text:
Spins spin coupling on proton NMR for an aldehyde?
On a proton NMR, does the hydrogen in the CHO group cause coupling, or get coupled itself? Specifically, I a considering 2-methylpropanal. I cannot decide whether the H in the aldehyde group would be split into a doublet or remain as a singlet, and I am unsure as to whether the single proton on carbon number two would be a heptet (from the 6 adjacent protons), an octet ( from the 6 protons plus the proton in the aldehyde group), or if it would be split into some combination of a heptet of doublets, or a doublet of heptets...
Any advice on dealing with spin spin coupling in proton NMR would be much appreciated!

I cannot decide whether the H in the aldehyde group would be split into a doublet or remain as a singlet […]

What would be the argument against $^3J$ coupling between these protons? The aldehyde proton will appear as a doublet.
Don't forget that coupling works both ways! $\ce{(CH3)2C\color\red{H}-}$ will (at least) split to a doublet too. Coupling with this proton will also  split the signal for the methyl protons to a doublet.
What does this mean for the proton at C-2?
Formally, a doublet (from coupling with the aldehyde proton) of a septet (from coupling with six magnetically equivalent methyl protons) is to be expected.
In really, you will observe a smaller number of lines! I suggest to try this with a pen on squared paper to figure out what happens when the two coupling constants are identical or one is twice as big as the other.


text:
Atomic number $Z$
If electrons were spin-$\frac {3}{2}$
instead of spin-$\frac {1}{2}$ ,
what would be the atomic
number $Z$ for the first noble gas ?

The spin multiplicity will be $2S+1=4$. This means that the first atomic orbital can hold $4$ electrons with four different quantum numbers.
So, the atomic number for the first noble gas is $4$.

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text:
Titration of C₄H₆O₂
$\ce{C4H6O2}$ (10.75 g) was dissolved in water (250 cm3) and then a 20 cm3 sample was titrated against an alkali metal hydroxide solution. This required 17.35 cm3 of the alkali. What was the concentration of the alkali?
The water was evaporated from the alkali solution (100 cm3) to leave a white solid (3.23 g). Identify the alkali metal present.
My working already is that I started by dividing 10.75g by 0.25dm3 to get 43g/dm3. Then divided 43g/dm3 by the relative formula mass of 86 to get 0.5mol/dm3. I did 0.5mol/dm3*0.01735dm3 to get 0.008675mol then divided that by 0.02dm3 to get 0.0434 mol/dm3 as the concentration of the alkali. I'm not sure how to proceed from there or know what I have already done is correct.

What have you tried and where specifically are you experiencing difficulty.  Note, that we are not a homework help site.

I started by dividing to 10.75 by 0.25 to get 43g/dm3. Then divided 43 by the relative formula mass of 86 to get 0.5mol/dm3. I did 0.5*0.01735 to get 0.008675mol then divided that by 0.02 to get 0.0434 moldm3 as the concentration of the alkali. I'm not sure how to proceed from there or know what I have already done is correct.

@Zara Edit the question to include your calculations that you have shown in the above comment.

Attach the units with your numbers. It makes it easier to spot errors.

My working already is that I started by dividing $10.75g$ by $0.25dm^3$ to get $43g/dm^3$. Then divided $43g/dm^3$ by the relative formula mass of $86 g/mol$to get $0.5mol/dm^3$.

Up to here, you have it correct. The next step is to take $20 mL$ from $250 mL$, which would decrease the amount of acid to titrate. Then you would find the concentration of the base.

I see my mistake, so 0.5mol/dm3 * 0.02dm3= 0.01mol of acid. Then 0.01mol/0.01735dm3 gives 0.576mol/dm3, the concentration of the alkali.

So now you go on to the next part. How many moles of salt are in the precipitate and determine the molecule mass from 3.23g of solid.

How do you find the mol of the salt, wouldn't it be 0.01mol?

Yes, that's correct.

So I divided 3.23g by 0.01mol to obtain the rfm, but that seems wrong as the answer is 323 which seems too high

of the product of the reaction? its a solution of 100 cm3 which evaporates to 3.23g of solid. there was 43 g/dm3 originally

My mistake; I misread the question. The solid (3.23 g) comes from 100 mL of just the alkali solution (I thought it was the salt that was formed). So the solid (3.23 g) comes from 100 mL of 0.576M solution. How would you find the molecule mass of the substance?

so 0.576M * 0.1dm3 gives 0.0576mol, 3.23g/0.0576mol gives 56

Good, now identify the metal ion from the molecular mass.

Wouldn't the product be C4H5O2 and the metal ion, but without the metal ion it already has an RFM of 85

C4H5O2 and the metal ion is in the other solution that you titrated. The 56 g/mol is from the alkali solution.

That's what I got.

I would suggest to solve a titration question by equating the meq of acids and base for a neutral condition.
$$moles \space of \space \ce{C4H6O2}=0.5=\frac{10.75*1000}{250*86}$$
 $$meq \space of \space \ce{C4H6O2}= 500$$
$$meq \space of \space \ce{C4H6O2} in \space 0.02 l=500*0.02=10$$
$$\therefore$$
$$meq \space of \space alkali= 10$$
The alkali is sodium, as it give a white ppt and the base was NaOH
$$moles \space of \space \ce{NaOH}=0.01=conc*volume$$
$$conc \space of \space \ce{NaOH}=\frac{0.01}{17.35}*1000=0.576$$
$$\therefore$$
The concentration is 0.576M

Thank you, that makes a lot more sense now, do you have any idea how to identify the alkali metal as I have no idea where to start

Your first line has a mistake. You are not finding the moles of acid, but the concentration of the acid. Please attach units and you will see that it is a concentration, not moles..

Yeah it's the molar concentration or the number or moles in 1l don't worry it's perfectly correct

The numbers are correct, but without the units, it's wrong.

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text:
Can we prevent metal corrosion by creating a vacuum around it?
I have a device in a box which has parts of metal. So can I prevent corrosion by creating a vacuum around the metallic object?

The word corrosion has a range of meanings, but it is generally taken to indicate the chemical reaction of a material with its environment. For example iron corrodes by reacting with oxygen in the air (or water if it's underwater).
If you put the metal in a vacuum then it doesn't have anything around it to react with, so in this case yes putting the metal in a vacuum will stop it corroding.


text:
Contact lenses come in two polymer classes, ionic or non-ionic. Whats the significance of having both?
As a contact lenses user, I was looking for information about why manufacturers produce soft contacts using polymers from both classes. Common knowledge would suggest that something placed in contact over the eyes for corrective vision should be inert. Ionic usually means more reactive. So why do they even make ionic contacts?

I have no knowledge on this topic, but am a fellow contact lens user. What brands offer ionic vs. non-ionic lenses?

I just edited my question taking your comment in consideration .

I don't know, if you had some specific question about the properties of these polymers or something, you might get some good answers, but I'm not sure the question of what is "better" for contact lenses is on topic here.

'Better' means 'more healthy'. I would like to be an informed buyer. This information is not available for the consumer. I just edited my question taking your comment in consideration .

Um, unfortunately, now your edit renders this question off-topic. We can't safely answer questions related to health. That's what doctors are for.


text:
How to define custom force in quantum chemistry packages?
I don't know if this is really applicable but I want to add a custom force (say, $F(x,y,z)$) to some atoms in my geometry optimization. Is there any way to add such forces to the DFT or molecular mechanics calculations? (preferably GAMESS or Gaussian for QC or Avogadro for MM).
It can be also done as Molecular Mechanics (MM) as I just need an approximate geometry after applying the force.
Let me give an example. Say I have molecule A and molecule B and a force that changes in space as $F(x,y,z)$ but is only applied to molecule B. Now I want to calculate the new geometry of A and B.
I can think this is like adding a $V_F$ to the external energy in the Schrödinger equation, but I don't know how to do that in QC packages:
$$\left[-\frac{1}{2}∇^2 + V(r) +\color{red}{V_F(r)}\right]ψ(r) = Eψ(r)$$
Update:  The idea of making a unit cell and making shrinkage on it looks great (suggested by Geoff Hutchinson here), but the problem is I couldn't figure out how to install CP2K on MacOS and also I am looking for something with a simple interface like Avogadro. I even tried Maestro by Schrödinger software, but couldn't make unit cell shrinkage for that.

Can you give a bit of insight into why you need the custom force - what you're trying to do here?

@GeoffHutchison I tried it but as I said I am not familiar with Linux command line. It needs SVN which is already a big issue to get it work and then its installation is another problem itself.  On the other hand I can not see any extension in Avogadro that can make an input file for CP2K.

Are you wanting to apply forces to the electron cloud, or just to the nuclei? Or, both?

The short answer is not. You cannot define a constant force for an atom or set of atoms in a system.
If you still want to modify force constants, this is the way to do it:

Gaussian, you can modify through the keyword GEOM(IHarmonic=n), where n is part of: $\frac{n}{10^6}$ which indicates the respective force applied to the system. Its units are $\frac{Hartree}{Bohr^2}$. To more info, please refer to this link.
If you want to introduce a specific force to an atom or set of atoms in MM, the best way to do it is by parameterizing your own system into your desire environment. The most important part is that you have to have in mind that the right part of the principal ec. of any MM describes the parameters of the specific force field of your desire system or set of systems under specific conditions (i.e. aminoacids in water like OPLS)

$$\vec{F}=\frac{\partial V}{\partial x}$$

Moreover, when you talk about forces in QM you are talking about the forces acting on electrons. Not on atoms.
But, if you still want to know a way to know how the forces of electrons (From DFT) and, at the same time, nuclei are, I recommend you Car-Parrinello. Please, refer to this link
If you have a particularly interest on simulating systems in QM with specific forces acting on electrons, the best way to do it (As far as I know) is to develop your own density functional.

Update
I Gaussian input file was requested to figure out how the keyword and input is used: Geom(iharmonic=n). I did not find the respective file, instead, I give a try twice by crating two input files with different values of n. The used Gaussian input files were:
chk=Test.chk
#p opt b3lyp/6-31g geom(iharmonic=n)

Test

0 1

H
H 1 0.9
O 2 1.2 1 120.

and the used values of n were 2 (On the left) and 20 (On the right). The respective Energy and RMS profile of both simulations are:

Also, in Gamess, you can use these keywords:  Statpt, Force and its related keywords like Zmt. More info, Please see this link at page 2-121 and 2-131.

I may want to give a try to IHarmonic concept.hartree/bohr^2 dimension is energy/dimension^2 = force/dimension. It is not force and that's pressure. If you look at my comments with Geoff, pressure is exactly what I want to apply to the system. Can you give a little more explanation or even better a worked example of such Gaussian input file please?

@Aug, thank you for commenting my post, please see the update of my post and have a good day :)

Wow !  That's really amazing !   I have just a little confusion about the physical meaning of this initial harmonic.  Is this equivalent to the external pressure ( as this external pressure is what I want to deal with ) ? If so, as I see 1 hartree/bohr^2= 1557 Pa. By setting n=20, we are dealing with 20 x1556x 10^-6= 0.031 Pa which has not really a physical justified meaning ( 1 atmosphere is 101325 Pa ) . It means even at 1 atm pressure we need to set n= 6.4 x 10^7 !! which is not really applicable .  Is there anything I am doing wrong here ?

