Asymmetry in trigonal bipyramidal geometry I teach an MCAT course in chemistry. I like to explain VSEPR by saying.. first, imagine arranging electron pairs around the central atom so they are maximally distant from each other, and uniformly arranged. If there are two pairs, they'll just be across from each other. Three? You get a triangle, and so on.

This rhetorical device fails with five pairs. Five is strange because the electron pairs are most definitely not equidistant from each other, with the equatorial pairs being 120 degrees from each other, but 90 degrees from the axial electron pairs.

My question: is this a simplifying geometry for high school / college classes, and the geometry is actually more equally arranged in 3D space among the pairs? Alternatively, if 'trigonal bipyramidal' is actually the geometry, why? What induces the asymmetry? And does that imply that molecules with the same formula and connectivity, but with an atom equatorially versus axially located correspond to distinct stereoisomers?

Yes, 5-coordinate compounds are always a bit confusing.

The best way I've answered this question about VSEPR for students is to do a little demo with balloons. Blow up balloons (i.e., electron pairs) and tie them around a central knot. It will definitely give a trigonal bipyramidal shape - it has nothing to do with chemistry, and everything to do with "fitting 5 things around a central point."

I disagree with @Martin that there's no theoretical basis to the shapes. The theory is purely geometric.

There are multiple possible shapes, including square pyramidal, but these are higher in energy because you have more ~90-degree angles. As discussed in the comments, if all 5 atoms are in a plane, the angles are 72 degrees, which is clearly bad. You could in principal have the top of an octahedron, in which case all angles are 90 degrees, but that's also clearly worse. (In practice, square pyramidal structures have the central atom slightly above the plane, increasing some angles slightly beyond 90.)

So I do the demo with the balloons, talk about square pyramidal and how this alternative is clearly slightly higher in energy.

The Berry pseudorotation, as mentioned above, does convert between the shapes, and it's known that at room temperature the axial and equatorial positions scramble quickly.

Update:

I decided to do some quick calculations using Avogadro and MOPAC on $\ce{PF5}$ in square planar versus trigonal bipyramidal using the PM7 semiempirical method. This took longer to write than to run the calculations (seconds).

Here's the optimized square pyramidal geometry:

The phosphorous isn't in the plane, it's a bit above, to increase four of the bond angles and minimize electron-pair repulsion. Note that the largest angle is ~103 degrees, but the F-P-F angles in the basal plane are all ~87 degrees.

Here's the trigonal bipyramidal geometry:

Now the angles are exactly what we expect from VSEPR. We see 90 degree angles between the equatorial plane and the axial F atoms. And the F-P-F in the plane are all 120 degrees.

Just from the angles we can guess that trigonal bipyramidal should be lower in energy.

For reference, the difference in energy between these two geometries using PM7 is ~2.76 kcal/mol. So there's not a huge difference in energy, but it's there.


Why do spin isomers of hydrogen (ortho and para hydrogen) change their nuclear spin with temperature variance? My book says that ordinary dihydrogen contains 75% ortho and 25% para forms of hydrogen, while at significantly lower temperatures (like 20K) ortho and para hydrogens are 0.18% and 99.82% respectively. It also says that pure para form can be obtained at very low temperatures.

So what is the relation between the nuclear spins and temperature, and also why do the nuclear spin changes with temperature variance?

The effect is due to the symmetry properties of the rotational energy levels and those of the nuclear spin. The change with temperature is, as usual, traced back to the Boltzmann distribution.

In homonuclear diatomic molecules such as $\ce{H2}$ it is necessary to consider the spin of the overall wavefunction to exchange of nuclei. If this is $\Psi = \psi_t\psi_v\psi_R\psi_n$. The translational wavefunction $\psi_t$ depends on the coordinates of centre of mass, the vibration only on separation of nuclei. Thus we need only consider the product of rotational and nuclear wavefunctions, $\Psi = \psi_R\psi_n$. As there are identical nuclei the degeneracy is $g=2s_n+1$ with nuclear spin $s_n$ ($=1/2$) for hydrogen atoms.

As the product wavefunction $\Psi$ is neither odd nor even, it is necessary to ‘symmetrise’ by making the symmetric form in terms of states $\psi_1$ and $\psi_2$ for particle $(1)$ and $(2)$, as these also satisfy Schrodinger’s equation.

$$\psi_S=\psi_{1}(1)\psi_{2}(2)+\psi_{1}(2)\psi_{2}(1)$$

and asymmetric form

$$\psi_A=\psi_{1}(1)\psi_{2}(2)-\psi_{1}(2)\psi_{n}(1)$$

The number of symmetric functions is $g_n(g_n+1)/2$ and asymmetric $g_n(g_n-1)/2$ making a total of $g_n^2$.

The rotational wavefunctions are symmetric for even J rotational states and asymmteric for odd rotational quantum numbers J. As the product $\psi_R\psi_n$ must be asymmetrical for a molecule with an odd mass number for even J only asymmetrical nuclear states are accessible and  and vice versa.

The partition function for odd mass number is
$$Z= \frac{g_n(g_n-1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n+1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) $$

and for $\ce{H2}$ the partition function is
$$Z= 1\sum_{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) + 3\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) $$

From the equilibrium distribution equation the ratio of number of ortho- to para- molecules is therefore

$$\frac{N_{ortho}}{N_{para}} = \frac{ 3\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT))  } {   \sum_{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT))} $$

As the temperature changes so does this ratio; as the temperature tends to infinity the ratio tends to a limit of 3 and to zero as the temperature tends to zero.

For any spin degeneracy the ratio in the limit of very high temperature is
$$\frac{N_{ortho}}{N_{para}}=\frac{g_n+1}{g_n-1}$$
and for $\ce{D2}$ with spin of $1$ and $g_n=3 $ the ratio is $2$.

The figure shows the ratio for hydrogen.


Precautions and tips for Qualitative Inorganic Analysis of Coins I'm a highschool student and as part of an investigatory project, I have decided to analyse which metals are present in our everyday coins.

I chose this because in my previous academic year, I had learnt how to analyze the cations present in inorganic salts and I employed the same systematic analysis method that I had learnt to test the presence of metal cations in some coins.

Obviously, I cannot expect the coin analysis to go exactly like the salt analysis. I would have to ignore certain tests like the 'flame test'.

Could you please tell me what precaustions I should take for this experiment? I wish to write up a clear and systematic process by which any student / coin collector can easily do in a laboratory with only common chemicals and salts.

All advice, resources and tips will be most appreciated.

1) The coins must be dissolved (at least partially). There is a good amount of metals that won't dissolve in common  acids (hydrochloric acids), some will require nitric acid, some will require aqua regia and some will require even worse, though that's unlikely. So, most likely you will have to deal with concentrated nitric acid, that produce nitric oxide (fox's tail) when react with metals. Given that you must have fume hood. In case something undissolved remains, you can try aqua regia, and if even that fail, you have something rare, like nioubium, with no common method of wet analisys.

2) Most metals used in coins have toxic salts. So, be cautios, and avoid skin contact. Latex gloves and protective goggles are not vital, but certainly recommended unless you believe you can be accurate enough. Usual safety requirements for work with conc. acids applies as well.

3) In case of non-destructive analisys, X-ray fluorescent spectrometry may be employed. It will give accurate analysis in almost any realistic case, but requires quite costly equipment. It is not extremely costly per analysis, though.


Carbonyl bond vs alkene bond Why is that carbonyl bonds are stronger than those of alkenes? It's a basic question but I've never really thought much about it.
I've read several times that carbonyl bonds are shorter and you need to apply a higher amount of energy to break them. Why is that so?

Is it due to the electronegativity of oxygen, which pulls the bonding electrons near it, thus attracting the carbon as well?

It's stronger because of a dipole moment. oxygen is more electronegative than carbon, so there is a stronger attraction between C and O than a C=C bond.


Difference between lattice energy and thermal stability For an ionic compound, is it always true that the greater the thermal stability, the greater the lattice energy. E.g. for 2 ionic compounds MX and MY, if MX has a higher thermal stability than MY, does MX necessarily have a higher lattice energy?

1) Lattice energy is the measure of the stability of an ionic solid.

It is the energy required to separate the ions in one mole of a solid ionic compound into gaseous ions.

The weaker the bond between the ions, the easier it will be to break that bond and separate the ions into gaseous ions and the lesser will be the lattice energy.

Therefore, the stronger the bond between the ions, the more is the lattice energy of the ionic compound.

2 )Thermal stability depends upon the amount of heat required to break the ionic compound. Clearly, the stronger the bond between the ions is, the more heat will be required to break the bond and as a result, the thermal stability of the compound increases.

Therefore, stronger the bond, more is the thermal stability of the ionic solid.

In general, we can say that as thermal stability increases, the lattice energy also increases, but they also also depend upon the crystal structure of the solid and other factors.

So, you cannot say with certainty that if the thermal stability is high, then lattice energy will be high too.

Hence, the answer is no. If MX has a higher thermal stability than MY, then it may or may not have higher lattice energy than MY.


What are oxidation states used for? When I took high school chemistry many years ago, considerable effort was spent on teaching us to compute oxidation states of atoms in various compounds, following a set of rules that looked somewhat arbitrary to me at the time. As far as I remember, we were never told what benefits (other than passing tests) knowledge of oxidation states would give us.

They were used as part of a convoluted procedure for "balancing redox reactions", but I never saw an example of that where it wouldn't give the same results simply to require that the number of nuclei of each element, as well as the total charge, must be the same on both sides of the reaction, and solve the resulting linear Diophantine equations.

Nevertheless, the concept must be useful other than for setting homework exercises -- I see encyclopedias and other sources use much space on classifying compounds based on the oxidation state of this atom or that. Still I don't recall seeing any case where the oxidation states are used for anything (other than computing other oxidation states).

What is this concept actually used for? Just a basic example or two where knowing the oxidation states helps produce a meaningful prediction.

It doesn't seem that "oxidation state" actually encodes any particular configuration change inside the atom that tends to be preserved across reactions. Or does it somehow require effort (energy?) to change an atom from
one oxidation state to another? In which case, what actually changes?

Sorry if this is too basic a question. It may be something "everybody knows". I've tried looking e.g. at the Wikipedia article, but that too seems to be entirely focused on how to determine oxidation states and doesn't explain why one would desire to know them. And all the oxidation-state questions here seems similarly to be about determining the oxidation states, and it is left unsaid why the asker wants to know ...

Edit: I'm looking for applications of oxidation numbers, by which I mean rules, laws, tendencies, rules-of-thumb or the like of the general form:

  When you have (condition that involves oxidation numbers) then (prediction that can be verified without knowing what oxidation numbers are).

Of course they are useful. Perhaps you have not come across them yet, but being able to determine the oxidation state of an atom allows us to understand the properties of chemicals and how redox reactions work.

Let's just give two very simple examples. There are tons more. (I don't even want to go into organometallic chemistry, where being able to determine the oxidation state is incredibly important.)

Potassium permanganate, $\ce{KMnO4}$, has manganese in a +7 oxidation state. This means that it would have an electronic configuration identical to the noble gas argon: $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}$, and therefore its $\mathrm{3d}$ orbitals are empty.

