text:
Iodometric titration of copper – standard electrode potentials
Upon inspection of the standard electrode potentials for the half reactions for both $\ce{Cu^2+}$/$\ce{Cu+}$ and $\ce{I2}$/$\ce{2I-}$, it becomes apparent it should be the iodide ions that act as the oxidising agent, as the standard electrode potential for the iodine half-cell is much higher.
$$\begin{alignat}{2}
\ce{Cu^2+ + e- \;&<=> Cu+}\quad &&E_\text{cell} = +0.159\ \mathrm{V}\\
\ce{I2 + 2e- \;&<=> 2I-}\quad &&E_\text{cell} = +0.54\ \mathrm{V}\\
\end{alignat}$$
To explain this, many websites that I’ve looked at suggest it’s the weak solubility of copper(I) iodide, increasing the $\ce{Cu^2+/Cu+}$ potential to around $+0.88\ \mathrm{V}$.
Such websites include:

http://www.titrations.info/iodometric-titration-copper
http://chemistry.tutorvista.com/analytical-chemistry/iodometric-titration.html

What I’d like to know, is how this value of $+0.88\ \mathrm{V}$ has been determined, and a little of the theory behind as to why it has increased from $+0.159\ \mathrm{V}$.

Let's consider the following redox couples:
$\ce{Cu^{2+} + e^- ->Cu^+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_1}(\ce{Cu^{2+}/Cu^+}=0.17\, \ce{V})$
$\ce{Cu^{2+} + e^- + I^- ->CuI\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_2}(\ce{Cu^{2+}/CuI}=?)$
As the chemical species $\ce{Cu^{2+}}$, $\ce{Cu^+}$ and $\ce{CuI}$ are together in the aqueous solution, the two redox couples  $\ce{Cu^{2+}/CuI}$ and $\ce{Cu^{2+}/Cu^+}$  are in equilibrium and that means: $$ E_1(\ce{Cu^{2+}/Cu^+})=E_2(\ce{Cu^{2+}/CuI})$$ Using Nernst equation for each couple:$$ E_1(\ce{Cu^{2+}/Cu^+})=E^0_1(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}]}}$$
$$ E_2(\ce{Cu^{2+}/CuI})=E^0_2(\ce{Cu^{2+}/CuI})+\frac{RT}{F}\ln{[\ce{I^-}][\ce{Cu^2+}]}$$
We can write then:
$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}][\ce{Cu^2+}][\ce{I^-}]}}$$ We rewrite the above equation after simplification:$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{[\ce{Cu^+}][\ce{I^-}]}}$$
$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{K_{sp}(\ce{CuI})}}$$
$$E^0_2(\ce{Cu^{2+}/CuI})= 0.17+\frac{8.314 \times 298}{96500}\ln{\frac{1}{10^{-12}}}=0.88 \ce{V}$$


text:
Any popular experiment in chemistry that digital signal processing played a crucial role in?
Is there any famous experiment in chemistry where digital signal processing played an important part? I don't mean using a machine that relies on such techniques (they all do) but an experiment where signals were first recorded (e.g. temperature/pressure/voltage/etc. as a function of time) and then analyzed with some sophisticated methods (e.g. time-frequency analysis with wavelets).

You mean something like Spectrograph

No, I don't mean that.

I do not understand. How would you imagine doing an experiment with signal recording without a machine?
Do you mean some pioneering experiment, where a guy measured some curve, which had no sense. Than he applied signal processing by hand and voila, nice results?
And it was so nice, he build it into automated machine with DSP.

Many spectroscopic experiments record analog data, which is subsequently Fourier transformed to give a frequency-domain spectrum. For example, in a nuclear magnetic resonance experiment, the signal emitted by an excited nucleus is picked up as a voltage in an RF coil for 1-2 sec. This time-domain signal is processed then converted to the frequency domain by FT.
As for 'popular', Richard Ernst won the 1991 Nobel in Chemistry for his contributions to this field of research.

Many mass spectrometers (ion cyclotron and also "orbitrap" instruments) now also work by Fourier transform.

How about the patch-clamp technique in neurochemistry?  This technique allowed the study of ionic currents through single molecules of a voltage-gated ion channel.  It won the Nobel prize in 1991.
The signal processing required to simply observe the single-molecule opening and closing events is fairly minor.  However, processing single-molecule ion-current data for multiple experiments when inhibitors are titrated, or to compare ensemble-average kinetics to single- molecule kinetics, can be fairly complex.
Here is a paper from 1998 on single-molecule enzymology; it's an extension of the patch-clamp experiments developed for the specific case of single-molecule ion channels: http://www.sciencemag.org/content/282/5395/1877.full  Read the paper for a better description of some of the signal processing involved.

source: Stack Exchange

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sort: votes


text:
Principle of potentiometric titration
In potentiometric titration between ascorbic acid and iodine, when we plot the graph of EMF with volume of iodine consumed, we get a sudden decrease in EMF at equivalance point--why? It is also negative--why?
Using Nernst equation at equivalence point we get
$$E_{cell}=E_{STP}~~~(voltage ~at ~STP)$$
What is its physical meaning?

In potentiometric analysis, there are two electrodes, an indicator electrode (platinum here) where reduction takes place and reference electrode (calomel electrode) where
oxidation takes place.
So here the reference is not $E_{STP}$ cell but the $E_{cathode}$ that is the calomel electrode, so there is drop to $-241mV$ after the equivalance point. Hence,
 $E_{cell} = E_{cathode}$

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text:
Can there be active THC be trapped in the evaporator?
In relation to the vaporization of THC, when smoked, is it possible that there is any active THC residue that becomes trapped inside the evaporator after the flame goes out?
Wouldn't some of the vapor get stuck inside because the instrument is not a hollow tube?

I have whittled this down to a minimum and toned the language down a bit to be more professional.

It's a site on which we discuss science. "Blunt" is fine to use with your friends, but it is not very professional. If anything, this edit actually helps deter stereotypes of said culture, so you should be happy about that. There's no need to be insulting.

Everything you just said takes away from what science means.
Stereotypes? Who cares? Aren't we scientists? Professionalism? Are you so concerned with political correctness that you would stop using something's proper label? Since when did colloquial labels take away from the value and meaning of information in the realm of science?

I'm not going to discuss this with you any further.

The words of a man who knows he is in the wrong.

When burning a blunt, joint, doobie or whatever, THC evaporates (bp of tetrahydrocannabinoles is around 155 °C). Depending on

the length of the "object" , related to the temperature gradient between the front and the mouth piece, and
the type of the "filling" and its adsorptive properties

a part of the vaporized material might condense and/or get absorbed in a colder region of the filling. Similar effects, such as the enrichment of nicotine in the stump of cigarettes have examined.
It is however unlikely that any of the transported and reabsorbed components can be inhaled from a cold "reactor". The relevant components are no longer in the vapour phase and the vapour pressure of the tetrahydrocannabinoles will be too small.


text:
What does it mean in a Ni(II)complex if the HNMR spectrum appears unbroadened and unshifted & what does this tell you about the coordination geometry?
I am trying to work out the an unknown Ni(II) complex. I don't known the geometry and I have read that broad lines on the nmr spectrum indicate large compounds with symmetric environments. Does this mean that the complex would have tetrahedral geometry as square planner would be symmetrical and octahedral too large?

If the proton spectrum is unbroadened and not shifted out of the ~0-10ppm range, then I would have to say it is a diamagnetic compound.
It could be low spin square planar.  Tetrahedral and octahedral d8 cannot be diamagnetic.

Image source: http://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png
Could also be two nickel atoms in the complex that are antiferromagnetically
coupled for a net spin of zero.


text:
Why would there be differences in UV/Vis spectrum?
$\ce{[NEt4]2[NiBr4]}$ dissolved in $\ce{MeCN}$ and $\ce{[NEt4]2[NiBr4]}$ dissolved in $\ce{H2O}$ have different UV/visible spectra in the visible region. How would they differ, for example would one of them have more peaks? Also how would the molar extinciton coefficients differ and what do these numbers tell you?

This seems a very strange molecule/complex to me. Assuming bromine has a -1 oxidation state and further assuming $\ce{[NEt_{4}]^{+}}$ it would leave nickel in a -1 oxidation state. I was unable to find this compound, could you add a reference or synthesis? The only compound that I found referenced in Greenwood/Earnshaw  with Ni (-I) was $\ce{[Ni2(CO)6]^{2-}}$.

$\ce{H2O}$ does not absorb the radiation almost at all, so it will not interfere in the absorbance of the analyte we are analysing, thats why we use water almost always as a solvent in instrumental methods of analysis.Therefore $\ce{MeCN}$ wil absorb some of the radiation so then of course the absorbance of the analyte will also change ( decrease ) and the spectrum will be different ( the peaks of the analyte will be lower ).

I haven't looked for the spectrums for the compound in the different solvents, so I can only theorize what the differences would be. Since the nickel ion has the energy levels for visible light, then the differences would be due to what surrounds the nickel ion. In water, the nickel ion would be surrounded by water, where as in acetonitrile, the molecule $\ce {[NEt4]2[NiBr]}$ would stay together.

would one of them have more peaks?

No, since it is the nickel atom that influences the visible spectrum.

Also how would the molar extinction coefficients differ and what do these numbers tell you?

I cannot predict the difference in the molar extinction coefficient. The coefficient would tell you how easy the transition took place in the atom. The larger the number, the easier the transition.


text:
Organizing my "chemical" cabinet
I'm reorganizing my garage (I do metalworking and some machining), and now I have an empty white cabinet that I'd like to dedicate to chemical stuff. I'm not a chemist, and I don't have the typical assortment of chemicals that a chem lab would have; the stuff I'm referring to is typical stuff you'd find in a garage: adhesives, lubricants, solvents, paint, etc. (household cleaners are stored elsewhere).
I'd like to organize things by functional groups (like those listed above). Is there a convention/standard for organizing shop chemicals?

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The only suggestion for organizing the cabinet would be based on safety. Put things which could potentially react dangerously far apart, would the container leak or break.
If this is not the case, than the safety comes from convenient and easy use, so organize at your will.

@ssavec - My knowledge of chemistry is very weak. Of the types of chemicals I mentioned, are there any that stand out as "KEEP THESE APART"?