First, Pascal is not a quantum unit. Second, you can find a description of the relationship between pressure, force and harmonics (From the quantum point of view) in here: http://physics.stackexchange.com/questions/147128/pressure-in-harmonic-oscillation  See at the largest answer

I thought this initial harmonic subject is worth to have a separate discussion.  Would you have a look at it please : http://chemistry.stackexchange.com/questions/29143/what-is-the-relation-between-surface-tension-and-initial-molecular-harmonic

@Aug Your new post is an interesting one. For the present post, hope you have found the most complete answer to this question. Thank you, have a good day and don't forget to choose your most complete answer  :D

This is certainly possible in most MM packages and hence it's possible via QM/MM as well. I had to do a bit of digging to see how you might do this in each program (hint, it's not easy).
For Gaussian, your best bet is to read the extremely long section on molecular mechanics paramters.
For GAMESS, you'd need to install the Tinker program and use the QM/MM integration.
Avogadro and Open Babel don't currently allow custom forces, although it wouldn’t be too hard to code.
I think the other alternative, depending on why you need the forces, is to use a geometry constraint during optimization. For example, ensuring certain distances are retained.
The problem is basically, that without care, your geometry optimization will be forced to use numeric derivatives, since the QM or MM program won't know much about the functional form of your custom force.

Thank you Geoff.   Actually I want to simulate the behavior of some gas compound materials under pressure. Here I am trying to add the total energy of  P.ΔV to the system.  First I thought I may use the idea of freezing the distance of some atoms as you said. I should say the results are really promising and compatible with the experimental data but I am thinking when I want to publish the results, I may face critics as the whole system may not collapse uniformly under the pressure.  Thats why I want to do something more sophisticated.

If you want to simulate pressure, your best bet would be to use a code with periodic boundary conditions and change the unit cell size (i.e., volume). Since n, R, and T are constant, P and V are inversely related.

Moreover, for such simulations, you might want to consider MD, e.g., using CP2K, which will let you perform both unit cell calculations and QM dynamics.

I really like the idea of unit cell. Is it possible in Avogadro ?( I am more comfortable with Avogadro and also I have Mac and seems CP2K is just for linux ) .   I tried making a larger unit cell from crystallography menu, ran UFF for two methane molecules as a test and then tried to make the unit cell smaller . but the whole molecular system burst !

At the moment, the UFF optimizations in Avogadro and Open Babel do not respect unit cells. You can, however, use Avogadro to create the unit cell and then use CP2K or other program for the simulation. For publication-quality results, that would be my suggestion (i.e., LAMMPS, Tinker, GROMACS, CP2K, etc.)

I have hard time getting CP2K or LAMMPS running on my Macbook. They look like work as command line executives and that's what I am really bad at !  Is there any app that can be installed on OSX or windows with a user friendly interface ( like Avogadro) that I can simply copy-paste the unit cell into it and run the shrinkages ?

Hi,  Finally I could install CP2K on linux cluster ( should say I went into the hell and came back !!). Now I can not make its input file with Avogadro . I made another post because that is a separate question: http://chemistry.stackexchange.com/questions/29184/how-to-make-a-cp2k-input-file-with-avogadro

If you are wanting the capability to apply forces to nuclei, to alter the equilibrium geometry, I do not know of any software with "a simple user interface like Avogadro" that offers built-in functionality to do what you seek. You are probably facing either a programming/scripting project or a great deal of manual iteration.
My recommendation would be that you investigate ORCA. ORCA incorporates the ability to perform geometry optimizations on systems where energy and gradient information is computed externally (ExtOpt mode; see Section 6.13.5 of the ORCA v3.0.3 Manual, downloadable from the forum linked above).  I am actually planning to develop something rather like this, though with functionality not nearly as general as what you describe [arbitrary $F(x,y,z)$].  I plan to write a Python application that calls ORCA in this ExtOpt mode, and then within each ExtOpt cycle:

Makes a normal call to ORCA to compute the single-point energy and gradient in the absence of the external force field
Calculates the contributions of the external force field to the energy and gradient
Merges the QM energy and gradient computed by ORCA with the force field energy and gradient
Passes the composite energy/gradient back to ExtOpt ORCA

ORCA supports semi-empirical methods as well as full quantum methods, so even large systems should be tractable in a reasonable time frame, once you have built the execution framework to carry it out.
EDIT: For your application, where you're trying to simulate increased pressure, the 'constrained volume' approach would probably work pretty well using a piecewise potential function, where $V = 0$ within the molecular volume calculated for a particular pressure, and then ramps to a very high value at the borders of the box.  The task then becomes appropriate definition of the box boundary, a problem which I suspect is similar to that involved in defining the cavity in implicit solvation models (COSMO$^1$, for example).
$^1$ Klamt and Schüürmann. Perkins Transactions 2 2: 799-805 (1993). doi: 10.1039/P29930000799

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Conductivity and Molar Conductivity of Equimolar Solutions of LiCl and KCl
If one took equimolar solutions of $\ce{Li}$ and $\ce{K}$, which would have higher conductivity? Is it the same for molar conductivity?
Say, $\ce{LiCl}$ and $\ce{KCl}$, of a litre each. Also, as Molar Conductivity = Conductivity / Molarity, are molar conductivity and conductivity directly proportional? Why or why not?

Since Li is highly solvated due to its small size, it's mobility is reduced and hence conductivity decreased.

I'm fine with this answer, but it would really pay if you be a little more elaborate. It might give some people a comment-like feeling.

Hmm, don't remember it now. :) I think I did click looks OK, but a better answer garners more votes, ya know. :)

Oh well, just saying. I think only a sentence can't help the OP (and they might be too shy to ask in their own post as a comment) You can leave this as is, or you can add a canonical and a comprehensive answer so that everyone else will be benefited. Oooh man! Now that will be helping. :)

@MARamezani :D I was earlier going to post Since Li has less mobility, it has less conductivity. I think it's enough and galdful that I added due to...


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MO for $N_2^{2-}$
Could someone please show me the MO diagram for $\ce{N_2^{2-}}$? What I did was since N has 5 valence $\ce{e^{-}}$, then $\ce{N_2}$ must have 10, and $\ce{N_2^{2-}}$ must have 12. And $\ce{N}$ has $s$ and $p$ orbital mixing. So when I draw out the MO it has all the orbitals filled. And the bond order is $\frac{1}{2}(6-6)=0$, but this is apparently wrong. Because according to the lewis structure the bond order should be $2$ I think.

This an image of the molecular orbital of $\ce{N2}$. If we add 2 more electrons, they will enter singly into the 2 antibonding pi orbitals and since bond strength is defined as bonding electrons minus nonbonding electrons over 2, or as I prefer, the net number of bonding/ nonbonding electrons times half, we would subtract 1 from the original bond order of $\ce{N2}$ to get 2.
Here is the mathematical calculations:
$\frac{1}{2}(8 (bonding)-4(antibonding))=2$
Antibonding orbitals are shown by an asterisk. Note that you will need draw in the 2 electrons on your own.


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Ring contraction in a carbocation due to ring strain and back bonding

Today our teacher told us that the following carbocation rearrangement occurs due to back bonding. I could not really follow what he meant. Can someone please explain what is actually happening during the following carbocation rearrangement, and why it happens?

The rearrangement has little to do with ring strain and a lot to do with resonance stabilization.
The ring strain in a cyclobutane ring is ~26.3 kcal/mol, while the ring strain in a cyclopropane ring is ~27.5 kcal.mol (reference).  This isn't much of a difference, certainly not enough to drive a reaction one way or the other.
The reason that cyclobutyl carbocations generally rearrange to cyclopropylcarbinyl carbocations is due to resonance stabilization.  Here are some drawings that may help, if you build a model and look at the model that will help even more. The $\ce{C-C}$ bonds in a cyclopropane ring are approximately $\ce{sp^{3.74}}$ hybridized (see here), said differently, they have a lot of p-character in the bond.  In fact, the p-character is so high that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.

here's another representation

Typically we discuss two conformations of the cyclopropane ring with the carbocation p-orbital, a bisected and a perpendicular conformation.  In the bisected conformation the cyclopropane $\ce{C-C}$ bond lies in the same plane as the carbocation p-orbital.  The carbocation p-orbital and the $\ce{C-C}$ orbital which resembles a p-orbital (because it is high in p-character) overlap!  In the perpendicular conformation, the cyclopropane $\ce{C-H}$ bond is in the same plane as the carbocation p-orbital, the $\ce{C-C}$ bond is not - there is no carbocation stabilization by the cyclopropane ring in this conformation.  Molecules where the bisected conformation can be attained show stabilization due to the stabilizing overlap between the carbocation p-orbital and the p-like $\ce{C-C}$ cyclopropane ring bond.  Sometimes we use resonance structures such as the following to denote this stabilization.


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Does the third law of thermodynamics apply to all compounds and elements?
According to wikipedia, the third law of thermodynamics is "The entropy of a perfect crystal, at absolute zero (zero kelvins), is exactly equal to zero.".
Then, theoretically, could all substances, including molecular compounds, attain absolute zero? Another way of thinking of this question is whether all compounds could become perfect crystals.

Entropy would be zero but it can't be.

@Mithoron So we can assume the law applies to all compounds?

Laws of thermodynamics are universal, applicable beyond chemistry...

see Absolute Zero

No, because it is possible that perfect crystals cannot be formed for statistical reasons. A simple example:
The absolute Boltzmann entropy is defined as $S = k_\text{B} \ln \Omega$ per molecule, where $\Omega$ is the number of possible microstates.
Consider a crystal of carbon monoxide at absolute zero. Now consider exchanging the position of all C atoms with those of all O atoms. This structure will be equivalent in energy so $\Omega = 2$. Thus $S = k_\text{B} \ln 2 > 0$.

Are you suggesting that if carbon monoxide is 2, monatomic atoms like He would be ln1, or 0? So it would be reasonable for elements in atomic state but not compounds? Or are you simply choosing 2 as a random number, but overlooking the electron cloud and its movement and even elements in atomic states would be unreasonable? I am curious to what cases the law would apply to.

Yes, He would be theoretically be able to form a perfect crystal as there would only be one possible microstate. At absolute zero all the atoms/molecules are in their ground quantum states so there is no entropy associated with excitement.

I strongly suggest you read this. From there:

The laws of thermodynamics dictate that absolute zero cannot be reached using only thermodynamic means, as the temperature of the substance being cooled approaches the temperature of the cooling agent asymptotically. A system at absolute zero still possesses quantum mechanical zero-point energy, the energy of its ground state. The kinetic energy of the ground state cannot be removed.

And:

An even stronger assertion is that It is impossible by any procedure to reduce the temperature of a system to zero in a finite number of operations. (≈ Guggenheim, p. 157)

And:

Perfect crystals never occur in practice; imperfections, and even entire amorphous materials, simply get "frozen in" at low temperatures, so transitions to more stable states do not occur.


