Manufacture an artifact with compression Room-temperature superconductivity has finally been demonstrated, as reported by Quanta (and others), with the caveat that it requires compression of the substrate at nearly 2 Mbar.  Obviously, this is an impractically high pressure, but if a different material were found that only required a much lower pressure, would it be feasible to manufacture wires under compression?  If so, what is the current state of the art?  I have not been able to find any examples of such a product.  Most apropos to chemistry, what is a practical upper-bound on tension/compression we can expect from currently known materials?
A trivial example would be to put a spring in a box, which only requires compressing the spring and sliding it into a box.  For the wire, one could perhaps imagine a strange "cylindrical press" (not sure if there is such a device) and a jacketing material with extremely high tensile strength into which the wire is thus threaded (or, more likely, the jacket is somehow rolled onto it).
A more sophisticated system would be a compressive jacket which is applied in a relaxed state, possibly at high temperature, and then cools/shrinks into a high-tension state which applies the necessary compression.

Jon Custer helpfully noted (in a comment) two common production processes that result in compression of the final product.  I did some research to discover the typical pressures involved.
Tempered Glass
Glass may be tempered by cooling the outer layer quickly, while letting the inner layer cool slowly.  As the inner layer cools, it shrinks; but because the surface layer is already cool, it cannot continue shrinking, so it experiences compression instead.  One can think of glass tempering as producing an end result in which the core glass is literally pulling the surface inward.  Very cool!
According to J.H. Nielsen, tempered glass retains compressive stress on the order of 100 MPa, if I am reading Figure 12 correctly.  This is equal to 1 Kbar.  That's 3 orders of magnitude short of the Mbar pressures created by the diamond anvil, not to mention the difficulties entailed by embedding wires in molten glass, and yet, this is at least a plausible way forward.
Shot Peening
The surface of a cold metal can be further compressed simply by banging on it really hard.  Most commonly this is accomplished by firing small metal shot at the surface at high velocity and volume.  This induces plastic deformations which have a higher density than the underlying metal, and thus a compressive strain within the surface which helps strengthen various industrial products like springs, crankshafts, and jet engine blades.
According to this source, shot peening can induce compressive strain on the order of 800-1000 MPa, according to Figure 1. However, it should be noted that the compressive layer is only about 100 μm deep.  So, this gets us up to 10 Kbar or so which is even better than glass tempering, and doesn't require high temperatures.  Very interesting!


Why does insufficient data lead to random errors? I've read up on the causes of random error and noticed insufficient data to be one of the causes of random error. However,  I'm confused as to why insufficient data could possibly lead to random errors. Could someone provide some context for why this might be?

More datapoints (if everything is done correctly) lead to more trustworthy results. With less datapoints, the possible (statistical) error grows, until at one point, when you only have as many datapoints as you have parameters, it looks like you have no error at all. (That is of course stupid, you should always have derived the error range of the individual data points.)

Now what is "insufficient"? Is that when the statistical error grows too large for your feeling? Or when you only have two points to draw a straight line through? Or when you only have two points to draw a parabola? That's your prerogative.

(note that I have also above used the word "error" in the same sloppy way your books seem to do. The "errors" are in your data, and they have a systematic component (hopefully small) and a distribution. The derived values (e.g. the slope of a linear regression) have a confidence interval.)


What is wrong in this argument that dG must always be zero? Under constant temperature and pressure, the change of the Gibbs free energy can be written as

$$\mathrm dG_\textrm{sys} = \mathrm dH_\mathrm{sys} - T\,\mathrm dS_\mathrm{sys}$$

And in the textbook that I have, when deriving this equation, they use the equation

$$\mathrm dS_\mathrm{univ} = \mathrm dS_\mathrm{sys} - \mathrm dH_\mathrm{sys}/T$$

I can understand to this step but the question occurs. Can't we write $\mathrm dS_\mathrm{sys}$ as $\mathrm dH_\mathrm{sys}/T$ also? If this T and the T that is used in the equation

$$\mathrm dS_\mathrm{univ} = \mathrm dS_\mathrm{sys} - \mathrm dH_\mathrm{sys}/T$$

are the same, than shouldn't $\mathrm dG$ always be zero? At which point am I wrong?

Your textbook's derivation is done under the assumption of constant $T$, which means $T_{sys} = T_{surr} =T$.  However, this does not mean $dG_{sys}$ is always zero.  Let's start with the following:

$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T_{sys}}+\frac{\text{đ}q_{rev, surr}}{T_{surr}}$$

Since heat flow always affects the surroundings reversibly (the surroundings are in the limit of being infinitely large, and thus heat flow affects them only infinitesimally), the reversible heat flow into the surrondings equals the actual heat flow, which is the negative of the actual heat flow into the system.  And since $T_{sys} = T_{surr} =T$, we have:

$$\frac{\text{đ}q_{rev, surr}}{T}=-\frac{\text{đ}q_{sys}}{T}$$

Which gives:

$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T}+\frac{\text{đ}q_{rev, surr}}{T}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{\text{đ}q_{sys}}{T}$$

Under the restrictions of constant $p$, and no non-$pV$ work,

$$\text{đ}q_{sys}=dH_{sys}$$

Hence:

$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T}+\frac{\text{đ}q_{rev, surr}}{T}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{\text{đ}q_{sys}}{T}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{dH_{sys}}{T} = dS_{sys} - \frac{dH_{sys}}{T},$$

which is what your textbook provides.

Note, however, that

$$\frac{dH_{sys}}{T} = \frac{\text{đ}q_{sys}}{T} \ne \frac{\text{đ}q_{rev,sys}}{T}=dS_{sys},$$
unless the process is reversible, i.e., non-spontaneous, in which case we could replace the "$\ne$" with "=", and $dG_{sys}$ would equal $0$.

This connects directly to the great beauty of using $\Delta G_{sys}$ at constant $T \text{ and } p$, when there is no non-$pV$ work.  Because, if the system is at constant $T$*,
$$\Delta G_{sys} = \Delta H_{sys} - T \Delta S_{sys}$$

[*$G=H-TS => dG = dH - d(TS) = dH -TdS -SdT$; at const $T, dG= dH -TdS $; note that a constant-pressure restriction is not required to write this.]

Where we need the additional restrictions of constant $p$, and no non-$pV$ work, is for the following:

$$\Delta H_{sys} = -T \Delta S_{surr}$$

This in turn gives:

$$\Delta G_{sys} = -T \Delta S_{surr} - T \Delta S_{sys} = -T \Delta S_{univ}$$


Why do experiments not verify the law of mass action? I read the answer to Rate Law vs Law of Mass Action and it's written in the answer:

    In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not
    be equal to stoichiometric factors. These are generally responsible
    for determining the order of reaction.

  Both the laws are correct because both have a wide area of
  application. Rate law is although quite experimental in nature and is
  generally considered to be more appropriate.

Why do reactions not obey the law of mass action? And if that's the case, then how does it even have any applications? Shouldn't all problems be solved using the rate law only?

The rate of mass action is true: the rate of reactions is proportional to the activity of its reactants.

On the other hand, from a macroscopic point of view, it is also true that reaction rates may not correspond to apparent stoichiometric proportions.

To understand this disparity, we need to refer to the concept of mechanisms.

The rate of mass action applies to elementary reactions, that is, to reactions that happen the way we write them at the microscopic level.

If the molecular, physical reaction consists on two molecules colliding and reacting, and we write it as

$$\ce{A + B -> C}$$

The rate of that reaction will, per the law of mass action, be

$$r = k · a_A · a_B$$

However, most macroscopic chemical processes aren't that simple, and in fact are the product of multiple steps; for instance, in the previous example, perhaps A undergoes a transformation into an intermediate which then goes on to collide with B to form C:

$$\ce{A -> D}$$

$$\ce{D + B -> C}$$

Macroscopically, we still have $\ce{A + B -> C}$, but the kinetics of the reaction will be very different.

For instance, imagine that the second step ($\ce{D + B -> C}$) is much faster than the first, so that any amount of D created reacts with B almost immediately to produce C. Then the first reaction will act as a bottleneck, and the overall reaction rate will be roughly the rate of the first step:

$$r = k_1 · a_A$$

If we do not know the underlying mechanism, this will be an apparent contradiction between the law of mass action (from which we could naively conclude that $r = k · a_A · a_B$) and the actual rate law ($r = k_1 · a_A$).

Much of the work on chemical kinetics consists in proposing mechanisms, testing these proposals (for instance, looking for experimental evidence that D exists as an intermediate) and deriving rate laws that are consistent with the law of mass action and experimental observation (which applies to elementary steps).


How much hydrogen or oxygen will be produced in the electrolysis of water? Is there a way to exactly calculate the quantity of hydrogen or oxygen in a water electrolysis? I am thinking of a plain and simple electrolysis like an anode and cathode in salted water separating oxygen and hydrogen.
I am looking for the relation between the amounts of hydrogen $n(\ce{H2})$ or oxygen $n(\ce{O2})$ and the amounts of water and salt, current, voltage and time of electrolysis.

I'm not sure it's appropriate to explain the relationship between all the variables above, here, as there are a lot of concepts that would need to be explored. However, fundamentally (assuming there are no other reactions apart from the electrolysis) the charge passed through the cell is proportional to the amount of products formed, described by Faraday's laws of electrolysis. Here is a useful rearrangement:
$$n = \frac{Q}{Fz}$$
Where $Q$ is the charge passed, $F$ is the Faraday constant and $z$ is the number of electrons transferred per redox reaction. For example, $z = 2$ for hydrogen because it takes two electrons to convert $\ce{H+}$ to $\ce{H2}$ as $\ce{2H+ + 2e- <=> H2}$. Therefore, for every coulomb of charge, we get: $$\begin{align}
  n &= \frac{1\ \mathrm{C}}{(96.485\ \mathrm{kC/mol})(2)}\\
    &= 5.18\ \mathrm{µmol}
\end{align}$$ of hydrogen gas produced. So if we can measure the charge (or current integrated over time), we can calculate the amount of product produced. This is true no matter what voltage is applied or what the electrolyte content of the solution is, however if more than one reaction is occurring at an electrode, you won't be able to separate the contributions from each reaction with only the charge as information. There will also be a minor contribution from charging the double layer capacitance when the potential is first applied.

Where things become less straightforward is understanding how changing the potential or the solution resistance affects the amount of charge that flows, but the charge itself is directly related to how much product is being produced.


Preparation of sulfuric acid through Contact process In the preparation of $\ce{H2SO4}$ through the Contact process, $\ce{H2SO4}$ is added to hot sulfur trioxide to form oleum. This oleum is then diluted by adding water to form $\ce{H2SO4}$.

$$\ce{H2SO4(l) + SO3(g) → H2S2O7(l)}$$

$$\ce{H2S2O7(l) + H2O(l) → 2 H2SO4(l)}$$

In a preparatory reaction, different reagents are used to obtain the desired product but here the product is itself a reactant and is used to make the product. Indirectly, $\ce{H2SO4}$ is being used to make $\ce{H2SO4}$.

Would this process be called as a preparation process?

You can call it whatever you like. Personally I would say you are preparing sulphuric acid because you end up with more than you started with - 1 mole is used to form 2 moles.