@ScottSmith: Out of the things you mentioned I do not see any problem. Household cleaners are separate, as you mentioned. Other problem could be garden chemicals (pesticides, fertilizers).
As a dirty guide: If you are not afraid working with it without thick gloves then it is reasonably safe.

As mentioned in the comments, safety is the main priority.  Take particular heed to the warning labels and information on the chemical label - if unsure, look up the chemical details MSDS (or SDS) = (Materials) Safety Data Sheet.
One of the most important thing to do (and is often overlooked) is to keep an inventory of what chemicals you have.  The safety data sheets can be obtained from the manufacturers or by looking up the chemical SDS itself, or from general websites such as MSDS.com - an example is this one for mineral turpentine - these contain considerable information about storage, disposal and first aid.  (It is always a good idea to have this readily available and updated).
Personal protective gear is a must - gloves, safety glasses etc - usually the label or the SDS will have this information.
According to the document NERC Guidance on Safe Storage of Laboratory Chemicals (2010), has information that is pertinent to a workshop environment.  There are 3 main principles in regard to storage of chemicals explained in the document:

Segregation, from the article:

The key incompatibles to segregate from each other are strong acids from
  strong bases and strong oxidisers from organic or flammable materials.

The NERC document has a list in its appendix (too long for this reply)

Separation, related to the segregation,

In a laboratory situation adequate separation can be achieved by means of
  storage cupboards which physically divide incompatible classes of hazardous
  chemicals. The cupboards may need specific properties or provide
  separation by means of distance. They will also need to provide secondary
  containment (eg spill trays or bunded shelves) and security (eg locks / bolted
  to wall).

Ventilation (from personal experience, this one is often overlooked).

Ventilation is often an essential requirement for safe storage of hazardous
  chemicals. Its main function is to allow dilution and extraction of vapours or
  gases that may escape / seep out from containers during storage so they no
  longer present problems from the viewpoint of noxious smell, hazardous
  personal exposure or creation of an explosive atmosphere.

Also consider an appropriate fire extinguisher, note that water is probably the worst one.  Here is an example of a fire extinguisher type guide - this guide suggests that powder based extinguishers may be best (but, make sure you study it and seek advice if unsure).
Some of this may seem 'overkill', but it is better to have some preparedness for if anything goes wrong, than be unpleasantly surprised by it (from personal experience), also, depending on where you are, there could be a legal requirement for a degree of proper storage, handling and protection.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].|| I have already approved your other edits, I can recommend the more advanced guides then: http://meta.chemistry.stackexchange.com/q/86/4945 and http://meta.chemistry.stackexchange.com/q/443/4945


text:
Why does water change phase so easily?
I did some undergraduate level Chemistry as part of my degree. It left me with the distinct impression that the polarity of water is responsible for its unusual phase properties: it's the only everyday substance that can exist in all three states under easily replicable conditions.
My daughter is learning about states of matter at school, and I thought this might be a fun "extra" for her homework, so I sat down to explain it to her. At which point I realised that I couldn't explain it myself. Polarity explains why water solidifies relatively easily, and why ice floats/expands - but I couldn't see/remember why it might have anything to do with transition into a gas.
So I taught her about phase diagrams instead, as a nice visual thing to take to class.
But it left me with the question: does water's polarity help explain its unusual phase properties? Or is it something else? Or are those phase properties perhaps not as unusual as I seem to remember?

It's all about hydrogen bonding. I find @ssavec's answer most adequate.

I would say the reason water is the only substance that can exist in all three states under everyday conditions is that we happen to live on the one planet in the Solar system whose surface conditions are near its triple point. (Which isn't necessarily that much of a coincidence, given the necessity of liquid water for life as we know it.)

does water's polarity help explain its unusual phase properties?

Not just its polarity, but its ability to form hydrogen bonds, together with the size and shape of its molecules.
Hydrogen bonds give water an unusually high boiling point for a molecule its size; many molecules of similar size that can't hydrogen-bond are gases at room temperature. They also give water a relatively high heat capacity and high enthalpies of fusion and vaporization. These latter two properties mean that ice has to absorb more heat to melt and liquid water has to absorb more heat to boil than would otherwise be necessary without hydrogen bonding.
Water's shape lets it form 4 tetrahedrally arranged hydrogen bonds. This lets it hydrogen-bond into networks of chair-shaped hexagonal rings in ice. The voids in the center of the rings make ice less dense than water. This is unusual; the solid form of a compound is most often denser than the liquid form.

Vapor pressure of water is quite small, thanks to H-bonding in the liquid. Therefore more energy is needed to evaporate (break H-bonds with neighboring molecules) a water molecule.
If you compare water to "nearest" non-H-bonded analog, dimethylether, $\ce{CH_3-O-CH_3}$ you'll find, that vapor pressure of ether is much higher (500 kPa) than water (2.3 kPa).


text:
How to activate charcoal
I readily admit to being a pyrophiliac and can generally be found near fire when there is fire. (Fire good, tree bad.)  One of the byproducts of fire, naturally, is charcoal.  I was looking into ways to activate charcoal instead of buying it over the counter - because how cool would it be able to just make your own - but my research so far isn't very promising.
Apparently, I require intense heat and very little oxygen in order to turn the home and garden variety of charcoal into activated charcoal (as one might find in Norit).
Is it possible to activate your own charcoal (preferably in one's backyard), and if so, does anyone have any ideas or suggestions that I can use to turn my meagre amount of (regular) charcoal into activated charcoal?

Welcome to chemistry.SE! If you had any questions about the policies that govern our community, please visit the help center or take a tour.

Thanks, @MARamezani!  The warm welcome is appreciated, though I am uncertain whether I should read your comment as a hint that my topic is not quite what you people generally cover? (I've checked the tour page, but I couldn't really make out whether that was what you intended either..)

Don't worry. I'm accustomed to leaving those links as a sign of feedback to new users. You understand that not all questions are as on-topic as your question is. Yours isn't off-topic at all: Asking for a home experiment involving a chemical reaction. It's sheer chemistry alright. However, my research too hasn't shown any safe ways of doing what you intend to do.

@MARamezani Oh, whew. :)  And yeah, the safety is one of the reasons I haven't yet tried it on my own.  Self preservation wins out every time.  I guess I'll never be a successful inventor at this rate, but I also won't be statistic #x that blew their own sorry ass to kingdom come. XD

@spoorlezer - you can get activated charcoal by cutting open an ordinary water filter. EDIT: oops, I see that you want to make your own! Well best of luck :D!

Note that like most of the colleagues on this site with practical lab experience, I've used activated charcoal once in a while, but I never bothered to make it myself.
Supposed that you live on the countryside with enough space around, you can make your own charcoal in a earth-covered pile or an oven - unless local environmental regulations ban that.
Activation, even when starting from ordinary charcoal is different. The gas activation requires heating (and partial burning) of the charcoal at up to 1000 °C in a stream of air, carbon dioxide or water vapour. I'd rather not do it in my backyard.

can't bunsen burnders achieve temperatures of 1000 C?

Sorry for reviving this question, but I just saw it today.
Since you already have charcoal, I guess you don't need the process required for making charcoal. But, you know, just in case:
making charcoal
activated charcoal from charcoal
The only extra thing you need is calcium chloride. There's a nice video explaining all the steps in great detail, if you need it. I used these sites myself for making activated charcoal and they worked pretty well.
Note: this will not produce high quality grade charcoal but it's effective.

Welcome to Chemistry.SE! Find out how to beautify your posts in the help center and take the [tour] to learn more about this site. One thing - we prefer if people give a brief summary of linked text in case the link breaks. Could you please summarize these two pages?


text:
Why is it obligatory to cool down the container of a sample to measure its mass in a lab?
We were doing an experiment about hydrated crystals and more precisely how to determine $n$ in $\ce{CuSO4.nH2O}$. After we heated the crucible we were to cool it down using a Desiccator. Then this question came to my mind: Why must we cool it down? By doing researches in the Net I were only able to come up with these suggestions, but none seem very applicable for me:
Using Einstein's formula:
His famous equation:$$E = mc^2$$
This means that matter is able to be converted to energy. Thus, more energy should mean more mass because $C$ is constant. However, I argue that since $C$ is a very large number, normal and observable-in-normal-labs changes in energy aren't able to modify the net change of the mass dramatically (to a degree that the balance is able to detect)
A gut feeling, no backup research:
Anyone sane, and with regards to the value the balance has in the lab, would give slightest chances that the hot container would harm the surface of the balance and therefore, damage it. But chances are that I'm wrong too.
So, why must we cool the crucible in order to measure the mass of the sample in it using a balance?

I'm confused why you accepted this answer. The below answer is good and your answer is not bad. I don't want to tell you which answer is better and You can accept which one you want.

Fred's answer is very informative, but it adds points to what ron and I said. And ron's answer was simple and short, while I went on telling grandma stories! :)

A container that is not at ambient temperature will generate air currents around it.  If you place such a non-ambient container in a balance, air currents will develop around the container as it heats or cools to ambient.  These air currents will cause the balance to read incorrectly.

The other two answers here are fine; it's true that as the hot object you're trying to weigh warms the air around it, the air will rise, creating air currents that will give you an unsteady reading.
I'd like to add three things they didn't mention:

The balance pan will heat up, too, causing metal parts in the balance to expand. This will also contribute to an error in your reading, and as the metal cools, the reading will change.
Another contribution to the error is that the density of a hot sample is less than the density of a cold sample, so its buoyancy is different. The object will appear to gain a tiny amount of weight as it cools!
Don't worry about $E=mc^2$ in chemistry unless you've got some serious nuclear reactions going on.

And a fourth thing: if you put a hot object into an air-tight dessicator, the air inside will contract as the object cools---and you might have some trouble getting the lid off again!