Now, most transition metal compounds are said to be coloured because of so-called "d-d* transitions". However, this can only occur if the d orbitals are partially filled - if there are no electrons in the d orbitals, like in $\ce{KMnO4}$, then there are no d-d* transitions available.

As such, the intense purple colour of $\ce{KMnO4}$ has to be explained via a different mechanism. In this case, it is explained by ligand-metal charge transfer.

Moving to organic chemistry. Let's say you have an acyl chloride, $\ce{RCOCl}$, and you want to convert it to an aldehyde, $\ce{RCHO}$. If you calculate the respective oxidation states of the carbonyl carbons, you get +3 and +1 respectively.

So, that tells you what kind of reagent you need to effect this transformation: you need a reducing agent. Hydrogen gas, $\ce{H2}$, is one such example of a reducing agent. Why? Well, that's because of oxidation states again. $\ce{H2}$ has hydrogen in an oxidation state of 0, and when hydrogen forms a bond to carbon, it has an oxidation state of +1.

Oxidation states are key to understanding why redox reactions work and why certain species are reducing agents (e.g. $\ce{LiAlH4}$: H(-1)) and why others are oxidising agents (e.g. DMSO, $\ce{(CH3)2SO}$, in the Swern oxidation).


Gas Electrode Working How does a SHE electrode work?

Does the $H_2$ gas get absorbed on the platinum electrode? How is it reduced? How are electrons transferred from the adsorbed gas to the solution? How is $H^+$ liberated or how does the $H^+$ in the solution take electrons from the adsorbed $H_2$?

If platinum doesn't adsorb the $H_2$ gas, what is its use? How does it provide a surface for reduction/oxidation of hydrogen?

Also in a concentration cell, why is it, in evaluating Q(reaction quotient)
you take pressure of a gas numerically and put it directly as a concentration?

And in evaluating log(Q) in nernst equation, Q isn't always unitless. But inside log, you can put only a unitless number, isn't that so?

And in concentration cells, if [c1] > [c2], then c1 acts as cathode. In Gases, considering Pressure=k[C], where k is a constant fixed at a particular temperature. then in gas concentration cell opposite happens if pressure 1 > pressure 2 then pressure 1 part acts as anode.

What is P? is it pressure of the gas above the solution?

In net, how does a gas electrode work?

Standard Hydrogen Electrode

A SHE consists of four parts: a platinum lead wire, a platinum foil contact, an acid solution, and a hydrogen gas flow. The word standard refers to the fact that all concentrations and pressures are at standard state, thus the acid is 1M and the gas flow is 1 atm. At these conditions, the electrode has a potential of 0 V (this is how potential is defined). If there are any deviations from these values, the electrode is no longer considered a SHE. The potential can then be calculated using the Nernst equation.

The electrode is one half of an electrochemical circuit. The other half cell determines whether the SHE will behave as the cathode or the anode. If it is the anode, the electron flow comes from the external system to the platinum contact. 2 electrons reduce 2 acidic hydrogens from the solution to form one molecule of hydrogen, as in $2H^+ + 2e^- → H_2$. The molecular hydrogen then bubbles up above the solution. If the electrode is cathodic, hydrogen gas is absorbed on the platinum contact plate and oxidized to two acidic hydrogens which remain in solution. The electrons then flow to the external circuit. Electrons do not flow from hydrogen gas to the acid, or vice versa.

Reaction Quotient

The reaction quotient is indeed unitless. While it is common to use concentrations and pressures to calculate Q, it is more properly calculated by using activity of each species, which are also unitless. The activity is the apparent concentration or pressure of a species (can be viewed as a correction to concentration or pressure). Because in most chemical applications activity is very close numerically to concentration, using concentrations or pressures directly for calculations is a close approximation.


Chemical water treatment: Acid Base reaction I was thinking of a hypothetical waste system that uses nanotechnology to its advantage.

At the individual level, waste just passes through into pipes when flushed like it normally does.

But once it reaches the water treatment plant there is an initial filter to catch solid waste(in other words poop) and divert it to a different part of the system where it gets dried and eventually ground into a biosolid that can be used as fertilizer.

Meanwhile all the urine, vomitus, blood, and any diarrhea that didn't get caught by the initial filter is passed through a series of filters to catch smaller and smaller waste products from cells all the way to molecules like urea being blocked with a nanofilter. Only a few compounds besides water are left. Yes there are still small ions like potassium and this might actually pose a problem.

But the real threat is that $\ce{HCl}$ is still in the water. Pepsin and other enzymes have been blocked by the nanofilters but $\ce{HCl}$ is still small enough to pass through all of the filters. A nanofilter that can block $\ce{HCl}$ is so impractical that it would most likely block the water itself until either the water treatment system or the microfilters and nanofilters burst from pressure and you end up with a critical failure either way.

So there has to be a chemical reaction that either deactivates the $\ce{HCl}$ and thus renders it harmless or a reaction that leaves no trace of $\ce{HCl}$.

I was thinking that to render the water safe, $\ce{NaOH}$ should be added to increase the amount of saltwater and totally get rid of the $\ce{HCl}$. But then I realized that with potassium and other ions besides sodium and chloride still in the water this could make the water toxic due to toxic salts forming.

Potassium chloride is the least harmful of all the salts that could form besides sodium chloride. Other than the bitter taste, there is basically nothing wrong with using potassium chloride. Similar for calcium chloride. Magnesium chloride though could cause diarrhea due to the magnesium(I know at least part of that diarrhea is from magnesium because the magnesium supplement I take helps to prevent constipation). I am not as worried about hydroxides because those would react with hydrogen ions to produce water.

But those salts are not the only problem. The other problem is how much $\ce{NaOH}$ to add.

I mean to get $\ce{HCl}$ down to 0 every time, there would have to be sensors to not only know the volume of water but also the concentration of $\ce{HCl}$ and a relatively complicated math problem like this:

$$[\ce{HCl}] * x = a mol \ce{HCl} = a mol \ce{NaOH} = [\ce{NaOH}] * y$$

What is complicated about it? Well for one how would you sense the exact concentration of a strong acid or base without damaging the sensory equipment? And unless you regularly go to the water treatment plant with this nanotechnology, you won't know what the concentration of $\ce{NaOH}$ is. And the concentration of $\ce{HCl}$ will most likely vary day to day depending on factors such as these:

How often any given person vomits
Hydration level before vomiting
Amount of bile in the vomit
Dilution of stomach acid
Amount of blood in the vomit if any
Prevalence of vomiting from any cause
Total Volume of vomit

So basically my question is how can I have the part of the system after the filters be constantly sensing the exact concentration of $\ce{HCl}$, doing the math, and adding the amount of $\ce{NaOH}$ that will get the concentration of $\ce{HCl}$ down to exactly 0?

The question is long and detailed, and deserves a long answer, but I can only hit on the highlights, and urge others with some expertise in this area (which I lack) to weigh in.

First, the science of water treatment is mature and has addressed many of the issues raised here, primarily through biological and chemical processes which are very good at breaking down and dissolving most of the different sized particles you refer to.

As far as the composition of sewage is concerned it contains a lot more water than human waste, and a lot more normal waste (urine and fecal matter) than vomit, which, since most people are sick only a tiny fraction of their lives, and nauseous an even smaller fraction, likely is a vanishingly small contributor to the content of sewage. The composition of sewage at the treatment plant intake varies much less, and more slowly, than you might think, since it is averaging the inputs from people all over a city or region. Still, it may vary seasonally due to changes in temperature or water usage, so the treatment must vary accordingly, as you have surmised.

Finally, you concern about neutralizing every last iota of HCl is misplaced. Adjustment of pH is commonly used in sewage treatment and is done using automated equipment and pH sensors that can withstand extended exposure to sewage and to pH fluctuations. The glass membrane electrode is a workhorse that is stable to extremes of pH well beyond what would occur in sewage.

But I conclude my answer by saying that some of your ideas are very interesting, and there may be many applications of nanotechnology in water treatment along the lines you mentioned, that could have practical benefits for peoples' lives. I urge you to continue your studies of wastewater chemistry with your full vigor and an open mind.


Why doesn't EDTA complex with alkali metal ions? EDTA complexes with all the other metal ions in the periodic table except those from the group 1. Why is this so ? What is the coordination chemistry behind this ?

If you're going to be picky that a "complex" doesn't form without covalent coordination bonds, then that's your answer. Alkali metals will tend to form cationic salts with EDTA.

Personally, I agree with the other comments. EDTA does form complexes with alkali metals.

I'd classify the disodium salt of EDTA as a complex. It has a metal and an organic ligand, and there's no direct metal-carbon bond (which would make it organometallic). As discussed in another question, it's a pretty arbitrary distinction between an ionic bond and a covalent one:

  What happens if the electronegativity difference is exactly 2.1?

This distinction is particularly cloudy in the case of metal complexes, since most metals readily gain positive charge.


Chemically Speaking, What's Bad About Inhaling Plastics? I hear about things like dioxin, but I really don't see what's dangerous about plastics-shouldn't you be ble to just breathe it out, or pass it out of your system? How would it cause cancer?

The primary dangers from plastics themselves come when the monomer, and any compound put in to make the polymer matrix flexible, separate from the polymer and are released (leached) into whatever the plastic is containing. There are several such compounds which are known to be problematic to human biochemistry:

Phthalates - These are aromatic diesters, with a structure incorporating a phenyl ring with two ester linkages, typically to alkanes like ethyl or butyl groups. Phthalates are used in certain plastics to link the monomers, such as in poly-ethyl terephthalate (PET aka polyester), and in others as a plasticizer to make the normally rigid plastic flexible (PVC). Phthalates mimic certain hormones in the human body and so can cause issues with growth and maturation, as well as feminizing characteristics in males (reduced body hair, weight gain, moobs). PET is OK for packaging and storage of cold foods (it is, for instance, the plastic used for soda bottles), but you don't want to heat it because that will degrade the plastic to release the phthalate.
The ones used in PVC are especially problematic because they're not well-bonded; instead of being an integral piece of the polymer chain, they form weak hydrogen bonds between the polymer chains (thus the flexibility, even in relatively thick layers of the plastic), and so these phthalates easily leach out into water or "off-gas" into the air, even without heat. These phthalates are a big part of the "plastic smell" of new plastic products, and of water stored in them; the more flexible, and the longer the plastic has been kept enclosed, the more intense the smell. There was a big stink (NPI) in the early 90s when PVC was being used for water supply lines in home construction, and then they found all these problems with the stuff and a lot of homebuilders and chemical companies went bust paying for all this piping to be replaced with copper.

Bisphenols - These are variations of 2,2-bis-(dihydroxyphenyl)propane (which is the infamous Bisphenol A or BPA). Bisphenol A is the monomer for most of the original formulations of high-impact polycarbonate, such as Lexan. Again, when heated, the polymer breaks down and leaches the water-soluble monomer, which is structurally similar in many ways to phthalate and again has hormone-mimicking, endocrine-disrupting properties, which is why newer polycarbonates intended for food storage are made with other bisphenol monomers (most of which have similar endocrine properties, but because they aren't Bisphenol A they're still legal for now and allow the manufacturer to label the product "BPA-free").

