text:
What are the differences between diesel and unleaded gasoline?
I am curious about the flammability of diesel fuel and its properties, as I don't deal with it on a regular basis (at all).
In particular, I want to know how the fuel works when powering a diesel engine.
Things I know about gasoline engines:

Compresses gasoline and air
Ignites compressed gasoline and air with spark from spark plug
The ignition pushes the piston back turning the crank shaft

I have heard that diesel is less flammable than gasoline and that diesel engines don't require spark plugs. How does this work?

Welcome to Chem.SE! I think this is a great question for the site, though you should probably add some more details.  For example, what precisely do you know about gasoline that you'd like to know also about diesel?  What do you know about how gasoline engines work, and how they use gasoline as fuel?

maybe something along those lines? And thank you for the welcome

I'm not clear if you're asking about diesel fuel or diesel engines, or both.  You're right that diesel engines don't have spark plugs.  They rely on spontaneous ignition of a compressed air / fuel mixture.   The compression stroke has to (probably not quite) adiabatically heat the air/fuel mixture to the point of combustion before the power stroke begins.

@CurtF. I would say both. I am not so clear on how a diesel engine produces so much energy without the use of a spark to ignite the fuel.  and I am unfamiliar with the term adiabatically I will have to look this up.

Diesel engines vs. gasoline engines
Diesel engines don't rely on spark plugs, but they still work by igniting the fuel to generate a force that moves a cylinder in the engine.  The air/fuel mixture in a diesel engine is ignited by the compression of the same cylinder during the "compression" stroke.  That is, like most gasoline engines in cars, diesel engines are four-stroke engines.
The compression stroke, in a gasoline engine or in a diesel engine, causes heating of the cylinder contents as the compression happens.  Diesel engines are designed to have a more severe compression of the cylinder contents during the compression stroke.  This results in much higher heating of the contents; in fact the temperature becomes so high that the ignition of the fuel and air becomes spontaneous.
You may ask, where does the energy to power the compression stroke come from?  It comes from inertia in the engine as it is running.  Another way to say it is that the previous power stroke provides the energy for the compression stroke.  (The power stroke is when the exploding air fuel mixture pushes the cylinder out, doing useful work.)  There is so much work released in the power stroke that there is enough to move the car forward as well as provide inertia for the next compression stroke.
Here are some more web links with helpful explanations:

Wikipedia:

The diesel engine (also known as a compression-ignition or 'CI' engine) is an internal combustion engine in which ignition of the fuel that has been injected into the combustion chamber is initiated by the high temperature which a gas achieves when greatly compressed (adiabatic compression). This contrasts with spark-ignition engines such as a petrol engine (gasoline engine) or gas engine (using a gaseous fuel as opposed to gasoline), which use a spark plug to ignite an air-fuel mixture.
  The diesel engine has the highest thermal efficiency of any standard internal or external combustion engine due to its very high compression ratio and inherent lean burn which enables heat dissipation by the excess air.

How stuff works:

In theory, diesel engines and gasoline engines are quite similar. They are both internal combustion engines designed to convert the chemical energy available in fuel into mechanical energy. This mechanical energy moves pistons up and down inside cylinders. The pistons are connected to a crankshaft, and the up-and-down motion of the pistons, known as linear motion, creates the rotary motion needed to turn the wheels of a car forward.
  Both diesel engines and gasoline engines covert fuel into energy through a series of small explosions or combustions. The major difference between diesel and gasoline is the way these explosions happen. In a gasoline engine, fuel is mixed with air, compressed by pistons and ignited by sparks from spark plugs. In a diesel engine, however, the air is compressed first, and then the fuel is injected. Because air heats up when it's compressed, the fuel ignites.

Be sure to watch the animation at How Stuff Works.  It's quite helpful!
Diesel fuel vs. gasoline
In diesel engines, premature combustion can be a problem.  This happens when the fuel ignites "too early" in the compression phase, before maximum temperature and thus maximum efficiency can be obtained.  This means that by design, diesel fuel is a bit harder to ignite than gasoline.  It's made of slightly less volatile hydrocarbons.
As noted by @steveverrill in the comments, the above paragraph was incorrect.  A better explanation (straight from his comment) is

Diesel is harder to ignite at ambient temp due to its lower volatility / higher flash point (approx 52 °C vs -43 °C). This means it's difficult to form an explosive mixture which could be ignited by a spark (which is a local area at much higher than ambient temp.) But inside an engine both diesel & gasoline are fully evaporated. Under these conditions diesel is easier to ignite due to its lower autoignition temperature (approx 265 °C vs 280 °C). Autoignition temperature is tuned by additives to help ensure diesel can be ignited just by compression, whereas gasoline cannot ignite without a spark.

Wikipedia again

Petroleum-derived diesel is composed of about 75% saturated hydrocarbons (primarily paraffins including $n$, $iso$, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\ce{C12H23}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.

Wikipedia on gasoline

The bulk of a typical gasoline consists of hydrocarbons with between 4 and 12 carbon atoms per molecule (commonly referred to as C4-C12). It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes).

Thus, diesel fuel contains mostly hydrocarbon molecules that have on average 12 or so carbon atoms, while gasoline contains hydrocarbons that contain on average 8 or so carbon atoms.  The heavier, bigger molecules in diesel fuel means it evaporates more slowly at ambient conditions, requires more heating (i.e. more compression) before it spontaneously ignites.

Diesel is harder to ignite at ambient temp due to its lower volatility / higher flash point (approx 52C vs -43C). This means it's difficult to form an explosive mixture which could be ignited by a spark (which is a local area at much higher than ambient temp.) But inside an engine both Diesel & gasoline are fully evaporated. Under these conditions diesel is easier to ignite due to its lower autoignition temperature (approx 265C vs 280C). Autoignition temperature is tuned by additives to help ensure Diesel can be ignited just by compression, whereas gasoline cannot ignite without a spark.

Diesel fuel
The difference between gasoline and diesel fuel lies in the length of carbon chains.
Gasoline and diesel fuel are examples of hydrocarbons, molecules made up purely of hydrogen ($\ce{H}$) and carbon ($\ce{C}$) atoms. Simple examples include $\ce{C_3H_8}$ (propane) and $\ce{C_8H_18}$ (octane).
Gasoline contains compounds with carbon chains containing a certain amount of carbon atoms; their formulas are typically in the range of $\ce{C_7H_16}$ to $\ce{C_11H_24}$. Diesel fuel, on the other hand, has carbon chains somewhere in the area of $\ce{C_13}$ to $\ce{C_18}$.
However, in order for some set of atoms to qualify as a chain, it must have a certain structure. You could not somehow make random hoops of carbon and call them carbon chains.
Diesel engines
I hope you read the Wikipedia article on diesel engines. It explains that compression is used to ignite the fuel, rather than a traditional spark (from a spark plug).
There are a few different reasons why diesel fuel is better than gasoline in diesel engines. One major one (referring to Engineering Stack Exchange) is that diesel has a lower autoignition temperature - which is perfect, because diesel engines don't have spark plugs like those used in gasoline engines.

It is worth mentioning that Diesel tends to consist of larger linear hydrocarbons but gasoline tends to consist of shorter unsaturated and branched hydrocarbons. This is an important reasons for the different combustion behaviours.


text:
Relationship between electric dipole moment and polarization?

$$\vec{p}_\text{ind} = \alpha \vec{E}$$
The induced dipole moment is the polarizability times the electric field vector.
$$\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}(\omega)$$
The polarization is proportional to the susceptibility times the electric field vector.

In spectroscopy we used the dipole approximation. Unfortunately, I often mix up the different formulas. Is there a relationship between both formulas, or do both equations tell us exactly the same thing?

In fact, the first equation is applicable on a microscopic scale  (for an atom or a molecule):
  $\space$ $\vec{p}_\text{ind}=\alpha \vec{E}_\text{local} $
While the second one is applicable on a macroscopic scale (for the bulk):
$\space$ $\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}_\text{ext}(\omega) $
So, the polarisability is a microscopic quantity, while the susceptibility is a macroscopic quantity.
You have to notice that the electric field in the first equation is the local field, while it's the external field in the second one.
For more details, please see this page.

Indeed, the microscopic / macroscopic versions carry forward also for $\beta$ and $\chi^{(2)}$, $\gamma$, etc.

I agree with @Geoff Hutchison that the microscopic / macroscopic versions carry forward also for (β and χ(2)), ( γ and χ(3)) etc.

@YomenAtassi I like your answer and as well the additional page. However, I've still confused, because in Spectroscopy both have been used and we only took a look into small molecules, not bulk. I also have no idea to which of the equations (eq. 1 or eq. 2) the formula for the dipole approximation $\hat{V}=-\hat{\mu}\overrightarrow{E}$ is best assigned. (related to my two weak old, but still unsolved question: http://chemistry.stackexchange.com/questions/33170/why-do-we-say-approximation-in-the-dipole-approximation-in-spectroscopy)

In the scope of our lecture of Spectroscopy (atoms, small molecules, aggregates) we have restricted to linear spectroscopy which means there is no dependence of quadratic terms in $\overrightarrow{E}$ in eq(2). I'm not sure but may be it's possible, you get also quadratic dependency in $\overrightarrow{E}$ for eq(1) if we deal with non-linear spectroscopy?

Regarding your second comment, you are right! there is also quadratic dependency in E for eq(1) if we deal with non-linear spectroscopy.

Regarding your second comment, you are right! There is also a quadratic dependency in E for eq(1) if we deal with non-linear spectroscopy. As for your first comment, the formula for the dipole approximation is best assigned to the first equation. You said" in Spectroscopy both equations have been used and you only took a look into small molecules, not bulk":  I  need to look more in depth to the use of these equations in your course in order to reply. But I think they use the second eq when they talk about the emission or absorption of the studied medium (as a bulk). First eq for single molec.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Deviation of real gases from ideal behaviour
From the Amagat's curve we can see two types of result for real gases.

If we increase pressure,the value of $PV$ decreases initially and then the value starts increasing for the gases like- carbon dioxide, nitrogen, oxygen, methane, etc..
If we increase pressure,the value of $PV$ starts increasing initially for the gases like - hydrogen, helium, etc.

Now please explain what're the reasons behind this kinda behaviour of real gases?

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

@Bon this isn't necessarily homework. I see it more as [tag:erratum]. It does deal with something in the textbook, just that it doesn't look for an error.

I think it's just bad tag 'cause it doesn't look like erratum.

@Mithoron I scratched my head for the tags. I, am the guilty fugitive who added [tag:erratum]. Can you please add better tags?

