At what voltage does the electrodeposition of the metal start? The question is as follows:

  An aqueous solution containing each of $\ce{Au^{+3}, Cu^{+2}, Ag^{+}\;and\;Li^{+}}$ ions each at the concentration of $1 \mathrm M$, is being electrolysed using inert electrodes. The value of the standard potentials are:
  $$E^{\circ}_{\ce{Ag^{+}|Ag}}=0.80\,\text{V} = E_1 \\
E^{\circ}_{\ce{Cu^{+2}|Cu}}=0.34\,\text{V} = E_2 \\
E^{\circ}_{\ce{Au^{+3}|Au}}=1.50\,\text{V} = E_3 \\
E^{\circ}_{\ce{Li^{+}|Li}}=-3.03\,\text{V} \\ $$
  With increasing voltage, the sequence of deposition of metals on the cathode will be:

  Ans: $\ce{Au, Ag, Cu}$

I understand with the increasing value of the standard reduction potential (SRP), the tendency of the metal to get reduced increases, and so obviously, the metal with the highest SRP is deposited first. So we get the increasing order as the answer. But, when I tried to justify this by trying to calculate the specific voltages when electrodeposition should start, I ran into some problems. What follows is my attempt at that. (The following part has been added to the question by me as an edit).

Let us assume we use a battery of $V$ volts, with the polarity as shown in the diagram. Edit 2: The $V$ here is the minimum voltage required for electrodeposition to start. Now, as the negative end of the battery supplies electrons, so the silver will deposit on that electrode, and so (using the comment by f") oxygen is evolved at the other electrode. We have the electrode reactions as:
$$\ce{O2 + 2H2O + 4e^- -> 4OH^-}\qquad E^\circ = V_1 = 0.401\,\mathrm{V}$$
$$\ce{Ag^+ + e^- -> Ag} \qquad E^\circ = V_2 = 0.80\,\mathrm{V}$$
So I reasoned as follows: the hydroxide ions are in the solution, and according to the equation (and its standard reduction potential), the solution is at a higher potential than the the "sink" of electrons, that is the anode. Similarly, the solution is at a lower potential compared to the source of electrons, that is the anode. As both $V_1$ and $V_2$ are positive, using Kirchhoff's Law, we have:
$$V+V_1+V_2=0$$
So here we have a contradiction. Both $V_1$ and $V_2$ are positive, but so should be $V$, because of our explicit assumption of how the terminals are connected. But this equation tells us that $V$ is negative!

So what is wrong with this line of reasoning?

Thank you.

We can't write a relation for V, V1 and V2, because we can apply virtually any V, and still have V1 and V2 constant :) Though, if V signifies the minimum applied potential for starting reactions, then it's ok.
Instead, V may be compared to total potential of cell, which is V2-V1 or V1-V2 (see [3] below about the sign), but not V1+V2, as you assumed.
From wikipedia:
E°cell = E°cathode − E°anode.
The exact sign of (V1-V2) is hard to remember and a subject to mess, so I would better remember the processes involved:
positive voltage means lack if electrons, i.e. oxidation. But (citation)
since the electrode potentials are conventionally defined as reduction potentials (literally: ability to be reduced), the sign of the potential for the metal electrode being oxidized must be reversed when calculating the overall cell potential.
In order to have a spontaneous reaction (ΔG° < 0), E°cell must be positive (see link above). The gold would deposit on the cathode spontaneously if you short external circuit and provide conditions for anode reaction with lower E°. This may be oxidation of OH- (alkaline solution) to O2 on inert electrode. Why do you wonder? Au3+ is powerful oxidizer.
Other note: for high-ionic mix (1M + 1M + 1M + 1M + anions, probably such a mixture is not possible at all) the real redox kinetic will easily be different from simple potential calculation because of polarization effects and other factors. For example, 1-molar Au3+ is probably stable only at low pH, which shifts redox potentials.


Is there a theory behind selecting elements that may be successful in potential superconductors? Looking at something like $\ce{YBa2Cu3O7}$ which was one of the first cuprate superconductors to be discovered, I'm always curious how the selection of these substances as likely superconductors comes about?

Does it have to do with optimal crystalline structure, unpaired electrons, or is it similar to the way semiconductors are designed with doping to create "holes"?  Or are these superconductors simply determined experimentally based on small changes of what has worked in the past?

As a side question: I've always been curious as to how they stumbled upon Yttrium for this particular case.  It's not something I think of as being carried in ready supply.  Would this have been after YAG lasers were already around?

Suffice to say, nobody really knows, and a Nobel prize surely awaits the first people who can explain the relationship between chemical composition and high-temperature superconductivity.


What chemicals react with alcohol? I am curious if there is any chemicals that react with alcohol to either produce a visible colour change or an odour someone could smell.

When you say "alcohol", I am not sure if you are referring to ethanol (drinking alcohol) or alcohols in general so in my example I will use ethanol. When an alcohol is combined with a carbocylic acid and a catalyst, they will produce esters (fragrances). For example, ethanol and ethanoic acid would react to produce ethyl ethanoate (a fruity smelling substance) and water.

There are a plethora of different reactions from all different types of alcohols and carbocylic acids that produce countless different smells. The fragrances also come in a variety of colors as well. If you want to find out more, I would suggest looking up ester synthesis. Hope my answer helps.


Is ln(K2/K1) = (∆h°rxn/R)(1/T1 - 1/T2) even when ∆h°rxn varies strongly with temperature? In all of my physical chemistry books I find the same expression for estimating the equilibrium constant of a reaction at a non-standard temperature. The derivation starts with $R \ln K_\text{eq} = -\Delta g^\circ/T = -(\Delta h^\circ/T - \Delta s^\circ)$. Then one takes the temperature derivative of each side assuming that $\Delta h^\circ$ and $\Delta s^\circ$ can both be considered constants, which eliminates the entropy term. The equation
$\mathrm{d}\ln K = (1/R)(\Delta h^\circ/T^2)\mathrm{d}T$ is then integrated between $T_1$ and $T_2$ to give
$\ln(K_2/K_1) = (\Delta h^\circ/R)(1/T_1 - 1/T_2)$.

Here's the question: $\Delta h_\text{r}^\circ$ is really NOT a constant, but a function of temperature. Assuming constant heat capacities, $\Delta h_\text{r}^\circ$ at $T = \Delta h_\text{r}^\circ + \Delta c_p(T - T_0)$. Likewise $\Delta s_\text{r}^\circ$ at $T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$. Where $\Delta c_p$ is the difference in the heat capacity of the products and the reactants.

I have tried to insert these expressions into the typical derivation but am getting wonky results. Can someone help me with the derivation or point me to a source where $\ln K$ vs. $T$ is derived with temperature-sensitive reaction enthalpies and entropies.

Edit 4/19/2015: Would it be more effective to invoke $\mathrm{d}\Delta g = \Delta v\mathrm{d}P - \Delta s\mathrm{d}T$ which at constant pressure, and in the case of free energy of reaction, gives $\mathrm{d}\Delta g_\text{r}^\circ/\mathrm{d}T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$ ?

Thanks in advance for any help.

For each species, the quantities $g^0$, $s^0$, and $h^0$ are defined as the free energy, the entropy, and the enthalpy of the pure species, respectively, at 1 atm. pressure and temperature T. So, for constant pressure of 1 atm., $d\Delta g^0=\Delta s^0dT$.  But, $\Delta s^0=\frac{\Delta g^0-\Delta h^0}{T}$. So $\frac{d\Delta g^0}{dT}=\frac{\Delta g^0-\Delta h^0}{T}$

So, $\frac{d(\Delta g^0/T)}{d{(1/T)}}=\Delta h^0$


Solvent for tire rubber I want to apply tire rubber on surfaces, for various reasons - including protection against water. I think of tire rubber, as it is so widely available - helping in reducing pollution (infinitesimally) at the same time.

I am interested in a solvent which will temporarily transform the tire rubber into a (viscous) liquid - in such way that after some time, when the solvent evaporates, the tire rubber becomes solid again.

I do not need any special properties of the new rubber, there will be nothing really safety critical. If it is similar enough to the original, it is good.

Ideally, the solvent should be easy to find and I should be able to buy it without any special permits.

Think of this:

there are some granules of salt;
I add some water - and get salty water;
after some time the water evaporates and I get 1 block of salt.

Now, the salt is actually pieces of tire rubber, and the water is... what? :)

I searched on the net, and the only solvent I recognized was acetone. Does anyone have any experience with this?

As it was mentioned in the comments, finished rubber is a cross-linked polymer, and this is the main reason why getting vulcanized rubber liquefied without destroying it seems to be a nearly impossible task.
Also, there might be other additives, e.g. tires are most likely going to contain powdered charcoal, so that even though one can dissolve the rubber by destroying the cross-linking $\ce{(-S-S-)_n}$ bonds, charcoal is likely going to form another phase, and who knows what might happen to other additives during this process.
Finally, at the end you have to somehow recreate cross-linking for the rubber to retain it's elasticity, and this doesn't happen on its own by just evaporating the solvent (see vulcanisation).

In general, the majority or rubber types can be affected by:

strong oxidizing agents (to the extreme of being irreversibly destroyed);
aromatic and chlorinated hydrocarbons (both have a tendency to soften rubber).
Visually it appears as if rubber is swelling; this can happen if one keeps rubber in chloroform or xylene for an extended period of time – for instance, this property is sometimes used to fit rubber tubing in place.

What I suggest is to use a so-called rubber mix or rubber compound: a semi-finished mix of [natural] rubbers, vulcanizing agents, plasticizers, fillers and additives (carbon, chalk, kaolinite etc.).
Rubber mix is used ubiquitously in the industry to produce all sorts of rubber products, including composite materials such as rubberized fabric.
Rubber mix is a versatile material which is highly viscous at the room temperature, but softens upon heating to approx. 100 °C and can be dissolved in volatile organic solvents (even benzene and gasoline).
The drawback is that at the end you still have to use vulcanization to crosslink the rubber.


Why is OH group activating towards electrophilic aromatic substitution? In phenol, the OH group has -I effect due to electronegative oxygen. So, OH group should deactivate the benzene ring towards electrophilic substitution.

[P.S.: I am not talking about the o,p-directing ability of OH, which appears due to lone pair. Also please do not suggest me the question at the link Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution?, because that is not what I am looking for.]

The oxygen in the $\ce{-OH}$ group in phenol has unpaired electrons that are capable of being donated to the ring. Although oxygen is electronegative and does exhibit some I- effects, the electron sharing that oxygen is capable of (shown by when drawing the resonance structures) has a stronger effect and results in $\ce{-OH}$ being a good ring activator. I've included the resonance structures below.


How do I convert this ion exchange capacity into water softening capacity? I am researching for an extended experimental investigation into methods of water softening. One such method is the use of ion exchange resin. I have contacted the supplier of the resin for its exchange capacity, but I'm not sure how their answer relates to the removal of magnesium and calcium ions from water.

They say the exchange capacity is "1.0 mEq/mL min of 99% in H form."