Your gut feeling is right: Very hot crucibles can cause damage to the balance. However, this isn't the main reason for why chemists cool their samples down before measurement.
We are all familiar with convection currents: When air is heated the molecules will collide more than its previous state; resulting in an increase in either volume or pressure depending on the "flexibility" of the environment of effect. Thus, the density of hot air decreases, and it goes up and "replaces" the cold air. If the process of heating the air continues, hot air will continuously go up in what we call a "convection current".
The system of the crucible is thermodynamically open, so in contact with air it will lose some of its heat to its surrounding gas, which is normally atmosphere air. If we put the hot crucible for measuring its mass, convection currents will occur around the crucible, where it's in contact with the surrounding air. The motion of the hot air upwards forces your crucible upwards, but in small amounts.
The balance has to be precise enough to detect these differences, though. Most balances in the labs can measure up to 0.01 or 0.001 gram, and will sense the difference. Since you usually need (or are made to write) exact measurement results up to 1 miligram, these differences are going to be irking pests in measurement.
This is a very useful source from which the answer came from. It also has useful recommendations on practical problems in the lab.

While these are all very good answers, I have a much simpler one. These far overthink the question. The crucible has to cool in the desiccator to avoid adding moisture to it. The experiment you're doing deals with hydrates, therefore getting as much moisture out as possible is important as it will reduce your percent error.

I'd agree with that in general, but the question is asking why it must be cooled, not why it must be cooled in a desiccator.

You must let the crucible cool before measuring it because the heat from the crucible warms the surrounding air, which rises, then that air cools down and falls. This rise and fall of surrounding air is called a convection current and will give you an unsteady reading that is rising and falling. These currents will make it hard to find the weight of the anhydrate(CuSO4), thus making it hard to find n, the moles of water per mole of CuSO4.

This was already told in other answers.


text:
Why can't bicyclic structures undergo E2 elimination?
Bicyclic structures like norbornane don't undergo E2 eliminations, but why is this the case? I understand that it can't do E1 as it can't form a stable carbocation (the bridgehead carbons that is) but I don't see the issue for E2. Can you explain?

Elimination reactions (including E2) are not forbidden on bicyclic molecules, they are just forbidden at the bridgehead (due to geometry issues that prevent effective orbital interactions).

I am aware of this - apologies that I didn't make it clear. I just don't understand why these carbons can't undergo E2 eliminations.

for E2 mechanism to work, the pair of leaving groups (commonly hydrogen and halogen) must be in antiperiplanar conformation (i.e. when loonking along axis of the carbon-carbon bond that would be double after elimination, they must be at 12 and 6 o'clock). In small bicyclic systems assuming such conformation is problematic. Also, formation of 2-nornbornil cation is well documented and E1 elimination for 2-nornbornyl species is known. Also, it is prohibited to form double forms at bridgehead, as the resulting system has too much angular strain.

Firstly, you have to specific the position of elimination. Like Ben Norris said, double bond can not be formed on bridge head in strained bicyclic structure simply because the geometry does not allow a planer double bond. For E1, cation can be generated but double bond can not be formed, it end up a SN1 reaction. For the concerted E2, it simply can never happen because the product can not be formed.

Elimination via a E2 mechanism cannot take place at the bridgehead carbon of a bicyclic compound due to the lack of the required trans geometry between the leaving group at the bridgehead carbon and any of the six β-hydrogens.
On the other hand, as you have correctly observed, a E1 mechanism has to be ruled out as well due to the strain at the bridgehead carbocation. In fact bicyclic structures prevent the tertiary bridgehead carbon becoming planar, which implies that is very high in energy because the nonplanar structure forces the cation to be an empty "sp3 orbital" instead of an empty "p orbital". This observation was made for the first time by Bredt, therefore is generally known as "Bredt's rule".


text:
How is the Rate law of a reaction measured?

What is the Rate law of the reaction below?
$$\ce{4 HBr + O2 -> 2 H2O + 2 Br2}$$

How do we approach finding the rate law of a reaction? Can this be found out only through experiments?

Preamble
For a reaction with no intermediate steps, you could derive the rate law from the reaction equation. For a generic equation
$$\ce{$a$A + $b$B -> C}$$
the rate is given by
$$r = k\cdot c^{a}(\ce{A})\cdot c^{b}(\ce{B}).$$
Rigorously, this is only possible for elementary reactions. A prominent example for a strictly unimolecular reaction is radioactive decay. Also quite commonly known is the photolytic or thermal cleavage of hydrogen or bromine:
\begin{align}
\ce{H2 &->[\Delta T] 2H. }\\
\ce{Br2 &->[h\nu] 2Br. }
\end{align}
An example for a bimolecular reaction would be
$$\ce{2NO2 -> NO3 + NO}$$
Reaction mechanism of the combustion of hydrogen bromide
This seems to be a reaction with a fairly complicated mechanism, involving multiple species, most of them radical. I would assume, that it is somewhat similar to the combustion of hydrogen, that I have described here.
A quick search on some forums will turn out a reaction mechanism, which I think is wrong. But I will show it here anyways, since it gives you a first clue, about how complicated this reaction might be. (Note: These are supposed to be gas phase reactions.)
\begin{align}\ce{
HBr + O2 &-> HOOBr\\
HOOBr + HBr &-> 2 HOBr\\
HOBr + HBr &-> H2O + Br\\
}\end{align}
The following mechanism is taken from Lilian G. S. Shum and Sidney W. Benson Int. J. Chem. Kin., 2004, 15 (4), 341-380. I think this is a much more suitable and credible approach to the problem. (Note: Most of the reactions are in equilibrium.)
\begin{align}\ce{
HBr + O2 &~<=> HO2 + Br\\
HO2 + HBr &~<=> H2O2 + Br\\
H2O2 + 2HBr &~->T[wall] 2H2O + Br2\\
Br + Br + M &~<=> Br2 + M\\\hline
4HBr + O2 &~<=> 2H2O + 2Br2\\
}\end{align}
Unfortunately I was unable to determine what $\ce{M}$ refers to in this case. It might be any inert gas used as a solvent in the reaction or the wall.
Rate law of the combustion of hydrogen bromide
As a consequence of the mechanism you cannot find the rate law from theory, but only from experiments. It is important to realize, that most of the reactions are in equilibrium and the whole system is dependent on the concentrations of each of the involved species. Therefore the kinetics of this reaction may be expressed as a complicated system of coupled differential equations.
However, Shum and Benson conclude, that the first step will be the rate-determining and the rate expression is given through

$$\frac{\mathrm{d}[\ce{H2O}]}{\mathrm{d}t} = 2k_g[\ce{HBr}][\ce{O2}].$$

Therefore it can formally be regarded as a bimolecular reaction.

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text:
How to interpret the instruction of the dilution of hydrogen peroxide?
I am required to prepare $\pu{120cm3}$ of hydrogen peroxide to each of one of my students tomorrow. Some material is given to me on the method of preparing the solutions. Within the method of preparing $\ce{H2O2}$, it says:

Prepare from freshly purchased $100$ volume ($30\%~\text{w/w}$, $\pu{8.3 mol/dm3}$) or $20$ volume ($6\%~\text{w/w}$, $\pu{1.7 mol/dm3}$) $\ce{H2O2}$. Dilute $\pu{200 cm3}$ of $100$ volume solution to $\pu{1 dm3}$ of $20$ volume solution, then dilute $\pu{59 cm3}$ of $20$ volume solution to $\pu{1 dm3}$ of $\pu{0.1 mol/dm3}$ solution.

Can someone please explain the quoted part?

Take $\pu{200 mL}$ of the $30\%$ solution, add water to a final volume of $1$ liter. Then you have a $6\%$ solution. Next, take $\pu{59 mL}$ of the $6\%$ solution and add water until it is $1$ liter.
After the first dilution it's not exactly $6\%$, because the density of $30\%$ $\ce{H2O2}$ is $\pu{1.135 g/cm3}$. It's more like $6.6\%$, but the instructions are referring to it as $6\%$. If you want exactly $6\%$ you should dilute by weight instead of volume.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Intrinsic Reaction Coordinate calculation-- both sides of TS lead to same minimum
I am currently trying to find an addition transition state and verify it using an IRC calculation. I have found a good candidate, but the IRC calculation gives the same minimum on each side of the transition state. In my mind, this does not seem any different from a vibrational mode.
I think the TS is submerged, and I am sure this is complicating things, but it does not make sense that an IRC calculation would give the same minimum in both directions.
Any thoughts on what this signifies or how to remedy it?
Thanks

Could you tell us more about the reaction you are looking into?

An information about the software package you use and the relevant pieces of the input file would also help the discussion.

Be sure you have really obtained TS, that means stationary point, i.e. all gradients are zero. Than, you should have one imaginary frequency corresponding to the motion along the reaction coordinate. If this is true and you still get both IRCs going to same minimum, your assumptions regarding the shape of Potential Energy Surface (PES) could be wrong.
It is easier to find the minima first, so be sure you have well optimized reactant, product, and if by any means reasonable, also intermediates. It is possible that there exist some high energy intermediates, which complicate the whole story (if you are not aware of them).

It could also be the case (impossible to tell, as you haven't told us what molecules you are using) that the two minima are equivalent by symmetry.

When I generate an IRC from a TS - using GAMESS (US) software - I calculate the IRC for the forward direction and the IRC for the backwards direction using MacMolPlt.  Is it possible you've stitched together two forward (or two backward) direction IRCs by accident?
Also, I once tried to calculate and IRC from a TS but both the forward and backward IRCs went in the same direction.  I don't know why ... I got around it by cribbing someone else's code.
A good place for code is the GAMESS (US) user group on google groups.
Hope the above helps...

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. I have no idea if this answer is acceptable here or not, so all I can do is give you a warm welcome!


text:
Cohesive forces in solid solutions
I have searched through printed books and the Internet, but I cannot find a definitive answer on an issue that fascinates me: what forces keep the various compounds belonging to a solid solution together?
If I correctly understand, a rock is a solid solution, and I know that the atoms of a single mineral species are bound by its particular covalent and ionic bonds, but what keeps different minerals together in a rock? Do they mix by sharing "neighbouring" cations and anions? I am not sure whether intermolecular forces such as Van der Waals forces are present, but I would definitely expect stronger forces in a solid solution or between the minerals of a rock.

Pure compounds and solutions are held together by the same types of forces.

No need to thank it's very general answer, you can see https://en.wikipedia.org/wiki/Solid_solution for some details, and generally rocks aren't solid solutions.

@Mithoron Nevertheless, I suspect that the forces keeping rocks together are the same that keep solid solutions together: am I wrong? Thank you again!

