Wavy baseline in carbon NMR I realize anything NMR-related is essentially physics, but I figured this was a better place to ask this. I have this issue fairly regularly (with one instrument in particular) where my baseline ends up looking like the one in the image below. The (inverted) sine wave-shape is what I'm referring to, if it's not obvious.

Is there something specific that causes this baseline shape?  Is there a way to fix it? Preferably I'd want to use something that fixed the signal/input, but if I have to resort to a processing method, I'd settle for that.

This baseline roll seen in this spectrum is an example of acoustic ringing, and is quite normal for many low frequency NMR experiments that are recorded over wide sweep widths. It occurs as a consequence of slight breakthrough of the excitation pulse, as it rings down to zero. Ideally, you would like to start collecting signal at your receiver immediately after your excitation pulse, however it does take a small amount of time (microseconds) for the pulse to decay to zero. If your pre-scan delay (typically configured for your spectrometer during the configuration process) is too short, you can get this ringing artifact. It is more pronounced for weaker samples, but usually nothing to worry about.

To correct this as an experimental acquisition parameter, you would make your pre-scan delay longer; for Topspin, this is the parameter DE in your acquisition parameter list. It is normally defined during spectrometer configuration, and needs to be as short as practically possible to maximise your signal detection - typically less than 10 microseconds. For some nuclei, much of the signal has already decayed during this pre-scan period, and require some backward linear prediction to help. I'd speak to your local friendly NMR spectroscopist before mucking about with changing parameters like DE.

It is perfectly normal, and you should be able to get rid of it by applying a simple baseline correction algorithm - for Topspin you should use the commands abs or absd. This should work for your spectrum. As a word of caution when applying baseline corrections to your 1H spectrum, this will also apply an automatic integration for you, overwriting any integration you may have done previously. You can avoid this by using the optional n switch (abs n). For severe cases of baseline roll, apply a partial basline correction using absf between the limits of absf1 and absf2. Your Topspin manual can provide a fuller description, including other baseline correction modules.


What is the role of POCl₃ in the reaction to form (NPCl2)3 (hexachlorophosphazene)? I did a reaction to form $\ce{(NPCl2)3}$. The reagents I used were:

phosphorus pentachloride $\ce{PCl5}$
ammonium chloride $\ce{NH4Cl}$
chlorobenzene - the solvent
phosphorus oxychloride $\ce{POCl3}$

The reaction is $\ce{3 PCl5 + 3 NH4Cl -> (NPCl2)3 + 12HCl}$

Where the chlorobenzene is the solvent.
But what role does the phosphorus oxychloride $\ce{POCl3}$ have in this reaction?

Is it some sort of catalyst? This reaction also needs to be performed under more acidic conditions, so it is the phosphorus oxychloride that creates the acidic conditions?

If your goal was to prepare hexachlorophosphazene $\ce{(NPCl2)_3}$,

then I'm not sure why $\ce{POCl3}$ was added.  The synthesis involving $\ce{POCl3}$ certainly works, but hexachlorophosphazene can be prepared more simply by heating phosphorous pentachloride and ammonium chloride in chlorobenzene.  Careful sublimation affords the desired product, pure and in high yield.  This route is more economical and less toxic since it avoids the use of $\ce{POCl3}$.

$\ce{POCl3}$ is typically added to the reaction when one is interested in preparing higher polymers of $\ce{(NPCl2)_{n}}$.  In this case, use of $\ce{POCl3}$ allows you to prepare and isolate $\ce{Cl3P=N-POCl2}~(1)$ in high yield and purity (ref. 1).  The purity is important in obtaining high molecular weight polymers of $\ce{(NPCl2)_{n}}$.  Condensation of $(1)$ to produce hexachlorophosphazene typically occurs when reaction temperatures are maintained well below $\pu{200^\circ C}$, while high molecular weight polymers are produced at temperatures over $\pu{200^\circ  C}$.

Reference:

Synthesis and Characterizations of Poly(organophosphazenes) By Roger De Jaeger, Mario Gleria. See page 26 here


Is Gibbs energy change during osmosis positive? If so then why is it spontaneous? For ideal solution, μA=μA* +RT
  lnxA

ln function is an increasing function, means that the higher the xA the higher the μA......(1)

Now consider osmosis. From Atkins' Physical Chemistry Ninth Edition, page 173,

  osmotic pressure, π, is the pressure that must be applied to the solution
  to stop the influx of solvent (into solution across a semi-permeable
  membrane).

Here is my interpretation and thought process:

Had you not exerted π on solution, the μ(solution)< μ(solvent) based on (1) because lnxA is negative for xA<1, so by Second Law, the osmosis is spontaneous and solvent should move from pure solvent (high μ) to solution (low μ). Sounds right.

However, during this process, μ(pure solvent) unchanged (μ is intensive, not affected by amount) but μ(solution) increases due to dilution. Based on (1), for the solution, since xA increases μ(solution) must also increase. Although the process stops when μ(solution)=μ(pure solvent), the effect is that μ increases. μ increases would in turn imply ΔG>0 which means non-spontaneous, but as far as I know osmosis is spontaneous, so where is my misconception?

Let the subscript A refer to the solvent and B refer to the solute, and let the subscript 1 refer to the pure solvent container and  2 refer to the mixture container.  Then the free energy of the combined system is given by:
$$G=n_{A1}\mu_A^0+n_{A2}\left(\mu_A^0+RT\ln{\left[\frac{n_{A2}}{n_{A2}+n_{B_2}}\right]}\right)+n_{B2}\left(\mu_B^0+RT\ln{\left[\frac{n_{B2}}{n_{A2}+n_{B_2}}\right]}\right)$$If the number of moles of solvent in container 2 increases by $dn_{A2}$ and the number of moles of solvent in container 1 decreases by $dn_{A2}$, the change in the free energy of the combined system changes by dG.  If we do the math correctly, we obtain:
$$dG=RT\ln{x_{A2}dn_{A2}}\tag{1}$$where $x_{A2}$ is the mole fraction of solvent in the mixture container at any point:  $$x_{A2}=\frac{n_{A2}}{n_{A2}+n_{B_2}}$$Eqn. 1 indicates that the change in free energy is negative (as expected).


Mercury into mercury oxide How can I turn mercury metal into mercury oxide? I know that many metals will react with oxygen if enough heat is supplied in the presence of oxygen. If I heated up mercury enough and also supplied oxygen from my oxygen tank, would that result in mercury oxide? Also how about lead? Thank you

If you give enough oxygen to let it react to mercury, it still won't work just like copper and sliver. Because they are unreactive metals.

However, it will form mercury oxide when mercury and oxygen react at 350°C as the chemical equation below.

$\ce{2Hg + O_2 -> 2HgO}$

P.S: The mercury oxide will decompose back to mercury and oxygen at 400°C.

Lead is a rather more reactive subatances. So, try burning it. If it don't work, means you need a higher temperature.


Do all salts taste salty? Recently, I am learning the production of soluble and insoluble salts. My friend and I have done this experiment at the school lab.

We wanted to taste them to see whether they are salty are not. The teacher luckily stopped us from doing that.

So without tasting them, I would really like to know whether all salts are salty.

No. There are sweet, bitter, and various other salts. (Likely, there are tasteless salts too). Pure salty taste is as far as I know exclusive for table salt, though I wouldn't bet on it.

Lead and Beryllium salts are said to be sweet, though toxic. Epsom salt, $\ce{MgSO4}$, is bitter. $\ce{CuSO4}$ has an incomprehensible, persistent metallic taste. (Based on personal experience. Copper salts are slightly toxic, but not extremely, so I survived with no consequences.)

Salts with hydrolysing cation (various alums) are acidic in addition to other notes.


Why do gases liquify under pressure? To my understanding, an increase in pressure (of gases in a container) is proportional to an increase in gases' kinetic energy, and hence temperature. However I read that gases liquify under pressure... However, gases liquify in colder temperatures, so how come gases liquify under pressure, where the temperature increases?

Thanks to all responses.

Think of it simply as spheres in a box. The spheres are your molecules. The higher the temperature, the higher the energy the molecules have and the more they move around. So when they have high enough energy for the molecules to move around a lot and stay away from each other they are a gas. (This is a simplified answer). If you decrease the energy they have (cool the system down), they move about less and there's a certain point the molecules aren't moving far away enough from each other to stay as a gas, so they liquify. As Raditz_35 said, look at a phase diagram. So when you put a gas under pressure, the molecules are getting "squished" together and they don't have a lot of room to move around, so they can't move away from each other enough and thus liquify. This is how they store Hydrogen as fuel. It's too dangerous as a gas and they put the Hydrogen under pressure to store it as a liquid. Hope this helps.


Law of Multiple Proportions: What is the significance of small whole numbers? I'm new to Chemistry and in my textbook, it describes the Law of Definite Proportions (aka Proust's Law) and then goes on to describe the Law of Multiple Proportions.

The example they give is carbon monoxide and carbon dioxide, where the mass ratio of oxygen to carbon in carbon dioxide is $2.67:1$ and the mass ratio of oxygen to carbon in carbon monoxide is $1.33:1$.

I understand the above example, and I understand that when you compare them in the following way:

$2.67~\mathrm{g}$ (proportion of oxygen to $1~\mathrm{g}$ carbon in carbon dioxide)$/ 1.33~\mathrm{g}$ (proportion of oxygen to $1~\mathrm{g}$ carbon in carbon monoxide) $= 2.00$, a small whole number.

$$\frac{\text{Mass oxygen to 1 g carbon in carbon dioxide}}{\text{Mass oxygen to 1 g carbon in carbon monoxide}} = \frac{2.67~\mathrm{g}}{1.33~\mathrm{g}} = 2.00$$

My textbook doesn't describe what a small whole number is, or what it means … just that, in this example, $2.00$ is a small whole number.

What is the significance of the $2.00$?

I understand that carbon dioxide has double the oxygen. Is this where the $2.00$ plays a role?

The law of multiple proportions is largely obsolete these days because we all believe that atoms exist.  Prior to about 1900 this and similar laws were used to show that, for example, the ratio of oxygen in $\ce{CO}$ to oxygen in $\ce{CO2}$ is 2:1.

Of course, that's trivially obvious when we write the formulas, but before we believed in atoms the "small whole number" ratios were used as an example of an experimental situation (the 2:1 ratio) could be explained by atoms (i.e. quantizing atomic masses.)

Today, the law causes more confusion than it cures.  Atoms are here to stay.


Is the elimination reaction between 1 1 dibromo cyclopropane and phenyl lithium feasible in alkaline medium (and some heat) I found this question in one of my question papers for the JEE ,
My sir gave me an explanation that due to resonance between lone pairs of the bromide and π bond formed.

This reasoning doesn't seem to be satisfying because of the cyclopropyl ring strain which is not lost even at the end of reaction.
Please give an alternate pathway of the above product is not correct.

