Order of elements in a formula? Possible Duplicate:
  Why is methane's molecular formula conventionally “CH4”, while water is “H20” (among others)?

If I have a formula containing 4 hydrogen atoms and 1 carbon

Should it be written $\small\ce{CH4}$ or $\small\ce{H4C}$?

In general, how do I know the order to put elements in, in a chemical formula?

There's something called Hill System. This is the system of writing chemical formulas.
In this system the carbon atoms are first, then hydrogen atoms and then other in alphabetical order. When the formula contains no carbon or hydrogen atoms, then all the elements are sorted alphabetically.

source: http://en.wikipedia.org/wiki/Hill_system

So it should be: $\ce{CH4}$

However, the Hill system is not the only method of ordering elements in a chemical formula. We can instead choose to list the elements based on their electronegativity. ManishEarth explains this in his answer to this question.


Distance between successive tetrahedral voids in FCC What is the distance between successive tetrahedral voids in a FCC
crystal structure?

The distance between a face centered atom and a tetrahedral void is $\sqrt{\frac{3a}{4}}$ if the cube's edge length is $a$. Now If I know the angle made, between the line joining a tetrahedral void and the face centered atom, and the successive tetrahedral void and face centered atom, I can find the distance.  But how do I calculate that distance since I don't know the angle.

As shown in figure, if we divide a FCC unit cell into 8 small cubes, then each small cube has 1 Tetrahedral void at its own body centre.
Thus, there are total 8 Tetrahedral voids in one unit cell.
It can also be seen from the figure that the nearest distance between two Tetrahedral voids is a/2.


Write SMILES to SDF in RDKit I have a file with smiles and I'd like to convert the SMILES to SDF and tag the properties of each molecule, so I can use the SD file in another tool. Here is my code:

pp = pd.read_csv('anti.smiles', names=['Smiles', 'BA'])
pp =  PandasTools.AddMoleculeColumnToFrame(pp,'Smiles','Molecule')
PandasTools.WriteSDF(pp, pp_out, molColName='Molecule', properties=list(pp.columns))

And the error I get is this:

NameError: name 'pp_out' is not defined

Kindly assist


How many dipeptides are possible by the reaction of glycine and alanine? How many dipeptides are possible by the reaction of glycine and alanine?

Glycine has no chiral centre. Alanine has one chiral centre.
The products are alanylglycine and glycinylalanine (I not sure whether this is the correct name or not. Please excuse), and I got a total of 4 isomers for each, (using D,L notation and optical isomers) so I got a total of $4+4=8$ products.

But the answer is given as $9$.

Where is my mistake?

The number of isomers are ${9}$ as shown below.

AlaAla has ${4}$ isomers.
AlaGly has ${2}$ isomers.
GlyAla has ${2}$ isomers.
GlyGly has ${1}$ isomer.


How can we separate carbon monoxide from wood gas? I am thinking about using open source wiki designs from allpowerlabs to construct a wood gas generator.  The motivation came from experiencing a 10 day power outage after super storm Sandy.  The utility company could not provide power due to damaged transformers, power lines, and other grid infrastructure.  Most of the gasoline service stations did not have backup power to run the pumps and meters and those that did had huge queues to wait through.

Meanwhile there was tons of biomass available in my 2 acre yard.  I used some of the wood in our wood stove for heating and some cooking but did not have a way of getting the energy stored in wood and other biomass to electrical energy.

The solution is a wood gas generator that converts the stored solar energy in wood to a gas that can be combusted in a hacked generator to provide electrical power.  Allpowerlabs.com has a turn key solution all ready to go but it is beyond my budget, but they offer the open source build it yourself GEK (Gasifier Electricity Kit), mentioned earlier, that I would like to build.

Instead of using the wood gas to run a hacked Briggs and Stratton  generator, I would prefer to route it to a BIY (Build It Yourself) fuel cell.  Wood gas consists of hydrogen, carbon monoxide, carbon dioxide, water vapor, and small amounts of other.  For fuel considerations hydrogen(H2) and carbon monoxide(CO) are usable.

While molten carbonate and solid oxide fuel cells can use CO, the operating temperatures are too high to be practical and we want to sequester the carbon instead of releasing to atmosphere.  I would like to separate the carbon oxides from the wood gas and use the H2 directly into a fuel cell or store it for later use.

I know that we can remove CO2 from the wood gas or exhaust with low cost or Brew it yourself sodium hydroxide (NaOH), via caustic scrubbing.

My question is: How can we cost effectively separate the CO from the wood gas?  I am not sure how NaOH reacts with hydrogen gas and CO.

This master's thesis by Dahiru Rufai Ahmed at the university of Oulu provides a comprehensive background into the removal of CO2 from wood gas.

On pages 36-37 it mentions the water-gas shift (WGS) reaction as a method of adjusting the hydrogen/carbon monoxide mixture or removing CO entirely to get pure H2.  "The WGS reaction is an exothermic reaction
(∆Hr = -41.1 kJ/mol) that converts excess carbon monoxide into hydrogen."

CO + H2O -->  CO2 + H2 + heat

"At low temperature the reaction can be catalyzed by catalysts based on copper, aluminum and zinc oxides. (Basu, P. (2010). Biomass gasification and pyrolysis: Practical Design and Theory. 1st ed., Elsevier Inc. 365pp. ISBN: 9780123749888.)

However, in recent times, interest has been shifted to a catalyst-membrane system that incorporates the WGS reaction and H2 separation via a hydrogen selective membrane (Mendes, D., Chibante, V., Zheng, J-M., Tosti, S., Borgognoni, F., Memdes, A. and Madeira, L. M. (2010). Enhancing the production of hydrogen via water-gas shift reaction using Pd-based membrane reactors. Inter J Hydrogen Energy. 35 (22), 12596–12608.)."

This article describes a membrane based catalyst for the WGS reaction.  The catalyst in the membrane uses Palladium alloys with Silver and Copper due to their higher selectivity and permeability to Hydrogen.

Since Palladium is an expensive noble metal and since catalytic converters for automobiles already use Palladium for CO oxidation, I am considering using an automobile catalytic converter (CAT) for this purpose.  The CAT uses oxidation rather than WGS to convert CO, so I am concerned that it would also oxidize the hydrogen.  However, I think the CAT would perform WGS if the input gas mixture is lacking oxygen which would be the case for wood gas.  Any useful comments and feedback appreciated.


Distinguishing between octahedral and tetrahedral holes I know the question sounds a bit silly and there are many similar answered questions but my 3D visualization (imagination) is very bad. There are some images of octahedral and tetrahedral holes but still there's no easy explanation. Can someone help me?

To visualize holes in closest packing of spheres, you need 6 balls (marbles, table tennis balls, etc) of equal size. 3D closest packing can be constructed from 2D layers of closest packed spheres as such:

The light blue layer is on the bottom and the dark blue layer is on top. The holes, tetrahedral or octahedral, are in between layers.

To find a tetrahedral hole, take three balls and arrange them in a triangle to form part of the lower layer. Then place one sphere in the center on top (you have to keep the lower level together somehow or the structure will collapse). Your tetrahedral hole is in the center of the tetrahedron formed by the 4 balls (marked T in the picture).

To find an octahedral hole, you need a helper or glue. Take three balls and arrange them in a triangle. Take another three balls and arrange them in a triangle. Place the two triangles on top of each other, twisted against each other by 60 degrees. This is a octahedron on its side, and in the center is the octahdral hole.

In the picture above, these positions are marked by T and O. You just have to imaging that they are in between the layers, and then you might be able to count nearest spheres. The picture below shows the arrangement of the nearest spheres, only.

If you don't have 6 balls handy, but have two hands handy, you can also do the following. Place thumb, pointer and middle finger together so that the finger tips form a triangle. Point at the center of the triangle with the pointer of the other hand. That's the tetrahedral hole. Now, form a triangle with the finger tips of your other hand. Bring the 6 finger tips together with a little twist so that they "lock in". At the center of the 6 finger tips, you have the octahedral hole.

So, whether you are a visual or a haptic learner (not that that would be a thing), this will do the trick, I hope.

  There are some images of octahedral and tetrahedral holes

Actually, I only found images of the spheres representing the atoms, no images of the shape of the space in between. If you squeeze some blue play dough between two layers of marbles, here is what you get (see also https://leancrew.com/all-this/2019/06/cheesegraters-and-crystals/):

Within a layer, each indentation is surrounded by six others. The arrows point at the six locations with the largest voids (in the center of triangles formed by the sphere and two of its nearest neighbors in that layer). If you look closely, each indentation has three holes, the contacts with the layer below. The voids take on different shapes depending what is underneath. If a sphere is directly underneath, it is a tetrahedral hole. The shape of this is a tetrahedron with "dented in" faces (far right). If a center of a triangle is directly underneath, it is an octahedral hole. The shape of this is a cube with "dented in" faces (near right; cube and octahedron are complementary archimedian solids: the centers of the faces coincide with the other's vertices and vice versa. A tetrahedron is complementary to itself).

If you shrink the spheres around the atoms to points, you get a space-filling arrangement of tetrehedra and octahedra called a tetrahedral-octahedral honey comb.

Teaching this material

Stefanie Lenzer, Bernd Smarsly and Nicole Graulich recently published a paper in J Chem Ed on using clear spheres to explore crystal packing and voids.


Why does Bismuth only create a +3 ion instead of a +5 ion? It has 5 valence electrons, but only loses 3 of them to make a +3 ion. Why does this occur?

I believe it may have something to do with how losing 3 electrons leaves you with the p sublevel full, but I do not understand why Bismuth is okay with that.

Isn't it just that +5 is a lot of (positive) charge for a single cation? Which other (transition) metals do form a $\ce{M^{5+}}$ cation? Just out of my mind, I can't think of any.

Typically, these high oxidation states exist in the form of oxo-anions ($\ce{CrO4-}$, $\ce{MnO4-}$, $\ce{VO4^{3-}}$, etc.) and bismuth is no exception here: $\ce{Bi(V)}$ exists as bismutate ($\ce{BiO3-})$.


Products of decarboxylation of sodium carboxylate My textbook says that the decarboxylation of sodium acetate with sodalime produces methane. But if we replace sodalime with barium oxide, pure methane is produced.

What is the difference between pure methane and the methane produced in the first case?

The first reaction is:
$$\ce{CH3COONa + NaOH -> Na2CO3 + CH4}$$
According to Encyclopedia of Inorganic Chemistry vol. 2:

The obtained methane in this process is not pure and may contain as
much as 8 % hydrogen and 10% other hydrocarbons such as the chloride
(to remove  acetylene) and concentrated sulfuric acid (to remove
ethylene and moisture). The purified  methane still contains hydrogen,
which may be removed as water by boiling at 110 °C the  excess oxygen
is absorbed concentrated sulfuric acid.