In the H form there are one plus charged ions ($\ce{H+}$) being exchanged in the water softening process. But the species being exchanged for the $\ce{H+}$ ions are doubly charged ($\ce{Mg^2+}$ and $\ce{Ca^2+}$), so two $\ce{H+}$ ions must be exchanged per $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ion.

So, "1.0 mEq/mL min of 99% in H form" could equivalently be read as:

The exchange resin has the ability to exchange $\pu{0.5mmol}$ of $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ions per $\pu{mL}$ of resin.


During the decomposition of sodium bicarbonate lab, the mass of the final solid I received was less than expected. Errors? The lab procedure that my class used can be found in the photo attachment.
The purpose of the lab was to determine the equation for the decomposition of sodium bicarbonate. The two possible equations were:

$\ce{2NaHCO -> Na2CO3 + CO2 + H2O}$
$\ce{2NaHCO -> Na2O + 2CO2 + H2O}$

I understand that the correct equation is supposed to be the first one where sodium carbonate was produced. However, through calculations using the ratio between reactant and solid product, my experiment showed that equation 2 where sodium oxide was produced is supposed to be correct.
I know that this is caused by the mass of product I produced which was 0.716 grams rather than the expected expected 1.187 grams. However, what I do not understand is what errors could have potentially lead to this less than expected mass?

I understand that the correct equation is supposed to be the first one where sodium carbonate was produced. However, through calculations using the ratio between reactant and solid product, my experiment showed that equation 2 where sodium oxide was produced is supposed to be correct.

So now you're faced with a decision. You can either:

Massage the result of your experiment until it comes up with the answer you think is "supposed" to be correct.
Report the result of your experiment as it actually happened.

If you take the first route, I'm sure we can come up with some convincing explanations. Maybe your starting sodium bicarbonate contained water (so it wasn't 100% pure) and that water was driven off in the heating. Maybe you spilled some as you put it in the crucible. Maybe some of the mixture in the crucible splashed out during the reaction. Maybe you read the scale wrong. Maybe there is another reaction that also happens, perhaps involving the air, or the crucible, or the gas from the bunsen burner.

But I think it would be much more honest to report your result as it actually happened. Your data supports the hypothesis that the sodium bicarbonate decomposed to sodium oxide. In fact, the agreement with that hypothesis is quite strong, and the agreement with the supposedly correct hypothesis is terrible. It could be that this is due to an error you made... or maybe the stupid sodium bicarbonate didn't get the memo that, according to the teacher (or the text or Wikipedia or whatever), it was supposed to decompose to sodium carbonate, so it just went ahead and decomposed all the way to sodium oxide.

I realize you're probably worried about being marked down on your lab report because you came to the "wrong" conclusion. Any decent teacher, though, would give you much more credit for correctly interpreting your data and coming to a conclusion based on the data, even if it's wrong, than for massaging your data to support a preconceived hypothesis.


Remove salt from dirt I have a mixture that contains out of dirt/sand and salt. I want to extract and measure the salt. What would be the best way to extract the salt from this mixture?

Assuming you mean NaCl - the common "salt" (chemists call lots of things salt!).

If you extract the salt by physical means, it's a physics question.

Assuming that the "dirt" is not (or poorly) soluble in water, I would simply dissolve the salt in water, filter the liquid, then recrystallize (by evaporation of the liquid) and weigh the resulting crystals. Not sure if you call that chemistry or physics... it's physical chemistry. It is well described in text books and online - see for example this Scientific American page

Depending on how accurate you want to get, there are a slew of precautions to take. That would be the realm of the chemistry.stackexchange sister site.


(Computationally) finding similarity between two organic compounds I have been struggling to find an acceptable answer for this question for my purposes.

There are many ways to find similarity between two organic compounds - some of which are particularly popular in chemoinformatics. The seemingly most popular way is to use fingerprints of molecules, which then somehow correlates to the structure/function of various parts of the molecule. This approach seems very good when you're looking for general similarity between molecules. It also has the added benefit of making it fast to compare huge amounts of molecules, as each molecule can be encoded separately (so it's generally speaking $O(n)$), rather than comparing every pair of molecules (which would be $O(n^2)$).

However, for some purposes, this fingerprinting approach is not very good. I need a function of two molecules with the following properties:

If $A$ and $B$ are two molecules, then $f(A, B) \in [0, \infty).$
$f(A, A) = 0$
$f(A,B) + f(B,C) \geq f(A,C)$

These are similar to the basic requirements for a metric space. (The big omission is that I don't necessarily need $f(A,B) = f(B,A),$ although this certainly couldn't hurt!)

The reason that I want such a function is because I'm designing a neural network that essentially outputs a molecule, and I want to have some sort of error on the output.

I have played around a bit with Python-RDKit, and its similarity module, but haven't really been able to form a good "error" function from the output.

I've also experimented a bit with an algorithm I created that looks for the largest identical connected subgraphs of two molecules, and essentially finds matches for each part of the query molecule. The final output is then how many different parts the molecule needs to be split into to find a match.

For instance, if the "true" molecule is ethylbenzene, while the query molecule is m-xylene, then the algorithm would find that the m-xylene needs to be broken into 3 parts for each part to find a perfect match: a benzyl group missing a hydrogen at the third carbon of the aromatic ring, the methyl group that was formerly attached at the third carbon, and the hydrogen that remains from the benzyl group.

However, this algorithm suffers from several problems:

If the query is a subset of the true answer, then the algorithm will always give the answer as 2 (one part is the subset, and the other part is the hydrogen that caps it - try seeing it with something like query - methane, true - acetic acid). This isn't that big of a concern, as you can simply run the algorithm twice - once comparing the query to the true molecule, and once in reverse. That way, if the two molecules are indeed far apart, it's not hard to see (the superset molecule may need to be broken into many pieces to be identical).
This algorithm is slow. Don't really see a way to speed it up. It searches for maps from subsets of the query molecule to the true molecule, then gradually builds up from there. It also can't pick maps randomly as being the best, then growing from there, as picking the wrong direction can easily make the "best" map drastically short. So it has to do all possible maps at the same time. Which is slow.

In short, this is a somewhat open-ended question that boils down to:

How can we put a (loose) metric on the set of organic molecules?

Using graphs to represent molecules, would it be reasonable to compute the distance between graphs A and B?

In a graph representation of a molecule, nodes are atoms and edges are bonds.

Example of an algorithm to compute distance between graphs: http://www.xavierdupre.fr/app/mlstatpy/helpsphinx/c_graph/graph_distance.html


Selecting between redox and neutralization reactions for a mixture of HCl, H2SO4 and NaOH I know that you can have 2 acids reacting with 1 base. These 4 reactions are an example of this:

$\ce{3 HCl + H2SO4 + NaOH -> 3 H2O + SCl2 + O2 + NaCl}$

$\ce{6 HCl +  4 H2SO4 + 2 NaOH -> 8 H2O + 2 S2Cl2 + 5 O2 + 2 NaCl}$

$\ce{HCl + H2SO4 + 3 NaOH -> 3 H2O + Na2S + 2 O2 + NaCl}$

And

$\ce{HCl + H2SO4 + 3 NaOH -> 3 H2O + Na2SO4 + NaCl}$

I checked and all these equations are balanced. But will these reactions forming sulfur chlorides and ionic compounds with sodium and sulfur as well as $\ce{H2O}$ and for 3 of them, $\ce{O2}$, actually occur without any added heat or are they endothermic acid base reactions and thus require heat or will the stronger acid(in this case $\ce{HCl}$) be the one that reacts and the weaker acid (in this case $\ce{H2SO4}$) just make the pH of the solution lower?

As Klaus and I mentioned, the first reactions seem very strange from a chemical perspective, even if they are correctly balanced. This is where a chemist needs to exercise their intuition and experience.

Balancing reactions is nothing but solving a linear system of equations, so it is entirely possible to propose reactions which balance perfectly, but which are based on questionable chemistry. For example, in all but the last reaction, the sulphur atoms in $\ce{SO_4^2-}$ anions are being reduced to lower oxidation states. While this is possible in some conditions, it takes quite strong reducing agents, and none of the reagents shown are known to possess significant reducing character. Reactions which release gaseous oxygen are also not particularly common. Furthermore, two reactions propose the formation of sulphur chlorides, which would actually hydrolyse in contact with water.

Disregarding intuition, it should be possible to pick out which reaction is the most favourable by simply comparing the Gibbs free energy of reaction in each case; the most exergonic reaction, with the most negative value of $\boldsymbol{\Delta_\mathbf{r}G}$, should dominate (assuming no kinetic factors bar it from happening, which is unlikely in this situation). I did a very quick and dirty calculation, out of curiosity, by taking the standard Gibbs free energy of formation $\Delta_\mathrm{f}G^\circ$ for each compound and summing them together to obtain the standard Gibbs free energy of reaction:

$$\Delta_\mathrm{r}G^\circ = \sum\limits_\text{products}\Delta_\mathrm{f}G^\circ - \sum\limits_\text{reagents}\Delta_\text{f}G^\circ$$

This is only meant as a qualitative way to compare the reactions, as I am assuming standard state conditions for the reactions, which are not necessarily true. Furthermore, for a few species, I didn’t find the correct values in these two sources [1] [2]. $\ce{SCl2}$ and $\ce{S2Cl2}$ probably can’t be found tabulated for the aqueous phase because of their propensity to hydrolysis, so I used the free energies of formation for the gas and solid phases, respectively. I also didn’t find a source for aqueous $\ce{Na2SO4}$, so instead I used the data for the solid decahydrate $\ce{Na2SO4 . 10 H2O (s)}$. The results I got were:

$$ \small
\begin{array}{lcc}
\hline
\text{Reaction} & \Delta_\mathrm{r}G^\circ / \mathrm{kJ\ mol^{-1}}\\
\hline
\text{A} & +440  \\
\text{B} & +1940  \\
\text{C} & +590  \\
\text{D} &  \mathbf{-2610}  \\
\hline
\end{array}
$$

As you can see, even though the calculation was very rough, the huge disparity in free energy values suggests that the first three reactions are completely overwhelmed by the last one, so that of the four possibilities listed, only the last has any chance of occurring. Strictly speaking this doesn’t mean the last reaction happens, only that it’s much more favourable than the others presented; there could be an unlisted reaction which is even more favourable than all four mentioned in the question. However, empirical data confirms that the last reaction, a simple neutralisation reaction, is the major event when mixing $\ce{HCl}$, $\ce{H2SO4}$ and $\ce{NaOH}$ in practically any conditions.


Is Hydrogen Bonding a Type of Dipole Dipole Interaction? I understand that dipole dipole forces is due to the attraction of the different partials charges of atoms in different molecules due to their different electro-negativities.

For hydrogen bonding, my understanding of it is that they are just a special case of dipole dipole forces. In certain molecules in where the difference in electro-negativities between the different atoms is high enough, then even stronger dipole dipole forces will be form which are known as hydrogen bonding. For example, in $\ce{H_2O}$ where the difference in electro-negativity of H and O is 1.24, hydrogen bonding occur. This is why hydrogen bonding can only occur between hydrogen and nitrogen, fluorine or oxygen.