There are two pathways that phenyllithium could follow - it could deprotonate, but there really not any reactive protons, or it could do lithium-halogen exchange with one of gem-bromines. This second pathway will lead to a carbene by loss of $\ce{Br-}$ and the carbene (a highly reactive intermediate) will rearrange to give allene $\ce{H2C=C=CH2}$.

This method is documented:

In Page 275: H. Maskill, "Chapter 1: Three- and Four-Membered Carbocyclic Rings," In Specialist Periodical Reports - Aliphatic, Alicyclic and Saturated Heterocyclic Chemistry, Volume 1, Part II: Three‐ and four‐membered rings (carbocyclic and saturated heterocyclic); W. Parker, Ed. (Senior Reporter); The Chemical Society: Burlington House, London, England, 1973, pp. 3-292 (ISBN: 0-85186-502-X).
In Pages 1492-1493: Armin de Meijere, Editor, Carbocyclic Three-Membered Ring Compounds, Houben-Weyl Methods of Organic Chemistry, Volume E 17 b - Cyclopropanes: Synthesis; Georg Thieme Verlag: Stuttgart, Germany, 1997,  p. 1492 (ISBN: 978-3-13-798204-3).


Why does ammonium carbonate smell? Why is it that $\ce{(NH4)2CO3}$ smells while other salts with ammonium like $\ce{NH4Cl}$ does not smell? Does it have something to do with the equilibrium reaction? I know it has nothing to do with the coefficient of $\ce{NH4}$ in ammonium carbonate. I am pretty sure the water in our nose causes it to dissolve. I do note that $\ce{CO3^2-}$ is basic along with ammonium is acidic while the chloride ion is neutral. So does it have to do with the fact that $\ce{(NH4)2CO3}$ is more neutral than $\ce{NH4Cl}$?

There are two reasons.

As mentioned by Ivan Neretin in the comments, carbonate is the conjugate base of a much weaker acid than chloride is the conjugate base of.  In other words, carbonic acid is less acidic than hydrochloric acid, and carbonate is more basic than chloride.  This means that carbonate solutions will have higher pHs than chloride solutions, in general.  At higher pH, the equilibrium between $\ce{NH4^{+}}$ and $\ce{NH3}$ is shifted more towards $\ce{NH3}$, which is volatile and smelly.
Carbonate is also potentially volatile.  If you leave ammonium carbonate, even in solution, around forever in an open container, it will eventually completely disappear.
$$\ce{(NH4)2CO3(s) -> H2O(g) + 2 NH3(g) + 2CO2(g) }$$This also favors more vaporization of ammonia.  In contrast, in ammonium chloride, chloride is essentially completely non-volatile.  Thus ammonium could not evaporate without a buildup of negative charge in the solution, highly unfavorable.


2nd Electron Affinity of halogens I have seen questions regarding EA but they were all about the 1st EA so this question is not duplicate.

I came across a statement which goes like "2nd EA of halogens is almost zero". How is this possible? How can energy gained/released on adding an electron be almost 0, particularly when we are adding an electron to a noble gas, because, according to me, a lot of energy is required to do this.

Chart of Electron Affinities

The Electron Affinity can be thought of as the "electrical advantage" given by adding an electron to an atom.  So, if you have a halogen that gains an electron, it becomes more stable because now it has its octet, orbitals filled, etc.  This is why the 1st EA is large.

But if you want to add an electron, it needs to go to a whole new energy level, which will not be very desirable in terms of the electronic structure of the atom.  So the difference in electric stability or potential formed is nearly 0.  In other words, the halide ion has "0" want for another electron, since that would require adding another energy shell.

Another way to think about it is like this: the 2nd EA of a halide is basically the same as the 1st Ionization energy of the alkali metal to which it is isoelectronic (since adding an electron releases as much energy as removing that electron will take).  Since the 1st Ionization energy of a Group 1 metal is normally very small because the electron is so far away and shielded from the nucleus, so is the 1st electron affinity of a noble gas, and thus, the 2nd electron affinity of a halide.

Furthermore, now that the halogen has a negative charge, it experiences repulsion with the electron soon to be added. (Why does oxygen not like to be doubly anionized?)


High/low density water solution values I would like to know what min and max values of densities are achievable with water solutions.
There are some conditions:
1. Normal or close to normal temperature;
2. Non-radioactive solutions/components;
3. Acids are good as long as they can be safely contained.

Thanks for your help!

Highest known is about 4.25 (all
densities in g/cm${}^3$) at room temperature:
http://en.wikipedia.org/wiki/Clerici_solution

If you're willing to allow heating it comes close to 5 because the solubility of the salts increases with temperature:
http://books.google.com/books?id=kaa2qeFRXmUC&pg=PA99

My favorite runner-up is the Thoulet solution mentioned at
http://en.wikipedia.org/wiki/Potassium_tetraiodomercurate(II) but it
only reaches around 3.20.

Lowest would depend on what you mean by "water solution". If you mix a
drop of water with a gallon of a very light liquid such as diethyl
ether (which is not very miscible with water but does dissolve some
-- hence the need to dry it in the lab over sodium or molecular
sieves), then the answer is the density of the liquid (ether = 0.73).
If water has to be the majority component I'd expect the
minimum to be very close to the density of water, since mixtures tend to decrease a bit in volume. (The density of water is 1.0 at 4${}^\circ$, close to 1.0 at room temperature, 0.94 near the boiling point).


Determination of the integrated rate equation for a reaction We were given this equation to determine its integrated rate equation:

$$\ce {A + B <=>[$k_1$][$k_2$] C +D ->[$k_3$] E + F}$$

We were given these assumptions to simplify the derivation:

$\displaystyle \frac{\mathrm d[\ce{A}]}{\mathrm dt} = \frac{\mathrm d[\ce{B}]}{\mathrm dt}$, and $\displaystyle \frac{\mathrm d[\ce{C}]}{\mathrm dt} = \frac{\mathrm d[\ce{D}]}{\mathrm dt}$
At any time, $[\ce{A}]=[\ce{B}]$, and $[\ce{C}]=[\ce{D}]$, and $[\ce{E}] = [\ce{A}]+[\ce{B}]$
$\ce{F}$ is continuously removed from the system such that $[\ce{F}] = 0$ at any given time

The answer given to us was:

$$\ln[\ce{A}] = \frac{1}{2} \ln \left( \frac{k_2+k_3}{k_1}\right) + \ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2 [\ce{A}]\right)$$

However, I refuse to believe that an integrated rate equation can be without the time variable. Our teacher argued that the time variable was removed by using steady state approximation, and that $\frac{\mathrm d[\ce{C}]}{\mathrm dt} =0$. However, I think there is something wrong with using the rate of formation of an intermediate in deriving the integrated rate equation for a certain reaction. Can anyone please help me?

Edit: Here is how our professor got the answer:

$$\begin{align}
\frac{\mathrm d[\ce{C}]}{\mathrm dt}=\frac{\mathrm d[\ce{D}]}{\mathrm dt} &= k_1[\ce{A}][\ce{B}] - k_2[\ce{C}][\ce{D}] - k_3[\ce{C}][\ce{D}] \\
\frac{\mathrm d[\ce{C}]}{dt} &= k_1[\ce{A}][\ce{B}] - (k_2 +k_3 )[\ce{C}][\ce{D}]
\end{align}$$

$\ce{A}$ and $\ce{B}$ are the only species initially present such that:

$$[\ce{A}]_0 + [\ce{B}]_0 = [\ce{A}] + [\ce{B}] + [\ce{C}] + [\ce{D}] + [\ce{E}] + [\ce{F}] $$

Since $[\ce{C}] = [\ce{D}]$ and $[\ce{A}] = [\ce{B}]$ and $[\ce{F}] = 0$ and $[\ce{E}] = [\ce{A}] + [\ce{B}]$,

$$\begin{align}
[\ce{A}]_0 + [\ce{B}]_0 &= 4[\ce{A}] + 2[\ce{C}] \\
[\ce{A}]_0 + [\ce{B}]_0 - 4[\ce{A}] &= 2[\ce{C}] \\
\frac{[\ce{A}]_0 + [\ce{B}]_0 - 4[\ce{A}]}{2} &= [\ce{C}]
\end{align}$$

$$\begin{align}
\frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 \\
\frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )\left(\frac{[\ce{A}]_o}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2
\end{align}$$

By steady state approximation (SSA), $\frac{\mathrm d[\ce{C}]}{\mathrm dt} = 0$

$$\begin{align}
\frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2 = 0 \\
k_1[\ce{A}]^2 &= (k_2 +k_3)\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2
\end{align}$$

Applying natural logarithms to both sides,

$$\begin{align}
2 \ln(k_1) + 2 \ln[\ce{A}] &= \ln(k_2 +k_3)+ 2\ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \\
2 \ln[\ce{A}] &= \ln(k_2 +k_3 ) - \ln(k_1)+ 2\ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \\
\ln[\ce{A}] &= \frac{1}{2}\ln\left(\frac{k_2 +k_3}{k_1}\right) + \ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)
\end{align}$$

As you can see from the solution, he was able to eliminate time from the final equation by using the steady state approximation. Is this a valid move to get the integrated rate equation? Please help.

After consulting the Lindemann mechanism, here is my attempt for an answer:

$$\frac{d[\ce{C}]}{dt}= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 = 0$$
$$[\ce{C}]^2=\frac{k_1}{k_2 +k_3}[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +k_2[\ce{C}]^2$$

Substituting $[\ce{C}]^2$,

$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +\frac{k_2k_1}{k_2 +k_3}[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{dt}= \frac{-k_2k_1[\ce{A}]^2 + -k_3k_1[\ce{A}]^2 +k_2k_1[\ce{A}]^2}{k_2 +k_3}$$
$$\frac{d[\ce{A}]}{dt}= \frac{-k_3k_1[\ce{A}]^2}{k_2 +k_3}$$

Let $k'=\frac{-k_3k_1}{k_2+k_3}$, then

$$\frac{d[\ce{A}]}{dt}= k'[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{[\ce{A}]^2}= k'dt$$
$$\int_{[\ce{A}]_o}^{[\ce{A}]} \frac{d[\ce{A}]}{[\ce{A}]^2}= \int_0^t k'dt$$
$$-\left(\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o}\right) = k't$$
$$\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o} = -k't$$
$$\frac{1}{[\ce{A}]}= -k't +\frac{1}{[\ce{A}]_o} $$

I wonder how is it correct when I also used the steady state assumption in my solution. How is it different from the answer posted by my professor? Which is more correct and corresponds to the principles of chemical kinetics? Why would the other one be wrong and the other correct? (That is if my current derivation is correct :) ).


Defining enantiomer of morphine by R/S designation How to know if the following chemical structure of morphine is an R or S? What I basically need to know is: which chirality center do we use to make the decision?

$\hskip2in$

The structure that you are showing is neither R nor S.

As it has five chirality centers you need to look at all of them.