The second reaction doesn't make much sense to me with $\ce{BaO}$. If this reaction is happening in an aqueous medium, then $\ce{BaO}$ would probably have reacted to form $\ce{Ba(OH)2}$*. So I think you could justify representing the second reaction as:
$$\ce{CH3COONa + Ba(OH)2 -> CH4 + NaCO3 + BaCO3}$$
This is also how the aforementioned book represents the reaction, stating:

Heating sodium acetate with barium hydroxide, instead of sodium hydroxide forms nearly pure methane:
Sodium Acetate + Barium Hydroxide $\rightarrow$ Methane + Sodium Carbonate + Barium Carbonate
$$\ce{CH3COONa + Ba(OH)2 -> 2CH4 + NaCO3 + BaCO3}$$

So I guess the increased purity, in this case, is just due to the different amount of other products produced during the reaction.


In preparing sulphonilamide, why is aniline acylated if the acyl group has to be removed later? Our professor talked about the six stages of the synthesis of sulfa-drugs, which are:

nitration of benzene
reduction of nitrobenzene
acylation of aniline
reacting acetanilide with chlorosulfonic acid
treating the product with a primary amine
subsequent hydrolysis of acetyl groups

If we skip step 3, and let aniline react with chlorosulfonic acid, and then go till step 5, could sulfonamide be synthesized? Why do we perform acylation of aniline if we just have to remove the acyl group by subsequent hydrolysis later?

The main problem encountered during electrophilic substitution
reactions of aromatic amines is that of their very high reactivity.
Substitution tends to occur at ortho- and para-positions. If we
have to prepare monosubstituted aniline derivative, how can
the activating effect of –NH2
 group be controlled ?

This can be done by protecting the -NH2 group by acetylation with acetic
anhydride, then carrying out the desired substitution followed
by hydrolysis of the substituted amide to the substituted amine

The lone pair of electrons on nitrogen of acetanilide interacts
with oxygen atom due to resonance as shown below:

Hence, the lone pair of electrons on nitrogen is less available for
donation to benzene ring by resonance. Therefore, activating
effect of –NHCOCH3
group is less than that of amino group.

Moreover, in the
strongly acidic medium, aniline is protonated to form the
anilinium ion which is meta directing.

However, by protecting the –NH2
 group by acetylation reaction
with acetic anhydride, the reaction can be controlled
and the p derivative can be obtained as the major product.

source: word to word from
NCERT(INDIA) CLASS 12 chemistry ;)
http://ncert.nic.in/NCERTS/l/lech204.pdf


How can we accurately carbon-date samples from La Brea? TL;DR - given a sample of bone which has been encased in asphalt for tens of thousands of years and has become saturated with the asphalt, how can you get an accurate age for the specimen through carbon dating?

This past weekend I took my kids to the Page Museum at the La Brea Tar Pits in Los Angeles. While there we saw an astonishing number of fossilized Pleistocene animals. The interesting thing is that all of these fossils have a lovely (to my eyes, at least) brown color which is staining from the asphalt in which these bones have lain for thousands of years.

While there, we saw a presentation which explained that the asphalt seeps into the porous bones and it was explained that we know how old the specimens are through carbon dating. Which got me thinking - if the specimens are all highly contaminated with asphalt, how can they know that the sample they're testing contains carbon from the bone and not from the asphalt (which is presumably much older than the bone)? I did try asking one of the docents, but they're really not equipped to handle such a technical question.

According to Radiocarbon dating of petroleum-impregnated bone from tar pits at Rancho La Brea, California. Science  30 May 1969:
Vol. 164, Issue 3883, pp. 1051-1052 :

  Extensive bone dating was not possible before a method was devised for the
  removal of the impregnating petroleum...Bone is now routinely dated from its collagen


Why "RCA clean" solution has to be prepared every time (and not stored long-term)? "RCA clean" solution is a 5:1:1 mixture of DI water, ammonium hydroxide and hydrogen peroxide.

Why the ready 5:1:1 solution is not stored for long term, but instead has to be prepared each time immediately prior to usage?

Hydrogen peroxide is not a very stable substance. Over time, and however you store it or however carefully you prepare it, it will decay. So solutions of a given concentration need to be prepared fresh when you want to use them. You could try to store them but you won't know whether the concentration is still suitable for whatever you are going to do with it at some later time. And, the decay may well generate gas and other side-products of oxidation what may interfere with the use and may be unsafe.

Hence you would make the mixture fresh every time when you want to use it.


How can water exist in three states in freezing point? Freezing point is temperature on which vapor pressures of solid and liquid are equal. This is definition from my book. How can water ice and vapor of those co-exist when water that is only possible in triple point? How can vapor of those even  exist when in freezing point (melting point) there is equilibrium between solid and liquid? Is it when phases co-exist there is always  equilibrium between them question is only is it placed more towards one phase or the other?

If you only have a single component, water, present, then the liquid, vapor, and solid can only exist at the triple point.  If there is a second component present, say air, the phase rule tells us that there is an additional degree of freedom, pressure.  So, at a total pressure of 1 atm, the freezing point is 0 C (slightly lower than the triple point temperature), and you have liquid water, solid water, and, in the gas phase, water vapor mixed with air.


Why are some molecules unable to absorb infrared radiation? Methane $\ce{CH4}$ is symmetrical and non polar, and so is oxygen $\ce{O2}$. But $\ce{O2}$ doesn't absorb IR. Why is that? People say that if a molecule can change its polarity then it will absorb IR, but I don't understand how a molecule changes its polarity while absorbing IR.

Please try to keep your answer simple — I don't have a strong background in infrared spectroscopy.

In basic terms, for a molecule to absorb radiation there has to be an oscillating dipole being produced. This can occur by nuclear motion (vibrations, rotations) or electronic motion to produce electronically excited states. You ask about IR radiation, which does not have sufficient energy to produce electronic excitation, so this is ignored in the text below.

The restriction on producing a moving dipole exists because radiation consists of oscillating electric and magnetic fields and to allow energy to go from the radiation into the molecule, there has to be an interaction between radiation and the molecule and the origin of this is the oscillating molecule's dipole. (Of course the radiation also has to have the correct energy to bridge the gap between any two energy levels, but this is just another way of saying that the molecular dipole and the radiation oscillate at the same frequency, i.e. are in resonance.)

Thus in any homonuclear diatomic there is no IR absorption because these molecules have no  intrinsic dipole or one caused by vibrations. Also as they have no permanent dipole there is obviously no dipole generated when molecules rotate. Heteronuclear diatomics, HCl for example, all have a permanent dipole and one that is changed  by vibrational motion, thus there is a vibrational and rotational spectrum in all heteronuclear diatomics.

All homo-nuclear diatomics, $\ce{N2}$, $\ce{O2}$ for example, are transparent to IR radiation and this is of prime importance in global warming, since if these molecules obtain energy, say by collision with water or carbon dioxide that has been excited by IR radiation, they cannot radiate the energy away and so must heat the atmosphere by collisions with other molecules.

In a molecule such as $\ce{CO2}$ although it has no overall dipole it does have moving (oscillating) dipoles due to the way it vibrates. One mode is to stretch one bond at the same time as the other contacts (an asymmetric stretch), and so an oscillating dipole is formed and this interacts with radiation and $\ce{CO2}$ has an IR vibrational-rotational spectrum. The bending vibration is also IR active.

A similar effect occurs in methane, the way the molecule vibrates develops oscillating dipoles and these interact with the radiation and so absorb energy. Molecular group theory is usually used to sort out which of the $3N-6$ ways molecules can vibrate can have IR spectra.

There are complicated higher order effect even in homonuclear diatomics which may generate absorption but these effect are absolutely tiny and can be ignored for most every purpose. The relative magnitudes of electric dipole, quadrupole and magnetic dipole intensities are $I_\mathrm{dip}:I_\mathrm{quad}:I_\mathrm{mag} \approx 1:10^{-5}:10^{-5}.$

We can measure the vibrational frequency of homonuclear as well as heteronuclear diatomics by using Raman spectroscopy, which is a scattering effect (no photons are absorbed), and depends on the shape of the 'electron cloud' in the molecule. This is more properly called its polarisability.


Kinetic theory of gases The intermolecular distance between the molecules are much greater than the diameter of the gas molecules so we neglect their volume compared to the volume of container. "The rapidly moving particles constantly collide among themselves and with the walls of the container. All these collisions are perfectly elastic. This means, the molecules are considered to be perfectly spherical in shape, and elastic in nature". How then are molecules thought to be perfectly spherical in shape although its volume is neglected.

The basic explanation that you've recited for the Kinetic Theory of gases is trying to present a simple complete and consistent model which explains the observed behavior. Thus to give a consistent perspective it is easiest to think of gas molecules as elastic spheres. Of course, the assumption about the negligible size of the molecules does indicate that the shape of the molecules is irrelevant to the volume of the gas. But if you consider the molecules as having edges, or different moments of inertia about the three axis then the "simple" model gets too complicated for "simple" mathematics. Think of the all the additional variables that would be needed.

The basic model neglects other factors of course. In the real world gravity plays a part. The atmosphere decreases in density as the altitude increases. So the basic model is trying to illustrate the key factors necessary to understand the kinetic theory of gases, not all the factors.

Edit - The basic gas law PV=nRT has been modified by a number of expressions where correction factors have been used to make the equation more accurate and to give the equation more dynamic range. See some of the equations in the following wikipedia article.


Dipole moment of borazine What is the dipole moment of borazine? I tried googling it but can't find a reliable source. I searched in JD Lee too. It might have it but I could not find.

This NIST publication lists the dipole moment as zero, while drugfuture.com (I've never heard of this site) reports a value of 0.50 as measured in benzene at $\pu{25^oC}$.


Does the volume change when the reactions occur? I have this reaction (all substances are gaseous):

$$
\begin{align}
\ce{A + 4.5 B &→ C + 2 D + 2 E}\tag{R1}\\
\ce{C + 7.5 B &→ 8 D + 2 E}\tag{R2}
\end{align}
$$

Does the volume change when the reactions has occurred?
Because if so, the concentration changes too, right?

A volume change from a reaction occurs if the difference in mole of gases does not equal to $0$

In order to find the answer for your question, we will have to add both reactions together, resulting in:
$$\ce{A + 4.5 B + C + 7.5 B -> C + 2 D + 2 E + 8 D + 2E}$$
We can shorten the reaction by adding up all duplicates and removing both $\ce{C}$ from each side, resulting in:
$$\ce{A + 12 B -> 10 D + 4E}$$
Now if we calculate the difference in mole gasses we get:
$$\Delta n = \pu{(10 + 4)mol} - \pu{(1 + 12)mol} = \pu{1 mol}$$
Based on the positiv difference in mole of gasses we can say that the reaction will change in volume, that is it will expand.


Why are moles used in the ideal gas law and not simply mass? The crux of my question is, why is just the amount of particles included in $PV=nRT$ and there is nothing to account for the mass of the individual molecules? Or is mass accounted for in temperature, the average kinetic energy which includes mass in $\tfrac{1}{2}mv^2$.