However, my school teacher says that hydrogen bonding in fact is not a type of dipole dipole force as she says that hydrogen bonding occurs differently to dipole dipole forces. She said that hydrogen bonding occurs to the attraction of the lone pairs in oxygen, fluorine and nitrogen atoms to atoms with a positive partial charge. Hence this is why hydrogen bonding is much stronger than conventional dipole dipole forces.

But this doesn't make sense because if hydrogen bonding is due to the attraction  of H to the lone pair of electrons, shouldn't hydrogen bonding also form between H and Cl or I or any other atom that has a lone pair and is more electronegative than H?

Edit 1

From the comments below, it appears that the actual answer is more complicated than the simple explanation of hydrogen bonding given above. After doing some research, I have became aware of the Advanced Theory of the Hydrogen Bond which states that the hydrogen bond actually contains 10% covalent nature. However none actually explain how they are formed. Could someone please explain how hydrogen bonding actually in reality form.

Edit 2

So in response to the excellent answer by jheindel, is the reason why hydrogen bonding is considered partly covalent in nature is because the let say in water, oxygen's MO partly overlaps with the MO of the hydrogen atom in the other molecule. Also this is why they are much stronger than conventional dipole dipole forces?

Well, it turns out that this is a very active area of research. I will only summarize what I understand to be true about the covalent nature of the hydrogen bond, so I'm sure the explanation could be more detailed and potentially more accurate in some places (I hope someone gives a more detailed answer), but here's what I've got.

As you said, it has been found that the hydrogen bond (specifically in water) is about 90% electrostatic and 10% covalent. See this link for a sort-of-review/overview of hydrogen bonding and some relevant research.

Upon following the paper trail, I found the following research article: E.D. Isaacs, A. Shukla, P.M. Platzman, D.R. Hamann, B. Barbiellini, C.A. Tulk, J. Phys. Chem. Solids, 2000, 61, 403-406. (Mirror)

Essentially, they use some method which I don't really understand and find quite conclusive evidence of the (word of the day) anisotropy of hydrogen bonds. That means that the bond is indeed directionally dependent. Bond direction is one defining characteristic of covalent bonds.

That raises the question of what is going on with the orbitals that allow for hydrogen to participate in more than one covalent bond at a time.

A simple explanation of the orbital interactions for a water dimer is to view the oxygen 2p,2s hybridized orbital as having superposition with the hydrogen 1s orbital and then interacting covalently with that same 2p,2s hybridized orbital on the other oxygen in the dimer. This superposition which includes the hydrogen then explains the way that hydrogen is participating in one covalent bond and another partially covalent interaction.

Furthermore, this explanation of what the orbitals are doing fits well with what we would expect from a hydrogen participating in two different covalent interactions, which is that the energy curve has two minima where one is much deeper than the other.

Further evidence for the covalent nature of the bond is the fact that the O-H bond elongation and O---H hydrogen bond length always change together. That is, as the O-H bond lengthens, the O---H interaction length shortens. This implies a shift in electron density towards hydrogen which is most easily explained by a delocalization of electrons participating in a covalent interaction.

So, I hope that's a good start toward understanding the covalent nature of the hydrogen bond--it was for me!

But to answer the simple part of the question which you clearly already know is that yes dipole-dipole is a significant part of hydrogen bonding, and it certainly should be, but as is always true of chemistry, there's a lot more going on than what one might initially think.

EDIT (06/01/2016)

Since I posted this answer, I think I have developed a better understanding of the situation, and feel obliged to update my answer. First, let me split hydrogen-bonding into two obvious parts, electrostatic and covalent characteristics. This is an obvious split, but it is not completely clear yet what we mean by covalent, so let us try to define that by looking at some examples. Furthermore, we must be careful when talking in about hydrogen-bonding because in every hydrogen bond there is a donor and acceptor, and these two roles are often played best by different chemical species.

Hydrogen-Bonding in thiols

Here we go breaking yet another rule. They told us that sulfur couldn't hydrogen bond. Well, that's not exactly true. Okay, but that hydrogen-bonding has got to be basically negligible right? Not exactly.

The first thing we need to do here is get some information. First, sulfur has an electronegativity of 2.58 on the Pauling scale. This is roughly the same as the electronegativity of carbon which we do not consider to hydrogen-bond well. So, if there are hydrogen-bonds involving sulfur, we should expect that they will be primarily covalent as any dipole interactions will be pretty small.

As it turns out, sulfur is a much better hydrogen-bond acceptor than it is donor, and at times is a comparable acceptor to oxygen. Here is a table comparing certain dimers from a great source you can reference at the bottom of the page$^3$.

As you can see, when the sulfur in these dimers (which are all named in the caption of the table) acts a hydrogen bond acceptor, the shift in vibrational frequency of the $\ce{O-H}$ and the binding energy are comparable to when an oxygen is the acceptor atom.

Looking at other research$^4$ involving sulfur hydrogen bonds reveals that approximately 70% of the interaction energy between sulfur and oxygen when sulfur is the acceptor comes from dispersion. That is the key physical insight in all this. Despite the loss of electronegativity, sulfur gains something in its greater dispersion interactions which makes it a good hydrogen bond acceptor. It should be noted, however, that the electronegative component of these hydrogen-bonds maintains their directionality.

Also, to compare, according to that wikipedia article I linked, the interaction energy of a water dimer is only 24% dispersion. In water electrostatics dominate and induction and exchange play meaningful roles.

Conclusions

I have more I could say, but it's perhaps a longer explanation than is worth going into. To conclude, what is meant by covalent interactions in hydrogen-bonding varies based on the species you are considering. In something like water, we can think about this in the usual way by saying that there must be some orbital overlap and hence electronic exchange which will be stabilizing. With other species, we can see that what we mean by covalent interactions might be different. That is, there is still some orbital overlap, but we must also consider non-electrostatic interactions like dispersion.

3 Noncovalent Forces-Chapter 2

4
Wennmohs, F., Staemmler, V., & Schindler, M. (2003). Theoretical investigation of weak hydrogen bonds to sulfur. The Journal of chemical physics, 119(6), 3208-3218.


Why do transition metals form particularly stable complexes? The pre-transition metal ions can form complexes but the ligands are weakly bound unless they are multidentate and thus benefit from the chelate effect. For example, crown ethers and cryptands.

However, transition metal ions can form a vast array of stable of stable complexes with monodentate and multidentate ligands. Why is there such a difference between the pre transition elements and the transition metals in this respect.

My initial thoughts are that the transition metal ions will have a higher charge and smaller radius so the electrostatic attraction is larger. I also thought about the d-orbitals but the pre transition elements have d orbitals available, they're just not filled. As a result, is increased stability related to exchange energy or perhaps the lower energy d orbitals? Are there any more factors? Are my suggestions valid?

To illustrate my answer, I am going to use a modified version of the picture of a simplified molecular orbitals scheme Professor Klüfers (LMU Munich) uses in his internet scriptum to the coordination chemistry I course.

On the left you can see the metal’s orbitals, on the right the group orbitals of six $\unicode[Times]{x3C3}$ donors (assuming no $\unicode[Times]{x3c0}$ interactions of any kind). $\mathrm{e_{g}}$ and $\mathrm{t_{2g}}$ correspond to the metal’s d-orbitals, $\mathrm{a_{1g}}$ is the 4s orbital and $\mathrm{t_{1u}}$ are the 4p-orbitals.

You can clearly see that the 3d-orbitals are similar in energy to the ligand group orbitals, and that they have the correct symmetry to mix. The greatest stabilisation comes from the $\mathrm{e_g}$ orbitals.

Empty d-orbitals can thus be thought to help with coordination by stabilising the coordination compound. Completely filled d-orbitals (e.g. $\ce{Zn^2+}$) are contraproductive and therefore zinc’s coordination chemistry is much like sodium’s.


Find moment of inertia around two axes of sulphur dioxide Problem:

  Find the moment of inertia for 2 axes of $\ce{SO2}$. The rotational constants are  given in $\pu{MHz}$. The answer is required to be $10^{-46}\rm~ kg\cdot m^2$, 7 digit accuracy.

I used $$I = \frac{h}{8\times\pi^2\times B}$$ but the answer is marked wrong. I do not understand what I am doing incorrectly. I thought maybe that $\ce{SO2}$ is asymmetric so that formula will not work.

This is an online course that has  minimal discussion assistance. I have scrounged on Google & physical chemistry textbook to see what the alternative could be. Did not see another formula. Where did I go wrong?

Without explaining some more its hard to help, but the most likely error is in using the wrong units. The rotational constant is usually given in wavenumber units (cm$^{-1}$), e.g. 2.02736 cm$^{-1}$. In a linear molecule the equilibrium value is

$B_e = \frac{h}{(8\pi^2cI)}$

with $B_e$ in wavenumbers and c is the speed of light in cm/sec. In your case since the SO$_2$ is not linear (its planar bent at the S atom, C$_{2v}$ point group) there are three rotational constants usually called A$_0$, B$_0$, C$_0$.  (You can check the units by converting Joules in to base units, m, kg, s etc . The moment of inertia is in kg m$^2$ and $10^{-46}$ is of the correct order of magnitude. Seven decimal places seems too many as the rotational constants are not usually known to more than 5 places.)


Chemical potential of mixtures I feel that there are very few textbooks that treat the chemical potential of mixtures in an understandable clear way, which is why I wanted to ask here about certain things. Although I do not have a single question, but 5 of them, I think that it is merely impossible to separate them thematically, which is why I hope to meet an expert here.

If we have an ideal gas then if we know the chemical potential for a particular $(p,T)$ we have:

$$\mu(p_2,T) \approx \mu(p_1,T) + \left(\frac{\partial \mu}{\partial p} \right)_T(p_1,T) (p_2-p_1) = \mu(p_1,T) + \frac{V}{N}(p_2-p_1).$$

Similarly for changes with respect to temperature we have
$$\mu(p,T_2) \approx \mu(p,T_1) + \left(\frac{\partial \mu}{\partial T} \right)_p(p,T_1) (T_2-T_1) = \mu(p,T_1) - \frac{S}{N}(T_2-T_1).$$

Now, if we have either a gas or a liquid (Question 1: is this true, does this law hold for both fluids?) $\ce{A}$ and put another material $\ce{B}$ into it, then we have by Raoult's law:

$$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_A(p,T) + RT\ln\left( \frac{N_A}{N_A + N_B} \right)$$

Question 2: Now, is the chemical potential of the total fluid given by $A$ and $B$ then the sum $\mu_{\mathrm{A+B}} = \mu_{\mathrm{A,~with~mixed~B}}(p,T) + \mu_{\mathrm{B,~with~mixed~A}}(p,T)$?

