Why is there a need for resonance? Why are we able to show structures of compounds with different position of electron, but with same position of them? Shouldn't the structure become unstable due to this?

Let's take benzene for the example of resonance.We know that in reality benzene has no pure single or double bond. How can this be possible?

The bond length for $\ce{C-C}$ and $\ce{C=C}$ bonds in benzene is same which is $1.39~\mathring{\mathrm{A}}$. This is due to partial single and double bond between carbon atom, which is due to resonance. Why is there a resonance? Why can't benzene be happy with three alternate single and double bond between each carbon atom?

As I mentioned in my comment, the Wikipedia article on resonance provides a rather nice description of the subject where the Resonance in quantum mechanics section is of particular interest for the question asked.

However, I decided to somewhat clarify the matter emphasising few important points. And I will start by quoting the OP.

  We know that in reality benzene has no pure single or double bond.

Now please, take a deep breathe and repeat with me: there are no bonds. In reality we do have electrons and nuclei that constitutes a molecule, but a chemical bond is merely a theoretical concept introduced in the framework of a scientific theory. In other words, chemical bonds do not exist in the same sense as electrons and nuclei do.

For instance, the concept of resonance was introduced by Linus Pauling in the framework of the so-called valence bond theory, according to which

  a covalent bond is formed between the two atoms by the overlap of half
  filled valence atomic orbitals of each atom containing one unpaired
  electron.

So, this is the way the concept of chemical bonding enters the scene in the valence bond theory: a chemical bond is used to describe overlapped atomic orbitals of two atoms in a molecule. Note that atomic orbitals also do not exist in the same sense as electrons and nuclei do, but an orbital has at least some strict mathematical and physical meaning, while a chemical bond, as introduced above, does not. Read the above quote carefully one more time: it is not even a definition! What we just say is that whenever half filled valence atomic orbitals of two atoms overlap a covalent bond is formed, but we do not say what exactly is a covalent bond. It is just something; something which is formed whenever orbitals overlap. It is a concept, purely theoretical construction. It is practically helpful in description of a reality, but no one will say that it is actually a part of it.

Do not be afraid of all this business, because it is the way science generally works. Scientific theories basically tell us how nature could be described, and they do so in the language of mathematics, since it is the only language nature can unambiguously "speak" with us. However, it is often so tempting to force it to speak with us in a different language, in a language of our theoretical concepts (such as, for instance, the concept of a chemical bonding), no matter how fuzzy they are. This is a very dangerous road, because, after all, who are we to tell nature which language to speak? Nature will likely refuse to "speak" with us in that language, and that would be it.

That is what actually happened with the valence bond theory as it was originally formulated. Indeed, all carbon-carbon bond lengths are equal in benzene, but the valence bond theory were not able to account for that simple fact. Nature refused to "speak" with us in the language of chemical bonding. At that point chemists had two choices: either abandon the valence bond theory or introduce another theoretical concept into it, introduce a new word in our language making it possible to communicate with nature in this language. The concept of a chemical bonding was already so tremendously useful in chemistry that the choice was actually predetermined, and in 1928 Linus Pauling introduced the concept of resonance between several valence-bond structures of a molecule.

What is resonance? Is it a physical process? No way! It is just a concept, a word, used to describe

  delocalized electrons within certain molecules or polyatomic ions
  where the bonding cannot be expressed by a single Lewis formula.

The word "resonance" was probably not the best choice by Pauling, since it gives an unwanted feeling that the molecule actually do oscillate back and forth between the structures, which is, as I said, not the case. But putting aside its somewhat misleading name, the concept of resonance is as good as the concept of chemical bonding and many other concepts in chemistry and other sciences.


What happens to the crystal structure of ice as it gets colder and colder Not sure if this belongs on Chemistry or Physics stackexcahnge.

As water freezes into ice, it turns into a solid, and expands as it forms a crystal structure.  But what happens to this crystal structure as it continues to cool?  For example, if you take ice and cool it down to -50C, how does the crystal structure alter?  -100C?

I would assume that it contracts, similar to any other crystal as it cools.

I'm familiar with the different phases of ice, which require different amounts of pressure and such to create.  For my example above, I'm talking about taking a cup of water, and slowly freezing it over time with roughly no change to the pressure in the room.

Well, the easiest way to address this question is take a look at the phase diagram for ice. Ordinary ice as we know consists of lots of hexagonal rings of water molecules. First off, as temperature is decreased well below $0~^\circ$C, there will be a slight contraction of the lattice due to decreased thermal contribution to vibrational frequencies. This will cause the average $\ce{OH}$ bond length to decrease which will cause a subsequent contraction of the average $\ce{OO}$ distance.

It is not obvious in any way that there should be another phase transition at low temperatures and pressure for ice Ih, but it turns out there is. One can understand why there is a phase transition by observing that in the low temperature limit, entropic effects will make a diminishing contribution to the free energy of any state, and the ground state will in the low temperature, low pressure limit will simply be the global minimum of the potential energy surface.

Now, experiments demonstrate that while the oxygen atoms in ice Ih are all arranged in these hexagonal rings, the hydrogen atoms are essentially distributed randomly subject to the conditions that two hydrogen atoms are covalently bound to each oxygen atom, each oxygen receives two hydrogen bonds, and no two hydrogen atoms point towards each other. These are known as the ice rules (or sometimes Bernal-Fowler rules). This, however, is not the most stable structure that one can imagine as the protons are disordered, and hence there is not optimal cooperativity in the direction of the hydrogen bonds.

For this reason, people thought there might be a low-temperature phase of ice which is proton-ordered. That is, you have the same lattice as in regular ice Ih, but the protons all point in the same direction when forming hydrogen bonds. This is easier to see, so I have put a picture of the lattice below.

This phase of ice is called ice XI. I think they are numbered based on the order they were discovered, which is very annoying.

In practice, even though this is essentially the ground state of bulk water, it is very uncommon to find in nature, and very difficult to synthesize in a lab. This is because at such low temperatures, the rates of proton rearrangements are very slow. These rearrangements occur by proton transfer in ice, or by rotating to rearrange which oxygen the hydrogen bonds are donated to. These processes individually have high barriers, and hence have low rates even at relatively high temperatures and pressures. So, going from ice Ih to ice XI occurs much more slowly than most phase transitions.

You can read about how ice XI is prepared in practice on the wikipedia page for ice XI, but basically you introduce defects which act as catalysts for the rearrangements for the hydrogen atoms.

TL;DR:
To be really pedantic, the answer to your question depends on how long you are willing to wait as you keep cooling down the block of ice. That is, if you have an ice cube and cool to e.g. 5K, then nothing will happen besides a slight contraction of the lattice. If you keep it at that temperature for, literally something like 1,000 years (evidence from Antarctic ice cores), then your ice cube will look exactly the same, but microscopically, the protons will be organized much more neatly. How pleasant.


Can 32% hydrochloric acid be stored in a milk jug (HDPE)? Can $32\%$ ($20$-degree Baume) hydrochloric acid (typically found as 'muriatic acid' in hardware stores) be stored in a milk jug?

I have read that acids can be safely stored in polyethylene bottles. The milk jug I have has the HDPE resin ID code on it. However looking at Wikipedia, it is clear that HDPE is used to make a wide range of products with different "strengths". Given the "squishiness" of milk jugs, I would assume it is on the lower end of the "strength" scale. Does this make it any less suitable for holding diluted hydrochloric acid?

Context: I plan on etching some circuit boards as per this tutorial. And I am looking for places to store the resulting copper (II) chloride solution.

$$
\ce{Cu + 2HCl + H2O2 -> CuCl_2 + 2H2O}
$$

I am assuming that if the container can hold hydrochloric acid, it can hold the resulting copper(II) chloride solution.

There is more than just material composition to container selection. Yes HDPE is compatible with hydrochloric acid, but you have to consider rigidity and failure probability. Think: would you want to store Concentrated Hydrochloric acid in a 2 mil bag? Probably not because the probability of puncture and resulting loss/cost from the spill would be sufficiently high. Though not as extreme your milk jug is along those lines as it is softer and less rigid as the original container.

There is also the issue of legal considerations. In the US I'm sure that milk jug would not meet the DOT requirements for transport, and OSHA prohibits using containers typically used for food as being used for chemical or waste storage.

That said for home use only, you will probably be okay to store it, certainly not recommended, but this is the real world. I would at least recommend that if you do use a milk jug, you at least use a basin that can hold all of the contents for secondary containment in case of a leak which milk jugs occasionally do  AND certainly label the contents.


How can I balance the following equation atomically and electrically? $$\ce{C2O4^2- + MnO2 -> Mn^2+ + CO2}$$

I think that the half reactions are

$$\ce{C2O4^2- -> CO2}$$
$$\ce{MnO2 -> Mn^2+}$$

I am supposed to balance these by adding water, $\ce{H+}$ atoms and by adding $\ce{e-}$’s, but I’m just not sure on the method to do this as we’ve covered it extremely quickly.

First, I found the oxidation numbers for the overall equation, and I think that $\ce{C2O4^2-}$ is the reducing agent because $\ce{C}$ is losing charge from +3 to +4, I just don’t know how to use that to balance this. Any help would be appreciated.

Deal with the two half equations separately and then combine them.

Starting with the oxalate equation: $$\ce{C2O4^{2-} -> CO2}$$

Balance the atoms: $$\ce{C2O4^{2-} -> 2CO2}$$

Now balance the charge by adding electrons: $$\ce{C2O4^{2-} -> CO2 + 2e-}$$

Now for the manganese dioxide reduction: $$\ce{MnO2 -> Mn^{2+}}$$

We can balance the half equations by adding water, hydrogen ions or electrons. Since water is the only one which contains oxygen we should add this first: $$\ce{MnO2 -> Mn^{2+} + 2H2O}$$

Now add hydrogen ions to balance the hydrogens: $$\ce{MnO2 +4H+ -> Mn^{2+} + 2H2O}$$

Finally add electrons the balance the charge: $$\ce{MnO2 +4H+ + 2e- -> Mn^{2+} + 2H2O}$$

Now combine the equations to cancel all the electrons. In this case there are two on both sides so we don't need to multiply any of the equations by anything.

$$\ce{MnO2 +4H+ + C2O4^{2-} -> Mn^{2+} + 2CO2 + 2H2O}$$


Acidic conditions, high heat, and return to carboxylic acids Apparently carboxylic acid derivatives such as cyanohydrins can be taken back to "daddy" - a carboxylic acid - under the right conditions. This involves acid, water, and "high heat." What, however, is the mechanism of a cyanohydrin going back to daddy?

In the cyanohydrin reaction aldehydes or ketones can react with HCN to form a cyanohydrin.  Inorganic cynanides and acid can be used in place of HCN, to generate the required cyanide anion.   The reaction proceeds by typical addition of the nucleophilic cyanide anion to the carbonyl carbon to form the tetrahedral intermediate, which can ultimately be protonated to yield the corresponding cyanohydrin.