Vinyl Chloride - The monomer of poly-vinyl chloride or PVC, this is typically a hazard in production facilities for PVC and when old, aged PVC begins to break down. The vinyl chloride monomer is a known carcinogen, which is the other reason PVC is no longer used for food packaging or water supply, and is being phased out from wastewater lines in favor of nylon.

As far as answering the question "shouldn't we just be able to pass it out of our system", the same could naively be asked of a whole host of chemicals that are bad for us. The biggest problem with that statement is the First Law of Toxicology; anything, in sufficient amounts, is harmful, even if it's absolutely essential for life in the right amounts. Selenium is an essential trace mineral in human biochemistry, required in certain protein synthesis mechanisms, but the LD50 blood concentration is about 3 mg/kg, and harmful effects such as stunted growth begin to be observed at concentrations within foodstuffs of as low as 4 to 5 ppm. Heck, you can die from drinking too much water, faster than you could from lack of it (it causes a condition called hyponatremia where the added water dilutes electrolytes to the point the body can no longer function).
What causes many chemicals to be harmful is in fact our body's own affinity for them as a beneficial chemical. Either the chemical is beneficial to us, or it looks like one that is, and the chemical is normally so rare in our diet and environment that our bodies grab all of it that they find, without an "off switch" to regulate intake.
Which leads to the second problem with that statement; sometimes our bodies don't efficiently metabolize and dispose of chemicals. Vitamin A, for instance, is an essential nutrient, and our body absorbs all it can, and gets rid of it only very slowly. As a result, it "bioaccumulates", and eventually becomes toxic. That's why beta carotene is the preferred "form" of Vitamin A in supplements; it's in fact an intermediate that our bodies can use to synthesize the vitamin as needed, but being water-soluble unlike actual Vitamin A, any unused excess of beta carotene is simply flushed out via the kidneys.
That process in itself can cause problems; the latest round of drug recalls the lawyers are running commercials on are for drugs that did exactly what they were supposed to, but were then metabolized into carcinogenic compounds that concentrated in the urine and caused various cancers of the excretory system.
And the last problem, tying back into the case in point, is that even if the body selectively absorbs it, efficiently processes it, and it isn't acutely toxic in any form or intermediate along the way, as long as there's an input of the chemical into your system, you'll always have some of it. Plastics are ubiquitous in consumer products, and especially food prep and packaging. As a result, virtually every resident of the industrialized world has a detectable amount of artificial phthalates in their blood, because it is simply impossible not to come into contact with them. Even USDA-certified organic foods have phthalates in them, because the cows eat food that's been stored or served in plastic bins and it shows up in their milk and meat, or the vegetable was fertilized with compost stored in a plastic bin or watered from a PVC irrigation system (even the farmer's market items from someone's backyard were likely grown using water collected in a plastic rain catchment and/or dispensed from a hose or a drip irrigator system that leaches plasticizer). What you store it in or prepare it with once it gets to you is immaterial; the damage is done.


Electrons split in Stern-Gerlach experiment but not in Thompson's experiment? So when electrons are fired through a magnetic field in Stern-Gerlach experiment they split into two beams depending on their spin but where was this phenomena when Thompson was calculating charge to mass ratio of electron? Shouldn't his cathode ray beam have split up too?

As has already been suggested by some of the comments to your question, there is a fundamental difference between a Stern-Gerlach-type experiment and a Thompson-type experiment, and this difference lies in the type of interaction of the particles.

First of all, as Ivan Neretin points out, a Stern-Gerlach experiment is performed with neutral species (the original experiment involved silver atoms), while Thompson's experiment employed a beam of charged electrons.

The Coulombic or Lorentz interaction between charged particles (or charged particles and fields) is rougly four orders of magnitude larger than the interaction between neutral particles and fields. In the latter case, one can have interaction between a magnetic dipole moment and a magnetic field and/or interaction with a electric dipole moment and an electric field. These interactions are referred to as Zeeman and Stark effects. The important thing to realize is that the sign and magnitude of the interaction depends on the quantum state of the particle (or in other words on the relative orientation of its angular momenta). It is therefore a quantum-state specific interaction. As a consequence particles in one state may be attracted to a region with high fields, while particles in another state may be attracted to regions with low fields, as if you have two different compasses; one that points to the North pole and one that points to the South pole.

Electromagnetic interactions between charged particles and fields are so strong that the quantum-state specific effects are much harder to see (and the Coulomb and Lorentz forces dominate).

Let's go back to the Stern-Gerlach experiment. In their experimental setup, Stern and Gerlach used an inhomogenous magnetic field with a well-defined gradient. The silver atoms they employed have one unpaired electron that could be oriented with respect to the external magnetic field "up" or "down", these different quantum states experience different forces and are therefore separated.

In Thompson's experiment charged electrons are send through a magnetic field where the force they experience only depends on their charge and velocity vector with respect to the magnetic field. This force is identical for all electrons, so no splitting is observed.


number of optical isomers of a cyclic molecule What is the number of optical isomers for this compound?

Also, there seems to be no clear-cut boundary between geometrical isomers and optical isomers. Referring to a post in StackExchange,

is (RS)- and (RR)-form of 1,2-dichlorocyclohexane considered optical isomers or geometrical isomers? The isomers exist due to restriction of rotation caused by the ring, you can also called the two geometrical isomers, can't you?

1-Bromo-3-chloro-5-fluoro-7-iodoadamantane would be an interesting example of central tetrahedral chirality without the actual (physical) center. Exchanging halogen atoms will result in only two stereoisomers.

Due to its rigidity, its stereochemistry is related to tetrasubstituted tetrahedrane (tricyclo[1.1.0.02,4]butane; or rather its imaginary geometrical model, because it would be very strained and unstable), whose stereochemistry is related to bromo(chloro)fluoro(iodo)methane.

To your additional question: It depends how (far) you'd like to go with removal of the rotation restriction, anyway the resulting compounds will already not be original 1,2-dichlorocyclohexane.

I mean yes, (1R,2R)-1,2-dichlorocyclohexane and (1R,2S)-1,2-dichlorocyclohexane can probably be considered geometrical isomers (in addition to optical), they can even be named trans-1,2-dichlorocyclohexane (which has two enantiomers, the other one being (1S,2S)) and cis-1,2-dichlorocyclohexane; but this isomerism is not caused by rotation restriction, but by different arrangements on chiral tetrahedral carbons.

But note that geometrical isomerism is usually used in relation with double bond cis/trans isomerism; and together with optical isomerism are both deprecated terms.


How can I safely carry out an electrolytic etch on stainless steel without producing relatively harmful by-products? I propose to carry out an electrolytic etch on a stainless steel plate like so:

One of electrodes must be the stainless steel plate which is being etched and the surface layer must be removed.

I'm aware such an electrolytic etches can produce harmful by-products such as Chlorine (NaCl electrolyte), Chromium compounds and more.

If it is possible, What electrolytes and electrodes can I use to prevent the production of these substances?

Would it be possible to create safer Chromium compounds?
If I can't, what other non-electrolytic options do I have?

Sand or shot blast it instead.  Rub it with crushed aluminum foil and saltwater.

Everything is harmful.  Voltage determines products.  Use the minimum voltage (slow way to travel).  Don't use a chloride electrolyte.  (TEST FIRST!) Use hydroxide, carbonate, (nitrate - Homeland Severity and OSHA will beat on you), sulfate, electrolyte (note solubility of metallic byproducts after reaction with electrolyte).  The larger the charge on the ion, the slower the kinetics.  Vent the chlorine plus inert air through lye solution to make bleach, or through a solution of reducing agent.

Cr(III) is OK, Cr(VI) is a carcinogen. Reduce spent electrolyte with sulfite, Vitamin C, zinc granules, FeCl2, shredded scrap iron, steel wool, etc - and have a way to test for Cr(VI) residual.

Before you vent or sewer anything, talk with your Safety Officer.  The EPA and OSHA are carnivorous.  it's always a good idea to be a good upstream neighbor.


Second ionization potential of Gadolinium I notice that Gd has a second ionization potential which is significantly higher than would be expected from the general trend in the lanthanides (see this paper p. 945 for a graph). What is the reason for this effect ?

The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} => [Xe] 4f^{7} 5d^{1}$. I understand that the 5d is occupied to prevent inter-electron repulsion between antiparallel f electrons, and this can be used to rationalize why the correspnding Tb transition requires less energy, but I'm not sure why the potential should be significantly higer than that for Eu.

The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} \rightarrow [Xe] 4f^{7} 5d^{1}$.

Well actually, the second ionization energy being plotted in the graph is only the energy to lose the second electron so the transition is

$[Xe] 4f^{7} 5d^{1} 6s^1 \rightarrow [Xe] 4f^{7} 5d^{1}$.

For all the other elements Pr through Yb, there is no 5d electron, and instead an additional 4f electron. In all cases, it is the 6s electron that is being lost in the second ionization.

The more the nuclear charge is screened, the easier it is for the 6s electron to be lost.

The 5d electron is less effective than a 4f electron at screening the 6s electron from the nuclear charge, because 4f electron density is concentrated much closer to the nucleus (see Fig. 2 of below reference).  Therefore it takes more energy to loss the 6s electron for gadolinium.

For more information on nuclear charge screening see Atomic Screening Constants from SCF Functions. II. Atoms with 37 to 86 Electrons J. Chem. Phys. 47, 1300 (1967).


Why doesn't the Bessemer process oxidize the iron? The Bessemer process for making steel involves blowing air or oxygen thru hot iron. Impurities and carbon oxidize away. Why does the iron itself not oxidize?

I thought you asked a great question, one that I have wondered about myself.

The answer, as best I could determine, is that the iron does oxidize.  A good description of the Basic Oxygen Steelmaking process, today's descendent of the Bessamer process, says so anyway.  The BOS process uses pure oxygen instead of the original Bessamer processes's air.  It reports that some iron does oxidize.

The oxidation of carbon to form carbon monoxide is more thermodynamically favorable than the oxidation of iron, so it occurs preferentially.  The page I found (once again the link is here) mentions using on the order of 1800 scf of oxygen per ton of metal.  A ton of steel is about 18 kmol of iron.  1800 scf of oxygen is about 2 kmol of oxygen.  So there is not enough oxygen to oxidize even close to all the iron in the average BOS run.  That fact, combined with the fact that carbon, silicon, and other impurities in the pig iron oxidize preferentially, mean that not much iron is oxidized during BOS or Bessamer.

But "not much" does not mean zero.  My guess is that iron oxides are more soluble in the slag than they are in molten iron and so are also removed with the slag after BOS.


Whats is the difference between Tris and Tris-acetate? I would like to run the following protocol to detect plasmid presence in bacteria.
Would be possible to replace 'Tris- acetate' with just Tris?

We don't have 'Tris- acetate' in the lab.
What is the role of acetate here?