Diamagnetic vs Paramagnetic As far as I know, to know whether a substance is paramagnetic or diamagnetic, we check if that particular species has got unpaired electron or not.

But my textbook lists-

Examples of Diamagnetic Materials - Copper,Lead,Sillicon,etc

Examples of Paramagnetic Materials - Aluminium,Sodium,Calcium etc

The valence shell configuration of Copper is 3d10 4s1 . So, shouldn't it be Paramagnetic?

The valence shell configuration of Calcium is 4s2 . So, shouldn't it be Diamagnetic?

You're right that Cu0 has one unpaired electron in the 4s orbital. If you were to observe a single Cu0 atom in the gas phase, you would observe some paramagnetic behavior.

However, in metallic copper, the valence orbitals are engaged in metal-to-metal bonding. The 3d and 4s orbitals interact in constructive overlap with neighboring copper atoms to produce bonding molecular orbitals.

When two 4s orbitals (each containing 1 unpaired electron) combine to form a sigma bonding MO, the electrons in that bonding MO become paired, per the image below.

Note: this MO diagram depicts the bonding in H2, but I'm just using it to illustrate the principle of partially filled atomic orbitals coming together to form a filled bonding MO

The process of deriving the MO diagram for a metallic solid is obviously more complicated than this, but the answer to your question is that a single Cu0 atom is paramagnetic, but Cu2 would be diamagnetic. By extension, metallic copper is effectively Cu$_\infty$ and also diamagnetic.

As you can tell, the process for determining whether an atom or ion is para/diamagnetic is a very different question than whether the bulk substance (metal) is para/diamagnetic. You can't do the latter by electron counting alone; you need to know more about bonding in the substance. As another example of this, you mention that lead is diamagnetic. Lead is another example of a paramagnetic atom (6p2; the 6p orbital contains two unpaired electrons) - but lead metal is diamagnetic because those electrons become paired in the bonding MOs.


Is it possible that atoms with 120 protons are possible, but that atoms with 119 protons aren't possible? We currently know that there are atoms with atomic number up to 118 are possible.

Is it possible that atoms with 120 protons are possible, but that atoms with 119 protons aren't possible? Or are there theoretical arguments or maybe heuristic arguments why this can't happen?

I suspect the answer to be yes, because technetium and promethium have no stable forms while the elements surrounding it in the periodic table do have stable forms. A similiar thing might happen here. The more general question: Does there exist $k > l > 118$ such that atoms with $k$ protons are possible, but that atoms with $l$ protons aren't possible?

It is not a duplicate of The last element's atomic number. I wonder whether there could be 'gaps' in the periodic table, not what the last element's atomic number is.

According to the Wikipedia page element 119 synthesis has been attempted indicating physicists think it is possible.  There is some speculation as to whether our current technology could even detect such short lived elements such as elements 119 or 120.  There is also speculation that the technology today may not be able to produce element 120.  It may be possible for such gaps (who knows maybe element 562 is stable and element 561 cannot form?) but at the current time no gaps are known with certainty.

One of the bigger problems for higher atomic number atoms is that as the nucleus becomes more charged the electrons orbiting the nucleus must go faster and faster to stay in orbit.  However nothing can exceed the speed of light and thus if an atom were to have electrons near the speed of light (say 99.99999% of c) at rest. The mere revolving of the earth around the sun would require the electron to move faster than light which presents and interesting case/thought experiment. Because of this it is reasonable to deduce that there is a penultimate element atomic number where anything larger would required electrons to exceed the speed of light though I cannot say what the magic number would be.

I will say though your question has a quite lot of merit as the Oddo-Harkins rule holds that the abundance of elements with odd atomic numbers in the universe is far less than that of elements with even atomic numbers except for hydrogen of course.  Further radioactive nuclei with even atomic numbers tend to be more stable and have a longer half life than those with odd atomic numbers.


Balancing using oxidation number method Balance the following equation using oxidation number method:
  $\ce{PbS + H2O2 -> PbSO4 + H2O}$

Attempt:

I have written down the oxidation states of all atoms and we can observe reduction of peroxide and oxidation of sulphur.

$\ce{Pb^{II}S^{II-} + H^{I}_2O^{-I}2 ->Pb^{II}S^{VI}O^{-II}_4 + H^{I}2O^{-II}  }$

The specific problem I am facing with this question is that oxygen is going from $-1$ to $-2$ in both the reaction products. How do I handle that?

For sulfur we can start by identifying the number of electrons lost in the oxidation half reaction:

$$\ce{PbS^{2-} ->PbS^{6+}O4 +8e-}$$

Now we can balance the oxygens using water, which also leaves us with an excess of protons:

$$\ce{PbS + 4H2O -> PbSO4 + 8e- +8H+}\tag1\label{Oxidation}$$

Now consider reduction half reaction:

$$\ce{H2O^{1-}2 +2e- -> 2H2O^{2-} }$$

It will obviously require a proton source, so we balance with protons:

$$\ce{H2O2 + 2 H+ +2e- -> 2H2O }$$

Multiply by a factor of 4 to match the number of electrons needed by the oxidation reaction:

$$\ce{4H2O2 + 8 H+ +8e- -> 8H2O }\tag2\label{Reduction}$$

Combine the half reactions $(1)$ and $(2)$:

$$\ce{PbS + 4H2O2 -> PbSO4 + 4H2O}$$


What is the difference between carboxylate ions and groups? I would like to know the difference between carboxylate ions carboxyl groups? An example is is $\ce{COOH}$ and $\ce{COO}$.

Carboxyl

The carboxyl group $\ce{-COOH}$ is an organic functional group consisting of a carbon atom double bonded to an oxygen atom and single bonded to a hydroxyl group.

Compounds with a carboxyl group are called carboxylic acids (e.g. acetic acid, $\ce{CH2COOH}$).

Carboxylate

A carboxylate $\ce{-COO^-}$ is a salt or ester of a carboxylic acid. For example, acetic acid reacts with sodium bicarbonate to form sodium acetate.

Compounds with this group have the carboxy- prefix or -ate suffix (e.g. sodium acetate $\ce{CH3COONa}$). When these salts are ionised (e.g. dissolved in water), the anion with the carboxylate group is called a carboxylate ion.

The carboxylate ion is the negative ion (anion) formed as the conjugate base when a carboxylic acid loses a proton.

$$\ce{R-COOH -> R-COO^- + H+}$$

Being its conjugate base, sodium acetate mixed with acetic acid makes a useful buffer solution for many biochemical reactions where a 'mildly acidic' pH of 4-6 is desired.


How to predict SN2 reaction when comparing nucleophilicty and leaving group ability In a lab, a common experiment is to perform SN2 reactions involving a primary alkyl bromide and sodium iodide in acetone (polar aprotic solvent).

The result is that the iodide ion is a good nucleophile, so the iodide will displace the bromide from the primary alkyl bromide, and sodium bromide will precipitate from the solution.

However, without the presence of the sodium cations, would the reaction still take place?

I'm somewhat confused, because iodide is both a better leaving group and a better nucleophile than bromide, so what is the determining factor in the reaction?

Is the strength of the nucleophile more important or the ability of the leaving group be displaced?

This is very famous "Finkelstein Reaction" that involved the exchange of one halogen to another through SN2 mechanism. The driving force for the reaction is the solubility of salts (NaI, NaBr, KF) under equilibrium.
There are different factor that effect the equilibrium position of a reaction, firstly, nucleophilicity of anion, presence of  good leaving group, and stabilization of anion in a given solvent.

first part of your question, I don't think with the absence of sodium cation the reaction would take place, because the driving force is the precipitation of salts of NaBr and NaCl in acetone.

In the second part of your question, you are right iodide is better leaving group compared to Bromide because iodide is weaker base compared to bromoide. Apparently, iodide can't replace bromine but here the reaction is driven toward products by "mass action" due to the precipitation of the insoluble salt as mentioned above.


Chemicals to clean natural stones I picked up some stones from a hiking trip. What can I use to thoroughly clean the mud and other hard deposits on the surface of the stones? I have only washed them with plain water.

Removing all of the debris from the stones without damaging them can be difficult, the following is what I've done in the past to clean geodes with debris and deposits.  Your mileage may vary, especially since we don't know exactly what is on your stones.

Soak the stones in soapy water for a day or so, then scrub them thoroughly with a stiff brush to clean dirt and most debris from them.  Be cautious if the stone has delicate structure that you're trying to preserve.  If some debris remains that appears to be washable, you can try scrubbing with a toothbrush and toothpaste to get rid of it.
If the stone is acid safe, soak it in vinegar for a couple of days to dissolve carbonate deposits.  You can check to see if it will react with vinegar by adding a few drops of vinegar to the stone's surface and waiting.  If it hasn't started bubbling within 10 minutes or so, it should be safe.
If the stone has rust deposits and is acid safe, you may be able to remove them with something like CLR Rust Stain remover.  If you need to go this route, skip the vinegar step, as this will also clean carbonate deposits.
Finally, it's possible that your crystal has something like granite attached.  If that is the case, nothing practical, short of cutting it off, is likely to clean the crystal.

Once you've got the crystal to the cleanliness you're happy with, you can polish it with powdered polish and a cloth, given time.

If instead of nice, angular crystals you're just trying to produce something like a smooth, rounded stone, after cleaning with water you can use a series of sandpapers (start at 60-80 grit and work up to ~400 grit) and polish to gradually smooth, then polish the stone by hand.  This will clear away any deposits, but it will also change the surface of the stone itself.  That might be acceptable to you if you've got something like a quartz river rock, and it will make a very nice end product.  Take care to keep the stone wet while working, you don't want to breathe in rock dust if you don't have to.


What does the label of stability: 0 mean on an MSDS sheet? I was reading an MSDS sheet on bleach. There is a section named NFPA. Under that is are the words,

  Health Hazard: 3, Flammability:0, Stability:0

Does that mean that Bleach is highly unstable?

According to the Wikipedia site for the NFPA 704   identification system, a Stability rating of 0 means:

  Normally stable, even under fire exposure conditions, and is not
  reactive with water.

The low concentration of the bleach, in other words it's mostly water, is the reason that it fits the "0" category.

Note that the bleaching component of the bleach solution is sodium hypochlorite (see your MSDS for bleach).  Pure solid sodium hypochlorite has a NFPA reactivity rating of "1".


ChemDraw: Automatic Caption with Number for Molecules When working on some reaction schemes in ChemDraw, I sometimes forget to add text/captions with numbering and align it with my molecules before placing all the molecules in a reaction scheme and aligning that.

Now that I already have a scheme, adding the numbers to the molecules will either mean I have to place the numbers by hand (which would mean they are not exactly centered) or aligning the captions to the molecule. The latter option would work, but would also mean I have to redo a lot of the work aligning my scheme.

The question is then: Is there a way to automatically place a number under a molecule, so I do not have to worry about realigning everything?

There is no fully automatic method for creating labels. However, you can manually generate property labels, which are automatically placed and centred without disturbing any existing alignment, as follows.