Once formed, the nitrile group of the cyanohydrin can undergo acidic (or basic) hydrolysis by the same mechanism as any other common nitrile.  The only difference in the case of a cyanohydrin is that the "R" group contains a hydroxyl substituent on its alpha carbon.

A useful variant of cyanohydrin formation is the Strecker amino acid synthesis.  Instead of forming a cyanohydrin, an aminonitrile is formed and the nitrile is hydrolyzed to the corresponding alpha-amino acid.  A detailed mechanism can be found in the above link.


How come one of the hydroxides in the phosphate group are more acidic than the other? Three phsiologically relevant $\mathrm{p}K_\mathrm{a}$ values are associated with psilocybin: 1.3, 6.5 and 10.4. Suggest the appropriate charged species and mark the $\mathrm{p}K_\mathrm{a}$ values next to the corresponding protons in your structure.

How come one of the hydroxides are more acidic than the other in the phosphate group? Is because of electron repulsion? Can the substituents on a phosphorus group not move around? Why are the two hydroxide groups on the phosphorus atom not the same functional group?

Because the rendering is not really accurate.  It would have been better to render the two dissociation constants without trying to associate them with specific hydrogen atoms.

Given the neutral phosphate group with both hydroxyl hydrogens attached, either one has an equal probability of being removed as a proton, corresponding to the lower of the two $\mathrm{p}K_\mathrm{a}$ values.  The drawing simply picked one of the protons at random.  Then with that proton removed and the functional group now negatively charged, the second hydroxyl proton has become harder to remove and that is what the higher $\mathrm{p}K_\mathrm{a}$ represents.


What does washing and drying a solution mean? After combining two solutions, a direction in a 1970s British chemistry book says to wash the new solution with a mixture of baking soda, table salt, and water. Then it says to dry it with anhydrous magnesium sulfate.
What does washing and drying mean? Do the chemicals get physically mixed together? Doesn't that contaminate the solution?

This is standard for purifying substances.
To wash means to add your product solution to an aqueous solution (or just water, but frequently a saturated solution) to a separatory funnel. After shaking, you drain the lower layer (which is usually aqueous). This process removes water soluble impurities. This is frequently repeated.
Drying is accomplished by taking the organic solution of your product and mixing it with a desiccant (usually anhydrous sodium sulfate or anhydrous magnesium sulfate). This removes any water that was left from the washing in the above step. The desiccant may then be removed by filtration.


How to determine an unknown metal based on the reaction products with oxygen and fluorine? The following question is, as it will probably be obvious, a homework question, however I have absolutely no idea how to progress with it beyond what I have already done. This is the question:

  $\pu{10 g}$ of an unknown metal reacts with oxygen to produce $\pu{16.58 g}$ of the metal oxide. $\pu{10 g}$ of the same metal reacts with fluorine to produce $\pu{25.63 g}$ of the metal fluoride. The valency of the metal is identical in the metal oxide and fluoride produced. Identify the metal after finding the mass of one mole.

My attempt:
\begin{align}
  m(\ce{O2}) &= \pu{ 6.58 g} & \implies&& n(\ce{O2}) &= \pu{0.206 mol}\\
  m(\ce{F2}) &= \pu{15.63 g} & \implies&& n(\ce{F2}) &= \pu{0.411 mol}
\end{align}

I also made the equations, where X was the unknown metal:
\begin{align}
  \ce{X + y/2 O2 &-> XO_y}\\
  \ce{X + z/2 F2 &-> XF_z}\\
\end{align}

Any help with where to go next would be massively appreciated.

Your answer is Magnesium. How you get there requires you to first realize that the number of moles of Fluorine is double the number of moles of Oxygen. Since Oxygen will accept 2 electrons and Fluorine will accept one electron, you can deduce based off oxidation states and the number of moles that this metal has a 2+ charge when it is a cation. From this, you can write your chemical equations out X+1/2O2-> XO and X+F2->XF2 . Moving on, you can do some basic  stoichiometry to determine you have .412 moles of X present and use that fact that X's mass is 10g to determine the molar mass of 24.272 g/mols. Then look at your Periodic Table to find what corresponds to 24.272 g/mol and Magnesium does. Hope this helps


Why does Cl- corrode aluminum containers? I heard that $\ce{Cl^-}$ will corrode aluminum containers. Why? Is it because it would form a galvanic cell or produce coordination compounds?

In general, it should not. Under the right conditions, i.e. acidic conditions at high temperature on a freshly exposed (e.g. scratched) $\ce{Al}$ surface, you could get some oxidation to $\ce{AlCl3}$.  One industrial process for producing $\ce{AlCl3}$ is via the reaction:

$$\ce{2 Al + 6 HCl → 2 AlCl3 + 3 H2}$$

at $\mathrm{600-800^oC}$. It would of course happen much more slowly, if at all, at lower temperatures and lower acidity.

The necessity of a freshly scratched surface is to expose the aluminum metal itself, as aluminum pretty rapidly forms a thin, inert oxide layer upon exposure to air or water.

In other words, in the absence of this perfect storm of conditions, aluminum cans should be very resistant to corrosion by $\ce{NaCl}$.


How to find the empirical formula for an nitrogen oxygen compound from the given mass of nitrogen and the product? $5.00~\mathrm{g}$ of nitrogen is completely converted into an oxide of nitrogen. The mass of the oxide formed is $19.3~\mathrm{g}$.
  The empirical formula of the oxide would be?

My working:

$$\ce{N2 + O_2 ->  N$_x$O$_y$}$$

$m(\ce{N2}) = 5~\mathrm{g}$
$n(\ce{N2}) = 5/28~\mathrm{mol}$
$n(\ce{N$_x$O$_y$}) = 19.3~\mathrm{g}$

How do I approach this?

$\ce{x/2 N2 +y/2 O2 \rightarrow N$_x$O$_y$}$

We have two unknowns $x$ and $y$. So, to determine them we need two independent  equations:

1 - The conservation of nitrogen mass: $5=14x$. So,  $x= 0.35714$.

2- According to the stoichiometry of the above balanced chemical equation: The number of moles of  $\ce{ N$_x$O$_y$}$ equals $n= \frac{5}{28}\times\frac{2}{x}$.

On the other hand, the number of moles of  $\ce{ N$_x$O$_y$}$ equals   $n= \frac{19.3}{14x+16y}=\frac{10}{28x} $.
By substituting $x= 0.35714 $ into the last equation, we find $y=0.89375$.
We find $\ce{ N_{0.35714}O_{0.89375}}$. We divide by $ 0.35714$, We obtain  $\ce{ NO_{2.5}}$.i.e. the formula of nitrogen oxide is $\ce{ N_2O_5}$.


A defining feature for acids / bases I know of 3 different definitions for acids and bases including Arrhenius, Brønsted–Lowry and Lewis. Can anyone tell me what is the common feature among these 3, that we have to label the compounds under those definitions as acids or bases? Is there any defining physical or chemical feature that guides this?

tl;dr version: You can only define an acid or a base in context of a reaction. Individual compounds are not acids by themselves. They are acids in reactions in which they lose protons (or gain electron pairs).

The Lewis definition is the broadest acid-base definition. Each of the others are specific subsets.
The Arrhenius definition is the simplest:

Acid = something that creates $\ce{H+}$ ions in water.
Base = something that creates $\ce{OH-}$ ions in water.

Thus any compound that can ionize in or react with water, like $\ce{HCl}$ or $\ce{HCO2H}$ is an acid and any compound that contains hydroxide ions or reacts with water to produce them, like $\ce{NaNH2}$, is a base.
However, the Arrhenius definition breaks down in non-aqueous media.
The Brønsted-Lowry definition is broader and more inclusive.

Acid = proton donor - All Arrhenius acids are Brønsted-Lowry acids
Base = proton acceptor - All Arrhenius bases are Brønsted-Lowry bases

The Brønsted-Lowry definition allows us to extend our discussion of acids and bases to, say, solutions in methanol $\ce{CH3OH}$. Note that the Brønsted-Lowry definition requires a proton transfer reaction to be occurring to properly identify acids. For example, water is amphiprotic - it can react as either an acid or a base. Many other compounds are similar. There is no absolute "this is an acid always". Instead, you have "this compound is an acid in this reaction". For example, water is an acid in the first reaction and a base in the second.
$$\ce{H2O + H- -> H2  + OH-}$$
$$\ce{H2O + H2SO4 -> H3O+ + HSO4-}$$
The Lewis definition is broader and more inclusive still. It allows extension of acids and bases into aprotic media and situations:

Acid = electron pair acceptor - all Brønsted-Lowry acids are Lewis acids
Base = electron pair donor - all Brønsted-Lowry bases are Lewis bases

Again, the Lewis definition requires a reaction to occur in order to classify compounds.
However, in all cases, there are compounds that behave as acids much more frequently than they act as bases, and vice versa. For example, $\ce{HNO3}$ is almost always an acid, but not exclusively. The first step of the mechanism of aromatic nitration is considered to be a proton transfer step between sufluric acid and nitric acid in which nitric acid is the base.
$$\ce{HOSO3H + HONO2 <=> HOSO3- + O2NOH2+}$$


What causes "stringy" melted cheese (and similar behavior in other materials)? I was wondering about this. What is the physics underlying the "stringy" consistency of melted cheese? That is, when the melted cheese is hot, and it is pulled apart, it forms strings of cheese.

This behavior also occurs with a number of other fluid substances. It seems to happen with polymer substances (like cheese which is made of proteins like casein, also plastics when melted), gums, etc. What is the technical term for this kind of behavior and what is the general mechanism by which it happens?

When cheese is heated to 140 degrees fahrenheit, the proteins line up in rows.  Soaking the cheese in a salt solution before pulling it helps to stiffen it and keep it pliable, as the salt neutralizes acids which could cause the long lines of protein to break.

Polymers, similar to cheese, can be stringy because their molecules are very large (macromolecules), and tend to entangle with one another.  @IanF1 is correct in saying this may be a chemistry question.  Here is a good explanation of polymerization:

http://courses.chem.psu.edu/chem112/materials/polymers.html


Can the degree of dissociation of a weak acid be more than 1? My common sense dictates that it can't be more than 1 because after all it is the number of moles dissociating per mole of a substance.
But I was reading about Ostwald's law of dilution and came across this formula:
a (degree of dissociation) = √K/√C
So if C is less than K by about 4 times, a becomes 2.
I'm sensing there's an extremely simple solution to this question but can't realise what it is.