The ideal gas law can be derived using kinetic molecular theory by making two very important assumptions that are not true for real gases:

There is no attractive force between gas molecules
The gas molecules have no volume

Think about how the PV curve would change when you remove one or both of these conditions. Then think about which types of molecules the assumption of "no attractive force" would be more important for.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What are good books about MO theory that don't require a lot of math background?
Are there any books about MO theory that explain the theory to a certain detail but don't rely on a lot of math background or explain the math within the book?

What level of math are you comfortable with? Ultimately, you need to be familiar with differential equations, preferable in spherical (i.e. non-Cartesian) coordinates. Fundamentally, it needs a certain level of math.

@JonCuster I am familiar with basic calculus.

my personal opinion is this: to 'get' it properly will need more than calculus.  Given the differential equation and the solution, you might be able to differentiate the solution to show that it solves the equation. But, you won't really understand the equation without quantum mechanics, and you won't be able to understand how to get a different solution when the equation changes a little bit.  However, just knowing that there is more underneath it all is important - as you learn more, you also learn that there is still more...

I teach the qualitative parts of MO theory to sophomore biology students using almost no mathematics. It can be done, but I don't have a book that I can point to.

From a very qualitative point of view I think this book by Fleming can teach you a great deal about MO theory and how to use it to predict reactivities and such. It is a good read (I only know the first edition but the second edition might be even better - it certainly has more pages). Fleming uses quite little math.

@JonCuster I will keep this in mind.

I'd suggest the seminal Orbital Interactions in Chemistry by Albright, Burdett and Myung-Hwan Whangbo.
The second edition came out in 2013 and I'm sure there are many copies of the first edition around as well.
They focus a lot on pictoral MO theory - indicating how orbitals of different shapes and symmetry combine.
To truly understand MO theory at a fundamental level, it really helps to study some quantum mechanics (and thus mathematics) but you can grasp a lot from the approach used in this book.


text:
A monocyclic 6 carbon ring with 6 double bonds
I would like to know if this molecule has ever been synthesized, if not why not, and what would its IUPAC name be?

There are several monocyclic $\ce{C6}$ isomers.  Three of them are pictured below

A - the compound you've drawn above, cyclohexahexaene
B - cyclohexatriyne
C - cyclohexa-2,5-diyne-bis-1,4-ylidene

While all of the carbons in these molecules are (more or less) $\ce{sp}$ hybridized, the molecules are different from one another due to differences in bond lengths and bond angles.  For example, all of the bonds in A are exactly the same length due to symmetry, whereas the triple bonds in B will be shorter than the adjacent single bonds.  Therefore, A-C are isomers, not resonance structures.
Ideally the atoms attached to an $\ce{sp}$ hybridized carbon form a linear arrangement (180° bond angle) with the central $\ce{sp}$ carbon, acetylene and the central carbon in allene serve as examples.

(image source)

(image source)
However in the cyclic $\ce{C6}$ compounds A-C, the $\ce{sp}$ hybridized carbons are forced by the ring to deviate from the desired 180° arrangement to something down around 120°.  Therefore, quite a bit of angle strain is incorporated into A-C making them extremely strained and unavailable through laboratory preparation to date.
As mentioned above, A-C have different bond lengths and bond angles therefore they are isomers, not resonance structures.  A fascinating question is,

Do A-C each sit in a potential well, separated from one another by an energy barrier and if so, which one is the most stable (energy minimum)?

or

Is there another monocyclic species lower in energy than A-C? Perhaps a molecule that could be described by resonance structures A'-C', where the prime denotes that all of the bond lengths and angles have been adjusted and made equal in A, B and C.

I would opt for A'-B' benzene heksaradical with only small in plane orbital overlap but symmetry and aromaticity intact.

This is a type of cyclocarbon, specifically Benzotriyne. Processes using it or synthesizing it have been described in Towards the total synthesis of cyclo[n]carbons and the generation of cyclo[6]carbon, among others.
It has not yet been successfully synthesized. Wikipedia states (without citations) that

Recent investigations have concluded that benzotriyne is unlikely to exist due to the large angle strain.

Not really with six carbon atoms, but with eighteen. Cyclo[18]carbon, preferred IUPAC name Cyclooctadeca-1,3,5,7,9,11,13,15,17-nonayne, is predicted to be the minimal stable carbocycle. According to the WP article, it has been synthesized at low temperature by decomposition of an organic precursor[1] and "claimed"  in the solid state[2].
References
1.
François Diederich; Yves Rubin; Carolyn B. Knobler; Robert L. Whetten; Kenneth E. Schriver; Kendall N. Houk; Yi Li (8 September 1989). "All-Carbon Molecules: Evidence for the Generation of Cyclo[18]carbon from a Stable Organic Precursor". Science 245 (4922): 1088–1090. Bibcode:1989Sci...245.1088D.
https://doi.org/10.1126/science.245.4922.1088. PMID 17838807.
2.
Kaiser, Katharina; Scriven, Lorel M.; Schulz, Fabian; Gawel, Przemyslaw; Gross, Leo; Anderson, Harry L. (15 Aug 2019). "An sp-hybridized molecular carbon allotrope, cyclo[18]carbon". Science 365 (6459): 1299–1301. arXiv:1908.05904. https://doi.org/10.1126/science.aay1914.

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


text:
Difference between oxygens in a carboxylic acid?
I've been reading about MO theory and am wondering if the traditional model of carboxylic acids is correct. Usually, how it's presented is that the carbonyl carbon is doubly bonded to one oxygen and singly bonded to a hydroxyl group. This would mean that one oxygen is $sp^3$ hybridization and the other is $sp^2$ hybridized, and the hydrogen is attracted to both (one through a covalent bond and the other through a hydrogen bond).
My book (Ansyln and Doughtery) says that

Hybridization isn't just limited to $sp^2$ and $sp^3$ and that bonding isn't localized between only two atoms but between the molecule as a whole. (paraphrased)

How does this affect carboxylic acids? Is there actually clearly defined '$sp^2$' and '$sp^3$' oxygens, or are they chemically equivalent? It seems like carboxylic acids react as alcohols (see: thionyl chloride). But does some sort of spontaneous symmetry breaking take place when a reactant approaches?
I can't find anything detailed/reputable about this on the internet.
The question What is the hybridization of the carbonyl oxygen in a carboxylic acid? does not address my concern.  I get what the classical hybridization of the carbonyl oxygen in a carboxylic acid is. I'm just wondering if there's a better non-classical description where the oxygen atoms are equivalent until approached by another molecule or something of the sort.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

There is no hydrogen bond in carboxylic group.

possible duplicate of What is the hybridization of the carbonyl oxygen in a carboxylic acid?

No, I get what the classical hybridization of the carbonyl oxygen in a carboxylic acid is. I'm just wondering if there's a better non-classical description where the oxygen atoms are equivalent until approached by another molecule or something of the sort.

@cwagen I have inserted your comment into the question. Please have a look and make further corrections if necessary.

First of all I would like to put my usual disclaimer about hybridisation: It is just a mathematical concept, nothing too real and it should always taken with a grain of salt. In the linked post I have already discussed certain aspects of that. Furthermore it has been discussed on this site in various occasions.
Going a bit further, especially for delocalised systems, the hybridisation picture starts to break down. In acetic acid (or any other carbon acids) this is not yet a big problem and we can still argue a lot with the involved concept. But I hope the following explanations will demonstrate some of this break down.
Now let's get back to the problem at hand. Here is the structure of acetic acid in the gas phase at 0 K, calculated with Gaussian09 at the DF-BP86/def2-SVP level of theory.

As you can see the carbon oxygen bonds are not equal at all. In the Lewis picture, we would say there is a single bond between carbon 2 and oxygen 1, which is a somewhat valid approximation. On the other hand we would say that between carbon 2 and oxygen 2 there is a double bond, which is also a somewhat valid approximation. In the molecular orbital picture we have three π-type orbitals (10, 13, 15). Therefore we have partial double bond character in the carboxyl unit due to delocalistion. Note that there is also an hyperconjugation effect involved. The main difference, however lies in the the molecular orbital 15, which is bonding with respect to C2-O2 and antibonding with respect to C2-O1. This orbital is basically the key, why the Lewis structure is a valid approximation.

Now we can unitary transform these orbitals to localised orbitals, therefore representing the bonding structure with hybrid orbitals. I am using the Natural Bond Orbital theory for this. The following summary gives the bonds of the carboxyl subunit, the numbers in brackets refer to the orbital number in the picture. Also note, that since these orbitals are no eigenfunctions of the Hamiltonian and they do not have an energy eigenvalue, hence there is no canonical order.
-----------------------------------------------------------
#  (Occupancy)  (Type)    (orbital)  || Contributions below
-----------------------------------------------------------
1. (1.99578)    BD ( 1)   C2 - O1   [6]
            ( 30.48%) C(2) s( 26.60%)  p 2.75( 73.10%)
            ( 69.52%) O(1) s( 35.03%)  p 1.85( 64.91%)

2. (1.99728)    BD ( 1)   C2 - O2   [5]
            ( 33.43%) C2   s( 33.28%)  p 2.00( 66.51%)
            ( 66.57%) O2   s( 42.68%)  p 1.34( 57.24%)

3. (1.99005)    BD ( 2)   C2 - O2   [14]
            ( 31.97%) C2   s(  0.00%)  p 1.00( 99.64%)
            ( 68.03%) O2   s(  0.00%)  p 1.00( 99.87%)

4. (1.98518)    BD ( 1)   O1 - H4   [7]
            ( 74.58%) O1   s( 19.52%)  p 4.12( 80.41%)
            ( 25.42%) H4   s( 99.60%)

5. (1.80107)    BD ( 2)   O1 - H4   [15]
            ( 99.93%) O1   s(  0.00%)  p 1.00( 99.93%)
            (  0.07%) H4   s(  0.00%)  p 1.00(100.00%)

6. (1.97613)    LP ( 1)   O1        [10]
                           s( 45.45%)  p 1.20( 54.52%)

7. (1.97395)    LP ( 1)   O2        [8]
                           s( 57.37%)p 0.74( 42.62%)

8. (1.83220)    LP ( 2)   O2        [16]
                           s(  0.00%)p 1.00( 99.91%)
------------------------------------------------------------

The total compliance with the Lewis bonding picture is 98.3%. Note the artefact in line 5, which should be a lone pair of oxygen.
From here we can see, that O1 is best described with an sp1.85 contribution towards carbon, an sp4.12 contribution towards hydrogen,
an sp1.20 and a p lone pair.
The O2 is best described with an sp2 contribution towards carbon, an π orbital towards carbon, an sp0.74 and a p lone pair.