Thanks in advance!

The overall notion is that gases exhibit "ideal" behavior so that the identity of the particular gas molecule doesn't matter. Thus regardless of what compound the gas molecule is, or if the gas is a mixture of different compounds, a mole of gas molecules at standard temperature and pressure (STP, 0 °C and $1.000\times10^5$ Pa) occupies a volume of 22.7 liters. So the equation for an ideal gas is

$$PV = nRT$$

In addition, there is the Maxwell–Boltzmann distribution which is for the speed of molecules. The gist is that regardless of the compound, gas molecules have the same kinetic energy spread. Since kinetic energy is given by $mv^2/2$, that means that lighter molecules have faster speeds than heavier molecules.

If you try to define the relationship by the molecular weight of the gas, then the value of the gas constant, $R$ would be different for different compounds and/or mixtures. Now in reality the ideal gas law works reasonably well, but there are empirical equations which fit the relationship between P, V, and T better. For those empirical equations the extra coefficients are fitted to the experimental data for particular gas compound under consideration.


Explanation for bond lengths in trans-hexatriene Hexatriene is an unsaturated hydrocarbon with six carbon atoms and five carbon-carbon bonds, three of which are double bonds.

However, the bond lengths of the $\ce{C=C}$ bonds are not the same. The middle $\ce{C=C}$ bond has a length of 137 pm while the $\ce{C=C}$ bonds at the end of the molecule have lengths of 134 pm, the length of a standard $\ce{C=C}$ bond. The two carbon-carbon single bonds are 146 pm long, also off from the standard 154 pm length of carbon-carbon single bonds.

Clayden's organic chemistry hints that the explanation has to do with the molecular orbits formed and the conjugation system in the molecule. However, I do not fully understand this explanation.

Why do these carbon-carbon bonds show this unusual bond length behavior? A thourough explanation using MO theory would be appreciated.

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

If you derive the π-type molecular orbitals of hexatriene, the three lower-energy MOs which are filled would look something like this (image from p 33 of Fleming's Molecular Orbitals and Organic Chemical Reactions, Reference Edition):

I suspect what Clayden is getting at is that in the second MO, there is some antibonding character between C3 and C4, whereas the C1/C2 and C5/C6 interaction is purely bonding.


Why is Cyclodextrin cavity hydrophobic? Actually this is going to be a list of interconnected questions. Please bear with me.

Wikipedia article on cyclodextrin (CD) says that it is hydrophobic from inside.

(1) How is it hydrophobic when there are so many hydroxyl groups on the interior surface ?

(2) This image
    labels the interir cavity as apolar rather than hydrophobic. I guess
    the apolarity is because of mutual cancellation of the dipole
    moments.
Am I right on this?

In any case I don't understand how the cavity excludes water.

There are reports on trapping organic substances such as naphthalene, toluene and crude oil in the CD cavity. Another report says that nitrosamime is trapped in CD by interacting with the hydroxyls.

(3) How does this trapping exactly occur ?

(4) Do the hydroxyls play a role in restricting the diffusion of the trapped organic molecules out of the cavity ?

All of this, I started investigating because of a recent news about an alcoholic beverage in powder form- palcohol. Initially I didn't believe it but later I found that alcohol is absorbed in CD to make a dry powder.

(5) What prevents alcohol from adsorbing on the exterior surface and not make the powder sticky ?

(6) Other studies say that there is a 1:1 ratio of trapping but palcohol claims that it is 58% alcohol by weight. Even this old patent says that CD absorbs a lot of alcohol. How does that happen ? In all it seems that CD behaves differently with different molecules.

Cyclodextrin hydroxyls are the external surface (hydrophilic and hydrogen-bonding).  They circle the larger open end.  The carbon and hindered ether skeleton is internal (hydrophobic).  A cyclodextrin looks like a bonded hollow detergent micelle in water.

Before you believe advertising hype, remember that underivatized cyclodextrins are nephrotoxic, crystallizing upon concentration, probably as the cholesterol endohedral complex.  Look it up.

Ethanols insert their hydrocarbon tails into the center, then hydrogen bond their polar ends with each other and the rim.


Why is there no salt bridge in electrolytic cell? I was just wondering why is there no salt bridge in electrolytic cell? It is said that electrodes are in a common electrolyte solution without salt bridge.
However, when I searched the image of the difference between electrolytic cell and galvanic cell, the salt bridge are in both of the cells.

The reference website http://mcat-review.org/electrochemistry.php

Electrolytic and galvanic cells have separation of electrolytes for different reasons. In galvanic cell you wouldn't have current if you mix electrolytes from both cells. In electrolytic cell you can dip both anode and cathode in the same solution. Then yes, you don't have a salt bridge because you use the same electrolyte.

Salt bridge passes ions at a certain rate that is not too high in order to prevent electrolyte mixing. In electrolysis you can allow mixing so you don't need a bridge.

In electrolytic cell you have two transports. Lets say you have Zn/ZnSO4 and Cu/CuSO4

(1) Electrons are flowing from more active (Zn) to more passive metal (Cu). This is essential for creating the electric electric current in the cell. In solution without a salt bridge you have constant concentrations of anion (SO4$^2-$). However, concentration of Cu2$^+$ is decreasing (Cu deposits on metal) and concentration of Zn${2+}$ increases (Zn dissolves). This creates charge imbalance that will quickly compensate difference in potentials of Zn/Zn${2+}$ and Cu/Cu${2+}$.

(2) You cave a salt bridge. It has to pass ions at a right speed. Not to slow and not too fast. If ions can travel too fast (no wall) then diffusion will mix electrolytes in two cells, copper will precipitate on both sides and there will be no current. If ions travel to slow (solid wall) then the charge difference on both sides of the electrolyte will prevent further current.

historically, it was just a piece of fabric between two compartments. Diffusion through it is slow, but drift along the electric potential is fast.


Would solar panels produce energy if uniform temperature in the universe? If the temperature on the surface of the sun and on the surface on the earth were the same, and the temperature of the universe were uniform, would the solar panels produce energy?

Solar panels work on the principle of photovoltaic effect. The only criteria for a voltage difference to be affected is that the light be incident on the solar panel. The energy of the incident light needs to be around that of the band-gap.
According to the question if both sun and the solar panel were of the same temperature, a solar panel on which light rays are incident would indeed generate a voltage. The band gap on the other hand depends on the temperature of the panel. Higher the temperature, smaller the band gap. Therefore, if the light incident on the panel is energetically close enough to the band gap (which depends on the temperature) the panel will generate a voltage difference.


Regioselectivity of electrophilic aromatic substitution in imidazo[1,2-a]pyrazine I am supposed to rationalise the regioselectivity of bromination of 8-chloroimidazo[1,2-a]pyrazine, as depicted in the scheme.

The left ring kind of looks like a pyridazine. If it's anything like a pyridine it'll be bad at electrophilic substitution without activation, so bromination at C–5 or C–6 is out. I guess it has to be one of the carbons in the five-membered ring. But why is it C–3 and not C–2? I am trying to draw out resonance structures and reason that way.

You are on the right track. Pyridine and similar nitrogen analogs of benzene are deactivated with respect to the electrophilic aromatic substitution reaction. You are also right that a resonance structure can help explain the regioselectivity of the reaction on the five-membered ring.

Electrophilic attack at C–2 (bottom line) directly gives an intermediate with non-aromatic resonance structure 2a. Moving electrons around, we are able to draw an aromatic resonance structure 2b, but this structure also contains a non-octet atom (carbocation), repulsive neighboring positively charged atoms, and charge separation.

On the other hand, electophilic attack at the observed site gives an intermediate 3 that maintains aromaticity in the six-membered ring and complete octets in all atoms.

We can predict that the intermediate arising from attack of the electrophile at C–3 would be more stable than the corresponding C–2 substitution intermediate. Since this step should be endothermic, the rate of this step will be influenced by the stability of the resulting intermediate (Hammond's postulate). From there, we can proceed through the usual addition/elimination mechanistic sequence to get to the product.


What is a secondary shell in an enzyme? I've searched the internet for a definition of what the different shells in enzymes/proteins are and haven't found a good answer. This Nature article, which is about an enzyme that catalyzes Kemp Elimination, has the sentence:

  The Val12Met and Phe77Ile mutations, also in the second shell, cause changes to internal cavities that cause a slight rotation in the backbone at Trp50 that also favors the alternative configuration.

What is the second shell, and how many different shells are there?

The authors mention three residues that they consider "second shell", 12, 77, and 102. They are shown in the figure below in green:

Residues Trp50 and Glu101 are directly involved in ligand binding, shown as space-filling in yellow. As you can see, they are part of the central beta barrel. The residues in green are also part of the beta barrel, but the side chains are not involved in ligand binding because they point to the outside. For that reason, the effect of mutating these residues on ligand binding must be indirect by repositioning the residues that point inside.

  What is the second shell, and how many different shells are there?

This is not a well-defined concept. The term is more common for hydration of ions (first shell of water molecules makes direct contact, second shell contact first shell and so on), or in the context of coordination spheres (https://en.wikipedia.org/wiki/Coordination_sphere). When researcher write papers, they come up with descriptions that fit their structures. A famous example is to describe polymerase active site secondary structure as hands, with a palm, thumb and fingers, see e.g. https://marlin-prod.literatumonline.com/cms/attachment/c9a1e727-5b4b-45ce-81c8-8a24c1aea2ef/gr1.jpg. Some of these descriptions become common (such as the beta-barrel mentioned above) and some do not (my research advisor once described an enzyme as hat-shaped - it did not stick, maybe because hats come in many different shapes and forms, and barrels are more defined).


What qualifies to be considered as a lone pair-lone pair repulsion? Referring to, NCERT Chemistry Part I, Textbook for Class XI[1].

Here the text puts forward three structures of $\ce{ClF3}$.
My teacher said to me that (b) structure is unstable due to lone pair lone pair repulsion while (a) is the most stable of the given structures as there are minimal lone pair and bond pair repulsions and no lone pair lone pair repulsion.
That got me thinking what actually qualifies as a lone pair lone pair repulsions. Is it when the lone pairs are at an angle equal to or less than the 90° or is it something else?
Reference:
[1]: Chemical Bonding and Molecular Structure. NCERT Chemistry Part I, Textbook for Class XI, First Edition(October 2019 Reprint); Uppal, Shveta, Banerjee, Binoy, Eds.; National Council of Educational Research and Training: New Delhi, India, 2006; pp 113.

