Electrical double layer I am seriously confuse with the sign of electric field in the electrical double layer. Say zinc is dipped into the zinc sulphate solution. Then why the electrical double layer has positive sign in the electrode side of EDL and negative sign in the electrolyte side of EDL.

I am confused that since zinc has more tendency to dissolve, then zinc should lose two electrons and retain those 2 electrons in the electrode side of the metal-electrolyte interface. This will give eletrode negative sign. Also a layer positive charge of $\ce{Zn^2+}$ develops parallel to electrode.
Where I am wrong here?

The phrase "retain those 2 electrons" is wrong.
Metal surface has positive potential in solution,
but this doesn't mean that lost electrons are concentrated somewhere.
Electrons don't form layers. You can think of electrons as much more "volatile" things than ions, they may discharge everywhere depending on potential, for example, to reduce Zn ions in solution.
So, surface is charged positively and hence first ionic layer is negative.


Elimination of HCl from isomeric chlorocyclohexadienes Which of these is most easily dehydrohalogenated?

a. 5-Chlorocyclohexa-1,3-diene (1)
b. 3-Chlorocyclohexa-1,4-diene (2)
c. 1-Chlorocyclohexa-1,4-diene (3)
d. Both 1 and 3

I encountered this question a while ago.  The given answer is option (b) (compound 2). But the dehydrohalogenation of compound 1 produces benzene which has high heat of formation. So why isn't that the answer?

That's a fair question,

Assuming that you use a strong base for dehydrohalogenation you will be deprotonating the most acidic $\ce{H}$ resulting in formation of a carbanion which has the highest stability in compound number 2

You can see that the carbanion formed in compound numer 1 will have resonance with only one of the double bond and hence will be less stable then the carbanion in the compound 2 which has resonance from both the double bond.

Mechanism can be as follow :-

Also;


Is this analogy explaining the difference between enantiomers and diastereomers correct? I tried writing an analogy to explain the difference between enantiomers and diastereomers. However, I'm not entirely sure if it's actually a correct representation of the differences. It might also be more confusing than it has to be:

To use an everyday object as an illustration of the difference, imagine two forks with four tines each. If fork 1 has its first and third tine longer than the other ones, and fork 2 has its second and fourth tine longer than the other ones, then the two forks would not match completely if you tried to stack them on top of each other. However, if you held fork 2 up to a mirror, then you would see it now has both its elongated tines in the same place as fork 1. These forks would be enantiomers of each other because they are the same thing, but mirror images of each other
If you changed fork 2 to have its first and second tine longer than the other ones, instead of the second and the fourth, then the two forks would instead be diastereomers. This is because now they both cannot match completely when stacked on top of each other, and holding either one up to a mirror will not make them look any more similar. This is exactly how diastereomers work too, because it only partially mirrors its counterpart. As you can see, fork 2 now has its first elongated tine actually match the first elongated tine of fork 1, while also retaining the effect that holding it up to a mirror will have its second elongated tine match that of fork 1. That is, in the mirror fork 2 will have its third tine elongated, just like fork 1 does when not viewed through a mirror. In other words, fork 2 has its first elongated tine match fork 1’s, but its second elongated tine is the mirror image of fork 1’s. This makes them diastereomers because they are the same thing, but neither structurally identical nor complete mirror images of each other.

So, is there anything off about my analogy, or does it correctly and comprehensibly explain the difference between the two?

Enantiomers are non-superposable mirror images (all the signs of one spatial coordinate are reversed).  Diastereomers are  formed by inverting some but not all stereocenters.

http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
 pp. 11-13.  Excellent overall.

Have some fun with your prof!  An acyclic, aliphatic, unstrained, tetrahedral carbon atom is $\ce{sp^3}$-hybridized.  It is an uncharged, valence-saturated singlet.  It bears four freely rotating, rigorously identical substituents, e.g., $\ce{CR4}$, with identical bond lengths to that central carbon.  Said carbon is a chiral center.  It has no ${S_n}$ improper axes of symmetry.

1) Show how this can be true.
2) Use any system of nomenclature to name left-handed and right-handed stereoisomers.

Acyclic, unstrained, etc., trigonal planar $\ce{sp^2}$-hybridized carbons ${R2>C=C<R2}$ are each identical stereocenters.  Each carbon and the olefin overall have no ${S_n}$ improper axes of symmetry.  The olefin is resolved chiral.  (1) and (2) as above.


Can a tertiary enamine tautomerize to an imine? Which of the following compounds does not show tautomerism?

The correct answer is (a), and I'm curious if this is because to show tautomerism, there can't be any net charge.

For instance, ketones can tautomerize to enols, and more importantly, enamines can tautomerize to imines.  In (a), we would get an imine-like thing, but the nitrogen would be positively charged.  Does this still count as tautomerization?

EDIT: Thank you everyone for your helpful clarifications, but I'm still a bit confused as to why, mechanistically, we need a hydrogen on the nitrogen.

When I Google the mechanism for imine tautomerization, what if find is this:

Where the nitrogen lone pair pushes down and the double bond reaches out to pick up a proton.  There isn't a transfer of hydrogen from the nitrogen to the solvent in the case shown below, yet the molecule still tautomerizes.

Tautomerization is a chemical reaction whereby constitutional isomers interconvert. The interconversion process involves the shift of a hydrogen atom accompanied by the shift of a double bond.  Here are a few examples.

In molecule (a) pictured above, there is no hydrogen attached to a nitrogen (it has been replaced by a methyl group), so there is no hydrogen atom available to shift.

Response to OP's edit:

The picture you posted in your edit is incorrectly labeled.  The charged molecule on the left is an iminium ion, not an imine.  So this is not an example of an imine-eneamine equilibrium.  You are correct that in this equilibrium (the one pictured in your edit) there is no hydrogen transferred to nitrogen, but again, this is not an example of an imine-eneamine tautomerization.  I think this error is the source of your confusion.

Here is a picture of an imine-eneamine tautomerization.

As you can see, a hydrogen is transferred between nitrogen and carbon.  This hydrogen transfer can occur with or without solvent molecules playing a role.  That is to say, the hydrogen can be transferred directly between the carbon and nitrogen atoms, or it can be transferred from one of the atoms to a solvent molecule which can then transfer it back to the other atom involved in the tautomerization. The image source also provides a nice explanation of tautomerization.


Why does the Co³⁺/Co²⁺ couple have such a high reduction potential? Cobalt is changing from $\mathrm{(3d)^6}$ to $\mathrm{(3d)^7}$ electronic configuration. What's so stabilizing about that? Just by reduction potentials this couple is more oxidising than hydrogen peroxide and that just sounds amazing and crazy at the same time.

The electronic configuration has nothing to do with it. The reduction potentials of $\ce{Ni^3+}/\ce{Ni^2+}$, $\ce{Cu^3+}/\ce{Cu^2+}$ and $\ce{Zn^3+}/\ce{Zn^2+}$, if they have been/could be measured, would be even greater.

The reduction potential for $\ce{M^3+}/\ce{M^2+}$ is most dependent upon the third ionisation energy. If $I_3$ is large then it will be difficult to oxidise $\ce{M^2+}$ to $\ce{M^3+}$ (hence a high reduction potential). Likewise if $I_3$ is small then the reduction potential will be smaller.

$$\begin{array}{c|cc} \hline
\text{Metal} & I_3\text{ / eV} & E^\circ(\ce{M^3+}/\ce{M^2+})\text{ / V} \\ \hline
\ce{Sc} & 24.76 & (-2.6) \\
\ce{Ti} & 27.48 & (-1.1) \\
\ce{V}  & 29.31 & -0.26 \\
\ce{Cr} & 30.96 & -0.41 \\
\ce{Mn} & 33.67 & +1.60 \\
\ce{Fe} & 30.65 & +0.77 \\
\ce{Co} & 33.50 & +1.93 \\
\ce{Ni} & 35.16 & (+4.2) \\
\ce{Cu} & 36.94 & (+4.6) \\
\ce{Zn} & 39.72 & (+7.0) \\ \hline
\end{array}$$

Data from Weller et al. Inorganic Chemistry 6ed; Johnson Some Thermodynamic Aspects of Inorganic Chemistry. Numbers in parentheses indicate predicted values. Graph by me in Microsoft Office 365.

So your question is essentially: why is $I_3$ of $\ce{Co}$ so "large"?

Going across the 3d block, effective nuclear charge increases which leads to the general trend of IE3 and $E^\circ(\ce{M^3+}/\ce{M^2+})$ increasing.

Any stabilisation derived from "special" electronic configurations e.g. the half-filled subshell in $\ce{Fe^3+}$, or the ligand-field stabilisation energy of $\ce{Cr^2+}$, merely lead to relatively small deviations from the general trend.

Moral of the story is: look at the big picture first. Those $\mathrm{d^5}, \mathrm{d^{10}}, \cdots$ configurations can be relevant, but they aren’t responsible for the overall trend, and certainly shouldn’t be the first thing you consider when trying to rationalise this set of data.


Why is wavenumber used in IR spectroscopy rather than wavelength? In IR spectroscopy, the $x$-axis is used to represent wavenumber, in $\mathrm{cm^{-1}}$. Why is wavenumber, equal to $1/\lambda$, used in place of wavelength, which is simply $\lambda$?

Sources I’ve already found explain why it was chosen rather than energy of waves, but the conversion from wavelength to wavenumber is never explained. Below are two relations from Wikipedia, which explain how it can be used in equations, but in all of these cases, $\lambda$ seems to be a choice that’s easier to work with.

What explanations are there, if anything other than “historical reasons”, for why $1/\lambda$ is favored over $\lambda$?

A spectroscopic wavenumber $\tilde\nu$ can be converted into energy per photon $E$ via Planck’s relation:

$$E = hc\tilde\nu$$

It can also be converted into wavelength of light via

$$\lambda = \frac{1}{n\tilde\nu}$$

where $n$ is the refractive index of the medium.

Not only in IR spectroscopy. Wavenumber is unit of energy and therefore you can directly deduce the difference of energy between states.

In addition, humans like to think in acceptably small numbers (0.01 - 10,000). Wavenumber allows this for IR and conveniently supplements the eV unit in small energy separations range. Admittedly, the conversion factor of 8,065.73 won't win beauty contest.


Crazy glue vs Nail glue So I have a nail repair kit, consisting of nail glue(ethyl cyanoacrylate), powder and a nail buffer. I'm almost of out glue yet stick have a bunch of powder. One of the crazy glue tiny bottles has almost the same name(ethyl 2-cyanoacrylate). Can I refill the nail glue with the crazy glue?

Short version -
Nail glue - ethyl cyanoacrylate VS ethyl 2-cyanoacrylate - super glue
Are these the same?

they are the same chemical yes - (the 2 just refers to the position of the side branch.)
I Would not expose the glue to air unnecessarily as any moisture will make it harden . I can not say for you using it on your nails for insurance reasons ( but scientifically they are the same )


What is this brown liquid that I got when I boiled glycerol and water solution? I boiled a solution of glycerol and water at about 160 degree Celsius and got a brown liquid much less in volume (boiled off the entire water I think; also not so sure about the accurate temperature).