It seems you were on right path - for example common feldspars are solid solutions and they are mixed because they share cations, while they easily substitute one another, otherwise they exsolve - become separate phases. Also you're right that van der Waals forces aren't important here as they are ionic compounds.


text:
Solvent interaction with thermal paper
If you place a piece of paper created by a laser printer, e.g. fax paper or some receipts, on top of a piece of normal paper you have drawn on with a Bic Marking 2300 permanent marker, the image transfers to the fax/receipt paper? Do any non-toxic solvents heat up as they evaporate? If so, would that explain the reaction between the paper and ink?

I'm not sure if I fully understand your question. But I know one thing: Evaporation is an endothermic process, so anything that evaporates has to gain heat in some way in order for it to evaporate.


text:
Is there a pressure dependent function for freezing points of water and carbon dioxide?
I'm trying to determine whether or not certain compounds form solid ice at certain atmospheric pressures. These pressures vary significantly, from 0.001 atm to 800 atm. I understand that there is no equation relating the freezing points of all elements, but is there one or two that relate to carbon dioxide and water?
Ideally I'd like a Python function (or library) or equation where the input is atmospheric pressure or kPa.
Or, failing that, if someone could point me in the direction of a phase diagram for both compounds where I can input specific atmospheric pressure values and read a specific temperature for freezing point, that would be helpful as well.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].|| I do not have the time to do more research, but maybe this page can help.

I did a lot of research already but I couldn't find a very detailed phase diagram. The link you posted's phase diagram for CO2 is broken.

Sorry about that. Maybe this helps for CO2 and this might help for water

Diagrams on https://en.wikipedia.org/wiki/Carbon_dioxide

@Martin That link does not load.@Mithoron that's the first place I looked. Not detailed enough.

Both links to phys rev lett work for me.

It works now, strange. Still, it's behind a paywall.

For ice I melting pressure as a function of temperature:
$$\pi = 1- 0.626000 \times 10^6 (1- \theta^{-3}) + 0.197135 \times 10^6 (1- \theta^{21.2})$$
where
$\pi =$ (pressure in MPa)/0.000611657 MPa
$\theta =$ (temperature in kelvins)/273.16K
Source is International Equation of the Pressure along the Melting and along the Sublimation Curve of Ordinary Water substance (no paywall).
For carbon dioxide:
$$\frac{p_m}{p_n} = 1 + a_1 (\frac{T}{T_t}) + a_2 (\frac{T}{T_t})^2$$
Where:
$T$ is temperature in kelvins
$T_t =$ 216.592K
$p_m$ is the melting pressure in MPa
$p_n =$ 0.51795 MPa
$a_1 =$ 1955.5390
$a_2 =$ 2055.4593
Source is A New Equation of State for Carbon Dioxide... (no paywall)
Check original sources before using equations in case I made typo(s).

Thanks! What exactly is P_sub_n? If I had an atmosphere of 0.006 atm or 0.00060795 MPa, would P_sub_n be 0.00060795? If I solve for T, would that give the temperature that C02 freezes at at at that pressure?

I'm not a Chemist, and unfortunately beyond Wolfram Alpha I don't know how to use these equations.

@anarchy8 p_n is the triple point pressure, the pressure at which gas, liquid and solid are in equilibrium.  P_m is the melting point pressure, so 0.00060795 in your example.  The co2 equation you could solve for t/t_t using the quadratic formula or a program like wolfram, or you could construct a table of values and interpolate the table.


text:
Electron Pair Geometry in Ibuprofen (Electron Pair Geometry in Organic Compounds)
I am trying to determine the electron pair geometry (and molecular geometry) of some atoms in ibuprofen. The first one I am trying to do is the double bonded oxygen. Ibuprofen is pictured below:

I counted the number of electron domains(steric number) to be 3. I counted the double bond as one and each lone pair as one. It has two lone pairs and one double bond. So this means the electron geometry should be trigonal planar.
But I am stuck on molecular geometry. There is no molecular geometry I have found in any chart that has two lone pairs and one bond. Am I making a mistake somewhere?

I assume you are trying to apply the VSEPR model. In that case you are correct in assuming two lone pairs, a sigma and a pi bond for oxygen, giving a trigonal arrangement in the surrounding of oxygen. Be aware, that this model is very crude and the bonding situation is much closer to  $\sigma-\ce{C_{sp^2}-O_{sp}}$, $\pi-\ce{C_{p}-O_{p}}$, $\sigma-\mathrm{LP}-\ce{O_{sp}}$ and $\pi-\mathrm{LP_{inplane}}-\ce{O_{p}}$.

You are right! There is no molecular geometry in any chart that has two lone pairs and one bond, because the only possible geometry with one neighboring atom is linear.
The central atom in a chemical moiety must have at least two neighboring atoms in order to predict the geometry around this central atom (according to VSEPR model).
Let's take the case of oxygen atom of the hydroxyl group (in Ibuprofen). The oxygen atom here, as a central atom,  has the steric number: 4 ( It has two lone pairs and two simple bonds with carbon and hydrogen atoms). So, The molecular geometry around oxygen atom is bent.


text:
Is it possible to make metal fulminates?
Is it possible to make fulminates such as: magnesium fulminate, copper fulminate, etc.
by the metal + $\ce{HNO3}$ + $\ce{EtOH}$ reaction?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour]. At least copper fulminate seems to be straight forward.

the wikipedia article on potassium fulminate mentions preparation by reacktion of potassium amalgam with mercury fulminate, which is slightly soluble. I'm pretty sure, other metals that can reduce mercury should work as well. However, it MUST be noted, that synthesis with slightly soluble explosive salt may take months, as stirring is obviously not an option.

In The Preparatory Manual of Explosives  (paperback at amazon), the author, Jared Ledgard, provides a procedure for the preparation of copper fulminate that can be summarized as follows:

To 100 mL of 70% $\ce{HNO3}$, 30 g of anhydrous $\ce{Cu(NO3)2}$ is added under stirring. If the copper nitrate doesn't dissolve properly, add water dropwise until it does. (NOTE: Adding water to concentrated acids usually isn't the best idea!) The solution is stirred for 24 hrs at room temperature and then heated to roughly 80 °C. Under stirring, 120 g of 95% ethanol are added. After addition of the ethanol, the heat source is removed and the mixture is alllowed to reach room temperature. The precipitate is rapidly filtered off, washed with several hundred mL od cold water and dried in a desiccator over magnesium sulfate. The product, copper fulminate, should be stored in a desiccator over sodium sulfate in a refrigerator.

NOTE

To my opinion, the procedure leaves out significant information, such as the exact temperature of the aqueous solution while adding the ethanol. It is not clear whether and how much ethanol is supposed to evaporate. While the author suggests to carry out the reaction in a beaker, a setup with a two-neck flask, reflux condenser and a dropping funnel might be a better idea.
Handling fairly concentrated nitric acid requires safety precautions: A fume hood, eye protection, gloves and a lab coat are mandatory!
The reaction product is a primary explosive! While less shock sensitive than mercury fulminate, it still will go off on impact (or upon heating above ~200 °C).

In the case of these or similar sensitive materials, there is a good rule of thumb: If you have to ask on a Q&A site - Don't touch it!

The best part of it is I think the insane amounts they use, absolutely "blown" out of proportion. (+1) for the safety advice.

@Martin the manual is NOT for academic science, it is for applied chemistry, when you need a lot of staff. Basically, it offer mostly upscaled procedures, usable to produce a large loads of staff. For example, a kilo or so of priming mixture for capsules, a good firework and so on. Some procedures mentioned in the book operates literally kilos of reagents (and litres of solvents)

@Martin And yes, it does contain a procedure involving close to kilo of mercury fulminate as a raw material.


text:
Why does symmetry have to be maintained in molecular orbitals?
Using the example of $\ce{XeF4}$:

What is the physical explanation enforcing the symmetry of the $\ce{1b_{1g}}$ orbital on the fluorine atoms? Why isn't the symmetry of a nonbonding orbital arbitrary?  If it's going to be nonbonding anyways, why can't we, for example, have a Fluorine p-orbital arrangement facing towards Xenon with three positive p-orbitals and one negative p-orbital?
To elaborate:
If I imagine a free Xenon atom in space, and the approach of four individual fluorine atoms, I would expect the bond formation to be randomized with respect to the orientation of the fluorine p-orbitals, and therefore for some arrangements to not be perfectly symmetrical, such as 3 positive p-orbitals, 1 negative, facing inward.  I understand bonds can't be made without symmetry between the Xenon and Fluorine orbitals; that makes physical sense because we can argue it by looking at orbital overlap that dictates bonds can only occur with appropriate symmetry.  But in a nonbonding case, such as $\ce{1b_{1g}}$ above, I don't understand why symmetry is also required.

Source for pictures:

https://scilearn.sydney.edu.au/fychemistry/calculators/make_mo.shtml?type=year1&theMolecule=xef4
http://www.chem.mun.ca/homes/cmkhome/SALCS&MOs.pdf

Source for pictures:

https://scilearn.sydney.edu.au/fychemistry/calculators/make_mo.shtml?type=year1&theMolecule=xef4

http://www.chem.mun.ca/homes/cmkhome/SALCS&MOs.pdf

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

The orbital and geometrical symmetry are closely related. You know that $\ce{XeF_4}$ is square planar, therefore $\ce{D_{4h}}$ symmetric. That also means that the four fluorine atoms are indistinguishable.
So if you perform some manipulation of the molecule, e.g. rotation by $\ce{90^{\circ}}$, you must end up with the same picture (or just with opposite sign), therefore ruling out our suggestion of 3:1 different orbitals. This holds for all orbitals, not only bonding.
One could possibly argue, what is the physical meaning of the unoccupied orbitals, but this is out of scope of the question.

Hi, thanks for the prompt answer. To clarify:
"So if you perform some manipulation of the molecule, e.g. rotation by 90∘, you must end up with the same picture"
Why must you end up with the same picture? I understand there's a pattern of symmetry, but I don't understand any physical explanation why.  If I were a Fluorine orbital, why do I care how the Fluorine orbital opposite me looks, in particular if neither of us are involved in bonding?

You as the fluorine play hide-and-seek with poor observer and you should be indistinguishable from other 3 fluorines, as symmetry dictates.
Would the molecule be non-symmetrical, this request is lifted and you can do whatever you wish. But you cannot.