Hot alkaline method for all plasmid sizes and bacteria (Kado & Liu,1981), modified

Centrifuge 2-3 ml of culture, resuspend pellet in 1 ml of solution containing 0.04 M Tris- acetate, pH 8.0 (adjust pH with glacial acetic acid) and 2 mM EDTA
Add 2 ml of lysis buffer (0.05 M Tris, 3% SDS, pH 12.50, adjusted with 2 N NaOH) and mix
Incubate at 60-68°C for 30-45 min (strain dependent)
Add to hot samples 6 ml of phenol/chloroform (1:1) and mix gently to complete emulsification
Separate phases by centrifugation at 10.000 x g for 15-20 min at RT and transfer the upper aqueous phase carefully (avoid interphase which contains debris) to new tube containing 1 volume of chloroform. Mix and centrifuge again for separation of phases

"Tris" simply means "three" and is used in a number of different contexts. However, in biochemistry it's normally used to refer to the common buffering agent tris(hydroxymethyl)aminomethane. You have to be a little careful, though, as there are other uses of "Tris" in biology, which may not refer to tris(hydroxymethyl)aminomethane. (For example, there's two other buffers called "bis-tris methane" and "bis-tris propane", each of which have very different properties from tris(hydroxymethyl)aminomethane.)

Your situation is referring to the standard tris(hydroxymethyl)aminomethane, though. However, when specifying a buffering system, it's often insufficient to specify just the buffering agent used. A buffer consists of a weak acid or weak base in near-equal concentration to its conjugate base or acid. Tris is the weak base being used, but to have a decent buffer you need to also have the conjugate acid form.

As this conjugate acid is charged, there's an obligate counter ion. Different buffering systems may use different counter ions. That's what the acetate is doing here - acting as a counter ion for the conjugate acid of tris(hydroxymethyl)aminomethane.

That's what "Tris-acetate" is specifying. It's using a Tris buffering system with an acetate counter ion. This buffering system is normally obtained by using enough Tris base to get the desired concentration, and then slowly adding enough acetic acid to obtain the desired pH.

Alternatively, you could use something like the Henderson–Hasselbalch equation to calculate the appropriate proportion of base and conjugate acid for your desired pH, and then specifically mix those two forms together. This isn't typically done with Tris-acetate, as the acetate counter ion form of Tris isn't commonly sold. However, Tris hydochloride is, so if you had a system which used a chloride counter ion, you could use this approach. (Though in practice you'll want to aim a little on the basic side, and adjust to the desired pH by titration - the numbers which come out of the Henderson–Hasselbalch equation are theoretically pure, and don't account for non-ideal deviations in behavior and things like weighing errors.)


Unexpected reaction during acid hydrolysis of cellulose Recently I tried to hydrolyze some cellulose:

I added some acid to it ($\ce{H2SO4}$) then noticed nothing. Even after a few days there was no visible change and after doing some tests for glucose, those were negative too.

So at the end - expecting nothing would happen - I added some sodium hydroxide ($\ce{NaOH}$) to the solution to neutralize the acid and then the most unexpected thing happened, the solution began to warm up and there was an immediate color change (turning from white (from the cellulose) to yellow/brownish) I suspect that color came from the dehydration of the glucose  that was produced at that moment due to instantaneous hydrolysis of cellulose. I finally diluted the solution with more water and the reaction stopped

So somehow there was no interaction between the sulfuric acid and the cellulose, but when sodium hydroxide was added, the solution did more than just neutralize the acid!

Does anyone have any idea why this is?

We miss some data to make a good hypothesis.

The concentration of the $\ce{H_2SO_4}$ solution.
The temperatures.

These are the main two factors that affect cellulose hydrolysis. Usually are used temperature above 200 °C and pressure above 25 MPa. For me its strange that the exothermic reaction between SA and $\ce{NaOH}$ could lead to an increase of temperature such that.  However some studies show that the reaction can occur at lower temperature with concentrated acid.

Another possible solution is the formation of alkali cellulose.


Why 6.023x10^23 is not written as 6023x10^20? We know that 1 mole is equal to $6.023\times10^{23}$ atoms. But why it is not written as $6023\times10^{20}$ ?

Dissenter's answer is not quite right, since, strictly speaking, as defined there, the scientific notation will work for positive numbers only. Obviously, it can be easily fixed to work for negative as well, but the case of 0 will continue to be problematic. Just try to follow this rule to understand why.

All in all, it is the matter of definitions, but usually, scientific notation is defined as just a way of writing numbers in the form $a \cdot 10^b$ without any restrictions on $a$ which can be any real number. So, both $6.023 \cdot 10^{23}$ and $6023 \cdot 10^{20}$ are considered to be numbers written in scientific notation.

However, in the normalized scientific notation, indeed,

  the exponent $b$ is chosen so that the absolute value of $a$ remains at
  least one but less than ten ($1 ≤ |a| < 10$),

and thus, only $6.023 \cdot 10^{23}$ is written in this notation. Normalized scientific notation  is just one special case of scientific notation. And yeah, you could not write down zero in this notation, which is kind of awkward.

Another special case of scientific notation is the so-called engineering notation in which the exponent $b$ is restricted to multiples of 3. Neither $6.023 \cdot 10^{23}$ and $6023 \cdot 10^{20}$ are written in engineering notation, while $602.3 \cdot 10^{21}$ is. Although engineering notation is a special case of scientific one as defined above, is rarely called scientific notation to avoid an ambiguity.


What are the products in the reaction of sulphuric acid and Group 2 metals? When group 2 metals, says magnesium, is reacted with nitric acid, oxide of nitrogen is produced beside hydrogen gas. The standard electrode potential values agree with this:

$$
\begin{align}
\ce{Mg^2+(aq) + 2 e- &<=> Mg(s)} &\quad E^\circ &= \pu{-2.38 V} \\
\ce{H+(aq) + e- &<=> 1/2 H2(g)} &\quad E^\circ &= \pu{0 V} \\
\ce{NO3-(aq) + 4 H+ + 3e- &<=> NO(g) + 2H2O(l)} &\quad E^\circ &= \pu{+0.96 V}
\end{align}
$$

Does sulfate ion from sulfuric acid is reduced too when reacted with group 2 metals? The standard electrode potential values suggest it is possible, but I can’t find any source from internet stating $\ce{SO2}$ is produced from reaction of sulfuric acid with, says, magnesium.

$$
\begin{align}
\ce{Mg^2+(aq) + 2 e- &<=> Mg(s)} &\quad E^\circ &= \pu{-2.38 V} \\
\ce{SO4^2-(aq) + 4 H+ + 2 e- &<=> SO2(g) + 2 H2O(l)} &\quad E^\circ &= \pu{+0.17 V}
\end{align}
$$

If sulfate ion can’t be reduced by magnesium to sulfur dioxide, then what’s the reason?

We should not extend the idea of electrode potentials for very high concentrations such as concentrated sulfuric acid which is on the order of 18 M. The interesting property of these acids (HNO3 or H2SO4) is that they behave as oxidizing agents or even as dehydrating agents (H2SO4) especially at high temperature and at high concentrations. This is not to be predicted by tabulated electrode potentials. These tabulated values are reliable for dilute solutions (~ 1 M or less).

So the short answer is that in dilute H2SO4, and Mg, we would see a classical reaction of Mg forming magnesium sulfate and hydrogen. Under concentrated conditions and heat, Mg, may be able to able to reduce it to SO2, just like copper.


Deriving heat capacity in terms of internal energy U and natural variables S & V My lecture notes have the following equation regarding heat capacities:

$$
C_p = C_v + T\left(\frac{\partial p}{\partial T}\right)_v\left(\frac{\partial V}{\partial T}\right)_p
$$

(just to be clear, all terms after the brackets in this question indicate a constant process of that term)

It then states that by using internal energy $U$, volume $V$ and entropy $S$, you can express heat capacity at constant pressure, $C_p$, just in terms of these 3 variables:

$$
C_p = \frac {\left(\frac{\partial U}{\partial S}\right)_v\left(\frac{\partial ^2 U}{\partial V^2}\right)_s }{\left(\frac{\partial ^2 U}{\partial V^2}\right)_s\left(\frac{\partial ^2 U}{\partial S^2}\right)_v - \left(\frac{\partial ^2 U}{\partial S \cdot\partial V}\right)^2}
$$

with no inbetween steps given.

From a previous derivation of $C_v$ I know that:

$$
C_v = \frac{\left(\frac{\partial U}{\partial S}\right)_v}{\left(\frac{\partial ^2 U}{\partial S^2}\right)_v}
$$

and I think I understand at the least how to express the $\left(\frac{\partial p}{\partial T}\right)_v$ term, because you can express $p$ and $T$ in terms of $U, S, V$ like so:

$$
\begin{align*}
-p &= \left(\frac{\partial U}{\partial V}\right)_s \\
T &= \left(\frac{\partial U}{\partial S}\right)_v
\end{align*}
$$

The real killer is the second term $\left(\frac{\partial V}{\partial T}\right)_p$, because the only thing I know I can do with this equation is use the Maxwell relation:

$$
-\left(\frac{\partial V}{\partial T}\right)_p = \left(\frac{\partial S}{\partial p}\right)_T
$$

but this still leaves a constant temperature term $T$ which is not present in the final equation. So, any help with this would be really appreciated.

Usually this part of thermodynamics is not presented in the most efficient way. To derive the expression for, e.g., $C_P - C_V$, one goes through a series of steps involving Maxwell's relations and the triple product rule. Figuring out what to do in each step is done mostly by guesswork.

However, all relations of these sorts really come from just a single mathematical relation that more generally holds. That is to say,

  Consider a twice-differentiable,  two-variable function $f(x,y)$ and its Legendre transform with respect to $y$, defined by $g(x,f_y) := f - f_y y$. 1  Here, $f_y$ denotes the derivative of $f$ with respect of $y$ with other independent variables held fixed. Then, the following identity is satisfied:
  \begin{equation}
\boxed{g_{xx}  - f_{xx}  = - \frac{f_{xy}^2}{f_{yy}} = \frac{g_{x f_y}^2}{g_{f_y f_y}}}.
\end{equation}
  Proof. We have
  \begin{equation}
\bigg(\frac{\partial f}{\partial x} \bigg)_{f_y} = f_x + f_y \bigg(\frac{\partial y}{\partial x}\bigg)_{f_y} = f_x - f_y \frac{f_{xy}}{f_{yy}}.
\end{equation}
  On the above, the first equality holds because of the chain rule [for the change of variables $(x,y)$ $\to$ $(x, f_y)$], and the second equality because of the triple product rule. Then,
  \begin{equation}
g_x = \bigg(\frac{\partial f}{\partial x} \bigg)_{f_y} - f_y\bigg(\frac{\partial y}{\partial x}\bigg)_{f_y}   =f_x,
\end{equation}
  and
  \begin{equation}
g_{xx} = \bigg(\frac{\partial f_x}{\partial x} \bigg)_{f_y}= f_{xx} + f_{xy}\bigg(\frac{\partial y}{\partial x}\bigg)_{f_y} = f_{xx} - \frac{f_{xy}^2}{f_{yy}}.
\end{equation}
  Also, using the chain rule for the change of variables $(x,y)$ $\to$ $(x, f_y)$, we get
  \begin{equation}
g_{f_y} = \bigg(\frac{\partial f}{\partial f_y} \bigg)_{x} -\bigg(\frac{\partial y}{\partial f_y} \bigg)_{x}f_y - y = -y.
\end{equation}
  Therefore, Legendre transform of $g$ with respect to $f_y$ is simply $g - g_{f_y} f_y = f$. Then, performing a similar analysis as above for $g$ gives
  \begin{equation}
f_{xx} = g_{xx} - \frac{g_{x f_y}^2}{g_{f_y f_y}}.
\end{equation}

Now, all that remains to be done is to choose the variables and functions $x$, $y$, $f$ and $g$. Below are some examples.