Before mixing glycerol with water it was a colourless liquid, now it is golden brown! Why would that be? Also, I smelled burnt sugar during the last couple of minutes of boiling. So I'd like to know what that gas was. So, do I have glycerin in my hand or any other byproducts?

If your mixture reached 160 degrees C then you must have boiled off all the water. If there was some water left, it would stay at around 100 degrees C and all (most of) the heat would be used to evaporate water.

The boiling point of glycerol is about 290 degrees C and so you didn't boil that off. I disagree with @Skeptical Chymist about you making the caramel (for the sake of god, don't eat it to test it out). Not all that gets brown is caramel and there were no sugars there I presume. If i had to guess I'd say that at 160 C you've managed to completely obliterate any complicated organic structures and what's left is glycerol, charcoaled bits of any organic impurities that you happened to have in the starting material and possibly some acrolein which is a common product of dehydration of glycerol (you can recognize it by it's smell).

Overall; it's hard to tell what one gets, when they heat the hell out of something.


How to test oxalic acid in potatoes How could you test that there's oxalic acid in potatoes.For example,

  what chemical reacts to oxalic acid and then what will happen if it
  reacts?

If you want to determine the total content of oxalate you would need to extract it from the matrix with hot HCl 1M, for instance, since it will dissolve calcium oxalate crystals. Afterwords you can precipitate oxalate again at pH=4-5 with calcium nitrate, thus forming white crystals of calcium oxalate. You can redissolve them by adding sulfuric acid and identify oxalate by taking advantage of its reducing properties - if you add potassium permanganate it will be reduced from Mn +7 (purple colored) to Mn +2 (brown colour) and oxalate will be oxidized to carbon dioxide. But of course you start with a complex matrix, the easiest thing would be to use chromatography to separate and identify oxalate.


Activity of Materials in Equilibrium constant I learnt that we use activity to define the equilibrium constant . But while calculating the equilibrium constant we plug in their concentration instead of activity .

For a general reversible reaction
$$\ce{aA + bB <=> cC + dD}$$
$$K=\frac{[C]^c [D]^d}{[A]^a [B]^b}$$
We put their concentration instead of activity.
Why is that done ?

While calculating concentrations is trivial (divide the amount of substance by the volume), activity is a complete different story. Activity is usually calculated by multiplying the concentration by a dimensionless number called activity coefficient($\gamma$):
$$a = c \times \gamma$$

The problem here is that $\gamma$ is not only different for every substance, it also varies with temperature, concentration of the substance itself, the diluent used, and so on. If the substance is a salt that dissociates then the calculation gets even more complex.

Since having the values of the activities would be quite a challenge, equilibrium constants are often written using concentrations, which are more readily available. It's not the most correct way of represent it, but it makes your life easier.


Thought experiment for evaluating total energy Atomization Energy ΔHat is the energy required to disintegrate a molecule into isolated atoms that are infinitely away from each other. The question is about atomization energy and its relation to the total energy (E) in Schrodinger equation (Eψ=Hψ).

If we disintegrate a molecule one step further to isolated electrons and nuclei, will the overall energy spent be equal to negative of the total energy of Schrodinger equation? I am assuming an energy reference will come into play as well as thermal energy. So let's say we are conducting the thought experiment of disintegration at 0 K (zero Kelvin) and we correct the values by assuming a reference energy point. Then, is the total Energy E equal to disintegration energy of the molecule into isolated electrons and nuclei?

If we disintegrate a molecule one step further to isolated electrons and nuclei, will the overall energy spent be equal to negative of total energy of schrodinger equation?

By convention, yes.

Energy as a absolute quantity is not well-defined, energy must be defined with respect to an energy zero.  By convention, when we perform quantum chemical calculations our energy zero is free electrons and nuclei infinitely far apart, which we define to have zero potential energy as they cannot interact.

Also, quantum mechanically, we consider only the case at $0$ K.  We can however add on corrections to the free energy due to entropy and internal motion at finite temperatures, as well as vibration zero point energies.

How is this different from the atomisation energy?

We can see that to get from the products of atomisation to the products of "disintegration" (as you refered to it), is to remove all the electrons from the neutral atoms.  That energy is given by the sums of all the ionisation energies for each atom in the system, removing each electron one by one.  This allows us to construct a Hess cycle.

Finite temperature corrections are added to the quantum mechanical energies as the potential energy and free energy are different at finite temperature.


Mechanism involved in bromine water test $$\ce{CH3CH=CH2 ->[Br2] CH3-CHBr-CH2Br}$$

In the reaction mechanism, why does the nucleophile attack at the 2nd carbon atom? Shouldn't it attack at the 1st one, so that in the transition state, the carbanion is relatively more stable?

There are no carbanion in this reaction.
In bromine water, $\ce{Br2}$ is partially made of $\ce{Br+}$ and $\ce{Br-}$ ions. Then $\ce{Br+}$ attacks the double bond, producing a carbocation, with $\ce{Br}$ at the outer $\ce{C}$ atom, and the + charge on the central $\ce{C}$ atom. This is the most stable carbocation. Then it will react with the $\ce{Br-}$ ion which will be attached on the central $\ce{C}$ atom.

It also should be mentioned that a secondary reaction may happen where $\ce{H2O}$ reacts with the transition state carbocation. As a result, the molecule $\ce{CH3-CHOH-CH2Br}$ is produced.


Thymol Blue - molar absorptivity I did a lab about spectrophotometry and right now I need the molar absorptivity(ε) of thymol blue in acidic solution and also in basic solution. I have tried looking online but I haven't found the data I need.

What is the molar absorptivity of thymol blue in acidic solution and in basic solution?
 information
The specific wavelength is 540.27 nm for the acid and 601.88 nm  for the base.

First of all, molar absorptivity is dependent on wavelength, so you either need to specify a wavelength or you need the entire spectrum.

When searching for absorptivity, it's often effective to search for the older name 'extinction coefficient'. Using that search term in Google, I found an absorption spectrum for thymol blue here (pdf).


Will supercritical nitrogen stay supercritical if we raise temperature and pressure high above critical values? Nitrogen's critical temperature is -146.96 °C and critical pressure 33.96 bar. If we maintain nitrogen at 220 bar and 400 °C, what will be the state of $\ce{N2}$? Will it be still supercritical? I know that phase diagram says after critical temperature and pressure we have a supercritical phase, but with enough p and T, more exotic states of matter occur.

If we maintain nitrogen at 220 bar and 400 °C, what will be the state of $\ce{N2}$? Will it be still supercritical?

Yes, it will still be supercritical.  The only reasonable definition of "supercritical" is something along the lines of "a fluid phase without surface tension wherein the fluid pressure and temperature exceed the critical point".

Thus, compressed air is a supercritical fluid.  The same is true of compressed nitrogen.

You may be wondering how "supercritical" air is different from regular, non-supercritical, gaseous air.  The answer is, it isn't meaningfully different.

If you research "supercritical fluids", it will be very hard to discover this fact.  That is because the "exotic" properties of "supercritical fluids", such as having densities "in between that of a gas and a liquid", are really only properties of supercritical fluids that are near (i.e. slightly above) the critical point.

It isn't really much of an exaggeration to say that supercritical fluids are really gases.  They have no surface tension.  They expand to fill their container.  Just like gases.  Near the critical point, supercritical gases can have densities that are very unusual for a gas. At these high densities, the gas is very non-ideal, and intermolecular interactions in the gas lead to all sorts of interesting behavior.  Those high-density intermolecular interactions give rise to the interesting properties that people refer to when they say "supercritical fluid".  The strength and nature of those interesting intermolecular interactions depends strongly on density, which is why people refer to "supercritical fluids" having "tunable" properties.  Small changes in density (by changing temp. or pressure) can give rise to large changes in those interactions.

At conditions way past the critical point, those intermolecular interactions don't matter very much any more, and the "supercritical fluid" again starts to behave more and more like an ideal gas.

  I know that phase diagram says after critical temperature and pressure we have a supercritical phase, but with enough p and T, more exotic states of matter occur.

Yes, other answers and comments have done a good job of discussing this point.  If you increase the pressure enough, eventually you'll have a black hole!  Of course, long before that happens, you can get other less exotic phases of matter such as various solid phases.  In general, predicting the high-pressure phase behavior of even small molecules like nitrogen is very difficult, and as such, experimental data is required.  Unfortunately, measuring the high-pressure phase behavior of materials is also very difficult, at least past a few hundred gigapascals.  So I'm not sure where the high-pressure "exotic" phases of e.g. solid nitrogen start to form.  But it's not at 220 bar and 400 ° C, under those conditions nitrogen will just be a gas, albeit a "supercritical" gas.


Deuterated solvents vs. regular solvents Working on dissolving aspartame and i was wondering if it dissolves well in methanol will this mean that it dissolves well in deuterated methanol ?

In general, it is safe to assume that solubilities in non-deuterated and deuterated solvents are much the same, and this would be the logical solvent of choice if looking for a NMR solvent.

However, there is a small difference in relative solubilities between H/D solvents, and you can gauge potential for differences based on the dielectric properties of the two. By memory, I think H2O/D2O have the largest difference (which is still pretty small).

Another factor that can cause differences in solubilities comes from the relative purity of the two solvents. For instance, many prepackaged deuterated solvents are often drier than bulk bench solvents, and this can sometimes cause significant solubility differences, especially in hygroscopic solvents like DMSO. I've seen countless examples of people claiming a compound was readily soluble in DMSO, only to have it sit like a brick when trying to dissolve in a freshly cracked vial of DMSO-d6.


How fast do electrons move around the nucleus? Electrons, as we all know, are incredibly small. Smaller things do tend to move faster, right? So exactly how fast considering how small they are? Also, does the electonegativity between two atoms alter the speed of the electron?

The ratio of the speed of an electron traveling in the first Bohr orbit to the speed of light is given by the handy equation

$$\mathrm{V_{rel}=\frac{[Z]}{[137]}}$$

where Z is the atomic number of the element under consideration and 137 is the speed of light in atomic units, also known as the fine structure constant.  Consequently a 1s electron in the hydrogen atom will travel at approximately 0.7% the speed of light. In silver (Z=47) the 1s electron will travel around 34% the speed of light, while the 1s electron in gold (Z=79) will travel at about 58% the speed of light.

Once we get up to around silver, the electrons are traveling at relativistic speeds and this can dramatically impact properties of the atom. For example, the relativistic mass of an electron is given by

$$\mathrm{m_{rel}=\frac{m_{e}}{\sqrt{1-(V_{rel}/c)^2}}}$$

where $\ce{m_{e},~ V_{rel}~ and~ c}$ are the electron rest mass, the velocity of the electron, and the speed of light respectively.  The following figure provides a graphical representation of how the electron mass increases as the electron velocity increases.