Concerning the intramolecular hydrogen bond, one can analyse the electron density with the Quantum Theory of Atoms in Molecules (QTAIM). This shows that there is no bonding interaction between H4 and O2.

(Laplacian distribution, solid blue lines indicate charge depletion $\nabla^2\rho<0$, dashed blue lines indicate charge accumulation $\nabla^2\rho>0$, red spheres are bond critical points, black lines are bond paths, dark red lines are zero flux surfaces. Hydrogens H2 and H3 omitted for clarity.)
One could argue that there is an electrostatic interaction, causing the slightly smaller (105°) CHO angle, but since the H4-O2 distance is only slightly shorter than the sum of the van der Waals radii (H = 1.20 A, O = 1.52 A) in vacuum this might already be a stretch.
In solution there are many more factors to be considered, i.e. dimerisation, protonation/ deprotonation, intermolecular hydrogen bonds, and possibly many more.


text:
The proper explanation for the general formula of the DU?
I found a formula for the degree of unsaturation (DU) in Clayden’s Organic Chemistry. It explains, simply, the unsaturation with the difference in H atoms. Then I stumbled upon the general formula:
$$DU=1+\frac12\sum n_i\left(v_i-2\right)$$
$n_i$ is the number of atoms with valence $v_i$.
Can someone explain from where this equation was derived? And how come the valence term was introduced?

Does this answer your question? http://chemistry.stackexchange.com/q/2819/4945

My answer is based on V. Pellegrin, J. Chem. Educ. 1983, 60 (8), 626 which might be inaccessible for you by a pay wall.
Starting with Eulers characteristic for planar graphs wich states that adding the faces and vertices of a planar graph and substracting all the edges must be 1 $$F+V-E=1$$ which can be rewritten as $$F=1+E-V$$
Now in more chemical words we can redefine this by saying that edges are bonds and vertices are atoms and the $F$ will be our $DU$ from now on. If we know the structural formula we could stop now, count all atoms and bonds and calculate what we want to know. Those are two examples from the paper:

But what if all we have is a sum formula from let's say mass spectrometry?
Then we have to calculate how many bonds this molecule will have based on the knowledge of how many atoms the molecule contains and how many valence electrons each atom has.
Each molecule consists of $$N=\sum_i n_i$$ atoms of element $i$ whereby $n_i$ denotes the amount of each element. Each atom has to be assumed in it's lowest valency state and has $v_i$ valence electrons which would make in total $$\sum_{i=1}^N n_i~v_i$$ valence electrons. As every single bond contains two electrons, the amount of all possible bonds is half the amount of all valence electrons, namely $$\sum_{i=1}^N \frac{n_i~v_i}{2}$$
Let's try it with the mass formula $\ce{C9H7NO}$:

The molecule consists of $N=18$ atoms.
The atoms are grouped in $4$ groups containing the elements $i=\{C,H,N,O\}$
Each element is contained $n=\{9,7,1,1\}$ times.
The lowest valency of each contained atom is $v=\{4,1,3,2\}$
The sum of all valence electrons is $\sum_{i=1}^N n_i v_i=9\cdot4+7\cdot1+1\cdot3+1\cdot2=48$
The sum of all possible bonds is $\sum_{i=1}^N \frac{n_i~v_i}{2}=48/2=24$

Now let's insert those numbers into the equation $$DU=1+E-V=1+24-18=7$$ Because I knew the moleule that I have chosen we can compare

and see that the calculation was correct ($5$ for the double bonds + $2$ for the ring closures).
As it was correct, we can combine it and derive your equation.
$$\begin{align}
F=&~1+E-V\\
DU=&~1+\text{number of bonds}-\text{number of atoms}\\
DU=&~1+\sum_{i=1}^N \frac{n_i~v_i}{2}-\sum_{i=1}^N n_i\\
DU=&~1+\sum_{i=1}^N \left(\frac{n_i~v_i}{2}-n_i\right)\\
DU=&~1+\frac{1}{2}\sum_{i=1}^N (n_i~v_i-2~n_i)\\
DU=&~1+\frac{1}{2}\sum_{i=1}^N n_i~(v_i-2)\\
\end{align}$$


text:
PTFE (Teflon) combined with Silicon Polymers
I wanted to know if it would be possible to combine basic silicon polymers with PTFE in order to create a stable, non reactive silicon rubber that won't react to butane or other hydrocarbons in particular. Is this doable? If so what would it take to accomplish this?

Do you mean as in a copolymer rather than a formulation of the two polymers ?

Yes I believe as a copolymer. Whichever will be better as I intend to combine the two for use as a flexible chem resistant mold.

I would suggest looking up "fluorosilicone" polymers, Kalrez plus silanes, then go from there. I suspect the organic portion of the siloxane will always be susceptible to something, just like PDMS or poly(diphenylsiloxane). To prevent this, you may need an alternating TFE /dimethylsiloxane sequence or as good as, rather than a block copolymer. Possibly this suggests a living polymerisation? I couldn't say.


text:
Is the pH of a solution (orange juice) affected by the presence of a metal?
Is the pH of a solution (orange juice) affected by the presence of a metal (aluminium, tin or iron for example)? I know that the metal will corrode in the presence of an acid (citric acid in this case), but will the metal have any effect on the actual acidity of the solution?

Is there some reason / context behind this question? Anyway, if the acid does react with the metal, then the amount of $\ce{H+}$ ions will decrease and pH will increase, leading to a less acidic solution (as you would expect, since the acid is reacted away).


text:
What determines the elastic limit of a substance?
from what I know, the elastic limit of a substance is the maximum amount of stress/force/pressure (a bit of confusion on my behalf here) a substance can withstand and still return to its original shape afterwards. my question is what factor(s) determine this limit?
also, I have noticed that some plastics go opaque when under stress and have found the reason for this effect in  this question however, what I do not understand is why some plastics and other synthetic materials do this before the elastic limit (and turn clear upon shrinking) and others do this after the elastic limit of the substance (and the effect is permanent). why can this process of turning opaque be reversed in materials that turn opaque before the elastic limit of the substance.

First let me clarify that there is no way to calculate the elastic limit of a substance - it has to be measured. Viscoelastic theory is pretty well-developed for only being about 150 years old, but it's not that well developed.
To answer your question, though, lots of factors affect elasticity in general and the elastic limit in particular.
It would probably be the most helpful to think about what is happening on the molecular scale. In any viscoelastic material, there is a range of deformations (strain, or the response to stress) and timescales over which molecules will go back to pretty much the same relative position that they were in before the deformation occurred. That's elasticity. What makes this happen? Each atom or molecule sits in a little potential energy well that exists because of its position relative to other atoms in the substance. Atoms like to be close to each other, but not too close. As a result, when given the opportunity they will arrange themselves so that as many atoms as possible are the optimal distance apart(resulting in the lowest energy).
When a material is deformed, we are moving atoms relative to each other. When one atom moves, its energy increases. This also changes the potential energy surface seen by its neighbors. If it moves a little bit, then the change is not usually great enough to cause the neighbors to move by much. On the other hand, for big movements, the effect can be great enough that the other nearby atoms rearrange themselves in order to find the next lowest energy level.
When the relative motion of atoms due to mechanical deformation is small enough that most of them go back to their original positions once the applied stress is removed, then we say the stress/strain was within the elastic limit.
When the atoms are permanently rearranged, the shape of the material is changed and we say that we have exceeded the elastic limit.
The ease with which atoms can move out of position (you can think of this like the depth of the wells), and their tendency to move back to the lowest energy (which can be thought of as the "steepness" of the wells) are the two main factors that affect how much stress/strain is needed to permanently deform a material. There are lots of things that can affect each of these, which is why there is no comprehensive theoretical model that can predict the elastic limit of materials based on molecular structure alone.
I can summarize a few important factors here, though:

degree of crystallinity in polymeric materials
type and number of dislocations in crystalline materials
length of time that the stress/strain is applied, in conjunction with the temperature of the material (this related to a phenomenon called time-temperature superposition)
structure of the molecules making up the material (e.g. linear vs. crosslinked or branched polymers, fcc vs. bcc crystal structure, or type of atoms in an alloy)
mesoscopic structure (e.g. crystalline grain size and shape, aggregation of nanoparticles, or long-range order in polymers)
macroscopic structure (e.g. fiber orientation in composite materials)

There are many more! And each of the ones I listed is practically an entire field of its own in materials science.
To answer the second part of your question - the cloudiness you are referring to is called "crazing" in polymer science. It is basically caused by molecules pulling apart from each other, without covalent bonds breaking. It can be reversed when it happens before the elastic limit is reached because the molecules can still go back to their original positions. If it doesn't happen until after the elastic limit, then the material is permanently deformed, and the molecules cannot return to their original position.


text:
Identifying functional groups

I've spent a few days googling functional groups and reading my text assigned on organic functional groups. I'm understanding very little and I just need a little clarification. I understand that at the top right and bottom left, the hexagon structure signifies an aromatic, right?
Top Left: My first guess was carboxylic acid, but with further googling, I now think it's an amide, because it contains nitrogen and oxygen?
Top Right: Since the hexagon is aromatic, I know it can only be the last three choices, but I'm going with C because the $\ce{-NH2}$ is not connected to an oxygen.
Bottom Left: Because of the $\ce{Br}$, I know it's a bromide and, assuming that my thinking is correct with aromatic, the answer should be B.
But the bottom right, I am not sure what exactly to look at here to even guess, besides that it is not an aromatic ring.
To show that I've tried the last one, all I can find is that an alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$ and, the aromatic being the hexagon structure, I'm unable to grasp/understand which one the last group would fall into.

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

@alyssa Is it multiple choice test?

No, it's not a test or homework, these questions are coming from my book, so I can be more prepared for our final in a month. The functional groups just make no sense to me.

Basically anything in this format is categorized as homework. It doesn't need to be homework itself. Self-study questions, puzzles etc. also count as homework. Regarding your edit, I'm really inclined to vote to reopen, but you've shown no evidence of your research. You shouldn't say I Googled but I couldn't find anything. that isn't helpful at all. You should say I Googled for blah blah, but it said bladiblah. so I can't understand the di in bladiblah. can you help me understand? Do as so and I'll vote to reopen.

https://en.wikipedia.org/wiki/Functional_group

This is much improved and it is clearer where your concerns are, asides from what @Mithoron linked to, here is another http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_01%3A_Chapter_1%3A_Introduction_to_organic_structure_and_bonding_I/Section_1.4%3A_Functional_groups_and_organic_nomenclature

Thank you guys. I'm still just extremely confused on the last one, as I can't find anything provided on either of the links shared to give me an answer that I can understand and know is correct. I'm leaning towards Alkane, because it's closest, but I feel like I'm missing something.