$\hskip3in$

You can see in the above figure that there are three R and two S centers in the configuration of morphine that you posted.


What are these upside-down glass egg-shaped containers for? I took this picture from a chemical vendor, but I've seen these things many times already. How are they called and why are they upside-down?

This is called an inverted specimen jar, sometimes called a sample jar, for obvious reasons intentionally designed to stand on the stopper (it is not a standard one to close the other jars in the lab).  Their use was something more popular around 1900's, when showcasing dyes and other visual appealing chemicals, especially towards the general public. Your example invites to look at the granular character of the product.

A more traditional design is this one, more conveniently used in collections / museums.

(source)


Covalent bonds other then 2-center, 2-electron bonds I've heard of single bonds, double bonds, triple bonds, and quadruple bonds between two atoms, but in each of those bonds the two atoms contribute the same number of electrons from each of their valence shells.

My question is, why can't the atoms contribute an uneven number of electrons? For example, why can't there be a situation where a carbon atom contributes two valence electrons while a nitrogen atom only contributes one?

Surely the sigma bonds arising from the overlap of two valence orbitals allow one valence orbital to have two valence electrons while the other only has one. (I don't know about pi-bonds, and I don't know if it is possible for a carbon atom and a nitrogen atom to give two and one valence electrons, respectively, but maybe there is another pair of atoms that would allow this).

One example to consider is benzene, where each carbon contributes 1 electron to a pi-bonding system of 6 electrons in 3 orbitals. The C-C bond lengths in benzene are intermediate between standard single and double bond lengths. Also take a look at the various boranes such as diborane (B2H6) which contain three-centered (or higher?) bonds involving boron and hydrogen.


The "hazardousness" of artificially made sweteeners In school I learned of the "possible hazardity of artificial sweeteners." But that was some time ago, now. I think that science should have caught up by now.
My question is as follows.

Which chemical properties in artifical sweeteners could be potentially cancerogenous?

(This is to say, if there exists cancerogenous properties in artificially made sweeteners from the start).
Please enlighten me.
Examples of sweeteners:

Fig 1 Acesulfame Potassium

Fig 2 Aspartame

From the NHS website in the UK:
The US National Cancer Institute conducted a study of nearly half a million people, comparing those who consumed drinks containing aspartame with those who did not. Results of the study found aspartame did not increase the risk of leukaemia, lymphoma or brain cancer."
Also:
In 2013 the European Food Safety Authority (EFSA) conducted a comprehensive review of the evidence (there is a pdf on the website) and concluded that aspartame was safe for human consumption, including pregnant women and children.
Cancer Research UK also have stated that artificial sweeteners are safe.
Rigorous testing is carried out before a sweetener can be approved for use in the EU.
So, no evidence of cancer in humans. Rats, maybe. Humans no.


Dissociation rates: trans-effect rule in square planar complexes In lecture, the trans-effect was described.

A ligand $L^t$ with a higher trans-effect as $L$ (cis to $L^t$) leads to a faster dissociation of ligand $L^d$ (trans to $L^t$). I would expect that the attacking ligand $B$ is needed to make a stronger bond as the leaving ligand $L^d$ did.

In order to see if these assumptions are always valid in planar quadratic systems I've set up following problem according to the given trans-effect order:

$\ce{CO}$ has a stronger trans-effect as $\ce{CH3-}$ and therefore the ligand trans to $\ce{CO}$ will be exchanged faster as trans to the $\ce{CH3-}$. But the leaving group $\ce{I-}$ would also have a higher trans-effect as $\ce{Cl-}$ so I wouldn't expect a substitution in b). Are these statements correct?

First off, there are two nice articles on the trans effect on Wikipedia and LibreTexts that are worth reading.

You need to remember that there are two factors leading to the trans effect:

Weakening the $\ce{Pt-X}$ bond trans to another group (i.e., a structural effect)
Stabilizing the 5-coordinate transition state, since substitution reactions in square planar complexes are generally associative. (i.e., a kinetic effect)

(a) I don't think the $\ce{I-}$ will substitute for the $\ce{NH3}$ group. Instead, I think the $\ce{CO}$ "wins" the trans effect "competition" and so the trans $\ce{CH3-}$ will be replaced by the $\ce{I-}$.

Next, substitution with the $\ce{Cl-}$ should replace the $\ce{I-}$.

Now your second question seems to be if you'll actually get a substitution with $\ce{Cl-}$ since $\ce{I-}$ has a greater trans effect. Yes, you will. All of these are in equilibrium, so if you generate a lot of $\ce{I-}$ that can substitute back (and get starting material). But you'll definitely get the Cl-substituted product too if you separate the reaction mixture.

Indeed, the prototypical example for the trans effect is synthesis of trans-platin or cis-platin, starting from either $\ce{[PtCl4]^{-2}}$ or $\ce{[Pt(NH3)4]^{+2}}$. Even though $\ce{NH3}$ has a smaller trans effect, you can synthesize cis-platin from $\ce{[PtCl4]^{-2}}$ and substituting two $\ce{Cl-}$ with $\ce{NH3}$.


Why Benzene Cannot be chlorinated using HOCl? Why HOCl doesnt do the job please , can anyone show me the mechanism ?

One possibility:  When you chlorinate with, let us say, $\ce{Cl2 + FeCl3}$, you generate a "leaving group", $\ce{FeCl4^-}$, that comes off easily because it is highly polarizable and only weakly basic.  The $\ce{HOCl +H^+}$ combination would have to displace a more strongly basic and less polarizable water molecule.

Chlorination can be effected with hypochlorous acid by adding silver ion as another catalyst/co-reagent.  See here.


Why is the equilibrium constant given in litre per mole? At $800\:\mathrm{K}$ a reaction mixture contained $0.5\:\mathrm{mol}$ of $\ce{SO2}$, $0.12\:\mathrm{mol}$ of $\ce{O2}$ and $5\:\mathrm{mol}$ of $\ce{SO3}$ at equilibrium. $K_\mathrm c$ for the equilibrium $$\ce{2SO2 + O2 -> 2SO3}$$ is $833\:\mathrm{L/mol}$. If the volume of the container is $1\:\mathrm{L}$, calculate how much $\ce{O2}$ is to be added at this equilibrium in order to get $5.2\:\mathrm{mol}$ of $\ce{SO3}$ at the same temperature.

The answer comes out to be $0.34\:\mathrm{mol}$.

I have understood the whole question and I am getting the answer. I just have a confusion why we are taking $K_\mathrm c$ as $833$ and not $1/833$ which makes it $\mathrm{mol/L}$. I think $K_\mathrm c$ should be in $\mathrm{mol/L}$.

$K_\mathrm c$ is defined as (for more information see here)

\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} = \frac{\prod \limits_{i \in \text{products}} [c_{i}]^{|\nu_{i}|}}{\prod \limits_{j \in \text{reactants}} [c_{j}]^{|\nu_{j}|}}
\end{equation}

where $\nu_{i}$ and $c_{i}$ are the stochiometric coefficient and the concentration of the $i^{\text{th}}$ component in the reaction, respectively.

If you use this definition for the reaction

\begin{equation}
  \ce{2SO2 + O2 -> 2SO3}
\end{equation}

you get

\begin{equation}
 K_\mathrm c = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^{2} [\ce{O2}]}
\end{equation}

where the notation $[i] = c_{i}$ was used.
This equation tells you that $K_\mathrm c$ has the unit

\begin{equation}
 \text{unit of }K_\mathrm c = \frac{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2}}{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2} \left(\frac{\mathrm{mol}}{\mathrm{L}} \right)} = \frac{\mathrm{L}}{\mathrm{mol}} \ .
\end{equation}


Nucleophilic addition of grignard reagent Here why isn't the phenyl group attacking the carbonyl carbon? Isn't the first step in addition of grignard reagent the attack by the carbanion on the carbonyl carbon ?

The attacking of the Grignard reagent in these kinds of $\alpha$-$\beta$ unsaturated carbonyl compounds depends on two factors mainly. $1$) Hardness of the acid centre $2$) Steric repulsion between the surrounding groups and incoming nucleophile.When the attacking nucleophiles are Grignard reagents,as per your equation, they are generally hard bases.According to Pearson's HSAB Theory, Hard bases prefer to bind to the Hard acid centre and soft bases prefer to bind to the soft acid centre.  In your compound i.e in $\alpha$-$\beta$ unsaturated carbonyl compounds, the carbonyl carbon centre is the Harder acid centre, whereas the $\ce{C=C}$ double bond centre is the softer acid centre.So, Grignard reagents try to first bind with the carbonyl carbons unless there is no other destabilising factor.The second governing factor is the steric crowding, which can sometimes make the attacking nucleophile attach to the least sterically crowded region even though it is violating HSAB theory. So, the addition compound actually depends on these two competititive factors.  Now, in your given compound the attacking nucleophile is $\ce{PhMgBr}$.It is undoubtedly Hard Base. But if it were to bind to the carbonyl carbon i.e. $\ce{C=O}$ double bond, there will be huge repulsion between the lone pair of oxygen and $\pi$-electron cloud of Phenyl ring which actually makes the previously stated second factor to dominate. So, due to this huge steric repulsion, The attacking nucleophile has to attack the less sterically crowded(Note: The repulsion is also there at the double bonded carbon centre , but repulsion between methyl groups and phenyl rings are only due to inefficient spatial orientation which is definitely smaller than inter electronic strong columbic repulsion forces in close vicinity in the case of $\ce{C=O}$) softer basic $\ce{C=C}$ centre which infact gives the product.


Why is STP mol/liter so specific? How do we know that one mole of an ideal gas occupies (exactly) 22.4 Liters? Why is it so precise a value? Many other constants/conversions in science have multiple decimal places, so why is this conversion terminated so soon?

Yes, I have googled it, and searched it up on StackExchange, and flipped though my chemistry textbook. I just want a nice clear explanation for the exact value of the the mol/liter conversion.

Finally, please don't use the ideal gas constant $\ce{R=8.314 J/K\cdot mol}$ to justify your answer, because we get $R$ by using this conversion fact.

Thank you for any information or insights!

EDIT: This has been suggested as a duplicate of What volume does one mole of an ideal gas occupy?, which does not answer my question because it derives the liter/mol conversion from the ideal gas constant.

How do we know that one mole of an ideal gas occupies (exactly) 22.4 litres?

It doesn't.

The current recommended value for the molar volume of an ideal gas at a temperature of $T=273.15\ \mathrm K$ and a pressure of $101.325\ \mathrm{kPa}$ is $V_\mathrm m=22.413962(13)\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$ (source). Note the given uncertainty; i.e. it is an experimental value  that is not exact.

Also note that these conditions do not correspond to STP. According to current IUPAC recommendations, STP corresponds to a temperature of $T=273.15\ \mathrm K$ and a pressure of $100\ \mathrm{kPa}=1\ \mathrm{bar}$. At this state, the molar volume of an ideal gas actually is $V_\mathrm m=22.710947(13)\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$ (source).