The following equation relates the ratio of the relativistic radius of the first Bohr orbit $\ce{R_{rel}}$ to the normal radius $\ce{R_{o}}$, to the relativistic velocity of the electron

$$\mathrm{\frac{[R_{rel}]}{[R_{o}]}=\sqrt{1-(V_{rel}/c)^2}}$$

As the relativistic velocity of the electron increases, the orbital radius contracts (the above ratio becomes smaller).  For silver, the first Bohr radius contracts ~6%, while for gold the contraction is ~ 18%.

Look at these earlier Chem SE answers to see the interesting physical effects that atoms can exhibit when their electrons travel at relativistic speeds.

francium
gold
mercury


Comparing van der Waals constant for gases How can I compare the van der Waals constant $a$ (the liquefication constant) for different compounds based on my knowledge of bonding in the compound? Let's pick the following sample set:
$$
\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Compound} & &\hspace{-1em}a \\
\hline
\text{Helium} & \d{0}{0346} \\
\text{Argon} & \d{1}{355} \\
\text{Xenon} & \d{4}{25} \\
\text{Water} & \d{5}{526} \\
\text{Ethanol} & \d{12}{18} \\
\text{Acetone} & \d{16}{02} \\
\text{Acetic acid} & \d{17}{71} \\
\text{Benzene} & \d{18}{24} \\
\text{Toluene} & \d{24}{38} \\
\hline
\end{array}
$$
A general trend that I can see is $a$ is proportional to molecular mass. Am I wrong to compare on the basis of intermolecular forces like hydrogen bonding, which is clearly very good in the water but the $a$ value isn't so promising?
Is it my misconception that $a$ depends on intermolecular forces? As far as I know, the stronger the forces, the more easily it will be liquefied, and greater should be $a.$

The reason that in your sample set water seems to have a low value of $a$ despite having hydrogen bonding is because most of the molecules that you chose were of much higher molecular mass. You should compare water with some gas of approximately equal molecular mass, then you'll see that value of $a$ in water is quite high. Your other observations are generally correct.


For bond dissociation energies (BDEs) of double and triple bonds, is number shown in tables for that second or third bond? Or all of them? Simple, 'dumb' question....   Is the kJ/mol number given in tables of bond strengths or bond dissociation energies , when it comes to double and triple bonds, just for the 'last' bond or for both (or all three) together?

They're for the entire bond (so for a triple bond, all 3).

From Wikipedia, it can be defined as

  The standard enthalpy change when [a bond, be it single, double, or triple] is cleaved by homolysis to give fragments A and B, which are usually radical species

So you get two fragments as your final state.


Electron Transfer in Organic Reactions: Why Does Brown Iron Oxide Improve Burn Rate of Rocket Candy? On wikipedia, this statement exists about rocket propellant aditives:

  Metal oxides have been found to increase the burn rate of sugar propellants. [...] Most often used are iron oxides. Red iron oxide is used most often as it is somewhat easier to obtain than the yellow, brown, or black versions. Brown iron oxide exhibits unusual burn rate acceleration properties under pressure.

I have no idea how would already oxidized compound increase burn rate. I have failed to research which compound is brown iron oxide. All I could find was red iron oxide, iron monoxide and magnetite.

So how do oxides help combustion generally? And what makes brown iron oxide so excellent for that purpose?

I would appreciate help with adding proper tags to this question. Couldn't think of any.

First of all, you're thinking of one of the two definitions of oxidation—breaking bonds in molecular oxygen and rejoining them in a different way in another molecule.

We need the electron transfer definition of oxidation however.

First off, to see what is happening we need to look at how many electrons are being transferred. In glucose the oxidation state of carbon is $\ce{0}$, whilst in carbon dioxide it is $\ce{+4}$. So during our rocket burning process, each carbon gives up four electrons. But what is it giving them to?

If we use molecular oxygen as our accepting species—our oxidisation agent—we have the following:

$$ \ce {6O_2} + \ce {C6H12O6}  \ce{->} \ce{6CO2} + \ce{6H2O} $$

With oxygen here going from an oxidation state of $\ce{0}$ in $\ce{O_2}$ to $\ce{-2}$ in both $\ce{H2O}$ and $\ce{CO2}$.

So if each $\ce{O2}$ molecule  is picking off two electrons and this is all it's doing, then why can't we replace this with two lots of $\ce{Fe^3+}$ to form $\ce{Fe^2+}$?

Before I write this in equation form, it's worth noting that organic electrochemical equations approximate organic matter (OM) with formaldehyde, $\ce{CH2O}$ since OM is biologically equivalent to a polymer (albeit complex ones!) of $\ce{CH2O}$. Note that glucose is therefore equivalent to six lots of this monomer. This simplifies writing balanced chemical equations a lot:

$$ \ce{2Fe_2O_3} + \ce {CH2O}  \ce{->} 4\ce{FeO} + \ce{H2O} + \ce{CO2} $$

Now molecular oxygen is by far a better oxidisation agent than $\ce{Fe_2O_3}$ however in fine powdered form, these iron oxide molecules are in immediate contact with your OM; we do not need to wait for molecular oxygen to be used up and for more to diffuse in.

So I'm warranting that red iron oxide ($\ce{Fe_2O_3}$) can be used as well as brown iron oxide (I think this means $\ce{FeO}$). The difference here is that in the former reaction each $\ce{Fe^3+}$ only captures one electron from your OM to form $\ce{Fe^2+}$, whilst in the latter case you're moving from $\ce{Fe^2+}$ to $\ce{Fe}$ metal which captures two electrons—so you need less. Note that formation of iron from $\ce{Fe^2+}$ will need really high temperatures or pressures, which is what you note in your question.

This is a super question in terms of getting people thinking about redox not just as a process which happens with ions, but with entire organic species. Such organic reactions can be characterised by a reaction voltage, just like the other types of electrochemical reaction you've been looking at. This is actually what happens in the field of biogeochemistry, where scientists look at which types of oxidisation agent are responsible for degrading OM in different environments. More can be found about this in this paper here.


Using a modified Arrhenius plot to find activation energy I have a question on how to determine activation energy from a plot using the Arrhenius equation. I am working as an assistant with general chemistry II students and their lab is the iodination of acetone. I am familiar with using a plot of $\ln(k)$ vs $1/T$ to get the slope of $E_\mathrm{a}/R$ to solve for $E_\mathrm{a}$. Their lab manual though tells then to plot $\ln(t)$ vs $1/T$ to determine $E_\mathrm{a}$. I was wondering if anyone could shed some light on the derivation of this equation.

The use of $t$ only makes sense if $t$ is defined as the time it takes for the reaction to progress a certain relative amount (and if we are talking about first-order reactions).  For example if $t$ is the time it takes for half (or 90% or 99% or whatever, as long as it is specified) the reactant to be consumed, then the idea makes sense.  In that case it would be more helpful and clear to say $t_{1/2}$ (or $t_{90\%}$ or whatever) instead of just $t$.

Assuming $t_{1/2}$ is what is meant by $t$:

Arrhenius equation:

$k = A_0 e^{\frac{-E_a}{RT}}$

$k = A_0 e^{\frac{-E_a}{R}\frac{1}{T}}$

$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T}$

First-order kinetics:

$C = C_0 e^{-k t}$

$\ln \frac{C_0/2}{C_0} = - k t_{1/2}$

$k = \frac{1}{t_{1/2}} \ln 2$

$\ln k = - \ln t_{1/2} + \ln \ln 2 = -\ln{t_{1/2}} + \ln \ln 2$

Combining:

$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T} = -\ln{t_{1/2}} + \ln \ln 2$

$\ln t_{1/2} - \ln \ln 2 = \frac{E_a}{R} T^{-1} - \ln A_0$

$\ln t_{1/2} = \frac{E_a}{R} T^{-1} - \ln A_0 + \ln \ln 2$

So if $t$ means $t_{1/2}$, then a plot of $\ln t_{1/2}$ vs. inverse temperature has a slope of $+\frac{E_a}{R}$, instead of $-\frac{E_a}{R}$ if you plot $\ln k$ vs. inverse temperature.

N.B. If $t$ just means time, then the idea doesn't make any sense, because then instead of $\ln \ln 2$ you would have $\ln \ln \frac{C(t)}{C_0}$, which is a term that depends on time, so instead of being an explicit equation that is clearly a line, you would have an implicit equation, non-linear in $t$, that would need to be solved before the plot was made, and the result would not be linear and would be impossible to use to determine $E_a$.


Is there any difference between Meprobamate and Equanil? My textbook mentions Meprobamate and Equanil separately, under Anxiolytic drugs and even provides two (not-so) different structures to back their claim.

However, Wikipedia would beg to differ:

  Meprobamate — marketed as Miltown by Wallace Laboratories and Equanil by Wyeth...

Which would naturally imply, that Meprobamate and Equanil are...the same thing. ಠ_ಠ

A quick search with Google Images got me this:

Now this particular textbook I use has, over the years, gained notoriety among several generations of us students as the Mater Omnium Errorum. So in all likelihood, my textbook's the one that's wrong here.

But since I'm not a Pharmacologist, I'd appreciate it if someone could verify if Meprobamate and Equanil are the exact, same, chemical.

I tried Google, but that endeavor wasn't particularly helpful.

Pharmaceuticals with meprobamate as the active compound are marketed by different companies under different names.

Meprobamate is the International Nonproprietary Name (INN) for the drug.

Whether two pharmeceuticals with the same active components acutally have the same kinetics does depend on additional factors, such as the type of the filler.


Explain the nomenclature of 7-amino-4-methylcoumarin Can someone explain the nomenclature of 7-amino-4-methylcoumarin, pictured below (from the Sigma-Aldrich website)?

The alpha-carbons of the methyl- and amino-groups here are separated by a minimum of three C's, whereas the numbers 4 and 7 are separated by two digits. What am I missing?

This is one of the tricky parts in the nomenclature of fused cyclic systems. Excerpt from page 180 of the 1998 IUPAC Nomenclature of fused and bridged fused ring systems
(I have added bold to the piece of importance to answer your question):

  The numbering of the peripheral atoms in a particular orientation starts from the uppermost ring. If there is more than one uppermost ring the ring furthest to the right is chosen. Numbering starts from the non-fusion atom most counterclockwise in the ring selected, and numbering proceeds in a clockwise direction around the system, including fusion heteroatoms. Fusion carbon atoms are not included; each fusion carbon atom is given the same number as the immediately preceding position modified by a roman letter a, b, c, etc. If
  heteroatoms are identified by replacement nomenclature all atoms retain the numbers of the parent
  hydrocarbon.

If one considers coumarin, this is not the IUPAC nomenclature but the numbering rule remains the same. Below is an example of the numbering scheme for coumarin which summarizes why the substituents are numbered by this manner on coumarin.


What reaction happens when carboxylic acids are treated with calcium hydroxide and distilled? What reaction takes place when $\ce{Ca(OH)2}$ reacts with $\ce{R-COOH}$? What is obtained after distillation?