Alkanes have only single bonds alkenes also double, alkynes triple, so what you think? Otherwise your answers are quite ok.

Basically, on the last one, list out what you see. As Mithoron stated, if you ONLY see single bonds, then alkane (CH4, C2H6, C3H8 etc...). If you see anything else, then it is usually classified otherwise since other functional groups are usually more meaningful for chemical reactions. Do you see any other bonds? Do you see a ring?

Okay, so based off of both of your explanations, it isn't an Alkane? Because of the H3C - CH?

Do you see the double bond? Two lines. I'll bring up some examples of alkenes. Search up examples such as propene, butene etc. In fact, the very compound is butene. The key is to look at the whole structures and list out what you see. Yes, there is the alkane PORTION but the fact that there is a double bond transforms its classification into alkene. it's sort of like the saying "one bad apple makes the whole lot rotten". Usually, in noting functional groups, you ignore the simple, saturated carbon chain (as in the alkane portion with only single bonds and CH bonds, no double)

The thing that looks like an equal sign? I was reading this as H3C - CH = CH2.. so, now I'm guessing that's wrong too. I told you guys the functional group is impossible for me to grasp!

Yes, the thing that looks like an equal sign signifies that it is a double bond. Three lines means triple. One line= 1 bond. 2 line = 2 bond. 3 line = 3 bond. And That's it. I doubt you'll every see 4 and more

Also, a trick for identifying alkane, alkene, and alkyne if you are more mathematically inclined is to use the general formula: CnH2n+2 for alkane, CnH2n for alkene, and CnH2n-2 for alkyne (related to the so-called degrees of unsaturation). This does not account for multiple bonds, cyclic structures etc. and should be taken with caution As you see 4. is C3H6. Also, you should note that alkenes and alkynes require at least 2 carbons

So the correct answer for this is Alkene because of the =?  I was basing the rest of my answers off what I previously said, so could you check those too?

Let us continue this discussion in chat.

Some pointers:

the hexagon structure signifies an aromatic

Not quite. The hexagon structure depicted with alternating carbon-carbon double and single bonds is aromatic. It is called benzene and is very commonly encountered in organic chemistry. It can be represented by two resonance structures as shown below and is sometimes also represented with a circle inside a ring of singly bonded carbons to show the delocalised $\pi$ system.

However, the are other molecules depicted as hexagons which are not aromatic such as cyclohexane.

Additionally, there are other molecules which are aromatic which are not benzene, and many which are not hexagonal but you probably won't come across these at an introductory level.

My first guess was carboxylic acid, but with further googling, I'm now thinking it's an amide, because it contains nitrogen and oxygen

You are correct that it is an amide. The amide functional group contains a nitrogen adjacent to a carbon-oxygen double bond. The carboxylic acid group is similar but the nitrogen is replaced by an $\ce{-OH}$ group. Amides can be primary, secondary or tertiary depending on the number of carbons that are bonded to the nitrogen.

alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$

These are the simplest example of alkanes, alkenes and alkynes. They are called methane, ethene and ethyne respectively. However, there are more complicated molecules involving these functional groups. In general, alkanes have only carbons-carbon single bonds and carbon-hydrogen single bonds. Alkenes are distinguished by having a carbon-carbon double bond denoted by two lines between the two carbons which look like an equals sign. Alkynes have a carbon-carbon triple bond which is denoted by three lines between the carbons. Therefore, your example is an alkene.

If the left compound of the three molecule you have draw had a halogen atom in the carbon chain could we still call it amide? I mean do we name a functional group by looking locally or by looking in the whole molecule

The bottom right compound is an alkene.
There are no atoms other than $\ce{C}$ and $\ce{H}$ present, so you can rule out most functional groups straight away. The double bond makes the compound an alkene. The name of the compound is propene.


text:
Chemicals in Urinal Odor Eaters
What chemicals are used in so-called urinal "odor eaters"? Do they actually destroy odors, or do they just mask them? Also, are these chemicals subject to any environmental laws or regulations?

What chemicals are used in so-called urinal "odor eaters"?

There are 1,4-dichlorobenzene-based and non-PDCB products. In the latter, the chemical mixture from the linked MSDS is:
SODIUM SULPHATE, SODIUM DODECYL SULPHATE, COCAMIDE MEA, DIPENTENE, CITRAL, LINALOOL,BENZYL ALCOHOL

Do they actually destroy odors, or do they just mask them?

They mask odors.

Also, are these chemicals subject to any environmental laws or regulations?

Yes, as detailed in the MSDS sheets linked above. Please note that my knowledge of urinal cakes and related products is not exhaustive.

could anyone (other than the manufacturer) have an exhausting knowledge of urinal cakes?

The MSDS of the non-DCB product says: HARMFUL TO AQUATIC ORGANISMS, MAY CAUSE LONG-TERM ADVERSE EFFECTS IN THE AQUATIC ENVIRONMENT.
DO NOT EMPTY INTO DRAINS, DISPOSE OF THIS MATERIAL AND ITS CONTAINER AT HAZARDOUS OR SPECIAL WASTE COLLECTION POINT.
ENVIRONMENTAL PRECAUTIONS:
DO NOT ALLOW PRODUCT TO REACH SEWAGE SYSTEM OR ANY WATER COURSE.
INFORM RESPECTIVE AUTHORITIES IN CASE OF SEEPAGE INTO WATER COURSE OR SEWAGE SYSTEM.

Of course, the sad envrionmental toxicological irony of the urinal cake is that is it mostly soluble in water and designed to end up "down the drain".

@iad22agp The caps are killing my eyes. Could you try and avoid them in future.

Sorry - I copied and pasted the text directly from the online MSDS. Must be a holdover from the days of the teletype.


text:
Average or individual molar heat capacity?

In finding enthalpy change $\Delta H$ during adiabatic, reversible expansion of  mixture of two ideal gases $n_1$ mol of $\ce{A}$ ($C_{\mathrm{m},v}=\frac{3}{2}R$) and $n_2$ mol of $\ce{B}$ ($C_{\mathrm{m},v}=3R$) taken in a container and during this temperature change is $T_\mathrm{i}$ to $T_\mathrm{f}$. Then what should be the value of $\Delta H$?

It can be $$\Delta H=\left(n_1\right)\frac{5}{2}R\left(T_\mathrm{f}-T_\mathrm{i}\right)+\left(n_2\right)4R\left(T_\mathrm{f}-T_\mathrm{i}\right)$$ taking individully for each gas. Or let say average molar heat capacity of gaseous mixture at constant pressure be $C_{\mathrm{m},p,\mathrm{avg}}$ then it can be $$\Delta H=\left(n_1+n_2\right)C_{\mathrm{m},p,\mathrm{avg}}\left(T_\mathrm{f}-T_\mathrm{i}\right)$$ so please help which is correct one. As per suggestion both are same.
$$C_{\mathrm{m},p,\mathrm{avg}}=\frac{\left(n_1\right)\frac{5}{2}R+\left(n_2\right)4R}{n_1+n_2}$$

They both are same.
If you plug $C_{\mathrm{m},p,\mathrm{avg}}$ in your second equation, you will get your first equation. The average $C$ is calculated as a mole fraction weighted average.


text:
Is there a way to dissolve glue?
Background: I have a journal from my childhood. I applied glue (uhu brand. this one) to the papers almost 10 years ago.
Q: Is there a way to dissolve the glue and make the pages readable?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Can you post some more information on the glue, such as manufacturer details etc.

You might try different solvents on a spot that does not have writing, to see if it removes the glue, and on glue with non-critical writing, to see if it removes or smears the writing as well as the glue.
Since it described as a clear synthetic resin, you might try the following sequence, from milder solvent to stronger:

Comparatively mild non-polar organic solvents, such as paint thinner or kerosene. If that does not dissolve the glue, try
Some Polar solvents such as acetone, methyl ethyl ketone or nail-polish remover. If that does not remove the glue, try
PVC pipe solvent (tetrahydrofuran).

N.B. All solvents listed are flammable, and some are fairly toxic. Use them in small quantities, closing the containers after getting a bit out, to prevent flash-back of flames, and use in well-ventilated areas.


text:
Complexity in computing the integrals involving higher angular momentum
I am quite familiar with Hartree Fock theory. However, writing it into program is an entirely different task, and that is where I have a couple of questions.
I wrote a Hartree-Fock routine in Python a couple of years ago. It involved a ultra-minimal basis: only one gaussian for two hydrogen atoms each. Kinetic energy term was pretty simple, so were the one-electron terms. Coulomb and exchange integrals were tricky, which involved evaluation of an erf. Now, my question is, if I included higher angular momentum basis functions (p,d,f,g type functions), what happens? Is it difficult to evaluate the J (Coulomb) and K (exchange) integrals because of the mathematical complexity? Or something else? Can someone give the math behind this and explain why it is difficult to evaluate?

FYI: the question might be more appropriate at CompScience.SE as noted in our Help Center. See specifically the note concerning Computational questions.

If it appears to be that you don't get any good answers to your question, you might want to actually consider migrating it the CompSci.SE as wildcat suggested. If you decide to do so, either flag the question with a custom flag or notify me here in the comments.

I think this is too specialized for Comp Sci, to be honest.  Adding to that, the actual question is too broad to get a concise answer.  You are certainly welcome to reask it there, but I will refrain from moving it at this time.

It appears that the integrals that are difficult are the two-electron repulsion integrals. There exists a recursive relationship that can turn integrals with high-angular momentum functions into integrals with only s-type functions. Refer to this lecture for details.
See the following articles for details concerning these recurrence relations:
M. Head Gordon and J. Pople, J. Chem. Phys. 89, 5777 (1988)
S. Obara and A. Saika, J. Chem. Phys. 84, 3963 (1986)

Could you summarize the link in your answer in case it becomes unavailable in the future. See the help centre for how to write a good answer.