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Equilibrium constants for lewis acids such as Ca2+
I need to calculate the theoretical $\mathrm{pH}$ in a $50~\mathrm{mL}$ water solution where $6.4\times10^{-4}~\mathrm{g}$ of $\ce{Ca^{2+}}$ is present. The reaction I base this on is the following
$$
\ce{Ca^{2+}_{(aq)} + 2H2O_{(l)} \rightleftarrows Ca(OH)2_{(s)} + 2H^+_{(aq)}}
$$
$$
K = \frac{[\ce{H+}]^2}{[\ce{Ca^{2+}}]} \Rightarrow [\ce{H+}] = \sqrt{K \cdot [\ce{Ca^{2+}}]}
$$
Is there a database of standard equilibrium constants for such Lewis acid reactions?
I appreciate any help!

Please avoid latex in titles due to issues with searchability. http://meta.chemistry.stackexchange.com/questions/253/should-we-revisit-the-url-slug-issue

I have never thought of that before. Thank you for pointing it out to me!

You're confusing terms here. $\ce{K_{a}}$ is the (Brønsted) acid dissociation constant. This suggests that $\ce{K_a}$ refers to the degree that a Brønsted acid dissociates into its conjugate base and hydrogen protons.
$\ce{K_{a}}$ is inapplicable to Lewis acids because Lewis acids are not proton donors.
Brønsted acids are Lewis acids in that Brønsted acids such as the hydrogen proton can accept a pair of electrons. But the converse is not true.
Lewis acids are electron acceptors. How do we measure the strength of Lewis acids? There is no universally accepted measure as there is for Brønsted acids. We could look at the enthalpy of reaction (Lewis adduct formation) between a Lewis acid and a Lewis base. This is a common method.

Duh, that was too obvious for me to think of! Thank you. Still, terminological errors aside, is there a place where I can find the equilibrium constant for the reaction?

@AndersMB - the best I can think of is using a bunch of already calculated K values to solve the unknown K value for your reaction. Try using Ksp of calcium hydroxide with Kw and any other applicable K values.

@AndersMB - try checking out this question to see what you might have to end up doing to solve for the unknown K value ...http://chemistry.stackexchange.com/questions/13624/why-may-k-expressions-be-manipulated-as-such

Oh, I am getting flashbacks to General Chemistry and Hess's Law! Thank you for your help.

You should be able to find the constant $K_\text{a}$ for the reaction
$$\ce{Ca^2+ + H2O <=> CaOH+ + H+}$$
I found a value of $\mathrm{p}K_\text{a} = 12.6$ (at 25 °C) in “Dissociation Constants of Inorganic Acids and Bases”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
Werner Stumm, James J. Morgan Aquatic Chemistry, 3rd Edition (1996), John Wiley & Sons, New York, NY gives a value of $\log K = -12.78$.


text:
Does the relationship equation between standard cell potential and equilibrium constant violate potential's intensive properties?
The equation:
$$E^{。}_{cell}= \frac{RT}{nF}\ln K_{eq}$$
We all know cell potential is intensive, not affected by the amount, Because: $volt=\frac{joule}{coulomb}$. Both joule and coulomb will be doubled altogether.
But as seen from the equation, cell potential is affected by the number of electrons transferred.

No, there is no violation of "intensivity".  The reason is that $K_{eq}$ depends on $n$, and changes in one cancel the other out.
For example, consider the electrolysis of water:

$$\ce{2H2O_{(l)} -> 2H2_{(g)} + O2_{(g)}}$$

The equilibrium constant for this reaction is $K_{1}=\frac{[\ce{H2}]^2 [\ce{O2}]}{1}$ and if you wrote out each electrochemical half reaction separately, $n$ would be 4.
Now consider this reaction:

$$\ce{4H2O_{(l)} -> 4H2_{(g)} + 2O2_{(g)}}$$

The equilibrium constant is now $K_{2}=\frac{[\ce{H2}]^4 [\ce{O2}]^2}{1}=(K_1)^2$.  If you wrote out the half-reactions for this reaction, $n$ would be 8, twice as big.  But the $\ln K_{eq}$ term would also be twice as big, since $\ln K_2 = \ln{(K_1)^2} = 2 \ln K_1$.

Wow, I should have spent sometime playing around with it first. But I was too lazy.

No worries.  Electrochemistry is confusing, mainly because physicists and engineers refuse to get rid of silly units like "volt", "coulomb", and "ampere".  Ever since Milliken's experiment, sensible folks have been wishing for those to be replaced by "J/mol", "mol", and "mol/s".

Is it more accurate to say standard cell potential is an intensive property?

Hello!! ^Last comment.

No, all cell potentials, standard or not, are intensive properties.

But if we change the concentration of a half cell, the reaction quotient would change and then by Nernst Eqn, the cell potential would change. Where am I going wrong?


text:
How to calculate the entropy change of benzene when it changes from vapour into liquid?

Given information:

The boiling point of benzene at atmospheric pressure is $353~\mathrm{K}$
the enthalpy of vaporization of benzene is $30.8~\mathrm{kJ~mol^{−1}}$ at this temperature.
The molar heat capacities of the liquid and vapour are $136.1~\mathrm{J~K^{−1}~mol^{-1}}$ and $81.7~\mathrm{J~K^{−1}~mol^{-1}}$,
  respectively, and may be assumed temperature independent.

Calculate the entropy change of the system, the surroundings and hence the universe when $1~\mathrm{mol}$ of benzene vapour at $343~\mathrm{K}$ and atmospheric pressure becomes liquid benzene at $343~\mathrm{K}$. Also, will this process occur spontaneously?

I know that $\mathrm{d}S = \frac{\mathrm{d}H}{T}$ therefore,
$\displaystyle\mathrm{d}S = \frac{-30.8 \times 10^3~
\mathrm{J}}{343~\mathrm{K}}=-89.8~\mathrm{J~K^{-1}}$ which is the entropy of the system.
How do I continue from there, utilising the molar heat capacities given?

I would use the clausius clapeyron relation to find the equilibrium constant at the lower temperature and then convert to gibbs free energy and then solve from delta S. There probably are many different ways to solve though.

@pikachugee -89.8 is the change in entropy of the surroundings. If that were the system, that would mean that there is a decrease in entropy if a liquid transitioned into a gas.

I can solve it if, you clear whether, heat capacity is for constant volume or constant pressure, $C_v$ or $C_p$

You just used $\mathrm{d}S=\mathrm{d}H/T$. Also you didn’t told whether the heat capacities are $C_{\mathrm{m},p}$ or $C_{\mathrm{m},V}$. Anyways I found that you are using $C_{\mathrm{m},p}$ from the benzene data page on Wikipedia. You need to do something like this:
$$\Delta S=\left(1\;\mathrm{mol}\cdot81.7\;\mathrm{\frac{J}{mol\cdot K}}\cdot\ln\frac{373\;\mathrm{K}}{343\;\mathrm{K}}\right)+\left(\frac{30.8\;\mathrm{\frac{kJ}{mol}}}{373\;\mathrm{K}}\right)+\left(1\;\mathrm{mol}\cdot136.1{\;\mathrm{\frac{J}{mol\cdot K}}}\cdot\ln\frac{343\;\mathrm{K}}{373\;\mathrm{K}}\right)\\\approx78.012\;\mathrm{J/K}$$

You already determined the $\Delta S$ for benzene at 353 K by dividing the $\Delta H$ at 353 K by the temperature 353 K.  Now all you need to do is determine the $\Delta H$ at 343 K using Hess' law?  Once you do that, you can get the $\Delta S$ at 343 K.

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


text:
How to calculate the required concentration for sodium sulfite to remove oxygen as an oxygen scavenger from a solution?
If I know the amount of oxygen in such a solution, how can I know the required concentration of sodium sulfite to be used and react as oxygen scavenger?

From the equation
\[\ce{2SO3^{2-} + O2 -> 2SO4^{2-} } \]
you could conclude that you need (at least) 2 moles of sodium sulfite $(\ce{Na2SO3})$ for each mole of oxygen in solution.


text:
Equivalent hydrogens in 2,3-dimethylbutane
So while calculating for the relative yield of monochlorination for 2,3-dimethylbutane, I was counting the primary hydrogens, I counted 6, but in fact it was 12. Why is it 12? How do you find equivalent hydrogens? Since the compound is symmetrical doesn't the hydrogens on both sides be equivalent to each other?

The hydrogen atoms in a methyl group are equivalent to each other. There are 4 methyl groups and therefore there are 12 equivalent hydrogens. 2,3-dimethylbutane is symmetric as you stated, and so on one side you have 6 equivalent hydrogens, this implies that there are 6 more equivalent hydrogens on the opposite side.

actually there are 2 pairs of 6 equivalent protons and 2 pairs of 6 enantiotopic protons.

@ron Ah, I see. Then that means she was correct in that there are only 6 equivalent protons?

Yes, but note that both the equivalent and enantiotopic hydrogens will not be distinguishable by an achiral reagent (a chlorine free radical in this case) .  So there are 12 primary hydrogens, it is just not the best terminology to say that they are all equivalent.


text:
Diluting acetic acid to obtain a solution of pH 5
What percent of a solution needs to be acetic acid for a particular $\mathrm{pH}$?  I have 100% acetic acid, and want $1~\mathrm{L}$ of solution with a $\mathrm{pH}$ of 5. I've done some research, but I still do not understand how to solve this problem.

The original question was fine as it is. But then someone edited and it seem's like that this is purely a homework question which it was not and was why I decided to answer.

This is a dilution problem. So in this instance, you want to add deionized water to your acetic acid. However, the question is, how much water do you add?
It is necessary first to find the molarity of the acetic acid. On PubChem, we see that the molecular weight is $60.05196 \ \frac{\text{g}}{\text{mol}}$ and density is $1.0446 \ \frac{\text{g}}{\text{cm}^3}$ @ $25\ ^\circ \text{C}$.
If we have 1 L of acetic acid, then
$$ 1.0446 \ \frac{\text{g}}{\text{cm}^3} \cdot \frac{1000\ \text{cm}^3}{1\ \text{L}} \cdot \frac{1\ \text{mol}}{60.05196\ \text{g}} = 17.394\ \frac{ \text{mol}}{\text{L}}$$
We know that,
$$ \text{pH} = -\log{\ce{[H3O+]}} $$
which implies that the concentration of hydronium ions to make the solution with pH equal to 5 is, $\ce{[H3O+]} = 1 \cdot 10^{-5} \ \text{M}$. We make our RICE table,
$$\ce{HC2H3O2 + H2O -> C2H3O2- + H3O+}$$
\begin{array} {|c|c|c|c|c|}
\hline
\text{Initial conc.} & x \ \text{mol} & - & \text{0 mol} & \text{0 mol}\\
\hline
\text{Change conc.} & -1 \cdot 10^{-5}\ \text{M} & - & 1 \cdot 10^{-5}\ \text{M} & 1 \cdot 10^{-5}\ \text{M}\\
\hline
\text{End conc.} & x - 1 \cdot 10^{-5}\ \text{M} & - &  1 \cdot 10^{-5}\ \text{M} & 1 \cdot 10^{-5}\ \text{M}\\
\hline
\end{array}
$$K_\text{a} = \frac{[\ce{H3O+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(0.00001)(0.00001)}{(x - 0.00001)} = 1.75 \cdot 10^{-5} $$
$$\frac{(1 \cdot 10^{-10})}{(x - 0.00001)} = 0.0000175$$
$$1 \cdot 10^{-10} = 0.0000175x - 1.75 \cdot 10^{-10} $$
$$2.75 \cdot 10^{-10} = 0.0000175x$$
$$x = 1.6 \cdot 10^{-5}\ \text{M} = [\ce{CH3COOH}] $$
Thus, you go from a $17.394\ \text{M}$ solution of acetic acid, to a $0.000016\ \text{M}$, which is a factor of $1.09\cdot10^{6}$. You can then dilute the solution by the appropriate amount by placing the amount of acid you want in a volumetric flask, and then successively dilute it to the required pH.
Two assumptions,

Deprotonation is small enough that the equilibrium concentration of the acid is approximately equal to the same as its initial concentration.
Autoprotolysis of water does not significantly contribute to pH.