Select the entire molecule, right-click the selected structure, then click Analysis, and then click one of the available options (Name, Chemical Formula, Exact Mass, Molecular Weight, m/z, or Elemental Analysis):

Use the Text tool (on the tool bar) to modify the text of the label (after your edit, you will receive the warning “This change to the text has caused auto-updating to be disabled.”):

You can use the Text menu, the Style toolbar, or shortcut keys (e.g. CTRL+B) to change the text format (structure labels are usually written in bold). Repeat this procedure for all relevant structures:


Mechanism for conversion of imide to ester with NaH What is the mechanism of this reaction? My proposed mechanism is shown below, would it work?

NaH is a strong base but a bad nucleophile. This is one possible mechanism.


Separating citric acid from sugar I'm trying to get as pure citric acid as possible from a mandarin. It's not because I desperately need citric acid, but because I want to achieve this.

I started by grating the mandarine and filtering the juice. After filtering overnight, I was left with slightly yellow liquid. I heated it while avoiding boiling and I reduced the volume to 50%. When volume was reduced, the solution started to become saturated and crystals appeared. At this point, I stopped.

Now, I'm looking for a way to remove sugar from the solution, as this is the most annoying part. It's sticky.

When crystals appeared while I was boiling the solution, I came to think, that perhaps that could allow to separate sugar. Citric acid has solubility in water 1477.6 g/L (20 °C), while sugar has 2000 g/L (25 °C). Wikipedia unfortunately doesn't provide values for the same temperature.

From the above it seems that I can have more sugar dissolved than citric acid. I think that means that citric acid will crystallise first. Does it work like this? Are the crystals I've produced the citric acid crystals?

Is there another way to separate the compounds? Does something react with sugar to produce precipitate that can be filtered?

Citric acid is relatively soluble in ethanol (62g/100g) whereas sucrose is relatively insoluble in ethanol (1g/170ml). There are many different kinds of sugars and sucrose is probably not the primary one in this extract; branched sugars... oligosaccharides (which have generally higher solubilities than sucrose) are likely, and I'm sure there are a few other organic acids involved with a crude, fruit extract, but the different solubility characteristics should be relatively helpful for extracting citric acid from sugars and some other organic acids.

So, I would try extracting the citric acid from the dried fruit or dried water extraction with 99%+ ethanol. I would then allow the alcohol to evaporate to help concentrate the acid.

Dilute citric acid (from a concentrated extract) can be precipitated with calcium hydroxide as practically insoluble calcium citrate, which is then reacted with sulfuric acid to form (food grade) citric acid and calcium sulfate (gypsum) as a recoverable by-product.

To really purify the stuff, an ion exchange resin or some other chromatographic method would be the last step.


Coagulation with Ferric Chloride I'm having a hard time understanding the use of coagulation wastewater treatment. I have read in literature that ferric chloride is used as a coagulant to remove COD, TSS and ammonia from landfill leakage. Their dose was $\pu{10 g/L}$ at a $\mathrm{pH}$ of 8. Since there is so much ferric chloride, it will neutralize all of the alkalinity in the leakage and then start decreasing the $\mathrm{pH}$. To maintain the current $\mathrm{pH}$ of 8 lime, $\ce{Ca(OH)2}$, is added so that the ferric chloride will neutralize that instead.

$$\ce{2 FeCl3 + 3 Ca(OH)2 -> 2 Fe(OH)3 + 3 CaCl2}$$

What I don't understand is what exactly is going on in this reaction? What do the products do in the leakage? What will happen to the reaction at higher or lower $\mathrm{pH}$?

According to this Wikipedia article:

  Iron(III) chloride is used in sewage treatment and drinking water production.$\mathrm{^{[1]}}$ In this application, $\ce{FeCl3}$ in slightly basic water reacts with the hydroxide ion to form a floc of iron(III) hydroxide, or more precisely formulated as $\ce{FeO(OH)−}$, that can remove suspended materials:

  $$\ce{[Fe(H2O)6]^3+ + 4 HO− → [Fe(H2O)2(HO)4]− + 4 H2O → [Fe(H2O)O(HO)2]− + 6 H2O}$$

1) Water Treatment Chemicals, Akzo Nobel Base Chemicals. 2007.


Why is PLGA so expensive in comparison to PLA? A Google search has lead me to believe that PLGA goes for anywhere between 8 USD and 100 USD per gram. Meanwhile, for PLA (3d printer grade), prices float around 50-100 USD per kilogram.
I understand there are differences in the two, but why is PLGA so much more expensive? Is it solely because of PLGA's application in medical-grade uses, while PLA has widespread consumer-grade application?

Poly(lactic-co-glycolic acid) or PLGA is a biodegradable copolymer of lactic acid ($\ce{CH3CH(OH)CO2H}$) and glycolic acid ($\ce{HOCH2CO2H}$). Poly(lactic acid) or PLA is, on the other hand, a polymer of lactic acid, which is also biodegradable. Since both polymers are linked together by ester linkages of their monomers, the resulting polymers can be categorized as linear, aliphatic polyesters:

PLGA is one of the most successfully biodegradable polymers, and hence it attracts considerable attention in the development of drug delivery systems due to its desirable properties and to its Food and Drug Administration (FDA) and European Medicine Agency (EMA) approval for parenteral administration. PLGA encapsulated drug delivery enjoys drug protection from degradation. In addition, PLGA also provides prolonged controlled release of encapsulated drugs. Therefore, as you have suspected, these medical uses (current and developing applications) needed GMP manufacturing of the product for the purpose of the use in Humans. This GMP manufacturing of drugs is always costly. The price of the compound also depend on the other factors such as the size of the polymer (molecular weight), percent enantiomeric ratio (Lactic acid has two enanthiomers), end groups, etc., which are critical in human use. Also remember, synthesis of copolymer with desired ratio is not a easy task.
PLA has become a popular material due to it being economically produced from renewable resources. In 2010, PLA had the second highest consumption volume of any bioplastic of the world. For example, PLA is the most widely used plastic filament material in 3D printing. This commercial success of PLA makes its bulk production (no need of GMP manufacturing processes). Thus, it is much cheaper than demanding PLGA. Yet, PLA can be expensive as well, depending on its use, which require careful synthetic processing. For example, D- and L-lactide copolymer is used in bone engineering (Ref.1).
Sigma-Aldrich listed series of PLGA and PLA with some different properties:
$$
\begin{array}{c|ccc}
\hline
\text{EXPANSORB$^®$} & \text{PLGA or PLA} & \text{L/DL Ratio} & \text{End group} & \text{Molecular weight} & \text{Price}/\pu{10 g}\\
\hline
\text{DL 100-2A} & \text{PDLLA} & 0/100 & \ce{COOH} &  \pu{10-25 kD} & \pu{501.00 USD}\\
\text{DLG 50-2A} & \text{PLGA} & 50/50^a & \ce{COOH} &  \pu{5-20 kD} & \pu{446.00 USD} \\
\text{DLG 75-9E} & \text{PLGA} & 75/25^b & \text{Ester} &  \pu{100-150 kD} & \pu{458.00 USD} \\
\text{DLG 75-2A} & \text{PLGA} & 75/25^b & \ce{COOH} &  \pu{5-20 kD} & \pu{446.00 USD} \\
\hline
\end{array}\\
^a: \text{D/L-Lactic : Glycolic} = 50:50; \ ^b: \text{D/L-Lactic : Glycolic} = 75:25
$$
The list was posted to show when synthesized for biological usage, either polymer can be expensive. For example, the entry 1 of the above table is essentially PLA (made by 100% D/L-lactic acid), but it is the most expensive one on the list.
References:

C. X. F. Lam, R. Olkowski, W. Swieszkowski, K. C. Tan, I. Gibson, D. W. Hutmacher, "Mechanical and in vitro evaluations of composite PLDLLA/TCP scaffolds for bone engineering," Virtual and Physical Prototyping 2008, 2(4), 193-197 (https://doi.org/10.1080/17452750802551298).


Physical meaning of thermal parameters dimension (Å²) In single-crystal x-ray crystallography both isotropic and anisotropic displacement parameters $U_{ij}$ of thermal ellipsoids have dimension of square  angstrom ($Å^2$) as follows from the definition of Debye–Waller factor ($T$ is dimensionless):

For isotropic approximation (conditions of Bragg's law):
$$T_\text{iso} = 8\pi^2U\left(\frac{\sin \Theta}{\lambda}\right)^2 \to [U] = Å^2 \tag{1}$$
For anisotropic displacement parameters (ADPs):
$$T_\text{ani} = 2\pi^2 \sum_{i=1\\j=1}^3{H[i]H[j]U_{ij}e^\star[i]e^\star[j]} \to [U] = Å^2 \tag{2}$$
where $H$ is reflex's index; $e^\star$ is length of the reciprocal lattice basis vector.

In both cases $U_1$, $U_2$ and $U_3$ ($U_{ij}$ in general) are basically showing contribution of oscillation along three principal (orthogonal) axes and are measured in $Å^2$. So, is there any physical meaning that can be attributed to square angstrom in this case? At first glance, it seems rather confusing that a parameter that seems to represent a linear quantity is measured in square units of length.

Model of a crystal and deriving the expected diffraction image

In single-crystal X-ray crystallography, the diffraction image is due to scattering of electrons in a crystalline sample. A convenient way of describing the electron density is to first specify crystal symmetry and mean atomic positions (coordinates), and then to describe in more or less detail how atoms deviate from these mean positions (modeled by atomic displacement factors). Combining this with a spherical description of the electrons "belonging" to each atom (modeled by atomic form factors), you arrive at an electron density.

This description is useful because in the model, the electron density is the convolution of the crystal lattice points with the coordinates in the unit cell with the distribution of atoms around mean positions with the distribution of electrons around the atoms. The convolution theory states that the Fourier transform of a convolution is the product of the Fourier transforms. So in this model, you can separately Fourier-transform the different levels of the model and then multiply them to arrive at the expected diffraction image (via the structure factors).

Atomic displacement and Debye-Waller factor

The atomic displacement around a mean position may be described through a probability density function $p_k(\mathbf{u})$, where $\mathbf{u}$ is the position (3D-vector) with the mean position as origin. The Fourier transform of this function defines the Debye-Waller factor (source: equation 1.4.8):

$$T_k(\mathbf{h}) = \int p_k(\mathbf{u}) e^{2 \pi i \mathbf{h} \cdot \mathbf{u}}d^3 \mathbf{u}\tag{3}$$

Here, $\mathbf{h}$ represents coordinates in reciprocal space (just as Fourier transform of a time series gives you information in frequency space, the Fourier transform connects real space - coordinate $\mathbf{u}$ - with reciprocal space - coordinate $\mathbf{h}$).

If the displacement is modeled by a Gaussian, the integral can be determined as (source: equation 1.4.10)

$$T_k(\mathbf{h}) = e^{-2 \pi^2 \langle(\mathbf{h} \cdot \mathbf{u})^2\rangle}\tag{4}$$

This reflects that the Fourier transform of a Gaussian is a Gaussian again, and the width of one Gaussian is the inverse of the other (i.e. if the atoms are widely distributed, the signal will diminish a lot with increasing resolution). Again using the time series as an example, if a sound is played for a very short time, the pitch will be ill-defined (or if an NMR FID signal decays rapidly, the peaks will be broad).