The $\sqrt K/\sqrt C$ ratio is only an approximation.  Let's do something more exact.  (Why do I say "more exact" instead of "completely exact"?  Read on, nature has a nasty curveball in store for us!)
Let $C$ be the nominal concentration of the acid, assumed monoprotic ($\ce{HA}$) for simplicity, and $K$ be its dissociation constant.  So, recognizing all ions are of course solvated, we have
$[\ce{H^+}][\ce{A^-}]=K[\ce{HA}]$
$[\ce{HA}]=C-[\ce{H^+}]$
$[\ce{A^-}]=[\ce{H^+}]$
So
$[\ce{H^+}]^2=K(C-[\ce{H^+}])$
$[\ce{H^+}]^2+K[\ce{H^+}]-CK=0$
From the quadratic formula
$[\ce{H^+}]=\dfrac{-K+\sqrt{K^2+4CK}}{2}=\dfrac{2CK}{K+\sqrt{K^2+4CK}}$
And our dissociation is
$\dfrac{[\ce{H^+}]}{C}=\dfrac{2K}{K+\sqrt{K^2+4CK}}=\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{C+K/4}}$
Now, if $C$ is large compared with $K$ then the terms in the denominator with $K$ become small compared with $\sqrt C$ and we simplify to $\sqrt K/\sqrt C$.  But if we dilute the acid, then the terms with $K$ in the denominator become dominant and we find a  limiting value of
$\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{C+K/4}}<\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{K/4}}=1$.
Thereby the dissociation remains limited to no greater than 100%.
There is one little thing we forgot.  If we keep diluting the acid so its concentration approaches the ion concentrations from water autoionization, then we also have to count the hydrogen ions from the autoionization into the equilibrium.  This extra source of hydrogen ions cuts down on the acid dissociation, so the limiting amount of dissociation is less than 100%.  For acetic acid it's still about 99.4%, but if we pick a really weak acid like $\ce{HCN}$ with a $pK_a$ above $7$ at ambient temperature, we never get more than a small fraction to dissociate unless we force the issue by reacting with a base.


Etching ITO on PET without damage I am planning to use some of this to construct a transparent touch panel for a prototype of mine. It is PET plastic coated in Indium Tin Oxide.

My plan is as follows:

Use some of this UV-cure dry film, a transparency, and a fluorescent light fixture placed very close to it for a while to cure it (not sure if enough UV there at the right wavelength...I may get one of those UV LED nail spa cure things).
Develop the film, remove excess with water.
Use some of this 32% HCl to etch the ITO.
Remove the film.
Enjoy my completed touch screen.

The problem is, I'm not really a chemistry person. My background is electrical engineering and my experience is limited to one semester in a cleanroom class where we did a lot of SiO2 etching (BOE) and some aluminum and nickel etching.

My questions are as follows:

Will dry film work well for this application? I worry that since it is designed for use with PCBs, made of the relatively stable FR-4, the etchant and remover provided with the kit I linked to will be too harsh and damage the PET or ITO
This paper seems to note that $\ce{HCl}$ is quite slow at $20^{\circ}$C for etching ITO. I estimate the molarity of the solution above is probably between 8-12M, giving me an etch rate of ~2-4angstroms/sec. I'm not sure what the thickness of the film is on the plastic, so I could be there a while. How can I determine the thickness of the ITO given its sheet resistance of 50 ohms? (What I'm really asking is how to determine the resistivity so I can solve for the thickness of the film.) I don't have access to anything fancy like a 4-point probe.
This chemical information sheet states that PET plastic does great with $\ce{HCl}$ at 10%, but terribly when "concentrated" (whatever that means). Am I going to destroy the PET during this process?
Is my plan fundamentally flawed? Is it likely not going to work? Do I need to use a different photoresist or etchant (aqua regia is terrible to PET according to that info sheet)?

I'm not too concerned with overetching or overdeveloping since the pattern doesn't need to be incredibly precise (this is a one-off thing and the contacts are around a half-inch square each).

Assume that I don't have access to a laser for ablation, fancy deposition equipment, or any complex measurement equipment (I've got a multimeter and oscilloscope). I can acquire basic glassware through amazon and such AFAIK.

My lab has a lot of experience with ITO-coated PET, but unfortunately I don't think anyone's tried etching it.
We typically etch ITO-coated glass with a mixture of hydrochloric acid/water/nitric acid (4:2:1 by volume) which is fairly fast at room temperature, going through a few µm in about 10 minutes and mask with a positive photoresist.

To your questions:

Will dry film work well for this application?

Because the resolution you need isn't great, it should work fine, so long as the film adheres properly to the ITO.
Underetching is a possibility, but if the features are big enough it should be okay.
The big issue here is trying to use a normal fluorescent light to expose the film.
This is probably hopeless; you're better off finding a proper UV lamp or even using the sun.
How can I determine the thickness of the ITO given its sheet resistance of 50 ohms?

The mixture I mention above is fairly quick, but either way, I wouldn't worry about the precise film thickness (which may actually vary somewhat depending on the batch/supplier).
The easy way to deal with this is to simply pull the sample out of the etchant occasionally and test the resistance with a multimeter in a region where the ITO should be etched away.
On PET, the conductivity should be negligible when the ITO is gone.
Am I going to destroy the PET during this process?

The datasheet you linked refers to the use of PET containers for storage of chemicals—it may actually last a few hours without significant damage, but I haven't tried it.
Is my plan fundamentally flawed?

I think it's worth a shot.
You will need to handle the ITO sheets fairly delicately—the ITO can crack if the sheet is flexed too much, but I don't see this method as being totally unworkable, though you may have issues with the quality of etching you'll get.

As a final note, I wouldn't worry about getting special glassware.
For measuring simple ratios and etching/developing baths, kitchen quality Pyrex dishes and measuring cups work just fine for this kind of thing.
We keep stacks of them in the cleanroom because they're nice and cheap.


Bromination of alkenes In regard to the mechanism of halogen addition, consider the situation when we add a halogen (say, bromine) to an unsymmetrical alkene about the double bond (say, propene) in the presence of $\ce{CCl4}$.

I knew the mechanism would be something of this sort:

  A cyclic bromonium ion would be formed with the two carbons of the double bond. The second bromide ion (acting as a nucleophile) attacks on the more substitued carbon from the side opposite to the bromonium ion to cause ring opening and formation of a vic-dibromide.

My teacher however tells me that other than ethene, no other alkene forms bridges like these, because of unstability of bromonium complex due to repulsion between substituents on carbon and the bridged element.

I have two questions in this regard:

Which of the two is the correct mechanism?
If the former is indeed the correct mechanism, then would the nucleophile  attack the least substituted carbon? (seeing as how the partial positive charge on the more substituted carbon is nuetralised to a greater extent due to $+I$ effect of attached methyl/ethyl groups).

The first mechanism is more likely unless specific conditions of the experiment dictate otherwise. In the presence of $\ce{CCl4}$, the addition of $\ce{Br2}$ takes place via the formation of a cyclic halonium ion as intermediate as follows:


Oxidation reaction of dimethylselenide / Selenourea with hydrogen peroxide For my paper,  I need to find the reaction between $\ce{(CH3)2Se2}$ and $\ce{H2O2}$ and between $\ce{SeC(NH2)2}$ and $\ce{H2O2}$. The mechanism below shows how I think it goes. The reaction should give an anion, an inorganic form of the selenide or selenourea but I'm unsure when the forms become anionic.

And

$$\ce{\overset{+IV}{Se}O3^2- + H2O2 -> \overset{+VI}{Se}O4^2- + H2O}$$

If anyone could give me a push in the right direction or can confirm if the reactions in the picture are correct, I'd appreciate it very much.
Thanks in advance!

I'm not familiar with selenium chemistry but am trying to help with the oxidation reaction of selenourea with hydrogen peroxide part. Here what I found on that reaction in literature (Ref.1):

  Abstract: Reactions of biological oxidizing agents, such as hydroxyl radicals ($\ce{^•OH}$), singlet oxygen ($\ce{^1O2}$), hydrogen peroxide ($\ce{H2O2}$), and peroxynitrite ($\ce{ONOO-}$) with selenourea were studied. The kinetics of the reactions was followed using time-resolved techniques, and the bimolecular rate constants were determined. In all these reactions, under aerated conditions, elemental red selenium was produced as one of the reaction products. The average size of the selenium particles could be controlled and stabilized in the range of $\pu{20−100 nm}$ with the addition of bovine serum albumin (BSA) or sodium dodecyl sulfate (SDS). The particles were characterized by dynamic light scattering studies (DLS), which revealed that the size and distribution of the particles depended mainly on the amount of selenourea undergoing oxidation. Other factors such as the nature of the oxidant and the concentration of the stabilizer also are important in stabilizing the particles. Nanoselenium-reduced $\ce{ABTS^{•-}}$ to colorless $\ce{ABTS^2-}$ ($\ce{ABTS} =$ 2,2‘-azinobis(3-ethylbenzothiazoline-6-sulfonate) and oxidized dichlorodihydrofluoresecein diacetate (DCFA) to fluorescent dichlorofluorescein (DCF) indicating its ability to participate in redox and free radical reactions. The reactivity of selenium nanoparticles with these systems varied linearly with the surface area of the particles. The studies demonstrate that selenourea undergoes oxidation with both one-electron and two-electron oxidants to produce elemental selenium, which, on stabilization to nanometer size, exhibits size-dependent redox activity.

The reaction mechanism is depicted in following Scheme, which is directly from Ref.1:

Hope this help.

References:

B. Mishra, P. A. Hassan, K. I. Priyadarsini, Hari Mohan, “Reactions of Biological Oxidants with Selenourea:  Formation of Redox Active Nanoselenium,” J. Phys. Chem. B 2005, 109(26), 12718–12723 (DOI: 10.1021/jp051328n).


How to write a cube file? This is a technical question. I have a real-space representation of a potential in a grid of points, however I have the grid in the format:

x  y  z  w  v

meaning: the coordinates of each point, the weight and the value of the potential.

I would like to visualize this potential but I need to re-write this potential in a cube file. I haven't found any particular software or script to do that.

I know how to get the header of the cube file. However I don't know how to re-write the points.

Is there a specific format to write the points?

Thank you

I would recommend using the functional PySCF cubegen script as a starting point for your own script. (https://sunqm.github.io/pyscf/_modules/pyscf/tools/cubegen.html)

If we have a grid defined with nx ny and nz points along the three axes, generating the cube file is just a nested loop over z, y, and x. (In that order) You should end up with nz lines of output each with nx*ny entries.


What is the difference between a ligand and a co-ligand? I've seen the word co-ligand pop up in several papers. Is there a particular difference between a co-ligand and a ligand?

I believe this refers to situations in tranistion metal complexes where there is more than one kind of ligand, and specifically, where one of the coligands could not ligate and endure by itself due to reactivity, such as a carbene being blocked from reactivity due to a coligands steric bulk


Why does radioactive decay occur? Is there a reason radioactive decay is exponential, as in after an extremely long time some of the original compound remains, or is this just an empirical model? Also, why does $\ce{^{238}U}$ have a much longer half life than $\ce{^{210}Pb}$, for example?

The simplified version of why radiation occurs is a balance between attraction of the protons and neutrons in the nucleus and the positive charge of the protons repulsing each other. Get the balance right, and you end up with a stable isotope, get it wrong, and the isotope is unstable and will decay.