(1) $x = S$, $y = V$, $f = U$, $g = H$, and $f_y = -P$:
\begin{equation}
H_{SS} - U_{SS} = -\frac{U_{SV}^2}{U_{VV}}.
\end{equation}
Note that
\begin{equation}
U_{SS} = \bigg(\frac{\partial^2 U}{\partial S^2}\bigg)_V = \bigg(\frac{\partial T}{\partial S}\bigg)_V = \frac{T}{T(\partial S/\partial T)_V} = \frac{T}{C_V},
\end{equation}
and that, similarly,
\begin{equation}
H_{SS} = \bigg(\frac{\partial^2 H}{\partial S^2}\bigg)_P = \bigg(\frac{\partial T}{\partial S}\bigg)_P = \frac{T}{T(\partial S/\partial T)_P} = \frac{T}{C_P}.
\end{equation}
Hence, we have
\begin{equation}
\boxed{\frac{1}{C_P} - \frac{1}{C_V} = -\frac{U_{SV}^2}{T U_{VV}} = -\frac{\Big(\frac{\partial^2 U}{\partial S \partial V}\Big)^2}{\Big(\frac{\partial U}{\partial S}\Big)_V \Big(\frac{\partial^2 U}{\partial V^2}\Big)_S}}
\end{equation}
which is equivalent to the relation OP wanted to derive.

(2) $x = T$, $y = V$, $f = A$, $g = G$, $f_y = -P$, and noting that $f_{-y} = -f_y$,
\begin{equation}
G_{TT} - A_{TT} = \frac{G_{T,-P}^2}{G_{-P,-P}} = \frac{G_{TP}^2}{G_{PP}}.
\end{equation}
Here,
\begin{equation}
A_{TT} = \bigg(\frac{\partial^2 A}{\partial T^2}\bigg)_V = -\bigg(\frac{\partial S}{\partial T}\bigg)_V = - \frac{1}{T}C_V,
\end{equation}
\begin{equation}
G_{TT} = \bigg(\frac{\partial^2 G}{\partial T^2}\bigg)_P = -\bigg(\frac{\partial S}{\partial T}\bigg)_P = - \frac{1}{T}C_P,
\end{equation}
\begin{equation}
G_{TP} = \frac{\partial^2 G}{\partial T \partial P} = \bigg(\frac{\partial V}{\partial T}\bigg)_P = V \alpha,
\end{equation}
where $\alpha := \frac{1}{V}\Big(\frac{\partial V}{\partial T}\Big)_P$ is the volume expansion coefficient,
\begin{equation}
G_{PP} = \bigg(\frac{\partial^2 G}{\partial P^2}\bigg)_T = \bigg(\frac{\partial V}{\partial P}\bigg)_T = - V\kappa_T,
\end{equation}
where $\kappa_T := -\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_T$ is the isothermal compressibility. Then, it follows that
\begin{equation}
\boxed{C_p - C_V = \frac{TV\alpha^2}{\kappa_T}}.
\end{equation}

(3) $x = V$, $y = S$, $f = H$, $g = G$, and $f_y = T$:
\begin{equation}
G_{PP} - H_{PP} = \frac{G_{TP}^2}{G_{TT}}.
\end{equation}
We have
\begin{equation}
H_{PP} = \bigg(\frac{\partial^2 H}{\partial P^2}\bigg)_S = \bigg(\frac{\partial V}{\partial P}\bigg)_S = - V\kappa_S,
\end{equation}
where $\kappa_S := -\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_S$ is the adiabatic compressibility. We have already shown that
\begin{equation}
G_{PP} = -V\kappa_T, \quad G_{TP} = V \alpha, \quad G_{TT} = - \frac{C_P}{T},
\end{equation}
whence we obtain
\begin{equation}
\boxed{\kappa_T - \kappa_S = \frac{TV\alpha^2}{C_P}}.
\end{equation}

1 To be more precise, one should first write down $f(x,y) - y f_y$ and then express $y$ as a function of $x$ and $f_y$. Also, for the Legendre transform to exist, $f(x,y)$ should be a convex or concave function of $y$.


Is there a relation between the ionization energy and the electron affinity? I am aware of the definition and equation of IE and EA. I was wondering if there is a direct relation between this two quantities with the energy of the neutral specie.

Okay, here's what I did (and I highly encourage everybody to do this along with this tutorial-style answer, just so you get the feel for prepping and looking at data from the world wild web).

Data Acquisition

First, I went and copied the data from the tables at Wikipedia: This one for the ionization energies and this one for the electron affinities.

Next I had to clean up this data somewhat. For the electron affinity data I first had to get rid of all the unnecessary data columns. This was quickly done with awk:

   awk '{print $1, $2, $4}' raw-affinity.dat > affinity.dat

Next, I filtered out all the lines containing "-" and "?" using inverse search with grep:

    grep -v "?" affinity.dat | grep -v "-" > affinity2.dat

And finally I got rid of all these pesky statistical uncertainties with either one or two numbers between parentheses with sed:

    sed -e 's/(.)//g' -e 's/(..)//g' affinity2.dat > affinity3.dat

Now the data file looks somewhat like this:

1 H  0.754195
3 Li 0.618049
5 B  0.279723
6 C  1.262118

Now for the ionization data file. Since we only want to look at the first ionization energy we can scrap all the other data. Again, we'll use awk for this:

    awk '{print $1, $2}' raw-ionization.dat > ionization1.dat

Next we'll filter out all the data we don't want in there (again with grep and inverse searches):

    cat ionization1.dat | grep -v "use" | grep -v "WEL" | grep -v "talk" > ionization2.dat

In this file we still have some nasty linebreaks... So I loaded the file in vim and substituted the line break + "CRC" with nothing and saved it as ionization3.dat:

    :%s/\nCRC//g
    :saveas ionization3.dat

I also corrected the value for At manually in vim while I was at it, because it wasn't entered like all the other data in the Wikipedia table.

The ionization data file now looks like this:

1 H  13.59844
2 He 24.58741
3 Li 5.39172
4 Be 9.3227

All right, so we have the data nicely prepared and we're ready for the next step, which is

Data Analysis Shenanigans

I did everything in python using the numpy and matplotlib packages. The code used can be found here.

I didn't really notice some large trend of something you might expect. The plot gets pretty messy somehow. The darker the scatter points, the heavier the element.

Maybe someone else notices a pattern that I'm missing right now.


What is the origin of the naming of deoxycorticosterone acetate? Deoxy- means the compound misses an oxygen comparing to the name without the prefix.
Acetate means ethanoate in IUPAC language. Many biologists still abide old rules and call it acetate. And finally, corticosterone. I just threw away thoughts and checked for the formula  online.
What is the law behind naming "corticosterone"? Is it relative to corticosteroids? Please answer in a detailed way!
Maybe it is corti+coster+one? Since I know that there exists cortisone and cortisol.

Naming complex chemical compounds in a straightforward systematic fashion is usually not very easy. And often there are trivial names, that existed way before this systematic naming scheme was introduced. Especially when it comes to naturally occurring compounds like steroids. The IUPAC defines steroids as the following:

  Naturally occurring compounds and synthetic analogues, based on the cyclopenta[a]phenanthrene carbon skeleton, partially or completely hydrogenated; there are usually methyl groups at C-10 and C-13, and often an alkyl group at C-17. By extension, one or more bond scissions, ring expansions and/or ring contractions of the skeleton may have occurred. Natural steroids are derived biogenetically from triterpenoids.

  (source: iupac.org)

There is even a quite lengthy article about the nomenclature of steroids: [1] G. P. Moss, Pure & App. Chem. 1989, 61 (10), 1783-1822.

The name Deoxycorticosterone acetate is itself derived from a trivial name. The underlying principle compound is corticosterone and its systematic steroid name is 11β,21-dihydroxypregn-4-ene-3,20-dione, see [1] Table 2. It can be derived from cortisol, which has an additional hydroxy group in position 17. The systematic name therefore is (11β),17,21-trihydroxypregn-4-ene-3,20-dione.

You assigned correctly, that the prefix de(s)oxy- refers to losing an oxygen containing functional group. In this case a hydroxy group at position 11. The systematic name for 11-deoxycorticosterone would therefore be 21-hydroxypregn-4-ene-3,20-dione.

Finally the acetate refers to an acetyl ester group at position 21 (in the sidechain). Interestingly the systematic name for cortisol acetate is 21-O-acetylcortisol. Since it has already a trivial name, see [1], 3S-4.3:

  When the steroid acid has a trivial name that already specifies the hydroxyl group that is esterified, the prefix must be O-acyl-.

Applying the same logic, the systematic name for Deoxycorticosterone acetate should be 21-O-acetyldeoxycorticosterone. Which makes little to no sense. A more systematic approach is described again in  [1], 3S-4.3

  Esters of steroid alcohols
  Esters of steroid alcohols are named by the appropriate steroid substituent group followed by that of the acyloxy group in its anionic form. The steroid name is formed by replacing the terminal -e of the hydrocarbon name by -yl, -diyl, etc. and inserting this before the locant with a Greek letter to designate position and configuration. When necessary locants should be used with the anionic part of the name. Examples: [...]
  The prefix form acyloxy-, with the appropriate locants, is used if there is a preferred functional group as the suffix (e.g. an acid or lactone). Examples: [...]

So in this case since you have a ketone, which is probably more important, I would choose the second part of this rule. Hence the systematic name becomes 21-acetoxypregn-4-ene-3,20-dione.

As you can see, naming is quite a big, often confusing subject and sometimes the origins are not clear at all. And your analysis was completely correct in every way. Trivial names are often like the vocabulary of a specific language - if you don't know it, you have to look it up.

Additional fun fact:
The name acetate is a retained name which can be used in a semisystematic way.


Does calcium nitrate decahydrate dehydrate or decompose upon heating to 100 degrees Celsius? Consider a sample of $\ce{Ca(NO3)2.10H2O}$. To remove water, I need to heat it to $\pu{100 ^\circ C},$ but will this high temperature decompose it or will I get the anhydrous salt?

Calcium nitrate exists under three forms : 1) $\ce{Ca(NO3)}_2 $ which melts at 361°C; 2) $\ce{Ca(NO3)}_2$$·\ 3 $$\ce{H_2O}$, which is not commercial and melts at 51°C ; 3) $\ce{Ca(NO3)}_2$$·\ 4 $$\ce{H_2O}$ which is commercially available and  melts at 43°C. It looses its water at 132°C and is deliquescent. The decahydrate that you mention does not exist.