Let's imagine that $A$ and $B$ would be gases, would this mean that

Question 3 : $$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_\mathrm{A}(p,T) + RT\ln\left( \frac{p_\mathrm{A}}{p_\mathrm{A} + p_\mathrm{B}} \right) ?$$

Question 4 : And what if $\ce{A}$ and $\ce{B}$ would be two two miscible liquids? could we still use the vapour pressures to calculate the chemical potential of the liquids mixture?

$$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_A(p,T) + RT\ln\left( \frac{p_\mathrm{A}}{p_\mathrm{A} + p_\mathrm{B}} \right) ?$$

Now, imagine that I have a liquid $\ce{A}$ and salt $\ce{B}$ in it. Also, I have a gas layer $\ce{C}$ of $\ce{A}$ above the liquid. What exactly is now in equilibrium: $\mu_{\mathrm{A+B}} = \mu_\mathrm{C}$, or $\mu_{\mathrm{A,~with~mixed~B}} = \mu_\mathrm{C}$?

Question 5: My last question would be: Imagine that you have a liquid (let's say water) and some gaseous water phase above (according to the vapour pressure). Now, you add oxygen to this. What would this mean for the chemical potentials of the liquid/gas water?

It's much simpler than all that.

Pure gas
$$\mu_g(T,P)=\mu^0(T)+RT\ln P$$
where $\mu^0(T)$ is the molar free energy of the pure gas at temperature T and pressure 1 Bar, and P is the pressure expressed in bars.

Pure liquid

If $P_{sat}(T)$ is the equilibrium vapor pressure at temperature T, then, at pressure $P_{sat}(T)$, we know that the free energy of the liquid is equal to the free energy of the vapor.  So, $$\mu_l(T,P_{sat})=\mu^0(T)+RT\ln P_{sat}(T)$$If we neglect the effect of pressure on the chemical potential of the liquid (because the specific volume is very small), then, for a pure liquid at temperature T and pressure P:$$\mu_l(T,P)=\mu^0(T)+RT\ln P_{sat}(T)$$

Ideal gas mixture

For species A in an ideal gas mixture, the chemical potential of species A is the same as the chemical potential of the pure species, except evaluated at the partial pressure of A in the gas mixture:$$\mu_{Ag}(T,P,y_A)=\mu^0_A(T)+RT\ln p_A=\mu^0_A(T)+RT\ln (Py)_A=\mu_A(T,P)+RT\ln y_A$$
where $p_A$ is the partial pressure of A (bars), $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, P is the total pressure of the mixture (bars), $\mu_A(T,P)$ is the free energy of pure A at the temperature and  pressure of the mixture, and $y_A$ is the mole fraction of A in the mixture.  So the chemical potential of A in the mixture is equal to the chemical potential of pure A at the same temperature and pressure as the mixture plus the correction term for mixing $RT\ln y_A$.

Ideal liquid mixture

By analogy to an ideal gas mixture, the chemical potential of species A in an ideal liquid solution is given by:

$$\mu_{Al}(T,P,x_A)=\mu_{Al}(T,P)+RT\ln x_A=\mu^0_A(T)+RT\ln P_{A,sat}(T)+RT\ln x_A$$
where $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, $P_{A,sat}(T)$ is the saturation vapor pressure of species A at temperature T (bars), and $x_A$ is the mole fraction of species A in the liquid.


Reaction of phenol and zinc dust There is a reduction reaction of phenol with zinc dust which is found in many books.

$\ce{Ph-OH + Zn -> Benzene + ZnO}$

However, a simple google search fails to explain anything about the reaction. So, I put forth my questions:

  1) What is the name of the reaction and who invented it?

I have already done a google search, but it yields no results. My textbook also mentions nothing about the inventor of the reaction, or any name of the reaction. I understand that every reaction might not have a name, but at least it must have an inventor. You can't say that the reduction of phenol by distilling it with zinc dust is known from time immemorial. Can you?

  2)What is the reaction mechanism of the reaction?

This and this page writes about a mechanism of the reaction. If drawn, I suppose it would look like this:

It would certainly explain the formation of biphenyl and hydrogen ( if they are indeed formed). But the problem is that no reference is given in the answers. There is at least one book or paper which wrote about the mechanism.

So, if you answer, please add a reference if possible. If you think that the above mechanism might be incorrect, then also add an answer. I want your insights on this problem.

[Note: There is another question at Reaction of phenol with Zinc dust which is similar. But I have already drawn a mechanism and I am asking a different question here. So I don't think this question is a duplicate.]

Ludwig Gattermann's The Practical Methods of Organic Chemistry (translated into English, 1896) has two references to reduction of aromatics with zinc dust:

The first one refers to Liebigs Annalen, issue 140, page 295 (1866):

Ueber die Reduction aromatischer Verbindungen mittelst Zinkstaub, Adolf Baeyer

This roughly translates to "On the reduction of aromatic compounds using zinc dust".
The second one refers to Chemische Berichte, issue 1, page 43 (1868):

Ueber Alizarin und Anthracen, C. Graebe C. Liebermann

This translates to "On alizarin and anthracen". Wiley, who now owns the Berichte, cites the page numbers as 49-51.
Name of the reaction
The reaction is given without name, even in modern textbooks. Maybe it should be called Baeyer reduction of aromatic oxygen-containing compounds.
Mechanism of the reaction
Maybe this would be a good research project because it does not seem like the mechanism is known.


Determination of pKa by absorbance and pH of buffer solutions I've been tasked (as part of a report) to calculate the $\mathrm pK_\mathrm a$ of 4-nitrophenol by measuring the pH and absorbance of six 'buffer solutions' containing the 4-nitrophenol:

$$\begin{array}{cccc}\hline
\text{Solution} & \text{Volume of base / }\pu{cm^3} & \text{Volume of acid / }\pu{cm^3} & \text{Volume of phenol / }\pu{cm^3} \\ \hline
1 & 2.80 & 7.20 & 1.00 \\
2 & 5.00 & 5.00 & 1.00 \\
3 & 6.20 & 3.80 & 1.00 \\
4 & 7.20 & 2.80 & 1.00 \\
5 & 8.70 & 1.30 & 1.00 \\
6 & 10.00 & 0.00 & 1.00 \\ \hline
\end{array}$$

We have to then plot a graph of $\mathrm{pH}$ versus $\log_{10}[A/(A_f-A)]$ and use that graph to calculate $\mathrm pK_\mathrm a$. $\text{A}_{f}$ is the absorbance of Solution 6 - we assume that the weak acid in the buffer is present only as its conjugate base - and A is the absorbance of the buffer solution.

My two questions about this are:
1) Where does the expression $\log_{10}[A/(A_f-A)]$ come from? I thought the derivation would be as follows, which clearly gives a different result ($\text{A}_{f}$ on the numerator rather than $\text{A}$):

  Using the Beer-Lambert Law,

  $$
\begin{align}
A &= \epsilon cl \\
c_\text{base} &= \frac{A_{f}}{\epsilon l} \\
\end{align}
$$

  And substituting into the Henderson–Hasselbalch equation:

  $$
\begin{align}
\mathrm{pH} &= \mathrm pK_\mathrm a + \log_{10}\left(\frac{[\text{conj. base}]}{[\text{acid}]}\right) \\
\mathrm{pH} &= \mathrm pK_\mathrm a + \log_{10}\left(\frac{A_{f}}{A_{f}-A}\right) \\
\end{align}
$$

2) Once I have the graph, how do I calculate $\mathrm pK_\mathrm a$ from it? Assumed it would be using the gradient of the line of best fit, but that won't work since $\mathrm pK_\mathrm a$ is given by the difference of $\mathrm{pH}$ and $\log_{10}[A/(A_f-A)]$ rather than the quotient.

Make the plot $10^{\mathrm{pH}}(A_{\ce{L-}} - A)$ versus $A$,
the adjusted equation will be of the form
$$
10^{\mathrm{pH}}(A_{\ce{L-}} - A)
  = 10^{\mathrm{p}K_\mathrm{a}}A
  - 10^{\mathrm{p}K_\mathrm{a}}A_{\ce{HL}},
$$
the slope is $10^{\mathrm{p}K_\mathrm{a}}$, that is, $\mathrm{p}K_\mathrm{a} = \log_{10}(\text{slope})$.


Van der Waals real gas equation Van der Waals corrected the ideal gas equation for real gases and gave the following equation:

$$\left(p+\frac{an²}{V^2}\right)\left(V-nb\right)=nRT$$

I’m very confused about one aspect of this equation, which is this: why add the pressure term but subtract the volume term?

Edit: If somebody could please just explain the physical significance of the corrected equation and also, exactly what Van der Waals was trying to do by correcting this equation, it would be tremendously helpful for me.

Let me begin my discussion by first introducing the notion, using some intuitive model building, and will try to derive, or perhaps guess the van der Waals equation of state.

The perfect gas equation, $ PV = nRT $ doesn’t allow for any interactions between molecules. The van der Waals equation is a refinement of this model, in the sense that it introduces the possibility of both repulsive and attractive interactions.

Imagine the gas molecules are small, hard, impenetrable spheres. This accounts for repulsive interactions, as now they can’t get arbitrarily close to each other. Consequently, they are no longer free to move about in a volume, $V$, and only have access to a much smaller volume $V-nb$ where $nb$ is roughly the volume taken up by the molecules.

The modified equation now becomes:
$$P = \frac{nRT}{V-nb}$$

Now to calculate the excluded volume:
$$V_\text{molecule} = \frac{4}{3}\pi r^3$$
where $r$ is obviously the radius.

Now note, a particle is surrounded by a sphere of radius $2r$ (two times the original radius) that is forbidden for the centres of the other particles. If the distance between two particle centres were to be smaller than $2r$, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do.

The excluded volume for the two particles (of average diameter $d$ or radius $r$) is: $\frac{4}{3}\pi (2r)^3$ or, $8V_\text{molecule}$. Divide this by two to get the volume excluded per molecule and we get $b \approx 4V_\text{molecule}$

Note: Of course, molecules are not infinitely hard, as van der Waals thought; there exists some “softness”. Thus, the factor four yields an upper bound; empirical values for b maybe lower.

Now turn on an attractive force between the particles.

Assume, notwithstanding the existence of this force, the density of the fluid is homogeneous.

And assume that a large majority of the particles are always surrounded by other particles, which attract each other (unless they get too close, when they repel; thus, the bulk of the particles do not experience a net force pulling them to the right or to the left. They feel that the container is infinitely large without any walls.

Now, the surface particles feel a net force from the bulk particles pulling them into the container, because there are no particles on the side where the wall is (and we assume here that there is no interaction between walls and particles, which is not true in reality).

This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density:
$$F = \frac{N_\mathrm a}{V_\mathrm m}$$

Assuming homogeneity, the number of surface particles is also proportional to the number density. So the total force on the walls is decreased by a factor proportional to the square of density.

Since pressure is force per unit area, it's decrease proportional to:
$$a'C^2 = a'\left(\frac{N_\mathrm a}{V_\mathrm m}\right)^2 = \frac{a}{V_\mathrm m^2}$$

Now, the modified equation becomes:
$$ P = \frac{RT}{V_\mathrm m-b}- \frac{a}{V_\mathrm m^2}$$

Derivation using Statistical Mechanics

For a system of $N$ particles, in 3 dimensions

The partition function:
$$ Z = \frac{1}{N!} \int \mathrm dp^{3N}\mathrm dx^{3N}\mathrm e^{-\frac{\beta}{2M} \Sigma p^2_n}\mathrm e^{-\beta u(x)} $$

This integral factorises into two, an integral over $p$, and an integral over $x$, i.e. a kinetic energy contribution and a potential energy contribution.