Computing overlap, kinetic, electron-nucleus attraction and electron-electron repulsion integrals is possible even for Cartesian gaussians of higher angular momentum. There are analytical formulas which gives you the solution of these integrals. In particular you can find them in Cook's book [1] (pay attention to the fact that the electron-electron formula is wrong! See this discussion for details.).
If you don't have access to Cook's book [1], you can find a detailed derivation of analytical formulas for overlap, kinetic energy and nuclear-electron attraction integrals in Refs. [3, 4, 5].
These formulas are quite straightforward to implement (apart from the computation of Boys function, see this discussion) but they are highly inefficient. This is the main problem!
This is why alternative methods have been developed. In Helgaker's book [2] you can find a good description of these alternative methods:

Obara-Saika scheme
McMurchie-Davidson scheme
Rys quadrature

If you just want to write a working Hartree-Fock program you can just use the analytical solution at first. However, if you need efficiency you will need to look at these methods.

A side note: personally I love Python but it is really, really unappropriate for computationally demanding calculations! My Hartree-Fock program in Python takes at least 2 minutes to compute two-electron integrals in the STO-3G basis set (a minimal one!) for $\ce{H_2O}$, while a Fortran version of the same program takes a bunch of seconds.

[1] D. Cook, Handbook of Computational Chemistry, Oxford University Press, 1998.
[2] T. Helgaker, P. Jørgensen and J. Olsen, Molecular Electronic-Structure Theory, Wiley, 2000.
[3] M. Hô, J. M. Hernández-Pérez, Evaluation of Gaussian Molecular Integrals: Overlap Integrals, The Mathematica Journal, Vol. 14, 2012.
[4] M. Hô, J. M. Hernández-Pérez, Evaluation of Gaussian Molecular Integrals: Kinetic-Energy Integrals, The Mathematica Journal, Vol. 15, 2013.
[5] M. Hô, J. M. Hernández-Pérez, Evaluation of Gaussian Molecular Integrals: Nuclear-Electron Attraction Integrals, The Mathematica Journal, Vol. 16, 2014.


text:
Electroplating metal inside polymer foam with acetone
In my research we have a method of making polymer open-cell foams with high tortuosity and pore sizes on the order of 50–200 nanometers. I can make thicknesses of 100 µm to 2 mm thick (and up to 5 inch diameter disk). I am looking for some cool things to do with this material, and one idea was to try to fill it with metal and then dissolve the polymer away to leave a nanoporous metal structure.
The problem is that so far I cannot get water to penetrate inside the pores, I tried up to 4 MPa pressure and it just won’t go through (small pores and hydrophobic) but acetone will get absorbed in very quickly. My question is, could there be a way to use acetone as an electrolyte and deposit metals (copper, nickel, whatever) to fill these open-cell pores?
I haven’t taken chemistry since intro chem back in undergrad so I really don’t know much of anything about electroplating, so I will be very grateful for any help on this experiment. Thank you very much, I hope I was clear in explaining what I am trying to do.

Do you know the polymer foam porosity?

What is the chemical structure of your polymer? Maybe somebody can suggest another solvent.

Thanks for the replies. The polymer is polyetherimide or polysulfone (both can be made porous) and depending on processing parameters these can be 10-60% void (air). I tried isopropyl and it goes in the pores although it takes longer to diffuse in, maybe 5-10x slower than acetone.

Electroplating might be possible, but getting good coverage of internal pores is tricky, even with aqueous electrolytes, as deposition will likely occur preferentially at the outer edge of the workpiece, leaving the internal passages lightly- or un-coated.
A more effective approach is likely electroless nickel deposition.  As long as you can find a suitable acetone-soluble reducing agent, it might work out.
I would also recommend testing other solvents for their ability to permeate the polymer structure -- if you could find a protic solvent that wets it (even something like 50/50 ethanol/water!), you'd have a much better chance of success at electrodeposition proper, and also at making electroless work for you, since (I think) most electroless recipes are aqueous-based.


text:
Diffusion and potential gradients for two similar atoms
Studying biochemistry I noticed that equilibrium is reached for K and Na separately (not taking into account electric potential), as would be the case I assume for any molecule or atom. This is of course entropy.
Eg: My understanding is, you have two types of atoms, and a membrane separating them, with different concentrations on each side. Each atom will reach equilibrium with itself across the membrane, as if the other type of atom doesn't exist (barring the influence of charge).
Why is it that, for instance, a K atom is so keen to reach equilibrium with another K atom in an area of lower concentration, and yet seemingly ignores the Na, even if the Na is in high concentration where the K is low? Both Na and K still have the same charge, and would also collide with each other I imagine? Why do identical atoms behave like this but non-identical ones with very similar properties don't?
Update #2:
My understanding was incorrect regarding K and Na having separate forces. The two atoms seem to be treated the same, except for selective permeability.
I'm still not understanding why the two atoms have separate gradient potentials, if they can be treated as the same. I made the following images to illustrate my point a little clearer.
There is a compartment with a left and right side, separated by a wall containing two selectively permeable channels that may open or close. Only Na can cross the green channel when it is open, only K can cross the orange channel when it is open.
First, the assumption: The main assumption behind all this is that, in answering this question, we can treat Na and K as being the same. They are effectively two marbles with identical physical and chemical properties, just with different colours. This is what I have come to understand based upon the answer by orthocrestol.
The cell on the left has a -70mv potential energy compared to the right compartment, caused by invisible negative ions which will always remain on the left side.

All that has been summarized by the discussed assumption is seen in figure 1.
Now in figure 2, we open the selective channel for Na. Since Na has a gradient potential, and there is also a potential difference, the positive Na flows into the left hand compartment.
In figure 3, an equilibrium is reached at an arbitrary value of -20mv. On average, no more Na will move to the left or the right, it's at equilibrium.
Figure 4, channels closed. Now this is where my understanding ends.

Apparently now, from what I gather, in figure 5 if we open the selective channel for orange (K+), since K+ has a gradient potential, it moves to the right. In the process, it takes its positive charge with it and restores more or less the original potential difference of -70mV on the left hand side.
But, if this is what happens, this must surely invalidate the marble analogy of the two atoms being the same. Because if they're the same, then whether we open the Na channel (which is already at equilibrium) or the K channel should make no difference, because once gain, they're the same atom with a different colour (except for being separately permeable via their selective channels).
The only difference that should be apparent is that if we leave both channels open, then the colours will eventually reach an equilibrium on both sides. But if an equilibrium is already reached for the charge (which it has in figures 3 and 4), I'm failing to grasp how opening then closing a different channel suddenly causes that equilibrium to change (as in figure 5 and 6). The explanation is the K potential gradient, but this is still the same marble as Na is it not? How come the potential difference changes?
Please direct me where my thinking has gone wrong, or if this is what even happens. And if it is, how is this possible? Many thanks.

There are a lot of different sub-questions embedded here. It might be prudent to try to split it into multiple posts.  For example, the mechanism of diffusion should really be a completely separate question from the requirement for both $\ce{Na+}$ and $\ce{K+}$ in action potentials.

I agree @Brian. Problem is though I thought I understood diffusion in terms of two atoms, as orthocresol correctly described, and yet in the neuron which I thought was just a biological application of this same system it seems actual charge transfer takes place (in depolarization/repolarization, absent the Na/K pump, which I imagine would play a trivial role in this discussion). My misunderstanding is lying either in the neuron or the diffusion, but I'm not sure which.

I'm insufficiently knowledgeable to say for sure, but I think the key aspect lies in the relative concentrations of $\ce{Na+}$ and $\ce{K+}$ in the system.  I'm pretty sure there's a lot more sodium than potassium, meaning that the overall potential is dominated by the instantaneous $\Delta C_{\ce{Na+}}$, but I haven't been able to turn up a good reference to confirm it.

Ah, nope - Molecular Cell Biology, 5th Ed, $\left[\ce{Na+}\right]$ and $\left[\ce{K+}\right]$ both around 150 mM in their high concentration zones in the resting state. But, I think it gives a good answer...I'll have to digest/relearn it a bit before I can write anything up, though.

I appreciate your time and help Brian. Specifically the phenomenon that I'm puzzled with is as per your first comment, why the relative concentrations (and their concentration gradients) lead to a different polarization state when their respective ion-channels open. If we treat Na+ and K+ as having the same physical/chemical properties, it seems they nevertheless reach separate equilibriums depending upon the charge state of the cell. Not sure why this is so? I think it may go back to the basics of diffusion.

The only way I could understand this currently is, after Na channels close, there is a net removal of Na/K from outside the cell, creating a negative potential difference, which when K channels open would then equilibrate. But nowhere have I seen this mentioned so I'm sure it's a more basic concept I'm not understanding.

On a high level, the differences in their movement and equilibria boil down to differences in the number and type of selective channels present for each ion, and how those channels respond to the dynamics of the ion movements and charge distribution. It's highly non-trivial. Will write it up as soon as I can, but it may be a day or three.

After sleeping on it, I think I won't write an answer after all. I recommend splitting the question (I've flagged for mod attention), leaving the diffusion mechanism portion here on Chem.SE but moving the action potential question to Bio.SE, as long as it is not a duplicate of another question there. Your having offered the bounty complicates this move, but hopefully a mod can make things happen smoothly.

It's much easier if you edit out the action potential portion and reask it on [Biology.se], though I think you'd find a few similar questions there already.  It's always best to tailor your questions for the target site.

Alright, I thought it was a fine line which target site to choose, as I noticed this site has a biochemistry tag, and the phenomenon probably dominated by diffusion and explainable via organic chemistry and electrochemistry. I'll edit it out later. To be honest all the biological sources, including my teachers, are unable to give a satisfactory explanation, leading me to question the pure chemical component itself. But the closest I can get to understanding that is facilitated diffusion, which doesn't explain the charge transfer.

I've now completely removed all references to biology and neuroscience and updated the question to better-fit this site. This will determine where my misunderstanding is lying.

Each ion has its own concentration gradient. The concentration gradient of K does not affect that of Na. The ONLY way that the diffusion of Na and K are linked in any way is via the electric potential difference between the LHS and RHS, which affects both Na and K equally. Btw, diffusion is an entirely statistical process.

In fig 5 - the electric potential difference makes Na and K both want to enter the cell, since the inside of the cell is more negatively charged. It is the concentration gradient of K - which, as you already know, exists mostly inside the cell (this concentration gradient is maintained by the Na/K pump) - that makes K want to exit the cell when the voltage-gated K channels are opened. These two factors, in this case oppose each other, and in this case, the concentration gradient wins out over the potential difference, leading to repolarisation.

In fig 4, if you opened the Na channels instead, nothing would happen. By virtue of the equilibrium that you established in fig 3, the concentration gradient of Na and the potential difference of -20mV cancel each other out precisely such that there is no net diffusion of Na.