As for the exponential decay: a nucleus falling apart is stochastic process, meaning it is only dictated by chance! When you take an unstable nucleus, after it's half-life there is a 50% chance that it will have decayed. Then, you can add more nuclei the chance of half of them decaying after it's half life is far, far larger than all of them being stable. The larger amount you start of with, the better this theory describes the decay. If you then measure a vast amount of atoms (a mole for example) the curve you find will describe an exponential decay.


Why is a C–D bond stronger than a C–H bond? I have heard that the carbon–deuterium bond is stronger than the carbon–hydrogen bond. What are the possible reasons for it? Is this also the reason that C–H bonds participate more in hyperconjugation than C–D bonds? Please explain.

Physics is better able to answer "how" questions than "why" questions, but here goes. The quantum mechanical description of the C–D system versus the C–H system gives the former a lower zero-point energy, which is the minimum energy the quantum system can attain. A good conceptual model of this is to consider the C–X system as two masses connected by a spring. When X = D, the system will vibrate more slowly than when X = H simply due to mass.

Since energy is proportional to the frequency of vibration, the energy of the C–D system is less (lower). This table lists the C–D bond dissociation energy as $\pu{341.4 kJ/mol}$ and C–H as $\pu{338 kJ/mol}$. Since this is the energy to break the bond, the C–D bond is stronger.

As far as your question about hyperconjugation; I'm surprised that C–H would have greater hyperconjugation than C–D, but it is not something that I know anything about. I guess I could speculate and wave my hands around and say that since the C–D bond is stronger, it keeps the electrons "closer" and so more confined to the sigma orbital, but that is just hot air.

Here's something which claims that the reason D does less hyperconjugation is because it vibrates less (see slide 15). So, it seems (if this is correct) that it's the smaller bond deformations that reduce the hyperconjugation for deuterated systems. Here is a pretty clear explanation of the kinetic isotope effect.


Why is lead(II) present in two groups of cation analysis? Well, I found that lead cation $\ce{Pb^2+}$ is present in both group 1 and group 2 of cation analysis. But why?

Simply because lead(II) fits both analytical groups:

it forms lead(II) chloride $\ce{PbCl2}$, a poor soluble chloride, which puts it in the 1st group;
it also forms lead(II) sulfide $\ce{PbS}$, insoluble in water and diluted acids, which also puts it in the 2nd group.

One reason for putting lead in both groups is that the chloride is somewhat of a borderline case in terms of being insoluble enough to appear in Group I.  Group II represents a more complicated analysis but is more sensitive for lead.


Is there any way to quantitatively find when a reaction has completed? So I'm designing an experiment for measuring how the reaction time of copper and ferric chloride is dependent on temperature; here's the reaction:

$$
\ce{FeCl3 + Cu -> FeCl2 + CuCl}
$$

The problem I have right now is how to quantitatively measure when the reaction has ended. I know you can visually see when a small copper rod has completely dissolved, but I wanted there to be a quantitative aspect to it rather than qualitative.

I've thought of using a pH probe to see if the PH changes after the reaction takes places and stays static, which I could leverage as an endpoint of the reaction, but the pH probes my school are damaged, I think, since the reading keeps jumping around by about 3 or 4.

If you have measured reaction time in general before, I'd love to hear how you did it.

Option 1: Spectrometry

Easiest method, in most cases accurate enough. If you know that there are no side-reactions, this is a safe bet. Simply measure some characteristic wavelength for one reactant (preferably $\ce{Fe^{3+}}$) and one product (I would go for the $\ce{Cu^{+}}$ there) and then make measurements of your reaction mixture on the fly.

Option 2: Titration

Not that recommendable if you need any reliable results, but if you are low on budget and/or want a didactic effect. Take a little sample from a big flask and do a $\ce{Fe^{2+/3+}}$-redox-titration.

Option 3: Electrolysis

Don't know how to implement that, but basically you would take the sample and measure the amount of electricity needed to convert all $\ce{Fe^{2+}}$ in that sample back to $\ce{Fe^{3+}}$. Probably more accurate than spectrometry as the measurements of currents can be done quite precisely

You could also do some weird stuff like quantitative chlorine-NMR, but I can not think of any benefit that you would have with that.


Why pipetting with a 30° angle? After seeing this youtube video, I was left to wonder why to use a $30^\circ$ angle between the flask and the pipette when emptying it. I can imagine it is important to touch the flask with your pipette when emptying, because the liquid prefers being attached to a solid surface due to capillary forces. When just emptying it from above the flask droplets would remain at the tip of the pipette. But why the $30^\circ$ angle? Is this an important standard or just arbitrary?

The angle of 30 degrees is not so important.

You have to make sure that the pipette is straight and the liquid runs along inside of the glass. You should not hold the pipette directly into the liquid. The drop in the tip has been included in the calibration. At the end you just have to wipe the pipette at the glass edge.


Using acetone instead of acetonitrile on reverse-phase prep-HPLC I've been using reverse phase prep-HPLC a LOT lately, and I'm growing increasingly concerned about the amount of acetonitrile needed. In an attempt to use a cheaper and more environmentally friendly solvent I had a read around and discovered you can substitute MeCN with acetone.

Has anyone ever tried this? How did it compare in a broad sense?

If you want to go completely green, you should use ethanol.
In many sectors it is considered the only green solvent (except for water of course).
Then, to be really green, you should make sure that it comes from fermentation, not from synthesis.
This quality exists.

The only problem with ethanol is the same as with acetonitrile, the price.
In this case it is due to the taxes that it normally has.
However, if you are working for some institution, it is probable that you can buy it tax free.

Another alternative is to use methanol.
It is not so green, but you can call it wood alcohol.
Methanol works very well in HPLC.
The only difference with acetonitrile is that the pressure in the system will be higher.
This is normally not a problem because the equipment is designed for it.

I prefer not to use acetone because it can give reaction with some compounds, or with itself to form isophorone or other condensation products.
As Waylander correctly points out in his comment, the UV cut off of acetone is also a problem in case you are using UV detection.

                 UV cut off (nm)
 acetonitrile        190
 methanol            205
 ethanol             210
 acetone             330

To change from one solvent to another you will have to change the gradient.


Comparing the nature of ionic compounds I want to know how to compare the strength or stability of ionic compounds.
Example : $\ce{Mg(HCO3)2}$ and $\ce{NaHCO3}$

I know that polarity and polarizability affect the ionic and covalent nature.

"Strength". Hmm, what firstly comes to mind is you want a comparison of lattice energies (or more precisely, lattice enthalpies)
Lattice energy is a [crude] way of estimating how strong the electrostatic attractions are between ions in an ionic compound. Let's take the formation of $\ce{NaCl}$ into account, as I see it's the most common "ionic" compound that teachers start teaching the lesson with.
According to Born-Haber cycle this is the reaction that results in the formation of pure table salt:
$$\ce{Na+(g) + Cl- (g) \rightarrow NaCl(s)}$$
Forming the salt crystals is exothermic and this is called the lattice enthalpy of formation.

Note that some textbooks mean to define lattice energy as the energy that is obtained by the thermodynamic system in the process of turning the solid compound to its gaseous ions.

To make a long story short, the more the lattice enthalpy is, the stronger "bonds" you get.
There are two main factors that affect the lattice formation enthalpy:

Charge of the ions
Radius of the ions

Charge of the ions:

Source
Look at the diagram above. $\ce{NaCl}$ is known to release approximately 822 kilojoules upon formation, while $\ce{MgO}$ releases 3890 KJ. And what does that mean? It means that Magnesium Oxide is way more "stable" than Sodium Chloride. So, complete the rest:

The more the charge of the ions is in an ionic compound, the _____ is its lattice enthalpy and the ______ its stability.

Radius of the ions:

Source
As you see, and could've imagined if you were on the right track, bigger ions (literally) mean weaker attraction. I wanted the answer to be fairly complete, so I did an addition to the main point that answers your question. Thus, there's no need of expanding this.
In your case:
You might as well have reached a conclusion. Mg ion has a 2+ charge, while Na has only a single electron missing. The anion is pretty much the same, $\ce{HCO3-}$ (hydrogen carbonate), and its priorly covalent nature shouldn't mislead you in doubting your answer. There is a case of comparison of the charge of the ions, and I would say Magnesium ion wins.
In case of additional study, you can visit:
http://www.chemguide.co.uk/physical/energetics/lattice.html
http://www.knockhardy.org.uk/sci_htm_files/08le.pdf
http://www.rsc.org/learn-chemistry/resources/gridlocks/downloads/answers/Born%20Haber%20NaCl%20ans.pdf

Addendum:
The charge and the radius of ions are by no means the only factors that affect the lattice enthalpy. It's just that the other factors are considered "trivial" by current textbooks.
The lattice enthalpy and lattice energy aren't the same. They do have close meanings, though.


Acetone absorbing water from the air Our company uses acetone and isopropanol in our manufacturing process to clean our product. Our operators get the chemicals from a large container and put it into small containers like this (http://www.soscleanroom.com/product/facilities/8-oz-esd-safe-solvent-dispenser-standard-pump/?gclid=CILep_rTzc0CFdBZhgodSWsEiA) to keep at their station.

Some times our operators say that it was more difficult to clean the product today and equate that to mean the chemicals were weaker that day.
Recently our operators stated that the chemicals seemed weak again but the next day they used the same chemicals and said they were fine now. This leads me to believe that the chemicals are fine and that there is a different problem leading to harder cleaning.

How fast does acetone or isopropanol absorb water from the air? How long will it take for the chemicals to be noticeably weaker if stored in the containers mentioned above?  Should we be dumping the chemicals everyday or would weekly dumping be fine?

The temperature and humidity is relatively constant at our factory.

For Acetone or IPA we are more than happy to keep it in normal squeeze bottles which are not air tight so the bottle you've got should be fine. I believe you're more likely to see it evaporate first than to see it get 'noticeably weaker' (I assume you're using technical grade acetone since it's just for cleaning). There should be no need to dump the acetone/IPA in the bottles on a weekly basis - although acetone is cheap its unnecessary to waste money like that!

Along the lines of trying to figure out what could be the cause of the experience of your staff, I doubt it has anything to do with the acetone itself because you say that they can use the same chemical on two different days and have two different experiences. So it's either they're imagining it (let's assume they are not), or it is the dirt that has to be cleaned that is the issue.

I've experienced it in the lab before that sometimes it does seem more difficult to wash out and clean glassware despite using the same chemicals - and this usually has something to do with how long the 'dirt' has been sitting in my flask or how it got there in the first place (slow precipitation/crystallization, crashed out, solvent dried out and left overnight, etc.). Without knowing what your product is and what it is you're trying to clean out, I can't offer much else other than checking out the consistency of the incoming product that needs to be cleaned. It could be a sign that the manufacturer's process is contaminated on some days.


What is the substance with the highest specific heat? I always learned that water has the highest specific heat, but I recently saw that Hydrogen has a specific heat as high as 14 cal/gC and helium has a specific heat of 5 cal/gC, which would be much higher than water? Is this true? and how can it be if there are so many "proofs" in nature , showing that water has the highest specific heat?