Displacement parameters $\mathbf{U}$

The displacement parameters $\mathbf{U}$ or $\mathbf{U^{jl}}$ in the OP's equation (1) and (2) are derived from the $\langle(\mathbf{h} \cdot \mathbf{u})^2\rangle$ expression in equation (3), either direction-independent in the case of an isotropic approximation (spherical distribution around the mean) or separately along the three lattice axes (leading to six anisotropic displacement factors). I am not explicitly showing this - it gets a bit complicated because the derivation requires transformation into the sometimes non-orthogonal coordinate system based on the crystal lattice (see here). In any case, the expression is a mean square displacement (analogous to a variance of a one-dimensional function). For that reason, the dimensions are length squared and the typical units are $Å^2$. As stated in the IUCR nomenclature document:

  the elements of the tensor U have dimension (length)2 and can be directly associated with the mean-square displacements of the atom considered in the corresponding directions

and

  If the atomic pdf is assumed to be a trivariate Gaussian, the characteristic function corresponding to this pdf - by definition, its Fourier transform - can be described by the second moments of the pdf, which in the present context are called anisotropic mean-square displacements.

Second moments (variance) have dimensions that are the square of the dimension of the quantity described by the probability distribution function (pdf). In this case, we are looking at the distribution of the position (dimensions length), and the second moment will have dimensions of length squared.


Would C bond with H or D? Is a C-D bond better than C-H bond?  I mean if carbon has a chance to either for a bond with deuterium or hydrogen what would it choose and why?  What if tritium competes too?

C-H would be the weakest bond, then C-D, then C-T would be the strongest.

This is due to the zero point energy of the bond and the fact the zero point energy is proportional to:

$$\sqrt{\frac{m_C + m_H}{m_C m_H}}$$

Zero point energy is the minimum energy the quantum harmonic oscillator can have.

Because the greatest mass isotope has the smallest zero point energy, the energy of the bond is greatest.


How to treat skin corrosion by sodium hydroxide? I'm under training of becoming floorman in the rig. We use a lot of chemicals and one of them is caustic soda or sodium hydroxide. In case it touches my skin can you tell me of ways to treat it? I have thought of neutralizing it by slowly going less basic by using for example household ammonia then hand soap then sea water.

We use a lot of chemicals […]

Before you start to use potentially hazardous compounds and products, get their Material Safety Data Sheets (MSDS) and read them.
Learn about storage, handling, disposal and emergency measures in case of

eye contact
skin contact
inhalation

For a $50\%$ sodium hydroxide solution, a typical MSDS looks like this and states:

  Skin Contact:

  In case of contact, immediately flush skin with plenty of water for at least $15$ minutes while removing contaminated clothing and shoes. Cover the irritated skin with an emollient. Cold water may be used. Wash clothing before reuse. Thoroughly clean shoes before reuse. Get medical attention immediately.

Forget about the ammonia! Use lots of water!


Viability of reduction of oil paint binder and solvent flammability upon disposal via addition of water As an artist, when disposing of a paper towel or cotton rag containing oil paint, turpentine, odorless turpentine substitute or mineral spirits at the end of a painting session, I generally pour water over the paper towel or cotton rag to reduce its flammability. Sometimes I rinse out the paper towel or rag in the sink before disposal to reduce concentration of the solvent. Does mixing water with the oil paint binder and solvent actually reduce its flammability? I understand mixtures can be categorized as suspension, colloidal or solution, and I would be curious to know whether or how this categorization pertains.

It really depends exactly on what you're mixing the water with. Soaking a rag or paper towel will reduce the flammability and is recommended by many safety organizations online (here is one). However, if you are washing off a large amount of solvents or flammables with water and the chemicals are immiscible in water, it can sit on the top of the water and still burn if exposed to some ignition source.
The best thing to do is to look at SDS or MSDS sheets if you are unsure what the best disposal procedure is for some chemical you are using. Additionally, a google search should offer additional information on disposal and hazards of the substances painters used.


Comparing basic strengths between pyridine and 1,2-dihydropyrazine Compare the basic strengths of above two compounds.

In the first compound (1,2-dihydropyrazine), the lone pair of one N will participate in resonance but the lone pair of other N will not, similarly in pyridine, lone pair will not participate in resonance.
I think that since the first compound atleast has $2$ nitrogens (and only one in pyridine), it should be slightly more basic than pyridine, but I came to know that it's the other way round. So how do we compare their basic strengths and what are their $\mathrm{p}K_\mathrm{b}$ values?

What I would do is I would compare their acid strengths. Due to inductive forces, deprotonating the 1,2-dihydropyrazine would result in a more stable compound than that in pyridine--the nearby nitrogen (electronegative) will share the burden of the negative charge. Therefore, 1,2-dihydropyrazine is more acidic and thus is less basic.


How to calculate Hirshfeld charges from Gaussian? I am new to Gaussian and want to calculate Hirshfeld charges from Gaussian 09 or 16.
I found this input file for Methanol:
%nproc=8
%mem=1600mb
# mPW1PW91/6-311+G(2df,2p) pop=Hirshfeld scf=tight
Methanol
0 1
8 -0.046153000 -0.751257000 0.000000000
6 -0.046153000 0.658469000 0.000000000
1 0.857779000 -1.062372000 0.000000000
1 -1.086394000 0.976180000 0.000000000
1 0.437376000 1.072716000 0.889654000
1 0.437376000 1.072716000 -0.889654000

but when I run it I get this error:
QPErr --- A syntax error was detected in the input line.
 p) Pop=Hirshfeld scf=tight Methanol 0 1
                            '
 Last state= "GCL"
 TCursr=      3523 LCursr=        51
 Error termination via Lnk1e in /opt/software/gaussian/g09.e01/l1.exe at Wed Aug 19 01:02:31 2020.
 Job cpu time:       0 days  0 hours  0 minutes  0.6 seconds.
 File lengths (MBytes):  RWF=      5 Int=      0 D2E=      0 Chk=      1 Scr=

Any suggestion which part is the problem?

%nproc=8
%mem=1600mb
# mPW1PW91/6-311+G(2df,2p) pop=Hirshfeld scf=tight

Methanol

0 1
8 -0.046153000 -0.751257000 0.000000000
6 -0.046153000 0.658469000 0.000000000
1 0.857779000 -1.062372000 0.000000000
1 -1.086394000 0.976180000 0.000000000
1 0.437376000 1.072716000 0.889654000
1 0.437376000 1.072716000 -0.889654000

will do. All I did was to introduce blank lines after Route section (the one starting with #) and Title section (Methanol). Also, make sure that there is a blank line after the Molecule specification part. Gaussian is usually a convenient entry point to computational chemistry and you can find a lot of advice for it. The Gaussian manual is not bad, either. Have a look at this to get an impression about the syntax of input files - this is something Gaussian isn't forgiving about.


How does the screening effect work for orbital in the same shell? Energy of an orbital in a single electron system depends solely on n, while energy of orbital in multi electron system depends on n and l, why is that? I understand that the screening effect is at work between two different shells, but how does it work in the same shell, is there any other reason?

Screening within the same shell is generally poor, which is why effective nuclear charge rises going across the Periodic Table: each added electron screens the others poorly from the added proton.  And, yes, as you suspect, the quality of screening varies with l.

In addition to the screening effect, in many-electron atom you have electron-electron repulsion, as well as quantum exchange-correlation effects between electrons.  Since these are interactions between electrons they depend on the distance between the electrons, which in turn depends not only on the distance between each electron and the nucleus (controlled by n) but also on the angular distribution of electrons around the nucleus (controlled by l).


Major product formed when 1,1,1-trichloropropane is treated with aqueous KOH What are (is) the product formed when 1,1,1-trichloropropane is treated with aqueous potassium hydroxide?

I think the major product should be propionic acid, but I'm not sure.

Figure: 1,1,1-trichloropropane

${\color{green}{\checkmark}}$ The above compound involves $\color{red}{\ce{C-H, C-C, C-Cl}}$ bonds, whose bond energies are $413, 348, 328 kj/mol $ respectively. So, initially $\color{orange}{\ce{C-Cl}}$ bond must break.

${\color{green}{\checkmark}}$ It may be noted that generally alcoholic $\color{blue}{\ce{KOH}}$ causes elimination (dehydrohalogenation reaction) in the molecule of haloalkane, while aqueous $\color{blue}{\ce{KOH}}$ leads to substitution. As in your reaction, aqueous $\color{blue}{\ce{KOH}}$ is used, substitution reaction is most preferable.

${\color{green}{\checkmark}}$ As primary carbon and strong $\color{orange}{\ce{C-Cl}}$ (compared to other $\ce{C-Br, C-I}$ bond) is involved in 1,1,1-trichloropropane, $S_{N2}$ mechanism is preferable.

From the above thought process, the most preferable reaction would be:
$$\ce{Cl_3C-CH_2-CH_3 + 3KOH(aq) ->[\text{heat}] (OH)3C-CH_2-CH_3 + 3KCl} $$

${\color{green}{\checkmark}}$ It may be noted that more than one $\color{blue}{OH}$ group cannot be present on the same carbon atom. In such a case, the compound will be extremely unstable and will readily lose a water molecule to form more stable aldehyde, ketone or carboxylic acid.

$$\ce{(OH)_3C-CH_2-CH_3 -> (OH)CO-CH_2-CH_3 + H_2O}$$

You are right, propionic acid along with water molecule is going to be the product.


Why do silver nitrate and sodium hydroxide react to produce silver(I) oxide? I did an experiment in class combining the aqueous solutions of $\ce{AgNO3}$ and $\ce{NaOH}$. I predicted that $\ce{AgOH}$ would form but it didn't, instead a brown precipitate. I found out it was silver oxide. But why does this happen?

I researched and found out that silver is a noble metal and is thus resistant to oxidation, so why is it oxidised in this case?

Also why is $\ce{AgOH}$ not formed? I looked at the reactivity series and apparently silver can't displace $\ce{H+}$ ions out of solution, is that why?

Silver in silver oxide is no more oxidized than in $\ce{AgNO3}$. So you should ask yourself the same question earlier, even before the reaction. Yes, noble metals are somewhat resistant to oxidation. But still they can be oxidized, and thus can form compounds, of which $\ce{AgNO3}$ and $\ce{Ag2O}$ are the examples.

Even gold can be oxidized, though that requires some extreme measures.

As to the formation of oxide instead of hydroxide, there is no clear-cut distinction between Ag and the rest of metals. Many insoluble hydroxides are not quite stable. $\ce{Cu(OH)2}$, for example, would slowly lose water and turn into oxide; likewise, $\ce{Fe(OH)3}$ turns into $\ce{FeO(OH)}$. Silver does the same, just a little faster.


Why does Benzoin give positive Tollen's Test? Certain alpha hydroxy ketones which can tautomerise to form an aldehyde will give a positive Tollen's Test. I was pretty clear with this idea until I came across this question and the answer given is A,B,C. I can't figure out why Benzoin gives a positive Tollen's Test as it can't give an aldehyde by tautomerisation.

What exactly does happen to Benzoin over here and why does it give a positive test? Is my idea about tautomerisation of alpha hydroxy ketones to aldehydes wrong?