What tools can be used to estimate binding energies? I want to estimate the binding energies of various molecules. What are the analytical and numerical tools I can use for this purpose? In particular I'm interested in bonding energies of dyes to substrates (example).

The steps for predicting a binding energy is relatively straightforward in computational chemistry but the process itself can become extremely difficult depending on

The types of molecules involved (size, complexity, etc.)
The desired accuracy of your final answer
The environment that the molecules reside in (gas phase, aqueous phase, etc.)

For very small molecules in the gas phase, the process is easy and routine assuming that electron correlation is trivial.  Aqueous phase environments, complex/large molecules and high accuracy quickly makes the problem more difficult if not impossible.

I have written up a previous answer which addresses how to predict an interaction energy ($E_{\mathrm{int}}$) which is completely different from a binding energy ($E_{\mathrm{bind}}$).  However, much of the concepts will remain the same.  I will try to outline here what in fact is the difference between an interaction energy and a binding energy.  Unfortunately, 'binding energy' is tossed around a lot in the literature and consequently grossly misused.  I hope to clarify the subtle differences between these two terms in my answer to your question.

I will focus my discussion around an elementary system of two water molecules to make this conceptually easier.  In the gas phase, an isolated water molecule will adopt a particular geometry (at 0K and in a vacuum).  We can determine what this minimum energy configuration will be at a particular level of theory by approximating the Schrodinger equation.  The geometry that gives rise to the lowest possible energy for an isolated water molecule will be the geometry that you will observe under those conditions.  Remember that the internal coordinates of water include things like the O-H bonds, and the H-O-H angle.  We don't need to guess at what these parameters will be in the end, solving the Schrodinger equation will determine that.

Now, we know exactly what an isolated water molecule looks like and what the corresponding energy of that water molecule will be.  Both of these components are equally important when it comes to determining a binding energy because (as the name suggests) we are trying to determine an energy.  So the energies that we start off with are crucially important.

We can now consider what happens when two water molecules interact (form a hydrogen bond).  What will be the binding energy of these two water molecules (i.e. what will be the strength of this bond that forms between then?).  It is common to approximate the strength of this bond by simply computing an interaction energy (its computationally easier to determine $E_{\mathrm{int}}$ rather than $E_{\mathrm{bind}}$).  How would we compute an interaction energy?  In this case, you wouldn't need to figure out what geometry or energy an isolated water molecule in its lowest energy configuration.  You would simply need to determine what the geometry of two interacting water molecules (i.e. dimer) would be (and corresponding energy) as well as the energy of each water molecule isolated from each other in the geometry that it adopts in the dimer.

I'll explain.  We can determine the lowest energy configuration of a single water molecule in the gas phase.  This geometry is inevitably different than a water that is not isolated... a water that is interacting with something else such as another water.  If you want to get an interaction energy, you only need...

Energy of the lowest energy configuration of the water dimer
Energy of each water (in the water dimer) isolated from each other but only at the geometry that the water takes on within the dimer.

In this case, you are NOT allowing each isolated water molecule to 'relax' into its lowest energy configuration.  You are simply going to neglect the contributions of this energy relaxation effect to the overall bond strength.  Therefore,

$E_{\text{int}} =  E_{\text{opt-dimer}} - (E_{\text{mon1}} + E_{\text{mon2}})$

where mon1 is the first water molecule in your dimer and mon2 is the second water molecule in your dimer. (opt-dimer means optimized dimer)

However, in the case of a binding energy, you MUST account for the relaxation effects that occur when you break the dimer apart and allow the waters to become isolated from each other (and everything else).  To do this, you simply carry out what I've stated above... you simply optimize each water and then take its corresponding energy.  (Determining an interaction energy means you don't have to do this extra optimization(s)... you would simply do a single point energy on each monomer).  Clearly for a water dimer, you would only do one optimization as each water will become identical when isolated.  For a heterodimer, you would have to optimize each piece separately.

To determine the binding energy, you would use the following equation:

$E_{\text{bind}} =  E_{\text{opt-dimer}} - (E_{\text{opt-mon1}} + E_{\text{opt-mon2}})$

Of course the problem becomes more complex when taking into account basis-set superposition error (you will need to perform counterpoise corrections though there exists a heated debate in the literature about the usefulness of this type of correction).  Also, zero-point vibrational energy corrected binding energies will require that the zero-point energy of the dimer and each monomer will need to be determined (which comes from taking the 2nd derivative of the wavefunction which is computationally demanding).

As far as tools go, you can choose from a number of available quantum software packages.  Gaussian 09 is a popular choice but very expensive.  Q-Chem (paid) has the benefit of having the most beautiful, gorgeous, well-laid out documentation of anything I've ever seen (makes for a good supplemental textbook to boot).  NWChem is opensource.  Google can help you find plenty of options.

A quick note: Basis set superposition error (BSSE) should NOT be referred to as an error as 'error' suggests deviation from the RIGHT answer. BSSE has been shown to actually make  non-corrected answers even WORSE.

One last important note:  When determining differences of energies (such as you do to get an interaction or binding energy), ALL energies being compared must be evaluated at the SAME LEVEL OF THEORY (method/basis set, etc.).


Will unused spare razor blades rust if stored for a long time? I am considering purchasing 16 razor blade spares for a reduced price, but am unsure if they will be good as new if I were to use them several months from now.

If left alone in the packaging (they come in plastic cartridge cases with one side exposed), will the quality degrade by the time I use the last several (if the blades were replaced monthly)?

Aside from storing a in dry location, you might use a vapor corrosion inhibitor (VCI) in a container with the blades. Amazon offers a similar product for ~US$11.

My experience has been that VCI is effective and does not leave a noticeable film that needs to be removed. Whether this is cost-effective for your intended use is another question (perhaps you can use the remaining VCI on other objects or share with ten other people).

And then, beards are even more cost-effective ;-)


Predicting the products of reaction between ammonia and hydrogen peroxide Like we do in organic chemistry is there any way to write the mechanism of the reaction between $\ce{NH_3}$ and $\ce{H_2O_2}$?

Can the products be predicted?

Majority of answer content came from this link and this link.

Strong solutions of $\ce{H2O2}$ with a few drops of $\ce{NH4OH}$ (aq. ammonia) is left for 24 hours without any nitrite formation occurring. But upon longer standing, even with a small amount of hydroxide then nitrite forms. Nitrite also forms when a dilute solution of $\ce{H2O2}$ is mixed with $\ce{NH4OH}$  and is evaporated over pure conc. $\ce{H2SO4}$ with a bell jar. $\ce{H2O2}$ forms (even in very dilute solutions) nitrite very rapidly, if the $\ce{H2O2}$ solution is mixed with a few drops of $\ce{NH4OH}$ and a little $\ce{NaOH}$  and this then boiled in a retort to a very small volume.

$\ce{2NO2- +  2H+(aq) <=> H2O + NO + NO2}$

$\ce{NO2- + NO3- + 2H+(aq) <=> H2O + NO2}$

Nitrogen dioxide is then able to attack ammonia.

$\ce{2NO2 + 2NO + 4NH3 -> 2NH4NO2 + 2H2O + 2N2~~~~~~~~~~~~~~~~~~~~(1)}$

In any case, solutions of ammonium nitrite decompose on heating to 60 to 70C.

Perhaps the very slow spontaneous reaction of aqueous ammonia with $\ce{H2O2}$ takes place the small equilibrium with amide anions, $\ce{NH2-}$.

$\ce{NH2- + H2O2 -> NH2- + 2OH-}$

Even in $\ce{NH4OH}$, there is a very slight equilibrium with amide ions.

$\ce{2NH3 <=> NH2- + NH4+}$

Amide rapidly and vigorously reacts with water, so the equilibrium of amide existence in water must be very low.

$\ce{NH2-+ H2O -> NH3 + OH-}$

In the base catalysed decomposition of hydrogen peroxide, the mechanisms is presumably:-

$\ce{H2O2 + OH- -> HOO- + H2O}$

$\ce{HOO- + H2O2 -> HOOOH + OH-}$

$\ce{HOOOH --> HOOO- + H+(aq)}$

$\ce{HOOO- -> OH- + O2}$

Also, ammonium nitrite decomposes into nitrogen and water.

$\ce{NH4NO2 → N2 + 2 H2O~~~~~~~~~~~~~~~~~~~~~~~~~~(2)}$

A combination of ammonia and hydrogen peroxide is often used in the process of bleaching hair. Hydrogen peroxide has the necessary properties to remove the color from hair - it oxidizes one of hair’s melanin pigments to a colorless substance. This results in the hair becoming lighter. However, it is not recommended by hair professionals (those who don't understand chemistry but want your cash), and can cause extensive damage to hair (even causing it to fall out).

Therefore, the overall reaction is:-

$\ce{2(NH3•H2O) + 3H2O2 ->[\Delta] N2 + 8H2O~~~~~~~~~~~~~~~~~~~~~~~(1) + (2)}$


What is the right way to express pH and concentrations? Which of the following expressions are used in English to express the pH of solutions and the concentration of solutions?  I think they are all clear to someone with knowledge of chemistry, but I want to avoid awkwardness in my English.

A. I prepared a solution such as below:

a solution of pH 3
a solution with pH 3
a solution with a pH of 3
a solution of a pH of 3

And

B. I prepared a solution such as below:

a 3 nM buffer solution
a buffer solution of 3 nM
a buffer solution of 3 nM concentration
a buffer solution with a concentration of 3 nM
a buffer solution of a concentration of 3 nM

I'm not a native speaker and confused. Especially, can I use A-1, B-2 or 3 expressions?

This is more a question of English usage than Chemistry, but the English Language and Usage people regarded it as “proof reading” and migrated it here. So I will move my original comment here to make it an answer (until the Chemists throw the question out).

First, you really should try to answer this yourself by looking at the style used in scientific journals relevant to your work in papers from English-speaking countries with authors whose surnames suggest they may be native speakers. Several of your alternatives are grammatically correct but unneccessarily long-winded and never used.

Having said that, my opinion (as a native English-speaking biochemist with many publications) is:

A: A-1 (“a solution of pH 3”) should be used.

The use of ‘with’ alone in A-2 is unusual, A-3 is ok in speech but too verbose for a paper, A-4 with two ‘of’s is clumsy, ugly and unnecesary.

B: B-1 (“a 3 nM buffer solution”) seems best to me — brief and correct.

There is no need to expand it as in B-2, B-3 (both of which sound clumsy and I advise you not to use), or B-4. B-5 has the ugly double ‘of’ again.

If for some reason you don’t like B-1 you can name the buffer in a phrase such as “3 nM Tris-HCl Buffer”.


Molarity Numerical on Water Gas Shift Reaction For the water gas shift reaction below, $K_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?
$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$

I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:
$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$
$$3.491 = \frac{x^2}{(1.526-x)^2}$$
Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$
$$3.491\cdot 2.32 = x^2$$
The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.