When applying VSEPR, imagine that the various electron pairs occupy "orbitals", in essence volumes shaped like balloons, protruding from the central atom (in this case chlorine). The electrons ("orbitals") repel each other.
Which repulsive interaction is strongest?
One assumes that, placed in equivalent relative positions (angles), LP-LP (LP=lone pair, BP=bonding pair) repulsion is strongest, followed by LP-BP and then BP-BP. The idea is that bonding electrons are more distant from the central atom, whereas nonbonding (LP) electrons are held tightly by the central atom and therefore represent a more "concentrated" charge, and therefore repel other electrons most strongly. The repulsion is also stronger the smaller the angle between the electron pairs.
The strength of repulsion determines the order in which to prioritize the interactions, but this can be a little tricky, as your example shows, because it is a "non-linear" problem. When more than one pair of LP is involved, the geometry should minimize LP-LP repulsions followed by LP-BP repulsions, but there is a tradeoff. While (c) places both LPs on opposite ends ($180^\circ$) and would seem to minimize the repulsion, there is a bit of a tradeoff between LP-LP and LP-BP repulsions. Geometry (c) results in a lot (6) of LP-BP repulsions at right angles ($90^\circ$), whereas geometry (a) has 4 LP-BP repulsions at right angles. This leads to positioning of the LPs in equatorial (!) positions, even though LP-LP distance is not maximized.
Such arguments are not transparent, they are based on a crude method that only approximates the "real" electron distribution. The experimental geometry is in fact a distorted version of (a), with the LP-BP angles slightly greater than right angles ($>90^\circ$).
My practical advice is to simply remember that according to VSEPR lone pairs are always placed in the equatorial plane in trigonal bipyramidal electron pair geometries (those with 5 pairs). See for instance the following table:


Calculating nearest neighbor distances For cubic materials, there are equations in which to calculate nearest neighbor (NN), second nearest neighbor, etc. Can this be done with tetragonal crystal structures? I want to calculate NN, 2NN, and 3NN of $\ce{TiO2}$ rutile with a tetragonal crystal structure but am unsure how to do it. Any suggestions/help would be appreciated!


Why do electrons jump back after absorbing energy and moving to a higher energy level? Electrons in a shell absorb energy and move to higher energy levels, but they release their energy and jump back to the shell they originally were in. Why do they jump back? Why can they not keep revolving around the nucleus?

This is a very fundamental question and for really understanding the "why" some advanced physics is involved. I will describe the process rather superficially.
As you might know, the level energies of atoms and molecules can be calculated (in principle) using quantum mechanics. The simplest system is the hydrogen atom as it consists of a single proton and a single electron. Ignoring higher order effects (such as interactions of electron and nuclear spins and QED effects), the quantum mechanical calculation gives the same result as the Bohr model, that is, the level energies of hydrogen are given by the Balmer formula, which you probably know.
The calculation does not predict that the excited levels fall back to the ground state. An electron in an excited orbital will, according to this calculation, always stay in this orbital if nothing happens to the system.
Because we know that excited states decay, something must happen to the system to induce the decay. It turns out that in our calculation we have ignored the interaction of the atom with the photon field. Atoms can absorb light and emit light and we have completely neglected this. If you do take this coupling into account, you will find that excited states have a limited lifetime and that an excited electron will fall back to a lower level eventually.


Experimental determination of pH I am trying to determine the experimental pKa for two weak acids that were titrated against 0.20M NaOH.

I have read elsewhere that you can take the point where the graph becomes steep and divide the value of base added by two the corresponding pH value would then be the pKa, but how do i choose which value since it may not be obvious which point the graph becomes steep.

Below I can see that the pka for acetic acid should be close to the theoretical value calculated of 4.76 and the Tris-HCl pka should be approximately 8.3 but there must be a better way than just guessing from a graph.

My textbook doesn't explain how to experimentally find pH just that it's the point where $[A^-]/ [HA]$. I am hoping someone can give me an equation to work with or guide me in the right direction.

Thank you,

So I think what you heard is about the right idea. The flat region is your buffering region, and isn't super helpful to deduce the pKa. Adding base consumes the weak acid, and because it is weak, you know you are mostly consuming the [HA] form, and equilibrating back to around the pKa. This is why your pH doesn't change much around the pKa.

As an example, if you had 10 mmol of HA to start, then at the pKa point you would have 5 mmol of HA and 5 mmol of A-. Now, if you keep adding base, at some point you will essentially consume the remaining 5 mmol of HA. Then, there will be negligible amount left and it can't buffer anymore - i.e. your pH will change rapidly because you are adding strong base. Here you will have ~0 mmol of HA, and ~10 mmol of A-. Note that you have twice the amount of A- now. The pKa will have been at the point where you had half of this.

Experimentally, I know two simple ways. Using a pH meter and no indicator, you have to measure the sharp region more carefully (i.e., drop by drop), because you want to be able to find the exact point where the pH changes the fastest. By approximating the derivative, i.e.

$$ \frac{d}{d(V_{base})} pH(x_i) \approx \frac{pH(x_{i+1})-pH(x_{i-1})}{x_{i+1}-x_{i-1}}$$
you can find the point of greatest change, or the equivalence point.  Obviously, better data in that region will give you more clarity. I used to do one titration quickly so that I would know the approximate region where I would need to be and then slowly titrate the rest.

The other method, i.e. using an indicator solution and then just very carefully reaching the equivalence point essentially gives you the same information, and I know people who can very quickly do this. But in general, this would not be as accurate.


What is density in density functional theory? This may be dumb question. Please bear with me.

Density functional theory (DFT) is a successful theory for electronic structure calculations of materials. In DFT, electron density is the fundamental variable that is proven by the Hohenberg-Kohn theorems. However, I find it difficult to understand how density is used to calculate the energy of the system.

In general, number of electrons per volume will provide the electron density. Okay. Now, I consider Hydrogen atom (for simplicity). If I calculate the electron density, in fact, it depends on how much volume is chosen. If I chose a unit volume, its charge density is 1e. If I double the volume, the charge density halves ($\frac{1}{2}e$). What decides this volume? How to chose this volume?

What is density in density functional theory?

The density in density functional theory (DFT) refers to electron density, which is the object of all DFT calculations. DFT differs from various other wavefunction-based methods which focuses on determining the ground-state $\ce {3N}$-dimensional wavefunction*, such as the Hartree-Fock (HF) method. Electron density, often denoted as $\rho$ or n(r), refers to the 3-dimensional function of 3 spatial coordinates describing the distribution of electrons in a given system. It is of interest in DFT as it allows us to determine all observable properties of the system without knowing the exact ground-state wavefunction, hence providing great computational ease. The following is the usual expression of electron density (for spin-restricted calculations, for a system of an even number of $\ce {N}$ electrons, which would have $\ce {N/2}$ different electronic energy levels since electrons in the same orbital are considered to have the energy, thus also meaning that we are summing only $\ce {N/2}$ of the different individual densities, and then multiplying by 2) that you would find in resources on DFT:

*$\ce{N}$ refers to the number of electrons in the given system. The wavefunction is a function of $\ce {3N}$ dimensions since each electron has 3 spatial coordinates. If spin is considered as a fourth coordinate, then the wavefunction becomes $\ce {4N}$-dimensional. Obviously, such a wavefunction would not be easy to solve for exactly.

  However, I find it difficult to understand how density is used to calculate the energy of the system.

This is where the functional in the name DFT comes in. The 1st theorem of Hohenberg and Kohn tells us that:

  Ground-state energy of the system is a unique functional of its electron density.

What exactly is a functional? A functional is a function of a function. A commonly used functional would be the definite integral. In Kohn-Sham DFT (which treats the system as comprising of non-interacting particles moving in an effective potential), notably the most widely-used version of DFT (another version being orbital-free DFT), the functional can be seen as comprising of three main contributions: 1) kinetic energy of electrons, 2) coulombic interactions between electrons and nuclei, and 3) coulombic interactions between pairs of electrons. However, there are some effects that we would need to take into account. For example, when we consider coulombic interactions of each electron with the electron density (in solving the Kohn-Sham equations), we also include a non-physical self-interaction (i.e. the interaction of the electron with itself). Particles in the system can also undergo exchange interactions and their motions are also correlated (i.e. they are not independent). Hence, another term called the XC (exchange-correlation) functional is also included into the overall density functional expression to account for all these additional effects that have not already been accounted for.

The following is an expression of the Kohn-Sham density functional taken from Wikipedia:

The first term is the electron kinetic energies, the second represents the electron-nuclei interaction energy, the third represents the electron-electron interaction energy, while the last term is the mysterious XC functional.

The site has more posts on the topic which you may find useful:

Are there any full worked examples of DFT calculations?

Density Functional Definition


What is the entalphy change with respect to pressure at Isothermal Process for *Real Gases*? How much enthalpy of argon changes when the pressure is isothermally increased by $\pu{1 atm}$? (Gas should be considered as real gas under specified conditions) ($\pu{1 mol}$ of Argon gas is available at $\pu{25 ^\circ C}$ and $\pu{11 atm}$ conditions)

-Firstly I took the total derivative of Enthalpy function of pressure and temperature:

$$dH =\left(\frac{∂H}{∂P}\right)_T \cdot dP + \left(\frac{∂H}{∂T}\right)_P \cdot dT$$

Then divided terms with $dP$ because as the question asks I need $\left(\frac{∂H}{∂P}\right)_T$. After this I get:

$$\left(\frac{∂H}{∂P}\right)_V = \left(\frac{∂H}{∂P}\right)_T + \left(\frac{∂H}{∂T}\right)_P \cdot \left(\frac{∂T}{∂P}\right)_V$$

Third term is enthalpy change with temperature at constant pressure means $C_p$, so the equation turns into:

$$\left(\frac{∂H}{∂P}\right)_V = \left(\frac{∂H}{∂P}\right)_T + C_p \left(\frac{∂T}{∂P}\right)_V$$
so I rewrite this like:
$$\left(\frac{∂H}{∂P}\right)_T =\left(\frac{∂H}{∂P}\right)_V-C_p\left(\frac{∂T}{∂P}\right)_V$$

Then I used Chain rule, which is: if $z=f(x,y)$:
$$\left(\frac{∂z}{∂y}\right)_x\left(\frac{∂y}{∂x}\right)_z\left(\frac{∂x}{∂z}\right)_y=-1$$

I used this rule for writing a derivative term, $\left(\frac{∂T}{∂P}\right)_V$, with constants, and my equation turns:

$$\left(\frac{∂H}{∂P}\right)_T = \left(\frac{∂H}{∂P}\right)_V - C_p\left(\frac{\kappa T}{\alpha}\right)$$
where $\alpha$: expansion coefficient; and $\kappa T$: isothermal compressibility coefficient (kappa t).

In this way, I came up to this point but I couldn't find the exact answer. I Lost myself in thermodynamic differentials. Can anyone explain me? Where did I mistake or is my solution method incorrect?

Edit:

$$\left(\frac{∂H}{∂P}\right)_T = V - T\left(\frac{∂V}{∂T}\right)_P$$

If i wasn't wrong rhs of the equation turns into = V - T ( ßV) which equals to partial derivative of enthalpy with respect to pressure at constant temperature. But i'm confused here. There are other terms that can be used for ßV. For example alphaV or Vkappa t. Which one should i use? What is partial derivative of volume with respect to temperature at constant pressure means? Question asked about isothermally increasing pressure so should i use isothermal compressibility coefficient or isobaric compression coefficient or thermal expansion coefficient or what? Really confused here..

$$\left(\frac{∂H}{∂P}\right)_T = V - TßV$$ this is the last point i came to

The equation for the partial derivative of enthalpy with respect to pressure at constant temperature is given by:  $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P$$From the equation of state for your gas, you can evaluate the rhs of this equation.


