Why is UuX used as a symbol for unnamed elements on the periodic table? Until recently, the last few elements on the periodic table had the chemical symbols (now they have formal chemical symbols!):

Uub, Uut, Uuq, Uup, Uuh, Uus, and Uuo

Why did they have these temporary symbols? Why the letter U? Is there any reasoning behind the third letter? Why not:

Uua, Uub, Uuc, Uud, Uue, etc...

I was just wondering some of the history of nomenclature and the thought process behind this specific nomenclature.

In 1978, the IUPAC Commission on the Nomenclature of Inorganic Chemistry decided that it would be necessary to have a systematic naming for the elements with atomic number greater than 100, even for those which had not been discovered. The recommendations are as follows:

  The name is derived directly from the atomic number of the element using
  the following numerical roots:

  $$\require{enclose}
\small
\begin{array}{lcc}
\hline
\text{Number} & \text{Latin Word} & \text{Greek Word} & \text{IUPAC Nomenclature} \\
\hline
\text{0} & \enclose{circle}{\text{nihil}} & \text{mēdén} & \text{nil} \\
\text{1} & \enclose{circle}{\text{ūnus}} & \text{en} & \text{un} \\
\text{2} & \enclose{circle}{\text{duo}} & \enclose{circle}{\text{dúō}} & \text{bi } \\
\text{3} & \text{trēs} & \enclose{circle}{\text{tria}} & \text{tri} \\
\text{4} & \enclose{circle}{\text{quattuor}} & \text{téttara} & \text{quad} \\
\text{5} & \text{quīnque} & \enclose{circle}{\text{pénte}} & \text{pent} \\
\text{6} & \text{sex} & \enclose{circle}{\text{héx}} & \text{hex} \\
\text{7} & \enclose{circle}{\text{septem}} & \text{heptá} & \text{sept} \\
\text{8} & \enclose{circle}{\text{octō}} & \enclose{circle}{\text{oktṓ}} & \text{oct} \\
\text{9} & \text{novem} & \enclose{circle}{\text{ennéa}} & \text{enn} \\
\hline
\end{array}
$$

  The roots are put together in the order of the digits which make up the
  atomic number and terminated by 'ium' to spell out the name. The final
  'n' of 'enn' is elided when it occurs before 'nil', and the final 'i' of
  'bi' and of 'tn' when it occurs before 'ium'.
  The symbol of the element is composed of the initial letters of the
  numerical roots which make up the name.
  The root 'un' is pronounced with a long 'u', to rhyme with 'moon'.
  In the element names each root is to be pronounced separately.$^{[1]}$

Though not in the original text, I went ahead and added the words from which the IUPAC names were derived. Though not obvious, bi- comes directly from di-, which is a result of duo, found in both languages.

$^{[1]}$ Recommendations for the Naming of Elements of Atomic Numbers Greater than 100, Pure Appl. Chem., Vol. 51, No. 2, (1979) 381-384.


How does grease become airborne? When cooking in the kitchen you usually use some oil or butter to bake your meat and potatoes in. After some weeks without cleaning (which happens every now and then) the surroundings of the stove will get a greasy, sticky layer. Since this only happens relatively close to the stove, say a 1-2 meter radius, I'm assuming that it is caused by my cooking.

My question is: how does the grease get so far away from the stove?

I know that sometimes oil will spray a bit when you add something water-containing to it, but the grease is actually also in places that are not in the 'line of fire' (on top of the ventilation hood for example). Therefore, the grease must be getting airborne somehow, but I don't understand how, because evaporation of the vegetable oil or butter seems highly unlikely to me. Does anyone know what is going on?

Let's reword it:

Traces of oil and grease are distilled off at temperatures below their boiling points.
The temperature in the frying pan is above 100 °C.
The food does contain significant amounts of water.

This sounds like the conditions for steam distillation, except that you're collecting the mixture of water and water-insoluble material everywhere in the kitchen and not in a flask.


The mass of a carbon-12 atom is 12 amu. What percent of the mass is located in the nucleus? The mass of a carbon-12 atom is 12 amu. What percent of the mass is located in the nucleus? and Why is the value so close to 100%?

I can only give a simple answer, so if others have better ones, great, but here's my shot:

An electron's mass is a lot less than a proton/neutron.
"When talking about relative masses, we use the term atomic mass unit (amu). Using this unit, a proton has a mass of 1.007 amu, a neutron a mass of 1.008 amu, and an electron a mass of 5.45 X 10-4 amu."

Basically, the proton and neutron is close to 1, and the electron is close to 0.

Carbon-12 means it's got a total of 12 protons and neutrons, and since carbon has 6 protons, that makes 6 neutrons. Since it's a normally charged carbon atom too, that makes 6 electrons.

Can you do the calculations yourself with that? Because now you know the masses.

Also this value should be really close to 100% because of how electrons are so much lighter than protons and neutrons; protons and neutrons are located in the nucleus and the rest of the atom is mostly empty space + some electrons, so that's why the bulk of the mass is there.


Permeability of containers to poop smells I do a lot of hiking trips that involve driving for an hour or two to the trailhead in the predawn hours. I'm a morning pooper, so I need to poop during the drive. Before COVID, I would usually plan a stop along the way at a Starbucks, get coffee, and use their bathroom. Then if necessary I'd also use a bathroom at the trailhead. Right now, during COVID, these bathrooms are all closed, so my new normal is that I pull over somewhere in the countryside, poop, collect my poop using one of my dog's poop bags, and bring it with me to a town or trailhead where I can dispose of it.
Sometimes this was stinking up my car, even though the bag was tied shut, so I tried putting the poop bag inside a large plastic (PETE) can that some nuts came in, and sealing it with the screw-on lid. Even then, there have been times when the smell seeped out within 30 minutes. Sometimes the smell has been like sour milk, sometimes like sulfur or eggs.
If I want to make this work better, what should I be focusing on? Is ~0.5 mm of PETE likely to be permeable to the smelly volatile compounds involved? Or is the issue more likely to be that there's a poor seal at the screw threads? Would I be better off trying something like the glass jars with clamping lids that people use for their coffee? A tupperware-style container with an o-ring seal? What kind of material should I be looking for in a gasket or o-ring?
(The other option I've tried successfully is to clamp the poop bag under a windshield wiper, so it's outside. But I'm worried this would, uh, catastrophically fail if I then drove at freeway speeds.)

The problem likely is not (so much) the plastic material of your container, either solid, or bag.*)  Assuming there is no contamination on the outer side of the container, most likely the smells are noticed because of closing the container by either knot or lid is not tight enough.
After some search on sibling sites LifehacksSE and OutdoorsSE I learned hikers found multiple solutions to this problem (example).  Assuming you stay on ground and don't climb, a) a poop tube may be suitable for you. In this tube, you carry the waste with you until you find back to a place connected to a waste management / waste water facility.

(credit, source describes both construction and use)
Step #3 of this guide includes an important clue how to reduce the smell.  To quote:

«Before you go out camping, hiking, or climbing, preload the airtight bags with 2-3 small scoops of kitty litter or powder detergent to remove moisture. This will also keep obnoxious odors from leaking out of your poop tube.»

b) The idea to catch liquids and thus smell in a bag filled partially with some of (advanced?) cat litter is commercialized elsewhere, too (example).
Maybe this avenue is working well enough that additional sealing of the lid by an O-rings of suitable diameter and material (viton is a material resistant to many chemicals, but I don't dismantle a vacuum deposition to check this field of application) is not necessary.
*) Although permeabilities of liquids and gases across plastics may be determined experimentally.


Formal charge and valence shell Is there a relationship between valence shell and formal charge of the atom......

Let's take an example.

  A carbocation for eg $methenium$ has a formal charge of +1,can we tell how many valence electrons it has using this.A carbon atom always forms 4 bonds but here it is only forming 3 bonds then surely there will be change in it's valence shell and the formal charge will be related to it?How about other atoms such as  ammonium ion.What will this positive charge signify....decrease in lone pairs or increase.Stuffs like that

I don't know much about formal charge so that's why I am asking such confusing questions.My teacher said to me that carbon always forms 4 bonds and since it is forming 3 bonds in cases like this , a positive charge will come to it.Same for nitrogen but he didn't tell me about the valence shell so that's why i am asking

Carbon doesn't always form 4 bonds. Consider neutral methylene, $\ce{H2C:}$ in which it forms two leaving two unshared electrons (:). Now suppose we protonate that (with a super acid?). We'd have $\ce{H2CH+}$ or $\ce{H3C+}$, a cation just as when we protonate $\ce{H3N:}$ we get the ammonium cation $\ce{NH4+}$. In both cases the bond to the proton is made from the unpaired electrons on C or N and not by sharing an electron from C or N with one from hydrogen.

In computing formal charge we assume that all bond electrons are equally shared. The formula for formal charge is V - N - B/2 in which V is the number of valence electrons in the neutral atom (4 for carbon), N is the number of valence electrons not participating in a bond (0 in the case of methenium) and B the number of electrons in bonds (6 in the case of methenium) so that the formal charge is 4 - 0 - 3 = +1.

It's a different way of indicating charge distribution. If we write methenium as $\ce{H2CH+}$ it implies that the carbon retained the electrons for the most part so that the bond to the proton is largely ionic. In the formal charge scheme we assume that this bond is totally covalent, that the + charge of the proton has transferred to the carbon because the proton got a half share in two electrons thus neutralizing its charge and write it $\ce{C^\oplus H3}$. It's a formality.


Can 1,2-dioxetanedione (active ingredient in glow sticks) undergo stimulated emission (and thus form a laser)? In common glowsticks, diphenyl oxalate reacts with hydrogen peroxide, forming 1,2-dioxetanedione. The dioxetanedione is unstable, but direct decomposition to carbon dioxide is a forbidden transition, so this can only happen in a reaction with another molecule. If the other molecule is a suitable fluophore dye, the dye molecule will become exited and then emit a photon of light. (The color depending on the fluophore used.)

Some types of lasers make use of forbidden transitions: atoms or molecules are brought into an excited state such that the transition to the ground state is forbidden and thus slowed down. This causes the exited state molecules to collect, forming a population inversion. The excited molecules can then undergo stimulated emission in a resonance cavity, producing the laser effect.

Now my question is if 1,2-dioxetanedione can also be made to undergo stimulated emission if the glow stick chemical reaction takes place in a suitable optical resonance cavity. (It looks to me like it should, but my chemistry knowledge is not enough for this.)

Alternatively, if the 1,2-dioxetanedione cannot undergo stimulated emission, could a suitably chosen dye molecule?

The activated complex 1,2-dioxetanedione can't undergo simulated emissions as it isn't the emitting molecule even in glow-sticks: it is the source of the activation energy for other dyes.

But most emitting dyes can be used in suitably configured lasers by providing energy as photons (=light) of a higher energy and a suitable wavelength in a suitable device. Whether they would make good lasers depends on a whole bunch of other factors (like how easy it is for them to absorb light, whether the subsequent transitions are forbidden or not (can't easily get population inversion if they are not) and various quantum efficiencies of the relevant conversions in the dye molecule.)