For benzoin, a similar mechanism of tautomerisation (of alpha hydroxy ketones tautomerising to aldehydes) happens , except for one step, where instead of the hydride ion (H-) migrating to the sp2 hydridized carbon, it is the Benzene carbanion (C6H5- or Ph-) that migrates to the sp2 hybridized carbon, since it is more stable than the hydride ion (an example of the stability of benzene carbanion over the hydride ion can be found in the pinacol-pinacalone rearrangement). Other than that, the rest of the mechanism is exactly the same.

Hope this cleared your doubt.


How many litres of water will be sufficient to generate 1600 psi when heated? Here's my setup. I have a pressure vessel with x liters of water. I'll be heating the water at 320 C. I want to achieve at least 1600 psi of force inside the pressure vessel.

So how do I calculate the minimum amount of water required to achieve that pressure?

First lets' state the Universal Gas Constant: R = (0.08206 x L x atm) / (mol x K)

And the Ideal Gas Law: PV = nRT

Now, lets find a handy calculator to do everything for us: http://www.ajdesigner.com/idealgas/ideal_gas_law_mole_equation.php

Now we can rearrange the gas law to solve for moles: n = PV/RT... and do some conversions... 320 °C = 593.15 K, and 1600 psig = 108.87 atmospheres (or did you mean psia?)... or, just plug the numbers into the calculator.

So, if the pressure is 1600 psi at 320 °C in a 20 L vessel you need 44.735912618737 moles of gas. Now, we multiply by the molar mass of water (18.01528 g/mol) to obtain more familiar values... it appears that you will need 805.93 grams of water (more or less). And let's use the density of water at room temperature to find the volume...  805.93g / 998.23 g/L = 0.80736 liters.

☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺

My advice, for what it's worth, is- if you are actually going to do this, you should probably use more water, and let the steam boil for a few minutes to evacuate the air from the vessel before locking it shut. Oh, and be sure to wear eye protection while you run for cover.

☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺

Edit based on comment:

Question: ...for how long would the pressure be mainted if [water vapor] is exiting also? I mean, I will start release pressure as soon as it hits 1600PSI, but I'll keep on heating the water too. So, for how long will I be able to maintain that 1600PSI exit force?

2nd Answer: The loss of pressure would be instantaneous... concurrent with the loss of water. To maintain the pressure you would need to increase the temperature as the water exits. The time it would last, as it empties and as the temperature increases, depends on the rate/amount of water being released and/or the rate of temperature increase. If you want to maintain 1600 psi, then you either need more water or more heat; if you use more water, then you could maintain the 320 °C because you would no longer have 20 L of "head-space" in the vessel. The amount of vapor in the head-space of the vessel would be less than 805.93g. Not that it matters, but to calulate the amount of vapor, you would subtract the liquid volume from the total vessel volume because water is practically  incompressible (maybe around 1% at 320 °C). The way to maintain 1600 psi would be to fill the tank (maybe 2/3 full), and use a pressure relief valve (like all pressure cookers do). A pressure relief valve would automatically control the pressure regardless of the change in temperature and volume (as long as it's sufficient or in excess- to supply the minimum amount of pressure to the valve).

Supposing that you had an infinite amount of heat on tap, the pressure would (theoretically) be maintained until the last molecule of water escaped the vessel. Now, let's forget that steel melts at about 1370 °C and water actually breaks down at around 2500 °C (says google- not sure if that chages because pressure would be 1600 psi) and steel burns at about 4000 °C; releasing one molecule of water per hour, the pressure would be maintained until you depleated 44.736 moles (so, multiplying by Avagadro's number: 6.022×10^23), you will have 2.694×10^25 hours.

To find the necessary temperature for 1 molecule of H2O to supply 1600 psi in 20L (the final temperature at hour 2.694×10^25), convert 1 molecule to moles and rearrange the gas law to solve for T.

n = 6.022×10^-23 : T = PV/Rn

T(final) = (108.87atm x20L)/(0.08206 x6.022×10^-23mol) = (about) 4.406×10^26 K. To convert K to °C, subtract 273.15 from 440,600,000,000,000,000,000,000,000 K.


Name of toxic liquid product that can be cleared with flocking agent I'm writing a paper rpg scenario, and even if we don't need to be 100% accurate about chemistry and physics in such games, one of my players has an advanced level in chemistry and I don't want him to cringe at the situation (and I can't ask him, obviously, or he'll be spoiled).
So, I need a river to be polluted with an invisible and odorless product, that slowly destroys the fauna. As it happens in Siberia, I thought about tritium first, but I don't think it's easy to clean. So I thought about a product that could be cleaned with a flocking agent. Problem is, when I was searching for "products reacting to flocking agent" on Google I didn't find any name of product that I could use.
So, which invisible, toxic and odorless product react to flocking agent to the point the river is cleaned in one or two months maximum (and eventually produced in the 1980's) ?
Thanks a lot, and sorry if it's not the SE to ask this question.

If your impurity is tritium, it can not be purified, whatever the method used. If your impurity is another substance, it can be purified by adding aluminum sulfate. Aluminum sulfate is slowly hydrolyzed in dilute solutions. And this phenomena produces a light and insoluble substance called aluminum hydroxide $\ce{Al(OH)3}$. The particles of this substance look a bit like snow flakes and have the property of fixing impurities on its surface, between its branches. So after a couple of hours, this aluminum hydroxide will be deposited on the ground of the river or the lake, with the impurities fixed inside its particles.


How to create a buffer using CH3COOH and CH3COONa that has a pH of exactly 3.75? I need to create a buffer using $\ce{CH3COOH}$ and $\ce{CH3COONa}$ that has a pH of exactly $3.75$.
A $\pu{50 mL}$ sample of your buffered solution will have to be able to withstand the addition of $\pu{25.0 mL}$ of $\pu{0.100 M}$ $\ce{NaOH}$ solution.
The buffered solution will break after the addition of no more than $\pu{35.0 mL}$ of the $\pu{0.10 M}$ $\ce{NaOH}$.

Here is what I have so far:
$$\mathrm{pH}= 3.75 \implies [\ce{H+}] = \pu{10^{-3.75} M} = \pu{1.78 \times 10^-4 M}$$
\begin{align}
\ce{NaCH3COO &-> Na+ + CH3COO-}  &&\text{(common ion effect)}\\
\ce{CH3COOH &<=> H+ + CH3COO-}
\end{align}
\begin{align}
K_\mathrm{a} &=\frac{[\ce{H+}][\ce{CH3COO-}]} {[\ce{CH3COOH}]}\\
1.8\times{10^{-5}} &=\frac{(1.78\times{10^{-4}})[\ce{CH3COO-}]} { [\ce{CH3COOH}]}\\
\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &=\frac{(1.8\times{10^{-5}})}{(1.78\times{10^{-4}})}\\
\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 0.1011 : 1~\text{(ratio)}
\end{align}
I found the ratio of $[\ce{NaCH3COO-}]$ to $[\ce{CH3COOH}]$, but I don't know what to do from here.
EDIT: "Break": The point where there's no acid left and the solution stops acting as a buffer. "Withstand": meaning that there can be a $\mathrm{pH}$ change of at most $1.0$. Thanks, guys!

In my understanding, there is considered $$\mathrm{p}K_\mathrm{a}=4.75$$ and therefore $\mathrm{pH}=3.75$ was chosen for easy calculation, as:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log \frac {c_{\ce{A-}}}{c_{\ce{HA}}}$$

and therefore
$\frac {c_{\ce{A-}}}{c_{\ce{HA}}}=0.1$

For the buffer change to $\mathrm{pH}=4.75$, at least $\pu{25 ml 0.1 M NaOH}$ has to be spent. ("withstand")

$$\frac {c_{ws,\ce{A-}}}{c_{ws,\ce{HA}}}=1$$

Calculating molar amounts to be equal for both acetic acid and acetate:
$$0.1\cdot c_{\ce{HA}}\cdot V_\mathrm{buf}+c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}\\=c_{\ce{HA}}\cdot V_\mathrm{buf}-c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}$$

From that, we calculate minimal concentration of acetic acid in buffer:

$$2\cdot c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}=0.9\cdot c_{\ce{HA}}\cdot V_\mathrm{buf}$$

$$c_{\ce{HA}}=\frac{2\cdot c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}}{0.9\cdot  V_\mathrm{buf}}=\frac{2\cdot 0.1\cdot 0.025}{0.9\cdot  0.05}=0.005/0.045=1/9 \mathrm{~ mol/l}$$

$$c_{\ce{HA},min}=\frac19 \mathrm{~ mol/l}$$
$$c_{\ce{A-},min}=0.1 \cdot \frac19=\frac1{90} \mathrm{~ mol/l}$$

All acetic acid in $\pu{50 mL}$ of the buffer must react with not more than  $35~\text{ml}$ of $0.1~\mathrm{M~}\ce{NaOH}$.

$$\begin{align}
n_{\ce{NaOH}}&\lt 3.5\mathrm{~mmol}\\
n_{\ce{CH3COOH}}&\lt 3.5\mathrm{~mmol}\\
n_{\ce{CH3COONa}}&\lt 0.35\mathrm{~mmol}\\
\end{align}$$

As the latter 2 amounts have to be present in $\pu{50 mL}$ of the buffer:

$$\begin{align}
c_{\ce{CH3COOH}}&\lt \pu{70 mmol/L}\\
c_{\ce{CH3COONa}}&\lt \pu{7 mmol/L}\\
\end{align}$$

Summary:
Concentration of acetic acid in the buffer must be at least $\frac19=0.1111\pu{ mol/L}$ and not more then $\pu{0.07 mol/L}$.

Such requirements is impossible to meet and such a buffer is not possible to prepare.

By other words, the buffer cannot have at the same time so big buffer ("withstand") capacity and so low "break" capacity.

If usage of acetate buffer was not mandatory, the goal would be achievable with buffers with pKa closer to 3.75 than the acetic acid has.  In such a case, the buffer capacity would be decreasing and not increasing between these pH values and the final break would be faster.


Are all exothermic reactions spontaneous? $\ce{GaAs (s)}$ is used in solar cells deployed on satellites found in the orbit of Earth. The best way to make these materials is through a technique called "molecular beam epitaxy", in which the following spontaneous chemical reaction takes place:
$$\ce{Ga(g) + As(g) -> GaAs(s)}$$
Pick the statement which best describes this process:
(A) The process is endothermic.
(B) The process is exothermic.
(C) The process leads to a decrease in the entropy of the universe.
(D) The process leads to an increase in the entropy of the system.
(E) The process leads to a decrease in the temperature of the system.

The key says the correct answer is B. I am wondering if this is because all exothermic reactions are spontaneous?

No, it is not the case that all exothermic reactions are spontaneous.  More generally, it doesn't even make sense to make a broad statement that a reaction is or is not spontaneous without also specifying the concentrations of reactants and products, since for a certain relative concentration of reactants and products, the reaction might be spontaneous in the forward direction, while for another set of relative concentrations it might be spontaneous in the reverse direction. [See illustration below.]

One could, however, make the general statement that chemicals will react spontaneously if, in so doing, the entropy of the universe increases.  This, however, is not one of the options you are given.