This may not be the absolute highest, but on a mass basis hydrogen gas has more than three times the specific heat as water under normal laboratory conditions.  Diatomic gases under ambient conditions generally have a molar specific heat of about $7\text {cal/(mol K)}$, and one mole of hydrogen has only $2\text {g}$ mass.  Thus $3.5\text {cal/(g K)}$ for hydrogen versus $1\text {cal/(g K)}$ for water.

Helium, contrary to the question, is $5\text {cal/(mol K)}$, not $7\text {cal/(g K)}$.


Relationship between thermodynamic reversibility and reactions Often we're concerned with changes in thermodynamic properties during reactions, like $\Delta H_p = Q_{reaction}$ (change in enthalpy for an isobaric process). What seems odd to me is that in the derivation of this relationship,

$\Delta H_p = \Delta U + P\Delta V=Q-W+P\Delta V$

we assume the work is given by $W=P\Delta V$, which requires that the process is reversible in the thermodynamic sense. Then the change in enthalpy simplifies to

$\Delta H_p=Q - P\Delta V + P\Delta V=Q$.

Supposedly, measuring the heat transfer during a reaction at constant pressure will give the enthalpy. Even though enthalpy is a state function in general, our case requires that the process is reversible.

To what extent does the restriction of thermodynamic reversibility apply to the reaction? Do we have to assume the reaction happens slowly and can be undone, leaving only an infinitesimal change in the universe? Or does it not matter?

Please correct me if any of my points are wrong. Classical thermodynamics usually doesn't mention kinds of matter, so I'm a bit confused when chemistry comes into it.

The process does not have to be considered reversible, and, if a reaction is occurring in the system, it is, of course, not reversible.  We are trying to find the change in enthalpy between the following two states:

State 1:  Reactants at temperature $T_0$ and pressure $P_0$

State 2:  Products at temperature $T_0$ and pressure $P_0$

We would like to measure the $\Delta H$ in such a way that, for the process path we select, all we need to do is measure the amount of heat transferred to the system Q.  The process path we select involves keeping the system in contact with a constant temperature bath at $T_0$ and controlling the motion of the portion of the system boundary at which the volume change occurs in such a way that the force per unit area is always constant and equal to $P_0$.  Since, in general, the work done by the system on its surroundings is determined by the integral of $dW=P_{ext}dV$ (where $P_{ext}$ is the force per unit area acting on the portion of the boundary at which the volume change is occurring), we have selected $P_{ext}$ such that $P_{ext}=P_0$; and the total cumulative work done by the system when it reaches its final equilibrium state will be just $W=P_0\Delta V$.

From the first law, it then follows that, for this irreversible path:
$\Delta U=Q-P_0\Delta V$.  So $\Delta U+P_0\Delta V=\Delta H =Q$


How to acid-wash sand? I need some acid-washed sand to smooth out boiling in diethyl ether synthesis. I aready have pre-washed sand for general filtration purposes, but it is not labeled as acid-washed.

Unfortunately, I have not found any reliable lab manual for acid washing of sand.

I found one reference instructing to wash 30 g of sand twice with 0.05 M sulfuric acid, then four times with distilled water.

However, other references instructing to use only hydrochloric acid and de-ionized water, but have not specified how much or the concentration.

Yet another reference says that sulfuric, hydrochloric or phosphoric acid can be used interchangeably.

I don't know how seriously should I take the acid washing step (for the ether synthesis from ethanol and sulfuric acid) and how to check whether it is properly washed (pH measurement?).

I know one can just purchase acid-washed sand, but I have excess of good quality sand and the professional acid-washed one is overly expensive.

I have just started working with sand columns and was also surprised that I could not find any valid and trustworthy guidelines or process breakdown for acid-washing sand.
However, I found this paper, which may answer some of your questions. In short, acid washing (or leaching) is supposed to remove impurities (mostly iron impurities) from the sand.

From the introduction of the paper:

  Iron is one of the most troublesome impurities in silica
  sand because it impairs the transparency of high-quality
  glass, e.g. for tableware, and optical glass; it impairs
  transmission in optical fibers; it adversely affects the production
  of pure silicon products, e.g. silicon metal for
  photovoltaic application, silicon carbide. One of the main tasks in the purification of silica sand
  is to further reduce iron compounds to a required level
  by a highly efficient method.

In table IV in the results section you can see a comparison between several acids with regard to their efficiency in removing iron and aluminum impurities. According to the data, sulfuric and phosphoric acids should be preferred. From what I understand, however, diluted HCl solution may be preferred by some, probably due to better price and availability, but that's just an assumption.

As for the procedure itself, I heard about adding a solution of 1% HCl to the sand, stirring and letting it stand for a few hours, and then washing with distilled water and decanting ~10 times, and then drying in an oven overnight.  I don't have any valid reference for it, though.


Two-electron integral algorithm I am writing a molecular integral code from scratch. Right now my code have the following structure:

loop over shellpair ab:
    loop over shellpair cd:
        loop over primitive ij in ab:
            calculate E_ij
            loop over primitive kl in cd:
                calculate E_kl
                calculate R_ij_kl
                calculate V_ab_cd_ij_kl
        contract V_ab_cd_ij_kl to V_ab_cd

The above pseudo-code have alot of implicit notation, so let me explain what it means before asking my question.

E_ij and E_kl is the expansions coefficients from the McMurchie Davidson scheme.
R_ij_kl is the Hermite integrals also used in the Mcmurchie Davidson scheme.
V_ab_cd_ij_kl is the cartesian integrals found by (note the equations below are not strictly correct):

$$V_{ab,cd,ij,kl}=\sum E_{kl}\sum E_{ij} R_{ij,kl}$$

And V_ab_cd is the contraction of the primitive to the basisfunction:

$$V_{ab,cd}=\sum_i\sum_j\sum_k\sum_l c_ic_jc_kc_lV_{ab,cd,ij,kl}$$

When calculating integrals of higher angular momentum (above $0$), I make sure to reuse common elements in E and R. I.e. when doing p-type integrals px, py and pz have the first element in common, and it is only calculated once in the recurrence relations of the McMurchie Davidson scheme.

For my implementation of the above pseudo-code even the (ss|ss) integrals are very slow. When doing benenze with 6-31G the (ss|ss) integrals uses 2.5 seconds, which is more than can be justified by the use of Python (accelerated with Numba) instead of a compiled language. I have thought about rearranging and saving more intermidiates in memory, i.e. doing something like:

 loop over shellpair ab:
    calculate E_ab (Here E_ab means all E_ij saved in memory)
    loop over shellpair cd:
        calculate E_cd (Here E_ab means all E_kl saved in memory)
        loop over primitive ij in ab:
            loop over primitive kl in cd:
                calculate R_ij_kl
                calculate V_ab_cd_ij_kl
        contract V_ab_cd_ij_kl to V_ab_cd

In the above by calculating E_ab and E_cd, recomputation of E_ij and E_kl is avoided. At this point I know that I lag some expertise in knowing what is worth it compared to memory usage.

My question is know, what is a proper way of implementing molecular integrals with regard of what intermediates to keep at which points. (I know thing like vertical recurrence relations exist, but these will for obvious reasons not speed on my (ss|ss) integrals).

There are many things to consider and improve your code. Many components
in an ab initio electronic structure calculations need supervision to achieve a
good performance. I am no expert on those calculations, but I will point you to some ideas and references. On the theoretical side, you could calculate E_ab for all shell pairs first. For more clarification on the order of operations and their scaling look at
Figure 3 in Multi-electron integrals. The article is written by a well-recognized researcher in the field that wrote a monumental book where you can find all needed expressions.
On the computational side, I would say that maybe Numba is not able to optimize your code. Writing it with Numba is not a guarantee of speed. I just took a glimpse at the code and, without going deep in the code, I saw many if statements. Only a straightforward observation. As general advice, it is all about how data is moved
and the flow of operations. Avoid allocations, try to rewrite as much as possible.
Profile your code to find the bottlenecks. Try to write it as vector or matrix operations to be able to use the large registers of the processor. There are many tricks to do these operations fast, but sadly they are buried elsewhere.
I would like to end with a reference to the recent literature on the topic. I just came up with an article that I managed to find again now: Fast Evaluation of Two-Center Integrals over Gaussian Charge Distributions and Gaussian Orbitals with General Interaction Kernels.


How does the process of nucleation work for boiling liquids? I have a somewhat clear picture how (on the molecular level) a gas turns into a liquids as it cools. When a gas has a certain temperature, its molecules on average have a high enough kinetic energy so that electromagnetic interactions cannot hold the molecules together and they are essentially free. When the gas cools, the kinetic energies are smaller and thus electromagnetic interactions between molecules start to pull the molecules together and the process of turning into a liquid begins. As more molecules lose kinetic energy, initially small numbers of molecules start to cluster up, forming "lumps" that continue growing as more molecules gather up into these clusters. I believe this process is called "nucleation".

I found this video that helps visualize the effect:

https://www.youtube.com/watch?v=wFT6G4CIL1o

But how does the reverse take place? Let's say I'm boiling water. I have a textbook that specifically states:

  During nucleation, small droplets of liquid form in gases or gas bubbles form in water as it starts to boil.

As I heat water, the average kinetic energies of the molecules increase. How does nucleation create bubbles in this situation? I understand that a bubble is a "pocket" of gas inside a liquid. If nucleation works the same way in the process of liquid turning into a gas as it works in a gas turning into a liquid, by analogy there should be some "lumps" of gas forming inside the liquid. But what would hold these gas "lumps" together as the molecules in the gas have a higher kinetic energy as the molecules around them (that are still part of the liquid)? And how do these molecules with higher energy "find" each other to start forming clusters/bubbles? Or do I have a completely wrong visualization as to how bubbles form?

Firstly, the average kinetic energy of the water has to be high enough (water obviously doesn't start to boil at room temp. and pressure the moment it is poured into a pan).

As the water gains more and more energy, small bubbles appear but don't fully rise to the surface as the energy of the gas particles is transferred to the surrounding water particles, eventually losing enough energy and becoming liquid again.

But if the energy of the gas particles is lost too slowly for this to occur they fully rise to the surface. This will only occur at a certain temp. difference i.e. the liquid has to be close to boiling temperature if not it will absorb too much energy and the bubble will become a liquid again.

A single bubble can increase in "size" (more molecules inside the bubble compared to when it started not talking about the diameter) as it rises, similar to the cluster effect you mentioned. This will only occur if the energy difference is high enough to vaporise the surrounding water molecules and not just increase their kinetic energy.

This can also not occur depending on how "violently" the water is being heated, or at which temperature the heating element is, therefore the temperature of the bubble being produced.