Dye lasers have been known since 1966. They have some advantages over more conventional lasers as they are often wavelength tunable (as dyes tend to have broad emission lines) and can often achieve very short pulses. Rhodamine 6G is a common laser dye but the dyes in a glow-stick may not be as good in lasers, though, in principle, there is no reason why not.


Why do the f-block elements ( inner transition metals) have an oxidation number of +3? The f-block valence electrons understandably include the outermost two from 6s. But why do the f-block elements have an oxidation number of +3, pulling an electron from an inner shell, not even a d-block electron?

First of all, lets define which elements we're discussing as f-block.  Generally the lanthanides are treated as f-block elements, though this introduces some ambiguity as La (which starts the lanthanides) has 1 d electron and no f electrons while Lu (which ends the series in some definitions) has 1 d-electron and all 14 4f electrons.  So one of these two is a d-block element but its not entirely clear which one.  Recall that in the f-block there can only be 14 elements due to the 14 electrons that fill f orbitals (just as the d-block can only have 10 elements).

Not all of the lanthanides have 5d electrons, only La, Ce, Gd and Lu do!  The rest fill the 6s and 4f only.  The 6s are heavily shielded and easily lost.  Only one 4f electron is lost because the 4f closer to the nucleus and more effected by losing electrons, ie once one is removed its difficult to remove another because the rest are far more attracted to the nucleus.

So the real question should be why are La, Ce and Gd exceptions to the filling order?  For La and Ce this is because they have a lower Z effective, they're not splitting the energy of the 4f and 5d as effectively.  For these two the 5d is lower energy and fills first, then the 4f fills.  For the later Ln with a higher Z effective the 4f (being in a lower shell) is pulled to significantly lower energies and fills first.

The reasons for Gd having a different filling order is slightly more complex but can be related to whats seen in the transition metals, recall that Rh and Ru don't actually fill the 5s despite not being at a half-shell closure, they fill one 5s and the rest in 4d.  Lu is not an exception.


Determine bond length via rotational spectroscopy Estimate the bond length of 12C16O (pure rotational spectrum) given J''=3 (15.356 cm-1) and J'=9 (38.356)

I understand that the first step is the calculate the rotational constant. So to do this I assume I would use one frequency value and one J value? And the equation to use would be

$$\nu = BJ(J+1)$$

If I do this with the numbers for $J=3$, then I get $1.28$ for $B$, which is apparently incorrect. I have seen someone else's attempt and they use

$$12B = (38.356 - 15.365)=1.916$$

And solved for B this way and they get the correct answer. Why is the way I thought not correct? Why is the second way correct and where did B come from?

The method they are using seems to be looking at the separation between two different transitions to determine $B$.

$J''$ denotes the lower energy state in a transition and $J'$ denotes the higher energy state. Here, you are looking at two different transitions: $$J_1'=4\to J_1''=3 \text{ with } \nu_{4\to 3}=\pu{15.365cm^-1}$$ $$J_2'=9\to J_2''=8 \text{ with } \nu_{9\to 8}=\pu{38.356cm^-1}$$

We can see in the figure below that $\Delta\nu=2B|J_2'-J_1''|$ or that the separation between transitions depends on difference between the $J$ of the higher state of the higher energy transition and the lower state of the lower energy transition. In your case, this gives that $\Delta\nu=2B|9-3|=12B$.

$\hspace{5cm}$

You could also get $B$ from the individual transitions, but you have used the wrong formula. $E_J=BJ(J+1)$ is the energy of the rotational state $J$, but you are given the energy of the transitions, which can be computed as $\nu=2B(J''+1)$. Checking with either transition should give the same result as using the spacing between the transitions.

One reason you might use the spacing to compute $B$, rather than individual transitions, is that if you have any systematic error in measuring transition frequencies, taking the difference will cancel out this error.


Effect of noble gas configuration on polarising power I don't understand this statement written in my textbook about the polarising power of cations:

If two cations have the same size and charge, then the one with pseudo noble gas configuration ( with 18 electrons in the outermost shell) has greater polarising power than the other with noble gas configuration (with 8 electrons in the outermost shell). This explains why $\ce{CuCl}$ is more covalent than $\ce{NaCl}$.

Why does this happen?

Recall the fact that the basic difference between an ionic bond and a covalent bond, is basically just charge separation. In ionic bonds, charges are well separated, and the bond arises due to opposite charge attractions. Covalent bonds, on the other hand, have the electrons shared in between them, and it is these electrons that hold the nuclei of the two bonded atoms together.

But sometimes, a covalent bond isn't completely covalent, nor is an ionic bond completely ionic. Even an ionic bond may have some degree of "covalentness" in them. You can compare the covalentness of two ionic bonds with the help of a few rules formulated by Fajans. A few of these rules would be:

Rule 1: Smaller the cation and greater the charge, more the covalent character.

Why is this true? If you consider a cation to be similar to a uniformly charged sphere, the electric field generated at it's surface would be:

$$E=\frac{1}{4\pi \epsilon}\frac{Q}{R^2}$$

Where $Q$ is the charge on the cation, and $R$ is it's radius. It's quite clear now that highly positive cations and smaller cations can generate higher electric fields, and can strongly pull the electrons of the anion into the space between the ions, which gives rise to a partial covalent character.

In your example, it was given that "cation with pseudo-inert gas configuration can polarize better than a one with inert gas configuration". This is very much true, and can be explained using screening effects of orbitals. In the case of $\ce{Na+}$, it's valence electrons are in the $p$ orbitals, while in the case of $\ce{Cu+}$, they are in the $d$ orbitals. It's a known fact that $d$ orbitals shield the electrons from nuclear attractions poorly compared to the $p$ orbitals (see Why do screening effect decreases due to d-orbital? for more details). As a consequence, the nucleus attracts the valence electrons much more strongly in case of $\ce{Cu+}$, leading it to have a smaller size, and more polarizing power.

Rule 2: Greater the size of the anion, more is it's polarizability and greater is the covalent character.

Now the first question that would arise is, what is polarizability and why does it affect covalent character? Polarizability refers to the ability of the anion to get polarized, which is charge separation within the ion. Larger anions allow more room for the electron clouds within them to move around, which causes the cation to polarize it better. See the picture below for better clarity:


Is sodium hypochlorite solution an ionic solution - does it conduct current? Does an aqueous solution of sodium hypochlorite conduct current?

Yes, it is a solution of an ionic salt. a table lists electrical conductivity at various concentrations.


What's a minimal yet chemically-meaningful kinetic system for an oscillating reaction? Oscillating reactions are a funny aspect of chemistry. I have tried to find various simplified kinetic models of oscillating reactions such as the Belouzov-Zhabotinsky, the Briggs–Rauscher or the Bray–Liebhafsky reactions in order to be able to study them. However, the models I have found so far are either too complicated (8 or more species considered), or loose chemical significance. For an example of the second case, Ball's 1994 model[1] corresponds to:

While this is useful and introduces a nice circularity in the model, creating the possible feedback, it has lost all chemical sense — by which I mean that no correspondance can be established between the species of this model and a real oscillating system.

So, my question is: what's the simplest known chemically-meaningful model of an oscillating reaction?

[1] Ball, P. 1994 Designing the molecular world: Chemistry at the frontier. Princeton, NJ: Princeton University Press.

I'm not sure if this is what you're looking for (it may well still be too abstract), but oscillations can occur in the Gray-Scott system, which is
$$
A + 2B \to 3B\\
B \to P,
$$
where $P$ is an inert product, and the reaction is assumed to take place in a flow reactor that provides a supply of A, giving rise to the dynamics
$$\begin{align}
\frac{\mathrm da}{\mathrm dt} &= f(1-a) - ab^2;\\
\frac{\mathrm db}{\mathrm dt} &= ab^2 - (f+k)b,\\
\end{align}
$$
where $f$ is a rate determined by the flow reactor, $k$ is the rate of the $B\to P$ reaction, and the rate constant of the autocatalytic reaction has been set to 1 without loss of generality by scaling $f$ and $k$ relative to it.

With the appropriate choice of the parameters $f$ and $k$, oscillation can happen because $B$'s concentration increases autocatalytically, but then it overshoots its food source (i.e. the concentration of A), which then builds up again, allowing the cycle to repeat.

You might not like the trimolecular step, but I've found you generally get similar behaviour if you split it up into something like
$$
A + B \to C + B\\
B + C \to 2B\\
B \to P.
$$
(Simply using $A+B\to 2B$ doesn't work, because its kinetics don't have the right sort of nonlinearity.)

I would say that this has an advantage over the Ball model that you posted in that it obeys the laws of thermodynamics. (At least from what you showed in your question, it looks to me that Ball's model only oscillates because the reverse reactions have been neglected, and if they weren't then it would have to go to equilibrium, because it's a closed system.) It makes explicit that you need a power source (the supply of $A$) for oscillation, and illustrates the connection between oscillatory behaviour and autocatalytic kinetics.


Why has nature evolved to put a porphyrin (i.e. chlorophyll) as the chromophore in a leaf? I'm stumped by this question:

  Why has nature evolved to put a porphyrin (i.e. chlorophyll) as the chromophore in a leaf?

I've thought that it might have something to do with how sun emits light at different wavelengths, in the green wavelength of 565 especially (as the sun emits green light), and therefore, through evolution, plants have evolved to incorporate chlorophyll as the chromophore in the leaf to absorb the wavelength of the sun?

Chlorophyll (chl) is involved in two ways in photosynthetic organisms firstly as a light gathering pigment and second in electron transfer once the light is captured. It is unique in this role.

In the first case, that of gathering light there are pigment protein complexes containing many chl (100's) and work in such a way that energy absorbed in one molecule is transferred among the others until it reaches the reaction centre where electron transfer starts. The feature that makes chl good for the energy transfer bit is that it has a large overlap in energy (equivalently wavelength) between its fluorescence spectrum and the absorption spectrum. This allows very efficient energy transfer between pigments, each step taking only picoseconds. Chlorophyll is among the best if not the best molecules known to do this.

(In some cases carotenoids may assist this energy capture process and in some antenna called phycobilisomes linear tetrapyrroles augment chl (they extend the wavelength range when the organism normally grows in low light conditions) but in these cases there are also chl pigment-protein complexes.)

The second role is that after energy is absorbed it is funnelled to a Reaction Centre. This is done by the energy transfer steps described above but slight changes in the chl environment change its energy levels slightly so that spatially the energy moves towards the reaction centre where electron transfer begins. The electron produced ultimately reduces carbon dioxide to carbohydrate and is replaced by dragging an electron of a water molecule, this is done in the protein in a sequence of complex steps called the dark reactions.

Returning to the Reaction Centre, this consists of a pair of slightly offset face-to-face chl molecules (the Special Pair)  that act as an energy trap and from here an electron is transferred in a couple of picoseconds to a nearby chl then to a pheophytin (a chl less the Mg atom) then to quinone then to an Fe atom all within a nanosecond. The crystal structures are all known and the photophysics of species involved also measured. The structure of these Reaction Centres can be found on the web.