Now just one last thing and please don't take it too personally because it's really not important, I just want to clear one thing up. Just to be nitpicky, the number -20 is not logical. During the action potential the membrane potential shoots up to about +30 mV. It has to be positive because again as you know, the concentration of Na outside the cell is much higher. Therefore the concentration gradient of Na makes Na want to enter the cell. For the potential difference to cancel out the concentration gradient, it has to force Na out of the cell, so it has to be positive.

Sorry, in my first comment above, I should have said - The concentration gradient of K does not directly affect that of Na, except for the fact that the electric potential difference itself is generated in part by the concentration gradient of K! which is what I was writing about in my answer a few days ago. If we were talking about neutral, uncharged species here, concentration gradients are entirely independent.

I've edited my answer. Hope it might help.

This answer has been edited quite a bit. New additions are put in blockquotes.

To avoid any potential (pun not intended) confusion, let me make one thing clear before I start: Any mention of "potential" in this answer refers to an electric potential, not a chemical potential. The definition of electric potential can be found here on Wikipedia.

Let's start off simple and say that we have a beaker with two compartments. The two compartments are separated by a membrane that is ONLY permeable to $\ce{Na+}$ ions. In the left compartment, we put 1 M of $\ce{K+}$ ions and in the right compartment, we put 1 M of $\ce{Na+}$ ions (with appropriate counterions). If diffusion of $\ce{Na+}$ entirely ignored the presence of $\ce{K+}$ as you say, then we would expect that after a while, we would have 1 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ on the LHS, and on the RHS we would have 0.5 M $\ce{Na+}$.
This does not happen. In actual fact, there would be a concentration gradient that favours net movement of $\ce{Na+}$ from RHS->LHS, and there would be a electric potential difference that favours net movement of $\ce{Na+}$ from LHS->RHS. We could model the concentration at any point in time with:

LHS - 1 M $\ce{K+}$ and $x$ M $\ce{Na+}$
RHS - $0.5 - x$ M $\ce{Na+}$

where $x$ is between 0 and 0.5.
When $x = 0$, there is no potential difference since both sides have the same amount of charge (overall both sides are electrically neutral). However, there is a steep concentration gradient, so $\ce{Na+}$ starts diffusing from RHS-> LHS, causing $x$ to increase. Now as $x$ increases, the concentration gradient becomes gentler and the potential difference starts to become larger. At one particular value of $x$, the concentration gradient will cancel out the potential difference and there will no longer be any net movement of $\ce{Na+}$ across the membrane. This is the point where we could say that diffusion has stopped.
Now, let me change the scenario and say that the membrane is permeable to BOTH $\ce{K+}$ and $\ce{Na+}$ ions. In that case, clearly the equilibrium concentrations will be 0.5 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ for both the LHS and RHS, because under these conditions there is neither a potential difference nor a concentration gradient.

It is important to note that the diffusion of $\ce{Na+}$ is NOT directly affected by the pre-existing concentration gradient of $\ce{K+}$. It is indirectly affected only because the concentration gradient of $\ce{K+}$ leads to a potential difference between the LHS and the RHS, which in turn affects the diffusion of $\ce{Na+}$. If we were to replace $\ce{K+}$ with an electrically neutral species, such as glucose, the concentration gradient of glucose would have absolutely zero effect on the diffusion of $\ce{Na+}$.

Now we can go on to explain the action potential in the neuron, which is not something I originally wanted to go into, but since the original question has been edited I feel that I have to address this now. Even though the original question now makes no reference to the action potential, it is describing exactly those events that occur during an action potential.
The inside and the outside of the neuron is separated by a cell membrane, which is selectively permeable to ions such as $\ce{Na+}$, $\ce{K+}$, $\ce{Ca^2+}$, and $\ce{Cl-}$. Of these four, sodium and potassium ions are the most important as they are directly involved in depolarisation and repolarisation when an action potential is generated.
The distribution of ions is not symmetrical across the membrane. When at rest, the extracellular concentration of $\ce{Na+}$ is much higher than the intracellular concentration of $\ce{Na+}$; the converse is true for $\ce{K+}$. Due to the membrane having different permeabilities to $\ce{Na+}$ and $\ce{K+}$ (for details of this, one can consult a neuroscience text), the resting membrane potential is -70 mV. Conventionally, the membrane potential is described as the intracellular potential with respect to the potential of the extracellular potential; this means that the inside of the cell is more negative.
Neurons react to certain stimuli by having $\ce{Na+}$ enter the cell, via various means which are not important to the current discussion. Since positively charged ions are entering the cell, this causes the membrane potential to rise. Once the membrane potential has risen to roughly -55 mV, voltage-gated $\ce{Na+}$ channels open, making the membrane permeable to $\ce{Na+}$.
At this stage, we have to consider the two factors that influence the diffusion of $\ce{Na+}$:

Since the potential difference is still negative, it favours influx of $\ce{Na+}$.
Since the extracellular concentration of $\ce{Na+}$ is larger than the intracellular concentration, the concentration gradient also favours influx of $\ce{Na+}$.

Hence more $\ce{Na+}$ enters the cell, leading to depolarisation of the membrane until the membrane potential reaches approximately +30 mV. At this stage, the voltage-gated $\ce{Na+}$ channels close and the voltage-gated $\ce{K+}$ channels open. Again, we need to consider the two factors that influence the diffusion of $\ce{K+}$:

The potential difference, which is now positive, favours efflux of $\ce{K+}$.
Since the intracellular concentration of $\ce{K+}$ is higher than the extracellular concentration, the concentration gradient also favours efflux of $\ce{K+}$.

Therefore $\ce{K+}$ leaves the cell, leading to repolarisation. There are further events such as hyperpolarisation and the refractory period, but they are not relevant to this discussion. After all this is said and done, the concentration gradients of both $\ce{Na+}$ and $\ce{K+}$ are not the same as they were at the very start of this discussion. They are restored by the $\ce{Na+}$/$\ce{K+}$ pump.
The most important thing to takeaway from this is that the diffusion of species X across a membrane is affected by two things: 1) the concentration gradient of X; 2) the potential difference across the membrane, if and only if X is a charged species. Other concentration gradients of other species, such as Y, Z etc. do not directly affect the diffusion of species X unless they contribute to the potential difference.
Prior to my edit you might have seen that I wrote that "diffusion does not ignore similarly charged species". This might seem to be a contradiction with what I say now, that the concentration gradient of Y does not affect the concentration gradient of X. The answer is that the concentration gradient of Y can only affect that of X when both X and Y are charged species, and this is only an indirect effect, in that the concentration gradient of Y affects the electric potential difference which in turn affects the concentration gradient of X. If Y was uncharged, its concentration gradient would not affect the potential difference; if X was uncharged, the potential difference would not affect its concentration gradient.
I think the marble analogy is inherently flawed in that there is no real analogous concept to a potential difference. You could modify it to have some kind of analogous concept, but in my opinion it wouldn't make sense any more.

After I finished typing out everything I reread your question and I think this might not really answer your question, so I apologise if this is the case. I think maybe my comment on the other answer might be more useful...

That's really helpful orthocresol, thank you. But just two points I'm not getting. First, if side A has 1M K, side b 1M Na, and a semi-permeable membrane for Na, wouldn't both these sides be in equilibrium in terms of their number of particles (atoms) ? If any Na did move to the left, wouldn't this break this equilibrium of 1M? I think there's something simple I'm not seeing in what makes the concentration gradient have potential like that. I assume this would also explain why in the neuron, if the Na channel is open we have an influx of Na, but the moment we open the K channel an efflux of K.

@user4779 The equilibrium that is established is Na (left) <=> Na (right), not for total particles. I could dump 1 M glucose and 1 M urea in the LHS and since both species are uncharged it wouldn't affect the diffusion of Na at all.

Really helpful explanation in the edit, thanks. So I see now that each atom does in fact have its own unique concentration gradient, provided it has no charge. This brings me to my original question. How could this be possible? Since there are so many more atoms, as you say, if we dump moles upon moles of some similar neutral atoms on one side, wouldn't those collisions near the membrane prevent the atoms on the other side from crossing? And even if not completely, surely moreso than if they weren't there in the first place?

If we have 10 marbles on side A, 10 marbles on side B different colours, shake them and they will be at equilibrium with enough trials, I'm fine with that. But put 1000 marbles on side B and 10 marbles on side A, how could they even get into side B? Is there some other analogy that could better describe this?

@user4779 yeah, theoretically that would have an effect, but I am not sure if it is possible to dissolve so many things in water such that that effect is non-negligible!

,I think I might be finally getting it. Let's say hypothetically, the channel was "one way" only into the cell. Every time K entered, it couldn't leave. Would this mean that we could keep adding K, and it would always increase in concentration inside the cell, because the chance of K leaving is now negated? Or would the potential difference prevent this from happening at some point, same as if the channel was two ways? Would this net concentration of K nevertheless be higher or the same inside the cell (as compared to the two-way system)?

Or in essence, even if we replaced all the Na inside the cell with K (in the one way system), this would make no difference. In effect the atoms would be treated the same.

@user4779 you're absolutely right. On top of that, as the extracellular concentration of K+ decreases, the rate of entry would also decrease. "even if we replaced all the Na inside the cell with K (in the one way system), this would make no difference" - yep!

Fantastic! Thanks for all your help. I finally get it now. The reason that each ion has its own concentration gradient is because, when one of the semipermeable channels opens, ions of that type are able to leave as well as enter. If their concentration is lower however, the equilibrium will still favour their entry overall due to probability of entry and exit (disregarding effects of potential difference).

Each atom does its own diffusion across the membrane. There is a probability that a $\ce{K+}$ or $\ce{Na+}$ ion will move in the "forward" or the "reverse" direction. The probability of getting through the membrane may differ between the types of ions (e.g. if the membrane pores allow one species through more easily than another because of ionic size), but, barring any external force, eventually each type of ion should come into equilibrium, with as many going forward as backward.
Imagine a box of marbles of two colors, with a partial separator between, and that the box is shaken. It is simple chance that a marble crosses the separator, and if many marbles of one color are on one side, there are fewer to migrate backward from the other.
See this online demo of diffusion across a semipermeable membrane.

So, this is what I"m failing to grasp. In a biological membrane of a neuron for example, if a sodium channel is open, the sodium ions will rush in. Then if it shuts and the potassium ion channel opens, the potassium ions will rush out. I can appreciate the channels being selective, but why is this particular behaviour observed? My understanding is because there is a lower concentration of Na inside the cell, lower concentration of K outside the cell. But why would the Na go in if there's already so much K in there? How can it differentiate between K and Na?

And if it's just a different colour of the same marble to use the analogy, wouldn't the marbles already be in equilibrium, and the colour that moves is then just random? If this is the case, how come a gradient potential exists for that specific atom then?