I write the result as
$$Z = \frac{z^N}{N!}$$
for an ideal gas, this result is
$$\frac{V}{\Lambda^{3N}}$$
where
$$ \Lambda = \frac{h}{(2\pi mk_\mathrm bT)^{1/2}}$$
(thermal wavelength)

For a van der Waals gas,

$$
U(r) =
\begin{cases}
\infty, & \text{when }r<d \\
-\epsilon\left(\frac{d}{r}\right)^6, & \text{when }r \ge d
\end{cases}
$$

basically attractive up to a certain distance, and repulsive when they come too close. r is the distance between the spheres and d is the distance where they touch. Also, each particle moves independently in an average potential field offered by the other particles

Also, for an ideal gas we integrate over all space i.e. the entire volume of the container, however, we cannot do that anymore. Instead our volume of integration is $V - Nb'$ (or $V – nb$, same as what we did before, $n$ is number of moles instead of number of particles).
$$b'= \frac{2\pi d^3}{3}$$

Now, assume that in a shell of thickness $\mathrm dr$ there are $ \frac{N}{V} 4 \pi r^2\mathrm dr$ particles, i.e. the position of the particles is averaged (reality, like always is different)

So for a single particle, the attraction felt is
$$\phi = \int_{d}^{\infty} u(r)\frac{N}{V} 4 \pi r^2\mathrm dr$$
and
$$z = \frac{(V-Nb')\mathrm e^{-\phi/2k_\mathrm bT}}{\Lambda^3}$$

I will omit the computations and quote the results,

$$\phi = -2a' \frac{N}{V}$$
where
$$a' = \epsilon (2/3) \pi d^3$$

going back and putting things together,

$$\ln Z = N\ln(V-Nb') + \frac{N^2 a'}{Vk_\mathrm bT} - N\ln(\Lambda^3) - \ln(N!)$$

Now,
$$ P = k_\mathrm bT \frac{\partial \ln Z}{\partial V}$$
which turns out to be,

$$ P = \frac{Nk_\mathrm bT}{V-Nb'}- \frac{N^2 a'}{V^2}$$

UPDATE: Physical Interpretation

I am writing this section because I get the impression (from the discussion the in the comments) that my answer hasn't adequately addressed this.

In the limit $P \to 0$ (or $V_m \to \infty $) we can expect perfect gas behaviour, i.e $PV = \text{constant}$

However, say at high pressures, we observe that real gases would have larger molar volumes than expected, so we remove a term from the volume.

Similarly, the pressure is lower than ideal gas pressure, so we add a term to it to compensate.

So, instead of saying $PV_m = \text{constant}$, we say $ (P +\text{"something"})(V_m - \text{"something"}) = \text{constant}$

It also accounts for critical behaviour, simply solve the following system to obtain critical constants (since the critical point, is a point of inflection on the PV diagram)
$$ \frac{\partial P}{\partial V_m} = 0 $$
$$ \frac{\partial^2 P}{\partial V_m^2} = 0 $$

Of course, it would be too optimistic to expect a simple, elegant model such as this to truly determine all properties of gases with complete precision; the van der waals equation of state does have its failings, but if nothing else, it improves at least the qualitative discussion of properties/behaviours of real gases


What is the role of dichloromethane in the ozonolysis / Criegee oxidation? Why is $\ce{CH2Cl2}$ used in ozonolysis of alkenes and alkynes? What role does it play in the reaction? Is dichloromethane a common solvent for organic reactions and used for its inertness, or is it particularly used in ozonolysis reactions? If it is the latter case, how does it enhance the ozonolysis reaction?

Sometimes the solvent choice is very important to enhance or decrease reactivity of either of the reactants.

Dichloromethane is a widely used organic solvent. Chemists sometimes say tongue-in-cheek that it can dissolve almost anything organic since it is rather mid-range in polarity. Other than that, it is also non-nucleophilic, non-electrophilic and generally does not take part in many reactions. (There are certain exceptions, however!) This means that it is a standard solvent choice for many reactions and the ozonolysis is no exception. Most reactions require a solvent to allow all reactants to get into close proximity of each other for the reaction to occur at all — not many reactions can also be conducted in solid state.

The standard mechanism for ozonolysis — the Criegee mechanism — is given for solvents such as dichloromethane, hexanes or any others that do not participate in the reaction in any way (that are inert). However, when using methanol or other alcohols the mechanism must be modified as also highlighted by Criegee in the original mechanism publication. I have included the mechanism for ozonolysis both in presence and in absence of methanol in the scheme below.

The resulting peroxy-methyl-hemiacetal is much more stable than the carbonyl oxide and thus various side reactions, e.g. formation of Staudinger ozonides with remote carbonyl groups are inhibited. Thus, in the case of the ozonolysis, adding methanol decreases the reactivity of the ozone in solution. Conduction the reaction in dichloromethane should be considered the standard case — unless one wants to say that it enhanced ozone’s reactivity.

These details can also be found on organic-chemistry.org and this page from Michigan State.


Does nitrogen deaeration need to be performed in a fume hood? I have a couple electrochemical cells I will need to sparge (deaerate) with nitrogen prior to and possibly during my experiment (the solution is .1 M NaOH). Does this need to be done in a fume hood?

From what I've read, it seems that $\ce{N2}$ gas isn't a huge danger, but if you're in a confined space it can cause a severe lack of oxygen. If that is the case, would a normal-sized classroom count as a confined space? Also, from what I understand about deaeration, the nitrogen bubbles escape the solution with impurities, but also with oxygen, so maybe in my case the lack of oxygen wouldn't be a problem.

This may be a simple question, but it has been difficult for me to find standard procedures for sparging a liquid with an inert gas, and I want to make sure I'm taken all the necessary precautions and steps.

As implied in the comments, you have nothing to fear by sparging your solution with nitrogen in the classroom. Even a good size tank only contains a few cubic meters of compressed gas, while your classroom likely contains a few hundred cubic meters of air. Thus, even if you could instantly empty the tank the effect would be negligible. And you will be using low enough flow rates that there will be no fear of a "localized" depletion of oxygen.

The safety issue that I need to mention, though you are almost certainly already aware, is regarding the sodium hydroxide solution. Even at $\pu{0.1 M}$ it's not something that you want in anyone's eyes, so protection against splatter from the sparging solution would be prudent.


What is the rotational symmetry number used to calculate the rotational entropy from an ORCA frequency calculation? The quantum chemistry software ORCA gives the following warning on frequency calculations:

CAUTION: The rotational entropy is not quite correctly treated here
         because it includes a symmetry number that is not yet correctly
         implemented in ORCA!
For a nonlinear molecule the correct rotational entropy is:
    S(rot) = R*(ln(qrot/sn)+1.5)
    R    = 8.31441 J/mol/K = 1.987191683e-3 kcal/mol/K
    qrot = 4922043.1563766
    sn is the rotational symmetry number. We have assumed 3 here
       if it is different for your molecule then you should correct
       the printed rotational entropy by manually evaluating the equation
       as given above

For convenience we print out the resulting values for sn=1 - 12:
 sn= 1  qrot/sn= 4922043.1564 T*S(rot)=    10.02 kcal/mol T*S(tot)=    34.72 kcal/mol
 sn= 2  qrot/sn= 2461021.5782 T*S(rot)=     9.61 kcal/mol T*S(tot)=    34.31 kcal/mol
 sn= 3  qrot/sn= 1640681.0521 T*S(rot)=     9.37 kcal/mol T*S(tot)=    34.07 kcal/mol
 sn= 4  qrot/sn= 1230510.7891 T*S(rot)=     9.20 kcal/mol T*S(tot)=    33.90 kcal/mol
 sn= 5  qrot/sn=  984408.6313 T*S(rot)=     9.06 kcal/mol T*S(tot)=    33.77 kcal/mol
 sn= 6  qrot/sn=  820340.5261 T*S(rot)=     8.96 kcal/mol T*S(tot)=    33.66 kcal/mol
 sn= 7  qrot/sn=  703149.0223 T*S(rot)=     8.87 kcal/mol T*S(tot)=    33.57 kcal/mol
 sn= 8  qrot/sn=  615255.3945 T*S(rot)=     8.79 kcal/mol T*S(tot)=    33.49 kcal/mol
 sn= 9  qrot/sn=  546893.6840 T*S(rot)=     8.72 kcal/mol T*S(tot)=    33.42 kcal/mol
 sn=10  qrot/sn=  492204.3156 T*S(rot)=     8.65 kcal/mol T*S(tot)=    33.36 kcal/mol
 sn=11  qrot/sn=  447458.4688 T*S(rot)=     8.60 kcal/mol T*S(tot)=    33.30 kcal/mol
 sn=12  qrot/sn=  410170.2630 T*S(rot)=     8.55 kcal/mol T*S(tot)=    33.25 kcal/mol

Now, I have got the following two questions:

What is the rotational symmetry number $s_n$, i.e. how is it defined?
What symmetry number $s_n$ would a molecule with no symmetry, i.e. $C_1$ point group have? The way I understand it, the rotational symmetry number reflects the rotational symmetry, e.g. $\ce{CHCl3}$ would have $s_n=3$ as there are three equivalent structures superimposed by the $C_3$ symmetry element. So that my guess is that $s_n = 1$ in that case. Is that correct?

Your #2 is exactly right, as far as I know, which answers your #1.

A key thing to remember is that the rotational symmetry number doesn't include reflections, so molecules in groups like $C_s$ still have $s_n = 1$ since there are no superimposable orientations achievable only with rotations.

For reference: Table II of the NIST CCCBDB introduction to thermochemistry is a nice, concise summary of the symmetry numbers for the common point groups observed in chemistry.


Detection of NaF levels in aqueous solution also containing NaCl Is it possible to determine the concentration of sodium fluoride, more specifically fluoride ions, in a solution that also contains sodium chloride?
I have looked into using an ion-specific electrode, but this seems very expensive.

Depending on the equipment you have available, you could buy some $\ce{CaCl2}$ de-icer, dissolve it in water, and add it to a known volume of your solution.  Highly insoluble $\ce{CaF2}$ will precipitate out; keep adding the $\ce{CaCl2}$ solution until no more precipitation is observed.

You can then filter the precipitated fluoride from solution, rinse and dry it, and weigh it.  The original fluoride concentration can then be back-calculated based upon the mass of obtained $\ce{CaF2}$, the percent by weight of fluoride in $\ce{CaF2}$, and the original solution volume.


Why does the tautomerism of purine favor 9H-purine? I tried to explain the equilibrium by comparing the basicity of different nitrogen atoms in a purine anion, but I can’t see any obvious difference. Seems like that the 9-N will be the most basic because it can have two adjacent imine structure while keeping the structure of pyrimidine untouched. Computational results will be appreciated.