Acetic acid (as an example of a typical carboxylic acid) reacts with calcium hydroxide to form calcium acetate. This is a simple acid-base reaction which forms a salt and water:

$$\ce{2CH3COOH + Ca(OH)2 -> (CH3COO)2Ca + 2H2O}$$

Upon heating, calcium acetate undergoes decomposition to acetone and calcium carbonate. The acetone can be distilled off from the mixture, leaving behind solid calcium carbonate:

$$\ce{(CH3COO)2Ca ->[\Delta] CaCO3 + 2(CH3)2C=O}$$

At even higher temperatures calcium carbonate itself can be decomposed to calcium oxide and carbon dioxide:

$$\ce{CaCO3 ->[\Delta] CaO + CO2}$$

References

https://www.tut.fi/ms/muo/vert/3_organic_chemistry/carboxylic_reactions.htm
http://www.docbrown.info/page04/OilProducts10.htm


Detecting trash (plastic) burning smoke with sensors I’m not a chemist, but a software engineer. I’m trying to build a device that could detect if someone is burning trash (mostly plastics and similar materials) vs wood or diesel. At my disposal I have the following sensors: https://www.mysensors.org/build/gas

What sensor or combination of them would be most reliable to detect smoke resulting from burning illegal waste?

The typical gas sensor marketed to hobbyists and enthusiasts (except for $\ce{CO2}$) are of the metal oxide semiconductor type, not to be confused with MOS as in MOSFET.
How do these work?
In these detectors, a semiconductor material made of the oxide of a metal, usually tin.
I'll quote from this question:

...sintered composite based on the semiconductor material $\ce{SnO2}$, which is the metal oxide. The different sensors have different admixtures and operating conditions to achieve different sensitivities to different gasses.
The principle of operation of MOS-type gas sensors is explained in Figaro sensors' Operating principle; MOS type.

The Figaro link shows the operating principle in the GIF below:
 GIF (Source: Figaro)
When the $\ce{SnO2}$ is heated by the (usually 5V but not always) heater supply, oxygen in the air will dissociate at the surface and oxygen ions will stick to the surface, depleting the semiconductor and cutting off the flow of current. When a carbon-containing gas (e.g. $\ce{CO}$ or $\ce{C_xH_y}$ (but not $\ce{CO2}$)) comes along, it will combine with the adsorbed oxygen and remove it from the surface. The more gas, the lower the equilibrium concentration of adsorbed oxygen and the higher resulting current flow through the semiconductor.
Not as gas-specific as you might think
By adjusting the amount and types of additives to the sintered $\ce{SnO2}$ and adding catalyzing metal clusters to its surface, the manufacturer can enhance sensitivity to some gases and reduce it to others, but there is no single-gas specificity with these sensors.
Since the carbon in $\ce{CO2}$ is already fully oxidized, it won't react with the adsorbed oxygen. Those sensors (e.g. MG-811) use a chemical cell sensor based on a solid electrolyte. They need very high impedance amplifiers that measure millivolts, a bit like pH sensors, in fact, a lot like pO2 sensors!
CO and VOCs
I would recommend you use a $\ce{CO}$ sensor (e.g. MQ-7, MQ-9, MQ-307A or MQ-309A) since burning garbage will likely be incomplete (not in an industrial incinerator) and a "Air Quality" VOC sensor like MQ-135.
You can experiment with others, some people like to build an "electronic nose" which is an array of a bunch of these, then they plot the data for different kinds of smoke and environments to see what happens, and try to use the pattern to identify the likely problem.
PM2.5
Inefficient combustion can also produce particles as soot, and there are more expensive "dust sensors" or "PM2.5 sensors" which measure small particles (2.5 microns or larger) in the air. Soot is a hallmark of incomplete combustion (think of the smoke from a badly tuned engine) and a huge problem from incomplete or low-temperature combustion from a garbage fire.
None of these will be conclusive, you'll have to look at your data and look at when there is a fire, and learn how the signals correlate to the events.
Ultraviolet light
There are UV light based fire detectors as well, but you'd need direct line-of sight access to the flames. If the sensor is designed to be solar-blind by adding a filter that removes all wavelengths from the Sun that make it to the ground, then it might be helpful here. However, they are usually designed to work indoors where indoor lighting may have less UV than outdoor sunlight. This would take some more investigation.
read the data sheets carefully
Finally, read each sensor's data sheet carefully and note carefully the specific voltages that each sensor needs. The "Air quality" sensors need you to run for a certain amount of time (about a minute) at one voltage, then a different amount of time at another voltage because the surface chemistry is a bit more complex.
Sensors can also be found here:

http://wiki.seeed.cc/search.html?q=gas+sensor
https://www.pololu.com/category/83/gas-sensors


Is Fe(OH)2 soluble in alkalis? I have just started inorganic chemistry in school and I have just learnt about metal aqua ions and how they react in water. I learnt that $\ce{[Fe(H2O)6]^2+}$ reacts in water by losing a $\ce{H+}$ acting as a Bronsted-Lowry acid producing $\ce{H+}$ ions. Due to this reaction we can test for metals such as $\ce{Fe^2+}$ by reacting it with $\ce{NaOH}$. This is because the $\ce{OH-}$ ions react with the $\ce{H+}$ ions produced pushing the equilibria to the right forming a insoluble precipitate. My questions is, why does the reaction stop there? The reaction stops once $\ce{[Fe(OH)2(H2O)4]}$ forms which is the insoluble precipitate but why doesn't it continue to form $\ce{[Fe(H2O)3(OH)3]-}$ and become soluble again?

This is a pretty interesting question. First of all, you are right, if you ever happened to see a yellow $\ce{Fe^3+}$ solution that was no pure aqua-complex but more likely $\ce{[Fe(H2O)5(OH)]^2-}$. The real $\ce{[Fe(H2O)6]^3+}$ for example, like it is formed in conc. $\ce{HNO3}$ or $\ce{HClO4}$ is actually somewhat purple. We call these compounds cationic acids. So the aqua-complexes will act as acids and become deprotonated in water. For your $\ce{Fe^2+}$ example it's pretty much the same. I often read in student reports that, if for example a hexaqua complex is treated with a base the $\ce{OH^-}$ would exchange for one water ligand. Many then concluded that $\ce{OH^-}$ had to be the stronger ligand in the spectrochemical series. This is, of course, wrong and not related to each other here. What is far more logical, at least to me is, that by coordination the water gets closer towards the highly charged metal center. But that means there is a repulsion from the $\ce{H}$ in $\ce{H2O}$ and the metal center since both have a positive (formal) charge. This makes it much easier to become deprotonated by a base.

But why does this stop at some points? I asked myself the same question for the example of $\ce{BeO}$ reacting with $\ce{OH^-}$ to form $\ce{[Be(OH)4]^2-}$. Why is Beryllium oxide amphoteric forming a berrylate while $\ce{MgO}$ to $\ce{BaO}$ do not show this ability? It took me quite a while to find an answer to this. There is an equation by  Cartledge that tries to describe this phenomenon:

$$\ce{Φ = z/r}$$

Where Φ is the ionic potential, z the charge and r the radius. If $\ce{Φ^{1/2}}$ is smaller than 2.2 the ion is considered basic, above 3.2 it's considered to be acidic. In the middle, it's considered amphoteric. What does that mean? It seems like there is a ratio of radius to charge. Small, highly charged ions seem to polarize the surrounding water molecules so much that they quickly lose a proton.

In your case, we are not specifically talking about oxides but I guess that something similar should be happening. The high charge on the metal and the fact that $\ce{Fe}$ is already in the middle of the d-row where atomic radii decrease along the row should be enough to actually cause even more deprotonation, this time perhaps not of surrounding water molecules like in the case of $\ce{BeO}$ but the ones that are already ligated to the metal center. At some point it will require additional help which happens when the pH is increased. And in fact for $\ce{Fe^2+}$ a  $\ce{Na4[Fe(OH)6]}$ seems to exist.

I guess the comparison with the $\ce{BeO}$ may perhaps be a bit far fetched but it would be an actual example where you can see how the charge to radius ratio acts on the way soluble hydroxo-complexes are formed.


Troubleshooting electrochemical batteries For my science classes, I have built multiple types of electrochemical batteries utilizing oxidation-reduction reactions, and I have demonstrated that they each produce potential difference across the different solid metals used using a voltmeter.  Unfortunately, I have been unable to harness any of this to light up a simple lightbulb.

For my first demonstration, I used a copper and zinc strip in vinegar solution.  There was a voltage difference across them of about 0.9V.  This was lower than the 1.1V I was expecting, so I hooked up two of these in series to produce 1.8V.  I attempted to light a simple light bulb (which works perfectly with a 1.5V D battery), but there was nothing.  No current flowed.

For my second demonstration, I made a voltaic pile with cardboard in vinegar, zinc discs, and copper discs.  Again, I had a great voltage prepared (about 3V or so), but I could get no current.

For my latest demonstration, I tried a 1.0M CuSO4 solution with a Cu strip in it connected to a 1.0M ZnSO4 solution with a Zn strip in it via a salt bridge of 0.2M KNO3.  The voltage difference was about 1.1V spot-on, but again, I could get no current to flow through the light bulb.

What could I be doing wrong?  Is there something fundamental I am forgetting here, or is there something perhaps wrong with the items I am using?  My original copper and zinc strips were about 1mm thick, if even that, so I replaced them with thicker strips in later demonstrations, but that did not fix anything.

Your experimental procedures seem to be perfect. There is just one minor error in your conclusion: you said that no current flowed. I disagree - current flowed, just not as much as you hoped. Harnessing one of these reactions to light up even the teeniest lightbulb would be similar to harnessing a bumblebee to a model airplane.

Getting a visible change upon connecting the battery is a reasonable goal. In one sense, you have already done it, by observing the needle move in the voltmeter. I know, you would like something more. A mechanical device might work well: say, a compass needle next to a coil of wire. When you connect the battery, some current will flow, some magnetic field will be generated, and the needle will move.

Another possibility: if you take a small battery operated clock (either mechanical/quartz or digital) and remove the battery, you should be able to operate it with one of your batteries (you may have to boost the voltage). These devices use very small amounts of current, in the range that your batteries can provide.


Adding concentrations to obtain mass I have a question that gives two concentrations and asks for the mass of HCl formed by the reaction.

I have two concentrations:

250mL of 4.00M $\ce{H2SO4}$, and

250mL of 1.00M $\ce{NaCl}$.

Here is the balanced reaction equation:

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

I know how to find the mass once I find the moles, molecular weights and then grams by multiplying the two.

However, how do I add those two concentrations? I assume the mixture will be at least 500mL, but how do I add the molarity?

Thanks!

Your working equation is correct.
$$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$

Find the amount of substance of protons of each solution via $n=c\cdot V$.

   $n(\ce{H+})=2~\mathrm{mol}$, $n(\ce{Cl-})=0.25~\mathrm{mol}$

What is the limiting agent?

   Chlorine

How many moles of hydrogen chloride can only be formed?

