equations represent redox reactions.
  B. Equation 1 represents a redox reaction but equation 2 does not.
  C. Equation 2 represents a redox reaction but equation 1 does not.
  D. Neither equation represents a redox reaction.

The answer key says that the correct answer is C. I picked A. But in equation 1, $\ce{H2S}$ donates a hydrogen to the $\ce{NH3}$ which would mean it's a redox as well. Both equations had donations and acceptances so I really don't understand why equation 1 is not a redox reaction.

Donation of a proton is not an indication of a redox reaction, rather an acid-base reaction (which may or may not be also involving a redox process). The only way to show the reduction/oxidation takes place is to demonstrate that the oxidation numbers do change:

$$
\begin{align}
\ce{\overset{+1}{H}_2\overset{-2}{S} (aq) + \overset{-3}{N}\overset{+1}{H}_3 (aq) &→ \overset{-3}{N}\overset{+1}{H}_4+(aq) + \overset{+1}{H}\overset{-2}{S}^-(aq)}\tag{1}\\
\ce{\overset{0}{Pb}(s) + 2 \overset{+3}{Fe}\overset{-1}{Cl}_3 (aq) &→ 2 \overset{+2}{Fe}\overset{-1}{Cl}_2 (aq) + \overset{+2}{Pb}\overset{-1}{Cl}_2 (aq)}\tag{2}
\end{align}
$$

As you can see, oxidation numbers only change in reaction (2), and are preserved in reaction (1).


What is the purpose of finding log k? I understand that k alone is the equilibrium constant and know how to find this number or when given it I can take the log of this number, but what is the purpose of this number? What does the log of k tell us in relation to the reaction? Why do we find log k?

$\log K$ relates to the thermodynamic changes in entropy and enthalpy of a reaction.
$$\Delta H - T \Delta S = \Delta G = \Delta G^\circ - RT \ln Q$$

At equilibrium $Q = K$ and:

$$\Delta G = 0 = \Delta G^\circ - RT \ln K \Longrightarrow  \Delta G^\circ = RT\ln K$$

$$\Delta G^\circ = \Delta H^\circ -T\Delta S^\circ = RT \ln K$$

and:

$$\ln K = \frac{\Delta H^\circ -T\Delta S^\circ}{RT} = \frac{\Delta H^\circ}{RT} - \frac{\Delta S^\circ}{R}$$


Formalin fallen out of solution Is there a way to bring back into solution a bottle of formalin that has crystallized? The bottle has been stored for several years at room temperature.
Thanks in advance.

The 'crystals' are polyformaldehyde, more commonly called paraformaldehyde.

Heat it up, with some base catalyst.


Extrapolating from a calorimetry lab to find a new delta T with different volumes and grams of substance So, we just finished an AP chem lab where the question was basically:

Given the results of the lab were that $\pu{5 grams}$ of $\ce{CaCl2}$ and $\pu{45ml}$ of water produced a $\pu{13.33 ^\circ C}$ change, find the new temperature change of $\pu{10 grams}$ of $\ce{CaCl2}$ in $\pu{40ml}$ of water.

Our teacher used a ratio strategy where basically $$\frac{\pu{5 grams}}{\pu{10 grams}}=\frac{\pu{13.33 ^\circ C}}{x}$$
Then, since x turns out to be $\pu{26.66 ^\circ C}$;
$$\frac{\pu{26.66 ^\circ C}}{45ml}=\frac{x}{50ml}$$
solving for $x$, the final temperature change is $\pu{29.6 ^\circ C}$.
My question is basically whether this is true, because I find the denominator in the first fraction of the second ratio sketchy - its saying $\pu{10 grams}$ for 35 ml of water, but, the $\pu{26.66 ^\circ C}$ is derived from finding $\pu{10 grams}$ with same amount of water. Thus, shouldn't it be 50, since in the first ratio we only changed amount of grams of solution, so water is still 40 ml, and now $\ce{CaCl2}$ is  $\pu{10 grams}$. $\pu{40ml + 10 grams = 50 grams}$.  Thus, shouldn't the first fraction be $$\frac{\pu{26.66 ^\circ C}}{50ml}$$
The final answer happens to be $\pu{26.66 ^\circ C}$ (which is pretty and convenient), but  I think that's just because we chose 5 grams to 10 grams which is a nice number to double and the volume decreases by 5, so everything cancels out nicely.
Kinda of a big segue and separate question but related:
What's wrong with the following logic to solve the problem?
$$\pu{50 g}\times \pu{4.184 J g-1 ^\circ C-1} \times \pu{13.33 ^\circ C}=\pu{2789 J}$$
Basically, using q=mcat (m is mass of water, 45, plus mass of solution, 5) to find q of 5 grams $\ce{CaCl2}$ and 45 ml water, and then, assuming if we double the mass we double the q, work backwards? (If my logic is wrong, I think it occurs in this assumption)
thus, since we need 10 grams and 40 ml in the end, the mass is still 50 grams, but the 2789 is multiplied by 2 (since mass is doubled) and 13.33C becomes an X.
$$\pu{50 g} \times \pu{4.184 J g-1 C-1} \times x = \pu{5578 J}\\
x = \pu{26.66 ^\circ C}$$
Basically, I think the answer is $\pu{26.66 ^\circ C}$ and my teacher thinks it's $\pu{29.66 ^\circ C}$.
I'm honestly very confused about q... Google does not seem to help answer this question, if amount of water in a solution/reaction (ie water and $\ce{CaCl2}$) changes, but mass of reactant stays the same, does q change?