Confusion in solubility expressions in Henry's Law While reading about Henry's law and solubility I frequently come across two relations:

$C = k_{h}P$ (c = concentration of a dissolved gas)
$P = k_{h}x$ (x = solubility/mole_fraction)

What is the difference in these two expressions, do they seem to contradict each other? I am not able to decide whether a gas with higher $k_{h}$ would be more soluble or less soluble.
Please help me out with the same I am new to this topic

Sander (Ref. 1) has compiled a useful review of Henry's law constants in water that includes an introduction showing notation and conversions.
There are two types of Henry's law constants:

Solubility constants convert from pressure to concentration in solution (solubility):

$$c=Hp$$

Volatility constants convert from concentration in solution (solubility) to pressure:

$$p=Kc$$
Usually it is possible to determine which constant is reported by inspecting the units.
References

R. Sander. Compilation of Henry’s law constants (version 4.0) for water as solvent. Atmos. Chem. Phys., 15, 4399–4981, 2015.
www.atmos-chem-phys.net/15/4399/2015/doi:10.5194/acp-15-4399-2015


Why shape of diamond is like diamond? I know that in diamond carbon atoms occur in 3d geometry but i want to ask that why it is in shape of diamond not like some other 3d geometrical shape like cube or cuboid?

Diamond has the shape of diamond for precisely the same reason why any other crystal has the shape it has. Some crystal faces have lower surface energy than others, so the crystal grows in these directions, as if it wanted to develop these faces and avoid developing others.

This, BTW, is not determined by crystal family alone. Think of pyrite: it also has cubic unit cell, but it mostly develops the face (001) (see Miller indices) and hence makes those nice cubic crystals. On the contrary, diamond prefers the face (111), which together with its symmetry equivalents produces an octahedron. And that's why natural diamonds are typically found in this shape (not counting scratches and fractures).

Because of its low energy, the (111) face is also the hardest face of a diamond, to the point that jewelers deliberately avoid it when cutting brilliants, for it can't be polished quite as good as any other arbitrary plane.

So it goes.


What is the difference between D and L configuration, and + and −? What is the difference between D and L configuration, and + and −?

My textbook says they are two different things.
It also says that the correct way to name glucose is D(+)-glucose.

Could someone please explain what D, L and +, − represent, and why they are different?

The D-L system corresponds to the configuration of the molecule: spatial arrangement of its atoms around the chirality center.

While (+) and (-) notation corresponds to the optical activity of the substance, whether it rotates the plane of polarized light clockwise (+) or counterclockwise (-).

D-L system tells us about the relative configuration of the molecule, compared to the enantiomers of glyceraldehyde as the standard compound. Compounds with the same relative configuration as (+)-glyceraldehyde are assigned the D prefix, and those with the relative configuration of (-)-glyceraldehyde are given the L prefix.

It's kind of another way to tell the configuration of molecules beside the Cahn–Ingold–Prelog convention (R/S system), with little difference. (D-L system labels the whole molecule, while R/S system labels the absolute configuration of each chirality center.)

In short, the D-L system doesn't have direct connection to (+)/(-) notation. It only relates the stereochemistry of the compound with that of glyceraldehyde, but says nothing about its optical activity. We may have compound with same relative configuration as (+)-glyceraldehyde (thus, it's given the D prefix), yet it rotates the polarized light counterclockwise (-), such as D-(-)-ribose.

And also, don't confuse the D-L system with d- and l- naming. d- and l- is the exact same with (+) and (-) notation.

Additional explanation

D-L system (also called Fischer–Rosanoff convention) is mainly used for naming α-amino acids and sugars. It compares the relative configurations of molecules to the enantiomers of glyceraldehyde. This convention is still in common use today.

Rosanoff in 1906 selected the enantiomeric glyceraldehydes as the point of reference[1]; any sugar derivable by chain lengthening from what is now known as (+)-glyceraldehyde (or named D-glyceraldehyde) belongs to the D series. In other words, we used a D to designate the sugars that degrade to (+)-glyceraldehyde and an L for those that degrade to (-)-glyceraldehyde.

In assigning the D and L configurations of sugars, we could direcly look for the OH group of the bottom asymmetric carbon in the Fischer projection. If it's located on the right, we designate it with D, and vice versa, since they would have the same relative configurations with glyceraldehyde for the bottom asymmetric carbon.

Reference

[1]: IUPAC and IUBMB. Joint Commission On Biochemical Nomenclature. Nomenclature of Carbohydrates. 1996, 7.


Can I completely dry air with calcium chloride? I'm trying to dry air by leading it through calcium chloride.

I'm measuring the resulting humidity with a simple capacitive humidity sensor. I do not see measurements below 10% relative humidity, but it is quite possible that it is a result of a calibration error (The sensor is not actually calibrated.)

But how dry can I get air by exposing it sufficiently long to calcium chloride dihydrate?

"Completely dry"? No, because molecules are always moving and there will be some exchange between desiccant and air. If you mean "below 10% relative humidity (RH)", there are a number of ways to do so:

Consider that another way to discuss humidity is the dew point. Use a cold trap: after drying the air with a desiccant, pass the air slowly through a U-tube (or better, a through a spiral condenser) cooled with ethanol/dry ice or similar bath. As the air emerges, the dew point, ~200 K for dry ice, remains that of the bath, but as it warms to room temperature, the RH would fall.
Dry the air with desiccant at room temperature, then heat the air (not desiccant) and the RH drops.
Synthesize air from ~80% liquid nitrogen and ~20% liquid oxygen. If you're a stickler for accuracy in the simulation, add a dash of liquid argon, and a smidgen of liquefied CO2 (not dry ice, which, being open to air, quickly condenses water). These liquefied gases can probably be obtained from a welding supply shop.

BTW, don't forget to bake your apparatus before use! Metal, and particularly glass, adsorb water quite well, and slowly release it at low RH or pressure.


Determining the product which is not possible in the reaction This is an objective type question I encountered in a objective test.

NBS (N-bromosuccinimide) is used for the allylic substitution of $\ce{Br}$ in a reaction. In this reaction, since the allylic carbon is the carbon adjacent to the carbon having double bonds, I thought that options B), C) and D) all can be the products but not A). However, the answer key gives the answer as D) and I can't understand why it should be?

@Pete
B1 is same as D1

But, D1 & D2 are not same.

@Aniline
I see some problem in the options while it was framed/typed/published

I have shown the 6 possible stereoisomers, which have have labelled independently as A, B, C, D, E & F

For the sake of clarity have generated their names with some software [not from my own] to be assured that these are actually different or same compounds.

Showing the mechanism, how there is a possibility of getting SIX by a free radical mechanism:

This one a redraw of the options A, B, C & D with their names:

The relationship of D1 & D2 as per question doesn't work to the the same:

In my opinion, there are six stereoisomers possible, but the correct option does not exist, in the question[options given, in fact].

Hope it throws some more light on your query.


How did the first amino acids appear on Earth? Is there proof or a theory that shows that the first amino acid syntheses on Earth were influenced by a change in heat, gravity, radiation, or something similar?

The universe cooled after the Big Bang. There was certainly an epoch when the universe’s overall temperature decreased from 200 °C to −40 °C throughout a very large volume. There is your test tube (more like a Florence flask) for abiogenesis and then the panspermia reservoir.

Before you suspect the preceding is clever, how and when (what ambient vacuum temperature) did elements heavier than boron arise?


Why a catalyst is not consumed in a reaction when all steps are reversible? It is said that a catalyst speeds up the rate of a reaction but is not consumed (assuming no side reactions take place). Suppose we have the following reaction:
$$\ce{A + B <=> C} $$
catalysed by $\ce{X}$ in the following steps:
$$\ce{A + X <=> A-X} \tag{1}$$
$$\ce{A-X + B <=> C + X} \tag{2}$$
Each step is reversible so every step has an equilibrium constant $K$. So an amount of $\ce{X}$ should be present as the intermediate $\ce{A-X}$ even after the catalysed reaction has reached equilibrium. That means that the amount of catalyst that we have added to catalyse the reaction is not the same after the reaction has reached equilibrium. So why we say that a catalyst is not consumed through the reaction?

By saying Catalysts are not consumed by reactions. is meant there is no stoichimetric ratio to reactants, consuming catalysts and forming from them catalytically inactive compound.
By other words, some of reaction steps regenerates the original form of a catalyst, consumed by a prior step, so the net consumption is negligible.
The transition between active states of a catalyst is not considered a catalyst consumption, as it does not affect its catalytic activity.
Catalysis means the catalyst is reversibly switching between its 2 – chemically different – active states, producing an alternative and faster reaction path. Each form catalyzes one of the mutually opposite direactions of the net reaction. It is like a kind of a ferry to move compounds between reactant and product shores, bypassing a faraway bridge. The ferry is not consumed if ends at the opposite shore.

Note that things get little complicated, if some of products or intermediates catalyse a part or whole net reaction. Not being particularly aware of the intermediate case, but for the product, the famous case is oxidation of oxalate by permanganate in acidic environment:
$$\ce{5(COO)2^2-(aq) + 2 MnO4-(aq) + 16 H3O+(aq) -> \\ ->10 CO2(g) + 2 Mn^2+(aq) + 24 H2O}$$
$\ce{Mn^2+(aq)}$ as the product acts as a catalyst, forming catalytic system $\ce{Mn^3+(aq)/Mn^2+(aq)}$ with a similar redox potential as $\ce{MnO4-(aq)/Mn^2+(aq)}$.
$$\ce{4 Mn^2+(aq) + MnO4-(aq) + 8 H3O+(aq) <=> 5 Mn^3+(aq) + 12 H2O}$$
$$\ce{(COO)2^2-(aq) + 2 Mn^3+(aq)  -> 2 CO2(g) + 2 Mn^2+(aq) }$$
The effective result is an autocatalytic reaction. There is sometimes applied a small amount of Mn^{+II} salt to speed up the reaction, or it is waited until initial addition of $\ce{KMnO4}$ creates enough $\ce{Mn^2+}$.


First order differential equation for transmembrane diffusion I want to model transmembrane diffusion using a first order reversible differential equation. The diffusion rate is known from literature. There are two volumes $V_{\mathrm{intra}}$ and $V_{\mathrm{extra}}$. The former is $10^{12}$ times smaller than the latter. $c_{\mathrm{intra0}}=0~\mathrm{mol~L^{-1}}$, and $c_{\mathrm{extra0}}=x~\mathrm{mol~L^{-1}}$, with $x>0$.

Thus far I am using this:
$$\frac{\mathrm{d}c_{\mathrm{intra}}}{\mathrm{d}t}=k_{\mathrm{diff}}\cdot c_{\mathrm{extra}} - k_{\mathrm{diff}} \cdot c_{\mathrm{intra}}$$

However, this does not account for the fact, that $V_{\mathrm{extra}}$ is much larger than $V_{\mathrm{intra}}$. I expect $c_{\mathrm{extra}}$ to be virtually constant. So far the the simulation results in the equilibrium being at $c_{\mathrm{extra0}}/2$.

Can someone suggest a method to account for the volume discrepancy?

Edit:

From the outlined scenario above, I expect in REALITY, my $c_{\mathrm{extra}}$ to not noticeably change from its initial value due to the large difference between $V_{\mathrm{intra}}$ and $V_{\mathrm{extra}}$.