Very good answer, in particular the use (and demonstration) of the RICE table

why use density?

@ADG I used density to easily find the molarity of 100% acetic acid. If you look at the chain-link calculations above, you can see I get the molarity that I wanted. For reference, the pH is roughly 1 with purely 100% acetic acid which is interesting since acetic acid is a "weak acid".

What percent of a solution needs to be acetic acid for a particular pH? I have 100% acetic acid, and want 1 L of solution with a pH of 5. I've done some research, but I still do not understand how to solve this problem.

You need to know that if you add water moles remian same so $\rm M_1V_1=M_2V_2$. You also need to know $\rm pH=-\log[H^+]$ form where you get concentration of $\rm H^+$. Now for weak acids the cubic equation used to determine pH gets some terms negligible hence we use $\rm [H^+]=\sqrt{K_aM_2}$ where $\rm K_a$ is very common and I remember it as $1.7\times10^{-5}\sim10^{-4.7}$.


text:
Why is a hydrocarbon tail non-polar?

Here it says that the hydrocarbon tail is non-polar. I understand that the $\ce{H}$ on top of the $\ce{C}$ and on the bottom have dipoles that cancel, but at the very end, with the $\ce{CH3}$ group, there is no opposing vector to cancel with that dipole. Why then is the hydrocarbon tail still non-polar?

This is a 3-dimensional structure - so the dipoles do NOT cancel anywhere... Rather, the dipoles created by C and H are weak, as the electronegativity difference is low - which is considered unpolar.

The hydrocarbon tail is considered nonpolar mainly due to

The Pauling electronegativity of Carbon is 2.55 and Hydrogen is 2.2. Due to the small difference, .35, and the definition of a polar bond, C-H bonds are considered nonpolar.

Indeed, the molecule methane, $\ce{CH4}$ has a dipole moment of 0 D (debyes). I think part of your confusion arises from the picture you provided. Skeletal drawings are 2-d in nature and do not convey information about 3-d. For instance, each carbon has it's hydrogens arranged in a tetrahedral structure, with bond angles 109.5 degrees. The symmetry of this cancels all the vectors contributing to polarity.

If you look at the picture of methane above, and consider the top most hydrogen as the next carbon in the hydrocarbon tail, you can see that in reality, the bottom 3 hydrogens do experience a net dipole moment towards the carbon, and this increased by having another carbon further up in the tail. Relatively speaking, this distance is rather large from carbon 1 to carbon 2. We can say that the vector that arises from this is negligible, and indeed, there is a net dipole moment anyways towards the oxygen at the polar head.

methane is CH4. Did you mean methyl?

@pikachu Oh, thanks for pointing that out. Edited.

The hydrocarbon that you have depicted (decanoic acid) does indeed have a polar head and a non-polar head, though it is not apparent in the form it is presented to you in. When viewing decanoic acid as a 3-dimensional, ball-and-stick model, it becomes more obvious:

The hydrogen on C-10 of this chain that sticks out in the plane of the rest of the carbon chain is balanced by C-9. Most importantly, it should be noted that there are no unbonded electron pairs in this region. Because the electron density along the chain remains consistent, the carbon chain is non-polar.
Each of the two oxygens in the carboxylic acid groups have two electron clouds that give these regions around the molecule a negative charge. The space around the hydrogen bonded to oxygen, and the space around C-1 that is oriented away from the oxygens are both more positive. These opposing charges result in the polarity of the carboxylic acid group.
Source: http://upload.wikimedia.org/wikipedia/commons/9/98/Decanoic-acid-3D-balls.png


text:
Viewing a protein or bio structure in Chembio3D in its natural conformation
I notice when using SMILES or a IUPAC name for many molecules and generating them in Chembio3D, and then running MM2 > Minimize Energy, the structures produced are most of the time not reflections of any conformation observed physiologically.
I'm wondering if there's a way I can import protein and other macromolecules into Chembio3D, perhaps from a database somewhere, where I can view their precise arrangement either as found in crystallography or in vivo in their usual forms at physiological pH. Most importantly, the proteins would need to be accurate (something I find lacking when generating structures by name in chemdraw/chembio3d).
I appreciate that the proteins and other molecules are always changing conformation and interacting with other molecules, but at least a reasonable prediction of some realistic conformations to view them in 3D would be nice, instead of conformations that might rarely/never exist in a cell. Is this possible? And if so, how would I go about this? What would be the specific steps I need to follow to import these proteins into Chembio3D?

A side question, can you alter bond angles in ChemBio3D?

The site you are looking for is the Protein Data Bank.
You can get to it with one of these links.
www.wwpdb.org
www.pdbe.org
www.rcsb.org/pdb
www.pdbj.org

Short and to the point. :) Might be better to link to the actual page, http://www.rcsb.org/pdb/home/home.do and not the Wikipedia page.

@Jun-GooKwak There are 3 links on the Wikipedia page, depending on the organization that accesses the information.


text:
Are these organo-metallic compounds?
My understanding says organo-metallic = organic + metallic compunds.
One already known to me is Grignard Reagent ($\ce{RMgX}$).
But would you consider $\ce{NaOMe}$ or $\ce{NaCCH}$ as an organometallic compound? Would you consider $\ce{LiOAc}$, $\ce{LiOMe}$, $\ce{LiN(Me)2}$ or $\ce{LiMe}$?

To my understanding, only $\ce{NaC#CH}$ and $\ce{LiCH3}$ are organometallic compounds.
According to the IUPAC definition of organometallic compounds, these must contain

at least one metal-carbon bond between an organic molecule, ion, or radical and a metal.

OK no metal (non)carbon bond that means.

Must the bond also be covalent? Or is ionic fine?


text:
Reaction with silver oxalate?
I was asked which geometrical isomer of $\ce{[M(NH3)2Cl2]}$, viz. cis or trans would react with silver oxalate. The possible reaction could be formation of silver chloride but since the coordination entity does not break up, any reaction thus, in my opinion can not happen. But the asker says that the answer is not none, but actually one of them.

[…] but since the coordination entity does not break up, any reaction thus, IMO can not happen.

Your opinion contradicts experimental findings. Two chlorido ligands in metal complexes can be replaced by (other monodentate ligands or) a bidentate chelating ligand, as demonstrated for the reaction of potassium tetrachloropalladate, $\ce{K2[PdCl4]}$, with diaminosuccinic acid (and its diethylester))(DOI).
When the binding atoms of a bidentate ligand are close together, as in oxalate, the ligand can't just somehow wrap around the central metal. Instead specific geometric requiries have to be met.
For a tetra-coordinated complex, such as $\ce{[M(NH3)2Cl2]}$, two major fundamental geometries are conceivable, namely

tetrahedral or
square-planar

Only in the latter case, cis-trans isomerism can be observed!
In the case of the trans-isomer the two square-planar complexes, the substitution of two chlorido ligands that point to opposite corners of the square is not possible: oxalate would have to bind all across the complex. Consequently, it is only the cis-isomer which can react here!
$$\ce{\mathit{cis}-[M(NH3)2Cl2] + ox^{2-} -> \mathit{cis}-[M(ox)(NH3)2] +2Cl-}$$
(For the abbreviation ox for oxalate in coordination compounds, see the IUPAC Red Book, Table VII, entry 118.)

Can you please elaborate with a diagram?


text:
Calculation of CFSE
I need to find CFSE for these:

$\ce{[Ti(H2O)6]^3+}.$ $\ce{Ti}$ is $\mathrm{(4s)^2(3d)^2}$, $\ce{Ti^3+}$ is $\mathrm{(4s)^0(3d)^1}.$ Afterwards it becomes $\mathrm{d^4sp^2}$ or $\mathrm{t_{2g}^5e_g^4}$ so CFSE is $\frac25\Delta_0$. And I'm given $\bar\nu_\mathrm{max} = \pu{20300 cm-1}.$ So, we can use $E=hc\bar\nu.$ And that doesn't seem to come close to answer, I think some conceptual mistake has occurred.
$\ce{[CoF6]^3-}.$ It's $\mathrm{sp^3d^2}$ or $\mathrm{t_{2g}^4e_g^2}.$ Here CFSE would be $4\frac25\Delta - 2\frac35\Delta = \frac25\Delta = \pu{4 Dq}$. This one's correct.

What transition do you think the max wavenumber corresponds to ?

@J.LS think?? what do you mean, please elaborate.

Well, what sort of electronic transition would that wavenumber correspond to ?

@J.LS $e_g\leftarrow t_{2g}$

That's correct; the only other band you would see would be LMCT at a higher wavenumber. So you know the energy of the delta parameter and thus the CFSE; be careful to work in SI units if you want to convert it.

@J.LS would you be helping enough to form an answer?

Calculation for the first part:
$\bar\nu_\mathrm{max} = \pu{20300 cm^-1},$ so $\Delta = \pu{4.03E-19 J}$ per molecule or $\pu{242.7 kJ mol-1}.$ $\mathrm{CFSE} = \pu{97.1 kJ}.$
$$\mathrm{CFSE} = \pu{6.6E-34 J s} × \pu{3E8 m s-1} × \pu{20300 cm-1} × \pu{6.02E23 mol-1} × \pu{E-3 J kJ-1} × \frac25 = \pu{97.1 kJ mol-1}$$


text:
How to determine which salt will precipitate from a solution containing multiple ions?
I have a solution containing $0.2\:\mathrm{mol}$ of $\ce{Br^-}$ ions and $\ce{SO_4^2-}$ ions each.
Now if I add excess  $\ce{AgNO3}$ in this solution,
what will precipitate?
 $\ce{Ag2SO4}$ or $\ce{AgBr}$ or both?
And if I add just $0.2\:\mathrm{mol}$ of $\ce{AgNO3}$, what will precipitate?
(I don't want to calculate anything. Just wanted to know qualitatively that which compound precipitates.)

You need to know that $K_\mathrm{sp}$ is used to determine the solubilities of sparingly soluble salts usually. You can get the values of $\ce{AgSO4,AgBr}$ and as $K_\mathrm{sp}=[\ce{Ag+}]^2[\ce{SO4^2-}]$ or $[\ce{Ag+}][\ce{Br-}]$. Now concentration of silver ion is constant so try calculating the other two anions and try comparing them.

So in excess both will precipitate eventually while in limited quantity, the one with lower Ksp will precipitate?