It's not so much that these are arbitrary arrangements or that symmetry is "conserved."
We take the 4 fluorine atoms as a set and consider the reproducible representation of all fluorine 2p orbitals.
Under the $D_{4h}$ point group of $\ce{XeF4}$, the 4 fluorine orbitals consist of an $a_{1g}$, $b_{1g}$ and $e_u$ representations. (There are other representations of the out-of-plane 2p orbitals too.) Four atomic orbitals, 2x1D 1x2D representation.
Now we try combining the representations of the fluorine orbitals.
It's a matter of math... the reason $b_1g$ is non-bonding is because there is no Xe atomic orbital with the same symmetry. If you consider the direct product of the fluorine $b_{1g}$ with any Xe orbital, you get zero - they are orthogonal and have no overlap.
Now, you ask why can't you have 3:1.. well, you have 4 identical fluorines. If you had $\ce{XeF3Cl}$ or something like this, you'd then have different symmetry and different orbitals.

Hi, thanks for taking the time to respond. I actually understand why the orbital is nonbonding. I guess my sticking point is that all the explanations so far approach from the angle of symmetry. Of course if we define a problem as symmetry, and then explain it using symmetry, it seems self-consistent. But nature itself isn't aware of a concept like symmetry. Nature is composed of fundamental forces dictating its behavior. In non-bonding, my confusion is what fundamental property of nature requires certain symmetries? In bonding, it's orbital overlap so that probabilities add, but nonbonding...

but non-bonding orbitals don't require orbital overlap to exist.  Why is a non-bonding orbital forced then to be in a non-arbitrary arrangement?

There's nothing that "requires" symmetry. You asked a question about $\ce{XeF4}$ which has a symmetric property. Because the molecule is symmetric, the orbitals will be symmetric. There's nothing arbitrary about it - it's the base property of the molecule itself. If I look out my window, I see snowflakes right now. The ice lattice dictates the symmetry of the crystal, much like the symmetry of the molecule (i.e., four equivalent fluorines) dictate the symmetry of the orbitals.

A 3:1 mixture will be possible under a different less-symmetric molecule. Under $D_{4h}$ such a mixture is a reducible representation, not an irreducible one.

If you haven't studied it, perhaps you should read about using projection operators to create the MOs.

Symmetry reduces the level of energy of any thing. Most of the things of nature has inbuilt symmetry with them. It is the property of nature to have different kinds of symmetry. So that it reduces the energy of the orbital or molecule


text:
How many delocalised electrons in gold?
How can we determine how many delocalised electrons every atom of gold contributes to the 'sea of delocalised electrons'? More generally, how can we determine the number of electrons any metal contributes? Must we have thermodynamic data to determine this?

According to the book Solid State Physics (Ashcroft, Mermin), the Drude model (electron sea model) as originally proposed gives no theoretical basis for finding how many electrons per atom in a metal are delocalized.

Briefly: valence orbitals on adjacent metal atoms interfere to form "bands" of closely spaced molecular orbitals that spread throughout the metal.
If you have N metal atoms with occupied s orbitals, they interfere to form an "s band" with N molecular orbitals. The lowest energy orbital is fully bonding; the orbital on the top of the band is fully antibonding. The spread of the band will be finite, and since N is a huge number, the molecular orbital energies will be very closely spaced, practically continuous.
If you have two valence s electrons in your metal, the s band will be full (each orbital can hold up to 2 electrons, you have 2N electrons, and N orbitals in the band). If you have only one valence s electron, the s band will be half full (the first N/2 orbitals will be filled, and the remaining orbitals will be empty).
Above the s band you'll have a similar p band. In gold, the p band is unoccupied, while the s band is half full (gold has 6s$^1$). However, IF the bottom of the p band overlaps with the top of the s band, the s electrons can easily move into the bonding MO's at the bottom of the p band. They can also move into any vacant orbitals in the s band that are just above the highest occupied orbital.  In either case, the s electron will be delocalized.
To figure out whether the valence electrons will delocalize, then, you'll have to decide what bands are present, where the highest occupied molecular orbital is, and what the width and the spacing of the bands are (does the unoccupied band overlap with the occupied band, or is there a gap between the two?)

Valance electron of s and p orbital take part in delocalisation over the positively charged gold atoms


text:
Dependence of steric hinderance on enol content
The following is extract from my book-

To illustrate this, let us consider the enol contents of acetyacetone and $\alpha$ - methylacetylacetone. Although both the enols are stabilized by H bonding the enol form of $\alpha$ - methylacetylacetone is destabilized to some extend by steric repulsion due to the presence of $\alpha$ - methyl group.

What kind of H bonding (sorry I know i am being stupid, the figure in my book showed a bond connecting H and oxygen and dotted line connecting another oxygen and the previous hydrogen. How can hydrogen form two bonds?) is being talked about and what is steric repulsion? How does the steric repulsion destabilize the enol form?

How can hydrogen form two bonds?

In the case of an enol, hydrogen does not form two full bonds.  What the drawing below is trying to show with the enol hydrogen is that one bond (the O-H with the solid line) is elongated, partially broken, while the other bond (the O..H with the dotted bond) is also elongated and only partially formed.  It's as if the partially broken OH bond and the partially made OH bond are each "half a bond", so all together the electron sharing is such that there is still just one full bond to hydrogen

How does the steric repulsion destabilize the enol form?

Look at the bottom part of the drawing where we have inserted an alpha-methyl group into the 1,3-diketone.  Notice the steric interactions indicated by the arrows.  The situation looks similar to that seen with cis-2-butene.  The methyl groups are large enough (larger than a hydrogen) and close enough to one another that their electron clouds "bump" into one another causing what we often call "steric destabilization."  Remember that trans-2-butene is more stable than cis-2-butene because the trans isomer does not have the adverse steric interaction present in the cis isomer.  It is that same type of steric interaction that occurs here with the enol when an alpha-methyl group is present, and just like in cis-2-butene, it destabilizes the enol.  Therefore less of the enol will be present at equilibrium when an alpha methyl group is added to the enol.

You said that there is one complete bond between OH and another partial bond between the other OH (dotted one). Hydrogen gas has one election and it get used up in forming the one complete bond. From where does the partial bond get electron?

No, I said one bond is elongated and partially broken and the other bond is also elongated and partially formed.  As the bond being formed strengthens and the bond length decreases, the other bond elongates and weakens.  There is not more than one full bond to hydrogen at any point in the process.  The transition state would look like it has two half-bonds between the hydrogen and the two oxygens.  The hydrogen is always contributing 1 electron to the bonds and each oxygen contributes between 0 and 1 electrons depending upon how strong\weak the hydrogen bond is to each oxygen.


text:
Dynamic equilibrium - effect of adding inert gas
Well consider the reaction $A + B ⇌ C$
Adding inert gas to a container where this reaction is taking place will increase the pressure of the system. In accordance with Le Chatelier's Principle, the equilibrium should shift in the direction which opposes this change in pressure, i.e. in the direction where fewer number of molecules are created. In this specific reaction, it happens to be in the forward reaction. However, I have been taught that adding an inert gas at constant volume has no effect on the equilibrium. Why is this?

In the case of, say, the equilibrium partial pressure of water vapor over water, the pressure of inert gas over the water is ignored.  This is because water is (pretty much) incompressible, so the inert gas pressure has no effect and Le Chatelier's principle results in no shift in equilibrium for any reasonable lab pressures. In many intro courses simple absolutes are presented - in the context of the introductory material they are fine, but the principles break down eventually, which is then the basis of the next courses...

@JonCuster Well that may  be applicable for the certain example that you have provided, but definitely not for all reactions. There are many equilibrium reactions which contain all reactants and products in gas phase.

At constant temperature, only changes to the concentration (technically activity) of a reactant or product, shift the equilibrium.
Adding inert gas does not change the concentration of any reactant or product, so it does not affect equilibrium.

Yes but changing the pressure doesn't have any affect on the concentration either. However, it does affect the equilibrium.

Changing pressure by decreasing volume changes concentration and affects equilibrium.  Adding inert gas at constant volume increases pressure but does not change concentration and does not affect equilibrium.

Is there no other way to change the pressure? So basically it is that when you are changing the pressure, you are changing the volume.

pressure can change by changing volume, temperature, or amount of one or more substances in the system.

Okay. So let's take that we increase the pressure by increasing the temperature. Consider the forward reaction to be exothermic, while the products have less molecules than reactants. In which direction will equilibrium be shifted?

For an exothermic reactions, increased temperature favors the reverse reaction.  For endothermic reactions increased temperature favors the forward reaction.  This is because the equibrium constant changes in accordance with the Van't Hoff equation http://en.wikipedia.org/wiki/Van_'t_Hoff_equation

Yes but wouldn't the change in pressure favor the forward reaction? As we are making less molecules and thus reducing the pressure?

No. Since volume does not change, concentrations do not change.  The equilibrium constant tells you that for a given set of [A] and [B], there is a unique solution for [C].  Only if the equilibrium constant itself changes, or a concentration changes, can the equilibrium shift.

The gas added works in both directions and the fraction of pressure is the same becaussed the total pressure increses but the inert gas do not affect the features of the others and exerts the same effect. If the number of molecules do not change, the concentration remains constant, and the principle of Le chatelier sais that in case of increase the pressure the reaction take the direction where there are less molecules because the pressure force them to agregate.


text:
Why can syn-periplanar E2 elimination happen in molecules like vinyl bromides but seemingly not in other molecules?
E2 eliminations occur when the leaving group and the proton are in the anti-periplanar conformation not syn-periplanar. This is beyond a merely a preference as leads to E2 reactions being stereoselective. However, when it comes to vinyl bromides for example (involving a double bond) the reaction can occur in the Z- and E- isomers (i.e both syn-periplanar and anti-periplanar). How can this be the case?

Elimination from vinyl halides occurs faster when the halide and proton have a trans relationship.  When the halide and proton are located on the same side of the double bond (cis), the elimination to form an alkyne is much slower.  The faster elimination from the trans arrangement is consistent with a preferred anti-periplanar arrangement in the E2 reaction.
The fact that elimination still occurs, albeit at a slower rate, when the groups being eliminated are arranged cis to one another suggests that a different mechanism is involved in the "cis" case.  Two possible mechanistic alternatives are

cis elimination from a higher energy (higher energy, therefore less preferred, slower) syn-periplanar conformation, or
an E1CB mechanism where the proton is removed in a first step and a carbanion is generated, and this is followed by halide ejection in a second step.