When it rains, it puddles. Spilled salt cycles between wet and dry with humidity. Is this akin to a phase change? I have some spilled un-iodized table salt (NaCl). It is very humid where I live, and when it gets very humid for a few days, the salt absorbs so much water that it becomes a puddle of (probably) saturated salt water. When the humidity drops for a few days, the water evaporates and I have dry salt crystals again. This is repeatable.

My questions about this process are:

What is this process called exactly? As far as I understand "deliquescence" is a property and not a process. Does the exchange of water between my pile of salt and the air have a specific terminology? Something that would go along with terms like evaporation, sublimation, condensation, etc.? Perhaps some mash-up of adsorption and dissolution?
Is this process akin to a phase change in some way? For example, if I had a sealed box of air with a dish of salt and a beaker of water, and I could ramp the  temperature slowly enough, would I see a plateau at a certain humidity as the salt hit a threshold and started absorbing water until saturated?

Some may be familliar with the Morton Salt logo and slogan "When it rains, it pours" meaning that other moisture-absorbing materials in the package will keep the sodium chloride as table salt from absorbing too much water and becoming stuck and un-pourable:

Source

What is this process called exactly? As far as I understand
"deliquescence" is a property and not a process. Does the exchange of
water between my pile of salt and the air have a specific terminology?
Something that would go along with terms like evaporation,
sublimation, condensation, etc.? Perhaps some mash-up of adsorption
and dissolution?

The property of salts to become wet as I mentioned in another post Scientific Reason for Salt Solution Gaining Volume is deliquescence. The corresponding verb is to deliquesce. It is a pretty old word. From OED (not open access):

To melt or become liquid by absorbing moisture from the air, as certain salts.
1756   C. Lucas Ess. Waters i. 14   They attract the humidity of the
air, and deliquesce, or run liquid.
1781   Philos. Trans. 1780 (Royal Soc.) 70 349   This
pot-ash..deliquesces a little in moist air.
1876   D. Page Adv. Text-bk. Geol. (ed. 6) xvi. 299   Pure chloride of
sodium is not liable to deliquesce.*

Regarding phase changes, there were many phases changes involved in that process. The water as gas became liquid. The solid NaCl phase become one phase with a liquid. Once it evaporated, you got the solid again.


Splitting of multiplets in ¹H NMR spectra? I'm currently taking VCE (Victorian Certificate of Education) Chemistry classes, and we're currently studying the interpretation of spectra produced by Hydrogen NMR (Nuclear Magnetic Resonance) spectroscopy.

When studying the spectra of High Resolution 1H NMR, the peaks representing the different Hydrogen environments are split into multiplets based on the protons surrounding these environments. There has been a considerable amount of confusion in my classes over the actual principles / rules of thumb on how to calculate the multiplets for a particular environment of a known chemical (i.e.; known structue), based off the 'n + 1' rule.
(i.e.; an environment with n neighbours will be split into n+1 multiplets).

We are absolute on the principles that;

Peaks of a particular Hydrogen environment are not split by neighbouring protons in equivalent environments.
OH does not, and is not split by, it's neighbouring environments.

However, immense confusion arised over whether the following principle was correct.

Peaks of a particular Hydrogen environment will only be split by the protons in neighbouring environments once for each type of neighbouring environment.
e.g.  The middle "CH2" environment in "CH3–CH2–CH3" will only have 4 peaks;
Although it has 6 neighbouring protons, they are two lots of the same environment (CH3).

As a class, we found numerous examples from different text books and sources that provide examples of ¹H NMR spectra which did not clarify the matter; Some considered all neighbouring protons as neighbours, others discriminated on the repeated neighbouring environments. For example, the CH2 in CH3–CH2–CH3 was sometimes split into 4 peaks or 7 peaks, depending on the source.

Many Chemistry teachers contradicted each other on the matter.
There was repeated self corrections made by the teachers, such that now nobody really knows whether this principle is correct or not.

So, is there anybody that has the correct information on the matter?
Is there a reasonable explanation behind this strange lack of correlation,
or is there a common misconception about multiplet splitting?

Ultimately; How many multiplets should the CH2 in CH3–CH2–CH3 have?

(Note that if there is a complicated explanation that I am currently only at Year 12 VCE level, so links to resources I can pursue would be extremely helpful! It's been established that VCAA (an authority for the education system in Australia) ensures that the chemicals featured in the exams for NMR analysis will not be of a structure so as to allow the ambiguity above.)

Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $\mathrm{^3J_{HH}}$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $\mathrm{^3J}$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $\mathrm{^4J}$-couplings or more.

So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way.

The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other.

In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C–C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7).

The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra.


How do I determine the crystal field splitting for an arbitrary point group? How can I figure out the degeneracy of the d orbitals for a site that has a given point group? Specifically I'm interested in $D_{3d}$ and $D_{3h}$, but it would be good to know how to do it in the general case.

If you already know the symmetry of your site then it is quite easy. In a lot of books (e.g. this one) and on this web site you can find the character tables of the point groups supplemented with two additional columns which show the transformation properties of the basis vectors (e.g. $\ce{p}$ orbitals), their rotations and their quadratic combinations (e.g. $\ce{d}$ orbitals). In the following picture I highlighted the important parts in the character table of the $\ce{D_{3\mathrm{h}}}$ group.
So in the $\ce{D_{3\mathrm{h}}}$ group the $\ce{d_{z^2}}$ orbital transforms as the irreducible representation $A_{1}^{'}$ and the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals transform as the irreducible representation $\ce{E^{'}}$.

After you have identified the irreducible representations of the $\ce{d}$ orbitals you can tell their degeneracy by identifying which $\ce{d}$ orbitals belong to the same irreduzible representation. If two or three $\ce{d}$ orbitals belong to the same irreducible representation they are degenerate. In the case of the $\ce{D_{3\mathrm{h}}}$ group shown below this means that $\ce{d_{z^2}}$ forms an energy level of its own, while $\ce{d_{x^2-y^2}}$ and $\ce{d_{xy}}$ (both belonging to $\ce{E^{'}}$) are degenerate and $\ce{d_{xz}}$ and $\ce{d_{yz}}$ (both belonging to $\ce{E^{''}}$) are degenerate. So, your $\ce{d}$ orbitals split into three energy levels, two of which are doubly degenerate, when your site has $\ce{D_{3\mathrm{h}}}$ symmetry.


Would the density and radius of your metal appear to be higher or lower if you forgot to tap the flask to release air bubbles? For this lab we had to measure out copper in flasks, but tap them out before so no air bubbles would affect the weight. We have this question at the end of the lab and so I wasn't sure if the density/radius would be higher or lower if I didn't tap them out and why.

We were trying to figure out the size on an atom of copper. To find the density of the sample, we had to add about 1/3 of copper to an empty flask and weigh it with the metal. Then after we had to add enough water to just cover the metal, then gently tap and rotate the volumetric flask to remove any air bubbles. Then fill to the calibration mark with water and reweigh the volumetric flask containing the water solid mixture.

In this experiment, you were determining the average volume that a copper atom occupied in your sample of copper.  In order to do this, you had to know the number of copper atoms and the volume that number of copper atoms occupied.

The moles of copper was measured by gravimetrically determining the mass of the copper (e.g. by weighing it). From that you determined the number of moles of copper using it's atomic weight, and from there you calculated the number of copper atoms using Avogadro's number.

The volume occupied by your copper sample was determined by measuring how much water the copper displaced. So, if the volume of water increased by 10 $\mathrm{cm^3}$, then you knew your volume of copper was $\mathrm{cm^3}$. But air bubbles occupy space also.  If, for example, 1 $\mathrm{cm^3}$ of the total displaced volume were actually due to air bubbles, then you would actually only have 9 $\mathrm{cm^3}$ of copper, but you would mistakenly have used a value of 10 $\mathrm{cm^3}$.

So, the bottom line is that because you were calculating the molecular volume as the volume occupied by the copper per molar quantity of copper, your calculated molar volume would have been proportionally higher than the true value.


Program that simulates basic reactions in organic chemistry To explain - a friend and I have been working on a program that would allow you to draw organic molecules, indicate the conditions, and predict the products.  This would be primarily aimed at basic, first year undergrad organic chemistry (SN1/2, E1/2, applications, potentially (retro)synthesis, etc).

We've realized that it is probably worthwhile to figure out if there are similar existing programs out there.  From what we've read there doesn't seem to be something that meets what we're planning - everything we've found seems to be aimed at professional/higher level chemists, not assisting a beginner.

Basically it comes down to a couple main points:

Are you aware of similar software?
Do you think that this sort of software would be used (either by yourself or students or whoever).
If/when we have a working version, would users here be interested in using/testing it?

I apologize if this doesn't meet the guidelines/standards for questions here, I was kind of iffy about asking it.

As you mentioned, there've been quite some attempts on CAOS (Computer-Assisted Organic Synthesis), such as

WODCA from the Gasteiger group
CHIRON from the Hanessian group
LHASA from the Corey group

I might be awfully wrong, but I think they were/are all closed source.

But you might want to have a look at OSET, an open source Java project on which Ivan Tubert-Brohman worked some ten years ago.


pH of amphoteric species For a solution of $\ce{HA-}$, I have seen the following approximations for the $\mathrm{pH}:$

$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})\tag{1}$$

$$\mathrm{pH} = \sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\tag{2}$$

How do they differ? Is the second approximation always more accurate than the first? I know that the second is derived from the mass and charge balance equations and the first is the same as the second, with several assumptions.

The former equation assumes
$$[\ce{H2A}]\simeq [\ce{A^2-}]$$
due reaction
$$\ce{ 2 HA- <=> H2A + A^2-}$$
The is possible with 2 simplifying conditions:

The concentration of oxonium resp. hydroxide ions originated from water dissociation is much lower than concentration of the basic resp. acidic ampholyte form.
$$[\ce{H2A}] \gg \sqrt{K_\mathrm{w}}$$
$$[\ce{A^2-}] \gg \sqrt{K_\mathrm{w}}$$
That assures the effect of ampholyte reaction with water dissociation products
$$\ce{ A^2- + H3O+ -> HA- + H2O}$$
$$\ce{ H2A + OH- -> HA- + H2O}$$
is negligible for the ampholyte  form concentration ratios and you can ignore water dissociation.
This condition is critical for low ampholyte concentration near neutral $\mathrm{pH}$.

The concentration of oxonium resp. hydroxide ions is much lower than concentration of acidic resp. basic ampholyte forms.
$$[\ce{H2A}] \gg [\ce{H+}]$$
$$[\ce{A^2-}] \gg [\ce{OH-}]$$
That assures the effect of production of oxonium resp. hydroxide ions to reach the final $\mathrm{pH}$
$$\ce{ HA- + H2O -> A^2- + H3O+}$$
resp.
$$\ce{ HA- + H2O -> H2A + OH-}$$
does not compete with creation of acidic/basic ampholyte forms.
$$\ce{ 2 HA- + -> A^2- + H2A }$$
This condition is not valid for very low /very high target $\mathrm{pH}$, as respective ions are produced in expense of the deviation from the same ratio of ampholyte form concentration.