Why, then, is "B" the correct choice?  Answer: It's because this Ga and As are forming a chemical bond, and bond formation is exothermic. In addition, there is a phase change here from a gas to a solid, and that is also exothermic.  Furthermore, since Ga and As are both monatomic gases in this reaction, no chemical bonds need to be broken to allow the Ga and As to combine. Thus, since we have chemical bonds being formed but not broken, and a phase change from gas to solid, the reaction is exothermic.

Here's a graphical illustration of the general concept articulated in the first paragraph:


Nitrogen Covalent Bonds Let me premise this with the fact that I really have zero knowledge of chemistry. So bear with me and my ignorance.  :)

Now, with that introduction out of the way, I've read up this site which presented in a very easy to understand way, how a Nitrogen atom bonds to yet another Nitrogen atom to make a fairly common element $\ce{N2}$.

Continuing, it says on that site that when you create $\ce{N2}$ from two Nitrogen atoms, that's as far as you can go for creating covalent bonds between Nitrogen atoms.

Armed with that knowledge, my question is as follows:

- Can you create further bonds utilizing another atom and still keep the strength of a triple bond and $\ce{N2}$'s natural inertness?

Since you have zero knowledge of chemistry, I will keep explanation as simple as possible. I have to emphasize that this answer is not fully correct and can be misleading. It is suited just for interested layman, but not someone trying to grasp deeper concepts of chemistry.

Nitrogen atom has 5 electrons in its outer shell. The most stable electronic configuration, by octet rule, corresponds to 8 electrons in outer shell. Nitrogen needs three electrons to make that shell full. It can achieve that configuration by bonding with another nitrogen. One nitrogen has his own 5 electrons and another lends him his 3 - now first nitrogen has 8 electrons, e.g. full shell. First nitrogen lends his 3 electrons to second nitrogen which also had 5 electrons, so now both of them have closed shell. Since both of them lended three electrons, and accepted three - they have three shared electron pairs, which corresponds to triple covalent bond. Closed shell configuration, which they both now achieved, corresponds to a state of the lowest energy. Since everything in nature strives to a state of lowest energy, you can see why $\ce{N}$ atoms prefer to bond, instead of being each on it's own. The molecule is inert, because it achieved what it strived to: energy minimum. By adding another (third) nitrogen in a game, you would necessary have to "break" closed shell configuration, which necessary leads to increase of energy, which is not favorable. That's the reason why you can't create another bond and keep inertness of $\ce{N2}$ molecule.


How does Potassium Hexfluoronickelate(IV) exist in a stable state? Nickel generally does not exist in a $+4$ Oxidation state. Wikipedia has the following to say regarding Nickel(IV):

  Ni(IV) is present in the mixed oxide $\ce{BaNiO3}$. ....... Ni(IV) remains a rare oxidation state of nickel and very few compounds are known to date.

This article directly contradicts the first statement, as it says:

  the photoelectron spectroscopy data suggest the description of $\ce{BaNiO3}$ in terms of trivalent Ni and oxygen anion holes.In terms of a local ionic model the idealized description of this high-valent oxometalate would correspond to $\ce{Ba^2+Ni^3+(O2)^2O^•-}$

Now, we end up coming to the curious case of Potassium Hexfluoronickelate(IV), which decomposes at 350°C, meaning that it is relatively stable. Not much literature is available on this compound online, except this article, which is unfortunately behind a paywall.

Could someone explain how nickel exists in a (IV) state in this compound instead of degenerating to $\ce{K4[NiF6]^2-}$? Also, are there any more Nickel compounds exhibiting the (IV) state? Explanations regarding the formation and stability of this compound are also welcome!

To quote from the comments: you can get things into an amazing variety of oxidation state given the right environment.
In the case of maximal oxidation states, particularly above $+2$ or $+3$ in the $3d$ transition series, the right environment (at least in simple species) involves complexation with ligands that are highly electronegative, capable of pulling electrons even from a strongly positive central atom, and resistant to oxidation themselves in an electron-poor setting.  Meaning, in practice, oxide or fluoride ligands.
We can therefore roughly classify high oxidation state complexes into oxo and fluoro complexes.  Oxo complexes are more likely to be seen with early transition metals, where the central metal ion contributes few $d$ electrons and there are a lot of vacant orbitals with which the strongly $\pi$-donating oxygen can form back-bonds.  The best known examples being chromate/dichromate ($\ce{Cr(VI)}$), manganate and permanganate ($\ce{Mn(VI), Mn(VII)}$), and ferrate (most commonly, $\ce{Fe(VI)}$).  Even cobalt reaches its apparent maximum $+5$ oxidation state with an oxide-ligand species, $\ce{[CoO4]^{3-}}$.
Later transition metals, which have lower maximum oxidation states due to increased electronegativity/ionization energy, have more $d$ electrons coming from the cationic shells and thus are no longer so stabilized by $\pi$ back-donation.  Then the fluoro complexes, with a relatively $\pi$-inactive ligand, take over.  In the $3d$ series cobalt, cited above as forming an oxide complex with the formula $\ce{[CoO4]^{3-}}$, also forms a +4 fluoride complex $\ce{[CoF6]^{2-}}$(1).  Fluoride complexes are also seen with both nickel and copper in oxidation states above $+2$, including the $\ce{[NiF6]^{2-}}$ ion mentioned in this question and the copper(III) species $\ce{[CuF6]^{3-}}$.  Zinc in the $3d$ series is limited to oxidation states of $+2$ or less, but if mercury has a tetrafluoride $\ce{HgF4}$ (disputed), it's no coincidence that this too is a fluoride.
Reference

J.W. Quail and G.A. Rivett, "Fluoride Complexes of Tetravalent
Cobalt",  Can J Chem 50:2447 (1972).


Is the conjugate base of pyrrole resonance stabilised? This is something I am conceptually struggling with, if I deprotonate pyrrole on the nitrogen in the ring, is it true that the conjugate base is not resonance stabilised, because the orbital is orthogonal to the ring? However, with a pKa of 23, I feel like it must have some resonance stabilisation.

There is like you assumed no reasonable amount of resonance stabilisation. However, the aromatic π-system contributes to the stability of σ-skeleton.

Below you can find the MO of pyrrole and its N-deprotonated anion (click it to enlarge). You can see that in MO 17 the electron density is shifted further into the centre of the ring, accounting for the larger negative charge on the nitrogen. It is a bit more difficult to see, but that is also true for MO 18 and 14 (pyrrole), which becomes 15 (anion). The most significant change is that the upon deprotonation resulting lone pair becomes much higher in energy (12 to 16), and it is as you note perpendicular to the π-system in a σ-type orbital. Note that the image only shows the trend and has no information of the actual orbital energies, apart from that they increase from bottom to top.


How to identify a compound as chiral or achiral I read my book and learned everything about enantiomers, racemization, $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions, but when I'm trying to identify molecules as chiral or achiral, I get confused.

Consider the two molecules below:

1.CH3-CH-OH
      |
      Br

2.CH3-CH-Br
      |
      Br

How can I tell if a compound is chiral?

Look for carbons with four different groups attached to identify potential chiral centers.

Draw your molecule with wedges and dashes and then draw a mirror image of the molecule. If the molecule in the mirror image is the same molecule, it is achiral. If they are different molecules, then it is chiral.

Here you're particularly interested in the C in the CH. However, take a look at the C in CH3. Why isn't it a chiral carbon?


"Scrubber" for storing HCl I'm in need of hydrochloric acid for cleaning parts before electroplating, but I'd rather not deal with the HCl fumes given off by it, as I'd rather keep my workshop contents from rusting.
Given that HCl basically from what I've understood escapes any container it's put in, I had the idea of putting the container inside a bigger bucket and putting something in the bottom of the bucket to react with the fumes give off by the hydrochloric acid.
My first idea was either sodium bicarbonate or lye in solid form, would these react with it or do I need to have the base in liquid form? I'd rather not have my hydrochloric acid container covered in a lye solution.

Inexpensive [about US\$0.30 / liter] marble or limestone scraps, $\ce{CaCO3}$, likely would serve well. You could make an outer bucket with polyethylene netting, so that the pickling bucket would be well below the level of surrounding limestone chips. Coarse netting would not work with powders. Make the neck just large enough to admit the parts.
Better would would be to draw air continuously from above the container through a scrubber stack filled with marble or limestone chips, and possibly vent outside.
Definitely don't use lye, $\ce{NaOH}$, which is perhaps more dangerous than $\ce{HCL}$, as well as hygroscopic! You'd wind up with a puddle of caustic lye solution at the bottom of the outer bucket. And $\ce{NaHCO3}$ would also make a mess, though non-toxic.
BTW, if using large amounts, be careful that a spill into the outer container doesn't release so much $\ce{CO2}$ as to displace air and asphyxiate someone.


Is there another name for compound "delta-7 sterine", does it exist alone? Is there a more common name for "delta-7 sterine"? Is this spelled correct?

I read all over the internet and this compound exists in many articles, but I cannot find any information on it at all, also it doesn't show any information it. Can you please direct me to more information on this compound on what it is, it's more common names, chemical makeup, etc. All I could find in articles is that it's a steroid found in pumpkin seeds.

I would like to buy this compound for experimenting but cannot find a supplier that recognizes this name of the compound, does it exist alone in this form?

"delta-7 sterine" seems to be an adaptation of German term "Sterine" (plural for "Sterin") used by German scientists who elucidated the structures of those steroids using NMR [1].
In English, it seems, the name should be written as Δ⁷–sterol and is not a single compound but a group of compounds (for more info I'd suggest searching Google Scholar).

References

Sucrow, W.; Reimerdes, A. Δ⁷-Sterine Aus Cucurbitaceen. Zeitschrift für Naturforschung B 1967, 23 (1), 42–45. https://doi.org/10.1515/znb-1968-0109. (Open Access)


What is the formula for evaporation rate of water? I did my own research, and according to The Engineering Toolbox,the formula for the evaporation rate of water is as follows:

$g_\mathrm h = \Theta A (x_\mathrm s - x)$

$g_\mathrm h$ = amount of evaporated water per hour ($\mathrm{kg/h}$)

$\Theta = (25 + 19 v)$ = evaporation coefficient ($\mathrm{kg/(m^2\ h)}$)

$v$ = velocity of air above the water surface ($\mathrm{m/s}$)

$A$ = water surface area ($\mathrm{m^2}$)

$x_\mathrm s$ = humidity ratio in saturated air at the same temperature as the water surface ($\mathrm{kg/kg}$)  ($\mathrm{kg}$ $\ce{H2O}$ in $\mathrm{kg}$ dry air)

$x$ = humidity ratio in the air ($\mathrm{kg/kg}$) ($\mathrm{kg}$ $\ce{H2O}$ in $\mathrm{kg}$ dry air)

However, later I realized that something is wrong with the formula. There is no variable for water temperature.

The formula for humidity ratio in the air is:

$$x = 0.62198 p_\mathrm w/(p_\mathrm a - p_\mathrm w)$$

$p_\mathrm w$ = partial pressure of water vapor in moist air (Pa, psi)

  $p_\mathrm a$ = atmospheric pressure of moist air (Pa, psi)

The formula for humidity ratio in saturated air is the same, except partial pressure of water vapor in moist air is replaced with saturation pressure of water vapor in moist air.