Writing a chemical equation with a reactant of unspecified chain length Would appreciate it if I got some general tips and hints and not the solution cause I'm trying to learn from this question :)
The question is as follows:

FA1 is a solution containing $\pu{5.00 g dm-3}$ of a dibasic acid, $\ce{HOOC-(CH_2)_n -COOH}$.
FA2 is a solution containing $\pu{5.00 g dm-3}$ of $\ce{NaOH}$.
$\pu{25.0 cm3}$ of the acid solution FA1 needed $\pu{17.0 cm3}$ of the sodium hydroxide solution, FA2, for complete neutralization.

Determine the value of $n$ in the formula of the acid.

The issue I have is in starting the question. I genuinely have no clue how to start a chemical equation with something like $\ce{HOOC-(CH2)n -COOH}$. Tips and hints would be greatly appreciated thanks.

Tips and hints would be greatly appreciated.

My tips and hints for you (I apologize for if my tips and hints direct to me treating OP as a novice):

Your dibasic acid is $\ce{HO2C-(CH2)n-CO2H}$. Thus, $\ce{A^2-}$ in the equation $(2)$ of Karsten Theis's answer is $\ce{^-OOC-(CH2)n-COO^-}$.
Molar mass of this dibasic acid is: $2 \times (12+32+1) + 14n = 90 + 14n$
Find the concentration of FA2 solution in $\pu{mol dm-3}$ (in other words, in $\pu{M}$. Hence, you can find the the amount of $\ce{NaOH}$ (in $\pu{mol}$) in FA2 solution ($\pu{17.0 cm3}$).
Determine the amount of dibasic acid (in $m_\ce{H2A} \ \pu{ mol}$) in FA1 solution ($\pu{25.0 cm3}$) using the equation $(2)$ of Karsten Theis's answer.
Now, you can find $n$ using following equation:
$$m_\ce{H2A} \ \pu{mol} \times \pu{(90 + 14n) g mol-1} = \pu{5.00 g dm-3} \times \pu{0.025 dm-3}$$

Since you have already calculated $m_\ce{H2A}$ in step 4, you can calculate the $n$ $(\pu{1.0 dm-3} = \pu{1000 cm-3})$.
Note that $n$ is an integer so final answer should be rounded to closest integer.


Successive ionization energies (work functions?) for metals Are there tables for successive ionization energies for metals (like for gases)? All I can find are work functions to remove one electron from the surface of a metal into vacuum (and to keep it there). However, I guess it gets harder to remove electrons the more electrons already have been removed, right?

Maybe there's a way to relate that energy to the average degree of ionization of the material. Is it possible at all?

Finally, is it harder to ionize a metal gas or a metal solid? Although most work functions are a little bit smaller than the (1st) ionization-energies of the same material in gaseous state but I still think ionization of solids is harder. Am I correct?

In an infinite metal solid, there are an infinite number of electrons, so each one removed will be at the exactly identical work function with no change in energy.

Intuitively, this may seem a bit strange because .. well, it should be harder to remove an electron once positive charges start to accumulate.

If I have even a micron-sized piece of metal, that's ~$10^{13}$ atoms. So you really have to do a lot of ionization before you're even going to get 0.1% of the metal ions.


Synthesis of phosphonate salt not working I have tried four times in the past few weeks to synthesise 2-ethoxyvinyl phosphonyl dichloride by reacting ethyl vinyl ether with phosphorus pentachloride in toluene.

I have been following the literature procedure (see below) successfully up to the point where it says that I need to bubble sulfur dioxide through the reaction mixture - at which point it will turn clear. The reaction turns a pale yellow color; the next step is where I'm at my wit's end - the literature says to obtain the 2-ethoxyvinyl phosphonyl dichloride by direct distillation and every time I've tried this, it's resulted in a black, opaque solution. The literature does not say what color the final product is although my supervisor has said that it should be a yellow oil. I'm thinking that it could have decomposed upon heating, although the boiling point according to the literature is 142 °C and the bp of toluene is 110. Previously I tried the same reaction in $\ce{CCl4}$, which distilled off at 76 °C and also turned black.

I would really appreciate help on this, since it's part of my final year project and I only have a handful of lab sessions left to finish this. Thank you in advance! (Also, I've attached an image of the literature procedure below if it might be helpful).

$%edit$

Looking at the procedure (and considering the structure of the desired product) it appears that its boiling point is 142C at 27 mm (not at atmospheric pressure). So you should try a vacuum distillation to purify your product. Heating it to try to distil it at 1 atm is probably causing decomposition. Also, be sure to protect the reaction mixture and the final product from atmospheric moisture.


Aldol Condensation with Cyclic Molecule How does one solve this mechanism? Why doesn't the negatively charged oxygen on the non-cyclic molecule remove an alpha hydrogen from the cyclic one?

It certainly could undergo the acid-base equilibrium you described, if there's enough energy in the system. Your problem is omitting an important piece of information, which is the temperature at which the reaction is taking place. Lithium enolates (the right compound) are usually formed at low temperatures, down to -78 C. The hope/assumption is that the aldol reaction happens faster than the acid-base reaction between the two reactants. This is certainly hard to control in this particular example, and thus the yield of the reaction is not great, according to this example mentioned in a review article on aldols:

Furthermore, it's likely that in doing this reaction, the experimentalist would form the lithium enolate separately before mixing it with the cyclohexanone. You can read about the importance of order of addition here: https://en.wikipedia.org/wiki/Aldol_reaction#Order_of_addition

Here's the mechanism, note that the problem didn't provide you with a viable source of H+.


Advantages of Travelling wave IMS and Drift tube IMS I am learning Ion Mobility spectrometry. I am sorry for a very fundamental question. After reading different articles, I really did not understand what is the advantage of Travelling wave IMS over Drift tube IMS (or vice versa). Can anyone explain very fundamentally, what are the adavantages/disadvantages of TW IMS over Drift Tube IMS (and vice versa)?

In a nutshell, since other sources probably would give much more accurate and complete answers:

Advantages of DT-IMS over TWIMS:

Direct access to absolute collision cross sections. This is a major advantage considering the difficulty to provide accurate CCS calibration scales for TWIMS measurement as can be seen in the recent litterature.
Separative resolution is usually reported as higher, although recent work by Dick Smith's group at PNNL seems to show that it is mostly related to separation device lengths which are different.

Advantages of TWIMS over DT-IMS:

Reduced ion losses and increased sensitivity. The stacked-ion ring basis leads to reduced drift towards the sides. This advantage has been reduced by the addition in several DT-IMS setups of ion funnel electrodes which focus the ion trajectories upon reaching the end of the drift tube.
No need for large potentials and thus large voltage generators and all the related problems. To reach a given electric field, the potential difference does not need to be applied to the whole drift tube.


Could neutralisation of safe chemicals produce dangerous ones? I'm basically thinking whether I could potentionally poison myself while neutralising strong acid/base by some weak base/acid when I accidentally spill it, touch it or when I'm cleaning reaction product.

Combinations of chemicals that are non-toxic in my opinion:

$\ce{HCl + NaHCO3}$
$\ce{H2SO4 + NaHCO3}$
$\ce{NaOH + CH3COOH}$

None of these are poisonous. I just thought these up, for example the first one produces table salt, $\ce{NaCl}$.

But could there be a combination of non-toxic acid and base that would produce some nasty salt when neutralised? Something worse than what I would be getting rid of?
Is there a combination where I should better just wash myself with water?

While not an acid-base reaction, an example of mixing two relatively "safe" chemicals to form dangerous ones is bleach (sodium hypochlorite) plus ammonia - both common household chemicals. This reaction yields chloramine, dichloramine, and nitrogen trichloride, all more toxic and hazardous than bleach or ammonia.


Why would sulfuric acid turn gray/black? About 2 years ago I purchased 2 liters of 98% concentrated sulfuric acid. It's in a standard poly bottle with leakproof lids, and originally was clear. Recently I noticed the acids have turned a dark gray; almost black color. They have both never been opened.

I suspect there was something in the lid that was eroded by the acid, but is there any other reason sulfuric acid would darken over time?

The only reason I can think of that would cause the solutions would blacken over time is if the solutions reacted with organic substances. Sulfuric acid is an extremely strong oxidizing agent, and dehydrates many organic compounds to produce carbon in the form of graphite, and water. For example, the reaction of sulfuric acid with paper (cellulose) is:

$$(\ce{C6H10O5})_{n}\ce{(s)} \xrightarrow{\ce{H2SO4}} {6n~\ce{C(s)} + 5n~\ce{H2O(l)}}$$

and can be seen in the following photo:

If a considerable quantity of organic material was reacted in your solutions, the $[\ce{H+}]$ in your solutions would actually decrease, so I would recommend titrating one of your solutions to see if this has changed since you bought them.


Peak Wavelength of Emission Spectrum of a Flame vs. Actual Flame Temperature I am looking at a lab test of two variants of mixed hydrocarbon gas (57.8% $H_2$, 16.9% $N_2$, 12.4% $CH_4$, 1.11% water vapor, and 11.8% "unknown" gases)  burning in oxygen which shows a discontinous emission spectra. The authors are claiming that since their graph shows a peak wavelength at 471 nm, the gas therefore has a peak flame temperature at 6132 K / 5859 °C / 10578 °F according to Wien's Displacement Law, which is absurdly higher than the adiabatic flame temperature of any known hydrocarbon gas.

Now, I know that the spikes shown correspond to the $C_2$ and $CH$ Swan Band. Can someone provide a detailed explanation as to why this is not an accurate measurement of flame temperature?

I should also mention the graph only ranges from 200 nm to 1100 nm, and that they fitted the emission spectrum to a blackbody curve. Could the peak wavelength actually be further into the infrared spectrum? Is fitting the emission spectrum to a blackbody curve even appropriate for a non-blackbody emitter such as a flame?

There seems to be a number of problems here.

spectrometers need to be calibrated in intensity on a wavelength-by-wavelength basis. Each part of the system has an efficiency which can vary with wavelength, but the grating response tends to be one of the strongest. Here is a random example of a grating efficiency curve from a Thorlabs catalog :

You can see right away that a grating efficiency curve can impose it's shape on the measured spectrum, generating a shape superficially similar to a blackbody, but totally instrumental. Without calibration for efficiency, you just have ADC counts vs wavelength raw data!

Fitting of a Plank shape should be done to the spectrum without ignoring the peaks. In other words, a good fitting program will fit the shape you are looking for (Plank) PLUS fit the peaks at the same time. And don't forget to calibrate intensity first (see 1.)!