Is there a difference between NaCl or table salt (processed) and salt from the Himalayas? So, I was doing some research on bonding and electronegativity and one of the examples used was NaCl (Sodium Chloride) which is salt. We don't buy processed salt, rather we buy salt from the Himalayas.

Apparently this is better salt because it is not processed, made, etc.
But, is there a difference between NaCl and Himalayan salt in regards to their molecular structure?

No salt will be pure NaCl.  Each will have some degree of other elements.  The fact that the salt isn't white confirms there are other elements present.

See Analysis of Gourmet Salts for the
Presence of Heavy Metals which investigates 14 salts including two Himalayan salts.

See especially "Table 3. Comparison of toxic elements in Table Salt".

A "Himalayan Pink Fine Mineral Salt" was found to have the highest level of cadmium and tied for first for the highest level of nickel.

NaCl isn't molecular, but instead is a lattice of Na+ and Cl- ions.  Other ions can lie within this lattice as impurities.


Why do the peaks in a radial distribution function graph have a probability >1 Why do RDF plots have the probability at a particular distance >1. How can probability be more than one?
I am asking this because I want to understand what do the peaks of different length but at the same position on the x axis mean
The below is an example of a RDF plot:

The radial distribution function isn't a probability or a probability density. The usual definition is $$g(r)=\frac{\rho(r)}{\rho_0}$$ where $\rho(r)$ is the number density at a particular distance $r$ away from some specified atom and $\rho_0$ is the bulk density of the material.
It can be larger than one if the local density is greater than the bulk (which you see in the first peak of your graph), so it can't be a probability. It also can't be a probability density, as the local density should eventually converge to bulk at long distance, so $g(r)$ converges to 1 (as you start to see on the right-hand side of your graph) so the integral would continue to increase as you increase the range of integration.
So what does the RDF tell you? Well if you integrate it over a spherical shell and multiply by the bulk density, you obtain the number of atoms in that shell. One particular shell we are often interested in is the first coordination sphere. Using the RDF, we can define the number of particles in the first coordination sphere as $$n(r')=4\pi\rho_0\int_0^{r'}g(r)r^2dr$$ where $r'$ is the location of the first minimum of the RDF. We can similarly determine the number of atoms in the $n^{\text{th}}$ coordination sphere by integrating from the$(n-1)^{\text{th}}$ minimum to the $n^{\text{th}}$ minimum.


Why does a catalyst increase the forward and reverse reactions equally With reference to the picture below, how is it that when there is a lower activation energy (due to the alternate reaction pathway provided by the catalyst) that the catalyst will increase the forward and reverse reactions exactly the same amount.

It only makes intuitive sense that when the $Ea$ is lowered, the forward reaction increases more than the reverse reaction because the decrease was an absolute amount (that is, the same decrease for both forward and reverse) and thus proportionately, the forward $Ea$ has been reduced far more than the reverse $Ea$. So why should the reverse reaction instantly be raised such that it is still equal to the forward reaction (if from equilibrium)?

The equilibrium is unaffected because the barrier to both forwards and back reaction is reduced, and even though the amount is constant it works out that this effect cancels out. This is a consequence of the fact that the population of energy levels is exponentially distributed from the Boltzmann distribution. This also leads to Arrhenius type rate expressions.

If the energy in your figure from reactants to transition state is $E_a$ and the difference in products to reactants $G$ and we lower the energy $E_a$ by $E_0$ then using Arrhenius type rate expressions the equilibrium constant is $K_e=k_{forward}/k_{back}$

$\displaystyle K_e=\frac{\exp(-E_a/RT)}{\exp(-(E_a+G)/RT)}$,  and for the catalyst  $\displaystyle K_e^C= \frac{\exp(-(E_a-E_0)/RT)}{\exp(-(E_a-E_0+G)/RT)}$  which is the same as $K_e$if you expand the terms in the exponentials.


Can a liquid turn to a solid by applying pressure? When pressure increases, volume decreases. If an extremely high pressure is applied to a liquid, is it possible to reduce it's volume to an extent where the molecules are very tight together, hence becoming a solid? Or do changes in pressure and volume not alter the physical properties of the substance and the liquid remains liquid?

It is possible to change a liquid into a solid, or in general change the phase a substance finds itself in by just altering the pressure.

You might want to refer to some graphs known as Phase diagrams that portray the effect of temperature and pressure on the physical state of a substance.

(source: purdue.edu)

In this example if you're to draw a straight line from the point B to D, it'll have a subtle positive slope. So there's a possibility to transform a liquid into a solid without making any changes to the temperature conditions.

You might have heard of solid $\ce{CO2}$ or dry ice. I might be wrong but a part of me says carbon dioxide is pressurised and refrigerated for it to form.

An interesting thing to note here is that in case of a gas turning into a liquid, there's a temperature known as Critical Temperature ($T_c$) above which the gas cannot be liquified no matter how much pressure is applied.


Strength of weak acids and bases: Is methane better base than ammonia? $\mathrm{p}K_\mathrm{a}$ of methane is $\sim 50$ and that of ammonia is $\sim 37,$ thus ammonia is a better acid than methane, which implies methane is a better base than ammonia.

But I don't see how: methane doesn't have any lone pair of electron, neither it has any $\ce{OH-}.$ How is it possible?

You got it wrong. Acidic and basic strength of a compound are unrelated, about as much so as the man's name and his weight. Within the definition you and I are currently using (Brønsted–Lowry), acidity is about having a proton on a polar bond ready to ionize, while basicity is about having electron pairs ready for accepting a proton. A compound can have one and not the other, or vice versa, or both, or neither. Why should the two be related? They shouldn't and aren't. Well, there is one obvious exception: a strong acid can't be a strong base at the same time, otherwise it would turn upon itself. Except that, everything is possible.

Indeed, methane is a very weak acid, about 13 orders of magnitude weaker than ammonia (as attested by the difference in their pKa). In fact, it is one of the weakest acids among all compounds. This is not unexpected, given its almost non-polar bonds and the lack of stabilization in the anion. Also, methane is a very weak base. Forcing it to accept a proton was no small achievement in itself (see Methanium). This is not unexpected either, given its lack of lone pairs or negative charge.

Now you must be thinking that I am pulling your leg, because every kid knows that strong acid means weak base and vice versa. It is written in every textbook, said by every teacher, and that with great confidence, too. Can they all be wrong? No, they are right. How so?

Here's the trick: strong acid does indeed mean weak base, but not in the same compound. Look again at that equilibrium:
$$\ce{HA <=> H+ + A-}$$

See that? It is $\ce{HA}$ that acts as an acid here. And now look at the reverse reaction: it is $\ce{A-}$ acting as a base. Now if the equilibrium is shifted to the right, then $\ce{HA}$ is a strong acid, which inevitably means that its conjugate base $\ce{A-}$ is a weak base.

So it goes.


Why are HgO and CdS compounds colourful? We know that the electron configuration in $3d$ orbitals of $\ce{Zn^2+}$, in $4d$ orbitals of $\ce{Cd^2+}$ and in $5d$ orbitals of $\ce{Hg^2+}$ ions are completely same.

Then why the compounds of $\ce{Hg(II)}$ and $\ce{Cd}$, such as $\ce{HgO}$, $\ce{CdS}$ radiate versatile colour where $\ce{ZnO}$, $\ce{ZnS}$ are white?

The band gap decreases as either electronegativity of the anion decreases (electronegativity of anion in CdS < CdTe , ZnS < ZnSe) or the overlap decreases (overlap in ZnS > CdS).

Also, note that colored compounds only result when 1.8 < Band gap < 3.0. Larger gaps give white/transparent & smaller gaps give a black colour to compounds.
(The band gap in ZnS is 3.54, the band gap in CdS is 2.45.)


Do phosphoryl thiocholines, their uncharged analogs, and other organophosphates, interact differently with Acetylcholinesterase? As asked in the title, would the various salts -such as hydrochloride or methyl iodide- of V-Agents interact any differently with acetylcholinesterase compared to the neutral compound? For instance, would the interactions between 33-SN and AChE differ in any way from those between 33-SN+ and AChE? Also, would the phosphoryl-thiocholine analog of Soman have different interactions with AChE compared to Soman itself? In short, would a set of organophosphate AChE inhibitors which only differ in whether the leaving group during the process of AChE inhibition is a fluorine, dialkylaminothiolate, or thiocholine, interact differently with AChE??

As a matter, of fact, yes, uncharged phosphoryl thiocholines do seem to interact with acetylcholinesterase differently compared to their charged analogs. For example, it has been reported that

Sundwall (1961) reported that P2S was considerably more effective as
an antagonist of the lethal action of 37 SN
+(isopropoxy-2-trimethylammonio-tiomethylphosphino oxide) in mice than it was of those of Sarin and 37 S-N
(isopropoxy-2-dimethylaminothiomethylphosphine oxide).

P2S refers to the methyl sulfonate salt of pralidoxime. This would seem to suggest that, at least in the case of 37-SN+, the adduct formed by the charged inhibitor is more susceptible to reactivation by oxime therapy as opposed to that formed by the neutral compound†.
Regarding the second half of the question, the answer is also yes. According to Boskovic, Acetylcholinesterase that has been inhibited by O-pinacolyl S-2-diethylmethylammonioethyl methyl phosphonothioate methyl sulfate undergoes aging two orders of magnitude more slowly compared to AChE that has been inhibited by Soman, and is also more susceptible to reactivation. The explanation proposed by the author is that the interaction between a phosphoryl-thiocholine and AChE results in the formation of an adduct in which the phosphoryl group is oriented differently compared to the adduct formed as a result of inhibition by an uncharged organophosphate. The author also points out that AChE which had been inhibited by the compound Ro 3-0411 is more easily reactivated compared to AChE inhibited by diisopropyl fluorophosphate.
In conclusion, it can be said that, for a set of organophosphorus compounds which only differ in whether the leaving group during AChE inhibition is a fluorine atom, a thiocholine, or dialkylaminoethylthiolate (And in which the other side chains are the same), each of these compounds would interact with AChE differently, which influences, probably among other things, their susceptibility to reactivation by oximes and the rate at which the inhibited enzyme undergoes aging.
†It should be noted that, under physiological conditions, some amount of these neutral compounds becomes protonated, although I do not know to exactly to what extent and how and if that affects their pharmacology.
References:
BOSKOVIC, B. (1981). The treatment of soman poisoning and its perspectives. Fundamental and Applied Toxicology, 1(2), 203–213. doi:10.1016/s0272-0590(81)80059-0
BošKović, B., Maksimović, M., & Minić, D. (1968). Ageing and reactivation of acetylcholinesterase inhibited with Soman and its thiocholine-like analogue. Biochemical Pharmacology, 17(8), 1738–1741. doi:10.1016/0006-2952(68)90237-2


What does it mean that an electron pair "spends more time" nearer an atom? In a book I am reading, it is said

  In the $\ce{HCl}$ molecule, the shared pair of electrons spend more time nearer the chlorine atom. In the $\ce{HF}$ molecule, the shared electrons spend more time nearer the fluorine atom.