And the formula for water vapor saturation/partial pressure is:

$$p_\mathrm{ws} = \mathrm e^{\frac{77.3450 + 0.0057 T - 7235 / T}{T\cdot8.2}}$$

$$p_\mathrm w = p_\mathrm {ws} \cdot HU$$

$p_\mathrm{ws}$ = water vapor saturation pressure (Pa)

$T$ = dry bulb temperature of the moist air (K)

$HU$ = humidity ratio (%)

So, according to this website, temperature-wise the formula is determined solely by the air temperature without considering the water temperature. Is it because the formula assumes the equivalence of the two temperatures for simplification or because I'm missing a variable? I searched for other methods, but so far no website is as comprehensive as this one, and I don't want to re-program everything again.

This is a different question from the existing ones. I've read many of them and the answers are not specific enough for my situation. Here I need to know how the temperature of the water plays a role in the formula for the rate of evaporation of water.

The temperature of water is implicitly covered by actual and saturated vapour content, or equivalent vapour pressures.

There is deterministic relation between relative humidity, air temperature and temperature of water, supposed to be in dynamic equilibrium.

In such a state, the cooling effect of water evaporation is balanced by the thermal transfer from warmer air.

This relation is used in meteorology/climatology as the standard method to determine humidity via the psychrometric tables of temperature of "dry" and "wet" precise thermometers.

Effectively, the wet bulb temperature is  somewhere between air temperature and the air dew point.

They are standard items in the white meteorological wooden box 2m above the lawn.

See psychrometrics particularly the sections dry and wet bulb temperature.

The main article Wet bulb temperature.


Xenon and the human body Reading this article on Wikipedia: Xenon Medical applications I see that Xenon can be used as an anesthetic, neuroprotectant and doping agent.

If it is a noble gas, and thus, chemically stable, how can it have those properties?

If Xenon does not react with any other substance, how does this happen, from a chemical point of view?

Chemist Neil Bartlett in 1962 discovered that although Xenon was a noble gas, it is able to form compounds with other substances even though it is chemically "inert."
Neil Bartlett, at the time in 1961, produced an unidentified red solid and discovered that the red solid was a reaction between gaseous fluoride, platinum hexafluoride ($\ce{PtF6}$) and oxygen that was oxidized. Finally, determining the chemical structure, they found that the red solid was $\ce{O^{2+}PtF6}$.
Oxygen is a powerful oxidizing agent, being consumed in many combustion reactions. However, Bartlett believed that the platinum hexafluoride ($\ce{PtF6}$) was a stronger oxidizing agent than oxygen, and oxygen is instead reduced. Bartlett then believed if the oxidation of oxygen was possible, he could apply this xenon.

Because my co-workers at that time (March 23, 1962) were still not sufficiently experienced to help me with the glassblowing and the preparation and purification of $\ce{PtF6}$ [platinum hexafluoride] necessary for the experiment, I was not ready to carry it out until about 7 p.m. on that Friday. When I broke the seal between the red PtF6 gas and the colorless xenon gas, there was an immediate interaction, causing an orange-yellow solid to precipitate. At once, I tried to find someone with whom to share the exciting finding, but it appeared that everyone had left for dinner!

The discovered new compound was xenon hexafluoroplatinate ($\ce{XePtF6}$). Bartlett's important contribution to chemistry is a series of noble gas compounds with specific and useful properties.
In regards to your question on how xenon can be so diverse, with over "100 noble gas compounds known today..."

Noble gas compounds have already made an impact on our daily lives. $\ce{XeF2}$ has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents. The reactivity of radon means that it can be chemically scrubbed from the air in uranium mines and other mines. Excimer lasers use compounds of argon, krypton or xenon to produce precise beams of ultraviolet light (when electrically stimulated) that are used to perform eye surgery for vision repair.

Material derived from a highly interesting article from the American Chemical Society.
http://www.acs.org/content/acs/en/education/whatischemistry/landmarks/bartlettnoblegases.html
Looking at the literature:

Neuroprotection

For the use of Xeon during mechanical ventilation, an article examining the neuroprotective effects of xenon,
Neuroprotection against traumatic brain injury by xenon, but not argon, is mediated by inhibition at the N-methyl-D-aspartate receptor glycine site.

Anti-Tumor Agent

Noble gas compounds have already made an impact on our daily lives. $\ce{XeF2}$ has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents.
The American Cancer states that

Fluorouracil belongs to the class of chemotherapy drugs known as anti-metabolites. It interferes with cells making DNA and RNA, which stops the growth of cancer cells.

Anesthesia

I'm still looking for literature to add here. Some of these cover very complex physiological processes, and this is not my area of expertise. I will try to find more pertinent articles from peer-reviewed journals with high enough impact and relevance. It will be up to the poster of this question to explore in-depth as much as they need by starting at some of the literature I provide.


Mustard oil test for distinguishing between primary aliphatic and primary aromatic amines? Can the mustard oil test be used for distinction between primary aliphatic and primary aromatic amines?

An alkyl isothiocyanate is formed when mustard oil is reacted with a primary aliphatic amine. But does mustard oil give a product with aromatic primary amines so as to make a considerable distinction?

No, aromatic primary amines cannot be distinguished from aliphatic amines using mustard oil test as it reacts with carbon disulfide to give phenylisothiocyanate.

Aniline is reacted with carbon disulfide in the presence of mercury chloride to give the isothiocyanate. The reaction is given in this site.

$$\ce{C6H5NH2 + S=C=S ->[HgCl2] C6H5NCS + HgS + 2HCl}$$.

Aliphatic amines also give the same products,  isothiocyanate, mercuric sulfide and hydrochloric acid and as such this test cannot be used to distinguish between aliphatic amines and aromatic amines.

Alternatively, it is also formed in the presence of ammonium hydroxide and lead nitrate.


Order of reactivity of hydrogen halides Solomon and Fryhle $11^{th}$ edition says :

Order of reactivity of HX in the acid catalyzed conversion of alcohols to alkyl halides follows the order : HI>HBr>HCl

We can account for this by saying that the acidity order is also HI>HBr>HCl
and hence the $-OH$ is easily protonated by the more acidic halide but the reaction is already acid catalyzed by a strong acid and the conversion of $-OH$ to  $-H_2O^+$ is fast so the acidity of $HX$ should make a negligible difference.

The slow step of the reaction is the formation of the carbocation and that does not seem to involve the halide so how can we explain this order ?

The halide ion is involved, indirectly.  The hydrogen halide has to protonate the alcohol first, and this requires dissociation of the proton from  the halide ion.  Chloride ion is more strongly basic than iodide ion, an effect we do not see in water (leveling effect) but can see it in the reaction with the alcohol.  So we can protonate more of the alcohol at any time, with hydrogen iodide, making more opportunities to then form the carbocation, than with hydrogen chloride.

Hydrogen fluoride would be way off the scale here, in the direction of less reactivity.  It is only a weak acid in water, let alone in this reaction mechanism.  It's one manifestation of fluorination being more difficult than other halogenations.


Empirical Evidence and Assignment of Orbital Energies (Electron Filling) I see a lot of questions on electron configurations from students at various levels here. Students mechanically remember how to fill orbitals and sorry to say many teachers do the same. Not a single general chemistry or even undergraduate physical chemistry book discusses the "why" part of orbital filling and what is the experimental evidence behind the electron configurations? Yes, hand-waving discussion is always present. It all comes from spectroscopy and so on.

A long time ago I asked a question here, on the physics forum and ResearchGate as how one would come up with an electron configuration from the raw spectrum of a non-hydrogen element. Experimental Assignment of Electronic Transitions in Atoms (Grotrian Diagram)  However, even trained atomic spectroscopists did not have an answer. Is it an art which has been lost and all we have now are tables?

The question is that what is the point in arguing and teaching that electronic configuration of copper is [Ar] 3d10 4s1 or [Ar] 3d9 4s2? What difference does it make in the world when the teacher and the student have no clue of experimentally verifying it except both are memorizing the rules and exceptions like a parrot? Suppose a very good undergraduate asks this question that how do we experimentally know that configuration of copper is [Ar] 3d10 4s1 or [Ar] 3d9 4s2 vice versa? Most of us may not have an answer.

I would be very happy to know a resource that talks about the experimental evidence and determination of electron configurations of heavier metals. So far my efforts have not been very fruitful.

A similar question has been asked here Evidence of orbitals but it does not fully address my old query.

As far as I know, it was a long and complex undertaking. You can't look at a single atomic species by itself. The understanding of level assignments to series slowly grew starting from the simplest cases.

Of course the first was hydrogen, where - at least initially - no series are discernible but one - there is almost perfect degeneration on l
. But that helped, as it allowed the first step, the one from wavelengths λ (or wavenumbers k=1/λ) to terms, later identified with energy levels. It was Ritz' combination principle:
k=Tm−Tn.
Already at this introductory level, a difficulty was to be overcome: the difference between emission and absorption spectra. The former is much richer. This was explained as a consequence of atoms in absorption initially being in the ground state alone, whereas in emission photons are emitted by excited atoms jumping to a lower state, not necessarily the ground one. For hydrogen this is especially notable, as there are no absorption lines in the visible region - the Balmer series has n=2

.

Then alkaline spectra were explored. In absorption spectra only lines Tn−T1
are observable, where T1 is what later would become the ground state energy (divided by hc). Note that here n=1, meaning the ground state, but this is not the principal quantum number of hydrogen-like classification, which is 2 for Li, 3 for Na, etc...

The main difference between hydrogen and alkaline spectra is that in the latter l degeneracy is broken. This required to separate terms in several series. Surely you know the origin of symbols S, P, D,... related to a different appearance of lines starting from different series towards the same final term. An instance is transitions S-P and D-P (in emission), In absorption only S-P transition is visible, giving rise to the principal series (therefrom P-terms).

But a second feature appears in alkaline spectra: the so-called "fine structure". It exists in hydrogen lines too, but is much less prominent, whereas the famous sodium doublet requires a modest resolving power to be seen. Its interpretation required the discovery of electron spin and of L-S coupling. The effect was a doubling of columns for all series, S excepted.

Well, this is a rough history, as I'm able to follow. Maybe a book could be of help: G. Herzberg, "Atomic Spectra and Atomic Structure". It's outdated as far as theory is concerned, but thanks to its publication date (first edition 1936) is nearer to the times when the studies were carried out.


Statistical model of ligand substitution Recently, I was told that in case of a particular step of a generic ligand substitution reaction:

$$\ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$

The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to

$$\frac{N - n}{n + 1}$$

by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?