Experiments in analogous molecules suggest that the E1CB pathway is most often followed in "cis" elimination from vinyl halides.


text:
Synthesis of 1-iodo-4-nitrobenzene
I was trying to synthesize 1-iodo-4-nitrobenzene in the lab today – however, the appearance of my product really confused me.
At first, I added conc. sulfuric acid to water and 4-nitroaniline.
This was followed by diazotising the amine sulfate with sodium nitrite and water. (This was done at temperatures of below $10\ \mathrm{^\circ C}$).
Then, I added the diazonium salt to a solution of $\ce{NaI}$ and water. A large plume/foam formed ($\ce{N2}$). After, I performed a filtration using a Büchner funnel. That’s where things went wrong.
I used water to wet the filter paper. I added in the iodonitrobenzene. A suckback occurred within a few seconds as the product was quite viscous. After resumption, the filtration went as planned. However, the product on the filter paper was a molten brown mush.
I tried to let it dry for a further 5 minutes but once turned off it remained a sludge.
Do you have any suggestions on how I could improve on this synthesis or maybe an idea of what went wrong?

How quickly\slowly did you add the diazonium salt to the NaI solution?  2) Did you maintain the reaction temperature below 10C during the addition?  3) Did you keep the diazonium salt cooled while adding it?  4) After addition was complete, did you let the reaction stir at 10C until nitrogen evolution ceased and then let the reaction slowly warm to room temp?  Here is a tested lab procedure, see where you differed.

Yes, I think a big issue was the fact that I didn't give enough time for the Nitrogen evolution to cease. You mention that I should have kept it at 10C even when the NaI was added. According to the manual I was following, this wasn't necessary.?

However, I didn't return it to the ice bath whenever I added the NaI. So, therefore, it didn't solidify. So, what was my brown sludge? Was it iodonitrobenzene, but in the wrong form?

Keeping the reaction cold during the addition is key.  If it warms up too much, the diazonium can react with itself to give colored by-products.

Here are some things to focus on:

How quickly\slowly did you add the diazonium salt to the NaI solution?
Did you maintain the reaction temperature below 10°C during the entire addition?
Did you also keep the diazonium salt cooled while adding it?
After addition was complete, did you let the reaction stir at 10°C until nitrogen evolution ceased and then let the reaction slowly warm to room temp?

Here is a tested lab procedure for the preparation of 4-iodonitrobenzene.  Compare it to your notes and see where you differed.
Keeping the reaction cold during the addition is key. If the diazonium salt warms up too much, the diazonium can react with itself (coupling) to give colored by-products.  The diazonium ion can also eliminate nitrogen to generate the extremely reactive phenyl cation which will react with any nucleophile around, water for example (see page 6, the Mechanism section in this reference).  Both pathways can  contaminate and reduce the yield of the desired product.

So the benzene ring of the diazonium could react with a Nitrogen triple bond? How could it ''react with itself''?   Would the Nitrogen dioxide act as a nucleophile on the nitrogen triple bond?

Yes, the terminal nitrogen in the diazonium can react with another activated aromatic compound (starting material, final product, by-product) to produce coupling products.  Your ring has a nitro group, so it is deactivated and less likely to couple.  More likely, your diazonium salt thermally decomposes (eliminates nitrogen) to generate the extremely reactive phenyl cation which will react with any nucleophile around, water for example.

@Maximilion  Feel free to ask follow-up questions. BTW, if the answer was helpful, please mark it as accepted, thanks!

Okay, so say if I decided to make 4-bromo or 4-nitrophenol or 4-nitrophenylhydrazine, would it be possible to use the Na- equivalent to NaI?

No, if you wanted to make 4-bromo-nitrobenzene from p-nitroaniline, you'd use CuBr (Sandmeyer reaction).  Iodide is reactive enough to react directly with the diazonium ion, bromine and chlorine need a little help from copper.  Water is also reactive enough that generating the diazonium ion in the presence of excess water is a standard route to phenols.

Thanks ron . How is it that the arenediazonium salt will lose its N2 group in the presence of I- , even when cold? An elimination reaction? But if it's an elimination reaction, then why doesn't it then react with water when cold?

N2 is a very good leaving group so it is easily eliminated.  Still, for the nitrogen elimination to occur the diazonium must be warmed up.  If you keep the diazonium below 10C appreciable reaction won't occur.  Above this temperature the diazonium will eliminate nitrogen and react with iodide, hydroxide or whatever nucleophile is present.

Yes, but why I- in favour over water?

When you're trying to make an iodo compound you have a lot more iodide present than water when you warm up the diazonium.

Okay, in the synthesis of 1,3,5 tribromobenzene, how could I make the initial aniline reactant?

Are you asking about what conditions are required to react aniline with bromine in order to get 2,4,6-tribromoaniline?

It's okay, thank you. Would H2SO4 and Na2SO3 suffice in the formation of 4-nitrophenylhydrazine?

@Maximilion  Yes, that method was used for the synthesis of 4-nitrophenylhydrazine

Wash the product (sludge) with 95% ethanol when filtering. The sludge is due to impurities.Your pure product should be light brown.


text:
What is the difference between IPA and Ethanol in terms of solubility?
I have read an article online that says that ethanol could be used in a solvent extraction of two solids, but it also says that isopropyl alcohol could also be used, but must be used in higher concentrations. Why?
From some preliminary internet searches, I have discovered that:
"Isopropyl alcohol is a slightly bigger molecule than ethanol (Methanol is $\ce{CH4O}$, Ethanol is $\ce{C2H6O}$ and Isopropyl alcohol is $\ce{C3H8O}$) and is often cheaper to make. It is not as poisonous as methanol (an ounce of methanol can kill you and a third of an ounce causes blindness). So it is useful as a solvent in the same way acetone is useful as a solvent."
So would this suggest that if one were to use IPA as a solvent, one could use the same amount of IPA as they would Acetone?
One final question. It would be best to use 99% pure solvents (mostly free of water), however for an ethanol replacement, I have seen a product used for ethanol fireplaces where the MSDS mentions 95% ethanol and 5% 2-propanol. Would this be an acceptable substitute for 99% ethanol? Thanks a lot.

This question is too broad to be answered well.

[...] ethanol could be used in a solvent extraction of two solids, but it also says that isopropyl alcohol could also be used, but must be used in higher concentrations. Why? [...]

They hopefully didn't write about the "concentration of isopropanol" but mentioned that a larger amount of the solvent has to be used.
It is not uncommon that organic compounds with polar substituents are less soluble in higher alkanols.
Vanillin is a typical example, the solubilities are in the range of 4.15 mol/L in methanol, 2.5 mol/L in ethanol and 1.8 mol/L in 1-propanol. (Data are taken from a freely accessible collection of experiments performed by students: OpenNotebook Science Challenge). I couldn't find any data for 2-propanol.

So would this suggest that if one were to use IPA as a solvent, one could use the same amount of IPA as they would Acetone?

No, not really. A ketone and a secondary alkanol, although looking similar on first sight, may have behave completely different as solvents. Polystyrene (foam), as an example is pretty much insoluble in isopropanol but highly soluble in acetone!


text:
Alcohol spectrophotometry
A couple of weeks ago we did a lab where we had a bunch of standard solutions of alcohol, dichromate, and sulfuric acid. We also had one unknown concentration.
We heated them so they would all react, and the initial orange dichromate turned slightly green (the ion $\ce{Cr^{3+}}$ formed). Then we placed each reacted soln in a calorimeter which can emit light of certain frequencies. I believe we chose 635, which should have emitted orange light through the solution. the absorbance/transmittence were calculated for each standard soln until we had a line of best fit. Then the absorbance of the unknown concentration was given, so we used to line to match it to it's concentration (of alcohol).
I'm confused as to why we had to use the color orange in our emission and why a reaction was even needed in the first place.

Since green is the absorption of orange light, you need to supply the orange light to measure how much is absorbed. As the alcohols are oxidized to the acids, the dichromate is reduced, so you can measure how much acid has been formed. BTW, what alcohol was used? Butyl alcohol oxidation has a distinct aroma that can be detected without spectroscopy.

authors: Архип Архипов


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Why does one bicyclic diastereomer react faster than the other one?
I have been asked to explain the following observation:

I have a number of issues with this:
1) The fast reaction must be a syn-elimination. But how can syn E2 elimination occur? If an E2 elimination is stereospecific because it must go through an anti-periplanar transition state surely syn-peri-planar elimination cannot happen.
2) How does the slow reaction even happen? The H and the Cl are never going to be in the same plane (syn or anti).
As a side question: does the boat conformation (which the bicyclic molecule must adopt) have equatorial and axial sites like the chair conformation does?

You might find this link helpful.  Do you know how to draw a Newman projection? If so, draw the Newman projections, sighting down the C2-C3 bond (C2 and C3 are the two chlorine containing carbon atoms), for both of your dichloro compounds.  This should make it easy to understand the arguments presented in the first link above.  Hope this helps.

What mechanism does the slow reaction undergo?

One could run additional experiments to find out, but without additional data I'm not sure, either E1 or E1CB.

Yes, the fast reaction is a syn-periplanar E2 reaction. If we are ranking the rates of E2 based on the orientation of the C-H and C-LG bonds, anti-periplanar would be the fastest, syn-periplanar would be slower, and perpendicular would be the slowest. The anti-periplanar arrangement is the fastest because it provides the best overlap of the (filled) bonding C-H orbital with the (empty) anti-bonding C-LG orbital. The syn-periplanar arrangement is slower because there is much less productive overlap between these orbitals. However, the small lobe of the bonding C-H molecular orbital does provide some overlap with the anti-bonding C-LG orbital. A perpendicular arrangement results in no productive overlap.

The slow reaction must occur by some other mechanism, such as an E1 or E2-conjugate base mechanism.

Why would E1 be slower? Is E2 absolutely impossible, they're not completely perpendicular are they? Could the structure be somewhat distorted? Also E1cb can't work because the lone pair isn't in the same plane is it?