P.S. I don’t think the question is a duplicate of the “purine nitrogen basicity” one because that one compares the basicity of nitrogens in a neutral purine molecule whereas my question compares the basicity in a purine anion.

The given purine has two structures as shown below.

In $\ce{9H–Purine}$ (figure 2), lone pairs are far apart from each other (3 and 7). Comparatively in $\ce{7H–Purine}$ (figure 1) lone pairs are very near each other (3 and 9) . Nearness of lone pairs in $\ce{7H–Purine}$ could leed to greater repulsions between them. This could make it energetically unfavorable in comparison to $\ce{9H–Purine}$.

This could be the reason why $\ce{9H–Purine}$ is more favored tautomer compared to $\ce{7H–Purine}$.


Electrophilic substitution reactions Why is the para position preferred more than ortho in the electrophilic substitution reactions of haloarenes?

Is it really? Always?

In an older article (J. Chem. Educ., 1988, 65, 367), the authors described an experiment that demonstrates the outcome of the alkylation of chlorobenzene (1) with 2-chloropropane (2) under kinetic control.

They first added 25 mL of 1 and 2 mL of nitromethane to 1.2 g anhydrous aluminium chloride, and stirred until the aluminium chloride was fully dissolved. To the cooled mixture (3-5 °C), a solution of 3 mL 2 in 20 mL of 1 was dropwise added within 10 min, while keeping the temperature of the reaction mixture below 10 °C. The reaction was stirred for another hour, allowed to come to room temperature.

After workup, gas chromatographic analysis showed that the o/m/p ratio for the resulting isopropyl chlorobenzenes was 63/5/32.

To cut the long story short, under the conditions describe above, the ortho isomer prevailed.


Why is a 7-endo radical cyclisation favoured over a 6-exo in this synthesis? Normally, according to Baldwin's rules, 3-exo to 7-exo cyclisations are preferred over endo cyclisations. But in the following example, a 7-endo-trig reaction is favoured over a 6-exo-trig and I am just not able to understand why. It's a real synthesis example (taken from Eur. J. Org. Chem. 2015, 2015 (15), 3240–3250). Please explain to me why the 7-endo product is favoured here.

I'm the co-first author of this paper. I want to thank Mathew Mahindaratne for the helpful and thoughtful discussion. I just want to add one more result when I was doing the synthesis, please see the attached. The beauty of total synthesis is that we never 100% sure what the product we get, but we can always find a way to explain the results.


Is it possible to improve electrolysis rate without increasing heat? I'm separating water into hydrogen and oxygen by connecting a DC current to 99.9% pure nickel electrodes in a solution of NaOH in deionized water.

It works great, except that any attempts to increase the rate of production by increasing current, increasing NaOH, bringing electrodes closer together, etc. also result in an increase in heat.

At the moment the bottleneck for my production rate is trying to keep the water from boiling or otherwise causing heat damage to the apparatus.

I've considered investing in a cooling system, but I'm also interested in knowing whether or not it is possible to somehow improve the efficiency of the electrolysis and thereby improve separation rate without increasing heat.

(In case you're wondering, I'm using hydrogen to fill balloons.  It works quite well, I can fill a normal sized balloon in a few minutes, but then I need to wait an hour for everything to cool down.  (Yes I know hydrogen and fire don't mix well.))

Consider the electrolysis apparatus as a resistor in series with an EMF. Ideally, the cell voltage should be just enough to overcome the EMF (potential needed to split water), but some over-voltage is inevitable to have a decent current flow. Much of the heating is due to this wasted over-voltage. Then, to increase the efficiency and reduce heating, it's necessary to reduce the series resistance.

$\ce{NaOH}$ is already highly ionized, so adding more, beyond a certain point, is not the solution (no pun intended;). Close spacing helps, but you must leave room to physically separate $\ce{H2}$ from $\ce{O2}$. The remaining variable is electrode area: make the electrodes as large as feasible. You can further increase area by using porous sintered, or at least coarsely-sanded electrodes.

Quantum Sphere Inc. claims to produce a more active electrode from $\ce{Ni-Fe}$ particles, though they're probably a bit pricey for filling hydrogen balloons. Perhaps they offer samples?

BTW, I assume you're using a moisture trap for the gas, otherwise bubbling hot caustic electrolyte would inadvertently wind up in the balloon, reducing lift, damaging the balloon and possibly causing severe chemical burns. $\ce{NaOH}$ is called "caustic soda" for good reason!


Sodium chloride + sulfuric acid Why does sulfuric acid displace more volatile acids from salts?
My textbook says that sulfuric acid can displace more volatile acids from metal salts.
How is $\ce{HCl}$, which is not even a reactant, 'displaced' from $\ce{NaCl}$, as there is no $\ce{HCl}$ to be displaced?

How is $\ce{HCl}$, which is not even a reactant, 'displaced' from
  $\ce{NaCl}$, as there is no $\ce{HCl}$ to be displaced?

Well, there will be some $\ce{HCl}$ due to a well-known chemical reaction which is used both in the lab as well as in production to get the hydrogen chloride:

$$\ce{NaCl(s) + H2SO4(s) → NaHSO4(s) + HCl(g)}$$

The reaction proceeds at room temperature, but note, that the reagents should be dry. If you heat it up above 200 °C reaction proceeds even further:

$$\ce{NaCl(s) + NaHSO4(s) → HCl(g) + Na2SO4(s)}$$

Hydrochloric acid is then prepared by dissolving hydrogen chloride in water.


Reason for double acylation of ferrocene only occurring with certain reagent combinations Can anyone explain why the acylation of ferrocene produces the following results:

Ferrocene + acetic anhydride + AlCl3 ==> mono-acylated ferrocene

but:

Ferrocene + acetyl chloride + AlCl3 ==> di-acylated ferrocene (an acyl group on each ring).

I have seen various statements confirming this, but not a single explanation for it! My current thoughts are that it must be something to do with the carboxylic acid produced from the reaction with the anhydride somehow hydrogen bonding/blocking the second site, but not really sure how it would work.

Any explanations greatly appreciated!

One factor:  since ferrocene is covalently bonded, the rings "communicate" with each other via the molecular orbitals passing through the iron atom.  Thus the acetyl group placed on one ring not only deactivates that ring, but to some extent deactivates the second ring as well.  The second ring would still be attacked preferentially in the second stage but may require a stronger electrophile, which acetyl chloride is compared with acetic anhydride.


Bond dissociation energy among halogens Among the halogens, why is it that the fluorine has the lowest bond dissociation enthalpy, considering the fact that fluorine is the smallest and the internuclear distance between the fluorine molecule is least and $2p-2p$ overlap should also very very strong ,the bond dissociation enthalpy should have been very high but it is lowest among the halogens. Why is it so??

The strength of the bond is due to 2 things : Difference in electronegativity and atomic radius of the elements .Halogens except fluorine , have a very big atomic radius so less bond energy.And they have 0 difference in electronegativity so even less bond energy.They both "want" to get the bonding pair of electrons closer to them , so because they cant both atoms are "unhappy".


Why do coefficients become exponents in the speed equation when dealing with elementary reactions In school, while studying chemical kinetics, we were learning about the speed law and equation for different chemical reactions. We learned that in general, the exponents of the concentration of reactants in the speed equation must be determined experimentally. However, earlier this week, we learned that if we are talking about basic elementary reactions, the coefficients of the reactants become the exponents in The equation. This puzzled me. Can anyone explain why the coefficients would become exponents? Thanks in advance

A reaction occurs when its reactives are in the same region of the space. If you have an elmental reaction of the form $a A + b B \rightarrow $, for the reaction takes place it is needed that $a$ molecules of $A$ and $b$ molecules of $B$ collide. How probably is this event? Well, if you only need one molecule of $A$ ($a=1$), the probability that an $A$ molecule be in this specific place (at some time) is proportional to the $A$ concentration. If $(a=2)$ we need another $A$ molecule, as the probability for a second molecule being in that space region (at some time, in particular at the same time) is independent, the probability of finding two $A$ molecules in the same space region (at the same time) is the product of the two individual probabilities (is like throwing a dice two times). In general, for the mixture the probability of the encounter is proportional to $[A]^a [B]^b$.
The probability of encounter is proportional to the number of encounters in some time interval. And the later to the reaction velocity.


Is there ANY chemical that can destroy PTFE, or Teflon? Polytetrafluoroethylene was discovered by accident. It now is an important material in the industry mainly because of its extremely high bonding energy, which prevents corrosion, halts reaction, and reduces friction (yeah carbon-fluorine bonds!)

And people would have definitely put it to the test, making it contain some of the most vicious and chemically diabolical substances ever created. There is a whole HOST of items it can contain that some chemists have gone so far as to say they were 'evil':

Dioxygen Difluoride
Known as the gas of Lucifer, there is a whole list of people blown up and killed while just trying to work with one of its components, fluorine. It ignites stuff at temperatures that most of the stuff that we breathe in would be in liquid form. No one really knows about its atomic structure (obviously).
Fluoroantimonic Acid
With a staggering pH of -25, it chews through stuff you might not even believe could be corroded; like wax or glass. It can even strip hydrogen off of methane

...There are a lot of other chemical demons it can contain, but this is not the point. Let this suffice: Chemical Resistance Comparison (Spoiler: Fluorine is good at this corrosion thing.)

With this kind of hyper-resistance to about anything chemically destructive, is there anything that can destroy Teflon through only chemical means? A chemical that reacts exothermically to release heat, which melts the PTFE does not count. You get the drift.

Also, I am very curious as to whether there is anything more resilient than Teflon? Polytetrafluoroethylene is made of many carbon-fluorine bonds in series. However, carbon-fluorine is second only to the Si-F bond. Is there an "overclocked" Teflon made of silicon-fluorine bonds that is even stronger?

EDIT: Now I know that some, but very few, solvents can make a mark on Teflon; but my question has not been answered: Is there any more resistant substances?

(More Teflon bragging: Here. Take that aqua regia)

Corrosion Resistant Products, Ltd., with the help of Dupont, has established this source of information on what can and cannot eat teflon.

Here's a list:

Sodium and potassium metal - these reduce and defluorinate PTFE, which finds use in etching PTFE
Finely divided metal powders, like aluminum and and magnesium, cause PTFE to combust at high temperatures

These reactions probably reduce PTFE in a manner that starts:

$$\ce{(CF2CF2)_{n} + 2Na -> (CF=CF)_{n} +2NaF}$$

The world's most powerful oxidizers like $\ce{F2}$, $\ce{OF2}$, and $\ce{ClF3}$ can oxidize PTFE at elevated temperatures, probably by:

$$\ce{(CF2CF2)_{n} + 2nF2 -> 2nCF4}$$

Similar things can occur under extreme conditions (temperature and pressure) with:

Boranes
Nitric acid
80% NaOH or KOH
Aluminum chloride
Ammonia, some amines, and some imines


History of octanol-water partition coefficient We are aware that one of the basic information about a material is the logP value it has. The logP is worked out from the partitioning between the octanol and the water phases. Does anyone know who was the first to introduce this octanol-water partition coefficient? I had a look through the literature but was not able to find the author of that rule.