" Is fitting the emission spectrum to a blackbody curve even appropriate for a non-blackbody emitter such as a flame?" Probably not unless you really know what you are doing. Flames are beautiful and complicated. There's thermal emission from gas in addition to peaks, but stronger is the blackbody-like emission from hot soot particles. Since there will be a distribution in temperatures within the 3D volume of a flame, you have a spectrum recorded from a significant fraction of the flame volume will include a variety of Plank spectra corresponding to different temperatures.

So the answer is: you have to do fitting, remove peaks by fitting them simultaneously, and more importantly, you must calibrate the spectrometer intensity, because grating efficiency curves look strangely Plank-like, althouth they fall to zero much faster above and below their useful wavelength range. Also have to deal with considering the sample volume of the flame and whether or not it's true black-body emission or not.

Those "peak" values, are closer to peak efficiency wavelengths of the spectrometer, and not really related to Wein's displacement law at all!


Acetal Formation Mechanism In this example, why is there is a ring formation with the attack of the OH from the intermediate itself rather than a continuation of addition? I know my proposed product is wrong, however I just want an explanation as to why this occurs.

Intramolecular reactions are always preferred over intermolecular.

Remember that these reactions are reversible so even if the intermolecular attack proceeded, the product would be unstable with respect to intramolecular attack by the free OH that is within the molecule.


Is the correlation energy of a system always negative? The correlation energy of a system is defined as the difference between the exactly energy and the energy in the Hartree-Fock method: $E_\mathrm{cor} = E - E_\mathrm{HF}$. In the case of an atom or a molecule, is it possible say that $E_\mathrm{cor} < 0$, since the interaction between the electrons is positive?

The answer to this question rests on the fact that Hartree-Fock is a variational method.
A variational method is one where you supply a guess wavefunction ($|\Psi\rangle$) with some parameter(s) ($\alpha$) and then use those parameter(s) to minimize the energy. Doing this, you are guaranteed that your resulting energy is an upper bound to the exact ground state energy.
$$
E_{\text{exact}} \leq \frac{\langle\Psi_{\alpha}|\hat{H}|\Psi_{\alpha}\rangle}{\langle\Psi_{\alpha}|\Psi_{\alpha}\rangle}
$$

So, if we are talking about the exact correlation energy (such as one given by full CI) then the answer is that the correlation energy must always be negative. This is  because Hartree-Fock always provides an upper bound to the exact ground state energy. However, if we use a non-variational method to approximate the correlation energy (such as coupled cluster) then we could find cases where we calculate a positive correlation energy, but this is distinct from the exact correlation energy.

(For an authoritative description of this in the context of quantum chemistry with examples and practice problems, I recommend the first chapter of Szabo and Ostlund's Modern Quantum Chemistry)


What is the correct net ionic equation for the neutralization of phosphoric acid? What is the net ionic equation for the reaction between phosphoric acid and sodium hydroxide?

I came up with
\begin{align}
\tag1 \ce{H+ + H2PO4- + OH- &-> PO4^3- + H2O},\\
\tag2 \ce{H+ + OH- &-> H2O}.
\end{align}

Or is it, since phosphoric acid is a triprotic acid, whose ionization is
$$\ce{H3PO4 <=>3H+ + PO4^3-},$$
therefore the net ionic equation will be
$$\ce{3H+ + 3OH- -> 3H2O}?$$

Phosphoric acid dissociates into hydrogen ions and $\ce{H2PO4^-}$ ions in water.  This then can dissociate further, but $\ce{H^+}$ and $\ce{H2PO4^-}$ are the primary ionic species you'd find in a phosphoric acid solution.

$\ce{NaOH}$ dissociates completely into $\ce{Na^+}$ and $\ce{OH^-}$ ions.

The reaction would then be between $\ce{H^+}$ and $\ce{OH^-}$, your option 2.

The difficulty arises because adding more $\ce{NaOH}$ to the reaction pulls off the second then the third hydrogen from the phosphoric acid.  In all cases it is an $\ce{H^+}$ reacting with an $\ce{OH^-}$ ion with the phosphate ions as spectators, but that really doesn't tell the story of the reaction well.  This is a case where the net ionic equation really isn't your best option.emphasized text


Predicting products of chemical equations? Consider the equation $\ce{2 NaOH (aq) +Cl2 (g) —> NaOCl (aq) + NaCl (aq) + H2O (l)}$. How would one know the products of this reaction? Is there a general way of doing this? I would have no earthly knowledge of predicting this. Is it that we would look at each molecule and decide which goes with which on the basis of octet? Also, why does it just so happen that 2 moles of $\ce{NaOH}$ react with 1 mole of chlorine? What would happen if it was a 1:1 ratio?

One would not "know" what the products of the reaction are, a priori. As mentioned by @IvanNerenten, the way you know is by studying chemistry. In fact, the reaction is different if you use cold NaOH vs hot NaOH. The reaction as written is with cold NaOH and is a disproportionation reaction.This is where the chlorine is both oxidized and reduced in the same reaction.


What is the "blue spirit" (or "red spirit") in mercury-free thermometers? I surmise it's some kind of alcohol, but is there a standard chemical used for these?

One of them just exploded in my students' lab! It was the kind with blue fluid in it.

As per the Wikipedia page on Spirit Thermometers:

  The liquid used can be pure ethanol, toluene, kerosene or isoamyl acetate, depending on manufacturer and working temperature range. Since these are transparent, the liquid is made more visible by the addition of a red or blue dye.

1,4-bis(butylamino)anthraquinone - commonly known as "Oil Blue 35" - may possibly be used as the blue dye.


Why do we assume 100% ionization for acetic acid (a weak acid)? This is about a titration. The base titrant used is sodium hydroxide.

You do not assume 100% ionization of acetic acid during titration. Acetic acid is a weak acid. Look at this equilibrium

$\ce{CH3COOH <=> H+   + CH3COO-}$

The base, $\ce{NaOH}$, keeps consuming $\ce{H+}$ during a titration, as a result acetic acids keeps on providing $\ce{H+}$ to maintain the equilibrium in water until all the acetic acid is converted into acetate ion.


How do phase transfer catalysts bring anions to organic phase? Many anions are quite stable in the aqueous phase, but not so much in organic solvents where its necessary to perform some reaction (usually nucleophilic substitutions). I have heard that crown ethers are good phase transfer catalysts that bring ions to the organic phase. 18-crown-6 is used to coordinate to $\ce{K+}$ ions and bring them to organic phases.

While the cation can enter the organic phase, how does the anion enter it then? Does the anion simply "tag along" with the crown ether complex?
 The anion gains no solvation stability in organic phase, so the anion moving from aqueous to organic phase seems unlikely.

So my question is: How does the phase transfer catalyst ensure migration of anions to the organic phase?

Actually, anions are not unstable in organic phases per se. There is no intrinsic physical property of organic phases that would make anions inherently unstable: the main problem is that most heteroatoms in organic solvents are more electronegative than carbon and therefore carry a slightly negative charge which attracts cations more easily than anions.

Most organic phases are best classified as ionophobe, however. They will not like cations or anions in their vicinity. Remember that organic phases are typically mainly hydrocarbons with the odd heteroatom, so they have a low polarisablility. Water, on the other hand, is a highly polarised dipole and able to stabilise both cations and anions easily with the appropriate side of the water molecule.

Now if we add a phase transfer catalyst to a biphasic mixture of water and an organic solvent — typical examples would be $\ce{Bu4N+ I-}$ or crown ethers — then these will readily transfer themselves into the organic phase. In the case of crown ethers, it is because they are basically already like the organic solvent: a hydrocarbon with a few heteroatoms. Crown ethers can, however, piggyback a coordinated cation at practically no extra cost: the cation is directly solvated by the crown ether’s oxygens. In the case of the tetrabutylammonium cation, the charge of the tetravalent nitrogen is somewhat shielded by the four butyl residues which interact much better with other organic residues than with water.

This means that we have a general positive charge in the organic phase due to the crown-ether solvated cations or tetrabutylammonium cations present in there. This in turn draws anions towards the cationic charge due to electrostatic interactions.

We can now also look at some basic principles explaining London interactions: generally, the more disperse the orbitals of a species are, the easier this species can be polarised and thus the better London interactions can be. Cations are species in which there is an overall deficiency of electrons so orbitals are drawn closer to the nuclei. Anions, on the other hand, have a surplus of electrons, meaning that these will repel each other more and orbitals will be more diffuse. Compare neutral atom radii to ionic radii to confirm this. The more diffuse anions will therefore have an easier time forming London interactions with the hydrocarbon solvent than the less diffuse cations.

Thus, in summary, one typically needs the ion pair to cross the phase boundary. An anion might transfer by itself but the cation needs help; thankfully, cations can be transferred easily with crown ethers or tetrabutylammonium compounds because these are more able to interact well with the solvent.


What is the origin of rotational energy barrier in ethane? What is the origin of rotational barrier in ethane?
In most organic chemistry textbooks, we learned that it's originated from steric repulsion between two methyl groups. But recently I heard that there also exists other perspective which explains rotation barrier using hyperconjugation model.
So I searched and read many papers about the origin of rotational barrier in ethane and I notice that there are two competing models for explaining this rotation barrier.
But there are no significant conclusions and it still remains controversy "what is the primary energy term in rotation barrier?".
Because many authors suggested many calculations to prove what is mainly contributed to rotation barrier and there results for supporting there argument, but each calculation is based on different approximations and theories. (e.g. NBOs are not optimal --> hyperconjugation energy is overestimated)
So my question is that "Is there no conclusion? What factor mainly contributes to ethane rotation barrier?"

As you say, this issue appears to be unsettled. I find the argument in favor of hyperconjugation compelling, but there is theoretical evidence that steric strain contributes as well.
This debate is about 20 years old. Here are two papers summarizing the opposing views:

Pophristic, V.; Goodman, L. (2001). "Hyperconjugation not steric repulsion leads to the staggered structure of ethane". Nature. 411 (6837): 565–8. doi:10.1038/35079036.
Bickelhaupt, F.M.; Baerends (2003). "The case for steric repulsion causing the staggered conformation of ethane". Angew. Chem. Int. Ed. 42 (35): 4183–4188. doi:10.1002/anie.200350947.

A more recent paper that has an interesting take:

Baranac-Stojanović, M. (2015). "Theoretical analysis of the rotational barrier in ethane: cause and consequences". Structural Chemistry. 26 (4): 989–996. doi:10.1007/s11224-014-0557-5.

I cannot read the entire article, but the abstract offers the conclusion that hyperconjugation is responsible for the initial energy rise during rotation, but that steric factors are needed to fully explain the high energy of the eclipsed conformation.


How to determine the molecular formula of a compound given its molecular mass and mass percentage contribution of the involved elements? Hydralazine is a drug used to treat heart disease. It is $59.99\%$ carbon, $5.03\%$ hydrogen, and $34.98\%$ nitrogen and has a molecular mass of $160.178$.

What is the molecular formula for hydralazine?