What does "spend more time" mean here? Has this been experimentally validated? If yes, how?

When you come to the quantum level (looking at one electron) things don't behave anymore like we normally think. To make it anyway understandable those expressions were invented such as "spending more time". Electrons are less described as a dot but as a smearing (electron density).
Chlorine and fluorine have higher electronegativities than hydrogen. The electron density therefore is not equally distributed between the two bonding partners but drawn towards fluorine/chlorine. To illustrate this, you can say "the electron spends more time at F/Cl".

This thinking was established as molecular orbital theory (s orbitals, p orbitals, ...). There have been other models before, e.g. you might know the shell model (k shell, l shell, ...). For experimental evidence you might want to take at look at this -- but most of all the citations therein: http://en.wikipedia.org/wiki/User:Chem507f10grp4/sandbox#Experimental_Evidence_Supporting_Molecular_Orbital_Theory
Basically, the shell model ran into a series of properties it did not calculate correctly or precisely. Molecular orbital theory was found to predict many properties and came to well calculated values.


Multiplet shape in proton NMR of morpholines I recently carried out a Buchwald–Hartwig reaction to attach a morpholin-4-yl group to an aromatic ring:

In the proton NMR of the (columned) product, I found the peaks corresponding to the morpholine protons. The multiplet at 3.86 ppm comes from the protons next to oxygen, and the protons next to nitrogen give rise to the multiplet at 3.04 ppm:

The spectrum was obtained on a 400 MHz spectrometer and the values of the "coupling constants" are shown in the second multiplet. (They are identical in the first one.)

Why do the peaks have the fine structure that they do? Based on a simplistic analysis (the "n+1 rule"), I would expect a 1:2:1 triplet for both of these peaks.

In my crude NMR (pre-column) there was some excess morpholine and those peaks had exactly the same fine structure.

Does it have something to do with the chair conformation of the six-membered ring and the differentiation between axial and equatorial protons?

It looks as if the NMR of morpholine is an AA′XX′ spectrum (the chemical shift difference is 0.80 ppm, or 320 Hz on your spectrometer, two orders of magnitude larger than the coupling constant).
Unlike linear AA′XX′ systems where the bonds can rotate, in the morpholine it very much has a fixed conformation.

In this fixed conformation, the protons are predominantly in a 'gauche' type arrangement due to the shape of the chair, this is where the appearance of the 'triplet' comes from in the NMR spectrum.
The image below is from Hans Reich's NMR notes, and shows some simulated NMR spectra depending on the gauche/anti ratio (Karpus equation), alone with a true spectrum of cyanomorpholine (which you can see looks basically like your spectrum above).

Source: Hans Reich (wisc.edu)

As a complete side-note. In TopSpin (which is what it looks like you used to process your NMR data) you can change the window function to use a Gaussian function, for strong spectra (which it looks like you have), this can help get a bit more resolution (at the expense of S:N- peaks on the baseline will disappear) such that what looks like a broad slightly messy spectrum actually gives resolved multiplets where the J values can be calculated.
In TopSpin, this can be done by entering the wm command; select "gaussian" from the drop-down list, and enter the two required parameters LB and GB (a decent starting choice is -2 and 0.2). The modified FID will now appear; this can be Fourier transformed using ft and automatically phase-corrected using apk to give the modified spectrum. [Even shorter is to sequentially type the commands lb -2, gb 0.2, and gfp into the command line.]
In MestreNova, the apodisation interface may be brought up from the Processing menu (or directly with the hotkey W) and the relevant parameters set in the same fashion.


How has iodine gained electrons? $$\ce{Cr2O7^2-(aq) + 14H+(aq) + 6Cl-(aq) -> 2Cr^3+(aq) + 3I2(aq) + 7H2O(l)}$$

Two chromium atoms gain 3 electrons each. They go from $0$ to $3+$

Similarily, 6 iodine atoms go from $-1$ to $0$, thus gaining one electron. However, my textbook says that iodine has lost electrons. How?

It went from negative to positive, i.e. from $-1$ to $0$, just like chromium from $0$ to $+3$.

Two points before we begin:

+ charge means deficiency in electrons.
- charge means excess of electrons.

Now lets come to the reaction you are talking about.

  Two chromium atoms gain 3 electrons each. They go from 0 to 3+

Moving from neutral to +3 charge means it just lost 3 electrons. (Electrons have negative charge.)

  Similarily, 6 iodine atoms go from -1 to 0, thus gaining one electron. However my textbook says that Iodine has lost electrons. How?

Your textbook is right and you're wrong here. Moving from -1 charge to 0 means that an atom just lost an electron (in other words, it lost the excess negative charge it had and now it is neutral).


Could micrograms of polonium really change the viscosity of a whole teapot? I was reading How radioactive poison became the assassin’s weapon of choice and it opens describing how

  The men eventually left, and Andrade cleared the table. As he poured the remaining tea away, he noticed that the consistency of the liquid that tipped into the sink was strange. Gooey. He couldn’t have known it as he puzzled over its weird yellow tinge, but the man who’d been sipping the tea was a 43-year-old Russian dissident called Alexander Litvinenko, and the tea itself, draining away into the London sewers, was lethally radioactive.

Later in the article it mentions the total amount he ingested was about a microgram.

Now, skipping past the part on how you wouldn't notice you are drinking gooey tea, could micrograms of polonium change the viscosity of a whole tea kettle?

It doesn't say how much he drunk, or the size of the teapot, so lets take some guesses. Lets assume he drank a very small amount of tea, say 10 mL to give us a decent upper bound on the concentration of the solution. So that would be 0.1 micrograms/mL or about 0.5 micromolar.

Could a solution in this concentration range change the viscosity of water enough for a casual observer to notice?

I think it is highly unlikely polonium makes water notably more viscous. The concentration is just too low, and radiation doesn't make water more viscous, controlrods are cooled using water and they produce a lot more radiation.

what can happen in the case of tea is that the radiation induces the formation of free radicals that in turn initiate polymerization or other reactions that form larger molecules which would make the solution more viscous, this still sounds highly improbable though.

there are multiple studies on radiation initiated polymerisation:
http://large.stanford.edu/courses/2012/ph241/weil1/


Can hydrogen alone be used as a fuel? I have always thought that fuel such as petrol, diesel, etc.. are getting burnt to move the pistons in the car (I am not bringing physics). Why not hydrogen be used as a fuel but I know the incident about the Hindenburg disaster but the same thing could happen if the Hindenburg was filled with petrol. So both are combustible and that's what we want. Petrol and hydrogen do the same job what does he want?. When hydrogen is burnt there won't be any smoke so it saves the environment. PS: This is just to make this question a bit understandable I have mentioned the word "Alone" in the question.

Hydrogen fuel cells which produce electricity from hydrogen and oxygen are over a hundred years old
https://www.thoughto.com/hydrogen-fuel-cells1991799

The recently introduced Toyota Mirai runs off a hydrogen fuel cell.
www.toyota.co.uk/mirai  as does the Hyundai ix35 and Honda Clarity

There aren't very many hydrogen filling stations in the UK which rather limits its appeal


What is the structure of nitrite? I know that $\ce{NO2}$ has 3 effective pairs,  so it is likely to have trigonal planar structure. In trigonal planar structure we usually see three atoms bonded to central atom at 120°. But in $\ce{NO2-}$, I observe that though it has three effective pairs,  but still, it is not having the desired structure,  with only 2 O atoms bonded to N atom. Can anyone explain the situation?

Nitrogen is actually surrounded by 3 effective pairs. But 2 of them are bonding and the third one is a lone pair. According to VSEPR theory, the structure should be  trigonal planar.

The angle is not exactly $120^\circ$, as the electrostatic  repulsion between non-bonding pair and bonding pair is higher than the repulsion between two bonding pair. So, the angle of $\ce{O-N-O}$ is slightly smaller than $120^\circ$.
The distribution of the three electron effective pairs is trigonal planar. But the molecule structure is bent (like letter V).


What is Fajans rule? when reading about polar covalent bond I came across the rule called 'Fajan's Rule' but could not find the explanation nor the derivation

Fajans' rules (note the difference) were formulated in 1923 by Kazimierz Fajans.

They are a method for predicting ionic vs. covalent that predates electronegativity (by three decades) and make use of ionic and atomic radius data that was becoming available through x-ray crystallography.

To use Fajans' Rules, assume your binary compound is ionic and identify the potential cation and anion.

By Fajans' Rules, compounds are more likely to be ionic if: there is a small positive charge on the cation, the cation is large, and the anion is small. For example, $\ce{NaCl}$ is correctly predicted to be ionic since $\ce{Na+}$ is a larger ion with a low charge and $\ce{Cl-}$ is a smaller anion.

Compounds are more likely to be covalent if: there would be a large positive charge on the cation, the cation would be small, and the anion would be large. For example, $\ce{AlI3}$ is correctly predicted to be covalent since it would have a small cation with a high charge and a large anion.