@pikachu i think you got it right.


text:
What is there in an atom between the nucleus and electrons?
As you can see the atom consists not only of the nucleus and electron but also of "empty space is the space empty or is their something else?

There is a question and answer on Physics.SE http://physics.stackexchange.com/questions/7615/what-is-in-the-space-between-a-nucleus-of-an-atom-and-its-electrons

It is not necessarily empty space.
For example the flux of neutrinos on Earth is about $\pu{3E12 cm-2 s-1}$.  At a given moment, a neutrino may or may not be within the atom or even within the nucleus of the atom. Similarly, other particles may be present within an atom at a given time.
Also, an electron may be inside the nucleus in accordance with the Fermi Contact Interaction.  So if you have a hydrogen atom, having only one electron, and the electron is inside the nucleus at a given moment, there will be no "between" for there to be space, empty or not.

Your doubt stems from the fact that you are using an incomplete view of what an atom is. That atomic model with the nucleus in the center and the electron zooming around in tilted circular(-ish) orbits, called the "planetary" model of the atom, is incomplete.
After the discovery of the fact that most of the mass (and the positive charge) of an atom was concentrated into a small point-like nucleus, chemists and physicists had to throw the currently prevailing model out of the window and were trying to come up with a new thing that incorporated the known facts about matter. This picture was the working idea, but it had not solidified into anything consistent because it had serious problems, mainly because they were describing the electrons with classical mechanics, since the discovery of quantum mechanics was yet to happen.
In order to resolve this apparent paradox of what exists between the electrons and the nucleus you have to shift into a quantum mechanical view of the atom.
Since you are using Rutherford's model as a starting point and because your question seem so innocent and non-professional I will assume you have no experience with the details of quantum chemistry, so I will provide an explanation based on images and allegories rather than equations and quantities. If you want me to be more detailed with the explanation, please say something in the comments and I can walk you through the bowels of the math in this particular problem.
So, it turns out even though electrons are pieces of matter, they are so small and so light that "quantum effects" start to take place. There is a lot of complicated things that go into "quantum effects" but the gist of it is that it becomes impossible to nail the electron down into one particular position and one particular classical orbit simultaneously. You can think of this as if the electron is smeared out. It becomes less like this:

And more like this:

This smearing effect turns the electron into some sort of diffuse cloud. And its dynamics, that is the way it moves and the way forces act on it to change its motion, changes accordingly.
So electrons stop being point-like things that occupy certain circular orbits around the atom and start being more like cloud-like things that have certain particular shapes. Since electrons are now described as cloud-like things, it is trivial to understand that there really isn't any empty space between the nucleus and the electrons. The cloud fills the entire space and the nature of the interactions between electrons and nuclei, along with the quantum dynamics required to describe the electrons, determine the actual shape of the "electron-cloud". I hope that is sufficiently clear.
P.S. The term "electron cloud" isn't very accurate and should not be used in a setting where accurate, formal description of quantum-mechanical systems is in order. It is just an allegorical way of conveying a complicated concept in a pictorial way that can be easily envisioned with the mind's eyes.


text:
Which intermolecular interaction is dependent on the inverse cube of distance?

The intermolecular interaction that is dependent on the inverse cube of distance between molecules is:

hydrogen bond
ion-ion interaction
ion-dipole interaction
London force

Inverse cube of distance I think means an ion-dipole interaction, since we know force due to a dipole on a charge is inversely proportional to the cube of distance. But I don't understand why the answer given is hydrogen bond. What is the distance dependence of the hydrogen bond? Shouldn't the answer be (3)?

2
The ion-ion or charge-charge interaction is given by
$$
V_{\text{cc}} = \frac{Z_{i}Z_{j}e^{2}}{4\pi\epsilon_{0}} \frac{1}{|\mathbf{R}_{i} - \mathbf{R}_{j}|}
$$
and is therefore proportional to $R^{-1}$.
3
The ion-dipole or charge-(permanent) dipole interaction is given by
$$
V_{\text{cd}} = \frac{Z_{i}e}{4\pi\epsilon_{0}} \frac{\mathbf{R}_{ij} \cdot \vec{\mu}_{j}}{R_{ij}^{3}}
$$
and is therefore proportional to $R^{-2}$.
4
"London force" is often meant to encapsulate any of the van der Waals forces (here, here), but in particular the dispersion force. All of the above have a $R^{-6}$ dependence.

Here is a general table that covers each of the interactions so far:
\begin{array}{lll}
\hline
\text{type of molecular unit} & \text{type of force} & n~\text{in}~R^{-n} \\
\hline
\text{ions}                        & \text{coulombic}               & 1 \\
\text{ion - polar molecule}        & \text{ion - dipole}            & 2 \\
\text{two polar molecules}         & \text{dipole - dipole}         & 3 \\
\text{ion - nonpolar molecule}     & \text{ion - induced dipole}    & 4 \\
\text{polar and nonpolar molecule} & \text{dipole - induced dipole} & 6 \\
\text{nonpolar molecules}          & \text{dispersion}              & 6 \\
\hline
\end{array}
1
From Wikipedia:

The hydrogen bond is often described as an electrostatic dipole-dipole interaction. However, it also has some features of covalent bonding: it is directional and strong, produces interatomic distances shorter than the sum of the van der Waals radii, and usually involves a limited number of interaction partners, which can be interpreted as a type of valence. These covalent features are more substantial when acceptors bind hydrogens from more electronegative donors.

This implies that hydrogen bonding interactions may not behave exactly as $R^{-3}$. Consider the water dimer:

which is a prototypical example of a hydrogen-bonded complex. From Stone, using density functional-based symmetry-adapted perturbation theory (SAPT-DFT) in the aug-cc-pVQZ basis,
\begin{array}{lr}
\hline
\text{Type of interaction} & \text{Energy (kJ/mol)} \\
\hline
\text{Electrostatic}                 & -33.4 \\
\text{Exchange-repulsion}            &  31.5 \\
\text{Dispersion}                    & -12.2 \\
\text{Induction}                     &  -8.0 \\
\text{Charge transfer}               &  -5.9 \\
\text{Exchange-dispersion}           &   2.3 \\
\text{Exchange-induction}            &   7.8 \\
\delta_{\text{HF}}~\text{correction} &  -3.4 \\
\hline
\text{Total}                         & -21.5 \\
\hline
\end{array}
the electrostatic component is clearly the largest contributor, but after taking the absolute value of all interaction terms, is about 32% of the total interaction. Another approach for decomposing the interaction energy is based on absolutely localized molecular orbitals (ALMO-EDA), which at the $\omega$B97M-V/def2-QZVPP level gives
\begin{array}{lr}
\hline
\text{Type of interaction} & \text{Energy (kJ/mol)} \\
\hline
\text{Electrostatics}  & -65.7 \\
\text{Pauli repulsion} &  65.0 \\
\text{Dispersion}      &  -7.7 \\
\text{Polarization}    &  -4.6 \\
\text{Charge transfer} &  -7.9 \\
\hline
\text{Total}           & -20.9 \\
\hline
\end{array}
SAPT-DFT and ALMO-EDA are in good agreement, though charge transfer is more important in ALMO-EDA primarily due to the lackluster definition of charge transfer within SAPT. Electrostatics are now about 44% of the total interaction. For reference, here is how the additional terms decay (Stone, Table 1.2):
\begin{array}{lccl}
\hline
\text{Contribution} & \text{Additive?} & \text{Sign} & \text{Comment} \\
\hline
\textbf{Long-range}~(U \sim R^{-n}) & & & \\
\text{Electrostatic} & \text{Yes}     & \pm & \text{Strong orientation dependence} \\
\text{Induction}     & \text{No}      & -   &                                      \\
\text{Dispersion}    & \text{approx.} & -   & \text{Always present}                \\
\text{Resonance}     & \text{No}      & \pm & \text{Degenerate states only}        \\
\text{Magnetic}      & \text{Yes}     & \pm & \text{Very small}                    \\
\hline
\textbf{Short-range}~(U \sim e^{-\alpha R}) & & & \\
\text{Exchange-repulsion}  & \text{approx.} & + & \text{Dominates at very short range} \\
\text{Exchange-induction}  & \text{approx.} & - &                                      \\
\text{Exchange-dispersion} & \text{approx.} & - &                                      \\
\text{Charge transfer}     & \text{No}      & - & \text{Donor-acceptor interaction}    \\
\hline
\end{array}
Answer
Although components of hydrogen bonding interactions may have a $R^{-3}$ distance dependence, there are significant contributors to the hydrogen bond interaction other than electrostatics that do not decay as $R^{-3}$; in general, it is a complex interaction and its definition is still hotly debated. For that reason, the best answer is (3), the ion-dipole interaction.
References

Masahiro Yamamoto, Kyoto University: charge-charge, charge-dipole, dipole-charge, dipole-dipole interaction
Stephen Lower, Simon Fraser University: Chem1 Virtual Textbook, Interactions between Molecular Units
Anthony Stone: The Theory of Intermolecular Forces
ALMO-EDA calculations: geometry is B97-D/def2-TZVP, EDA input as comment

Simple Google search led me to the content at Wikipedia page:

The energy of a Keesom interaction depends on the inverse sixth power of the distance, unlike the interaction energy of two spatially fixed dipoles, which depends on the inverse third power of the distance.

so this ion-dipole interaction comes under which one exctly?? keesom interactiom?? or two fixed dipoles?? i thought it comes under the interaction between a charge and a dipole in space

This answer does not seem to fit any of the given options.

@GaurangTandon only if hydrogen bond is considered dipole-dipole interaction...


text:
installation error of dalton 2015 in ubuntu 12

what are the answer for CONFIRMATION OF REGISTRATION when register in dalton forum ?
install got error, how to install dalton 2015

 wonder@wonder-VirtualBox:~/dalton/DALTON-Source$ uname -r
    3.8.0-29-generic
wonder@wonder-VirtualBox:~/dalton/DALTON-Source$ sudo apt-get install openmpi-bin openmpi-dev
    [sudo] password for wonder:
    Reading package lists... Done
    Building dependency tree
    Reading state information... Done
    Note, selecting 'libopenmpi-dev' instead of 'openmpi-dev'
    libopenmpi-dev is already the newest version.
    libopenmpi-dev set to manually installed.
    The following packages were automatically installed and are no longer required:
      libghc-regex-posix-dev libghc-http-dev libghc-stm-dev libghc-quickcheck2-dev
      libxmu-headers mesa-common-dev libghc-hunit-dev libghc-cgi-dev alex
      libghc-monadcatchio-mtl-dev libghc-syb-dev ghc-haddock happy
      libglu1-mesa-dev libdrm-dev freeglut3-dev libghc-html-dev libxt-dev
      libxmu-dev libghc-regex-base-dev libghc-fgl-dev gap-libs cabal-install
      libghc-opengl-dev libgl1-mesa-dev libghc-random-dev libkms1
      libghc-haskell-src-dev libghc-regex-compat-dev libghc-xhtml-dev freeglut3
      libghc-glut-dev libghc-parallel-dev
    Use 'apt-get autoremove' to remove them.
    The following extra packages will be installed:
      blcr-util libcr0 openmpi-checkpoint
    Suggested packages:
      blcr-dkms gfortran
    The following NEW packages will be installed:
      blcr-util libcr0 openmpi-bin openmpi-checkpoint
    0 upgraded, 4 newly installed, 0 to remove and 404 not upgraded.
    Need to get 239 kB of archives.
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    Processing triggers for libc-bin ...
    ldconfig deferred processing now taking place
    wonder@wonder-VirtualBox:~/dalton/DALTON-Source$ ./setup --mpi
Configuring system: Ubuntu
 FC=mpif90 CC=mpicc CXX=mpicxx cmake -DENABLE_MPI=ON
 -DENABLE_SGI_MPT=OFF -DENABLE_OMP=OFF
 -DENABLE_64BIT_INTEGERS=OFF
 -DENABLE_OPENACC=OFF -DENABLE_COLLAPSE=OFF
 -DENABLE_CSR=OFF -DENABLE_SCALASCA=OFF
 -DENABLE_VAMPIRTRACE=OFFDENABLE_TIMINGS=OFF
 -DENABLE_XCFUN=OFF -DENABLE_INTEREST=OFF -DENABLE_ICHOR=OFF
 -DENABLE_STATIC_LINKING=OFF -DENABLE_SCALAPACK=OFF
 -DCMAKE_BUILD_TYPE=release /home/wonder/dalton/DALTON-Source

CMake Error at /usr/share/cmake-2.8/Modules/CMakeTestFortranCompiler.cmake:40 (MESSAGE):
  The Fortran compiler "/usr/bin/mpif90" is not able to compile a simple test
  program.