The ratio strategy will definitely work, but I think going back to $q=m\cdot C\cdot \Delta T$ is easier to understand.

You've already figured out a bunch of these constants: $m = 50$, $C = 4.184$ (we assume that the heat capacity of the water doesn't change too much when adding the solid), and $\Delta T = 13.33$.

$$\begin{align}
q & =m\cdot C\cdot \Delta T \\
& = 50\cdot4.184\cdot13.33 \\
& = \ce{2789 J}
\end{align}$$

The heat produced by mixing $\ce{5 g}$ of $\ce{CaCl_2}$ is $\ce{2789 J}$, therefore,
the heat of reaction ($\ce{\Delta H_{rxn}}$) in $\ce{J/g}$ is $\ce{557.7 J/g}$.

When we mix $\ce{10 g}$ of $\ce{CaCl_2}$, the total heat produced will be $\ce{10 g \cdot 557.7 J/g = 5577 J}$.

Now we need to determine the heat change had we mixed in $\ce{10 g}$ of $\ce{CaCl_2}$. We can rearrange the specific heat formula to get $\ce{\Delta T = \frac{q}{m\cdot C}}$. We know $q = 5577$, $m=50$, and $C=4.184$.

$$\begin{align}
\Delta T &= \frac{q}{m\cdot C} \\ & \\
&=\frac{5577}{50\cdot4.184} \\ & \\
&= 26.66\, ^\circ \ce{C}
\end{align}$$

The reason the ratio works is the same logic as the gas laws:

$P_1\ V_1 = C$, $P_2\ V_2 = C$, therefore $P_1\ V_1 = P_2\ V_2$.

$q_1 / T_1 = m\cdot C$ and $q_2 / T_2 = m\cdot C$.

Therefore, $q_1 / T_1 = q_2 / T_2$ by the transitive property of equality


What is wrong with storing deionized water in glass containers? I have heard that in general storing deionized water in glass bottles is ill advised and that it is better to use plastic. Additionally this is especially important for water used in ICP. I have noticed that ICP standards (which use 1-2% nitric acid) all come in polyethylene bottles as well though the nitric acid used for rinsing (10%) is stored in a glass reservoir. Is there some reason that glass is unsuitable for ICP applications or is it another just economical to use plastic?

It is very difficult to make a container that does not introduce impurities to a liquid you place into it. As mentioned in the comments to the question, storing in a glass bottle will result in ions in your formerly pure water.

This is thought to proceed by a mechanism called ion exchange (source):

$$\ce{ Glass.Na+ + H3O+ -> Glass.H3O+ + Na+}$$

The Wikipedia article on dealkalisation states:

  Routine tests for surface dealkalization in the glass container industry all generally aim to evaluate the amount of alkali extracted from the glass when it is rinsed with or exposed to purified water. For example, dealkalization can be quickly checked by introducing a small volume of distilled water to a freshly made bottle and rolling the bottle gently to pass the water completely over its inside surface. The pH of the rinse water is then measured; untreated containers will tend to yield a slightly alkaline pH in the 8-9 range due to extracted alkali, while dealkalized containers tend to yield a pH that remains approximately neutral.

The article also mentions a technique called secondary ion mass spectrometry that can probe the composition of solid glass up to 1-2 nm depth below the surface. Leaching has been studied quantitatively in the context of storing nuclear waste in the glass state, see for example this PhD thesis (abstract in English, main text in French).

So storing in a glass bottle is not ideal because of contamination with ions. If you store it in a plastic bottle, on the other hand, you will find organic substance in your formerly pure water. Depending what kind of analysis you want to use your water for, different types of impurities will be more relevant (i.e. give spurious signals in the analysis). If you are analyzing samples with a GC/MS, glass bottles would be better (well, you would use an organic solvent, not water). If you are analyzing samples with atomic emission or absorption spectroscopy, plastic bottles would be better.

A first pointer would be the materials used in sample handling by the instrument itself.