E1 would be slower because carbocation formation is inherently slow. I think a concerted E2 is unlikely. You're right that the C-H and C-LG bonds aren't completely perpendicular, but they are close enough that it's unlikely that E2 can operate. I think E2cb is unlikely because the base isn't strong enough. However, if that deprotonation did occur, the elimination of chloride would be fast.

But the lone pair orbital would not be in the same plane as the C-LG bond in E2cb (same problem as E2) surely? If deprotonation were to occur that is.

This is a guess, but the anion would probably be high enough in energy to overcome the non-ideal orbital alignment.

Wouldn't the empty p orbital of the carbocation in E1 also be in the incorrect or orientation? Also, surely the base is too strong for E1

Similar to the E2cb situation, the orientation is likely less important to the E1 due to the high energy of the carbocation intermediate.

As for the strong base, E2 is usually preferred with a strong base because it can avoid the high energy carbocation. In this case though, the E2 will be slowed by the suboptimal stereo electronic arrangement making E1 competitive. Once a carbocation is formed, it's going to react rapidly with any base, including t-butoxide.


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What does it mean if a functional group has "higher priority" in terms of IUPAC naming of organic compounds?
What does "higher priority" mean exactly in the following statement from this wikipedia article (article name: "E-Z notation") on naming organic compounds in IUPAC nomenclature of organic compounds?

If the two groups of higher priority are on opposite sides of the
  double bond, the bond is assigned the configuration E (from entgegen,
  German: [ɛntˈɡeːɡən], the German word for "opposite").

Whether trying to figure out if a double bond is "E" or "Z", or if a chiral center is "R" or "S", the Cahn–Ingold–Prelog priority rules (see the section "Assignment of priorities") are used to prioritize the groups.  In the case of a double bond there are 2 substituents (a lone pair of electrons can be a substituent) at each end of the double bond.  Priorities (1 and 2) are assigned to the substituents at each end using the Cahn–Ingold–Prelog rules provided in the link above.  The first rule used to set priorities is based on the atomic number of the atoms attached to the double bond.  If there is a tie (the 2 attached atoms at one end of the double bond have the same atomic number), then we move out to the next atoms and compare these.  The process is continued until the tie is broken (eventually the tie must be broken if the molecule being analyzed is really capable of being described as "E" or "Z").  After the priorities have been assigned, if the two priority-1 substituents at opposite ends of the double bond are on the same side of the double bond, then the molecule is the "Z" (zusammen, together) isomer, otherwise it is the "E" (entgegen, opposite) isomer.
Example:

image source

I just want to nitpick, that the CIP rules in Wikipedia are of course missing a key component. The rules, as provided by the key publication are: (0) [...] (1) Higher atomic number precedes lower. (2) Higher atomic mass-number precedes lower. (3) [...] Cahn, Ingold, Prelog Angew. Chem. Int. Ed., 1966, 5 (4), 385-415. This is very important and the best example why is chiral acetate.


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Are there different types of melamine resin?
Background
I have a little knowledge of chemistry.
I was talking to a sales rep from a cleaning supplies company. They sell a scrub sponge that looks and functions just like Mr. Clean Magic Eraser. The sales rep was trying to scare me by saying that magic erasers have formaldehyde.
I know that the properties of a chemical compound are different than the properties of each component. I suspected that their product is actually the same as the magic eraser so I contacted the company directly, but got a less than satisfactory answer.
I asked if their sponge was made of formaldehyde-melamine-sodium bisulfite copolymer. They did not answer yes or no. They told me their sponge is made of melamine resin.
According to Wikipedia, melamine foam (a foam-like material consisting of a formaldehyde-melamine-sodium bisulfite copolymer) is a special form of melamine resin that is used mainly as an insulating and soundproofing material and more recently as a cleaning abrasive.
I would like to be more knowledgeable next time this sales rep contacts me. Hence the question.
Questions
Is there another way to make an abrasive foam sponge out of melamine resin that is not formaldehyde-melamine-sodium bisulfite copolymer? Is there a way to make a melamine resin without using formaldehyde?

Certainly the are many different types of melamine resin.
Is there a way to make a melamine resin without using formaldehyde?
Yes, a different aldehyde could be used such as acetylaldehyde.  Glucose, benzaldehyde, acrolein and other alternatives to formaldehyde have been studied also (see page 33 of Polymer Synthesis by  Sandler et al.).
Is there another way to make an abrasive foam sponge out of melamine resin that is not formaldehyde-melamine-sodium bisulfite copolymer?
Good sources of information about formaldehyde-melamine-sodium bisulfite copolymer and possible alternatives are the patent Resilient foam based on a melamine-formaldehyde condensate and related patents.
First the introductory portion of the patent explains:

It is true that foams based on melamine-formaldehyde condensates have been described in various patent publications, but they have not hitherto been employed industrially for heat or sound insulation in the construction industry. They are produced by stirring air into an aqueous melamine resin solution which contains an emulsifier and a curing agent. Such as described, for example, in German Pat. No. 870,027, have the serious disadvantage that they are very hard and brittle, and easily break or crumble when handled. It is alleged in French Pat. No. 1,073,642 that melamine resin foams can be produced by heating a resin powder in a mold under reduced pressure. This process, however, does not give useful resilient foams. U.S. Pat. No. 3,093,600 describes melamine resin foams which are said to have improved resilience and resistance to crazing as a result of the incorporation of triols, e.g. trimethylolpropane. However, it has been found that the resilience, and especially the recovery after compression, of such foams is inadequate for many applications.

Then in describing what is considered new about the invention, many different alternatives are included. Such as adding aldehydes other than formaldehyde, urea, and many other substances.


text:
Alcoholic proof and temperature
The US defines the "proof" of an alcoholic beverage to be twice the alcohol per volume percentage at 60° F (about 15.5° C)
As someone who does not have a chemistry background, I am wondering what the 60°F requirement means.
Given a mixture of alcohol and water that is 80 proof (40% alcohol by volume at 15.5°C), what is the alcohol by volume at room temperature 20°C?

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Briefly, since ethyl alcohol expands more rapidly with increasing temperature than does water, then measuring the mix by volume at a warmer temperature would give you less alcohol and more water, because the alcohol would be less by weight for the same volume.
Thermal Coefficient of Expansion (approximate, measured at room temperature)
Alcohol, ethyl (ethanol):   0.00109 / K
Water:                            0.00021 / K
Using these constants, you can figure out the answer to your question. But note the wording: given a mixture. If the EtOH/H2O is already mixed, changing the temperature does not change the concentration or proof, and you would have to separate the ingredients to measure their respective volumes. BTW the volume of the mix is slightly less than the sum of the volumes of its ingredients.
Moral: if your measuring your drinks by shot-glass rather than with a scale, you'll get a weaker drink in a warm bar ;-)

In all normal cases (tax purposes, food quality, etc...), it is the density which is measured, using http://en.wikipedia.org/wiki/Hydrometer#Alcoholmeter
This measurement is very temperature sensitive and therefore needs a reference point.

Good point... and for convenience, a refractometer may be used, though it also is temperature dependent. See http://web2.airmail.net/sgross/fermcalc/fermcalc_alcohol.html for correction calculation.


text:
How does an atom get rid of electrons?
I know that if an atom has 4 electrons in its outer shell it will share 4 with other atoms to have a full valence shell. But what if the atom only has 3? I understand that it will likely get rid of three instead of picking up 5 more to share, but how does it get rid of the three?
Thank you

You might wanna study some things about bonds.

Atoms with one outer electron (e.g. sodium), called alkali metals, or two outer electrons (e.g. calcium, strontium), called alkaline earths, and even the aluminum-scandium metals are not good at sharing -- they tend to lose their outer electrons given the opportunity, e.g. when a halogen is near. These form ionic compounds, such as table salt, in which case the Na+ has given up one electron, which stays close to the Cl- ion.
Elements in columns closer together don't get away with outright theft, and share their outer electrons more equitably in a covalent bond.
That said, it's an oversimplification, in that a bond is not necessarily fully ionic or fully covalent and may fall somewhere in between. See https://www.youtube.com/watch?v=OY2_q-1t2Q8 for a video on this.


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Vacuum good inside a box?
I have no clue whatsoever about Chemistry on a academic point of view.
We wish to transport some goods.
Let's say we vacuum some of those goods, by doing like so:
We place those goods on a box, we close the box, and then we vacuum seal the box (not the goods).
Can we consider the goods to be vacuum protected?
Does that mean, that no hight heat will affect those goods?
Thank you in advance, and sorry for not using specific chemical terminology.

Please attach simple schematics of your idea and add few more details on goods, heat and how would you vacuum seal.

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I have no clue on what schematics does you expect sorry. The thing is, we wish to ship something, for example, fresh herbs. Those herbs should not be exposed to high temperatures during the shipping process. By high I don't mean 300ºC or something. I presume it will never be hooter then 50ºC. What I'm trying to understand is: If we place the fresh herbs on a box, let's say, a regular plastic box, we close the box, and then we vacuum seal the box with some material, using a vacuum sealer bag, will the contents inside that box, be considered sealed?

Klaus's answer exactly shows my point. Is the box itself airtight? The herbs are freely lying on the bottom of the box? Would the herbs profit from being kept in vacuum, compared to air? (U would guess they will). Or is the thermal insulation the main point? Than see e.g. http://www.spacaky.cz/thermopack_eng.htm

Let's imagine that your transport box has a decently closing lid. It you place the herbs in the box, close the lid , stuff the box into a vacuum sealer bag, remove the air from the bag until it is shrunk around the box and finally seal it, you may consider it sealed.
This means:

According to "household standards", it is air-tight to the outside. No air from the environment will enter the bag, unless it is damaged.
There is however still air in the box. Whatever degradation, loss of flavour, etc. may happen to the herbs during transport still can, since there is oxygen.

However, if you vaccum-seal the box, or replace the air in the box with nitrogen, you will minimize (or fully prevent) degradation of the herbs.

I know this is an old post but I hope I can still help.  It would be a very bad idea to vacuum pack fresh herbs for shipping.  From a chemical stand point, the living herb are respiring and so releasing CO2.  Cold shipment will be needed to lower the rate of respiration, check the Q10 for the herbs you are shipping, or a not air-tight container.  Look at the packaging in a grocery store, they are not air-tight.
From a microbiological view, you want to place fresh herbs (may have some soil still on them), that most likely have not been irradiated (best way to kill microbes on food), into an anaerobic environment.  Research Clostridium botulinum and you will see why this is a bad idea.