The important property here is that the special pair of chl molecules has the right redox potential to donate an electron to a nearby chl and this the correct potential to transfer the electron to the pheophytin at the same time preserving as much energy as possible and preventing the electron from returning, i.e. making sure it goes down hill in energy but only slightly and thus preserving as much of the initial photon energy as possible. The special pair does not fluoresce to any appreciable extent and so does not loose energy this way.

Chlorophyll has a unique set of properties, excellent ability to transfer energy among other chl with out being quenched until it reaches the reaction centre where the special pair uses this energy to transfer an electron to nearby molecules and start the chemistry leading to reducing carbon dioxide using water as a fuel. All the variant of photosynthesis, algae, plants, bacteria work in essentially the same way, the bacteria have one photosystem, the plants two (PS1, PS2) that work together.

(You should be able to find lost pictures of xray structures of pigment protein complexes on the web but if not I can post some later)


Why do we need to assume that gas is made up of large number of small particles in kinetic molecular theory of gases? Among the five postulates or assumptions of kinetic molecular theory of gases (KMT), only the assumption that a gas is made up of large number of small particles doesn't seem to make any sense to me. The only reason I could of think of was due to lack of negligible pressure. I reasoned that if there aren't enough molecules in the container there will be no evidence of pressure, since most of collision would be within molecules and not between the wall of the container and the molecules. But there is something missing in this explanation or it is completely wrong, as it fails to adequately explain the above stated question.
Why do we need this assumption?

To echo and expand on Porphyrin's comment, the idea of having a large number of particles shows up in KMT because this is a statistical theory. That is, everything KMT predicts is some average property, or, to be rigorous, the expectation value of a distribution.

Thus, if one only had a very small number of particles, you would never arrive at something like the Maxwell-Boltzmann distribution because you would have to treat each particle individually rather than averaging over a phase-space.

So, to answer your actual question, we need this assumption because it justifies us in doing things like the derivation in this link where they take the derivative of the unknown function (eq. 9.8). That assumes this function is continuous over all space because no constraints are put on the resultant function. The only way we can do that and expect to have our theory match reality is if reality has an approximately continuous distribution of particle speeds, i.e. a lot of particles present.

As it happens, $10^{23}$ particles really isn't that many in practice, so when you get down to very low pressures and temperatures (that is the number density $\frac{N}{V}$ is small) the theory still works because something like $10^6$ probably still makes a pretty continuous function.

As a totally unrelated aside, you might also wonder if the MB distribution is wrong because it allows particles to have speeds ranging from $0\rightarrow \infty > c$... I computed the special relativistic correction one time and it's so laughably small as to actually annoy you that you even took the time to do it.

I know this answer focused mostly on the MB distribution, but that is very central to KMT so I thought it would make for a representative though not exhaustive answer.

EDIT:

To be critical of my answer, however, I'm almost certain that much of KMT is derivable while only thinking about a single particle sampling its whole phase-space (perhaps all at once) and once could simply argue that the above function I mention taking the derivative of is a hypothetical distribution in which each of the infinite states need not be occupied at any given time. So basically taking a time average of the phase-space over infinite time gives this continuous distribution. That idea also lines up well with ideas to do with the partition function where every state could be equally likely but we get a non-even distribution because some states are more likely to be sampled.

Then perhaps it is just more physically instructive to think about a large number of particles.


Can you meassure a negative pH value with a pH meter? I was checking the $\mathrm{pH}$ of $\pu{1 M}$ chromic acid solution, and I got its $\mathrm{pH}$ around -1.7. Yet, $\mathrm{pH}$ of $\pu{0.4 M}$ solution was around -0.6. Could these values be correct? The $\mathrm{pH}$ meter was calibrated before use.
I'm thinking not. I guess $\mathrm{pH}$ meters work in the interval 2-12 $\mathrm{pH}$.

You are most likely getting an inaccurate value. As anticipated in Poutnik's comment above is difficult to reliably measure a $\mathrm{pH}$ of very acidic solution but the $\mathrm{pH}$ scale is indeed open, and negative $\mathrm{pH}$ are a real thing.
It is easy and more meaningful to state a very high concentration or activity rather than reporting inaccurate $\mathrm{pH}$. This, together with typical figures found in textbooks, has spread the misconception that $\mathrm{pH}$ is bound in the range 0 - 14, but it is not.
A simple example of a solution with negative $\mathrm{pH}$ is in every lab, namely conc. $\ce{HCl}$, 37% by mass, $\mathrm{pH} = -1.1$.
Edit: For sake of clarity, as for its definition, pH can exceed 14, too. As an example, a saturated NaOH aqueous solution has pH = 15.


Aluminium Oxide as Dehydrating agent According to this paper on dehydration by $\ce{Al2O3}$,

  Dehydroxylation causes a surface defects as Lewis acidic centers (unsaturated $\ce{Al^3+}$ ions) and basic Lewis centers ($\ce{O^2-}$ ions) creation. Water molecules deactivate Lewis centers and create strongly acidic Brönsted’s centers. Alcohols dehydration reaction performs on Brönsted’s active centers where hydroxyl groups disconnect proton from alcohol molecule. During reaction alcohol molecule adsorbed on aluminium oxide surface reacts with proton.

But I am not able to understand what they are trying to convey here.

Alumina absorbs water to give aluminium oxide hydroxide, a solid which removes the water from the reaction environment.

According to the paper by Pethrick et al. [1]:

  Aluminium oxide exists in several forms which on addition of water give a complex range of oxide-hydroxide; boehmite, bayerite and gibbsite.

References

Pethrick, R. A.; Hayward, D.; Jeffrey, K.; Affrossman, S.; Wilford, P. Investigation of the Hydration and Dehydration of Aluminium Oxide-Hydroxide Using High Frequency Dielectric Measurements between 300 KHz–3 GHz. J Mater Sci 1996, 31 (10), 2623–2629. https://doi.org/10/chcqs8.


How does dimethyl sulfate react with water to produce methanol? I came across the reaction of dimethyl sulfate with water to yield methanol. I was curious how that works as $\ce{H2O}$ isn't a strong nucleophile.

From my understanding the $\ce{H2O}$ should pop off a $\ce{-CH3}$ and you have $\ce{[CH3OH2]+}$ which has a $\mathrm{p}K_\mathrm{a}$ of -2.5 so a relatively strong base. Does the $\ce{[CH3OH2]+}$ protonate the $\ce{H2O}$ to form $\ce{H3O+}$ and $\ce{CH3OH}$?

The mechanism of methyl transfer of $\ce{Me2SO4}$ is generally thought to be an $\mathrm{S_N2}$ reaction. Thus, water would attack the methyl group, releasing $\ce{MeO-SO3-}$, a good leaving group generating methanolium $\ce{H3C-OH2+}$. The proton is later transferred back to a methyl sulphate to give methyl hydrogen sulphate $\ce{HSO4Me}$ and methanol.

One could imagine the proton from methanolium to initially be transferred to water to generate hydronium. That’s actually pretty probable and it will be hard to tell where it will end up. Water, methanol and hydrogen sulphate (and therefore likely methyl sulphate, too) all have similar $\mathrm{p}K_\mathrm{b}$ values and are equally eager to displace it/non-eager to accept it. adding some base, e.g. sodium carbonate should provide a good ‘resting point’ for such an unhappy proton.

The question is why water can do this even though it is a lousy nucleophile. Well, it turns out that for one, $\ce{MeSO4-}$ is a great leaving group enhancing the reaction rate and for two, there is nothing else around if you mix just $\ce{Me2SO4}$ and water that would be a better nucleophile. ($\ce{OH-}$ is present in too low concentrations to actually count.) Thus, the two reagents have the choice of doing nothing or slowly doing something thermodynamically favourable (= reacting). Kinetics dictate that it must happen slowly, but it can happen.


What is the unit of Z effective? Z effective is the net positive charge experienced by an electron. But what is the unit of this net charge: coulomb (C) or microcoulomb (μC), or something else?

This is an example of terminology which should be taken with a grain of salt.
The term "effective nuclear charge" is often casually symbolized with $Z_\mathrm{eff}.$ This is a universally accepted simplification, but you should keep in mind that the effective nuclear charge is, strictly speaking, $Z_\mathrm{eff}e,$ where $e$ is the elementary charge.
Similarly, the full nuclear charge is $Ze,$ where $Z$ is the atomic number.
So, it would be more appropriate to refer to $Z_\mathrm{eff}$ as to effective atomic number since $Z_\mathrm{eff}$ is defined as a function of $Z$ $(Z_\mathrm{eff} = f(Z_i)).$
For example, the common form provided in every introductory physical chemistry textbook is
$$Z_\mathrm{eff} = Z − σ,$$
where $σ$ is the shielding constant of the nucleus (empirical dimensionless parameter).
Since $Z$ is dimensionless, so is $Z_\mathrm{eff}.$
However, effective nuclear charge $Z_\mathrm{eff}e$ is not, and can, for instance, be expressed in coulombs (in the SI system $e = \pu{1.602176634E−19 C}).$


Solubility of Lead in water from leaded pipes in Roman aqueducts Lead can undergo the following reactions when cold and hard water passes over it:
$$\ce{Pb^2+ + 2Cl- -> PbCl2}$$
$$\ce{Pb^2+ + SO4- -> PbSO4}$$

All these salts are sparingly soluble in water and have a tendency to stick to lead, rendering it passive and not allowing plumbosolvency to occur:
$$\ce{2Pb + 2H2O + O2 -> 2Pb(OH)2}$$

My question is:

How did lead dissolve in the water flowing through the aqueducts if most of it's salts are sparingly soluble?

I have considered erosion of the salt layer formed on the surface but according to Wikipedia, the aqueducts had a very gentle slope

  Vitruvius recommends a low gradient of not less than 1 in 4800 for the channel, presumably to prevent damage to the structure through erosion and water pressure. This value agrees well with the measured gradients of surviving masonry aqueducts. The gradient of the Pont du Gard is only 34 cm per km, descending only 17 m vertically in its entire length of 50 km (31 mi):

Wikipedia also has the following to say regarding lead concentration in the water:

  Where lead pipes were used, a continuous water-flow and the inevitable deposition of water-borne minerals within the pipes somewhat reduced the water's contamination by soluble lead. Nevertheless, the level of lead in this water was 100 times higher than in local spring waters.

Any insights would be much appreciated!

Let me give an extreme example of chemical analysis. It is an anecdote. Some analytical chemists were using a very sensitive technique to analyze metals in blood. It is called neutron activation analysis. What they saw was that a very small quantity of molybdenum (Mo) was present in small quantities persistently. Nobody expected that. It turned out after a lot of hard work that Mo was coming from the stainless steel syringe needle used for drawing blood. Lesson of the story: There is nothing which is perfectly insoluble in water. It all depends on the analytical technique. If you try to weigh a hair on a "weighing machine" it will tell you that the mass is zero kg, but if you use a fine analytical balance it will tell you that hair has a mass.

The same goes for lead pipes. Natural water has plenty of minerals, dissolved oxygen anions, cations and don't forget the microorganisms. The bacteria do a lot of chemical transformations in water. As a result, Pb would slowly leach into the water at parts per million or parts per billion levels. Don't assume that Romans were drinking lead salt solutions in high concentration ranges. Not only water lead was present in water, it was also present in alcoholic beverages.