   $n(\ce{HCl})=0.25~\mathrm{mol}$

Calculate the mass of hydrogen chloride via $m = n\cdot M$

   $M(\ce{HCl})=36.5~\mathrm{g/mol}$, $m(\ce{HCl})=9.1~\mathrm{g}$


Calculating exposure time to inactivate coronavirus by UV light Sorry for a layman question. I'm having difficulties understanding some basic math apparently.

According to a paper, for a coronavirus, the UVGI dose for 90% inactivation is 339–423 μW sec/cm2.

I have a UV lamp with the following specifications:

it's 15 watts
it takes 1h30 min to 90% inactivate coronavirus in a 20 square meters room with a 3-meter ceiling.

But that doesn't make any sense if you believe the paper's number.

This is how I calculate the necessary exposure time:

the surface of the floor, ceiling, and walls in the specified room is 94 square meters (or 940,000 square centimeters.)
15 watts is 15,000,000 microwatts
from the 339–423 μW sec/cm2 in the paper, take the highest number

So since our surface is 940,000 cm2, we multiply 423 * 940,000 and that gives us 397,620,000 μW sec/surface of the room.
Since our lamp is 15,000,000 microwatts, we divide 397 million by 15 million, and we get 26 (watts provided by our lamp) * sec/surface of the room.

So 26 seconds is the necessary time exposure, but it should be 5400 secs as specified in the manual. Is there an error somewhere?

yes, there are several errors in your calculation.
Germicidal lamps specify an effective power output in UVC. This lamp
https://www.osram.de/ecat/PURITEC%20HNS%20UV-C-UV-C%20lamps%20for%20purification-Ultraviolettstrahler-Industrie-Spezialbeleuchtung/de/de/GPS01_1028570/ZMP_1022255/
consumes 16W of electricity but has a UVC output of 4W. You need to calculate with the 4W value.
Then you need to calculate the intensity of your irradiation on the surface you want to inactivate the virus on. The distance from the lamp to the surface is key, as the 4W UVC power remains contant over the surface of the light cone that shines from it. In the lamp's data sheet it says that this is valid for up to 3m from the lamp. (after 3m it most likely decreases due to other effects.) So you need to calcuate the distance from your lamp, and you need to calculate your powerdistribution over the surface of the sphere/cylinder (depending on your geometry), with the lamp in the center. It does not do to calculate the surface of the room as a whole. Given the irradiation intensity on the surface of that sphere you can now calculate the time needed until the UVC output of the germicidal lamp manages to reach the 90% inactivation dose for a given virus. For corona viruses different values for that dose are given in different papers. Some say 10.3mJ/cm^2 for 99%, some quote 0.6mJ/cm^2 for 90% inactivation. This one gives a good overview: https://www.nature.com/articles/s41598-020-67211-2.pdf
I recommend to choose a simple geometry for calclation of the irradiation power. Living rooms are quite tricky. Spheres are better.


Is selenium disulfide polar or non-polar? So I saw in some internet site that $\ce{SeS2}$ is a polar molecule. When I drew the Lewis structure of the molecule, it showed up as a linear molecule like this:
$$\ce{S=Se=S}$$
The electronegativity difference of $\ce{Se-S}$ is just $0.01$ and the linear form shows that vectors cancel each other. Yet, some websites state that this molecule is polar, when everything to me indicates is a non-polar molecule.
Are they wrong or am I missing something here?

I understand your assumption of selenium disulfide being a simple molecular like that. It is similar to the mistake of silicon dioxide, which at first glance we might assign it to be a linear molecule like carbon dioxide. However the crystal structure of silicon dioxide is far more complex.
In selenium disulfide, the selenium and sulfur atoms form a 8-membered ring, similar to the 8-member rings you would find in certain allotropes of either elemental selenium or elemental sulfur. The ratio of $\ce{Se:S}$ atoms is not exactly $\mathrm{1:2}$ as the chemical formula implies (that would give $\mathrm{\frac{8}{3}}$ atoms of $\ce{Se}$ and $\mathrm{\frac{16}{3}}$ atoms of $\ce{S}$ per ring). Indeed the ratio is closer to $\mathrm{3:5}$ ($\mathrm{3:6}$ is $\mathrm{1:2}$ anyways). A better representation of the compound is $\ce{Se_{n}S_{8-n}}$.


Why is chromate stable in basic medium and dichromate stable in acidic medium? When I was studying the equilibrium existing between chromate and dichromate ions in the solution, I came to know that

  Chromate ion is more stable in basic medium while dichromate ion is stable in acidic medium.

However, I am not able to understand what is the governing factor to decide which species will exist to a greater extent and how the equilibrium shifts the direction on changing the nature of the medium.

Why is chromate stable under basic conditions and dichromate stable under acidic conditions?

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements
The preface of the book says:

The approach: These reactions represent important facets of the elements involved but are typically presented as no more than facts. (Why does boiling white phosphorus in alkali lead to hypophosphite and not phosphate?—Current texts make no attempt to address such questions.) Arrow pushing demystifies them and places them on a larger logical scaffolding. The transformative impact of this approach cannot be overstated. Almost to a person, students who have gone through our introductory course say that they cannot imagine how someone today could remain satisfied with a purely descriptive, nonmechanistic exposition of inorganic main-group chemistry.

Potential concerns: Given the plethora of advantages of a mechanistic approach, it’s worth reflecting why it has never been adopted for introductory inorganic chemistry. A plausible reason is that, in contrast to common organic functional groups, simple p-block compounds such as hydrides, oxides, halides, and so forth, tend to be much more reactive and their vigorous and even violent reactions have been much less thoroughly studied. As good scientists, inorganic chemists may have felt a certain inhibition about emphasizing an approach that has little grounding in experimental fact. This is a legitimate objection, but hardly a dealbreaker, in our opinion, for the following reasons.

Our ideas on main-group element reactivity are not taken out of the blue but are based on parallels with well-studied processes in organic and organoelement chemistry.
Second, it no longer necessarily takes a prohibitive amount of resources to test amechanistic proposal, at least in a preliminary way. Quantum chemical calculations, particularly based on density functional theory (DFT), very often provide an efficient and economical way of evaluating reaction mechanisms.
Third, and perhaps most important, it’s vastly better to be able to formulate a hypothesis on how a reaction might happen than to have no inkling whatsoever about the mechanism

What all of this basically means is that in this approach, you use the classical organic chemistry concepts of electrophiles and nucleophiles to write an arrow-pushing mechanism for the reaction. It will only be a probable mechanism, but often, it's not much far away from the reality, and this methodology can serve to be a pretty great teaching tool for inorganic chemistry, so long as you accept the fact that the mechanism you have made can only be proved by a research paper. But having an idea is also good.
Having got the introduction and disclaimer of this method out of the way, we can now start to explain your phenomenon by arrow-pushing diagrams.
The dichromate anion $\ce{Cr2O7^2-}$ has got a structure consisting of a $\ce{Cr-O-Cr}$ bridge, four $\ce{Cr=O}$ bonds, and two $\ce{Cr-O^-}$ bonds.
Here, the $\ce{Cr=O}$ bonds can be said to be structurally similar to the carbonyl group.
Hence, like the carbonyl group, it can be said that there will be high partial positive charge on the $\ce{Cr}$ atoms due to the electron-withdrawing effect of two $\ce{Cr=O}$ groups per $\ce{Cr}$. Hence, they can serve as the ideal electrophilic site for the attack of a nucleophile like $\ce{OH-}$ in basic medium, and lead to bond cleavage as follows:

The $\ce{HCrO4^-}$ formed can  again be deprotonated in the basic medium to form $\ce{CrO4^2-}$
In acidic medium, the $\ce{CrO4^2-}$ anion can be protonated to form $\ce{HCrO4-}$ , which can react with another $\ce{CrO4^2-}$ anion in the following manner to give $\ce{Cr2O7^2-}$ and expelling water as a leaving group:

Any alternate protonation pathways, for example, protonating $\ce{CrO4^2-}$ twice to form $\ce{H2CrO4}$ and then reacting it with $\ce{CrO4^2-}$ might also give the same product with a few extra proton transfers here and there, but the way I have presented thing would be the most probable way, as the reaction is progressing stepwise with each protonation.
Note: Remember that these are just probable pathways we have drawn to reach our product, but there's a good chance they resemble the reality( which can only be determined by kinetic experiments) and as the authors have stated, it is a good learning tool, and not too far off the mark.


What is the source of discrepancy in the calculation of Standard State Gibbs Free Energy change of the reaction of HCl (l) dissolving in water? I have calculated $\Delta_\text{r} G^o$ two different ways and got completely different results. Why? What assumptions have I made incorrectly, particularly in the first calculation shown below?

Okay, so everyone knows that liquid Hydrochloric acid is a strong acid so it completely dissociates in water according to the following chemical equation:

$\ce{HCl (l) + H2O (l) -> H3O+ (aq) + Cl- (aq)}$

If I used tabulated values of Standard State Gibbs Free Energy of formation $\Delta_\text{f} G^o$ to calculate $\Delta_\text{r} G^o$, standard State Gibbs Free Energy formation change of the reaction.

Standard State Gibbs Free Energies of formation:
$\begin{array}{lr}
\ce{HCl (l)} & \pu{-131.17 kJ mol^{-1}}\\
\ce{H2O (l)} & \pu{-237.19 kJ mol^{-1}}\\
\ce{H3O+(aq)} & \pu{-237.19 kJ mol^{-1}}\\
\ce{Cl-} & \pu{-131.17 kJ mol^{-1}}
\end{array}$

Source (I used this one because I couldn't find any that listed $\ce{HCl}$ as a liquid and not a gas):
http://www.vias.org/genchem/standard_ent...

Then, if you have one mole of each constituent according to the chemical equation, the standard state Gibbs Free Energy of the reaction = 0 kJ

by means of calculating ΔGoreaction = ΣΔGoformation of products - ΣΔGoformation of reactants

So ΔGoreaction should be less than zero if the reaction is favorable! But we know that HCl completely dissociates in water into Chloride ion and hydronium. What is going on here!?

If I calculate the standard state Gibbs Free Energy of the chemical reaction a different way.
That is that

$\Delta G^\circ_\text{reaction} = -R \cdot T \cdot \ln(\ce{Keq})$ at equilibrium, where

for the acid $\ce{HCl}$, $\ce{Keq = Ka} = \pu{1.3e6}$,

Then the Gibbs Free Energy change of the chemical reaction at standard state

$\Delta G^\circ_\text{reaction} = (\pu{-8.314 J K-1 mole-1}) \cdot (\pu{1 mole}) \cdot (\pu{298 K}) \cdot \ln(\pu{1.3e6})$

$= \pu{-34.88 kJ}$,

which is less than zero; therefore, the reaction is favorable. What is going on!? See the difference in the calculations!?