However, in the simulation I don't want to set $c_{\mathrm{extra}}$ as constant, because I want to apply a 'extracellular' decay rate later on. Basically, I want to be able to modulate the extracellular concentration over time and see how the concentration within the cell changes.

This is why I want to account for the volume. Maybe, one needs an entirely different approach for that?

The solution to your differential equation, subject to the prescribed initial conditions is $$c_{intra}=c_{extra0}(1-e^{-k_{diff}t})\tag{1}$$ under the constraint that $c_{extra}$ does not change from its initial value of $c_{extra0}$.  The true material balance constraint is $$V_{intra}c_{intra}+V_{extra}c_{extra}=V_{extra}c_{extra0}$$
which would yield a result virtually indistinguishable from Eqn. 1.

CASE WHERE $c_{extra}$ IS NOT CONSTANT

OK.  Here is the solution for the case where $c_{extra}$ is a function of time $c_{extra}(t)$, and $c_{intra}=c_{intra}(0)$ at time t = 0:

$$c_{intra}(t)=c_{intra}(0)e^{-k_{diff}t}+k_{diff}\int_0^t{e^{-k_{diff}\tau}c_{extra}(t-\tau)d\tau}\tag{2}$$
where $\tau$ is a dummy variable of integration and where $c_{extra}(t-\tau)$ represents the value of the concentration $c_{extra}$ at time $t-\tau$.  This solution reduces to the previous result if $c_{intra}(0)=0$ and $c_{extra}$ is constant.

So, to find the value of $c_{intra}$ at time t, you need to use Eqn. 2 and perform the indicated integration (usually numerically).


When does a piston stop in an irreversible process? Suppose we have an ideal gas filled inside a container fitted with a piston. The external pressure is equal to internal pressure.
If we suddenly decrease the external pressure, we say that the gas will expand until the internal pressure and external pressure become equal.
Considering an irreversible process, shouldn't the gas expand beyond that point because by that time the piston will have gained some velocity?

You're right.  The piston can gain kinetic energy and overshoot its equilibrium position, but then, the force of the gas on the piston will drop until the piston comes back the other way.  The piston will undergo an oscillation back and forth, but this oscillation will be damped as a result of viscous dissipation stresses in the gas.  Eventually, the oscillation will decay until the piston stops.  At this point, the gas pressure will be in equilibrium with the lower pressure outside.
If I do a force balance (Newton's 2nd law) on the piston, I get
$$F_\mathrm{g} - p_0 A = m \frac{\mathrm{d}v}{\mathrm{d}t},$$
where $F_\mathrm{g}$ is the force the gas exerts on the piston, $p_0$ is the outside pressure, $A$ is the piston cross sectional area, $m$ is the mass of the piston, and $v$ is the piston velocity.  If I multiply this equation by the velocity of the piston $v = \mathrm{d}x/\mathrm{d}t$, and integrate with respect to time, I obtain:
$$
W_\mathrm{g}(t)
  = \int\limits_0^{x(t)} F_\mathrm{g}\,\mathrm{d}x
  = p_0(V(t) - V(0)) + m\frac{v(t)^2}{2},
$$
where $W_\mathrm{g}(t)$ is the work done by the gas on the piston (its surroundings) up to time $t$, $V(t)$ is the gas volume at time $t$, and $m\frac{v(t)^2}{2}$ is the kinetic energy of the piston at time $t$. At infinite time, the piston is no longer moving, so its kinetic energy is zero at final equilibrium. Therefore, at final equilibrium, the work done by the gas is just
$$W_\mathrm{g}(\infty) = p_0\Delta V,
\text{ where } \Delta V = V(\infty)-V_0.$$


What is the mathematical meaning of the plus sign (+) in chemical reaction equations? Reactants and products are connected with plus signs in a chemical reaction equation, e.g.
$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}.$$
Is there a strict mathematical interpretation of what looks like sums before and after the reaction arrow? If so, what do the sums represent, and what are the mathematical objects we would use for the chemical species?
Often, when we use a chemical equation to solve a problem, operations other than addition come into play. For an equilibrium constant expression, concentrations of species are multiplied, not added. In stoichiometry, the chemical equation is interpreted as ratio of changes in amounts. In kinetics, we divide by the coefficients to be able to compare rates of changes of different species (and there is an equal sign, not a plus sign between the rate expressions of the different species.
So why are there plus signs between species in a chemical equation, and what is their mathematical interpretation?

I think we are over-interpreting the poor $+$ sign. In older books you will also see an $=$ sign instead of an arrow just like what Lavoisier did. We have to get hold of La Traité Élémentaire de la Chimie (1789).
We should look at the historical perspective and I always want to see the earliest usage in chemistry just to see what was the original person thinking. I think we can credit Lavoisier for the "+" sign in chemical equations and let us see what was going on in his mind. I quote from the History of Chemistry, by F.J. Moore 1910 (Full text in Google Books).

The cumulative effect of these researches led to the sudden conversion of most French chemists about $1785,$ and the new ideas were firmly fixed by the publication of Lavoisier's great text-book La Traité Élémentaire de la Chimie in 1789. The change was well called the Chemical Revolution, for it inverted completely the chemical point of view. The mysterious hypothetical substance, phlogiston, which did not obey the law of gravitation, and changed its properties arbitrarily as theoretical considerations dictated, was banished from the science and the law of conservation of mass vindicated once for all. This made possible quantitative analysis and the chemical equation. The latter is a convenience in form of expression which we owe to Lavoisier, and of whose assistance in stating and solving chemical problems he made immediate use. He writes: $^{1}$
If I distill an unknown salt with vitriolic acid and find nitric acid
in the receiver and vitriolated tartar in the residue, I conclude that
the original salt was nitre, and I reach this conclusion by mentally
writing the following equation based upon the supposition that the
total weight is the same before and after the operation.
"If $x$ is the acid of the unknown salt, and $y$ is the unknown base
$I$ write: $x+y+$ vitriolic acid $=$ nitric acid $+$ vitriolated
tartar $=$ nitric acid + vitriolic acid + fixed alkali. Hence I conclude: $x=$ nitric acid, $y=$ fixed alkali, and the original salt was nitre."
Nothing like this had ever appeared before in chemical literature and
what a fog of mystery and superstition it removed!

Update Surprisingly, I cannot find this quote in Lavoisier' book (can't read French) but + sign is obvious! The only equation in his 382-paged book is
moût de raisin = acide carbonique +  alkool.
grape wort = carbonic acid + alcohol
Anyway, plus sign is quite old. In the book titled Historical studies in the language of chemistry by Crosland, Maurice P, a 1744 text is cited showing

Readers should consult this book (Internet Archive, need to make an account for borrowing it) if the want to dig deeper. The author discusses plus sign at several places at length.


Why do we say "approximation" in the dipole approximation in spectroscopy? In the dipole approximation, the following relation holds:

$$
\hat{V} = -\hat{\mu} \cdot \vec{E}.
$$

When we say "approximation", I guess we want to point out that we are dealing with linear spectroscopy, where we have the polarizability of the molecule proportional to the electric susceptibility of a molecule:

$$
\vec{p}(\omega) = k\chi \vec{E}.
$$

If the wavelength of the light is not larger than the atom size, the dipole approximation is no longer a good approximation. This is probably also true for large molecules. I'm not sure if this has to do with large molecules no longer having symmetry, so transitions might be allowed which otherwise should be forbidden.

Is this correct?

As per your linked reference, it's an approximation because only the first term in the expansion (here, unity) is considered.

The issue with transitions that are allowed or forbidden arises as a result of using such an approximation: "Forbidden" transitions are not strictly forbidden, it's just that they are weak and, in the linear approximation of the transition dipole operator, are usually not observed.


Is silicon dioxide decomposition possible? I was thinking about this, I made Gibbs free energy calculations and I can't figure out if

$$\ce{SiO2(s) -> Si(s) + O2(g)}$$

is actually possible, even at a ridiculous high temperature.

Silicon dioxide is difficult to decompose directly, but it can be broken down in the presence of other reagents at more moderate temperatures.  In particular, the reaction of silica with elemental silicon has been known since 1905:

In 1890, the German chemist Clemens Winkler (the discoverer of germanium) was the first to attempt to synthesize $\ce{SiO}$ by heating silicon dioxide with silicon in a combustion furnace.[1]
$$\ce{SiO2 + Si ⇌ 2 SiO}$$
However, Winkler was not able to produce the monoxide since the temperature of the mixture was only around 1000° C. The experiment was repeated in 1905 by Henry Noel Potter (1869–1942), a Westinghouse engineer. Using an electric furnace, Potter was able to attain a temperature of 1700° C and observe the generation of $\ce{SiO}$.[2] Potter also investigated the properties and applications of the solid form of $\ce{SiO}$.[3,4]

The reaction may be used to remove silica from ores.
Cited references:

C. Winkler Ber. 23, (1890) p. 2652.

J. W. Mellor "A Comprehensive Treatise on Inorganic and Theoretical Chemistry" Vol VI, Longmans, Green and Co. (1947) p. 235.

U.S. Patent 182,082, July 26, 1905.

E. F. Roeber H. C. Parmelee (Eds.) Electrochemical and Metallurgical Industry, Vol. 5 (1907) p. 442.  Link


Why even the most volatile solvents leave a residue after evaporating? After a solvent evaporates off of a surface there's always some residue left in the form of a strain.

Why it is the case even for highly volatile solvents such as acetone, toluene etc?

If the surface material is clean and neither soluble nor reactant in any way with the solvent, then it is probably a solvent purity issue. If you have a good micro-balance, you could pour a known volume of the solvent into a pre-weighed light-weight, i.e. aluminum foil, container, dry it, and re-weigh. From that you can calculate the non-volatile solids content of your solvent and compare it to the specifications.


Is biphenyl considered to be aromatic? Biphenyl has 12 π-electrons and according to Hückel’s rule, it should not be aromatic because $12 \neq 4n+2$, but it nevertheless is aromatic. Why is that so?

The two rings in biphenyl are inclined at an angle of 44° to each other because of the large steric repulsion between the ortho hydrogens on each ring.

For Hückel's rule to be applied, the molecule must be able to be planar and it must be monocyclic.

Due to the molecule not being planar, the two rings are not strongly conjugated to each other and you cannot apply Hückel's rule to the whole system. Additionally, the whole system is not monocyclic. Hückel's rule can be applied to each ring individually and they are both aromatic.


Latent blood detection with fluorescein I need help with the reaction between fluorescein and blood in the method of detecting latent blood stains.

I find it hard to find any literature on this subject.
As far as I am concerned, the fluorescein is reduced into fluorescin in a stock solution with powdered zinc, distilled water and sodium hydroxide pellets.
$\ce{Fe^2+}$ (found in the latent blood) is oxidized into $\ce{Fe^3+}$ with hydrogen peroxide.
When fluorescin is reacting with iron ions in the haemoglobin, it will be oxidized back to fluorescein.