The most oft-cited publication is Partition Coefficients and Their Uses. A. Leo, C. Hansch, and D. Elkins. Chemical Reviews 71(6), 525-616 (1971). This manuscript was noted as a "Citation Classic" (see text here) in which A. Leo writes (in part):

  In recognition of the effort of designing
  a method of calculating log P (o:w) from
  structure, the early version of which was reported
  in this article, the Chesapeake section
  of Sigma Xi presented me with its Excellence
  in Science Award for 1980.”

Note bene: In the last article I've linked, mention is made that the phenomena was first investigated by Bertholet in 1872 and "put on a firm basis by Nernst in 1891." That said, the 1971 publication of Leo $\it{et\; al}$ is the first modern/comprehensive listing of octanol-water partition coefficients and the place where the symbology (and definition) were written.


Is the chemical structure of an amide bond (-CONH) or (-CONH2)? The Dictionary definition of an amide is

  an organic compound obtained by replacing the −OH group in acids by the −NH 2 group.

So from this, I deduced that the bond would be CONH2.

But when I searched around, I found some images that showed the bond only had one hydrogen atom, like so:

Which is the correct bond structure? I'm looking specifically at the condensation polymerization of diamine and diacarboxylic acid to form nylon.

And also, I see online that lots of bond structures (like NH2 and COOH) are prefixed with a dash, so they'd appear like -NH2 and -COOH. Is this just a personal preference, or is there a more elaborate reason behind this?

I'm currently doing the GCSEs.

Thanks

Amides are derivative of carboxylic acids. It consists of 2 parts, and the bond between carbonyl carbon and nitrogen of ammonia is called is amide bond.

The bond structure of 1° amides is as shown in the figure.

What you have shown in the question is bond structure of 2° amide, which can be obtained by removing 1 hydrogen from 1° amide.

Now if you observe structure of nylon, you will see there are n numbers of 2° amides.

  And also, I see online that lots of bond structures (like NH2 and COOH) are prefixed with a dash, so they'd appear like -NH2 and -COOH. Is this just a personal preference, or is there a more elaborate reason behind this?

The dash means it is Functional Group like -R means hydrocarbyl functional group, -COOH means carboxylic acid functional group, etc.


R and S Configuration Understanding So I have been trying to understand how to draw R and S configurations. I feel like I have understood how to draw them, but every time I do a question, I get a wrong answer.

So here is what I have understood,

The lowest priority goes to the back of the compound meaning on the dash line.
The highest priority is on the top.
You can play around with the 3rd and 4rth, so that one is on the
'out of page' and the other is on the line, so that you can
differentiate between the R and S.

But then there is this picture for example from my book,

I have been going crazy trying to understand it. The things I do not understand are these,

1) I understand the diagrams on the right side. That I have been doing as well

2) But then when we look at the diagrams on the left. The configuration is completely opposite. Even though it says S, the configuration is like of a R. Plus the highest one is going into the page and not the hydrogen. WHY? It is driving me crazy because they do that for all the examples. When they S, they actually make an R and vice versa. Whereas on the internet it is a different story. So I really need someone to tell me what I should do.

This is difficult to explain, even face to face, but I'll give it a try as I know how frustrating this can be to grasp.  Looking at the wedge and dash model, pick one of the four atoms/groups that ISN'T the lowest priority and ISN'T in the dashed (going away from you) position.  Doesn't matter which as long as it meets the two aforementioned rules.  Personally, I prefer to pick the wedged (going towards you) group.  Now, "freeze" whichever atom/group you chose, and by this I mean this group/atom will remain in the exact same position it's already in.  In fact, go ahead and draw the central carbon atom with the dashed line, wedged line, and both normal lines, but omit the groups/atoms for now (and draw it the same way it is pictured above on the left, it makes it much easier to visualize).  So you should have a C with all 4 bonds drawn out from it.  Now, go ahead and write in the atom/group you decided to freeze in the SAME position as it was before.  Now, the objective is to get the H atom (or lowest priority group) to be at the dashed position.  And remember, the frozen atom/group cannot be moved.  There are only two other positions the H atom could assume, as it cannot occupy the frozen atom/groups location, and you don't want it in the position it's already in.  So in the case of the first molecule, it can either occupy the position of CHO or OH (if you chose the wedged group/atom to be stationary).  Well, you want the H atom to be in the dashed position, which is currently occupied by OH.  So on your new molecule, go ahead and draw in the H at the dashed position where the OH was in the original drawing.  Now, youve effectively rotated the H into the position that the OH was.  Now the final rule you must follow, is that no group EXCEPT the one you chose to freeze can remain in the same position as it was in the original drawing.  This leaves only one location for the OH to be moved to, and one location for the CHO to be moved to.  The hydrogen was already moved to a new location, so now you must move the OH group and the CHO group to new positions, and it cant be either of the already occupied positions (the positions occupied by hydrogen and the group you chose to freeze, which in this example is the wedged position).  The OH must occupy the old position of CHO, and the CHO group must occupy the old position of H (why must they occupy these exact positions?  Because there are only two unfilled positions, and remember the newest rule, all groups/atoms that arent the frozen group MUST occupy NEW positions).  So go ahead and draw them in.  Now your molecule should be ready to evaluate for R/S configuration.

Be consistent when redrawing the molecule with the original drawing.  Don't draw the wedged bond on the top of the molecule if it was on bottom in the original drawing. Hopefully this helped, this is a confusing subject though and like I said earlier it's difficult to explain even in person.  I would definitely recommend getting a set of ball and stick models as it makes visualization much easier.  Practice enough and you'll get to where you dont need them anymore.

By the way to clear up any confusion, this step-by-step is essentially rotating the molecule (you aren't changing the structure or stereochemistry).  I simply laid out detailed directions so that if you were having difficulties visualizing rotating the molecule in your head, you wouldn't have to think about it.  As long as you follow the directions and rules I have laid out, it should result in a new drawing of the original molecule from a different angle, the angle in which the lowest priority group is in the dashed (going away from you) position


Combining solutions with multiple concentrations - Mixing NPK fertilizer ratios I'm trying to mix my own fertilizer to grow some peppers.
Fertilizers are labeled with NPK ratios, so mixing two volumes and finding the result is easy: Just multiply the concentration of each component by the ratio of the solution over the total volume. For example, $\pu{1L}$ of a $3:4:6$ solution plus $\pu{1L}$ of a $5:2:6$ solution yields $\pu{2L}$ of an $4:3:6$ solution.
Something like:
$$ NPK_\mathrm{final} = \left(NPK_{s1} \times \frac{V_{s1}} {V_{s1} + V_{s2}} \right)+ \left(NPK_{s2} \times \frac{V_{s2}} {V_{s1} + V_{s2}}\right) $$
However, if I have a desired ratio in mind, how can I solve for how much of each of 2 solutions to add? I can plug in different values till I get close, but I'm curious if there's a general formula. I figured there's probably a way to get there with linear algebra (but I never took linear algebra!).

Let consider ratio of N/P.
Let fertilizer A has this ratio N/P=a.
Let fertilizer B has this ratio N/P=b.
Let you want the ratio c, where a < c < b.
Then you need to mix A : B in ratio $$\frac{ b - c}{c-a}$$
In the Central Europe with German scientific influence, it is called the "mixing cross rule".

Note that you can chose just ratio of 2 from 3 elements and only in the range between these 2 source fertilizers. The ratio of the 3rd element is determined by your prior choice for the other two elements.
You would need 3 fertilizers to mix to independently choose ratio of all 3 elements, again within the range of these sources.

A graphical chart method(*) may be more practical than a numerical method.
One can draw a equal-side triangular ternary chart. Each each corner of the triangular ternary plot represents formally a fertilizer with the solely N ( or P or K ) component.
If we take e.g. the N corner, than lines  parallel to the side opposite to N corner represent N relative content to the sum N + P + K. The same for other 2.
Within this triangle, one can draw 3 points, each representing 3 different particular fertilizers.
Then one can draw similar chart between these 3 points for these fertilizers.
Available ratios than fall into this smaller triangle.
Then one can on the big chart choose N:P:K ratios and than read from the small triangle chart the ratios of all 3 fertilizers.

There is small catch in all this. The amounts of fertilizers to be mixed should be normalized for their total NPK content.
So mixing fertilizers A, B, C would not affect the total NPK amount per volume/mass, only the component ratio.

(*) It is also called a ternary plot on Wikipedia. See also the listed referenced there:

an online tool to create ternary plots)
simple Excel template
Trip-Plot Excel package

Or you can used PDF chart template to be printed, provided by @MaxW.

Alternatively, I have prepared on Google docs the spreadsheet,

using linear algebra, which provides the mixing rations for given 3 fertilizer compositions, checking forbidden combinations.


What is the structural formula of alkali hypohalite: MOX or MXO? There seems to be an inconsistency in the formula. Somewhere it is written as MOX and in some places, it is written as MXO. This is what Google gives when you search sodium hypochlorite:

I think the formula of alkali hypohalite should be MOX because since the negative charge bears over the oxygen atom, it accepts the electron coming from the alkali metal ion and forms a duplet  thus forming ionic bond.

Is my assumption correct? What is the correct formula of alkali hypohalite?

I felt tempted to close this as a duplicate of the sodium acetate notation question, because of my answer there, but the question in itself does not ask the same thing, so I’ll just rewrite the relevant parts of my answer.

In short, there is no correct formula. Depending on which point you want to bring across, you can use either, and depending on the context, $\ce{ClNaO}$ may also be acceptable.

Structually, you are correct with your assumption; the oxygen bears the negative charge. But that doesn’t include any implication that the formula should be written that way; compare $\ce{SO4^2-}$ where the charge is also on the oxygen atoms and yet they come last.

In fact, that leads us to our first clue. By convention, oxygen atoms are typically written last; compare $\ce{ClO3- , ClO4- , BrO3-}$ and others. You could probably explain that with electronegativity, the convention being that electronegative elements come last. This assumption argues for $\ce{NaClO}$.

On the other hand, the corresponding acid $\ce{HOCl}$ is often understood as derived from water $\ce{HOH}$, replacing a hydrogen with a chlorine atom. From this starting point, it makes sense to write $\ce{NaOCl}$, because then you replaced the other hydrogen; this time with sodium. It also includes a resemblance to $\ce{NaOH}$.

Both ways can thus be supported by the reasons given above. In practice, I think I have always seen the substance being referred to as $\ce{NaOCl}$ but I could simply have ignored the occurances of $\ce{NaClO}$.


Is there an example of a bond with negative bond enthalpy? My textbook says bond breakage is "generally endothermic." This implies there are some bonds out there that release energy when broken.
Is there a classic example? Perhaps diatomic Helium?

Bond enthalpy is like a measure that tells us how strong/stable a particular bond is.
Definition:

Bond enthalpy (also known as bond energy) is defined as the amount of energy required to break one mole of the stated bond.

To break a bond, energy must be provided to the bond. How is that energy provided?