It fails with the following output:
   Change Dir: /home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp

  Run Build Command:/usr/bin/make "cmTryCompileExec/fast"

  /usr/bin/make -f CMakeFiles/cmTryCompileExec.dir/build.make
  CMakeFiles/cmTryCompileExec.dir/build

  make[1]: Entering directory
  `/home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp'

  /usr/bin/cmake -E cmake_progress_report
  /home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp/CMakeFiles 1

  Building Fortran object
  CMakeFiles/cmTryCompileExec.dir/testFortranCompiler.f.o

  /usr/bin/mpif90 -c
  /home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp/testFortranCompiler.f
  -o CMakeFiles/cmTryCompileExec.dir/testFortranCompiler.f.o

  --------------------------------------------------------------------------

  The Open MPI wrapper compiler was unable to find the specified compiler

  gfortran in your PATH.

  Note that this compiler was either specified at configure time or in

  one of several possible environment variables.

  --------------------------------------------------------------------------

  make[1]: Leaving directory
  `/home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp'

  make[1]: *** [CMakeFiles/cmTryCompileExec.dir/testFortranCompiler.f.o]
  Error 1

  make: *** [cmTryCompileExec/fast] Error 2

  CMake will not be able to correctly generate this project.
Call Stack (most recent call first):
  CMakeLists.txt:18 (project)

-- The Fortran compiler identification is unknown
-- The C compiler identification is GNU
-- The CXX compiler identification is GNU
-- Check for working Fortran compiler: /usr/bin/mpif90
-- Check for working Fortran compiler: /usr/bin/mpif90  -- broken
-- Configuring incomplete, errors occurred!

wonder@wonder-VirtualBox:~/dalton/DALTON-Source$ ls -l /usr/bin/mpif90
lrwxrwxrwx 1 root root 24 Dec 12 22:51 /usr/bin/mpif90 -> /etc/alternatives/mpif90

I suggest you better try to search/ask on their own forum.

Or, wait. You know about the forum. If you have some problems with registering there, Chemistry.SE in obviously not the right place to ask about it.

But for me it looks like you don't even have Fortran compiler (gfortran) installed. Try install it by something like sudo apt-get install gfortran and run DALTON's setup after it.

I'm voting to close this question as off-topic.

i do not know the answer when fill in information when register in the forum

As pointed out by Wildcat have a look at these messages:
CMake Error at /usr/share/cmake-2.8/Modules/CMakeTestFortranCompiler.cmake:40 (MESSAGE): The Fortran compiler "/usr/bin/mpif90" is not able to compile a simple test program.
and
The Open MPI wrapper compiler was unable to find the specified compiler gfortran in your PATH.
mpif90 is a environment variable set by (older versions of) MPICH and is actually deprecated in recent versions of Open MPI.
So, it you didn't forget to install gfortran, what is the value of mpif90 on your system?

how to show the value of mpif90? maybe i forget to install gfortran, i will tomorrow


text:
How to calculate the concentration based on dilution of a stock solution?
One of my exam questions is about lipstick and calculating the amount of lithium in the lipstick by AAS. The determination took place by making standard solutions with known concentrations of lithium. For the stock solution $653.3~\mathrm{mg}$ lithium chloride, $\ce{LiCl}$, was dissolved in $100~\mathrm{ml}$.
$M(\ce{Li}) = 6.49~\mathrm{g/mol}$
$M(\ce{LiCl}) = 42.4~\mathrm{g/mol}$
From the stock solution the following volumes were pipetted into $100~\mathrm{ml}$ flasks: $3$, $5$, $8$ and $10~\mathrm{ml}$. Also $1~\mathrm{ml}$ $\ce{KCl}$ solution, $1.0~\mathrm{M}$, was added to prevent ionisation.
I calculated the concentration Lithium in the stock as follows:
The mass of Lithium in the weighed $\ce{LiCl}$ is equal to the molar mass of lithium divided by the molar mass of lithium chloride multiplied by the mass that was weighed. So in the form of an equation:
$$m(Li)=\frac{M(\ce{Li})}{M(\ce{LiCl})}\cdot m(\text{total})=99.9~\mathrm{mg}$$
Therefore the concentration of the stock solution is equal to the mass of lithium divided by the volume of the flask:
$$C(\text{Li, stock}) = \frac{m(\ce{Li})}{V(\ce{Li})}=\frac{99.9}{0.1}=999~\mathrm{mg/l}$$
The concentration of the standard solutions therefore would be:
$$C(\text{standard}) = C(\text{stock}) \times \frac{V(\text{stock})}{V(\text{flask volume})}$$
With the concentration of the standard in $\mathrm{mg/L}$ and the volumes of both stock and flash volume in $\mathrm{mL}$.
This would give concentrations of $29.9$; $49.9$; $79.9$ and $99.9~\mathrm{mg/L}$ to the standard solutions where $3$, $5$, $8$ and $10~\mathrm{ml}$ were pipetted into. The answer sheet however provides us with $300$, $500$, $800$ and $1000~\mathrm{\mu g/L}$. I suspect I've made a mistake in unit conversion somewhere, due to the factor 100 mistake in the answer, although the units don't correspond either.

I don't see anything wrong with the math. I calculate the same numbers you have.

@LDC3 Which is strange, because the answer sheet provides me with a different answer O_O. I'll send an email to my teacher.

Well, I'll start by saying I redid your math, and you have rounding errors.  Using correct sig figs, your stock solution has 100. mg of Li in it, giving a concentration of 1.00E3 mg/L.  Those dilutions therefore should be 30.0 mg/L, 50.0 mg/L, 80.0 mg/L, and 100. mg/L.
You're correct aside from the rounding errors, and the answer key is wrong.  The question as you've written can't have those answers, as 300 μg/L is 0.3 mg/L, which is not the answer.


text:
Experimental verification of the field effect
March's Advanced Organic Chemistry has the following paragraph

The other effect operates not
  through bonds, but directly through space or solvent molecules, and is called the field
  effect. It is often very difficult to separate the two kinds of effect, but it has been
  done in a number of cases, generally by taking advantage of the fact that the field effect
  depends on the geometry of the molecule but the inductive effect depends only on the
  nature of the bonds. For example, in isomers 1 and 2 the inductive effect of the chlorine
  atoms on the position of the electrons in the COOH group (and hence on the
  acidity, see Chapter 8) should be the same since the same bonds intervene; but the field
  effect is different because the chlorines are closer in space to the COOH in 1 than they
  are in 2. Thus a comparison of the acidity of 1 and 2 should reveal whether a field
  effect is truly operating. The evidence obtained from such experiments is overwhelming
  that field effects are much more important than inductive effects.

The diagram:
I simply do not understand how 2 is more acidic than 1. Since the field effect is greater in 1 , should not 1 be more acidic?

The inductive effect is transmitted through bonds; the field effect is transmitted through space.
The $\ce{C-Cl}$ bond is polarized due to the electronegativity difference between the two atoms (carbon 2.55; chlorine 3.16), the chlorine end of the bond being more negative than the carbon end of the bond. We could note this excess of electron density (partial negative charge) around the chlorines in your structures by placing a ${\delta^{-}}$ by the chlorine and a ${\delta^{+}}$ by the carbon.
Now, when the acid ionizes there is also a negative charge associated with the carboxylate group.  When the chlorines are on the same side of the bridge as the carboxylate (syn isomer) the (through space) field effect acts to destabilize the situation because we have placed the partial negative charge around chlorine closer to the negative charge on the carboxylate (placing the negative charges closer together is destabilizing since like charges repel). The destabilization results in a higher activation energy being needed to reach the transition state for ionization.
On the other hand, the anti isomer with the chlorines further from the carboxylate will be less destabilized because the negative charges are further apart.  This means that the anti isomer will have a lower activation energy for ionization compared to the syn isomer.  Therefore, the anti isomer will be more acidic (display a lower pKa) than the syn isomer due to the field effect.

Ahh , cool! So field effect here is destabilizing while inductive is stabilizing. Thanks a lot!


text:
Why can some redox half-reactions be combined and some cannot?
I mean by "combining" is to make a new half-reaction equation and not an overall equation for a reaction in whole.
For instance, I was trying to arrive at the following half-reaction:
$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$
by combining
$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ=  1.51 \,\mathrm{V}$
and
$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ=
1.7\,\mathrm{V}$
using Hess's Law. My calculation yielded +0.19 V, and not +1.23 V as I would expect.
Why caused this? I know that there are half-reactions that can be constructed in this way (as above, by combining two or more other half-reactions) that result in the correct potential being calculated.

I think there is a mistake in your middle equation. It is not balanced w/r/t H or O: $$\ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O}$$

Also, for folks who might expect that there is a mistake in the reduction potentials, here is a table with references. Hess's law should work here...

I think you got -0.19V. At least, that is what I get.

@LDC3 - I also got -0.19 V, but for the sake of preserving the original content of the post, kept the answer the OP got in there.