Read more here Lead in ancient Rome’s city waters

Lead in water is still a serious problem in several areas of the modern world. It is not an ancient problem.


What does the letter R represent in chemical structures? The letter $\ce{R}$ appears frequently in chemical structure illustrations. For instance, when I Google keratin, I find Wikipedia's Spanish article about it.

In the image illustrating part of the chemical structure, I see elements like $\ce{H, C, O, N, and S}$ - which I recognize - but I also see $\ce{R}$, which I do not. What does $\ce{R}$ stand for in this context?

Depending on the context, the letter R can stand for several things. From our friends teaching organic chemistry at UCLA:

  R group: An abbreviation for any group in which a carbon or hydrogen atom is attached to the rest of the molecule. Sometimes used more loosely, to include other elements such as halogens, oxygen, or nitrogen. R is an abbreviation for radical, when the term radical applied to a portion of a complete molecule (not necessarily a free radical), such as a methyl group. Should not be confused with R (the gas constant), R (the one-letter abbreviation for the amino acid arginine) or R (a designation of absolute configuration).

In the image from the Wikipedia article you have referenced, the letter R is as defined in the first section of the above quote - that is, it stands for "any group in which a carbon or hydrogen atom is attached to the rest of the molecule."


Claisen Condensation to Ketone Can someone explain (and preferably draw) the intermediates of this reaction?

I understand that it is a Claisen Reaction, and likely a condensation due to the heat, but I am not sure why there isn't an ester in the final product.

Step 1 is a Claisen ester condensation with loss of ethoxide from the ester acting as the carbonyl part. The product is the $\beta$-ketoester (2).

Upon acidification and heating, the ester is cleaved, the resulting $\beta$-ketoacid decarboxylates to yield the symmetrical ketone.


Why does the SnAr reaction proceed by the Meisenheimer complex? I am studying SNAr mechanism from this webpage. I am confused as to why this convoluted mechanism occurs. It seems as though we simply create an anion when the hydroxide attacks in order to quench it directly after by pushing out the chloride. Why does the hydroxide not push out the chloride in a typical substitution reaction? Also, what evidence do we have for the Meisenheimer complex actually forming?

Basic level explanation:
SN2 occurs only at sp3 carbons, where you can get backside attack.

Intermediate:
Addition to the double bond generates a highly stabilized anion. This is a lower energy transition state than the concerted SN2.

Advanced:
The reaction occurs at the site of maximum frontier molecular orbital overlap. The HOMO (highest occupied molecular orbital, the hydroxide nucleophile) adds to the LUMO (lowest unoccupied molecular orbital,  π* of the arene). To do SN2, one would need to interact with σ*, which is not the LUMO, and has an unreachable trajectory (points towards the middle of the ring).


Reference request on formate potash process Wikipedia article on Potassium Formate mentions a so-called "formate potash process" for production of potassium, presumably referring to elemental metal.

Concise encyclopedia chemistry by Mary Eagleson, page 888 is a given as a reference, but I can't find a PDF, hence this request.

The reference is correct, from [1, p. 888]:

Potassium formate: $\ce{HCOOK},$ colorless, deliquescent, rhombic crystals; $M_\mathrm{r}~84.12,$ density $1.91$, m.p. $\pu{167.5 °C}.$ P. is formed by neutralization of aqueous formic acid with potassium hydroxide or potassium carbonate. It is an industrial intermediate in the formate-potash process (see Potassium carbonate).

However, formate-potash process doesn't yield in elemental potassium, rather potassium carbonate [1, p. 886–887]:

Potassium carbonate, potash:
[…]
P. is also produced by the formate-potash process, in which potassium sulfate and lime are treated under pressure with carbon monoxide:
$$\ce{K2SO4 + Ca(OH)2 + 2 CO -> 2 HCOOK + CaSO4}.$$
The resulting potassium formate is oxidized to P. by heating it in air:
$$\ce{2 HCOOK + O2 -> K2CO3 + CO2 + H2O}.$$

English Wikipedia page claims formate-potash process is for “production of potassium” probably because the author attributed “P.” from the encyclopedia entry to potassium.
Whereas in reality it refers to the entry title, “Potassium carbonate”, which is a standard referencing method in data sources such as a handbook or an encyclopedia.
German Wikipedia page doesn't have this flaw and informs that “Formiat-Pottasche-Verfahren” is a technically insignificant method of producing $\ce{K2CO3}.$
References

Concise Encyclopedia Chemistry, English language ed., 2nd ed.; Scott, T., Eagleson, M., Eds.; de Gruyter: Berlin; New York, 1994. ISBN 978-0-89925-457-9.


Does battery self discharge when connected? I am an electric engineer, and I read about battery self discharge phenomenon. I was wondering whether a battery self discharge rate is the same when it is unconnected (i.e. in storage) and connected (i.e. when it is plugged into a device and being discharged "normally").
In other words, when I calculate battery life time for a product, should I take add the rated self discharge of a battery to the general current calculations or not? And also-is the answer similar to all types of battery?

Great question! I studied battery chemistry as part of my graduate work, and I think the answer depends on the mechanism of self-discharge.

I think that by discharging the battery through an external circuit, you will generally cut off the self-discharge mechanisms (depending on the discharge rate), so if you add the discharge rate and the self-discharge rate, you will probably underestimate the battery life. You should also keep in mind that the rate of self-discharge depends on the state-of-charge for the battery, and is generally worse at high state of charge. Thus, even the manufacturer specification for self-discharge rate isn't a constant.


Can the total amount of solution be found as a ratio between molar mass of a component and total mass of solution? I wonder whether the following relation is true:
$$n_\mathrm{solvent} + n_\mathrm{solute} = \frac{M}{m_\mathrm{solvent} + m_\mathrm{solute}},$$
where $M$ is the molar mass of the component, $n$ is the amount of substance and $m$ is the mass.
It was derived assuming $n = m/M,$ $n = n_\mathrm{solvent} + n_\mathrm{solute}$ and $m = m_\mathrm{solvent} + m_\mathrm{solute}.$
I don't think this is true, but I wanted to be sure before doing anything weird on a test.

To sum up the comments, only the following relation for the total amount of solution $n_\mathrm{tot}$ is universally true:
$$n_\mathrm{tot} = n_\mathrm{solvent} + n_\mathrm{solute} = \frac{m_\mathrm{solvent}}{M_\mathrm{solvent}} + \frac{m_\mathrm{solute}}{M_\mathrm{solute}}\tag{1}$$
The best you can do is to assume that $n_\mathrm{tot}\approx n_\mathrm{solvent}$ for the diluted solutions of small molecules. Also, if the molar masses are similar $(M_\mathrm{solvent}\approx M_\mathrm{solute}\approx \bar{M}),$ the expression can be lead to a common denominator:
$$n_\mathrm{tot} \approx \frac{m_\mathrm{solvent} + m_\mathrm{solute}}{\bar{M}}\tag{2}$$
This can be the case, for example, for the solution of ammonium nitrate $(M(\ce{NH4NO3}) = \pu{80.043 g mol^-1})$ in dimethyl sulfoxide $(M(\text{DMSO}) = \pu{78.13 g mol^-1}).$
Algebra aside, your mistake was also neglecting $M_\mathrm{solvent}.$
Keep in mind you can always check dimensions.
For your proposed formula

$$n_\mathrm{solvent} + n_\mathrm{solute} = \frac{M}{m_\mathrm{solvent} + m_\mathrm{solute}}$$

it doesn't work out well:
$$\mathrm{dim}~n_\mathrm{tot} = \mathsf{M}\cdot\mathsf{N}^{-1}\cdot\mathsf{M}^{-1} = \mathsf{N}^{-1} \neq \mathsf{N}\tag{3}$$
Illustrating with common units used in chemistry:
$$[n_\mathrm{tot}] = \frac{[M]}{[m_\mathrm{solvent}] + [m_\mathrm{solute}]} = \frac{\pu{g mol^-1}}{\pu{g}} = \pu{mol^-1} \neq \pu{mol}\tag{4}$$


How do I find the final temperature of a calorimeter using specific heat capacities? What is the final temperature of water and iron if a $\pu{30 g}$ piece of iron at $\pu{144 °C}$ was dropped into a calorimeter with $\pu{40 g}$ of water at $\pu{20 °C}$?
Specific heat of water is $\pu{4.184 J g-1 ^\circ C-1}$, and of iron is $\pu{0.449 J g-1 ^\circ C-1}$

Here is my work:
\begin{align}
Q &= mc\,\Delta T\\
Q_1 &= (\pu{30 g})(\pu{0.449 J g-1 ^\circ C-1})(x - \pu{144 ^\circ C}) \tag{Iron}\\
Q_2 &= (\pu{40 g})(\pu{4.184 J g-1 ^\circ C-1})(x - \pu{20 ^\circ C}) \tag{Water}\\
\text{Since,} \quad Q_1 &= -Q_2\\
13.47 (x-144) &= - (167.36)(x-20)\ \pu{J}\\
13.47x - 1939.68 &= -167.36x + 3347.20\\
180.83x &= \pu{5286.88 ^\circ C} \\
x &=  \pu{0.03420 ^\circ C}
\end{align}

This is giving me an answer that isn't correct according to my book. What did I do wrong, and how can I fix it?

All you did is essentially right, your only mistake is in the last step, as LDC3 already pointed out in the comments. However, I am encouraging you to use units all the way and when dealing with thermodynamics use Kelvin instead of Celsius.
\begin{align}
Q &= mc\Delta T\\
\end{align}
Now you can form the equations for each of the problem, while substituting $\Delta T$ with a temperature range, being $x$ the final temperature the whole system will end up on. Also note, that the iron will be cooled down, while the water will be heated. (I am using a different approach than you.
\begin{align}
Q_\mathrm{loss} &= m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x]\\
Q_\mathrm{gain} &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\
\end{align}

The transferred heat has to equal $$Q_\mathrm{gain} = Q_\mathrm{loss}$$

With this you can solve for $x$.
\begin{align}
m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x] &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\
%m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe})-m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x-m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})\\
% m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x+m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x\\
% m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&= [m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})]\cdot{}x\\
x &=\frac{m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})}{ m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})}\\
%
x &=\frac{30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}\cdot{}417~\mathrm{K} +40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}\cdot{}293~\mathrm{K}}{ 40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}+30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}}\\
x &= \frac{5616.99~\mathrm{J}+49036.48~\mathrm{J}}{167.36~\mathrm{\frac{J}{K}}+13.47~\mathrm{\frac{J}{K}}}\\
x &= \frac{54653.47}{180.83}~\mathrm{K} =302.24~\mathrm{K}\\
x &\approx 29~\mathrm{^\circ{}C}
\end{align}


Why does benzoyl peroxide homolytically cleave to give phenyl radical Benzoyl Peroxide cleavese rapidly to give out $\ce{CO2}$ and phenyl radical.

What I could not understand was why phenyl radical is formed because it is not at all resonance stabilized and nor any electronic effects are acting on it that could help in its stabilization then why is it being formed ?? Wouldn't it be too unstable ??