Oh, just to directly answer your question "But why would the Na go in if there's already so much K in there? How can it differentiate between K and Na?" - it's actually just all about statistics. Let's say you and I sit across a table and you have 20 red marbles and I have 20 blue marbles. We blindfold ourselves and spend the next 1 hour exchanging random marbles. At the end of the 1 hour we take off our blindfolds and guess what the most likely outcome is? Of course - the most likely outcome is that each of us have 10 blue and 10 red marbles.

Of course, there is a chance that maybe you have 15 blue and 5 red, and by implication I have 15 red and 5 blue. That's always possible and if we kept the blindfolds on, there would be no way for us to know that that happened. It's just that statistically speaking it's more likely that we have nearly equal numbers. The same is true for the neuron - where the numbers are much, much larger than 20. The entire concept of entropy is actually just statistics and probability. @user4779 I should have tagged you in the earlier comment to alert you, sorry.

Just regarding the marble analogy, so the net number of particles remain the same. But in the neuron for example, the depolarization will release a lot of Na into the cell. And then release K in the repolarization. If K and Na are just different colours, then is this to say that if we hypothetically left the Na channel open, depolarization would also occur with Na items then moving out the membrane just like K would through its channel? Since they'd be interchangeable?

@user4779 Not entirely, but partially. Because the concentration of Na+ outside the cell is larger, diffusion of Na+ would not be as fast as that of K+.

But they're still the same charge right? If it's just statistical probability that K will equilibrate with its concentration outside, then we could take any single K atom, and any single Na atom at a point in time (let's assume both their channels are open) and regardless of the concentrations, they have the same probability of crossing the membrane right? Since there's no actual force that makes K or Na more likely to move, just statistical probability.  And yet Na enters until the membrane potential is greatly raised, and K leaves until its greatly lowered, causing actual charge difference?


text:
Oxidation of Carbon atom in Alkene, Alkyne and Alkanes of functional groups?

Nomenclature priorities are in order of the degree of oxidation of the
  carbon atom.

I am not sure if it's advisable to just take it as it is and use it or try to understand the basis of the functional group prioritization. I guess the latter interest me. So looking at the priorities for functional groups,
Alkene has higher priority than Alkyne and followed by Alkane. I started reading some materials about Oxidative Cleavage of Double Bonds and rest of the reactions, which kinda went way over my head.
Can someone explain, how and why the carbon oxidation in Alkynes are lower than Alknenes? Or put in another word, why would Alkene carbon oxidate faster than Alknenes?

@bon why did you remove the functional group and hydrocarbon tag? Just curious to know.

@Have a look at the meta post about functional group tags.

There are two separate questions in the last paragraph. They are not the same question and I think the first is the one which is relevant to the rest of the post.

Nomenclature priorities are in order of the degree of oxidation of the carbon atom.

This may be true for the heteroatom functional groups, but not for alkenes, alkynes, and alkanes.
Consider that each $\ce{C-H}$ bond contributes $-1$ to the oxidation state and each $\ce{C-C}$ bon regardless of order contributes $0$ to the oxidation state of carbon.
Thus:
hydrocarbon   formula   functional group   oxidation number of C
ethane        CH3CH3    alkane             -3
ethene        CH2CH2    alkene             -2
ethyne        CHCH      alkyne             -1

Alkynes have a higher oxidation number than alkenes.

Alkenes react better in oxidation reactions than alkynes because alkenes are less oxidized than alkynes.
The organic chemistry textbook that I teach from explains that alkenes have higher nomenclature priority than alkynes because alkenes are more common. I cannot find a more authoritative reason/source.

The question that drives me a little confusing is, the order is: alkene, alkyne and alkane... so from the perspective of single, double, tripple bond, I really do not see an order at all... You have stated because alkenes are less oxidized than alkynes. Well how important to understand the underlying oxidation theory of these guys, from biochemistry and pharmacology aspect?


text:
Which mechanics or law is predominately seen on the nanoscale?
Nanotechnology and sciences are the study or the manipulation of matter on an atomic, molecular, and supramolecular scale especially at nano meter range. We know that at macro molecular level a system can be described using Newtonian/classical mechanics (eg: $F=ma$) whereas for an electronic level quantum mechanics are used. In nanotechnology and science which mechanics is seen predominantly; is it Newtonian, quantum or both and why?

Seems like a nice essay topic for HW... what are your thoughts?

I dont know whether this answer is correct but after searching a while came across this site and its about planck length. the scale at which classical mechanics become invalid and quantum mechnaics comes into effect.  So we can confirm that at nano scale its between Newtonian mechanics or an intermediate of newton-quantm mechanics

Quantum mechanics dominates by the very definition of the nanoscale. "Nano" is essentially just a buzz-word for "quantum". Classical methods can be used for supramolecular aggregates, but just as a rather crude approximation and only for the reason that the application of quantum mechanics to such relatively big systems is yet too expansive.

Its depence on the size of the system your are looking at. To understand spectroscopy of molecules, you usually use quantum mechanics. In molecular dynamics simulation, you'll need both: On the one hand, there is a potential needed based on QM-interactions, on the other hand, the classical equations of motion are good enought to move your atoms in a QM-force field.
Based on your problem, you have to decide how much classical mechanics you can justify. But in the nanoscale everything is made up with quantum mechanics and classical mechanics are used to approach a problem, which is not solvable in quantum mechanic way in an accetable periode of time.


text:
Why ginger reacts with sparkling water?
I think that the ginger ale is born this way.
Anyway, if you put the ginger in a glass and add sparkling water, small drops splash out of the glass, the water is fizzier, more effervescent.
Why? what makes the two of them react?

Just to clarify, are you adding ginger ale to the water or ginger (powdered, root or otherwise)?

It doesn't react chemically, but probably acts as nucleating agent for CO2

I'd agree with @Mithoron that the ginger acts as a set of nucleation points for the gas bubbles. Adding raisins, almonds, .. pretty much anything somewhat rough will do very similar things.

@bon, i'm putting pieces of the root in the glass and than I add sparkling water.

Anyway, thanks to @Mithoron and Geoff for the explenation!! So Anything rough will give this "side" effect?

According to this Wikipedia article:

Carbonated water (also known as club soda, soda water, sparkling water, seltzer water, or fizzy water) is water into which carbon dioxide gas under pressure has been dissolved. Some of these have additives, such as sodium chloride, sodium bicarbonate or similar, but seltzer water is almost always composed of water and carbon dioxide with no other additives.

So the only possible candidates for a chemical reaction from the sparkling water are $\ce{CO2}$, and water itself. The other salts present at low levels would not have sufficient reactivity to produce the relatively high-energy results observed. And water can be ruled out as a reactant because ginger itself of course is a plant that contains water (or contained water previously if dried ginger was used). For the sake of thoroughness, I went to my kitchen and added tap water to dried, ground ginger and observed no effervescence.
This leaves $\ce{CO2}$ as the only candidate for causing the observed reaction upon the addition of ginger to sparkling water. A chemical reaction can be ruled out from a couple perspectives. First, as $\ce{CO2}$ is a very stable, completely oxidized compound, ginger would have to contain some strong reducing agent for it to react. We know this is not the case as the ginger is produced and stored in an oxidizing environment (our atmosphere of $\pu{21\%}$ oxygen). The atmosphere also contains on the order of $\pu{0.04\%}$ $\ce{CO2}$, which is less than that found in sparkling water but would still have quenched any $\ce{CO2}$ reactive component, albeit in slower less dramatic fashion.
At this point we've determined the effervescence must be a result of the $\ce{CO2}$ in the sparkling water, but we've eliminated the possibility of a chemical reaction. Because the carbonation process leaves the sparkling water supersaturated in $\ce{CO2}$, the only thing preventing the $\ce{CO2}$ from bursting out all at once is the lack of sites for $\ce{CO2}$ bubbles to nucleate. And this is what the ginger provides, particularly if it were in a ground, high-surface area form, but even in the case of raw ginger slices there are many more surface sites available than a glass container.
Bottom line, TL/DR: There is no chemical reaction, rather the ginger just provided a relatively high surface area for the rapid formation of $\ce{CO2}$ bubbles from the $\ce{CO2}$ supersaturated sparkling water.


text:
Calculation of fluorescence quantum yield
The fluorescence quantum yield of a sample $\Phi$ is defined as the fraction of photons emitted per number of photons absorbed. The quantum yield is often calculated by comparing a sample with a reference substance of known quantum yield $\Phi_\mathrm{r}$ such that
$$
\Phi=\Phi_\mathrm{r}\times\frac{n^2}{n_\mathrm{r}^2}\times\frac{I}{I_\mathrm{r}}\times\frac{A_\mathrm{r}}{A},
$$
where $n$ ($n_\mathrm{r}$) is the refractive index of the sample (reference), $I$ ($I_\mathrm{r}$) is the integrated emission of the sample (reference), and $A$ ($A_\mathrm{r}$) is the absorbance of the sample (reference) at the excitation wavelength (see here or here for details).
It seems odd to use the integrated emission because it is not proportional to the number of photons emitted unless the emission peak is extremely narrow. In particular, the integrated emission is
$$
I=\int_{\lambda_1}^{\lambda_2}\mathrm{d}\lambda\,I_\lambda,
$$
where $I_\lambda$ is the spectral intensity. Because the number of photons for a given energy changes with wavelength ($E=hc/\lambda$) using a quantity that is proportional to the number of photons seems more sensible. In particular,
$$
I_\lambda= n_\lambda \frac{hc}{\lambda},
$$
where $n_\lambda$ is the photon density at wavelength $\lambda$. Consider the quantity
$$
N=\int_{\lambda_1}^{\lambda_2}\mathrm{d}\lambda\, \frac{I_\lambda \lambda}{hc}.
$$
It is directly proportional to the number of photons even if the emission peak is broad. Why is $I$ used instead of $N$?

After some more reading, it turns out that the protocols suggesting to use the integrated intensity are wrong and it is necessary to correct for the variable wavelength across the emission peak. Würth et al. (2013) provide a description of the correct method. First, they state that

all fluorescence spectra have to be converted to a (relative) number of photons per unit time

Second, they describe the conversion as discussed in the question:

the radiometric quantity has to be multiplied with $\lambda/(hc_0)$

Its probably not important now after a year of asking , but you should convert your spectra to be in wavenumbers and try to get a standard that emits in the same region as your compound because detectors etc have a wavelength dependence.


text:
Which substance has the highest temperature range between melting and boiling point
Which substances exist that are normally liquid and that will not freeze nor boil even at relatively low/high temperatures as compared to the freezing and boiling points of water? Or how can I identify such substances online?
If it is relatively non-reactive, non-toxic, cheap and has high thermal conductivity that is also a bonus!