In general, you cannot simply subtract electrode potentials like this to find the potential of another half-reaction. They have to be weighted by $n$ (the stoichiometric coefficient of $\ce{e-}$). In cases where your method works, it is only because $n$ is coincidentally the same for both half-reactions you are combining.
The foolproof way is always to convert electrode potentials to Gibbs free energy changes:
$$\Delta G^\circ = -nFE^\circ$$
So $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ is
$$(-5)(96485 \text{ C mol}^{-1})(1.51 \text{ V}) = -728.4 \text{ kJ mol}^{-1}$$
and $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}$ is
$$(-3)(96485 \text{ C mol}^{-1})(1.7 \text{ V}) = -492.1 \text{ kJ mol}^{-1}$$
Subtracting the second equation from the first, you will find that $\Delta G^\circ$ for the half-reaction of interest ($\ce{MnO2 + 4H+ + 2e- -> Mn2+ + 2H2O}$) is
$$(-728.4) - (-492.1) = -236.3 \text{ kJ mol}^{-1}$$
which gives:
$$E^\circ = -\frac{-236.3 \text{ kJ mol}^{-1}}{(2)(96485 \text{ C mol}^{-1})} = 1.22 \text{ V.}$$
I presume some rounding errors led to the slight discrepancy between my answer and the answer you want. You may leave out the Faraday constant in the calculations once you are more familiar with them; basically that means you work with the quantity $nE^\circ$ instead of $\Delta G^\circ$. The answer should be the same since they are directly proportional.
For more information you can refer to the redox chapters in any inorganic chemistry textbook (Shriver/Atkins, Housecroft, etc.). These calculations are dealt with under the sections discussing Latimer and Frost diagrams.


text:
Software for viewing a structure and altering bond angles
Does anyone know of any good software that would allow me to view the structure of a compound while also allowing me to change the bond angles in the structure (and give me numeric values for those angles)?
I'm looking for something freely available, but that's not a major requirement. Thank you so much!

Have a look at Avogadro!

build your own structures or import files in various formats.
optimize the geometry using different force fields
change and measure bond length, bond angles and dihedral angles

Once you've used it, you don't want to miss it again.

Geoff Hutchison, who cannot be praised enough for his child Avogadro, has given the right directions in his comment. I'm just adding some images to illustrate how easy angle adjustment is:

Draw a molecule in Avogadro
From the main menu, select Build -> Properties -> Angles or whatever the localized menu of your system shows (on my German machine, it is Ansicht -> Eigenschaften -> Winkeleigenschaften)
In the table that pops ups, select the angle you want to change. Note that the atoms of the selected angle are highlighted

Change the value for the angle. Note that the geometry of the molecule is immediately updated!

So again, I think that Avogadro is one of the best tools available and it is free and open source! Use it in education, use it in research and whenever you do, cite the Avogadro paper:
Hanwell, M. D.; Curtis, D. E.; Lonie, D. C.; Vandermeersch, T.; Zurek, E.; Hutchison, G. R.; Avogadro: an advanced semantic chemical editor, visualization, and analysis platform. J. Cheminf. 2012, 4 (1), No. 17. DOI: 10.1186/1758-2946-4-17.

With the latest beta (5) of Chem4Word 2020 we have introduced the ability to set (fix) bond angles to any value in the range -180 to +180 (measured relative to Screen North)

Ensure that ACME (our brand new editor) is selected as your editor in the Options dialogue
Import your structure (MOLFile or CML)
EDit your structure, then right-click on the bond you wish to rotate.

If the bond is a terminal bond only that will be rotated about its anchor point
Otherwise, the whole structure will be rotated.

The text box is free text and is validated to contain numbers in the range -180 to 180.  If this contains invalid numbers a save can still be performed so that other changes can be carried out.
You can then export the structure as MOLFile (SDFile) or CML.
Head over to Chem4Word Homepage to find out what else it can do.
Please read our beta testing guide Link on our download page before you start to use the beta.

fair point by @andselisk. With v3.0 using ChemDoodle Web specifying bond angles is not supported.
I will ask the team about whether this can be added to our own editor in V3.1

With the embedded version of ChemDoodle web, when adding a "C" bond the bond angle is snapped to 15 degrees (I think). With our new editor inserting a "C" bond or moving a "C" atom the bond angles are snapped to 15 degrees.

Let us continue this discussion in chat.

@andselisk I have updated the post to reflect new functionality in beta 5 of our new product.


text:
Find the equilibrium constant
In a container with volume 4 litres we have 2 moles $\ce{COCl2}$ and 4 moles $\ce{Cl2}$. Equilibrium is achieved when 60 % of the reactant $\ce{COCl2}$ is wasted 60 %. What is the equilibrium constant $\ce{CO + Cl2 <=> COCl2}$?
In my book it is explained the solution. I understand it when it says that $\ce{PCl5}$ is the substance that is wasted, but after finding the concentration of each substance to find the constant it uses the formula: $k=I$ understand that we get the concentration of $\ce{Cl2}$, $\ce{COCl2}$, but then when he finds the constant it divides $[\ce{CO}]\cdot[\ce{Cl2}]$ with $[\ce{COCl2}]$.
But I think that they should divide $[\ce{COCl2}]$ with $[\ce{CO}]\cdot[\ce{Cl2}]$.


text:
Which carbocation is more stable, the ethyl- or 1-propyl-carbocation?
$\ce{C2H5+ vs \ C3H7+}$
There are two conflicting trends here.
Inductive effect of ethyl will be higher than that of methyl so carbocation attached to ethyl (3 carbons in total) should be more stable.
On the other hand 3 hyperconjugation structures can be drawn for the carbocation attached to methyl (2 carbons in total) and only 2 hyperconjugation structure for the other.
Which effect wins? I searched a lot of books but didn't find anything.

Further reading https://chemistry.stackexchange.com/questions/139494/order-of-reactivity-towards-cesn1

The ethyl carbocation ($\ce{C2H5+}$) has a different structure than what you might have anticipated.  It is a non-classical ion with a bridging hydrogen.
For a description of just what a non-classical carbocation is see this answer.
The hallmark of a non-classical ion is 3 atoms with 2 electrons spread over them.  This is called a 3-center 2-electron bond (hypercoordinate bonding). In the case at hand the 3-atom 2-electron bond involves the 2 carbon atoms and the bridged hydrogen atom.

So there is no "methyl" group present in the ethyl carbocation that can inductively release electrons.  Additionally, the hyperconjugated resonance structure you mentioned has actually been replaced by a hydrogen that is partially bonded to the other carbon atom.
The n-propyl carbocation would likely have a similar non-classical structure - if it existed at all.  This carbocation has not been experimentally observed, it rearranges too rapidly to the much more stable 2-propyl carbocation.
So while we can't answer your question about whether inductive or hyperconjugative effects are more important in this series of compounds, we can assess which cation is more stable.
A variety of experiments suggest that the heat of formation of the non-classical ethyl cation is around 215 kcal/mol (reference, see pp. 68-72). This same reference points out that the heat of formation of the classical 1-propyl cation lies approximately 20 kcal/mol above that of the 2-propyl cation.  The heat of formation of the 2-propyl cation is around 193 kcal/mol.  Adding 20 kcal/mol to that value would place the classical 1-propyl cation around 213 kcal/mol. But, assuming the 1-propyl cation would also be a non-classical ion it would be lower in energy than the classical version, its heat of formation would therefore be less than 213 kcal/mol.
This analysis suggests that the 1-propyl cation would likely have a lower heat of formation than the ethyl cation.  In other words, the 1-propyl cation would likely be more stable than the ethyl cation.

But what is the reason behind the increased stability of the 1-propyl cation over the ethyl cation? With reference to an inductive argument, does a longer chain mean a "larger stock of electrons" to release to stabilise the carbocation?

You never defined which carbon is the carbocation in the case of the propylium carbocation. I will go through a comparison of all three possibilities: the case of the ethylium carbocation, the linear propylium carbocation with a terminal carbocation, and the linear propylium carbocation with an internal carbocation.
Thermodynamically: Ethylium < terminal propylium < internal propylium
Kinetically: Terminal propylium < ethylium < internal propylium
The difference in the kinetic and thermodynamic stability arises from the fact that the terminal propylium will readily undergo a 1,2-hydride shift to create the internal propylium ion.
As for the thermodynamic order, it gets a bit more complex. Remember that for alkanes, it is ALWAYS the case that the more substituted carbocation is more stable and in cases of ties, more substituted substituents make for more stable carbocations. As for why this is true, well it comes down to the fact that carbons are electron donating to carbocations (and other electron deficient carbons). There is a role from hyperconjugation but you are forgetting that you can have hyperconjugation from bonds other than $\ce{C-H}$. You also seem to think that more hyperconjugatory structures is better when really that is not the case. Hyperconjugation is NOT an example of resonance where the molecule is in a superposition of multiple states so there is no enthalpic benefit of more methods of hyperconjugation. There is an entropic effect from the additional degrees of freedom but this would be a very small difference and if I remember right, would actually benefit both propylium carbocations, not the ethylium carbocation.

Why isn't hyperconjugation an example of resonance, there is no nuclear movement in such structures?

I'm sorry for not making it clear but I mean the  linear propylium carbocation with a terminal carbocation.

Also all books I've read state that more hyperconjugation structures imply more stability. March in his book for example , gives this as the reason why secondary carbocations are more stable than primary ones.(Along with inductive effect explanation).

See http://chemistry.stackexchange.com/questions/4151/why-does-carbocation-stability-increase-in-the-order-1%C2%B0-2%C2%B0-3%C2%B0?rq=1 for another example.


text:
How can I balance the following equation atomically and electrically?
$$\ce{C2O4^2- + MnO2 -> Mn^2+ + CO2}$$
I think that the half reactions are
$$\ce{C2O4^2- -> CO2}$$
$$\ce{MnO2 -> Mn^2+}$$
I am supposed to balance these by adding water, $\ce{H+}$ atoms and by adding $\ce{e-}$’s, but I’m just not sure on the method to do this as we’ve covered it extremely quickly.
First, I found the oxidation numbers for the overall equation, and I think that $\ce{C2O4^2-}$ is the reducing agent because $\ce{C}$ is losing charge from +3 to +4, I just don’t know how to use that to balance this. Any help would be appreciated.

Deal with the two half equations separately and then combine them.
Starting with the oxalate equation: $$\ce{C2O4^{2-} -> CO2}$$
Balance the atoms: $$\ce{C2O4^{2-} -> 2CO2}$$
Now balance the charge by adding electrons: $$\ce{C2O4^{2-} -> CO2 + 2e-}$$
Now for the manganese dioxide reduction: $$\ce{MnO2 -> Mn^{2+}}$$
We can balance the half equations by adding water, hydrogen ions or electrons. Since water is the only one which contains oxygen we should add this first: $$\ce{MnO2 -> Mn^{2+} + 2H2O}$$
Now add hydrogen ions to balance the hydrogens: $$\ce{MnO2 +4H+ -> Mn^{2+} + 2H2O}$$
Finally add electrons the balance the charge: $$\ce{MnO2 +4H+ + 2e- -> Mn^{2+} + 2H2O}$$
Now combine the equations to cancel all the electrons. In this case there are two on both sides so we don't need to multiply any of the equations by anything.
$$\ce{MnO2 +4H+ + C2O4^{2-} -> Mn^{2+} + 2CO2 + 2H2O}$$

It might be a good idea to balance your half reactions first and then sum up:
\begin{align*}
\ce{C2O4^{2-} &-> 2CO2 +2e-\\
MnO2 + 4H+ + 2e- &-> Mn^2+ + 2H2O}\\
\hline
\ce{MnO2 + 4H+ + C2O4^{2-} &-> Mn^2+ + 2CO2 + 2H2O}\\
\end{align*}


text:
Difficulties understanding how salt bridge works
A picture for reference:
Difficulty 1: I don't get why a salt bridge is needed. People say it's to maintain the overall neutrality. But in both cases where  salt bridge is a tube of solution and a membrane, two half cells can be seen as one. So as long as no thing leaves either of the two half cells, the whole is always neutral.
Difficulty 2: as in the picture, cations go to the right cell, and anions go to the left. I can understand that anions go the left because the number of zinc ions increase as the reaction proceed. But in the right half cell, the electrons are being absorbed, its charge cancelled by that of the copper ion; I don't see any need for sodium ions to move to the right.

related,but not a duplicate http://chemistry.stackexchange.com/questions/12332/is-there-always-a-need-for-a-salt-bridge-to-let-a-galvanic-cell-work-continuousl

You might want to give a quick search on salt bridges because there are a lot of these questions.