I mean after first cleavage, (Before $\ce{CO2}$ leaves) it should be more stable due to resonance (and that too equivalent resonating structures)

The mistake you're making (which is an easy one to make) is that you're neglecting to consider the carbon dioxide product. Carbon dioxide is a stable molecule, and its release also increases the entropy of the system.

So even if the phenyl radical is less stable than the benzoyl radical intermediate, the combination of phenyl radical + $\ce{CO2}$ can still have an overall lower free energy than the benzoyl radical.


Making nylon experiment - Questions I made nylon in a lab with these procedures -

1) Collect 30mL of sebacoyl Chloride dissolved in hexane

2) collect 30mL of 1,6-hexanediamine dissolved in water .

3) Transfer the beaker containing sebacoyl Chloride in hexane and carefully pour the solution down the side of the beaker containing 1,6-hexanediamine in water , such that the hexane layer floats on TOP of the water layer , without mixing the layers.,

I got these procedures from the Internet .

Now , I got a few doubts and questions ..

In the past , dichloromethane was used as the Solvent for the sebacoyl Chloride reagent. However dichloromethane is not environmentally friendly solvent and hexane was chosen to replace it in the experiment ...

a) how does switching it to hexane change the experimental procedures ?

b) What physical and chemical properties of the Solent must be considered when switching ?

c) why is hexane a good alternative in this experiment ?

Thanks in advance !

Nylon reaction:

$\ce{nH2N(CH2)6NH2 + nClOC(CH2)8COCl → H2N[(CH2)6NHCO(CH2)8]_{n}COCl + nHCl}$

Looking at the reagents, 1,6-hexanediamine is dissolved in water as it can form hydrogen bonds with the water molecules. Meanwhile the sebacoyl chloride is non-polar and thus dissolves best in non-polar solvents. The two solvents (polar and non-polar) cannot dissolve each other, just like oil and water.

The hexane is poured so it floats on top to minimize mixing -- the reaction only happens at the boundary between layers. Then we pull the thread causing more of the reagents to mix and react at the boundary. If we poured the hexane solution in quickly, we'd end up with a large lump of nylon instead of a thread.

To help with your other questions, some questions...

I've given some details about the solvents above. What kind of solvents would you choose to replace the hexane and water with?
Looking at the densities of hexane (0.659 g / mL) and water (1 g / mL), and looking at the procedure, why was the hexane poured? Why not pour the water onto the hexane?
Following on, do some research into the density of dichloromethane. That will give you your answer to (a)
You've mentioned "not enviromentally friendly" for dichloromethane. Are there any other issues? A good starting point are the CLEAPSS student safety sheets.


What is the effect on humans of using dry ice and liquid nitrogen as food additives? I have seen on a TV show that some  guy uses dry ice (solid $\ce{CO2}$) to make ice cream and on an another TV show, liquid nitrogen ($\ce{N2}$) is used with tequila.  Both are edible.

I know that both gases are inert, but could these gases (perhaps in larger concentrations) be poisonous or create any side-effects in our body?  If these gases don't create any side effects, what mechanism(s) does the body use to remove these gases from the body?

Nitrogen has a boiling point of 77K (-196 °C), CO2 melts at 195K (-78 °C). You don't cool down the ice cream or whatever else you want to eat to those temperatures, that would be very unsafe. Both will evaporate or sublimate quickly when they are in contact with warmer material, so there won't be much left once you eat e.g. the ice cream.

Air is composed of 78% N2 and 0.04% CO2, nitrogen gas can only be dangerous when it replaces all the oxygen in the air. Carbon dioxide is toxic at around 7-10% concentration. Both conditions are not realistically achievable by the methods you describe.

The biggest danger I see would be ingesting a piece of dry ice shortly after it was put in a drink, which would be cold enough to hurt.


Why absorption/transmission spectroscopic data is normalized? A typical data processing step for acquired absorption/transmission data is normalizing, i.e. stretching the curve such that it is bounded between 0 and 100%.

However, the absorption/transmission curve changes with the density and thickness of the sample. Moreover, absorption is a non linear phenomena (for example Beer–Lambert law), so I would expect that the relative height of the various peak would not remain constant across several samples of the same material.

Why, therefore, is it justified to normalize acquired spectroscopic data? Why not present data with original abs./trans. values, while providing details about the sample itself (for example thickness)?

In addition, how can spectroscopy be quantitative if the relative height of the peaks changes with illumination/thickness/density/etc.?

There is confusion here about absorption at one wavelength and the absorption spectrum. At each wavelength the Beer-Lamber law applies $$I_{trans_{\lambda}}=I_{0_{\lambda}}\exp(-\epsilon_{\lambda} c l)$$ for a compound with extinction coefficient $\epsilon$ at wavelength $\lambda$ and concentration c and path length l.

When an absorption spectrum is produced it is normal to show it as optical density vs wavelength, i.e. $OD_{\lambda}=\log(I/I_0)_{\lambda}=-\epsilon_{\lambda} c l $ and in this case the ratio of peaks for the same compound is constant no matter what the concentration. (Naturally assuming that it does not absorb so much that the instrument fails).


Why is it called "bicarbonate"? I'm helping to teach some kids about the baking soda/vinegar reaction, and was hoping to point out how the chemical names are just a direct translation of the chemical formula.

However I was surprised to find there is no "2" in the formula for "sodium bicarbonate", $\ce{NaHCO3}$.

Can someone explain where the "bi" comes in?

From Wikipedia

  The prefix bi in bicarbonate comes from an outdated naming system and is based on the observation that there is twice as much carbonate ($\ce{CO3}$) per sodium in sodium bicarbonate ($\ce{NaHCO3}$) as there is in sodium carbonate ($\ce{Na2CO3}$).

The IUPAC name for it agrees with its chemical formula: sodium hydrogen carbonate.


Why are zinc shot and mossy zinc not considered toxic but zinc granules are? In our lab we have three forms of zinc. We were initially experimenting with these three forms to optimize our application. We have "zinc shot", "mossy zinc", and "zinc granules". The granules are ~20 mesh. I am tasked with creating chemical standard operating procedures (SOPs) for all reagents that are considered "particularly hazardous". Zinc granules are considered particularly hazardous by our University and is an "acute toxin to aquatic life with lasting effects" in a recent safety data sheet (SDS) while zinc shot and mossy zinc are not considered hazardous by our University or on comparable recent SDSs. I see zinc powder is also considered toxic. To me this suggests particle size dictates toxicity, but why? I understand surface area influences the rate of reactions but if the effects are long lasting, then why are all forms of zinc not considered to be an aquatic life toxin?

The key to toxicity is concentration, which is related to how rapidly the material disperses (or concentrates) in the environment. Which is also related to water type (fresh or salt) and temperature. MSDS sheets are a bit of a joke. They are worded in a way that fends off lawyers rather than gives detailed useful toxicity data. Consider this MSDS for Sodium Chloride
Bear in mind this useful advice when using it at home: "Eye Protection Wear safety glasses with side shields (or goggles) (European standard - EN 166). Hand Protection Protective gloves"


How does the litmus pH indicator work? I'm wondering about how does the litmus solution work as a pH indicator.

And another question:

When you put drops of litmus solution into an colorless acid (e.g. $\ce{HCl}$), it turns red. But then if you add colorless base (e.g. $\ce{NaOH}$) to the mixture, it turns blue. How come?

I thought that the litmus solution doesn't change color after it's changed color once already after a chemical change with an acid or a base. In this case, I thought that the mixture wouldn't turn blue after adding the $\ce{NaOH}$ into the mixture. But it seems that it still works well after used once.

How come?

This explanation is true for most organic indicators.  Litmus is a complex mixture of organic molecules.  The molecule responsible for the color change (the chromophore), is 7-hydroxyphenoxazone (image taken from Wikipedia, public domain):

See that $\ce{OH}$ group at the lower left?  That's where the so-called "acidic proton" is.  This molecule is actually a weak acid, meaning that in the presence of a strong base, the $\ce{OH}$ group can lose a proton, which reacts with an hydroxide ion ($\ce{OH^{-}}$) to form a water.  The 7-hydroxyphenoxazone is now a negatively charged ion: it looks the same except it is missing the $\ce{H}$ on the $\ce{OH}$ group, and it leaves behind an electron.

To think about the equation that describes what's going on, pretend that the entire 7-hydroxyphenoxazone molecule is represented by the formula $HLit$, where $\ce{H}$ is the hydrogen on the $\ce{OH}$ group and $Lit$ is everything else (this is common procedure: only the hydrogen matters in the acid-base equation).  Then the acid-base equation is:

$$\ce{OH^{-} + HLit <=>H2O + Lit^{-}}$$

As more strong base is added, more hydroxide ion is present, driving the equilibrium to the right and producing more $Lit^-$ ion.  And now, the color change: $HLit$ is red, but $Lit^-$ is bue.  So the more acidic your solution, the more the equilibrium is to the left and thus there is more $HLit$ than $Lit^-$.

Litmus is pretty good for checking rough acidity or basicity, because its midpoint pH is close to 7 (neutral).  Litmus is as red as it gets at a pH of 4.5, and is as blue as it gets at a pH of 8.3, so the midpoint is about 6.4, a bit more acidic than pure water.


Liquid that burns in a vacuum This may seem like a silly question, but is there a flammable liquid that could oxidise itself, so it could effectively burn in a vacuum?

Indeed yes, and it's been used in all kinds of airless situations, including torpedos and rockets and I think even submarines.  It's hydrogen peroxide, which when exposed to a catalyst will oxidize and reduce itself:

2 H2O2(l) -> 2 H2O(g) + O2(g)

You'll notice that the oxygens on the left are in oxidation state -1.  Half of them become reduced to the O in water (oxidation state -2) and half become oxidized to the O in O2 (oxidation state 0).


Gaseous state: Boyles, Charles and Lusaac's law We know that pressure is inversely proportional to volume (Boyles law) and volume is directly proportional to temperature ( Charles law). So then won't pressure be inversely proportional to temperature?

Actually, it's a big NO.

Why?

Because you need to consider the statement of the laws carefully. Boyle's law states that pressure is inversely proportional to volume at isothermal conditions. Boyle's law is not valid if temperature isn't constant. The same goes for Charles' law, where you must have isobaric conditions. The conditions required for both laws aren't the same, and so you can't really make a logical conclusion that way.


A Tool to atomize liquid in the milliliter range for some graduate project I'm trying to create a mist from some water based solution to apply it to a surface, basically the surface is solid but very gentle and I need to spray it with this solution as evenly as possible, I cannot apply the solution with a syringe and then spread it around since this will pattern the surface, also the volume applied to the surface is important, I need to spray 2ml+-0.5ml, spraying 4ml for example would be too much

I've looked into an air brush but I couldn't find a way to control the volume and I've looked into a spray bottle but the volume there is too big.
any advice or suggestion would be welcome for a tool that can do this would be great!

You might want to look into using an inkjet (salvaged from an old printer). They can produce a regular, directed stream of small droplets and have a capacity of at least several mL. You can power them and control their behavior electronically. There are probably inkjet printers available that print onto a fixed platen, so that would take care of the production aspects.