It seems that the Ka of strong acids like HCl is usually assumed to be too large for most websites to even bother displaying. But, I found a table of Ka's of strong acids here: http://depts.washington.edu/eooptic/links/acidstrength.html

Using the $\ce{\Delta{G^{\circ}}}$ value of liquid $\ce{HCl}$ ($-131.17$ kJ/mol) is wrong. Simply put, we are indeed interested in gaseous $\ce{HCl}$. The normal boiling point of $\ce{HCl}$ is $-85^\circ$ C. On the other hand, the standard state of Gibbs Free Energy is $25^\circ$ C for what you are calculating. (Having liquid $\ce{HCl}$ entails water being below that temperature as well, in which case it would freeze. Additionally, the equilibrium constant for the dissociation of $\ce{HCl}$ is also measured at $25^\circ$ C). Finally, though this is unrelated to your question, if $\ce{HCl}$ was liquid, it would not be included in the $K_a$ expression.

Obtaining the $\ce{\Delta{G^{\circ}}}$ value of $-95.30$ kJ/mol for $\ce{HCl}$ $(g)$, we find that $\ce{\Delta{G^{\circ}}}$ for the reaction is $-35.87$ kJ/mol. Applying the equation $K = e^{-\ce{\frac{\Delta{G^{\circ}}}{RT}}}$, we find that $K_a$ for $\ce{HCl}$ is $1.9 * 10^6$ (within magnitude of your given value).

What this looks like physically is that $\ce{HCl}$ gas is bubbled up into water, and it dissolves (dissociates) into the water as it mixes.


Why do you drop the charge in the crossover method? When I do the crossover method of crossing ions, my teacher always tells us to drop the charge, but won't tell us why. I was wondering if we add it back in later, and if not, why does it disappear?

Let's figure out the molecular formula for calcium chloride.

Calcium has a +2 charge. Chloride has a -1 charge.

So with the crossover rule we get

$$\ce{Ca_{-1}Cl_{+2}}$$

But for molecular formulas a -1 is impossible. The coefficients must be positive integers (usually...). Also have a 1 is superfluous, so it is just $\ce{Ca}$ and not $\ce{Ca_1}$.

Now for the chlorine the + is superfluous, so it is just $\ce{Cl_2}$ and not $\ce{Cl_{+2}}$.

I noted that the coefficients must be positive integers. A case where that is not true is when there is a double salt. Say a mixture of NaCl and KCl. The cations and the anions must balance so the formula can be written as

$$\ce{Na}_x \ce{K}_{1-x}\ce{Cl}$$

thus $(x) + (1-x) = 1$ and the charges balance.

Frankly I find this whole rule thing much to do about nothing. The net charge must be 0. Calcium has a +2 charge and Chlorine has a -1 charge. So you need two -1 charges to balance the +2 charge. Thus $\ce{CaCl2}$. Note that the coefficients must be the lowest common multiple, so it is $\ce{CaCl2}$ and not $\ce{Ca2Cl4}$ which also has a net 0 charge.

So to me the only difficult part is figuring out the charges on the ions. Once you know that, the rest is easy. Making a special rule just seems to be overkill.


How to make dichloromethane human-friendly and safe for storage? I use dichloromethane to polish 3D printed object as postprocessing step.
Most of it is stored in darkened glass on my balcony (now at $\pu{-15^\circ C}$).

What chemical reactions I need to make it human-friendly (non toxic)? Is it ok to use non-darkened polypropylene/polyethylene box to store it?

What I can do during the summer season? (Dichloromethane is volatile liquid)

According to A Comprehensive Guide to the Hazardous Properties of Chemical Substances at pages 445-446:

Disposal/Destruction
...
Small amounts may be destroyed in the laboratory by hydrolyzing with ethanolic solution of potassium hydroxide


Real life chemistry :boiling carrot dices This morning I was boiling carrot dices. And saw something quite interesting happening to the dices.

The water is not boiling. But before it boils, the dices started gathering at the center  of the pot.

Any one has any answer as to why this happened?

My guess would be natural convection. Hot water is less dense than cold water so when the heat source is at the bottom, the freshly heated water rises and the cooler water at the top sinks. The cooler water can't sink through the rising hot water, so the two flows run along side each other. Depending on the fluid properties, this can result in circular "rolls" of flowing water, known as convection cells. If you use a wide pot and small particles, you may be able to see many, but in a relatively narrow pot, you're more likely to observe a single recirculating flow around the centre, as shown in this simulation. The liquid loses more heat around the sides of the pot, so the descending flow rides down the sides and the ascending flow is in the middle where it can stay the hottest.

I'm guessing that your dice is too heavy to ride the ascending flow, but the flow from the edge to the centre pushes all the carrots towards the middle.


Calculating purity from NMR spectrum I just started organic chemistry in university. I have no idea what I'm doing — it's wonderful.

My assignment was to produce $\pu{10g}$ of fluorescein, the theoretical yield is $70~\%$ — I need at least $60~\%$ yield. I've made $\pu{11.07g}$ of product and my lab instructor analysed a sample of my product on the NMR spectrometer, which gave us the following result:

There are some impurities in my product (I know that f.e. $1.12$ is from the impurity and $3.42$ is from the fluorescein, same with $1.00$ and $3.46$) and I have no idea how to use that data to compute purity or actual yield.

Can somebody help?

A couple of points for starters:

to say anything about impurities you need to show the entire spectrum, typically from $0$ to $\pu{10ppm}$ for $\ce{^1H}$. Merely describing the existence of peaks is not enough.
you also need to make sure your baseline is a proper baseline as you need to use the integrals to calculate purity if you even can. Here again, $n_\mathrm{t} = 8$ is absolutely too little.

However, it still needs to be said that determining the purity in percent just from NMR spectra is a very non-trivial task. It can only be done, if you know exactly which impurities you have, if they all contain the corresponding NMR-active nucleus, if the NMR-active nucleus is sufficiently predominant for the read to be quantitative and if there is no mechanism by which a non-NMR-active nucleus may be introduced into your impurity. Especially the predominance of NMR active nuclei along with another few requirements for ‘nice’ NMR spectra means that purity via NMR can in practice only be measure for $\ce{^1H, ^19F, ^31P}$ and a few much less common elements.

The main deal breaker is typically the requirement to contain the NMR-active nucleus. You can have a beautifully clean $\ce{^1H}$ NMR spectrum but the actual substance can be as little as $1~\%$ pure: most notably if the impurities are inorganic salts and do not contain any hydrogen. In most non-carbohydrate organic chemistry, this is not so much of a problem, though.

For any substances that contain the NMR-active nucleus and are impurities, the first step is to find out exactly what the substance is. To be able to calculate anything from an NMR spectrum, you need to know the impurity’s molar mass as NMR readings can only give you relative amounts.

Finally, you also need to make sure that your impurity does not lose its NMR-active nucleus. For example, if you are measuring in $\ce{CD3OD}$ and the impurity happens to be $\ce{C3F7OH}$ (1,1,1,2,3,3,3-heptafluoropropan-2-ol), the reading of the final remaining signal (if it even remains) will not be accurate due to proton/deuterium exchange:

$$\ce{C3F7OH + CD3OD <=>> C3F7OD + CD3OH}\tag{1}$$

(The equilibrium is shifted towards the products due to the abundance of methanol as the solvent.)

This is why NMR analysis is typically only used to determine relative purity, e.g. if two diastereomers form during a reaction. These will have the same molar masses allowing for an easy comparison.

To actually perform this, you need to:

Identify the peaks belonging to an impurity, the impurity’s structure and the signals’ integrals.
Normalise the impurity’s integrals to a single proton equivalent (i.e. if your impurity is cyclohexane, divide the integral by $12$ since you have $12$ magnetically equivalent protons.)
Normalise your desired compound’s integrals so that a single proton has the integral 1.
Determine a ratio of amounts in moles. (e.g. $\pu{8mol}:\pu{2mol}$)
Calculate the corresponding masses of desired product and impurity.
Add up the masses you calculated to a total mass.
Divide your desired product’s mass by the total mass to get your purity. Multiply by $100~\%$ to get a percentage.

From the information you give and the spectrum you show it is absolutely impossible to tell you anything about the purity of your compound. Signal integrals themselves are also not helpful; if your impurity is huge in molar mass but only has few protons, or if it has multiple magnetically identical protons a simple integral reading will be way off.

The practical approach for an organic chemist to determine purity is to use chromatography (either flash or HPLC) until no signals can be detected that do not belong to the compound in question and then label it ‘pure’.


Does an airborne chemical stimulant exist? I recently read the creepypasta entitled "The Russian Sleep Experiment". In this story they lock five test subjects in a room with the intention of keeping them awake for 30 days by releasing a gaseous stimulant into the room constantly. I'll let you read the rest, it's an incredible horror story, but not very credible. Anyways I wanted to know if in real life such a gaseous stimulant exists.

During the Moscow theater hostage crisis Russian forces "pumped an aerosol anaesthetic, later stated by Russian Health Minister Yuri Shevchenko to be based on fentanyl". So it's possible for an opiate-based depressant can be made gaseous (technically aerosol).

So is there an airborne stimulant that actually exists?

Regardless of whether the stimulant is actually gaseous at standard temperature and pressure, it can be made airborne as an aerosol simply by spraying. If the room to be filled with vapour is sufficiently small then it is no problem to just spray a solution of your stimulant and pump it inside. But even gym hall sizes should be fine with adequate technique applied.


Hamiltonian 2nd positional derivative analogous to acceleration? In Quantum Mechanics, we learn that the Hamiltonian operator for an electron confined to a 1-D space is:

We learn in QM that many operators have analogous interpretations familiar to us from classical mechanics. My question is then: is there a classical mechanics interpretation for the second positional derivative? Is its physical significance at all related to the second positional derivative in classical mechanics (i.e. acceleration)?

It's related to kinetic energy
$$T = \frac{1}{2} mv^2 $$

which can be rewritten in terms of momentum $p$
$$ T = \frac{p^2}{2m} $$

Replacing $p$ with its quantum mechanical operator
$$ p = -i\hbar\frac{\partial}{\partial x} $$

gives
$$\hat{T} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} $$

I'm not sure why you bring up acceleration. Acceleration $a$ is the second time derivative of position, $$a = \frac{d^2 x}{d t^2}$$


Which intermolecular interaction is dependent on the inverse cube of distance? The intermolecular interaction that is dependent on the inverse cube of distance between molecules is:

  hydrogen bond
  ion-ion interaction
  ion-dipole interaction
  London force

Inverse cube of distance I think means an ion-dipole interaction, since we know force due to a dipole on a charge is inversely proportional to the cube of distance. But I don't understand why the answer given is hydrogen bond. What is the distance dependence of the hydrogen bond? Shouldn't the answer be (3)?

2

The ion-ion or charge-charge interaction is given by

$$
V_{\text{cc}} = \frac{Z_{i}Z_{j}e^{2}}{4\pi\epsilon_{0}} \frac{1}{|\mathbf{R}_{i} - \mathbf{R}_{j}|}
$$

and is therefore proportional to $R^{-1}$.