The reactants are either heated or they take up energy from the surroundings, that is, they take in energy in the form of heat from the surroundings, overcome their bond enthalpies and break the bond.
All bonds require energy to be broken.
A negative Bond enthalpy simply means that the bond is highly unstable and would not exist in the first place.
I highly doubt if any compound has negative bond enthalpy.{Not even Dimeric Helium}
From Wikipedia:

Based on molecular orbital theory, $\ce{He2}$ should not exist, and a chemical bond cannot form between the atoms. However, the van der Waals force exists between helium atoms as shown by the existence of liquid helium, and at a certain range of distances between atoms the attraction exceeds the repulsion. So a molecule composed of two helium atoms bound by the van der Waals force can exist.The existence of this molecule was proposed as early as 1930.


Quantify the effectiveness of baseline correction in Raman spectra I've produced several thousand Raman spectra using hyperspectral imaging techniques. These suffer from significant fluorescent interference which produces a curved, sloped baseline.
Using Matlab, I have attempted several baseline subtraction techniques but I can't say quantitatively which is performing best - beyond a visual inspection I'm not sure how to compare them.

Is there any quantitative way of assessing the performance of a baseline subtraction? Or is visual inspection really the best we can do?

Thanks for taking the time to consider this.

Edit:
I've added an example spectrum along with one with a corrected baseline.

Thank you for your suggestions as regards methods for baseline correction, however, I have already used several different methods successfully. My intention was to discover if there was a way of measuring the performance of various techniques in order to asses which is the most effective.

Thanks again for your consideration.

Edit 2:
I just wish to clarify that I am not looking for methods for baseline correction. I have used several which have given me varying results - these include polynomial fitting, piece wise fitting, ModPoly, subtraction of iteratively smoothed savitzky-golay baselines, and WPLS.
What I'm looking for is a way to quantify their performance, is there any metric that can be used to do this?

Could you also mention the correction method? Your auto-fluorescence background is really huge and your signal to noise ratio is low. One of the most cited methods in automated background correction is indeed based on polynomial fitting, albeit little but more sophisticated than just least square fitting. This article is worth reading.

Automated Method for Subtraction of Fluorescence from Biological Raman Spectra
https://journals.sagepub.com/doi/abs/10.1366/000370203322554518

Another simple option is that you can choose a spectral range where to apply a polynomial baseline. Say from 0 to 1200 you apply a polynomial of degree 5, but 1200 onward you choose polynomial of degree 6. However, you have thousands of spectra and it is not clear whether your background shape is similar in each case or not.

Addendum:  I am afraid there is no solid way to compare the baseline correction approaches. The reason is that you do not know the "true" model of the autofluorescence. This is also stated by Ed in comments. He is a leading spectroscopist. In such cases, you should simulate a baseline like these authors and do a comparison. Make a synthetic baseline by polynomials with say n=5, make it more complicated by mixing certain regions with polynomial  n=6, 3, 2, and then try your approaches. Don't add any peaks to the baseline, and just check which technique is showing least residuals. Is there a trend in the residuals? The trend is also important. The residuals should be scattered. What technique did you use in your case?

Alternatively, you can try a known fluorescent compound, Rhodamine G, and collect a fluorescence spectrum. In this case you will know the true shape of the spectrum. Use that as a true model and try your baseline corrections.

One general advice given to us by a professor, who was an author of Raman spectroscopy monograph, was that a better and better fit does not indicate reality. The model may be completely wrong but may have a correlation coefficient close to unity. Thus fitting a polynomial or any experimental baseline correction is just a utilitarian approach, which is changing the aesthetics and cosmetic appearance of a Raman spectrum. There is nothing fundamental going on.


Should equiatomic iron–cobalt alloy be written as FeCo or CoFe? I have an alloy made of 50% iron and 50% cobalt. I see occasionally in the literature this is written as FeCo alloy.
But shouldn't it be CoFe alloy since the chemical symbols are supposed to be arranged in alphabetical order?
Is there somewhere where these rules are written down?

Current IUPAC recommendations suggest following electronegativity ordering. The symbol for iron (a less electronegative element among the two) should be written first: $\ce{FeCo}.$
From the section IR-4.4 Sequence of citation of symbols in formulae [1, pp. 58–59]:

IR-4.4.2 Ordering principles
IR-4.4.2.1 Electronegativity
If electronegativity is taken as the ordering principle in a formula or a part of a formula, the atomic symbols are cited according to relative electronegativities, the least electronegative element being cited first. For this purpose, Table VI* is used as a guide. By convention, the later an element occurs when the table is traversed following the arrows, the more electropositive is the element.
[…]
IR-4.4.3. Formulae for specific classes of compounds
IR-4.4.3.1 Binary species
In accordance with established practice, the electronegativity criterion (Section IR-4.4.2.1) is most often used in binary species.2
[…]

For intermetallic compounds, earlier recommendations prescribed alphabetical ordering
rather than by electronegativity (see Section I-4.6.6 of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 1990, ed. G.J. Leigh, Blackwell Scientific Publications, Oxford, 1990).

Examples:
[…]

$\ce{Rb15Hg16}$
$\ce{Cu5Zn8}$ and $\ce{Cu5Cd8}$

Table VI as seen in [1, p. 260]:

Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.


What does it mean by " concentrations are (value) in relation to each other"? This is not a home work help question or something like that, these days I self learn chemistry and when I was doing a model paper I came across this question. And the question itself is not difficult but I don't understand this term.

More accurately, I am not familiar with this term,

that part of the question says,

  The cation Ag+(aq) and Hg2^2+ are in an aquesous solution.Their concentrations are 0.2 moldm-3 in relation to each other. Find which cation precipitates first as its iodide when a solution of NaI is added gradually to the above solution... and there are some other sub questions,

If you can explain to me if what it means by "Their concentrations are 0.2 moldm-3 in relation to each other", I think I can do the question easily, Thank you!

The cation Ag+(aq) and Hg2^2+ are in an aqueous solution. Their
  concentrations are 0.2 moldm^-3 in relation to each other. Find which
  cation precipitates first as its iodide when a solution of NaI is
  added gradually to the above solution

It is a poorly worded question. The author means that the ratio of Ag conc. to Hg conc. is 0.2. It is implied that the concentration of silver is lower. Which ion will ppt out first, if we add NaI?

The author should have written that the relative ratio of Ag:Hg concentration is 0.2.


What are the effects of the repulsive forces between electron on the potential energy graph for the dihydrogen molecule? This diagram is from my textbook. It states that as long atoms get closer, the energy decreases due to the attractive forces between the electrons and the nuclei. When the nuclei get too close, the nuclei’s repulsive forces drives the energy upward. However, what of the repulsive forces between the electrons? As the atoms get closer and closer from infinity, where are the effects of the increasing repulsive forces between the electrons?

First of all, you have to understand that the potential energy depicted on the graph in your question already accounts for repulsive forces between the electrons. This potential energy is the sum of two terms, namely, the sum of electronic energy $E_{\mathrm{e}}$ and nuclear repulsion energy $V_{\mathrm{nn}}$, where the first term contains contributions from kinetic energy of electrons motion $T_{\mathrm{e}}$, potential energy of their interactions with the nuclei $V_{\mathrm{en}}$, as well as potential energy of their interactions with each other $V_{\mathrm{ee}}$. This sum of electronic energy $E_{\mathrm{e}}$ and nuclear repulsion energy $V_{\mathrm{nn}}$ plays a role of potential energy for the motion of the nuclei, hence its name.

All these components of the potential energy, $E_{\mathrm{e}} = T_{\mathrm{e}} + V_{\mathrm{en}} + V_{\mathrm{ee}}$ and $V_{\mathrm{nn}}$, behave differently with respect to internuclear separation. For instance, as you have mentioned, the last term $V_{\mathrm{nn}}$ grows when nuclei get closer to each other. The same has to be true for $V_{\mathrm{ee}}$ term since the closer the nuclei are the closer are the electrons. However this effects of increasing the energy could be overcompensated by a decrease in $V_{\mathrm{en}}$ term thanks to attractive nature of the corresponding forces between nuclei and electrons.

When a decrease in $V_{\mathrm{en}}$ overcompensates an increase in $V_{\mathrm{nn}} + V_{\mathrm{ee}}$ the bond is formed. Not that an overall decrease in potential energy is accompanied by an increase in kinetic energy in accordance with the virial theorem. But the increase in kinetic energy is only half of the lowering in the potential energy, so that the total energy goes down when a bond is formed.

Here is the graph of all these terms for $\ce{H2}$ molecule obtained from quantum chemistry calculations1:

Wilson, C. W. & Goddard, W. A. The role of kinetic energy in chemical binding. Theor. Chim. Acta 26, 195–210 (1972). PDF from caltech.edu.


What about the aromaticity of cyclopropenylidene? Here's a compound which we have to tell whether it's aromatic or not.

Drawing pi-orbital for the 3-membered ring, I think there are only 2 pi electrons delocalised all over the ring. The orbital having non-bonding electrons on the carbon having sp2 hybridization(carbon no. 1) is not having the correct orientation with the pi-orbital of the ring so these electrons are not delocalised over the ring. 2 being a Huckel's number, I think that this compound should be aromatic.

But the answer key says that it is ANTI-AROMATIC! Where am I going wrong? Perhaps the ring is not planar?

EDIT: The carbon no. 1 doesn't have a negative charge so it is not an anion. It is neutral.

This is a problem of counting lone electron pairs as part of the delocalized pi system.  In pyrrole you count the lone pair on the nitrogen, in pyridine you don't.  What's the difference?
In pyrrole you have the lone pair left over after you have given every ring a ligand (the outer hydrogen atoms).  With the sigma bonding requirements thus satisfied, the lone pair goes into the pi system.  In pyridine the nitrogen is missing its ligand, so the lone pair must replace that in the plane of the ring and can't go into the pi system.
Which situation corresponds to your compound depends on what compound is meant, and there seems to be confusion on this point.  Cyclopropenylidene ($\ce{C3H2}$) and cyclopropenyl anion ($\ce{C3H3^-}$) differ only by a carbon-hydrogen bond, and the shorthand method commonly used to render organic structures thus fails to distinguish them.  The problem statement should be checked to ascertain which of these compounds is meant because they give different outcomes, and the answer key is right for only one of them.
If it is cyclopropenylidene
In cyclopropenylidene, the carbon with the lone pair is missing a ligand like pyridine, so the lone pair has to replace the missing ligand in the plane of the ring.  That leaves two pi electrons to be delocalized which makes the ring aromatic.  Cyclopropenylidene is not only aromatic by the electron count, it is a sufficiently preferred three-carbon structure to be seen in interstellar space.
If it is cyclopropenyl anion
Cyclopropenyl anion has all the ligands in place, so the lone pair goes into the pi system like pyrrole.  Unfortunately in this case, adding that lone pair gets you four pi electrons in the ring and that's bad.  In fact it turns out to be really had, even worse than larger (and thus more flexible) formally antiaromatic rings like cyclooctatetraene.
Cyclopropenyl anion goes through some serious contortions to get rid of this antiaromaticity.  See this question.