Preparation of Effexor I take venlafaxine (Effexor) as medication, and having wrapped up 2 semesters of Organic Chemistry I thought it might be fun to design a synthesis for it. A picture of venlafaxine:

I came up with the following preparation (beginning with phenol) and was hoping someone could check it for me:

React phenol with $\ce{NaOH}$ to form phenolate, followed by methyl iodide to form methoxybenzene.
This forms a moderately activated ring with ortho/para direction.
Friedel-Crafts alkylation: allyl bromide (3-bromo-1-propene) and $\ce{AlBr3}$ adds propene to the para position.
Ozonolysis with DMS to give p-(acetaldehyde)methoxybenzene.
Reductive amination with dimethylamine and triacetoxyborohydride.
Radical bromination should be selective for the benzylic position.
Finally, Grignard reaction with cyclohexanone.

You may not necessarily need to start your procedure beginning with phenol. You may start it with readily available (for cheap price compared to the trouble you are going to have by handling methyl iodide, allyl bromide, and $\ce{AlBr3}$ or $\ce{AlCl3}$) 4-anisaldehyde (4-methoxybenzaldehyde). That is the exact product you are getting after your step 3. Although, steps 4-6 are some very thoughtful organic synthesis steps, I doubt it will give you the expected product. Therefore, I suggest following steps to follow:

Buy 4-anisaldehyde (4-methoxybenzaldehyde) from Sigma-Aldrich (available in 98% purity.
Prepare Grignard reagent using chlorocyclohexane/$\ce{Mg}$ in diethyl ether.
React Grignard reagent (cyclohexyl magnesium chloride) with 4-anisaldehyde.
Perform dehydration of resultant $2^\circ$-alcohol to give double bond.
Perform epoxydation of double bond with 3-chloroperbenzoic acid.
React epoxide with nitromethane.
Reduce nitro group to $1^\circ$-amino group.
Finally, convert $1^\circ$-amino group to dimethyl amino group by reductive amination with formaldehyde.


The Lewis structure for NCO- I was having trouble drawing the Lewis structure for $\ce{NCO^-}$. I found this image online:

But why is structure number one preferred over structure number two? Why does carbon prefer to have a triple bond with nitrogen, rather than have a double bond with nitrogen and a double bond with oxygen?

If you count electrons and determine the formal charge on each atom, you find that in structure #1, the negative charge is on the oxygen.  Do the same exercise for structure #2 and you find that the negative charge is on nitrogen.  Since oxygen is more electronegative then nitrogen, the negative charge is more stable when its on the oxygen atom.  Therefore, structure #1 is more stable (preferred) than structure #2.

Structures #3-#5 are even less stable.  They either violate the octet rule (#4, #5) and / or have charges (sometimes multiple) on more than one atom (#3-#5).


Prove that a Change of Volume will not Change the Value of the Equilibrium Constant According to my book, a decrease in volume will result in an increase in pressure for an equilibrium mixture, which will in turn cause its equilibrium to shift to the side with the least number of gas molecules, given the reaction involves gases (and according to Le Chatelier's Principle).

It also says that Volume and Pressure changes will cause the position of equilibrium to shift, but the equilibrium constant remains unchanged...

If the equilibrium constant (K) depends on the concentrations of reactants and products...

and according to the molarity formula, Volume is inversely proportional with respect to concentration...

... then why does the equilbrium constant remain unchanged?

So far I have this, but I don't know what it means...

In your workyou have changed the "equilibrium constant" which is a fixed while keeping the number of moles constant of all species in both instances. Rather its the opposite

Consider the reaction;$$\ce{AB <=> A + B }$$ with equilibrium constant $K_c$ at temperature $T$ and volume $V_0$

Initially let $n(AB) = x_0$ , $n(A) = y_0$ , $n(B) = z_0$ where $n()$ refers to the number of moles.

then $$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}}$$

Now the volume is reduuced from $\ce{V_0\bond{->}V}$
Hence all the concentrations increase.
Then the reaction quotient at the moment the volume is changed;$$Q_c =  {\frac{{\frac {x_0}{V}}{\frac {y_0}{V}}}{{\frac {z_0}{V}}}}$$

Now $Q_c > K_c$ hence equilibrium shifts to the right reducing the number of moles of $A$ and $B$ increasing the number of moles of $AB$.Due to this effect $Q_c$ will gradually reduce and at acertain point will becaome equal tp $K_c$ with different number of moles.

$$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}} = {\frac{{\frac {x}{V}}{\frac {y}{V}}}{{\frac {z}{V}}}} $$

The only parameter remaining constant(under constant temperature) is the Equilibrium constant is a certain parameter(i.e. Volume here) is changed, other parameters will vary keeping Equilibrium constant a constant.


Formation of acid chlorides In Sn2 nucleophilic substitution reactions,the attacking nucleophile or leaving group is eliminated. The weaker base is eliminated. Then how do acid chlorides form as OH- is a stronger base than Cl–.

Acid chlorides are normally formed from carboxylic acids by the action of reagents such as thionyl chloride (SOCl2). This is not a SN2 reaction well not at carbon.

Normally a carboxylic oxygen attacks the sulfur of the thionyl chloride displacing an chloride from the thionyl chloride. After loss of a hydrogen ion from the carboxylic acid we will have

R-C(=O)-O-S(=O)-Cl

A chloride anion then attacks the carbonyl carbon as a nucelophile, the sulfur group acts as a leaving group. It will form sulfur dioxide and another chloride anion. In terms of the reaction at the organic part of the molecule it should be classified as a nucelophilic attack on a carbonyl compound which results in the substitution of one group for another. In terms of mechanism it is very similar to the formation of an ester from an alcohol and a carboxylic acid.

Here is the mechanism


Detection of oxygen by Lassaigne test I have read that compound with nitrogen and carbon atoms gives Prussian blue color when sodium extract of the compound is added with $\ce{FeSO4}$ . Will the test work even if the compound has oxygen atoms?

Sodium-Fusion test or Lassaigne test was originally introduced by Lassaigne in 1843 (Ref.1), and was subsequent modified by Jacobsen in 1879 (Ref.2). The major modification of Jacobson was the replacement of potassium by sodium metal as the active reagent. Since then, the test has altered very little, and basically is still in use today. The difficulties associated with the test have been reviewed in 1969 (Ref.3).
The sodium-Fusion test or Lassaigne test is a general test for the detection of halogens, nitrogen, and sulfur in an organic compound. These elements are usually covalently bonded to the organic compounds such as halo, nitro, amino, and thio compounds. In order to identify their presence, each covalently bonded element has to be converted into their ionic forms. This task is achieved by fusing the organic compound with melting sodium metal. The ionic compounds formed during the fusion are then extracted into aqueous solution and subsequently identified by simple chemical tests. The aqueous extract is called sodium fusion extract or Lassaigne's extract.
The reference 3 gives good procedural practice for preparation of Lassaigne's extract, thus I'm not going to elaborate too much unless it is necessary and going to address the question directly: Test for nitrogen usually involves boiling an alkaline sample of
the water-extract from the sodium fusion with ferrous sulfate and then cooling and acidifying with dilute sulfuric acid. The development of a blue coloration or a dark blue precipitate is confirmatory for nitrogen.
The theory: The carbon and nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide ($\ce{NaCN}$), which is soluble in water. This is converted into sodium ferrocyanide ($\ce{Na4Fe(CN)6}$) by the addition of sufficient quantity of ferrous sulphate ($\ce{FeSO4}$). Ferric ions ($\ce{Fe^3+}$) generated during the process react with ferrocyanide to form $\color{blue}{\text{prussian blue}}$ precipitate of ferric ferrocyanide ($\ce{Fe4[Fe(CN)6]3}$):
$$\ce{Na + C + N -> NaCN }$$
$$\ce{6NaCN + FeSO4 -> Na4Fe(CN)6 + Na2SO4}$$
$$\ce{3Na4Fe(CN)6 + 4Fe^3+ -> Fe4[Fe(CN)6]3 + 12Na+}$$
There are two issues with this test, according to Ref.3:

It seems that the development of a blue coloration and/or precipitate in a positive test depends on pH and on temperature. With $\mathrm{pH}$ only just below 7, the development of coloration is inhibited. It is, thus, necessary to ensure the presence of an excess of acid. Similarly, cooling of the solution is advisable because development of the blue coloration appears to he inhibited in hot solutions.

During the Lassaigne's extract, if the compound has too much nitro groups and too little carbons, it is advisable that to add a little paraffin or a hydrocarbon before heating with sodium. Carbon in them provide enough carbon to make cyanide with nitrogen. In this praise, there is a answer for your question: Nitro group contains oxygen but did not interfere with the test.

References:

J. L. Lassaigne, "Mémoire sur un procédé simple pour constater la présence de l'azote dans des quantités minimes de matière organique (Memoir on a simple procedure for confirming the presence of nitrogen in minimal quantities of organic matter)," Comptes rendus 1843, 16, 387-391.
Oscar Jacobsen, "Ueber die Oxydation der Parasulfamintoluylsäure (On the oxidation of 3-methyl-4-sulfamoyl-benzoic acid)," Berichte der Deutschen Chemischen Gesellschaft 1879, 12, 2316-2320.
R. P. Gower, I. P. Rhodes, “A review of techniques in the Lassaigne sodium-fusion,” J. Chem. Educ. 1969, 46(9), 606-607 (https://doi.org/10.1021/ed046p606).


How does steam prevent oxidation in tea leaves? I recently started branching out into different "flavors" of tea. In this endeavor I've been learning more and more about different varieties of tea and was wondering if anyone could offer an explanation as to how steam can prevent the process of oxidation in tea leaves.

According to Tony Gebely at Tea Epicure (https://teaepicure.com/tea-leaves-oxidation/), it's not steam per se, but rather heating that slows oxidation, by denaturing two enzymes (polyphenol oxidase and peroxidase) that are responsible for it.  Steaming is simply one way to do this:

Much of the oxidation process revolves around polyphenols, particularly the enzymes polyphenol oxidase and peroxidase. When the cells inside tea leaves are damaged and the components inside mix and are exposed to oxygen, a chemical reaction begins. [Specifically when polyphenols in the cell’s vacuoles and the peroxidase in the cell’s peroxisomes mix with polyphenol oxidase in the cell’s cytoplasm. From Michael Harney’s 2008, “The Harney and Sons Guide to Tea.”] This reaction converts the polyphenols known as catechins into flavanoids called theaflavins and thearubigins (which are also polyphenols).

Fixing works by denaturing polyphenol oxidase and peroxidase – the enzymes primarily responsible for oxidation using heat. Fixing is also commonly referred to as de-enzyming, denaturing or kill-green.

The term kill-green is derived from the Chinese term shaqing (杀青), which translates to killing the green. The tea leaves must be heated to approximately 150 degrees Fahrenheit to “halt” oxidation. Oxidation is further slowed by drying the leaves, but it never completely stops. At temperatures over 150 degrees Fahrenheit, oxidation continues to occur at an extremely slow pace.