3

The ion-dipole or charge-(permanent) dipole interaction is given by

$$
V_{\text{cd}} = \frac{Z_{i}e}{4\pi\epsilon_{0}} \frac{\mathbf{R}_{ij} \cdot \vec{\mu}_{j}}{R_{ij}^{3}}
$$

and is therefore proportional to $R^{-2}$.

4

"London force" is often meant to encapsulate any of the van der Waals forces (here, here), but in particular the dispersion force. All of the above have a $R^{-6}$ dependence.

Here is a general table that covers each of the interactions so far:

\begin{array}{lll}
\hline
\text{type of molecular unit} & \text{type of force} & n~\text{in}~R^{-n} \\
\hline
\text{ions}                        & \text{coulombic}               & 1 \\
\text{ion - polar molecule}        & \text{ion - dipole}            & 2 \\
\text{two polar molecules}         & \text{dipole - dipole}         & 3 \\
\text{ion - nonpolar molecule}     & \text{ion - induced dipole}    & 4 \\
\text{polar and nonpolar molecule} & \text{dipole - induced dipole} & 6 \\
\text{nonpolar molecules}          & \text{dispersion}              & 6 \\
\hline
\end{array}

1

From Wikipedia:

  The hydrogen bond is often described as an electrostatic dipole-dipole interaction. However, it also has some features of covalent bonding: it is directional and strong, produces interatomic distances shorter than the sum of the van der Waals radii, and usually involves a limited number of interaction partners, which can be interpreted as a type of valence. These covalent features are more substantial when acceptors bind hydrogens from more electronegative donors.

This implies that hydrogen bonding interactions may not behave exactly as $R^{-3}$. Consider the water dimer:

which is a prototypical example of a hydrogen-bonded complex. From Stone, using density functional-based symmetry-adapted perturbation theory (SAPT-DFT) in the aug-cc-pVQZ basis,
\begin{array}{lr}
\hline
\text{Type of interaction} & \text{Energy (kJ/mol)} \\
\hline
\text{Electrostatic}                 & -33.4 \\
\text{Exchange-repulsion}            &  31.5 \\
\text{Dispersion}                    & -12.2 \\
\text{Induction}                     &  -8.0 \\
\text{Charge transfer}               &  -5.9 \\
\text{Exchange-dispersion}           &   2.3 \\
\text{Exchange-induction}            &   7.8 \\
\delta_{\text{HF}}~\text{correction} &  -3.4 \\
\hline
\text{Total}                         & -21.5 \\
\hline
\end{array}
the electrostatic component is clearly the largest contributor, but after taking the absolute value of all interaction terms, is about 32% of the total interaction. Another approach for decomposing the interaction energy is based on absolutely localized molecular orbitals (ALMO-EDA), which at the $\omega$B97M-V/def2-QZVPP level gives
\begin{array}{lr}
\hline
\text{Type of interaction} & \text{Energy (kJ/mol)} \\
\hline
\text{Electrostatics}  & -65.7 \\
\text{Pauli repulsion} &  65.0 \\
\text{Dispersion}      &  -7.7 \\
\text{Polarization}    &  -4.6 \\
\text{Charge transfer} &  -7.9 \\
\hline
\text{Total}           & -20.9 \\
\hline
\end{array}
SAPT-DFT and ALMO-EDA are in good agreement, though charge transfer is more important in ALMO-EDA primarily due to the lackluster definition of charge transfer within SAPT. Electrostatics are now about 44% of the total interaction. For reference, here is how the additional terms decay (Stone, Table 1.2):
\begin{array}{lccl}
\hline
\text{Contribution} & \text{Additive?} & \text{Sign} & \text{Comment} \\
\hline
\textbf{Long-range}~(U \sim R^{-n}) & & & \\
\text{Electrostatic} & \text{Yes}     & \pm & \text{Strong orientation dependence} \\
\text{Induction}     & \text{No}      & -   &                                      \\
\text{Dispersion}    & \text{approx.} & -   & \text{Always present}                \\
\text{Resonance}     & \text{No}      & \pm & \text{Degenerate states only}        \\
\text{Magnetic}      & \text{Yes}     & \pm & \text{Very small}                    \\
\hline
\textbf{Short-range}~(U \sim e^{-\alpha R}) & & & \\
\text{Exchange-repulsion}  & \text{approx.} & + & \text{Dominates at very short range} \\
\text{Exchange-induction}  & \text{approx.} & - &                                      \\
\text{Exchange-dispersion} & \text{approx.} & - &                                      \\
\text{Charge transfer}     & \text{No}      & - & \text{Donor-acceptor interaction}    \\
\hline
\end{array}

Answer

Although components of hydrogen bonding interactions may have a $R^{-3}$ distance dependence, there are significant contributors to the hydrogen bond interaction other than electrostatics that do not decay as $R^{-3}$; in general, it is a complex interaction and its definition is still hotly debated. For that reason, the best answer is (3), the ion-dipole interaction.

References

Masahiro Yamamoto, Kyoto University: charge-charge, charge-dipole, dipole-charge, dipole-dipole interaction
Stephen Lower, Simon Fraser University: Chem1 Virtual Textbook, Interactions between Molecular Units
Anthony Stone: The Theory of Intermolecular Forces
ALMO-EDA calculations: geometry is B97-D/def2-TZVP, EDA input as comment


is every molecule either a base or acid I read today that water is a base because it can receive a proton when mixed with some acids. That got me thinking: is every single molecule capable of either donating or receiving a proton -- when mixed with just the right complementary molecule? Thus, is every molecule either a base or an acid to some degree?

In theory, any molecule that contains a H atom ("proton") can act as a Bronsted acid - it can donate that proton. How likely this is to happen is described by that proton's pKa value, the lower the value, the more likely this is to happen (the more stable the conjugate base that forms will be). You can look up pKa values for different types of proton on an inorganic or organic molecule by searching for pKa tables like this:

Different proton environments on the same molecule will have different pKa values:

The pKa describes the degree to which that proton will dissociate in a solvent (normally water, but other solvents are sometimes used) and high values mean that this essentially doesn't happen, or only to a tiny extent (in the figure above, these are labelled as "not acidic"). However, a very strong base could abstract these protons, so the molecule could still act as an acid.

To act as a base, a molecule normally needs to possess a lone pair. The Bronsted definition of a base only requires a molecule to accept a proton, but a lone pair is generally required in order for this to happen.

The example in the figure above shows ammonia acting as a Bronsted base, using its lone pair to bond to a proton. Any lone pair could (potentially) do this, but some functional groups (such as amines) have more propensity to do this. The pKb value for the functional group will quantify this propensity.

Molecules often contain both protons to donate and lone pairs, so can act as Bronsted acids and/or bases.

Lewis acids and bases are defined slightly differently. The definition here involves the electron pair, rather than the proton. So you can have Lewis acids (electron pair acceptors) that have no protons at all. These tend to be electron deficient species such as the $BF_3$ in this figure below:

Other examples of Lewis acids include small, highly charged cations that can accept coordinate bonds.

Some molecules can accept protons by using electrons from unsaturated moeities rather than lone pairs. An alkene is a classic example, which can use the pi electrons in the double bond to accept a proton, forming a carbocation.

So an alkene could be described as a base, but we don't normally categorize it in that way.

It is also worth looking at the distinction between acid/base and electrophile/nucleophile. There are some good posts on this site like this one


Why are noble gases stable I was recently asked the question "Why are noble gases stable? with the expectation of providing an answer beyond the general explanation of "they have full valence layers" and I couldn't think of one.

I would love to hear a full description of the this stability, I have a solid background in quantum mechanics, so feel free to speak of wave functions or the like if necessary.

Actually it is not necessary to dig deep into quantum mechanics. There are several reasons why noble gasses are stable (as gasses at room temperature).

First of all, there is the obvious full valence shell.
Trend in the periodic table make clear that the charge of the nucleus grows from left to right in every period. The attractive force towards the electrons therefore increases. (This holds also from top to bottom.) Electrons in lower orbitals ($n<\text{period}$) now shield the charge of the nucleus. This will be somewhat the same throughout the period. (The orbitals also shrink due to higher charge of the nucleus.) In noble gasses the valence shell is completely filled, providing quite a good shield for the next shell. Also the increase in the main quantum number means a significant increase in the energy level of the next orbital. This makes it unlikely, that a noble gas will accept another electron.

Due to the high charge of the nucleus it is also not easy to remove an electron from an orbital. However, this is possible from (at least) Argon on downwards. For example: $\ce{HArF}$ is stable in a matrix at room temperature. It has a strong covalent $\sigma$ bond ($\ce{H\bond{-}Ar+}$) and a not quite as strong ionic bond ($\ce{[HAr]+\cdots F-}$). Krypton already does some fancy chemistry, that is stable at room temperature. As Uncle Al has pointed out Xenon is well known for its reactivity.

But why the sudden change? It is fairly intuitive, that the maximum electron density for each orbital with increasing main quantum number is also further away from the nucleus. That makes the valence shell very well polarizeable. Also the electrons are better shielded by the previous shells. However, the natural occurrence of these elements are as (homonuclear) gasses.

However, if you bring these elements in close contact with each other it was found out, that they have a very small dissociation energy ($D_e<1~\text{meV}$). This is due to dispersion, and van-der-Waals Forces, which is the main interaction between these elements. However, in $\ce{He2}$ no binding mode was observed (instant dissociation).

But that still does not answer, why those elements are stable as gasses, as pointed out that there are in fact attractive forces. The reason is as simple as obvious: Entropy. If two noble gasses would form a molecule/adduct, the bond/association energy of this molecule would have to compensate for the loss of entropy (Two volume elements would become one, hence the gas would have to expand to cover the room previously occupied, which requires work energy).

The explanation by tschoppi also covers, why from a MO point of view there could not be a bond in $\ce{He2}$. Go ahead and ask yourself the question if this would be true for $\ce{He3}$. We also know that orbital overlap is only one component to the truth. There are many other more. If it was not for such a nice thing like dispersion and other weak chemical interactions we would not exist.


Electric field required to split water into oxygen and hydrogen Others sources typically say that the minimum voltage required to split water into hydrogen and oxygen is 1.229 V. Overvoltage is typically required in a practical device due to inefficiency leading to heat production.

However, suppose that two electrodes are separated by a large distance.  This will result in a very low electric field strength (generally, electric field equals $V/d$ where $d$ is the distance of separation).  Is there a threshold electric field level required for electrolysis to take place?

In part, there is no abrupt cutoff because the gradient of the field is not spread evenly across the electrolyte: the gradient is steep at the interface between electrodes and the electrolyte, where gas evolution actually takes place, not in the body of the electrolyte. Note where bubbles arise in this video.

However, if the conductivity of the electrolyte is low, whether due to electrode separation or low ionization, then electrolysis proceeds more slowly, perhaps imperceptibly, though there is no complete cutoff point.


What is the boiling point and density of LSD? I have tried to do a lot of research and didn't succeed to find the boiling point and density of LSD. I would want to know this information. In addition, I can't use Clausius–Clapeyron equation because I don't have the data that I would need to use that equation.