I think the easy way out is to invoke $S_\mathrm m = R \ln \Omega$. If we assume that for a generic complex $\ce{MA_{n}B_{$N-n$}}$,

$$\Omega = {N \choose n} = \frac{N!}{n!(N-n)!} \quad \left[ = {N \choose N-n} \right]$$

and that for the individual molecules $\ce{A}$ and $\ce{B}$, $\Omega = 1$, then the equilibrium constant $K$ for

$$\ce{MA_{n}B_{$N-n$} + A <=> MA_{n + 1}B_{$N-n-1$} + B}$$

is given by

$$\begin{align}
K &= \exp\left(\frac{-\Delta_\mathrm r G}{RT}\right) \\
&= \exp\left(\frac{\Delta_\mathrm r S}{R}\right) \\
&= \exp\left(\frac{S_\mathrm{m}(\ce{MA_{n + 1}B_{$N-n-1$}}) + S_\mathrm{m}(\ce{B}) - S_\mathrm{m}(\ce{MA_{n}B_{$N-n$}}) - S_\mathrm{m}(\ce{A})}{R}\right) \\
&= \exp[\ln\Omega(\ce{MA_{n + 1}B_{$N-n-1$}}) + \ln\Omega(\ce{B}) - \ln\Omega(\ce{MA_{n}B_{$N-n$}}) - \ln\Omega(\ce{A})] \\
&= \exp\left[\ln\left(\frac{\Omega(\ce{MA_{n + 1}B_{$N-n-1$}})\Omega(\ce{B})}{\Omega(\ce{MA_{n}B_{$N-n$}})\Omega(\ce{A})}\right)\right] \\
&= \frac{\Omega(\ce{MA_{n + 1}B_{$N-n-1$}})\Omega(\ce{B})}{\Omega(\ce{MA_{n}B_{$N-n$}})\Omega(\ce{A})} \\
&= \frac{N!}{(n+1)!(N-n-1)!} \cdot \frac{n!(N-n)!}{N!} \\
&= \frac{N-n}{n+1}
\end{align}$$

The reason for ignoring $\Delta_\mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $\Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.


What eco-friendly substance can I use to write and erase on a wall for UV glowing without detection in daylight? I rent an office and want an alternative to a whiteboard and continual dry-erase marker purchases.

I can shut out the light and use the surface under UV lamps only, that's fine.

Commonly listed organic glowing elements are blood, urine and semen. None of which, I think would be untraceable by eyes (or nose) during the day!

The surface I would like to use contains typical white interior paint, applied last year.

I think that in a wall you can't write and erase so easily. You can write and then re-paint the surface in this case you can use Zinc White (make sure that what you use is $ZnO$). It was a common white pigment before $TiO_2$ discovery and has a strong fluorescence under UV light so if your wall is white (best $TiO_2$ painted, $CaCO_3$ may have some fluorescence) you will not notice $ZnO$ area in visible light but only under UV light.


How does a Microbial Fuel Cell work? I am doing a science fair experiment about Microbial Fuel Cells and was wondering how exactly do they work?

I know the basics, that they work similarly to a normal Fuel Cell through oxidation and reduction but am still a little unclear on the full details.

A microbial fuel cell (MFC) is an electrochemical cell that converts chemical into electrical energy, and in which one redox half-reaction is catalyzed by enzymes of microorganisms. The fuel cell consists of anode and cathode compartments which are separated by a membrane that is only permeable for specific cations. At the anode, bioorganic fuel is oxidized to $\ce{CO2}$ and $\ce{H3O+}$. The gained electrons flow from the anode through an external electrical circuit to the cathode, where they are consumed together with protons by the reduction of $\ce{O2}$ to $\ce{H2O}$. The $\ce{H3O+}$ ions are transferred from the anode to the cathode compartment through the membrane.

The fuel is a medium rich in nutrients which can be oxidatively degraded by the microbes, like, for example, sugars (in the following formula, $\ce{C12H22O11}$ stands for a variety of disaccharides):

$$\ce{C12H22O11 + 13H2O -> 12CO2 + 48H+ + 48e-}$$

The electrons from this half-reaction are either transferred directly to the anode (mediator-free MFC) or via a mediator (mediator MFC), which integrates into the electron transport chain of the cell and transfers electrons to the anode by passing the cell membrane. The mediator gains and loses electrons in every transport cycle, and switches between its reduced form (laden with electrons from the cell) and oxidized form (has given up electrons at the anode).

The scheme below (image source: MFCGuy2010/Wikimedia Commons) shows a MFC which uses soil as biofuel. Nutrients from the soil, like acetic acid, are oxidized by microbes in the anodic biofilm to $\ce{CO2}$ and $\ce{H3O+}$. Protons are then consumed in the reduction of supplied oxygen at the cathode, and electrons flow through the external circuit from anode to cathode. In this case, the soil acts simultaneously as nutrient medium and proton exchange membrane.


Limiting the minimum percentage of oxygen in atmosphere maintained by an oxygen absorber / scavenger? I'm aware that oxygen in the atmosphere of a sealed package can be readily removed by using oxygen absorber/scavenger such as granulated iron and sodium chloride but my understanding is that these bring the O2 content all the way down from 21% only limited by the permeability of the packaging to oxygen and the ratio of scavenger to package volume to something approaching 0%.

My question is if there is a way to formulate an oxygen scavenger that would be limited by some other processes to higher floor of say 5% to 10%... Without having to measure and calculate appropriate ratio of scavenger to package on a per use basis ?

Oxygen absorbers have the purpose to prevent food spoilage by oxidation (of fatty acids, flavors, etc.), so the desire is to remove oxygen as much as possible and also to bind small amounts which may diffuse into the package upon storage.

In order to achieve a certain defined oxygen containing atmosphere, consider flooding the package with an artificial gas mixture, e.g. 5% oxygen, 95% nitrogen. Should be easy to obtain from scuba dive shops who are filling bottles with homemade nitrox blends if they are willing to give you a custom mixture (which in this case is not suitable for diving).

Or just mix air and nitrogen or carbon dioxide (from a soda machine) by adjusting the flow.


What will be the Ozonolysis product? Problem:

My Answer:

According to me, both the left side and the right side double bonds undergo cleavage and hence the product must be, two molecules: Keto aldehyde and Ester aldehyde. But since, such product is not given in options, let us assume that only one bond gets oxidized. But which one? Which bond is the most reactive? I don't know.

Answer given in the book:

  (B)

How come?

There are two types of control, that governs reaction path:

kinetic control (path with least activation energy) and
thermodynamic control (path with least energy of products).

Given temperature favors kinetic control.

Ozonolysis starts as 2+3 electrocyclic addition (with later regroup of the product), similar to Diels-Alder reaction . Please, read about this reaction in good organic chemistry textbook, especially about what controls geometry of products, it will help to understand what I'm going to say next.

Ozone has low energy of LUMO, so alkene part here provides HOMO, that is still lower then Ozone's LUMO. Oxygen from $OMe$ group donates some electronic density to $C-C$ double bound, so this bond becomes more active, then opposite one.


Nomenclature of ester with cyclic substituents I have to tell the name of the following compound

The name is cyclopropylmethyl cyclobutanecarboxylate. All of it makes sense but I don't understand why the suffix carboxylate comes? Why are we writing the single carbon left unaccounted for by using carboxyl ? Kindly clarify.

The names of esters are usually formed by placing the alcoholic component in front of the name of the acid component as a separate word. The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

P-65.6.1 General methodology
Neutral salts and esters are both named using the name of the anion derived from the name of the acid. Anion names are formed by changing an ‘-ic acid’ ending of an acid name to ‘-ate’ and an ‘-ous acid’ ending of an acid name to ‘-ite’. Then, salts are named using the names of cations, and esters the names of organyl groups, cited as separate words in front of the name of the anion.

Since the acid component of the ester that is given in the question is cyclobutanecarboxylic acid, the corresponding ester is named as … cyclobutanecarboxylate.
However, that does not completely answer your question why the extra carbon atom is named using the ‘carboxy’ part.
Simple carboxylic acids are named substitutively using the suffix ‘oic acid’ as follows.

P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

Thus, the ‘carboxy’ part is not used in such simple cases. For example, pentanoic acid is not named ‘butanecarboxylic acid’.
However, the acid component of the ester that is given in the question is not such a simple case since the $\ce{-COOH}$ group does not terminate the chain of the parent structure (the cyclobutene ring). In such cases, the suffix ‘carboxylic acid’ is used instead.

P-65.1.2.2 The suffix ‘carboxylic acid’ is used for all carboxylic acids not covered by P-65.1.2.1, except for benzoic acid, a retained name (see P-65.1.1.1). The carboxy group can be attached to any atom, carbon or heteroatom, of any parent hydride; (…).


Wavelength extention in AAS Theoreticly, the width of the spectral line in  AAS (atomic absorbtion spectroscopy) is 10^-5, but in fact there happens an extention of it and it becomes 0.002-0.005. There are some factors which lead to this extention. Most common types of extentions are :

Doppler extention
Pressure extention
Self-absorbance extention.

Does anyone know how to explain in a simple way why do these 3 kinds of extentions take place?

Do these deviations happen all the time in these measurments or are they some deviations that can happen in case the concentration of atoms is to high?

From this source broadenings occur due to

  Doppler effect
  This effect arises because atoms will have different components of velocity along the line of observation.

  Lorentz effect
  This effect occurs as a result of the concentration of foreign atoms present in the environment of the emitting or absorbing atoms. The magnitude of the broadening varies with the pressure of the foreign gases and their physical properties.
  Quenching effect
  In a low-pressure spectral source, quenching collision can occur in flames as the result of the presence of foreign gas molecules with vibration levels very close to the excited state of the resonance line.
  Self absorption or self-reversal effect
  The atoms of the same kind as that emitting radiation will absorb maximum radiation at the center of the line than at the wings, resulting in the change of shape of the line as well as its intensity. This effect becomes serious if the vapor, which is absorbing radiation is considerably cooler than that which is emitting radiation.

As Greg pointed out in the comment, You can find more detailed explanation on the topic here

Pressure broadening should be dependent on the concentration of atoms. The other type of broadening shouldn't be dependent on the concentration


How to turn oil into in a gel? I've seen some EPR experiments where they add gelatin to water with spin-probes in it, as it "hardens" it immobilizes the spin-probes so that you can check effect of tumbling rate on the EPR. Cool
I would like to do the same with asphaltene in crude oil.... So I'm looking for a chemical that I can add to oil that will turn it into a more-or-less solid black gooey mess. It shouldn't be something too reactive though, because I don't want it to change the chemical aspects of the asphaltene (e.g. not make it self-aggregate). Just lock it in place.
Thoughts?

Put your asphaltene source on a silica gel column and elute with hexane and then hexane/toluene mixture to remove paraffins (saturated compounds), aromatics, and resins (small polar compounds). At last, remove the asphaltenes (high molecular weight polar compounds) with pure toluene, evaporate to dryness (a brittle black solid), then with heat you can disolve/disperse the asphaltenes in a simple high molecular weight neutral material like petroleum jelly and let it cool. This will give a high and/or adjustable concentration of what you want to look at. https://en.wikipedia.org/wiki/Asphaltene


How to figure out which phase on a phase diagram will be the most dense? If I am given a phase diagram, how can I determine which phase will have more density?

What should I consider to solve these kinds of questions?

Does the answer have to do with the area that each phase occupies?  Or the more area a phase occupies, the more the density?

The solid phase is more dense than the liquid phase. The line that separates solid and liquids bends right.

If the solid phase is less dense than the liquid phase, the line that separates solid and liquids bends left.