The process of fixing requires precise control of the temperature and length of heating; each has to be adjusted depending on the size and thickness of the leaves and the amount being processed.

Note also that, for teas, oxidation is not a bad thing.  Rather it is a parameter that is precisely controlled in order to achieve the desired type of tea.  Green teas are produced by minimizing oxidation. Oolongs are produced by allowing oxidation to proceed part-way (and they are thus known as "semi-oxidized teas").  Black teas are produced by allowing oxidation to proceed completely, or nearly so:

Oxidation refers to a series of chemical reactions that result in the browning of tea leaves and the production of flavor and aroma compounds in finished teas. Depending on the type of tea being made, oxidation is either prevented altogether or deliberately initiated, controlled and then stopped. [emphasis mine]

When oxidation is prevented altogether, the catechins are left largely intact. The tea leaves keep their green color and vegetal characteristics.

When a semi-oxidized tea is being produced, some catechins convert to theaflavins and thearubigins, resulting in a slight browning usually along the edges of the leaves and yellower liquor. Lipids, amino acids, and carotenoids also begin to break down into flavor and aroma compounds.

When oxidation is allowed to run its course, the leaves exhibit an aroma and taste profile completely unrecognizable from a finished tea that was exempted from oxidation.


Is there a quantum chemistry package which calculates a second-order wavefunction? I am trying to generate a change in density which corresponds to a second-order wave-function in standard Rayleigh-Schrodinger perturbation theory. I would like to try out a simple test on an atomic dimer. However, I am not sure if such a method is implemented in any of the standard quantum chemistry software packages. Does anyone know of such a package?

EDIT: What I am after is the following wavefunction:

$\Psi = \psi^{(0)} + \psi^{(1)} + \psi^{(2)}$

As @Wildcat already mentioned, here is a Gaussian 09 input that will generate the MP3 density, which should correspond to the full second-order wavefunction, thanks to Wigner's $2n+1$ rule:

%chk=g09_mp3.chk
%mem=2gb
#p mp3(full)/sto-3g density=current pop=full output=wfn

wavefunction

-1 1
O     -2.0481500778    0.0447984581   -0.0601937089
H     -1.1514551262    0.4243541138   -0.0806303110
H     -1.8059176361   -0.8898508193    0.0217239645
Cl     1.2314805228   -0.2273431057   -0.0015462666

g09_mp3.wfn

The MO coefficients corresponding to the fully relaxed wavefunction $\Psi$ are also dumped to the *.wfn file in addition to the normal output, which I think can be parsed with cclib.


Rusting of an Iron nail at oil water boundary surface Recently at school, I did some experiments on corrosion. Everything mostly went as expected, but the result in one of the test was quite baffling.

This test had an iron nail in water with some of the nail protruding the surface. The tip of the nail was then covered by oil so that it was completely submerged. It was left as is for 4 weeks. At the end of this time, there was a bit of rusting (as I had expected), but most of it was right at the surface of the water, right where it and the oil meet. Is there any reason why this would be?

Edit
I had a beaker with the nail in it, leaning upright against the wall of the beaker. I put about 200 ml of water in, then about 50–100 ml of oil. The water didn't completely cover it, but the oil did. After 4 weeks, there was rust, mostly situated around (but below) the oil/water border. As for why it was done, I don't know. My lab partner decided to test this. I think it was kind of testing how much it would rust when starved of oxygen.

According to your description, this looks like localized corrosion. Due to differences in the availability of oxygen in the solution, the two half-reactions of the redox process take place on different areas of the surface of the corroding metal. The oxygen reduction reaction preferably takes place on the nail surface in the medium with higher concentration of dissolved $\ce{O2}$ (water).

$$\ce{O2 + 4e- + 2H2O -> 4 OH-}$$

The metal dissolution (oxidation) is then limited to the more oxygen-starved surface area, which is presumably right at the the phase border, where consequently the deposit of corrosion products is formed.

\begin{aligned}
\ce{Fe &->~ Fe^2+ + 2e-}\\
\ce{Fe^2+ + 2OH- &->~ Fe(OH)2 \rightleftharpoons FeO + H2O}\\
\end{aligned}

Since these two half-reactions are coupled, the amount of corrosion products formed depends on the oxygen concentration. When there is sufficiently dissolved oxygen available, $\ce{Fe^2+}$ is easily oxidized by to $\ce{Fe^3+}$, and thus the final product, which is commonly called rust, is a mixture of more or less hydrated iron(III) oxides and hydroxides.

$$\ce{Fe^3+ + 3OH- -> Fe(OH)3 \rightleftharpoons FeO(OH) + H2O}$$
$$\ce{2FeO(OH) \rightleftharpoons Fe2O3 + H2O}$$


Are there any general chemical rules for choosing initiators for a polymerization reaction? I'm working with some very basic free-radical initiated polymerization reactions. For the most part, I just follow a recipe I'm given (by a member of my group or the broader literature), it works, and I don't worry too much about the chemistry of why it works. Clearly, that will only work for so long, and I'd like to better understand the chemistry of what I'm doing.

  So what general rules can I use to predict what kind of initiator will polymerize a given monomer?

To give a concrete example of what I mean: I know that some persulfate-based initiators will polymerize some acrylamide-based monomers. Can I generalize that to all persulfate initiators and all acrylamide monomers? What about methacrylamides? What about styrenes? Essentially, at what point have I changed the reactants so much that the chemistry doesn't work anymore?

Odian's Principles of Polymerization (4th ed) reviews some things that have been done, but doesn't discuss any general rules. Are there other good sources for this information?

EDIT: I am looking specifically for chemical reasons. Are some initiators better at reacting with some monomers? Are there any issues like side reactions I should take into account?

EDIT(2): from @Ben Norris's answer and the ensuing comments, it sounds like, as long as I can get the initiator to radicalize, and everything is soluble in the solvent, then I will be able to use any initiator with any monomer. Is that correct? I've found another textbook on this, Polymer Chemistry: An Introduction by Malcom Stevens, and that seems to support this.

It's been my experience that initiators are chosen for: 1) initiation temperature, 2) solubility, and 3) cost.

1) Initiation temperature - Most radical initiators are stable at room temperature; they need to be heated to dissociate. For example, the half-life of benzoyl peroxide at room temperature is very long, but it is about 30 minutes at 92 °C and 1 minute at 131 °C. However, some monomers boil or decompose at temperatures lower than those temperatures. AIBN (azobisisobutyronitrile) decomposes above 65 °C, so less heat is needed. Both benzoyl peroxide and AIBN are suitable for bulk polymerization of most monomers, including acrylamides, acrylates, styrenes, etc.

2) Solubility - Benzoyl peroxide and AIBN are soluble in organic solvents, which include most monomers for bulk polymerizations. However, if you want to perform your polymerization in water, for example if you are doing an emulsion polymerization, then you need a water-soluble initiator, like a peroxodisulfate salt.

3) Cost - Every initiator has a price, even if you are only using tiny amounts. AIBN may have a lower activation temperature than benzoyl peroxide, but AIBN is also more expensive. There are other more specialized inhibitors that are even more expensive.


Why there will be only limited shells in a particular atom? What is the maximum possible principal quantum number for an atom(any atom not a specific one) ?

Imagine a universe that is not expanding, and contains only one large star and one small planet. It is technically possible to put the two objects into orbit no matter how far apart they are. At some very large distance, the gravitational attraction is so small that it is nearly negligible, but if given infinite time you will be able to trace the orbits for the planet-star system. If, however, you introduce more planets and stars close to the planet, then the influence of the star that is very far away will become irrelevant to the motion of the planet.

We can see a similar case when we look at an electron around an atom. Each time you excite it, you move the electron into a higher-level orbital that is more weakly bound. Eventually you get the electron into orbitals that are nearly degenerate and asymptotically approach zero binding energy. If you have a universe that contains only the electron and the atom, and get rid of the vacuum fluctuations, then yes, it is theoretically possible to put an electron into an orbital with arbitrarily high level of n.

In practice, at some point the electron becomes so weakly bound that electromagnetic forces in the environment will dominate and take the electron away from the atom, ionizing it.

It is experimentally possible to isolate an atom from the environment quite well and then excite the electron to a pretty high n without ionizing it. An atom that is excited in this way is called a Rydberg atom.


Choosing an electrolyte for a galvanic cell I was wondering what processes are used to select an electrolyte for a galvanic cell.

I know a zinc copper battery usually has zinc and copper sulfate electrolytes. How do these electrolytes interact wishing each half cell (i.e.: how does zinc sulfate interact with water, the anion from the salt bridge and zinc itself)?

The point of an electrolyte is to allow ion flow between the anode and the cathode in order to balance the charges on either side so the reaction can continue. What you are looking for in a good electrolyte is that it won't react with either of your metals (i.e. won't form a precipitate), and that the ions will readily move (that's what makes soluble ionic compounds a great choice).

With $\ce{ZnSO4}$, for example, you wouldn't want to have $\ce{Pb^2+}$ ions anywhere in your cell, since $\ce{PbSO4}$ would precipitate and your concentrations of your ions would change, therefore changing your overall cell potential.


What will be the product when 3,4-dibromocyclohexene is reacted with sodium followed by heating with selenium? I'm tutoring few Class 12 students for their chemistry exams. I'm stuck with few problems. I don't know, if the key is wrong or I am wrong.

According to me product (A) must be the cyclic diene, as given in option (A). Now, I don't know, what happens when you heat cyclic diene with selenium. But I do know that,  $\ce{SeO2}$  oxidizes allylic positions of alkenes into alcohols. But, none of the options given are fitting into that argument.

The answer given in the book is option (B).
How can that be explained?

Assuming your Product A is correct then product B should be benzene. Selenium act as a mild oxidizing agent at temperature 220 °C to 330 °C. In the above reaction selenium oxidizes cyclohexadiene to form benzene.

Below is a reaction in which 1-methyl-1,3-cyclohexadiene is converted into toluene.

you can also go to this link for more info


Acid/alkali + salt reaction? Why some acid/alkali can react with some salt, forming another acid/alkali? For example:

$\ce{2NaCl +H2SO4->Na2SO4 + 2HCl}$

$\ce{Ca(OH)2 +Na2CO3->CaCO3 +2NaOH}$

but not the reverse.
What determines this? I heard someone said the weaker acid/alkali produced is favourable, but HCl is stronger than $\ce{H2SO4}$. I also heard that only if the products are insoluble or lost from the solution, the reaction happens. Someone suggested that the problem is concentration, as sulfuric acid is 98% while HCl is 38% Is that true?

Hydrochloric acid solutions can only readily reach about $\pu{40\%}$ concentration. So say you add $\ce{NaCl}$ to $\pu{98\%}$ sulfuric acid. You quickly end up supersaturated with respect to dissolved $\ce{HCl}$ and it goes into the gas phase which thus limits the ability for that reaction to go in the reverse direction.

The situation is similar for the second reaction, except in this case rather than removing carbonate from the system via volatilization, you are forming the insoluble $\ce{CaCO3}$, preventing the reaction from happening in the reverse direction.