Why does burnt hair smell bad? When I use hot stuff like hair straightener on my hair, my hair begins to smell bad, which is very different from smell produced from burning other things. So what's the gas produced that is responsible for this smell?

Hair is largely (~90%) composed of a protein called keratin, which originates in the hair follicle.

Now, keratin is composed of a variety of amino acids, including the sulfur containing amino acid, cysteine. All these amino acids are joined to each other by chemical bonds called peptide bonds to form these long chains that we call polypeptide chains. In the case of human hair, the polypeptide that we're talking about is keratin. The polypeptide chains are intertwined around each other in a helix shape.

The average composition of normal hair is 45.2 % carbon, 27.9% oxygen, 6.6% hydrogen, 15.1% nitrogen and 5.2% sulfur.

(I got that diagram off of Google Images)

Now, there are a whole bunch of chemical interactions that maintain the secondary and tertiary structures of proteins, such as van der Waals forces, hydrophobic interactions, polypeptide linkages, ionic bonds, etc. But there is, however, one additional chemical interaction in proteins that contain the amino acids cysteine and methionine (both of which contain sulfur) called disulfide linkages. You can see that in the diagram above (it's been marked in yellow, which is fortunately, a very intuitive color when you're dealing with sulfur).

When you burn hair (or skin or nails... anything that has keratin in it for that matter) these disulfide linkages are broken. The sulfur atoms are now free to chemically combine with other elements present in the protein and air, such as oxygen and hydrogen. The volatile sulfur compounds formed as a result is what's responsible for the fetid odor of burning hair.

Quite a few of the "bad smells" we come across everyday are due to some sulfur containing compound or the other. A great example would be the smell of rotten eggs, which can be attributed to a volatile sulfur compound called hydrogen sulfide. Yet another example (as @VonBeche points out in the comments) would be that of tert-butylthiol, which is the odorant that is used to impart the characteristic smell of Liquefied Petroleum Gas (LPG).


How is the "quasi-steady-state" assumption of Michaelis-Menten kinetics different from the steady state assumption of ordinary kinetics? In Wikipedia's article for Michaelis-Menten kinetics, it titles the "[ES] = constant" section the "Quasi-steady-state approximation". Lehninger's Principles of Biochemistry, 4th Ed. pp. 203ff call this "steady state kinetics".

I know that steady state is an assumption which is used for various kinetic models that have an intermediate, and I believe the term "steady state" is used in other settings as well. Is Wikipedia's use of "quasi" in the title for the enzyme kinetics reflective of $V_0$ only being the measurement of the initial rate, and thus the further assumption that [S] is constant also adds more uncertainty to the steady state assumption? If this isn't the reason (and I'm not certain that it is), why add a further descriptor that denotes uncertainty or inaccuracy when we already know that steady state is an assumption?

Is Wikipedia's use of "quasi" in the title for the enzyme kinetics reflective of [...] the further assumption that $[S]$ is constant[...]?

No I don't think the "quasi-steady state" hypothesis is related to whether $[S]$ varies with time.

  [W]hy add a further descriptor that denotes uncertainty or inaccuracy when we already know that steady state is an assumption?

Different fields use different terms for the same concept.  For example, some folks call it the "pseudo-steady state hypothesis" and abbreviate it "PSSH".  Some people just say "steady state assumption".  It's all different vocabulary for the same thing.

I don't think there's a principled reason to prefer one term over another.  If you solve the full differential equations for kinetic models that would normally invoke the PSSH, you'll see one problem.  Before the reaction starts, the concentration of intermediate $I$ is zero.  But when the reaction is going on, we assume the intermediate is at steady state, i.e. that $\frac{dI}{dt}=0$.  But if $I$ starts at zero and has an initial concentration of zero, then it is everywhere zero, which violates the prinicple that intermediates actually occur in real-life chemical reactions.  So there are two phases to such models: the first is the "burst" phase, also known as single-turnover conditions, or unsteady state kinetics.  In that phase $\frac{dI}{dt}$ is not 0 but increases rapidly.  That phase ends quickly and is replaced by the pseudo-steady phase for $I$.  If you want to get precise about the strict definitions of the two phases, you'll need to invoke singular perturbation and asymptotic matching, as described in this thorough book chapter from Mathematical Biology by James D. Murray.

So you could argue that the "pseudo" or "quasi" is good to have in there to remind you of these complexities.  Or you could argue that all those extra letters aren't doing a whole lot of good.  Either way, it's just a semantic issue and is unrelated to differences in enzyme kinetics and more "traditional" chemical reactions.


Melting point vs Freezing point Is it possible to have Melting point different that Freezing point

I mean is there any element/molecule in the liquid status that freeze at point X Centigrade but when it become solid then you need a higher temperature to become liquid again like X + Y degrees in the same conditions?

That's very common, actually.

https://en.wikipedia.org/wiki/Polymorphism_(materials_science)#Ostwald's_rule

A substance melts, but the recrystallisation from the iostropic melt can be kinetically inhibited. If you cool further down, a less stable form of the same substance, with lower melting point, can crystallise, and re-form over time, into the original phase.

This (according to Ostwalds rule) happens practically every time a suitable, less stable polymorph exists. Or let's rather say there is a general tendency for it to happen.

Another, very common example (to which Ostwalds rule also loosely applies) are polymers. Because of their chain structure, crystallisation (if possible at all) is much inhibited and usually never occurs until you cool a good ten K or so below the equillibrium melting point.


Why is the only H NMR resonance of 1,4-dibromobenzene a singlet? I am working on a chemistry problem set and the answer key says all H's will show up as singlets. This does not make much sense to me as there are adjacent hydrogens to each carbon which should mean the H's show up as a doublet. What am I thinking incorrectly about this?

All four hydrogens are equivalent and therefore they do not display any coupling to each other. (They do couple, but the coupling cannot be seen in the spectrum.) See: Why don't equivalent hydrogens cause splitting in NMR? for more discussion of this topic.

Just to be extra nitpicky, I wouldn't write "all four hydrogens show up as singlets", since that implies that there are four different singlets. I'd write something like "one singlet is observed in the spectrum, corresponding to all four hydrogens".


Effect of impurities on melting and boiling points I am tutoring my nephew for his science GCSEs; and his chemistry textbook states that you can tell how pure any compound is by how close its melting and boiling points are to the "official" mp and bp for that compound. It says that impurities will always lower the melting point of any compound, but - paradoxically (intuitively speaking) - will always raise the boiling point. But it just states this as a fact and offers no explanation!
I found a simple explanation for why the melting point is reduced in the case of crystalline solids:

Foreign substances in a crystalline solid disrupt the repeating pattern of forces that holds the solid together. Therefore, a smaller amount of energy is required to melt the part of the solid surrounding the impurity. This explains the melting point depression (lowering) observed from impure solids.

... although this doesn't explain why impurities in solids that are not crystalline also lower the melting point.
But I can't find any easy to understand explanation anywhere for why impurities would always increase the boiling point of any compound. I've found explanations that are specific to water (and even then, the explanations I've seen only make sense if the impurity binds more strongly to water molecules than water molecules bond to each other, which isn't always the case); and I found one explanation that says it's because of increased solution phase enthalpy, without explaining what on earth that means!
So I emailed a Professor of Physical Chemistry about this and she replied

The reason for impurities lowering the melting point yet increasing the boiling point is because the impurities stabilise the liquid phase, making it more energetically favourable. This extends the liquid range to lower temperatures (lowering the melting point) and to higher temperatures (raising the boiling point). The liquid phase is stabilised due to the entropy increase when you have the impurity (or any solute) in the liquid.  Normally solids do not dissolve the impurities, so the melting point involves transition to pure solid (with impurity separating out or staying in the liquid state).

She made it clear she didn't have time to answer any follow-up questions, and I can't make full sense of her answer, so I am hoping someone here can understand her explanation and elucidate it.
I think I understand why impurities increase the entropy of a substance in its liquid phase by more than they increase the entropy of the same substance in its solid or gasseous phases - I presume impurities don't significantly affect the entropy of a solid because the solid can't rearrange itself into different configurations without absorbing a lot of energy; and they don't affect the entropy of a gas because the gas can disperse over an almost infinite distance to all intents and purposes, so it will be almost infinitely disordered even when pure; but they do increase the entropy of a liquid because there are many more ways of arranging the molecules within a liquid if there is more than type of molecule in that liquid, and the molecules in a liquid can rearrange themselves in all those different configurations without any input of energy.
First of all, have I got that much right?
But even if I have, I don't understand why increasing the entropy of the liquid by adding an impurity to it would make it energetically more stable - and therefore both more difficult to freeze and more difficult to vapourise.
Can anyone help to explain this?
Dave

All generalities are false (as is this one). Though impurities usually lower the melting point (m.p.) by disrupting crystallization on the atomic order, consider the phase diagram of the binary alloy (amalgam) HgxAg1- x: Pure Hg melts ~-39°C, and adding even a little bit of impurity raises the melting point considerably!
And without resorting to entropic explanations, consider that a high boiling point (b.p.) solute would "hold onto" the solvent, decreasing vapor pressure. $\ce{CaCl2}$ will absorb $\ce{H2O}$ from the air, driving down the vapor pressure (and humidity) even after the $\ce{CaCl2}$ liquefies.
That said, the answer they expect on the exam is that impurities decrease the m.p. and increase b.p.


How is spiro[4.4]nonatetraene antiaromatic How is spiro[4.4]nonatetraene antiaromatic? I know that spiro compounds exhibit homoaromaticity, but how do they exhibit antiaromaticity?

The orbital interaction between the 2 butadiene units in spiro[4.4]nonatetraene is referred to as spiroconjugation.  Note that the 2 rings attached through the common or spiro carbon are perpendicular to one another.  Each cyclopentadiene ring contains a 1,3-butadiene unit.  In the figure below, I've drawn in the molecular orbitals for the highest occupied molecular orbital ($\Psi_2$) of 1,3-butadiene.

It is a bit easier to analyze the situation using a Newman projection.  In the Newman projection on the right I only show the 4 carbons (and their P-orbitals) that are directly attached to the central spiro carbon, since these atoms and orbitals are the only ones close enough to interact between one ring and the other.  Notice how all of the through space interactions between nearby P-orbitals is stabilizing (indicated by arrows).  That is, all of the overlaps are construcive (bonding}, either + + or - -.  This indicates that $\Psi_2$ has the correct symmetry to interact with itself.  Look at the molecular orbitals (MO) for 1,3-butadiene and convince yourself that $\Psi_1$ and $\Psi_3$ do not have the appropriate symmetry to interact with $\Psi_2$ (e.g. there are no + + overlaps between the carbons attached to the spiro carbon, only + - overlaps that cancel out; and energetically, $\Psi_4$ is to far away to interact.  When the two $\Psi_2$ MOs interact, as usual, new MOs are formed; one is stabilized and one is destabilized and the destabilized orbital is always slightly more destabilized than the stabilized orbital is stabilized.  The net result in this case is that spiroconjugation produces a (small) net destabilization and the molecule is termed antiaromatic in a spiroconjugation sense.