Lysergic acid diethylamide doesn't boil. It decomposes. The density is $1.2 \pm 0.1~\mathrm{g~cm^{-3}}$.

Data can be found at  ChemSpider. There's a predicted boiling point of 541°C. However, lysergic acid decomposes at its melting point of about 240°C (CRC Handbook of Chemistry ad Physics, 64th ed.). This suggests that lysergic acid diethylamide does not have a real boiling point - at least at normal pressure.


effect of temperature on position of equilibrium I was wondering if there was an intuitive way of explaining how a change in temperature would effect the position of an equilibrium, without using le chateliers principle. Le Chateliers principle doesn't seem to explain why the equilibrium shifts the way it does, I was wondering if there was a way of explaining this.

Well, For endothermic reaction the Number of moles at the product side increase with increase in temperature.
And for exothermic, it decreases with Increase in temperature..
And exactly vice versa with respect to the reactant side..
It is in Accordance with the Le chaitliers Principle..


Why does lithium move on the surface of water during the reaction? If someone asks me to number the most notable "textbook experiments" I would definitely name "the comparison of the reactions of alkali metals with water" among the most favorites.

We just happened to do such a reaction with small pieces (approximately $\text{7.5 g}$) of Li, Na and K. As we were supposed to write down anything that we observe, I came across an interesting phenomenon.

It seems that the answer is so simple, I didn't get it through many resources.

  Li piece is moving rapidly (better call it spinning) on the surface of water as it is reacting with it and a subtle explosion sound is heard.

Why does the motion of Li on the surface of the water in the beaker take place? Could the heat released by the reaction be causing current in the water (I don't believe so, as the possibly hot water isn't involved in any convection currents)? Info you might need:

Our chemistry teacher cut a non-oxidized piece of well-preserved lithium. The piece of the reactant wasn't of any distinguishable geometrical shape. After repeating the experiment for several times, I sensed a headache that as I believe is because of release of the impurities of the sample.

What happens when you put lithium (or sodium or potassium) in water? What is the chemical reaction that is occurring?

The alkali metals react with water to produce alkali hydroxides and hydrogen gas:

$$\ce{2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g)}$$

The two products of the reaction of water and lithium are lithium hydroxide (which dissolves) and hydrogen. Hydrogen is not very soluble in water, and it is much less dense than air. The motion of the lithium on the surface of the water is created by the escape of the hydrogen gas, which provides upward force.


How are subshells and shells arranged in an atom? If $\mathrm{2s}$ and $\mathrm{2p}$ are in the second energy level, and $\mathrm{3s}$, $\mathrm{3p}$, and $\mathrm{3d}$ are in the next (3rd) energy level, how are these subshells arranged in space in relation to one another? Particularly, could this be pictured as the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals being relatively the same size (although different shapes) and superimposed over them being the $\mathrm{3s}$, $\mathrm{3p}$, and $\mathrm{3d}$ orbitals all of a relatively larger size than orbitals of the 2nd energy level?

I will start off by offering an apology for anything I might have misunderstood - if this is the case, just tell me in the comments below.

  If 2s and 2p are in the second energy level, and 3s, 3p, and 3d are in the next (3rd) energy level

This is, in general, not the best choice of words. The 2s and 2p orbitals are only degenerate, meaning that they have the same energy, in a hydrogenic atom: a species which contains one nucleus and one electron. The hydrogen atom itself satisfies this requirement. Other hydrogenic species include $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, ...

For any species that has more than one electron, the 2s and 2p orbitals have different energies (see this question, although it's a bit of a read) and therefore it would not be correct to refer to them belonging to the same "energy level".

What you can say is that 2s and 2p orbitals belong to the same principal quantum shell. All this means is that the number in front of them - called the principal quantum number - is the same, in this case, 2.

  How are these subshells arranged in space in relation to one another?

Orbitals are not actually containers, or tangible objects that occupy space. They are mathematical functions obtained by the solution of the Schrodinger equation. The pictures of orbitals that you are probably familiar with are just pictorial representations of these functions. Without going into too much detail, one quantity that is used in the functions is the radial distance between the electron and the nucleus, denoted $r$. $r = 0$ corresponds to the nucleus's point in space. What this means is that all orbitals - regardless of their shape or energy - are centred at the nucleus. The $\text{s}$ orbitals are spherical in nature, and the centre of the sphere corresponds to the nucleus. And the same goes for $\text{p}$ orbitals, $\text{d}$ orbitals, and so on.

Therefore, to put it very succinctly, everything is superimposed on everything. You could think of it as shining different colours and shapes of light onto a screen - the projected images can all simultaneously exist.

  3s, 3p, and 3d orbitals all of a relatively larger size

You're right about this - in general, orbitals with a larger principal quantum number tend to be larger, as long as we restrict the discussion to the same atom. For example, the $\text{3s}$ orbital is larger than the $\text{2s}$ orbital, which is in turn larger than the $\text{1s}$ orbital.

Now since I did say that orbitals are mathematical functions, this does raise the issue of how you define an orbital to be larger than another. If you are interested, you can read up about the radial distribution function, although I shall not expand on it as that would really be beyond the scope of this question.


I believe you have not correctly interpreted the first basis set, which should read:

O1 = 0.061 f(172) + 0.358 f(25) + 0.7 f(5.5)
O2 = -.39 f(3.66) + 1.21 f(.77)
O3 = 1 f(.195)

Note that these are only the s-orbitals. Using the coefficients in the right-most column, two triplets of p-orbitals are also present. By doing it this way for the second basis set, which is 6-31G, you would obtain 3 s-orbitals and two triplets of p-orbitals, making it nine AOs in total. Note that all orbitals in a tuple must have the same contraction scheme, otherwise, the calculation would not be rotationally invariant (to within integration accuracy).

Aside: Why does this basis specify s- and p-orbitals this way?
Depending on the integral scheme, one can reuse certain intermediates of some integrals for others and save some computational time. However, modern programs tend to work better with segmented basis sets (i.e. basis sets that do not allow for reuse, because the $\alpha$ exponents are all different). The results are also of higher quality. IMHO, the Pople basis sets are only of historical interest or for hobbyists with very limited resources.


Is it possible to express Gibbs free energy as following? We know $$\Delta G =\Delta H -T\Delta S$$
$$\Delta S= \frac{Q_{rev}}{T} $$
Hence $$\Delta G=\Delta H-Q_{rev} $$

Yes, but only if the temperature T in the initial and final equilibrium states of the system is the same and you are able to identify a reversible path between the initial and final equilibrium states over which all the heat transfer takes place at the temperature T.


Negative Kelvin Temperature I remember my Physical Chemistry Professor saying that very tiny negative Kelvin temperatures have been achieved on the quantum level. Is this true?

Yes.  But it is a bit of a joke of a sort.  In any system with energy levels, the lower energy levels never have a lower population than the higher ones.  In fact there is an equilibrium arrangement of the populations of such levels characterized by the temperature.  The higher the temperature, the more evenly populated the levels are.  There is even a formula for calculating the populations in this case.

The energy levels of some systems depends on a magnetic field.  You can let the populations of the energy levels come to equilibrium at some temperature $T$ and then quickly invert the magnetic field.  What was a low energy level now becomes a high one and vice versa.  You have an upside down distribution of populations.

Soon enough that upside down population decays to a right side up population.

But while the distribution is "upside down" one finds that it is given by the same formula used for an ordinary distribution, but with a negative temperature.


Reaction mechanism of a terminal alkene with an anhydride In a reaction (as given in my textbook),

$$\ce{R-CH=CH2 + (CH3-CO)2O  -> R-CH2=CH-COCH3}$$

an alkene reacts with acetic anhydride to give methyl alkenyl ketone.

I am not sure of how this proceeds, a search on google displays no links to any relevant matter, and I have not successfully found this reaction in any of the other books I possess.

According to me, it should have been an addition reaction (characteristic of alkenes) across the double bond. But this isn't the case here. So, what exactly is the reaction mechanism?

This is the  Friedel-Crafts acylation reaction of alkene. Catalyzed by Lewis acid. Addition-elimination mechanism.


Two reactions that catalyze each other Is there an example of two reactions that catalyze one another? More than two is also helpful, but I guess it will be hard to find one.

By "catalyze one another", I mean that one of the products of the first reaction acts as a catalyst for the second reaction.

An early synthesis of borazon from hexagonal boron nitride would be one possible example.  When other approaches did not work, Robert H. Wentorf Jr. added magnesium and voila!  The magnesium had reacted with some of the hexagonal nitride to form $\ce{Mg3N2}$, and the latter then catalyzed the hexagonal to cubic conversion.  (Today other methods have taken over.)


Unknown white precipitate in electrolysis of water I was electrolysing water using a copper anode and zinc cathode.  The water was from the tap and where I live is very hard.  I was using about 20V between the two electrodes.  As I applied the voltage, a white precipitate formed around the anode.  What is strange is that I can't replicate this using any other type of metal as the anode.  Any ideas on what this might be?   Or, I suppose a better question is: are there any ideas on how I can find out what this was?

The slight bluish tinge to that white precipitate makes it look to me more like copper(II) hydroxide than anything to do with the hardness minerals, especially since you haven't observed it with other anode metals thus far.

One way to test this would be to make a whole bunch of the precipitate, by running the current for a long time, and then taking a small sample of the liquid with precipitate in it and adding an equal volume of distilled vinegar (5% acetic acid).  If it is indeed $\ce{Cu(OH)2}$, I think it should dissolve to form a clear blue solution of copper(II) acetate.

The only problem might be if 5% acetic acid isn't able drop the pH far enough, in which case you'd have to track down a more concentrated acid. One choice might be $\ce{HCl}$, which is available at, e.g., aquarium supply stores under its historical name of muriatic acid. You'd then get a solution of copper(II) chloride, the color of which will vary depending on the $\ce{Cl-}$ concentration. Another option, per MaxW's comment, would be pickling vinegar (10% acetic acid) -- in fact, a quick search just revealed that Amazon has several vinegar options available in higher acidities.


Can I get ride of dichlorobenzene in honey? I have some honey that was stored near dichlorobenzene crystals & has taken on the smell. I'm wondering if I turn it into mead if it would remove the chemical?

The procedure would be to add 5-10 parts water to the honey. Then boil the solution for at least an hour. I would think this should rapidly remove the benzine component.

The second part of the process would to be to add a yeast culture that would not ferment until the chlorine has dissipated.

It would seem that I could tell the chemical is absent when the strong odor is gone.

Just theoretically, if you were to heat the honey (primarily fructose and glucose) to 373 K, the boiling point of water, it would speed up the evaporation of the dichlorobenzene (vapor pressure ~8,000 Pa). That might be enough to get rid of the odor, if kept at that temperature for a few days.

That said, the flavor of the honey would likely be destroyed, as well. Even though pasteurization of honey is not normally recommended, that only requires heating to 145° F (336 K) for 30 minutes, and it still changes the flavor.

So toss the honey, or use it as an embalming agent (see also National Geographic). The dichlorobenzene "moth flakes" should make this honey even more effective!