I did:
\begin{align}
59.99~\ce{C}/12.011&=4.994\\
5.03~\ce{H}/1.01&=4.9801\\
34.98~\ce{N}/14.007&=2.4973.
\end{align}

Then divided all $3$ by $2.4973$, and got 2 carbon, 2 hydrogen, and 1 nitrogen atoms i.e. $\ce{C_2H_2N}$. The online homework said that the stoichiometry was wrong. I don't know what I did wrong.

The percentages are based on the molar mass. You have to multiply the percentages with the molar mass and divide by the mass of the atoms.

\begin{align}
\ce{C}:&& \frac{0.5999 \cdot 160.178}{12.011} &\approx 8\\
\ce{H}:&& \frac{0.0503 \cdot 160.178}{1.01} &\approx 8\\
\ce{N}:&& \frac{0.3498 \cdot 160.178}{14.007} &\approx 4\\
       && \implies \ce{C8H8N4}\\
\end{align}


Does fluorine in FNO3 have +1 oxidation number? According to my textbook and also Wikipedia, fluorine nitrate is a compound which can be created (though it is unstable). Also my textbook says that fluorine has a $+1$ oxidation state. But how is $+1$ state of fluorine possible?

I read in another textbook that fluorine can only show $0$ and $-1$ oxidation states (even Wikipedia says so).

Can someone explain as to how fluorine nitrate exists with a $+1$ oxidation state of fluorine?

tl;dr The ambiguity is due to an unfortunate incompleteness of the previously (prior to 2016) existing rules. The oxidation state of fluorine in $\ce{FNO3}$ is $-1$ according to the present rules.

As a general reminder, it is important to understand, that oxidation states are a bookkeeping tool only, and that they hardly represent the general bonding in molecules. With that being said, it is a very simple and powerful tool in many situations. As it is very simple, one has to especially pay attention to its shortcomings or ambiguities.

Prior to 2016 the IUPAC recommended an incomplete set of rules, which I discussed in more detail in Electronegativity Considerations in Assigning Oxidation States.

The special case you have cited is very close to the case of $\ce{HOF}$, which has been previously discussed by me on two occasions already: Oxidation state of fluorine in HOF, What should be the oxidation state of oxygen in HOF (hypofluorous acid)?

According to the new rules, which are based on Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail), and of which the sort version is below adapted from the gold book, there is no more ambiguity about fluorine, which will be assigned $\text{OS}(\ce{F}) = -1$.

  oxidation state
  gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:
  OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

Applied to $\ce{FNO3}$, you would assign fluoride an oxidation state of $\text{OS}(\ce{F})=-1$, the terminal oxygens $\text{OS}(\ce{O})=-2$, the nitrogen will have its highest possible state $\text{OS}(\ce{N})=+5$, and the bridging oxygen will have $\text{OS}(\ce{O})=0$.

Contrary to that are the pre-2016 rules by the IUPAC according to the gold book (via the Internet Archive):

  A measure of the degree of oxidation of an atom in a substance. It is
  defined as the charge an atom might be imagined to have when electrons
  are counted according to an agreed-upon set of rules:

  the oxidation state of a free element (uncombined element) is zero;
  for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;
  hydrogen has an oxidation state of 1 and oxygen has an oxidation state of -2 when they are present in most compounds. (Exceptions to
  this are that hydrogen has an oxidation state of -1 in hydrides of
  active metals, e.g. $\ce{LiH}$, and oxygen has an oxidation state of
  -1 in peroxides, e.g. $\ce{H2O2}$;
  the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the
  oxidation states of the constituent atoms must be equal to the charge
  on the ion. For example, the oxidation states of sulfur in $\ce{H2S}$,
  $\ce{S8}$ (elementary sulfur), $\ce{SO2}$, $\ce{SO3}$, and
  $\ce{H2SO4}$ are, respectively: -2, 0, +4, +6 and +6. The higher the
  oxidation state of a given atom, the greater is its degree of
  oxidation; the lower the oxidation state, the greater is its degree of
  reduction.

You can already see that there is a problem in these rules since they are not definitive. The keyword here is most in bullet 3. They are at least incomplete since there is no mentioning of other elements than oxygen and hydrogen.

This will obviously lead to contradictory results. According to the deprecated rules we need to give oxygen an oxidation state of $\text{OS}(\ce{O})=-2$ in $\ce{NO3F}$. This will give us a total of $6$ electrons that need to come from the other elements. Since nitrogen only has five valence electrons, anything larger than $\text{OS}(\ce{N})=+5$ does not make any sense. Unfortunately that leaves one electron, which can only come from fluoride, i.e. $\text{OS}(\ce{F})=\color{red}{+1}$. It is obviously ridiculous that the most electronegative element has a higher oxidation state than one with a lower electronegativity. This is also in no way an accurate representation of the charge distribution in the molecule.

The above image shows the (NPA) charges of the molecule calculated on the DF-BP86/def2-SVP level of theory. It clearly shows, that fluoride is not positively charged.

One really has to understand that an oxidation state of $+1$ for fuorine does not make any physical sense, and that this set of rules is insufficient. Thankfully this has been addressed in the new definition, which is not perfect, but at least complete.


Standardization of sodium thiosulfate using potassium dichromate In this process first we need to liberate iodine to react with sodium thiosulfate.

First, sodium bicarbonate is added to a iodate-free solution of potassium iodide. The mixture is shaken until the salts dissolve, then conc. hydrochloric acid is added. The solution is shaken again, then covered and keept in a dark place. I have two questions here:

What is the role of sodium bicarbonate here?
Can conc. sulfuric acid be used to provide the acidic media instead of conc. hydrochloric acid?

I think your procedure is incorrect. Typically you would mix $\ce{KIO3}$ (limiting reagent) with access of $\ce{KI}$. Then you add $\ce{HCl}$ to have a reaction:
$$\ce{KIO3 + 5KI + 6HCl -> 6KCl + 3I2 + 3H2O}$$
This is a way to generate a known amount of $\ce{I2}$.

You would then dissolve $\ce{Na2S2O3}$ in a buffer (sodium bicarbonate). This is done because $\ce{Na2S2O3}$ is not stable under acidic conditions. Then you can titrate $\ce{I2}$ in presence of starch with $\ce{Na2S2O3/NaHCO3}$. $\ce{NaI}$ and $\ce{Na2S4O6}$ are formed in this reaction.
In any case you cannot replace $\ce{HCl}$ with other acids. $\ce{H2SO4}$ and $\ce{HNO3}$ are oxidizers strong enough to turn some more $\ce{KI}$ into $\ce{I2}$ and mess up your measurement.


Identification of zinc/aluminium/magnesium with copper sulphate solution We are doing an experiment and would like to be able to identify zinc/aluminium/ magnesium . Would a copper sulphate solution work ? I hear that it turns black on contact with zinc, stays clear on contact with aluminium and fizzes on contact with magnesium.

If you mean identification of the metals in their elemental form, then yes, it might be possible to distinguish between them with an aqueous $\ce{CuSO4}$ solution due to the different reactions.

Zinc in $\ce{CuSO4}$ will be oxidized to colorless $\ce{Zn^2+}$, while $\ce{Cu^2+}$ is reduced to elemental copper, because copper is the more noble metal. The $\ce{Cu}$ precipitate is normally not black, although it can appear dark in the background of the blue solution. It can also form a layer on the piece of zinc, and in the course of the reaction, the initially blue solution will become more and more colorless.

$$\ce{Zn + CuSO4 \rightarrow Cu\downarrow + ZnSO4}$$

Aluminium does not react at all because it is protected by its native oxide layer which is stable in this solution. Magnesium, however, is oxidized under evolution of $\ce{H2}$ gas, which fizzes off from the solution because it is poorly soluble in water. The reaction will gradually become slower when insoluble $\ce{Mg(OH)2}$ passivates the metal surface.

$$\ce{Mg + 2 H2O \rightarrow Mg(OH)2 + H2\uparrow}$$


How to calculate the mass of air that fits into a balloon of a certain size? A hot air balloon has a $\pu{500 m^3}$ volume.
  How much mass of normal air fits into the balloon, if the relative molar mass is $\pu{28.8 g/mol}$?

I am not sure if I should simply use the $PV=nRT$ formula and therefore
$$1\times500= m/28.8 \times 273 \times 8.31 .$$

I feel that there is something wrong, can anyone enlighten me with the right formula or the basic concept of solving this kind of questions?

You're right in using the ideal gas law, but the numbers are wrong. Assuming atmospheric conditions, the pressure is 101,325 Pascal and the temperature is likely to be higher than 0°C (273 K). So the mass of the gas (in grams) would be:

$$m = \frac{\pu{101325 Pa} \times \pu{500 m3} \times \pu{28.8 g mol-1}}{\pu{8.31 J K-1 mol-1} \times \pu{290 K}^{[1]}}$$

where you can choose an appropriate value for [1] for your location. All values (except for the molar mass) are in SI base units, as described here.


Which plastic can be partially broken by pressure easily? I need to create a container which hold ~150 to 500 ml water. My plan is to create something like a inverse cone. I will apply pressure (~20psi) on the top of the cone and I want to create a hole at and only at the tip of the cone.

I plan to make the tip of the cone thinner, so that it will  be easier to be break. However, I am new to plastic. Would anyone please give me some suggestion of material that I should use?

Thanks

If you need good control, put a spring-loaded or cantilever-loaded valve at the tip.  Rupture diaphragms typically have a needle point at the far end. The bubble deforms with pressure, contacting the point that then ruptures the diaphragm that is under tension.  It is much easier to control and reproduce deformation than it is to engineer catastrophic failure at maximum deformation.


What causes the smell of something being "musty"? You know the smell that arises when you leave a wet sponge out for too long? Does anybody know the chemical structure of the molecule that causes that smell? Similarly, what chemical process might be underpinnig that smell from occuring in the first place?

The "musty smell" is from mold, sometimes bacteria. Different compounds are responsible depending on the species. Some examples: trichloroanisole, tribromoanisole, dimethylmethoxypyrazine. Search "mold smell molecules" for many links.

Anisole is also called methoxybenzene, structure here: https://en.wikipedia.org/wiki/Anisole


Gas chromatography flame ionization detector (FID) - why hydrogen gas? Why is $\ce{H2}$ used in an FID, apart from the fact its combustion does not contaminate the flame?

In other words, is the temperature of the flame important or critical? Is the stability of the temperature?

The key to the answer is understanding how FID works.

The hydrogen flame has a minimal flame ionisation, what is needed for the low signal baseline.

Incoming organic molecules from the HPGC column create in the flame a lot of ions and increase the flame electric conductivity.

Using alternatives causing higher ionisation would decrease FID sensitivity that is not great even at ideal conditions, compared to e.g. ECD.


Glass with a low melting point? What kind of glass can I purchase that will melt in a Bunsen burner flame?

Is there any way to treat glass such that the melting point is lowered? I tried adding some sodium hydroxide but the glass shattered when heated.