Is there an electronic structure method which accounts for 3-, 4-, ..., n-electron exchange integrals? Something I learned recently which really helped me in understanding exchange is that the case of exchanging two electrons is really a special case of enforcing symmetry (or anti-symmetry) in which the permutations of labels in cycles out to length $N$ are used. $N$ is the number of particles.

Just to write it out, the two-electron exchange integral is defined as,
$$K_{ab}=\int d\textbf{r}_1d\textbf{r}_2\psi^*_a(\textbf{r}_1)\psi_b(\textbf{r}_1)r_{12}^{-1}\psi_b^*(\textbf{r}_2)\psi_a(\textbf{r}_2)$$
where everything means what it normally means and I took this integral and notation straight out of Szabo and Ostlund.

Now, this is clearly only a permutation of two electrons, and introduces the only correlation present in Hartree-Fock by allowing for exchange of spin-paired electrons.

Shifting gears, I've been reading Quantum Mechanics and Path Integrals by Richard Feynman and Albert Hibbs. In the chapter on Statistical Mechanics, they talk about liquid helium, which has a peculiar transition around $2K$ in which the heat capacity begins to increase. Feynman was the first to explain that this is due to the exchange interaction, and at this temperature is when permutations of the atom labels larger than just one (i.e. two particle exchange) become important. The following statements from the book are relevant to electrons (as opposed to helium because $\ce{^4He}$ is a boson!).

In a different context, the calculation of partition functions, the authors state:

  If we were dealing with Fermi particles, e.g., the isotope of helium which has three nucleons, we would have to include an extra factor of $\pm1$, positive for even permutations and negative for odd permutations. There would also be some extra features which depend upon the spin of the atom in our result.

And more directly about electrons,

  Consider the behavior of electrons in a solid metal. The mass of the electron is so much smaller than that of a molecule that the critical temperature is much higher. At room temperature, electrons in a metal are described accurately only by equations which include the exchange effects of these cyclic permutations. From this point of view, room temperature is very cold for electrons.

So, hopefully we have now reached a place where it is clear why I am a bit confused. Namely, to my knowledge, we only ever consider the exchange of two electrons of the same spin in our electronic structure methods. First, is this actually true? That is, are there methods which implement (in principle) $N$-electron integrals to describe all permutations of the labels?

Second, is the comparison I am making even strictly valid? That is, in the case of a partition function, which is what Feynman is talking about, permutations of even length tend to cause the electrons to stay further away from each other. This is the behavior we are used to with exchange. In contrast, for permutations of odd numbers, the contribution to the partition function picks up a minus sign. So, the total partition function for a system of fermions is an alternating sum of positive and negative terms in the path integral formalism. Can I simply substitute partition function for wavefunction and sum for integral and find what would happen in the electronic structure case? That is, will the 3-electron exchange integral (assuming this is real and I'm not mistaken) be a negative number, so that we can picture the wavefunction relaxing a little bit from the so-called "heaps" which form due to exchange?

Finally, if I'm not just delusional about this whole thing, is there evidence that permutations beyond the two labels are important for describing any known chemical processes?

References:

[1]: Szabo, A., & Ostlund, N. S. (2012). Modern quantum chemistry: introduction to advanced electronic structure theory. Courier Corporation.

[2]: Feynman, R. P., Hibbs, A. R., & Styer, D. F. (2010). Quantum mechanics and path integrals. Courier Corporation.

I am posting this answer not as the definitively correct answer, but as an example of the type of thing I am talking about.

azago's answer is correct in pointing out that the Slater determinant contains within it the correct permutational symmetry of the wavefunction required by electrons. Namely, it is antisymmetric upon interchange of two orbitals. Additionally, it symmetric upon interchange of three (or any odd number of) orbitals. This is the symmetry which must be present in higher cycles of exchanges.

What I have found upon further reading is that I could have been more precise in the wording of my question. Namely, these higher-order exchange interactions are not surprising if I had made it clear that these are necessarily three-center, four-center,... integrals. Then it is more clear that HF, while having the correct symmetry to adhere to Pauli exclusion, does not include these integrals as it only deals with 2-electron integrals at the most.

So, I will discuss three contexts I have found in which the types of integrals I am thinking of show up.

Hamiltonian for a Three-Atom System:

Following along with ref. [1], the Hamiltonian for the three atom system, $\ce{H3}$ is,
$$
H=\sum_i\left(-\frac{1}{2}\nabla^2_i-\sum_{\alpha}\frac{1}{r_{\alpha i}}+\sum_{j\gt i}\frac{1}{r_{ij}} \right)+\sum_k\frac{1}{R_k}
$$
where $i$ and $j$ are labels for the three electrons, and $\alpha$ is the label for each Hydrogen atom with locations $R_1$, $R_2$, and $R_3$.

It is clear from this Hamiltonian that we have the usual coulomb terms and the usual two-electron integrals. Something weird is going to happen with the exchange terms though. There are actually three possible configurations of the atoms, but the only point I'm making here is that the matrix elements for one of the atomic configurations can be written as,
$$
\langle\psi_1|H|\psi_1\rangle=4J+4K_1-2K_2-2K_3-4K_{123}
$$
I have deviated from their notation here for consistency with current convention. These integrals are written explicitly as:
$$
\begin{align}
J&=\langle abc|H|abc\rangle \\
K_1&=\langle abc|H|bac\rangle \\
K_2&=\langle abc|H|acb\rangle \\
K_3&=\langle abc|H|cba\rangle \\
K_{123}&=\langle abc|H|cab\rangle \\
\end{align}
$$
where now it is quite clear that $K_{123}$ is a double exchange. That is, two electron labels are changed, and it is clear this is distinct from the other exchange integrals which are the usual exchanges we are used to.

I do not think this has anything to do with the fact there are three atoms present, however. Only that there are three electrons. Thus, I believe that even a complete description of Lithium would include these three-electron exchanges. Thus, the reason we do not see these integrals in HF is that they are left out of the Hamiltonian.

Explicitly-Correlated Methods:

I'll be brief here because explicitly-correlated methods are pretty complicated and I haven't spent enough time studying them to really know what's going on, but one thing which does happen is that you get matrix elements of the following form[2]:
$$
Z_{\nu w,p}^{xy,q}=\langle w_{\nu w,p}|(1-\hat{p}_{13})r_{13}^{-1}+(1-\hat{p}_{23})r_{23}^{-1}|w_{xy,q}\rangle
$$
The only important in this horrible integral is that the exchange operator $\hat{p}_{ij}$ appears with the labels of three-distinct electrons. Now, I have no idea if it's valid to interpret this as the type of integral I'm talking about, but these kinds of integrals appear all over the place in explicitly correlated methods.

Three-Body Symmetry-Adapted Perturbation Theory

Now, two-body symmetry-adapted perturbation theory (SAPT) is quite a common method, but there is a 3-body version of SAPT which can be quite a useful tool at times.

In Ref. [3], the authors present the relevant equations for 3-body SAPT, but they note that they approximate the exchange term out to the fourth power of orbital overlap. That is, for the exchange operator they do the following:
$$
P\approx 1+P_2+P_3+P_4
$$
They describe this as follows:

  In this expression, $P_2$ represents the operator which interchanges
  one electron from $X$ with one electron from $Y$ and
  therefore produces terms proportional to $S^2$. The operator
  $P_3$ includes double exchanges between three monomers involving
  three electrons. This operator, absent for a dimer,
  generates terms proportional to $S^3$. The operator $P_4$ is a
  double exchange operator which interchanges four electrons
  among the monomers and generates terms proportional to
  $S^4$. The double exchanges produced by $P_4$ are of the
  $(XY)(XY)$ or $(XY)(XZ)$ types and can be thought of as two
  exchanges of the type produced by $P_2$ occurring simultaneously.
  In contrast, double exchanges of the $(XY)(XYZ)$
  type are included in $P_5$ and double exchanges of the
  $(XYZ)(XYZ)$ type are included in $P_6$, leading to terms proportional
  to $S^5$ and $S^6$, respectively.

The whole purpose of three-body SAPT is to be able to describe the 3-body non-additivities which arise from intramolecular interactions. It has already been pointed out that the electrostatic interaction is additive up to 2-body because Coulomb's law follows the superposition principle. This then makes it clear that these higher-order exchanges involving more than two electrons generate some of the non-additive forms of coulomb correlation.

Finally, the authors of ref. [3] provide the form of these three-electron cycles I was trying to describe which can be written as:
$$
P_{111}=\sum{i\in A}\sum{i\in B}\sum{i\in C}[P_{ij}P_{jk}+P_{ik}P_{jk}]
$$
which interchanges electrons in a cycle on atoms $A$, $B$, and $C$. This operator generates all possible three-electron exchanges between three atoms. There are no intra-monomer exchanges however.

[1] Porter, R. N., & Karplus, M. (1964). Potential energy surface for H3. The Journal of Chemical Physics, 40(4), 1105-1115.

[2] Hättig, C., Klopper, W., Köhn, A., & Tew, D. P. (2011). Explicitly correlated electrons in molecules. Chemical reviews, 112(1), 4-74.

[3] Lotrich, V. F., & Szalewicz, K. (1997). Symmetry-adapted perturbation theory of three-body nonadditivity of intermolecular interaction energy. The Journal of chemical physics, 106(23), 9668-9687.


How to identify lead crystal glassware? I would like to identify if a wine glass is made of lead crystal (a destructive method is allowed). Lead crystal is said to be eroded by dishwasher detergent. I don't have a dishwasher, what causes the stains? Could I quickly identify lead crystal by exposing it to hot caustic soda?

Since a destructive method is allowed, the other suggestions did not work for you (despite being excellent, especially the Archimede's scales suggested by @porphyrin) and you have hot caustic soda (aka $\ce{NaOH}$), I suggest this plausible method (which, thankfully, I have never had to use). First, crush a small amount of the suspect lead glass, trying to get it a finely crushed as you reasonably can. Second, mix that crushed glass with solid  $\ce{NaOH}$. You do not need much of the mixture: a teaspoon (or a couple of grams) should suffice. Third, do a fusion of that mixture. A crucible would be used in a laboratory, but you can improvise using a small cast iron skillet. They sell very small ones (4 or 5 inch diameter) at some hardware stores. Absolutely do not use aluminum because the $\ce{NaOH}$ will destroy it!

You need to do the fusion at moderately high temperature, such as can be obtained with a charcoal grill, gas grill or a campfire. You may need to be a bit inventive on this and take care to avoid having anything spatter onto yourself! This is where a grill cover is very helpful. Fourth, let the fusion cool down and add some white vinegar to neutralize the $\ce{NaOH}$. You can add an excess, but don't get carried away: dilution is not your friend. If lead was present, you should now have some 'sugar of lead' (lead acetate) in solution. Fifth, hardware stores usually sell inexpensive test kits for lead (from leaded paints). So use one of those to test your neutralized solution for lead.

What about false positives? Do the whole thing over with crushed glass that definitely does not contain lead. It would be best to do this 'blank' first, to avoid contaminating the new little skillet (which, by the way, can be cleaned, seasoned anew, and used for its orthodox purpose).

I have not tried this and I hand-wash my lead 'crystal' single malt glasses, so best of success!