Can anyone help me with this? Maybe even on a molecular level.

Fluorescin(left), a non-fluorescent compound, reacts with iron in residual blood samples to produce highly fluorescent fluorescein(right).

The reference$\ce{^{[2]}}$ from which the following excerpt is taken describes the chemistry, preparation of reagents, and application to forensics.

  Fluorescin is the oxidized colorless form of fluorescein. Fluorescin reduced with hydrogen peroxide is immediately oxidized to fluorescein in the presence of blood-associated proteins and iron ion found in hemoglobin. Fluorescein fluoresces when viewed using blue light and orange or yellow filters, whereas fluorescin will not fluoresce under the same lighting conditions.

Literature Resource

An excellent starting point to consider for searching for relevant literature is PubMed.

References

Image sources - PubChem
Google Books - Complete Crime Scene Investigation Handbook, Everett Baxter, Jr.
Latent-Prints.com - Fluorescein Detection of Latent Bloodstains
The Presumptive Reagent Fluorescein for Detection of Dilute Bloodstains and Subsequent STR Typing of Recovered DNA
A Comparison of the Bluestar and Luminol Effectiveness in Bloodstain Detection


How is ammonia used in production or processing of steel? There are some large tanks of ammonia in nearby iron works.

I wonder, what do they need it for? What is the application of ammonia in steel industry - what processes of making or processing steel require this much ammonia?

Likely, for nitriding http://en.wikipedia.org/wiki/Nitriding . Essentially, ammonia is used as nitrogen donor. Nitrogen atoms diffuse into steel, producing hardened, corrosion-resistant surface layer.


Stoichiometry Assignment $\ce{2Ca3(PO4)2 +6SiO2 +10C->6CaSiO3 +P4 +10CO}$

What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phosphorite sample is 75% $\ce{2Ca3(PO4)2}$ by mass

I am confused. How do I find a maximum amount of a reaction

Phosphorite is a mineral containing calcium phosphate. This particular 1kg sample has 75% percent of it by weight i.e. 750 grams. All others reagents are assumed to be in excess as you have to find the maximum p4 obtained which would be at a point where all of our mineral is used.


Explaining Ionization to kids I have a quest to explain a physical, chemical theory to kids of age 8-12.

The topic is "Ionization". We all understand from early experiments of Physics and Chemistry (the Millikan oil drop experiment) how a neutral atom in gas becomes a charged atom where radiation knocking the electrons out of the neutral atom.

Here's the story line I came up with. Can you guys please help me to make this a success and correct the wrong impressions, expressions and examples in the narrative?
I am open for any better examples.

There's a high-class (=high voltage) shopping mall area (=radio activity area metaphorically) in the country I am living right now. One particular shopping mall called ION. If you enter that shopping mall, you will shop and get excited (=charged) and spend your pocket money (=losing electrons) by the temptation (=radiation) from fancy shoppers around. Shopping means getting exposed to radiation.

Some shoppers are highly influential (=alpha rays with +2 charge) and there's a high tendency you will spend money (=lose electrons). e.g. Movie theater. You get excited even looking at it from far.

Some other shoppers are less influential (=beta rays with -1 charge). e.g. Junk food, a piece of cake you like to eat. You have to go near to see the food and then spend your money on it. The excitement of eating the food lasts for shorter period compared to the movie you watched. So the impact from movie is higher. You will talk about it in school for a week or two. Unlike the food you had and forgotten by the next meal.

The last type of shoppers are least influential (=gamma with 0 charge). e.g. Manga shop. You don't find it as exciting as watching a moving or eating favourite food. But you may go there as a matter of need (for a manga copy) or peer-pressure from your friends to buy manga and spend your money. You are kinda forced to get excited (=charged) to spend a buck on it.

So it's quite dangerous for you (=neutral atom) to enter or go near a shopping area (= radiation area) where you will be losing your money and get excited for nothing...

I think the key concepts that are explainable at this age level are:

Atoms are made of protons, neutrons, and electrons
Protons have positive charge, electrons have negative charge, neutrons have no charge
Opposite charges attract
Atoms can gain or lose electrons, but not protons or neutrons
Atoms hold onto their electrons more or less tightly depending on which atom it is (you won't be able to really explain why, but you could just show them a periodic table and explain that non-metals (the right side) like electrons more than metals (the left side))
Atoms can "steal" electrons from each other
Electrons can be "knocked off" of atoms by high-energy photons (light or EM radiation)

The difficulty with explaining these things by analogy to regular-life experiences is that there is no regular-life experience that captures the whole picture. This means you are limited to activity-type demonstrations, videos/pictures, or analogies for specific pieces of the puzzle.

I would recommend a mixture of all three approaches. Kids like hands-on stuff, so maybe you could explain the basic concepts with very simple analogies, then do an activity where the kids are given "electrons" (tennis balls, maybe) and are assigned to be metals or non-metals. Then you could have them give each other electrons and count charges as they go. If you had containers that could hold 8 electrons, you could even represent valence shells and the octet rule - metals could start with two or three electrons, non-metals with 5, 6, or 7, and the metals would need to empty their shells while the non-metals would need to fill them.

This would demonstrate a very important concept that is hard to teach in the abstract: Electrons don't disappear.

Since you are also interested in teaching about ionization by radiation, you could assign each student a number representing their ability to "hold on" to an electron. Here is where you might be able to use a shopping-type analogy - only instead of students giving up electrons as money, they would give them up in exchange for money. In this case, the money represents the energy, and you would be giving them the energy (money) needed to remove (sell) an electron. Atoms that charged a higher "price" (require more energy) would wind up with more money at the end of the exercise.

There are many variations on these ideas that you could use, but I think anything that covered the seven points I mentioned above while simultaneously developing an intuitive understanding that energy and electrons are conserved could be called successful.


SN2 with thionyl chlorides Grignards don't couple with alkyl halides. I know that thionyl chloride reacts with alcohols and other $\ce{-OH}$ group containing molecules to replace the $\ce{-OH}$ with a $\ce{-Cl}$. So we probably aren't making an alkyl halide from the alcohol with the thionyl chloride.

The book says that the starting product, which is an alcohol, attacks the sulfur in the $\ce{S=O}$ bond of $\ce{SOCl2}$. I got that the starting product should be an alcohol, but I did not expect the alcohol to attack the thionyl chloride.

Nevertheless, this generates an $\ce{R-O-SOCl}$ molecule, and the resonance-stabilized $\ce{O-SOCl-}$ ion is a good leaving group. This is the substrate that the Grignard can perform an SN2 type reaction on.

My question is: how common is it that alcohols attack thionyl chloride in this pattern? I'm guessing "not very common" because this would make thionyl chloride an undesirable reagent for making alkyl halides from alcohols. Is this reaction synthetically useful?

This generates an R−O−SOClR−O−SOCl molecule, and the resonance-stabilized O−SOCl−O−SOCl− ion is a good leaving group. This is the substrate that the Grignard can perform an SN2 type reaction on.

I could be wrong about this, because I always doubt everything that I do, but I don't think that's quite right.

Here is a good discussion about thionyl chloride works.  It should turn your alcohol into an alkyl chloride, and now that chloride is your leaving group for the SN2 reaction.

Knowing that, we can work retrosynthetically.

Your product is octane.  You get to the octane by reacting something with ethyl magnesium bromide (a great nucleophile), so that's a strong indication that you did a substitution reaction.  If you add on an ethyl group and get octane, then your substrate was an ethyl group shorter, so it would have contained a hexyl chain.  You got to that hexyl species by treating an alcohol (as it was described in the initial question) with SOCl2, and you know that SOCl2 turns alcohols into chlorides, so your hexyl species that you treated with EtMgBr was 1-chlorohexane, which you obtained by treating hexan-1-ol with SOCl2.


How to make 1,1-diphenyl-1-butene from benzophenone and 1-bromopropane? I would like to know the method of syntheses of making
1,1-diphenyl-1-butene $(\ce{C16H16})$ from benzophenone $(\ce{(C6H5)2CO},$ generally abbreviated $\ce{Ph2CO})$ and 1-bromopropane $(\ce{CH3CH2CH2Br}).$

I think that the carbonyl group which is a functional group composed of a carbon atom double-bonded to an oxygen atom $(\ce{C=O})$ of benzophenone will react to the alkyl halide.

Because oxygen is more electronegative than carbon, carbonyl compounds often have resonance structures which affect their reactivity.  This relative electronegativity draws electron density away from carbon, increasing the bond's polarity, therefore making carbon an electrophile (i.e. slightly positive).

But I have no idea how it reacts and proceeds to work. Can anyone help me understand the detail mechanisms and  reaction procedures to make 1,1-diphenyl-1-butene from benzophenone and 1-bromopropane?

Two steps:

Form the Grignard reagent with 1-bromopropane and magnesium metal. This can be down in a variety of ethereal solvents, THF or $\ce{Et2O}$ are most commonly used. This species is nucleophilic through carbon and will add to the carbonyl group.

Cool the Grignard solution in an ice bath under nitrogen with stirring, slowly add a solution of 0.9 eq of benzophenone. Allow it to stir until TLC shows consumption of the benzophenone starting material.

Add a solution of aqueous ammonium chloride to quench residual Grignard. Add organic solvent $(\ce{Et2O}$ or $\ce{EtOAc})$ and collect the organic phase, dry and concentrate. This will form 1,1-diphenyl-butan-1-ol.
Take the crude product from step 1 and dissolve in toluene. Add a catalytic amount of p-toluene sulfonic acid (PTSA) and heat under Dean-Stark reflux to remove the water. The PTSA will protonate the tertiary alcohol you have formed in step 1. This gives a cation stabilised by the two phenyl groups so it forms readily. This eliminates water to give the required product cleanly.

When the system has ceased to generate water allow it to cool, wash with diluted aqueous sodium carbonate solution, dry and concentrate. This should result in fairly clean 1,1-diphenyl-1-butene.

To produce cleaner material use chromatography on silica with hexane/ethyl acetate (0 to 10%).

Further reading on Grignard reagents: Master Organic Chemistry.


Can class D fire extinguishers be used for other fire types than metal fires? Burning metals like sodium should not be extinguished with standard fire extinguishers, but they require either sand or a special class D fire extinguisher.

Now, if there is a fire in the lab and the closest fire extinguisher is one of those class D extinguishers meant for metal fires, can those be used for other types of fires?

The main case where there might not be enough time to get specific equipment is probably when people are on fire. Can a class D extinguisher be used in this case, if it's the closest available one?

Unfortunately, as far as I can tell, specific class D fire extinguishers aren't even suitable for all types of class D fire, much less other types of fire.

The copper powder variety in particular has limited effectiveness against fires that aren't lithium-based, and the thermoplastic additives in the sodium chloride-based extinguishers may also catch fire.

Dry sodium carbonate-based class D extinguishers should be generally effective but suboptimal against many types of fire, but I'd say in general, even if you have to go a little further, you should go for the proper sort of extinguisher to minimise accidental horrible disasters.