On the other hand, consider spiro[2.4]heptatriene. Now the only interaction of consequence (e.g. where the interactions are of the ++ or -- type rather than +-; said differently, only $\Psi_2$ of the diene and $\ce{\pi_^{*}}$ of the ethylene have the appropriate symmetry to interact) occurs between the HOMO of the diene ($\Psi_2$) and the LUMO of the ethylene in the 3-membered ring.

(image source)

In this case, as the orbital mixing diagram indicates, there is a net stabilization of the pi electrons that results.  Spiro[2.4]heptatriene is said to be aromatic.

Spiroconjugated compounds containing 4n+2 pi electrons will turn out to be stabilized (aromatic), while those with 4n pi electrons will be destabilized (antiaromatic).  Remember, these effects are small, but they can be seen in UV spectral shifts and even in solvolysis reaction rates of related compounds.  If you google spiroconjugation or spiroaromaticity you can find links to the original articles that appeared in the late 60's, as well as some more recent information on the subject.


Delta G for Standard Hydrogen Electrode reaction I’m dealing with electrochemistry problems.
For the following reaction
$$\ce{2H+ +2e- <=>H2}\quad E^\circ=0\ \mathrm{V}$$
Does it make sense to say
$$\Delta G^\circ=-nFE^\circ=0$$
Which would mean the reaction is in equilibrium?

You cannot apply the $\Delta{G}$ equation to a single electrode potential. It can be applied to a cell though so if the hydrogen electrode is connected to another electrode (say copper dipped in copper sulfate solution) then you can find the free energy.

It's better to remember the formula as:

$\Delta{G} = -nFE_{cell}^o$

Where

$E_{cell}^o = E_{cathode}^o - E_{anode}^o$ (using all standard reduction potentials)

That way you won't accidentally apply it anywhere else.

In this case, for $\ce{\Delta G}$ to be zero you would have to take Hydrogen electrodes for cathode as well as anode.

A More Complicated Explanation

Just in case you're curious about the reasons behind this (Since I was also very confused about this when I first learned Electrochemistry. It's ok if you don't understand this),

It is very difficult to measure the absolute potential of a single electrode and so the standard reduction potentials you are using are actually based on a reference system.

This is actually very similar to the system of atomic masses. The atomic masses you use in calculations are not the actual masses of the elements in grams but they are measured in unified atomic mass units. One unified atomic mass unit is defined as 1/12th the mass of a carbon-12 atom.

[Note: Now of course, after the mole was made a fundamental SI unit, the unified atomic mass unit has become equivalent to $1 g/mol$. This is just for the sake of explanation.]

So if you come across a situation where you have to use the mass of the atom (maybe some nuclear chemistry calculations) you can't use those normal atomic masses but you would have to convert it into kg (One unified atomic mass unit is about $1.6\times10^{-27}$ kg).

Similarly, the standard reduction potentials are actually measured with reference to a standard electrode (Whose reduction potential is arbitrarily taken to be zero) which is the standard hydrogen electrode.

This standard reduction potential of an electrode can be measured easily by connecting the electrode to a standard hydrogen electrode (SHE) and then measuring the potential difference using a voltmeter and applying the $E_{cell}^o$ equation. (If $H^+$ is being reduced to $H_2$, then $E_{cathode}^o$ is zero otherwise $E_{anode}^o$ is zero).

So this just means that when you say $E^o$ of the SHE is zero, you are actually saying that "the potential of the SHE with respect to itself is zero" which is quite obvious. That doesn't mean that the potential of the electrode itself is zero.

That is why you cannot apply the first equation in this case. But this does not apply to cells. If we assume the absolute potential of the SHE to be some $x$, then,

$E_{abs}^o = E^o + x$

Then,
$E^o_{cell} = E_{abs}(cathode) - E_{abs}(anode)$

$= E^o_{cathode} + x - [E^o_{anode} + x]$

$E^o_{cell} = E_{cathode}^o - E_{anode}^o$

The absolute potential of the SHE cancels out in the difference.


Corrosion of iron: no electron pushing mechanism? Why does the oxidation-reduction reaction between iron, water, and $\ce{O2}$ gas to form rusts lack an electron pushing mechanism and a specific sequence of steps?
$$\ce{4Fe^0(s) + 3 O2(g) + 2n H2O(l) -> 2 Fe2O3·nH2O(s)}$$

I've noticed that most of the reactions described in undergraduate organic chemistry textbooks have detailed and widely accepted electron pushing mechanisms whereas most inorganic reactions (such as the corrosion of iron) lack electron pushing mechanisms. Why?

Also, which type of $\ce{O2}$ is involved in the corrosion of iron? The singlet excited state or the triplet ground state or doesn't matter?

$$
\underset{\text{triplet oxygen}~\ce{^3O2}}{
\ce{
·\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}-\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}·
}
}
\qquad
\underset{\text{singlet oxygen}~\ce{^1O2}}{
\ce{
\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}-\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}
}
}
$$

Arrow pushing mechanism is just a formal way of thinking how the reaction might be occurring. Unfortunately, organic chemistry textbooks present it as if all of these reactions were occurring in real time, step by step, and we are able to watch it like a movie. The reality must be far more complex. If you were truly investigating a reaction mechanism, we have to design very clever experiments (there are books on this topic). It is experimental data that determines the true outcome of an organic reaction not the arrow pushing formalism. Usually, when chemists find out the experimental products, then they postulate arrow pushing mechanisms.
Coming to the rust question. Rust has puzzled chemists and engineers for decades because it causes millions of dollars of losses every year.  Your equation is a very simplistic way of thinking that rust forms in a single step.
$$\ce{4Fe^0(s) + 3 O2(g) + 2n H2O(l) -> 2 Fe2O3·nH2O(s)}$$
It does not proceed in a single step. Our atmospheric chemistry is very complex, you have carbon dioxide, you have sulfur dioxide, nitrogen oxides, ozone, water vapors, sunlight, plenty of free radicals and plenty of minor trace components. Then there are different phases of rust too. You can have a quick look at Google Scholar and one such representative abstract [1].
In short, people still do PhD in this field and one can only imagine how complex corrosion science is.
References

Misawa, T.; Hashimoto, K.; Shimodaira, S. On the Mechanism of Atmospheric Rusting of Iron and Protective Rust Layer on Low Alloy Steels. Corrosion Engineering 1974, 23 (1), 17–27. https://doi.org/10.3323/jcorr1974.23.1_17.


The mole is used extensively in chemistry, why not elsewhere? In chemistry, the unit "mole" is used in many places, and is an important tool when working with quantities.

I have not seen it used elsewhere--for example, we never count people or TVs with the unit.

What makes the mole so useful in chemistry, and why isn't it used elsewhere?

Because the mole is not a convenient basis for people and TV sets (for instance, one mole of people would be approximately 86 trillion (short-scale) times the population of earth, or put another way the earth contains a human population concentration of approximately 12 femtomoles/planet), and because there's no connection back to atomic mass, which is what makes the mole useful in the first place.

This was not part of your original question, but addresses Manishearth's modifications:

Because atoms and molecules react in discrete units (cf. the law of definite proportions), the number of particles is more fundamental than the masses of those particles. The mole is useful in chemistry because it is defined such that the mass of Avogadro's number of particles (atoms or molecules) of a substance (the molar mass) is equal to the sum of the numerical values of the atomic masses of its constituent elements. The value of the mole is calibrated such that the molar mass of carbon-12 is exactly 12 grams per mole. On the periodic table you will note that the atomic masses given are non-integral (often markedly so, e.g. chlorine at 35.45 g/mol). This is because the atomic masses represent a weighted average of isotopic masses, and because the proton and neutron masses aren't quite equal.

The mole makes stoichiometric calculations particularly convenient as we can measure out, for instance, 16 grams (one mole) of $\ce{^{16}O}$ and 2 grams (two moles) of $\ce{^1H}$ and obtain one mole of $\ce{H2O}$. In the case of water, oxygen makes up 89% of the molecule by mass, but is only one of the three atoms present in the molecule. The mole is necessary to transpose stoichiometries involving single atoms and molecules (with their tiny masses) into the everyday world of grams and litres.


Buffer and strong acid If a weak acid is added to a neutral solution, it creates equal amounts of  $\ce{H3O+}$ and conjugate base.

So what then what happens to the original $\ce{H3O+}$ created by the weak acid when a strong acid is added to a solution.

From what I understand, some of the newly created $\ce{H3O+}$ from the strong acid will react with the conjugate base of the weak acid, preventing the pH from getting lower. But as I stated above, what about the original $\ce{H3O+}$ from the weak acid? For every molecule of the conjugate base that reacts with a $\ce{H3O+}$ from the strong acid, there must have been a $\ce{H3O+}$ created from the weak acid. So how is pH actually prevented from being lowered?

I have read your question several times to figure out what your problem is.
Below, I will try to give you a simple explanation without any fancy mathematics or complex equations.

Now, suppose we dissolve the weak acid "A" in water. The acid will react with water and form hydrogen ions and the corresponding base "B" of the weak acid. We further suppose that the pH of this solution is 3 at equilibrium.

We have the following equilibrium:

$$\ce{A + H2O <--> H3O+ + B}$$

The concentrations are supposed to be low, so we can state an equilibrium constant:

$K = [\ce{H3O+}]*[\text B]/[\text A]$

Now, we add a strong acid to this solution. What will happen?
The pH of the solution will drop. Suppose it will drop to pH = 2. We will get an immediate response from the equilibrium reaction of the weak acid system. Since the amount of hydrogen ions have increased in the solution by the addition of the strong acid, the equilibrium will be pushed to the left.

Pushed to the left? What does it mean?

Well, the new situation implies that the weak acid cannot give off  the same amount of hydrogen ions as it did at pH = 3. pH is now 2 and we have a completely new situation.

Now I will target on what I think is your main problem.

When you add a strong acid to the weak acid solution, the pH will drop.
The equilibrium of the weak acid will then reset into a new equilibrium determined by the actual pH and the equilibrium constant of the system.

At pH = 3, hydrogen ions were formed from the weak acid. These (or other) hydrogen ions will be forced to react with the corresponding base B, giving more A at pH = 2.
Thus, the net effect is that the hydrogen ion concentration related to the strong acid will be in principle unchanged.


Aromatic ring substitution: how are MNT and DNT further nitrated if nitro groups deactivate benzene rings? This is probably a common question because it involves TNT precursors, but if an $\ce{NO2}$ group deactivates benzene rings toward electrophilic aromatic substitution reactions, then how can toluene be tri-nitrated? I understand that the methyl group will donate electron density and activate the ring toward the first substitution, but after two nitrations I'm sure the nitro groups slow the third substitution. So what's the trick? How does toluene get three nitro groups?