Intermolecular interaction between aniline and dichloromethane Short background

As it is known there are interactions between:

Charges
Charge and dipole
Hydrogen bonding
van der Waals (VDW) forces

From stronger (1) to weaker (4).

VDW forces are divided into dipole-dipole, dipole-induced dipole, induced-induced dipole; also from stronger to weaker.

The problem

Reading about solutions I found interesting the intermolecular forces involved on aniline and dichloromethane mixing. Below both structrures are shown:

Aniline
 Dichloromethane

Question

What is the most important intermolecular force between those molecules and why? It might be interesting to see an AIM analysis of aniline, but I have no idea how to do that. Also, I'm not sure about what experiment could give valuable information on this case.

My attempt

Those previous interactions aren't mutually exclusive. $\ce{-NH2}$ might be a negative density region of the molecule (inductive attraction), but it is also donor by resonance, so the ring has partially a negative charge density. In consecuence, it is not easy to see the dipole. Dichloromethane, on the contrary, is an easy-to-see dipole.

It might result that London dispersion is the most relevant interaction here, but I can't very well justify it.

According to this NIST publication (caution: link leads directly to a biggish PDF file) the dipole moment in the gas phase of dichloromethane and aniline are quite similar (1.6 and 1.5 D respectively).  These are respectable dipole moments, and neither molecule is very big, so offhand I would say an argument can be made that the dipole-dipole interaction is most important, particularly because it's orienting.

Note by the way that rating dispersion forces as the weakest should be done cautiously. It works for small molecules, but since dispersion forces scale with the number of electrons in a molecule and no other interaction does, once you get above a fairly modest size of molecule dispersion forces (along with excluded volume forces) dominate.  That's why eicosane is a solid at room temperature while water is a liquid.


How is the electron "trapped in the crystal" in irradiated NaCl? Pennsylvania State University provides irradiated $\ce{NaCl}$ (180,000 rads of gamma radiation) to teachers for a very entertaining demonstration. The salt is orange after irradiation. When it is placed on a hot plate, you see a flash of visible light and the salt returns to it normal color. The explanation provided with the salt is that "The heat allows the electrons that were trapped in the crystal structure (when the energy was deposited) to return to their original position." This language evokes a picture of a particle electron in a crystal jail for me. The explanation goes on to say that the energy is released as a photon of visible light. (I understand this part of the explanation.) So, my question is, how are electrons trapped in a crystal structure? What keeps them in an excited state?

What you're describing is thermoluminescence, which is used as a paleochronology method (aka Trapped Electron (TE) dating, afaik). The wikipedia article has a good synopsis. Simply, electrons get bound to potential wells generated in the radiation-defected structure. When you heat the crystal up you allow it to anneal and return to a more regular structure. The defected crystal is stuck in an energetic local minimum.

You can also do this trick with X-rays. Calcite (or was it aragonite?) exposed to X-rays will turn light blue for a while.


Why are amino acids least soluble at isoelectric point? At isoelectric point amino acids exist as zwitterions. As a molecule it is electrically neutral, but still has a positive and a negative charge. Then why are amino acids still least soluble at isoelectric point?
The positive and the negative charges of the amino acid can interact with the negative and the positive charges of water and can dissolve, right? Where am I going wrong?

In order for molecules to precipitate out of solution, they need to aggregate together. Amino acids that have zero net charge can aggregate together much more easily than those that are charged.
Molecules that have net charge need counterions to aggregate with them to offset the charges or the electrostatic repulsion will be too great.
The concept of isoelectric point is more often applied to polypeptides and proteins than to individual amino acids. In those much larger molecules, the tendency to aggregate when there is no net charge can be quite high. This is especially true if there are any hydrophobic regions that can favorably associate with each other and reduce the interaction with water.
It is important to note, as mentioned in the comments, that this does not mean that all amino acids and proteins are insoluble at their isoelectric point, simply that they tend to be less soluble at that pH than at pH values where they carry net charge.


What type of reaction is this? (Modified organic nucleoside synthesis) What type of reaction is shown in step F of this patent image?
I tried looking it up in SciFinder, but I didn't get any results, and I have not seen this before.

Thank you!
Edit 1: I'm also wondering if there is a specific name for this reaction, so that I can look it up.
Edit 2 Yes I'm sorry I forgot the reagents. The patent number is WO2014078463A1 (Bennett et al.) and I included a screenshot below of the relevant description.

My thoughts are along the lines of @Andrew and @Waylander. Step F doesn't tell us much, @orthocresol! I prefer a cyclization, 1 $\rightarrow$ 2a, followed by hydrolysis of ester 2a to alcohol 2b. Alternatively, N-O bond homolysis with light would probably involve a cage mechanism. .