There are a few options for low-melting glasses that will easily melt at the temperatures of an air-gas torch (i.e. a Bunsen burner).  I suspect that either soda-lime or lead glasses would be the soft glasses of choice for your application (making Prince Rupert's drops).

Soda lime glass should be an adequate choice.  For some applications, it can be prone to cracking if not properly annealed, however.

Lead-based glasses are among those with the lowest melting points of any glass, and probably the lowest applicable to your application.  Annealing is simple and can just be done with the torch itself.  Lead glasses also have a large "soft" temperature range (the range of temperatures below the melting point at which they are still somewhat pliable).  This is seen as a disadvantage in many use-cases, where sagging prior to sufficient cooling/hardening is a problem.  For your work, I would think this would either be a non-issue or possibly could even expand the possibilities for making cool looking drops ;)

Given the availability and affordability of these solutions, further treatment to reduce the melting point just seems like a waste of good glass-working time.


What is the difference between balancing a redox reaction in acidic solution vs basic solution? I'm asked to balance the potassium permanganate and hydrogen peroxide reaction. I misread the problem and thought that the reaction was in acidic solution. This is the result that I've got:

$$\ce{2MnO4 + 5H2O2 + 6H+ -> 2Mn^2+ +5O2 +8H2O}$$
What do I have to change to balance the equation in a basic solution?.

You basically just have to add hydroxide ions to either sides of your equilibrium.

$$\ce{2MnO4 + 5H2O2 + 6H+ + \color{blue}{6OH-} -> 2Mn^2+ +5O2 +8H2O + \color{blue}{6OH-}}$$

$\ce{H+}$ and $\ce{OH-}$ ions can be collectively re-written as $\ce{H2O}$, so we have:

$$\ce{2MnO4 + 5H2O2 + \color{blue}{6H2O} -> 2Mn^2+ +5O2 +8H2O + 6OH-}$$

Finally, we can cancel out the excess water molecules which leaves us with:

$$\ce{2MnO4 + 5H2O2  -> 2Mn^2+ +5O2 +2H2O + 6OH-}$$


Electrophilic substitution on disubstituted benzene rings If I have a strong deactivating group (like nitro) and a weak activating group (like methyl) attached to the benzene ring, which group's preference will direct the next substitution on the benzene ring? For example, if a chloro group is attached to meta-nitrotoluene, where will it substitute?

If I have both strong activating and also deactivating groups in the benzene ring, which position will be preferred for the next electrophilic substitution? For example, if I want to attach a chloro group to the meta-nitroaniline, where will it substitute?

Electrophilic Substitution of Disubstituted Benzene Rings

When a benzene ring has two substituents, each substituent exerts an effect on subsequent substitution reactions.

  Electrophilic substitution of a disubstituted benzene ring is governed
  by the same resonance and inductive effects that affect
  monosubstituted rings. The only difference is that it’s necessary to
  consider the additive effects of two different groups.

If the directing effects of the two groups reinforce each other, the situation is straightforward. In p-nitrotoluene, for example, both the methyl and the nitro group direct further substitution to the same position (ortho to the methyl and meta to the nitro). A single product is thus formed on electrophilic substitution.

From above, the methyl and the nitro group direct further substitution to the same position (ortho to the methyl and meta to the nitro). This is similar case with chlorine.

If the directing effects of the two groups oppose each other, the more powerful activating group has the dominant influence, but mixtures of products are often formed.
Further substitution rarely occurs between the two groups in a meta-disubstituted compound because this site is too hindered. Aromatic rings with three adjacent substituents must therefore be prepared by some other route, such as by substitution of an ortho-disubstituted compound.

You may also wish to review this table showing positions where common substituents direct:

Basing on the above explanations,  it is easy to predict the products of  electrophilic substitution of disubstituted benzene ring, the only thing you have to know is the effect of each substituent whether it is activating or deactivating. In some cases, observe the positions of the substituents as point 3 mentions steric effects may come into play and substitution may not occur.

On the second note, concerning diazonium ion, I think you have to look at measured dipole moments, experimental data etc I doubt if there is a way of telling whether a substituent is strongly deactivating by merely looking at it

After explaining concepts now lets see what the answer for the question in comments will look like:

Reasons

the directing effects are opposing (nitro is meta while methyl is both ortho and para directing) but the methyl is more activating so we have a mixture:

meta to nitro and methyl
ortho to the methyl but para to the nitro

In this scenario since position 3 is not deactivated by nitro group hence there is also a possibility of a mixture here, also take note of how the nitro group is strongly deactivating at position 4 (more additive effect)

Hope this helps

References

McMurry Organic chemistry
Electrophilic Substitution of Disubstituted Benzene Rings


Balancing redox reaction of zinc and hot diluted nitric acid I need to balance the following reaction:
$$\ce{Zn + HNO3 -> Zn(NO3)2 + NH4NO3 + H2O}$$
I assigned all oxidation numbers:
$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{2+}{Zn}(\overset{+5}{N}\overset{-2}{O_3})2 +\overset{-3}{N}\overset{+1}{H_4}\overset{+5}{N}\overset{-2}{O_3} +\overset{+1}{H2}\overset{-2}{O}},$$
but I'm having troubles finding the half reactions. I know
$$\ce{Zn -> Zn^{2+} + 2 e-}$$
is one half reaction, but what about the other? Since ammonium nitrate has nitrogen atoms in two different oxidation states, what do I do with them? Do I add them to see if the nitrogen is or it is not balanced?

The oxidation state of $\ce{Zn}$ as calculated by you is incorrect. The correct oxidation state of $\ce{Zn}$ is $+2$ as it is associated with two mono negatively charged $\ce{NO3-}$ ions.
Thus the unbalanced reaction with correct oxidation states would be:
$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{+2}{Zn}(\overset{+5}{N}\overset{-2}{O_3})2 +\overset{-3}{N}\overset{+1}{H_4}\overset{+5}{N}\overset{-2}{O_3} +\overset{+1}{H2}\overset{-2}{O}}
$$
Now the half-cell reactions are as follows:
$$
\ce{Zn -> Zn^{2+} + 2e-}\\
\ce{NO3- + 10H+ + 8e- -> NH4+ + 3H2O}\\
$$
Thus the balanced iconic reaction would be:
$$\ce{4Zn + NO3- + 10H+ -> 4Zn^{2+} + NH4+ + 3H2O}\\\\$$
Now adding nine $\ce{NO3-}$ spectator ions to the ionic reaction gives us the standard molecular reaction.
$$\ce{4Zn + 10HNO3 -> 4Zn(NO3)2 + NH4NO3 + 3H2O}\\\\$$
What did you learn from this?
Always try to write the half-cell reaction in ionic form and only involve the ions which undergo oxidation state change. Don't bother yourself with the spectator ions at first. After you balance the equation in an ionic  form adjust it to the molecular form by adding relevantly needed ions. Identify the oxidizing and the reducing species is the key.


Cleaning up dispersed potassium hydroxide (KOH) spill? So I tried to replace a dead triple-A alkaline battery in a waterproof compartment; it had leaked fluid + gas which had built up a bit of pressure: When opened, the material (about a dozen tiny droplets, total volume less than a pea) spattered on my hands, clothes and desk :-[

I am assuming that it was pretty concentrated KOH given that it felt slippery / turned the skin greyish before washing it off with plenty of water (no visible chemical burn or aftereffects several days later, though).

Is it good enough to wash the contaminated clothes in a regular washing machine? I did do an additional rinse cycle afterwards.
Will any droplets that I haven't been able to find become neutralized over time, or do I have to tear my study apart and wipe/hose everything down?

Some documents state that KOH absorbs atmospheric water while others state that it reacts with atmospheric CO2 to form a stable carbonate - Surely it can't do both at the same time?

Sorry for the somewhat off-topic question, but it feels like all the MDS are quite unhelpful about how to remediate a spill beyond just 'wash skin with water, remove clothes'..

You just encountered a situation where you did something without sufficient protective gear; in this example, if you had known the battery contains corrosive material, you would have used protective gloves (for your hands) and a pair of safety glasses (for your eyes).

Your estimate that the Leclanché and alkaline battery contains lye is correct. The sensation of the skin turning slippery is expectable, since $\ce{KOH}$ started to degrade the protective coating of your skin (fat) into soap.  Rinsing promptly with plenty of water was a good action.

The lye equally attacked your cloths (one reason chemists wear lab coats), and as your skin, it is beneficial to rinse them right after the skin with plenty of water, too; and then to wash in a washing machine.  On contrast, bringing them as they are into the washing machine only will spread unnecessarily the contamination to other clothes.  Expect that the areas touched by the lye eventually will become brittle, especially if i) they "dried" for some time prior to rising or washing and ii) upon action of heat (e.g. ironing).

Regarding your study, I recommend to go the extra mile, and and clean all the possible surfaces; albeit hosing sounds a bit overkill (esp. inside an appartment).  You want to eliminate the pollution quickly, as it may attack wood and wood stain; and imagine you accidentally glide a book over such a forgotten droplet, it will be attacked just as your skin was attacked, too.  (Even more, as in contrast to human skin, the dead book does not have the ability to re-cover itself by the thin film of protective glycerides.)  In addition, droplets of lye do not dry away like water, $\ce{KOH}$ is hygroscopic and will constantly remain somewhat wet (and corrosive).


Can nonpolar molecules exhibit dipole-dipole forces? Dipole-dipole forces occur when the positive part of a polar molecule is attracted to the negative part of a polar molecule. In a nonpolar molecule, there may still be polar bonds, it's just that the dipoles cancel each other out. So why can't there be dipole-dipole forces between nonpolar molecules with polar bonds? There are still positive and negative parts of the molecule, so there can be attractions between them.

For example, in $\ce{CO2}$, there are negative areas near the oxygens and positive areas near the carbon.

Short answer: there are many electrostatic interactions between two non-polar molecules.

Beyond monopole (full charges) and permanent dipole moments (polar molecules), there is a full multipole expansion for the electrostatic potential around any molecule. (This is technically true for atoms and ions too, but higher-order terms are really only useful for molecules.)

So there are electrostatic potential energy interaction terms for charge-dipole, dipole-dipole, dipole-quadrupole, quadrupole-quadrupole, etc.

These terms are important - the quadrupole-quadrupole interactions dictate the orientation of the benzene dimer and $\ce{CO2}$ dimer in your example.1

The problem is that most of these interactions die off very quickly. The quadrupole-quadrupole term is:1

$$E(r)=\frac{-\Theta_1\Theta_2}{4\pi\epsilon_0r^5}\times\Gamma(\theta_1,\theta_2,\phi)$$

So roughly $1/r^5$, compared to $1/r^3$ for dipole-dipole interactions, or $1/r^6$ for dispersion forces like induced-dipoles.

When such molecules are close, the quadrupole moments (and other multipole electrostatic terms) can dictate packing and distances, but are not as strong or as long-range as dipole-dipole or charge interactions.