What do you exactly mean by $\ldots$ such substances online?

I mean, is there a tool somplace online that I can use to look up and sort substances by their thermal properties son I don't have to ask silly novice questions here.

Though not entirely adequate for your purposes, one of the substances (if not the substance) with the highest temperature range between melting and boiling at ambient pressure is gallium, which melts at 30°C and boils at 2400°C. That's an amazingly huge liquid range of 2370°C! Alloys with gallium can be made to melt below 0°C, I'm pretty sure.

What temperature range are you looking for? and what application? More info on what you want to achieve would be helpful.

@steverill This started as a crazy thought experiment if it would be possible to build an universal "constant temperature source". I would have a cooler and a heater operating on a circulating liquid and a digitally controlled mixing valve for transferring the constant temperature to the desired location.

@Lennard Rolland Interesting. We had something similar in a lab I worked in, basically a refrigerated bath with a heating element and a circulation pump in it, -40C to +100C, alcohol/water mixtures. BTW the easiest way to achieve constant temperature without instruments is to keep a fluid at its boiling temperature. for example water boils at 100C at atmospheric pressure. Use a mechanical valve to increase the pressure to 50 bar, and the boiling point rises to 265C. Handy if you have many heat exchangers on a common circuit and want to keep perfect distribution. Perhaps not for amateurs though

The polar aprotic solvents (for example dimethylformamide, mp −61 °C, bp 153 °C, or hexamethylphosphoramide, mp 7 °C, bp 230–232 °C) would be a place to start. Silicone oil is often used in heating baths – one product in the Aldrich catalog is advertised as having a working range of −40 °C to +230 °C.

Gallium melts at 30 °C but doesn't boil until 2200 °C.  If 30 °C is a bit too warm to count as "room temperature" or "normally" for you, I found an old paper that recommends tetralkyl silanes such as tetradodecyl silane as lubricants that are liquid over very wide temperatures.
Addendum:
Dowtherm A is a eutectic mixture of biphenyl and diphenyl ether.  According to its manufacturer:

These compounds have practically
  the same vapor pressures, so the
  mixture can be handled as if it were
  a single compound. DOWTHERM A
  fluid may be used in systems employing
  either liquid phase or vapor phase
  heating. Its normal application range
  is 60°F to 750°F (15°C to 400°C),
  and its pressure range is from
  atmospheric to 152.5 psig (10.6 bar).

[...]

The viscosity of DOWTHERM A
  fluid is low and changes only
  slightly between the melting point
  of the product and its top operating
  temperature.

Alloying can get the melting temperature down, at the expense of the boling point.

A good place for a general list is http://www.engineeringtoolbox.com/melting-boiling-temperatures-d_390.html
For high temperatures, here is a list of alloys https://en.wikipedia.org/wiki/Fusible_alloy Some of these have melting points below 0 °C and boiling points as high as you will ever need. Some are pretty expensive, though. Note that in general mixtures have a lower melting point than pure substances. The mixture whose percentage composition has the lowest melting point is called a Eutectic.
As a pure metal, tin is a good choice as it has low toxicity and a wide range. Hence it is used in the float glass process for window glass. Even though it is not particularly reactive, there is some potential for an explosive reaction of molten tin with water/steam if used in large quantities, partly due to the liberation of hydrogen.
The solar energy industry has used sodium and potassium nitrate (and previously sulfuric acid) but both are corrosive. Additionally, nitrates may be difficult to get hold of outside the agricultural industry in some countries as they are used in making explosives.
Then there are the organics. Silicone oil and Dowtherm A have been mentioned. Neither is particularly good for you. Cooking oil doesn’t have a bad range and is preferable to Diesel from both toxicity and flammability points of view. Among the alcohols, ethanol and methanol are probably too volatile for your use. The glycols (diols) are used as antifreeze additives with water. Ethylene glycol is quite toxic. Propylene glycol is much less toxic. Glycerol (a triol) is also nontoxic and higher boiling than propylene glycol, but it is highly viscous which will negatively impact its heat transfer properties.
It should be remembered that non-toxic organics can decompose into toxic ones, particularly when the heat flux applied to them is too high (as can happen with electric heating elements.)

Fun fact: Propylene glycol is much less toxic than ethylene glycol because the latter is metabolized to oxalic acid, whereas the former is metabolized to lactic acid, one of the most common substances in the body.

Galinstan is a substance similar to mercury in its basic properties, but has a liquid range at atmospheric pressure which is much larger than that of mercury's. It melts at −19 °C and boils at 1300 °C.
At the same time, it also has good thermal conductivity (16 W/(m K)) and is relatively non-toxic and unreactive.

From Wikipedia:
\begin{array}{c|c|c|c}
 \mathrm{Element}  &   \text{MP in K} &  \text{BP in K} &\Delta T \text{ in K} \\\hline
\mathrm{Neptunium} &  917 &4\,273 &3\,356\\
\mathrm{Uranium}   & 1\,405 &4\,404 &2\,999\\
\mathrm{Thorium} &  2\,115& 5\,061& 2\,946\\
\mathrm{Cerium}  &  1\,068& 3\,716& 2\,648\\
\mathrm{Plutonium} &  913& 3\,501& 2\,589\\
\mathrm{Praseodymium} &1\,208& 3\,793& 2\,585\\
\mathrm{Zirconium}& 2\,128& 4\,682& 2\,554\\
\mathrm{Lanthanum}& 1\,193& 3\,737& 2\,544\\
\mathrm{Protactinium}& 1\,841& 4\,300& 2\,459\\
\mathrm{Tantalum}&  3\,290& 5\,731& 2\,441
\end{array}
And that is just elements (top 10 delta).

These materials are not liquid at RTP.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.

@M.A.Ramezani This does indeed provide an answer to the question in the title.  But not to the question in the body of the question text.  It's hardly the respondent's fault if the OP asked two different questions.

@DavidWallace I disagree. I do think that it is the answerers responsibility to read the question completely.


text:
Can I successfully use a humidifier to distribute caffeine to wake someone up without the usual morning drowsiness?
Would I have any success putting coffee in a humidifier? Or perhaps water mixed with caffeine powder?
So I could wake up without a jolting alarm clock, feeling like I already had a cup of coffee?
I understand this will have to be a carefully controlled dose if working with caffeine powder, but is it feasible?

You would have to put the humidifier on a timer, so it starts about an hour before you wake (otherwise you might not sleep much). I don't know what the absorption of caffeine by inhalation is, but it might be quite low. The other problem is that you would end up with a lot of powder spread all over the room.

The are devices used to treat sleep apnea that are humidifier-like, but utilize a mask. You could look into that. It would solve he problem that @LDC3 mentioned of ending up with caffeine powder everywhere.

+1 for the weird question :D. I suspect that it won't work though as mentioned by others.

It looks like at least one person/organization has thought of this and patented a delivery mechanism ("Delivery of caffeine through an inhalation route").
That said, you might be able to couple the system outline in the patent documentation with an alarm clock of sorts to trigger the release at a specified time. I'd certainly buy one if the price was right.
Edit: The inhalation part has been investigated. As far as I can tell, the business about administering caffeine to a sleeping person and then having them awaken into a fully alert (or pseudocaffeinated) state has not.

I had a feeling that someone has jumped on this before!

If you scroll to the bottom of the page I linked, there are relevant patent and non-patent citations - there appears to be no shortage of investigation into this sort of thing!

Others have covered the practical parts a bit, so let's move on to the pharmacology of it.

Dosage

In a cup of coffee you can get ~120mg or more of caffeine. That quantity isn't really large so filling the whole room with caffeine droplets / powder could be practical. The body absorbs most of caffeine, whether given orally or by inhalation so it should be fine either way.

Rate of delivery

Here we may need to do some guessing. Let's say we want to deliver the 1 cup of coffee equivalent of caffeine (abbreviated cec as I made it up) over 30 minutes. Assume the room is sufficiently large and the air inside the room is well mixed, so the rate of delivery is mostly constant.
(inhalation is much faster than oral absorption, and we don't want to overdose our test subject, so a sufficiently slow delivery is necessary. Say the subject stayed in the room for 30 minutes after release of caffeine, the person will be dosed 1 cec instead of 6 if the delivery rate was 1 cec/5 min. Slower delivery rate gives better safety margin)

Pulmonary delivery

On average human pushes 5 L/min of air through the lungs. Pulmonary delivery of drugs could be much more complicated tho. If the caffeine particles are too large, most of it would just hit the airways and be trapped within the mucus. This greatly reduce the rate of delivery (but much less so for the total quantity delivered, as the mucus will likely get swallowed as it is swept towards the throat).
In normal design of pulmonary delivery, we want the particle size to be within a small range, as too small particles don't have enough time to settle and get absorbed. While the data available are mostly not applicable as we tell patients to hold breath when using normal inhalers, you may still want to maintain the particle size somewhere within the optimal deposition range for pulmonary delivery (somewhere between 2 and 5 micron).

Chemistry

With some guessing / calculation, you may want to have ~0.8 mg caffeine (free base) /L in the room. Now we commonly use caffeine as a salt, e.g. citrate. Caffeine is only weakly basic but ideally you should use a mix of salt and free base to make a more or less neutral product, as you're going to breath that every single day. This is mostly not important but I just want to mention it. So let's skip the calculation and assume we have 1 mg/L of caffeine in the room.

Indoor air quality (IAQ)

Let's assume all particle are 2 micron diameter, perfect sphere. That means they are around 4 μm3 in size. Caffeine density is 1.23g/cm3, so each particle weights about 5e-9 mg. To achieve 1 mg/L of caffeine in air, you want 200,000,000 particles in a liter of air.
Using Clean room as comparison is obviously not appropriate, but this value is much higher than FED STD 209E room air equivalent (or ISO 14644-1 class 9). From where I live in the IAQ standard is less than 180 ug/m3 of respirable particulates. Caffeine delivery obviously introduced a lot of particulates, potentially making the room hazy.

Glossary

You better ensure that the particulate generator, caffeine / salt / solution etc. that you're using is sterile and free of any other contaminants. Afterall, the lung is not designed for absorption of solid matters, and some carefully designed drugs turns out to up cancer risk by just giving insulin through the lungs.

Conclusion:

Your scientists were so preoccupied with whether or not they could,
  they didn’t stop to think if they should.

Jurassic Park