After vacuum cleaning of goods which you are mentioning.. there is still a chance of entry of gases during the packing time. If you are good enough to manage the goods vacuum protected until packing then your goods will be vacuum protected if you vacuum seal the box perfectly..


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Order of molecular mass of solvent; given boiling point

Given three solvents X, Y, Z. Order of boiling point of pure solvent X < Y < Z and same order of molal elevation constant ($K_\mathrm{b}$). Find the order of molecular mass.

So I got this question in my tuition test and the solution is unclear. This question is troubling me a lot.

I can see how this question is troubling.
It don't think there is a solution.
$K_b = \frac{RT_b^2M_w}{1000\Delta H_v}$
Where $T_b$ is boiling point, $\Delta H_v$ is the molar heat of vaporization, and $M_w$ is molecular weight of the solvent.
(see Colligative Properties by W. R. Salzman)
So you can see that knowing $K_b$ and $T_b$ is insufficient to find $M_w$, knowledge of $\Delta H_v$ is also required.
Experimentally, from Table of cryoscopic and ebullioscopic constants you can also see there is no solution.


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How are pencil marks "adhered" to the paper?

Pencils create marks by physical abrasion, leaving behind a trail of solid core material that adheres to a sheet of paper or other surface. Wikipedia

Normal paper is mainly composed of cellulose fibers, which were "funded" by the plants. And this "solid core material" is supposedly composed of graphite and a type of binder clay.

Image source
"Adhering" is usually caused by intermolecular forces. (e.g.: when sugar adheres to wet surfaces because of Hydrogen bonding). The questions are:

Is the case of a pencil mark some kind of intermolecular force too?
If so, what is it?

The graphite itself can adhere to paper fibers with London forces; it's not a strong adhesion because the marks can be erased easily. Some of the graphite is just mechanically stuck between the paper fibers; rubbing a mark from a soft pencil can easily blur it. Enough graphite is deposited that a line drawn by a pencil can conduct electricity.
The clay gives the lead structural integrity; an "H" pencil has a harder lead, and deposits less graphite; a "B" pencil is softer, and an "F" pencil can be sharpened to a finer point.
In my time as an art major, I made a lot of sketches using vine charcoal on newsprint. Vine charcoal has no binding clay, and the marks were easily smeared. On paper without much "bite", it was easy to lift the marks off completely with a piece of kneaded rubber, and if you wanted to keep a charcoal drawing (or a soft pencil drawing) you really had to spray it with a fixative.
So I'd say that it's possible that the binding clay also plays some role in binding the graphite or charcoal to the paper.

In the manufacture of graphite pencils, air-classified grades of Volclay sodium
  bentonite and micronized hectorite are regularly used to bind the graphite
  compound in pencil lead. --- American Colloid Company, Industrial and Household Applications

Bentonite and hectorite both have ionic surfaces that probably help bind it to OH groups on the surface of cellulose fibers.

So the clay (or any other additives) doesn't have any effects on strengthening the van der waals forces?

Without knowing anything about the structure of the clay it's hard to say. The fact that it's harder to erase marks made with a high-clay "H" pencil than with a low-clay "B" pencil suggests that it's possible that it affects the way the graphite binds to the paper. The clay probably has charged sites on its surface that can play some role in binding it to the paper, but I'm speculating.


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alpha-hydrogen atom and keto-enol tautomerism
The following is an extract from my book-

For an aldehyde or a ketone to exhibit keto-enol tautomerism, it is essential that it must have atleast one $\alpha$-hydrogen atom. Thus, acetophenone, butan-2-one and propionaldehyde all contain $\alpha$- hydrogen atoms and hence show keto-enol tautomerism. In contrast, benzaldehyde, benzophenone etc. do not contain $\alpha$-hydrogen atoms.

What is meant by $\alpha$-hydrogen atom? How does it affect keto-enol tautomerism?

In general, the "alpha" hydrogen is a hydrogen attached to a carbon that is "alpha" (adjacent to) the substituent.

image source
So in this case, an "alpha" hydrogen is a hydrogen attached to the carbon alpha to the carbonyl.

image source
The following drawing represents a typical keto-enol equilibrium.  If we removed the last alpha hydrogen and replaced it with a substituent (alkyl group, phenyl group, bromine, etc.) then we could not form a double bond with that alpha carbon, it would have 4 non-removable substituents.  If we can't form a double bond that means we can't form an enol.

image source

Why is the secondary alpha carbon more acidic than the primary alpha carbon? I thought primary carbanions were MORE stable than secondary/tertiary carbanions...

An $\alpha$ hydrogen atom is bound to a carbon next to some functional group, e.g. a ketone or an aldehyde. Without this hydrogen there is no hydrogen that can be removed by enolization. Other hydrogen atoms on the carbon chain are much less acidic due to the lack of an electron stabilizing group such as a keto-group. Benzaldehyde contain no $\alpha$ hydrogen atoms because of the $\ce{sp^2}$ hybridized carbon atom. On the other hand, acetone for example has 6 $\alpha$ hydrogen atoms.


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Why can't alcohols turn blue litmus red?
I came across the fact that alcohols cannot turn blue litmus red. The fact confuses me as they are acidic and acids do turn blue litmus solution red. What prevents alcohols from turning blue litmus solution red?

I'm not sure but I think because of two reasons: 1- Aqua in aqueous solution (the solution I assume you mean) is amphoteric. 2- Alcohols aren't very strong acids.

Litmus is a weak acid, with a $\mathrm{p}K_\mathrm{a}$ of about 6.5.
Alcohols are extremely weak acids, with $\mathrm{p}K_\mathrm{a}$'s (typically) around 17.
The litmus itself is a much stronger acid than the alcohol. Dissociation of the alcohol won't produce enough protons to shift the equilibrium between the two colored forms of the indicator, so you'll see no color change.

Alcohol in water solution is not acidic enough to change the solution pH. So, it doesn't turn blue litmus red.

The pH of ethanol ($\ce{C2H5OH}$) or pure alcohol is $7.33$. This means that ethanol is slightly basic. And the colour of litmus paper change at $\mathrm{pH} =7$.
Consider the ionization of indicator  in solution represented by this equation:
$$\ce{HIn  <-->  H+    +   In-}$$
And the acidic  colour  of  $\ce{HIn}$ is red, the basic colour of $\ce{In-}$ is blue.
According to Le Chatelier’s principle, if we put indicator in basic solution, $\ce{H+}$ reacts with base, more $\ce{HIn}$ ionized, more $\ce{In-}$ formed, so the colour of blue litmus unchanged.


text:
What kind of fire extinguisher for chemistry at home?
The are various types of fire extinguishers. They can be based on

water
foam
dry powder
carbon dioxide
wet chemicals
halon
etc.

Suppose one is learning chemistry at home. What type of fire extinguisher one should ideally get? It seems the dry powder option is the cheapest ($\approx 30€$), foam mid-priced ($\approx~60€$) and the one using carbon dioxide the most expensive ($\approx~150€$). I did not find solutions based on water/wet chemicals/halon yet, though I know such exist. If there's no differences in achieved safety, I would probably pick the dry powder one.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website . || When it comes to safety, I wouldn't think about expenses that much, especially when I'm the one conducting an experiment optionally. This is a nice question though.

This is highly dependent on what experiments you're going to conduct. Water is the cheapest extinguisher, but at times it's not effective, chemically harmful or even could make matters worse. I think it's better if you mention what "home experiments" you're going to try.

I'm studying the equivalent of two first courses of general chemistry (uni level). The topics vary a lot but you as chemists probably know what such 1st-year university courses typically contain.

A bucket of sand would be the most useful one for small experiments, it works on most reactions, doesn't cause much damage and is very cheap. A real fire extinguisher would still be useful for any larger fires.

You need to investigate the safety and handling of every chemical you use in your experiments individually. The handling and safety precautions can be very different depending on what substances exactly you use.
In terms of fire, the one thing you must know before each experiment is which methods won't work or even fail catastrophically. Trying to extinguish a sodium fire with water is a really, really bad idea, and that is something you must know in advance.
Metal fires are the most common kind of fire where you have to be careful about choosing the right method. Some metal fires can't even be extinguished by carbon dioxide. There are special fire extinguishers for this (class D), other methods might not work.
A bucket of sand is the most useful method for small fires in a lab. It works for almost everything and doesn't cause a big mess. But it is obviously useless if you cause a bigger fire and can't reliably cover it with sand anymore.
Powder extinguishers cause quite a mess, you'll spend a lot of time cleaning up the room afterwards. Obviously this is a lot better than burning down your home, but I wouldn't use them for small fires you could stop with some sand.
Foam and powder are probably the most useful general-purpose extinguishers. Carbon dioxide is a bit trickier to use from what I heard, I probably wouldn't recommend it in your case.
Thinking about a fire extinguisher is a good first step. Another thing I very strongly recommend is that you read up on general safety procedures in the lab. Simple things like not storing large amounts of solvent directly in your fume hood (or workplace) might prevent a small fire from becoming a dangerously large one. Knowing that you need to inactivate any reactive reagents before disposing them might save you from throwing away some oxidizing agent and accidentally setting your waste on fire. There is a lot more to chemical safety than just choosing the right fire extinguisher.

I would also suggest fire blanket. The carbon dioxide extinguisher is quite usable for small incidents. Quite often you just blow off the flame or fill some reasonably contained area (fume hood) with CO2. In terms of speed, blanket is fastest, costs close to nothing and you can have it just at the workplace. Also usable for extinguishing yourself.

Admittedly, I'm not a big fan of experimenting at home - particularly when (larger volumes of) flammable liquids are involved.
Home labs typically lack huge sinks with faucets high and huge enough to get a face under it. Neither do they have a fire shower at the exits, which is standard in German university labs. This aside and adding to the answer given by Mad Scientist:
Foam extinguishers work well even in the rare case when a person is on fire. This does however require a second person to operate it! With respect to this and other scenarios, where pain and panic prevents you from treating a mishap reasonably (acid and alkaline spills in the face, severe cuts due to breaking glas, strong burns, etc.) the golden rule for students in university labs therefore is: NEVER work alone!