Note that Fajans' Rules have been largely displaced by Pauling's approach using electronegtivites. However, the remnants of Fajans' Rules are found in Hard-Soft Acid-Base Theory, which predicts bonding properties based on polarizability (which is based on size and charge). Binary compounds having a soft acid and/or a soft base are often covalent.

$$\begin{array}{|c|c|c|c|}\hline
\mathrm{compound} & \mathrm{Fajans} & \mathrm{Pauling} & \mathrm{HSAB}\\ \hline
\ce{NaCl} & \mathrm{low + charge \\ larger\ cation \\ smaller\ anion \\ionic} & \mathrm{3.16-0.93 = 2.19\\ ionic} & \mathrm{hard\ acid \\ borderline\ base \\ ionic}  \\ \hline
\ce{AlI3} & \mathrm{high + charge\\ smaller\ cation \\ larger anion \\covalent} & \mathrm{2.66 - 1.61 = 1.05 \\covalent} & \mathrm{hard\ acid\\ soft\ base \\covalent} \\ \hline
\end{array}$$


Estimation of the bond angle of water We know from experimental data that $\ce{H-O-H}$ bond angle in water is approximately 104.5 degrees. If its two lone pairs were bonds (which is unfortunately impossible) also $\ce{O-H}$ bonds and a perfect tetrahedron resulted, then VSEPR theory would predict that the bond angle would be 109.5 degrees - this number can be easily derived using the geometry of a tetrahedron. However, how would people give estimates of the actual bonding angle of water, which is caused by a slightly greater repulsion by the lone pairs than there would be if they were bonds? What physics would be involved in the calculation? Thanks!

how would people give estimates of the actual bonding angle of water
What physics would be involved in the calculation

Background
That's a very good question. In many cases Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.
$$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

For example, in the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.
How can we use this approach to answer your question?
Unlike methane, water is not a perfectly tetrahedral molecule, so the oxygen bonding orbitals and the oxygen lone pair orbitals will not be exactly $\ce{sp^3}$ hybridized.  Since addition of s-character to an orbital stabilizes electrons in that orbital (because the s orbital is lower in energy than the p orbital) and since the electron density is higher in the lone pair orbital than the $\ce{O-H}$ orbital (because the electrons in the lone pair orbital are not being shared with another atom), we might expect that oxygen lone pair orbital will have more s-character and the oxygen $\ce{O-H}$ orbital will have less s-character.
If we examine the case where the $\ce{O-H}$ bond is $\ce{sp^4}$ hybridized, we find from Coulson's Theorem that the $\ce{H-O-H}$ angle is predicted to be around 104.5°. So indeed, in agreement with our prediction, removing s-character from the oxygen $\ce{O-H}$ orbital gives rise to the observed bond angle.
Note on reality
For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized.  The general description is that there are two equivalent O-H sigma bonds and two equivalent lone pairs orbitals.  The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the hybridization changes as described above and the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.
With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs).  Now, the hybridization of water is described as follows:

2 $\ce{sp^4}$ O-H sigma bonds
one lone pair in a p orbital
and the second lone pair in an $\ce{sp}$ orbital


Why are equations of state for a non-ideal gas so elusive? The ideal gas equation (daresay "law") is a fascinating combination of the work of dozens of scientists over a long period of time.

I encountered Van der Waals interpretation for non-ideal gases early on, and it was always somewhat in a "closed-form"
$$\left( p + \frac{n^2a}{V^2} \right)(V - nb) = nRT$$

with $a$ being a measure of the charge interactions between the particles and $b$ being a measure of the volume interactions.

Understandably, this equation is only still around for historical purposes, as it is largely inaccurate.

Fast-forwarding to the 1990s, Wikipedia has a listing of one of the more current manifestations (of Elliott, Suresh, and Donohue):

$$\frac{p V_\mathrm{m}}{RT} = Z = 1 + Z^{\mathrm{rep}} + Z^{\mathrm{att}}$$

where the repulsive and attractive forces between the molecules are proportional to a shape number ($c = 1$ for spherical molecules, a quadratic for others) and reduced number density, which is a function of Boltzmann's constant, etc (point being, a lot of "fudge factors" and approximations are getting thrown into the mix).

Rather than seeking an explanation of all of this, I am wondering whether a more "closed form" solution lies at the end of the tunnel, or whether the approximations brought forth in the more modern models will have to suffice?

It's been known since 1941 that the answer to your question is in the negative, i.e. that there will never be a closed form equation of state for a nonideal gas.

In 1941 Mayer and Montroll developed what is now known as the cluster expansion for the partition function of a nonideal gas whose particles have pairwise interactions. This cluster expansion provides a 1-1 correspondence between various integrals over the interaction potential and virial coefficients in the Kamerlingh Ohnes equation of state which has in principle an infinite number of terms in it. Therefore, it would not be considered a closed form equation of state. In practice the virial coefficients are known to decay so it is usually safe to truncate the expansion for practical calculations.

One could extend the cluster expansion to three-body and higher-order interaction potentials but this will not change the fundamental argument above.


Evidence of orbitals? How do we know that there are different types of orbitals? For example, what evidence is there for the existence of $\mathrm{p}$ orbitals instead of there being multiple $\mathrm{s}$ orbitals (for example, why isn't the electronic configuration of sodium $\mathrm{1s^1, 2s^2, 2s^2, 2s^2, 2s^2, 3s^2}$ instead of $\mathrm{1s^2 2s^2 2p^6 3p^1}$)?

Let me approach this another way than the others: orbitals are NOT physical objects! They do  not exist in physical sense, they are theoretical constructs, chemical concepts that help understand / visualize / etc. mathematical solutions of Schrodinger / Dirac / Kohn–Sham / etc. equations.

Orbitals are not unique: given linear combinations are equivalent with each other, and there is no "correct orbitals", one can choose whichever they like. Canonical orbitals, natural orbitals etc are all good to go.

What is the evidence they exist? They do not exist, they are just mathematical solutions for given equations, and it is a purely mathematical question if they are good solutions for those equations or not. The theories themselves are consistent with experimental data, e.g. spectroscopic properties, geometries, reactivity.


How are diazonium salts prepared without self coupling While preparing diazonium salts, what prevents the freshly formed diazonium salt from reacting with the still unreacted amines left in the solution and perform diazo coupling?
If this happened then there won't be an appreciable yield of the diazo salt.
So what's the reason?

During diazonium ion preparation, first, sodium nitrite is mixed with hydrochloric acid to produce nitrous acid. The nitrous acid can be protonated under acidic conditions, which undergoes the loss of water and produce the nitrosonium ion ($\ce{O=N+}$; this reaction is usually performed at or around $\pu{0 ^\circ C}$):

An aromatic amine can attack the electrophilic nitrosonium ion to form a nitrosoamine $(\ce{Ar-NH-N=O})$.
The nitrosoamine can undergo a proton transfer (tautomerism) to give corresponding imminol $(\ce{Ar-N=N-OH})$.
Under acidic conditions, this $\ce{N=N-OH}$ group can be protonated and left as a water molecule to give the diazonium salt, which is resonance stabilized $(\ce{(Ar-N#N)+Cl-})$.

The key word here is all of these happen under acidic conditions. Therefore, the aromatic nucleus of aniline derivative is not active because of protonation. Also, during the synthesis, the starting aniline derivative is used as the limiting reagent. In addition, many of diazonium salts are unstable and hence, they are always prepared as needed under acidic conditions with good stirring, kept at or near $\pu{0 ^\circ C}$ to minimize the side reactions such as the reaction with water to produce a corresponding phenol until used in the coupling reaction (usually immediately).
The diazonium salt reacts as an electrophile with an electron-rich coupling component (a phenol or an aniline derivative), through an electrophilic aromatic substitution mechanism under basic conditions (specially when coupling reagent is a phenol derivative). To complete the synthesis of some azo dyes, it is also required heating.


Finding large databases of sphere packings Part of my research is investigating sphere packings and their properties.I really would like a large database of sphere packings that I could investigate using a program I have written that tells me certain things about the packings. After doing some research I found some online databases but a lot of them haven't been taken care of so they're riddled with dead links.  I was wondering if anyone knows of any books or websites that have a lot of sphere packings that I could investigate.  Or if there is a way to interpret crystal structures as a sphere packing, seeing how there is a lot of crystal structure databases.

If anyone is confused as to what kind of data I'm looking I just need coordinates of the centers of spheres and their corresponding radii.  These need to form a packing where the spheres are touching like in this paper where the authors have found new bi-disperse packings.  Thanks!

http://pubs.acs.org/doi/abs/10.1021/jp206115p

I have found a partial answer to finding a large database of sphere packings, thanks to the comment from permeakra!

So far I have been obtaining .cif files from the large databases of crystal structures online such as, http://www.crystallography.net/.  Once I have the .cif files I use openbabel to convert those files to .xyz files.  From here I have a table that describes a crystal's structure by the x,y, and z coordinates of each particular element for example,

\begin{array}{c|c c c}
\ce{Mg}      &   1.01054     &   1.01054   &     1.01054\\\hline
\ce{Al}      &   1.01054     & 1.01054     &   1.01054\\\hline
\ce{Al}      &   4.04217     &   4.04217   &     4.04217\\\hline
\ce{Mg}      &   4.04217     &   4.04217    &    4.04217\\\hline
\ce{O}       &   2.11390     &   2.11390    & 2.11390
\end{array}

But my problem from here is that babel has failed to return the ionic radius which I suppose can be retrieved online via wikipedia or something.  I posted another question that is more specific to .cif files which expands on this problem which can be found here, Extracting "Crystal Radii" from a CIF file.\

Limitations to this approach include that I can only find 3D sphere packings and it's a rather long and clunky procedure for obtaining lots of sphere packings.


How can either reactants or products be "favored" at equilibrium? From Atkins' Elements of Physical Chemistry (6th ed.):

  $$\Delta G_\text{r}^\circ = -RT \ln K \tag{8}$$
  This fundamentally important equation links thermodynamic quantities – which are widely available from tables of thermodynamic data – and the composition of a system at equilibrium. Note that:

  If $\Delta G_\text{r}^\circ$ is negative, then $\ln K$ must be positive and therefore $K > 1$; products are favored at equilibrium.
  If $\Delta G_\text{r}^\circ$ is positive, then $\ln K$ must be negative and therefore $K < 1$; reactants are favored at equilibrium.

It says products/reactants will be favored at equilibrium, but I thought at equilibrium neither are favored?

At equilibrium the rates of the forward and reverse reactions are equal. This means that you can have more product than reactant, or more reactant than product. It just means that the rates of formation of each are equal and so there is no net change in the amount of each. The amount of products and reactants at equilibrium can be favored, their amounts just won't change once they reach equilibrium.
Depending on the temperature and pressure, the composition of the system at equilibrium can change.
Recall that in the reaction $$\ce{aA + bB<=>cC + dD}$$
The equilibrium constant is: $$K=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$
These concentrations refer to the concentrations of $\ce{A}$ and $\ce{B}$ at equilibrium.
Therefore in a situation where $\ce{A<=>B}$ and $\Delta G_\mathrm{r}^\circ\lt0$, then $K>1$, meaning that there is more of $\ce{B}$ than $\ce{A}$ at equilibrium (and vice versa).

If you are interested to know why $\Delta G_\mathrm{r}^\circ = -RT \ln K$ gives the standard change in Gibbs energy for a reaction, then you should note that the general form of the equation at any point in the reaction is given by
$$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + RT \ln Q$$
Where $Q=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$ at any point in the reaction. At equilibrium, there is no more Gibbs free energy, as the system no longer tends to the formation of products or reactants. Therefore at this point $\Delta G_\mathrm{r} = 0$, and the original expression given in Atkins is what you get. Also we define this special case of $Q$, saying that at equlibrium $Q=K$, the equilibrium constant.

Source
You might find the above image helpful. The y-axis shows the amount of Gibbs free energy, $G$. The reaction will want to tend towards the state where it has the lowest amount of Gibbs free energy (the local minimum). The slope of the curve can be therefore seen as $\Delta G_\mathrm{r}$. Therefore the point where the system has the least Gibbs free energy will be when $\Delta G_\mathrm{r} = 0$ could occur anywhere in the extent of reaction. This shows that the equilibrium concentrations need not be equal, but shows that the reaction will not progress any further when $\Delta G_\mathrm{r} = 0$.