On a industrial scale by using highly concentrated nitric acid known as fuming nitric acid (usually > 90%, but in the last step from DNT to TNT may also be anhydrous) along with oleum, which is a mixture of $\ce{SO_3}$ and sulfuric acid. Preparation in the lab may be achieved by more or less ordinary nitration at elevated temperatures. In a micro scale experiment [1] they used two different types of nitrating acid:

Nitrating acid I contains concentrated nitric acid 19% v/v, concentrated sulfuric acid 58% v/v and water 23% v/v.
Nitrating acid II is a mixture of concentrated nitric acid 25% v/v and oleum (30% $\ce{SO_3}$) 75% v/v.

Nitrating acid I was used for $\ce{Toluene -> MNT}$ and $\ce{MNT -> DNT}$ (65 °C). Nitrating acid II was used for $\ce{DNT -> TNT}$ (110 °C).

1 The Stepwise Nitration of Toluene: A Multistep Microscale Synthesis Based on an Industrial Process
Richard A. Russell, Robert W. Swltzer, and Robert W. Langmwe
University College, University of N.S.W., Northcotl Drive, Campbell ACT. 2600, Australia


What is this reference material in spectrophotometry? First, I should say I never have done any spectroscopies and I am just reading about it.

As far as I know, UV-Vis spectrophotometer has 2 beam samples: sample and reference. The sample has very diluted test material in a solvent and the reference is only the solvent.

I was reading a book about some experimental values for UV-Vis spectrophotometry and I noticed this page.  My question is about that sentence saying "The data were obtained .. against water reference" .   As the reference is the solvent alone, does it mean the samples are dilute solutions in water ?  if so, how they solved benzene in water?

Here is a snapshot of the text:

As the reference is the solvent alone, does it mean the samples are dilute solutions in water ? if so, how they solved benzene in water?

I come to a different conclusion.

To me it reads as follows:

sample: solution of benzene, acetone, acetonitrile, etc. in water
reference: water

Benzene is soluble in water: about $1.8\,\mathrm{g\cdot L^{-1}}$ at 25 °C.

Edit

As a reference (blank), a solution of everything except the sample is used. In organic (photo)chemistry this usually means: just the solvent. The situation is different in biochemistry, medicinal chemistry, etc. when complex aqueous buffer systems are used. Here, everything except the probe really means that the whole buffer is used as a blank.


What is the real pKa of water? What is the $\text{p}K_\text{a}$ of water? A simple google search yields the value $15.74$, but this site and this paper say it's $14.0$.

According to my understanding, the correct answer should be $14.0$:

$$\text{p}K_{\text{a}}= -\log([\ce{H+}][\ce{OH-}])$$

For $25~\text{°C}$:
$$[\ce{H+}][\ce{OH-}] = 10^{-14} = K_{\text{w}}$$

Thus follows $\text{p}K_{\text{a}}=14$.

Can you tell me which value is correct and why?

The ion product of water is usually expressed as
$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$
At a temperature of $25\ ^\circ\mathrm C$, its value is approximately $K_\mathrm{w}=10^{-14}$, or $\mathrm pK_\mathrm{w}=14$.

However, the ion product of water is not to be confused with the acid dissociation constant of water.

Generally, the dissociation constant for the simplified reaction
$$\ce{HA <=> A- + H+}$$
is defined as
$$K_\mathrm{a}=\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$$
Thus in case of water
$$K_\mathrm{a}=\frac{[\ce{OH-}][\ce{H+}]}{[\ce{H2O}]}$$
The concentration of pure water at a temperature of $25\ ^\circ\mathrm C$ is $c=55.345\ \mathrm{mol\ l^{-1}}$.
Therefore,
$$\begin{align}K_\mathrm{a}=\frac{[\ce{OH-}][\ce{H+}]}{[\ce{H2O}]}=\frac{10^{-14}}{55.345}=1.807\times10^{-16}=10^{-15.74}\end{align}$$
or
$$\mathrm pK_\mathrm{a}=15.74$$


Polycarbonate resistance to acids If polycarbonate plastic is a polar material with ester bond linkages which usually hydrolyze in acids, why is it resistant to both weak and strong acids (as polar substances)?

Looking at what has been published about hydrolysis of polycarbonate resins to recover bisphenol A, typical conditions involve water, heat, optional base catalyst (or more rarely, acid, which probably further degrades BPA), and very often a cosolvent such as tetrahydrofuran that itself is stable to hydrolysis. So your intuitions about whether polycarbonate should be susceptible to acid hydrolysis are correct.

There has been a lot of attention paid to whether polycarbonate in household items (e g. baby bottles) releases bisphenol A into foods, causing health concerns- which it certainly can.

On the practical side, such as preparing dilute aqueous acids or bases in a polycarbonate container at ambient temperature (e.g.for lab use), degradation would be very slow due to the density, hydrophobicity,  and perhaps cross-linking of polymer chains into a tight matrix. This prevents the aqueous acid or base from physically contacting the bulk polymer, although slow surface degradation is still likely.


Why does H+ attack alkene first in this reaction? The reaction is:

Why is the double bond attacked here even though it is in conjugation with the lone pairs of O thereby making it harder to extract. Also,don't the 2 lone pairs on OH have more electron density than the pi bond?

It's better to check the resonance hybrid whenever we have a confusion, so, in this case resonance takes place between lone pair of oxygen atom and the double bond and the electrons shift from oxygen towards the alkene. This creates a partial positive character on the oxygen atom and partial negative character on the alkene carbon which is farthest from the oxygen atom. Now we can easily decide that in between ether oxygen atom and alkene proton will tend to attack alkene carbon and that to on the carbon, which is farthest from the oxygen which results in the positive charge on the carbon next to ether oxygen (this cation is also stable because of octet completion due to resonance of oxygen (may also be called as backbonding)). Now deciding between hydroxy oxygen and alkene is a bit tougher task. Let us assume that proton attacked hydroxy oxygen and it formed oxonium ion, now it can be removed as water and a cation can be formed there. After formation of cation there is nothing to do and it will be waiting for the hydroxy component of water to get attacked by. But if the cation is formed at the alkene carbon it can go for rearrangement which increases the size of 4-membered ring to 5, which increases entropy of the system and also it will lead to formation of $\ce{C=O}$. That will tend the reaction to move forward (like pinacole pinacolone rearrangement). So,it's better to form cation at alkene site rather than forming oxonium ions.

Following is an arrow-pushing diagram of the above description:


Why is aluminium carbonate unstable? Can it be isolated at room temperature?  Or even at any temperature?

Aluminium carbonate reacts easily with water/water vapour to form aluminium hydroxide as you can see on the Wikipedia page.

$$\ce{Al2(CO3)3 + 3 H2O → 2 Al(OH)3 + 3 CO2}$$

So, if you want to store aluminium carbonate, you have to make sure it is free from moisture.

Why it is unstable:

Smaller and more charged metal cations (i.e. metal cations with high charge density) tend to distort the electron cloud of the carbonate ions (polarisation).

Polarisation eventually leads to abstraction of oxygen from the carbonate ion, making one of the $\ce{C-O}$ bond in the carbonate weak, which makes it very prone to attack by water and heat (similar to the decomposition of copper (II) carbonate when compared to alkali metal carbonates).

The reaction itself doesn't release a lot of energy; hence why it is rather unstable and not reactive in chemistry terms.


How to find the mass of dissolved AgCl when placed in a solution of NaCl? I think the following problem in my textbook is malformed:

  Solid $\ce{AgCl}$ is placed in $1\,\mathrm{L}$ of $0.55\,\mathrm{M}$ $\ce{NaCl}$. Find the mass of $\ce{AgCl}$ which dissolves.

I know the method to solve the problem. I think the problem is malformed: When we get an answer, say $x$ grams $\ce{AgCl}$ dissolves, but if we placed less than $x$ grams solid $\ce{AgCl}$ in the solution initially, then it would be impossible for $x$ grams to dissolve.
So we obviously have to make certain assumptions, but I am looking at the solution in my textbook and I don't see in which part of the solution is the assumption used (implicitly). And by the way, how do we solve the problem if we have place less than $x$ grams solid $\ce{AgCl}$?

The question is most likely badly worded and should have been

  What is the maximum mass of $\ce{AgCl}$ that can be dissolved?

As we all know, $\ce{AgCl}$ is damn insouble in aquaeous solutions — and due to the equilibrium constant it doesn’t get better at all if you add chloride ions (reducing the silver ion concentration). So without having done the maths, I’m going to assume that a small spec of nothingness is all that will dissolve. That spec is so tiny that you will have a hard time weighing less so you can actually perform the experiment you suggested.

Because it’s just a flick of dust, and because the authors likely

knew that, too
assumend you knew that, too

they may have gone with a less obvious wording. I mean it’s clear: You can’t dissolve what you didn’t put in, and to-be-dissolved silver ions aren’t going to magically appear with Alakazam.


Products which adhere to human nail while still being removable/reusable? I am at loss as to what I can use for my project. I require a product which can adhere to the human nail (not any damaging glues) while still being easily removable. There will be a degree of 'overlay' on the nail so it's going to need to cover the entire nail plus a bit more (maybe seen with fake nails?).

I've looked online and video tutorials on adding fake nails isn't convenient (while it could work). My biggest problem is I don't want to damage the nail so I can't use any popular practices often seen in beauty salons.

Try flour and water paste or washable school pva glue. Start with the flour. It would be an interesting experiment to try a 50/50 mix of both.
Another possibility is rubber cement (art gum). This will be quicker drying.


When is it permissible to use the stoichiometry of a reaction to determine rate laws? I have read in a general chemistry textbook that rate laws can only be determined by experiment. This seems intuitive because if a mechanism is not elucidated and tabulated reaction data would be one way of bypassing that info to reach the rate law. However, I have seen instances where the rate law is simply determined by using the reactants and raising them to their respective stoichiometric power...which I thought contradicted the idea of needing an experiment? I highly speculate that this is only acceptable if it a singular reaction that the reaction order can be determined experimentally? For example:

$$\ce{2A -> B_2}$$

Can be surmised as:

$$\text{rate} \  = k[A]^2$$

Is this true? My speculation is based on the notion that if one reactant is splitting to a product and that stoichiometric equivalent is needed for the conversion of reactant to product, then the rate law is determined by stoichiometric equivalent number of molecules hence the justification for the "raise to the power" method. Or is there another underlying factor when the exponents have been raised and no detail is given to rate laws determine by experiment?

Your reasoning is correct. When the reaction you are investigating is an elementary reaction step, you can write down the rate law using the stoichiometry of the reaction. Mathematically, this is done as follows.

Consider a reaction as

$$
\sum_{i} \nu_{i} R_{i} \rightarrow \sum_{j} \nu_{j} P_{j}
$$

Where $R$ and $P$ are reactants and products respectively and $\nu_{i}$ and $\nu_{j}$ are the stoichiometric coefficients.

If this reaction is an elementary reaction step, than the rate law in the forward direction is given by:

$$
r = k \prod_{i} [R_{i}]^{\nu_{i}}
$$

where $[R_{i}]$ is the concentration of reactant $R_{i}$ and $k$ is the rate constant.

For example, if you consider the reaction

$$
A + 2B \rightarrow 3C,
$$
then the rate in the forward direction is

$$
r = k[A][B]^{2}.
$$