Regarding Difficulty 1)  When the zinc bar (unconnected) is first immersed into the solution, some zinc ions immediately dissolve away.  Now the zinc bar has extra electrons and the solution has extra positive zinc cations.  This is entropically acceptable, however, because cations in the solution (both zinc cations and hydronium cations) can align close to the surface of the zinc metal (creating what's called the 'Helmholtz layer').
One might imagine that as soon as the zinc bar was connected, the extra electrons in the zinc would drain to the other side of the cell.  But they don't because the electrostatic pull of the zinc and hydronium cations in the left-beaker solution is holding them back.
There's an analogous situation in the right cell:  the copper metal can't accept any electrons because of the electrostatic repulsion of the sulfate and hydroxide ions in right-beaker solution.
Regarding Difficulty 2) When the cell is running, there is an electrostatic gradient across the membrane where the salt bridge meets the right-side solution (among other places, for more info see note 1).  Cations (such as sodium) are driven out of the salt bridge;  anions (such as sulfate) are drawn into the salt bridge.  Sodium cations are smaller and diffusion more quickly than the larger, slower sulfate anions. Overall, many more sodium ions will leave the salt bridge compared to the number of sulfate anions that will enter the salt bridge.  There is an analogous situation at the membrane where the salt bridge meets the left-beaker solution.
Note 1, Electrostatic Gradients) When the cell is running, the zinc bar has a slightly negative charge and the copper has a slightly positive charge.  If we let ourselves ignore the metal wire for the moment, this creates an electrostatic field throughout  the cell very much like the classic "electrostatic field around adjacent positive and negative charges in a uniform media".  Notice that in the diagram of that situation, there are electrostatic field lines pointing from the positive charge straight-across to the negative charge.  Another way to describe this situation would be to say that there is an electrostatic gradient between the two charges.  If you drop a much smaller, and otherwise free-floating positive charge any where in the neighborhood around those two charges, it will move down electrostatic gradient to the negative charge:  the same way rain-water flows downhill, (in that case due to a gravitational gradient).  Now imagine there was an imaginary plane exactly halfway between the two charges:  you could now also say that there is an electrostatic gradient across that plane, from the one side to the other. If you dropped a positive charge right beside the plane, but on the same side as the bigger positive charge, the electrostatic gradient would drive the positive charge across the plane.  Similarly, if you dropped a negative charge next to the plane,  on the same side as the bigger negative charge, then again the electrostatic gradient would drive the negative charge across the plane.
In the electrochemical cell problem, the electrostatic field lines are going to be more complicated because there are two materials (air and water / salt-bridge) with different permittivity.  Long story short: the electrostatic fields will actually be much stronger in the water than in the air.  The strongest part of the electrostatic field will go from the copper electrode, across the right-side solution, through the salt bridge, and across the left-side solution to the zinc.   So, among other places, when the cell is running, there will be an electrostatic gradient across the membrane where the salt bridge meets the right-side solution.

I have BIG trouble understanding you mean by electrostatic gradient. Still, my question remains that if the charge is already balanced, why do we need Na+ to go the right half cell?

Difficulty 1: I don't get why a salt bridge is needed. People say it's to maintain the overall neutrality. But in both cases where salt bridge is a tube of solution and a membrane, two half cells can be seen as one. So as long as no thing leaves either of the two half cells, the whole is always neutral.

Salt bridge is to:

maintain neutrality
complete the circuit

Your problem is that through wires electrons move from one compartment to another, but if there's no salt bridge and charge cannot build up, current simply does not flow.

Difficulty 2: as in the picture, cations go to the right cell, and anions go to the left. I can understand that anions go the left because the number of zinc ions increase as the reaction proceed. But in the right half cell, the electrons are being absorbed, its charge cancelled by that of the copper ion; I don't see any need for sodium ions to move to the right.

Yes you are right, since charge is cancelled, we need to get back the positive charge back in right compartment.
Similarly in left as zinc is increasing, positive charge is increasing, so to balance sulphate moves.
In both cases, the total charge in a vessel must remain constant (not neutral).

I'll put it in a different way for you, so essentially what we are doing is that one side is taking electrons, so it becomes more negative, and the other side becomes positive. And the salt bridge is to maintain this electroneutrality or in other words, to maintain one side from becoming too negative or positive.
And this is exactly why we need a salt bridge with the $\ce{Na+}$ and $\ce{SO4^{-2}}$
So at the end of the day, the positive cations go to the cathode and the negative anions move to the anode.


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Yeast Fermentation with YPD + Apple Juice
I am working on a project for my General Chemistry class.
I am trying to compare the EtOH production (and the rate of EtOH production) between seven types of yeast strains by fermenting the YPD media, apple juice, or both, if necessary. I can describe the difference between the yeast strains if asked, but I don't think it is necessary right now.
Right now, I have only tested plain YPD media, which is 1% yeast extract, 2% peptone and 2% d-glucose mixture. All strains yield around 8-9 mg/mL of EtOH, and the fermentation stops after ~24 hours. My next step is to test YPD + apple juice media (because the project asks for "fun" element), and I was wondering if I should use the same method for analyzing this YPD + apple juice fermentation.
The method I used for YPD media is:

Grow each strain (total of 7 strains) in ~5 mL YPD media at 30 degrees shaker ~250 rpm overnight.
Subculture each strain into new fresh YPD media so that the new media is 50 mL and has 0.5 OD at 600 nm. Use the "special" flask designed for anaerobic growth. This flask is a 250 mL flask with a special cap. The cap is plastic at the outer radius and rubber at the inside. This allows me to grow yeast anaerobically and poke a needled syringe and retrieve small amount of sample without letting the air in.
Flush all flasks with nitrogen gas for four minutes.
After 6 hour, 12 hour, and 24 hour growths, use syringe to retrieve ~1.5 mL of sample.
Measure OD600 of each sample. (This is used to get the number of yeast cells in the media)
Centrifuge samples for 8 minutes in 14,000 rpm.
Pipette 1 mL of the supernatant into a GC vial.
Use HPLC to obtain concentration of ethanol. I am using refractive index detector, ion exclusion column(?) (I forgot the type of column used at this moment), and sulfuric acid as mobile phase.

For YPD + apple juice fermentation, I was thinking of doing the same procedure with 50 mL YPD + 50 mL apple juice in each "special flask". Will this work? Will this be good for the same HPLC setting?

Thank you for an especially clear and detailed question!

For YPD + apple juice fermentation, I was thinking of doing the same procedure with 50 mL YPD + 50 mL apple juice in each "special flask". Will this work?

I think this would work but it seems like a better comparison to your earlier experiments to keep using 50 mL of total medium, i.e., mix 25 mL of YPD with 25 mL of apple juice.  That way the volume of the fermentation is the same in each experiment.

Will this be good for the same HPLC setting?

Probably, although it depends on the components of the apple juice you use.  Have you injected just apple juice (unfermented) into your HPLC?  Have you verified that this negative control injection contains no ethanol or other interfering peaks in the region of interest?  If not, those are things you should do to make sure the HPLC method you have developed for YPD medium will still work in apple juice-based media.


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In electrolysis, does it make any difference if the electrolyte is an molten, aqueous or dilute aqueous solution?
I'm an IGCSE Chemistry-extended student, and my question is: in electrolysis, does it make any difference if the electrolyte is in a molten, aqueous or dilute aqueous solution form?
An example mentioned in my book:

Explain  the difference in the electrolysis products of
a) molten sodium chloride;
b) dilute aqueous solution of sodium chloride.

In molten sodium chloride, the only available ions are $\ce{Na^+}$ and $\ce{Cl^-}$. With electrolysis, the products will be sodium metal and chlorine gas.
In an aqueous solution, the sodium chloride still moves the charges from one electrode to the other, but the potential for electrolysis of water is less than $\ce{Na^+}$ or $\ce{Cl^-}$. So the products of a solution of sodium chloride is hydrogen gas and oxygen gas.

I think adding corresponding reactions would be a nice addition (for clarity).

Depending on concentration, it may be chlorine instead of oxygen due oxygen overpotential


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Is the 'n' in Delta G = -nFE, ever negative?
Is it possible for $n$ in $\Delta G=-nFE$ to ever be negative? Trying to figure out where I went wrong on some thing and want to rule that out.

No. This result occurs from the Nernst equation.
The "n" is the number of electrons transferred. If no electrochemical reaction occurred, then n = 0.
Otherwise "n" is positive.
Having a negative number of electrons transferred would be impossible.

If no electrochemical reaction occurred, then n = 0. really?

The lowest limit is when no oxidation or reduction occurs. At that point, n = 0. I answered the OP's question, which is that it can never be negative.

And indeed, there is a direct relationship between Nernst and $\Delta G = -n F E$ as indicated in the link in my answer (i.e., here) or in many other sources.

@ADG I think you're being incredibly rude. I teach physical chemistry. I don't need to look this up on Google.

Could it be that both these standard equations can be derived from each other? Backtracing the chemwiki derivation?

Indeed, derivations can go both directions.

@ADG I've been responding to flags for 5 minutes and I've run into a multitude of rude, obnoxious, and otherwise non-constructive comments that you've left.  It's going to be a matter of shape up or ship out here.  Some radical changes need to be made immediately.  Every user here deserves your utmost respect and courtesy and no one needs your grandstanding.

@ADG, it's not wrong, because you can derive the Nernst equation without using that exact formula (here's one example, using the Boltzmann distribution.).

@J.LS OK. I never saw such a super high level proof. sorry

Is it possible for $n$ to ever be negative? Trying to figure out where I went wrong on some thing and want to rule that out.

$\Delta G$ is the maximum non expansion work and hence should be equal to external force under consideration times the distance for a reaction to be at equilibrium, hence we need to find that external force.
I know that from electrostatics that for a charge in a uniform electric field the force is $\bar F=q\bar E$. for $q=-1.6\times10^{-19}\ \mathrm C$. When we take our system to be $1\ \mathrm{mol}$ of electrons, we have a total of $-1.6\times10^{-19}\ \mathrm C\times6.02\times10^{23}\approx -96500\ \mathrm C$ and we have that $1\ \mathrm F=96500\ \mathrm C$. So total force is:
$$\Delta G=W_\text{max,add}=\bar F_\text{ext}\cdot \bar d$$
$$\Delta G=\bar F_\text{ext}\cdot \bar d=q_\text{total}\cdot E\cdot 1\ \mathrm m=(-n) _\mathrm F\cdot E _\mathrm V=-nFE_\mathrm J$$
Now think, can the number of moles of electrons transferred be negative?

$n$ cannot be negative.
$$\Delta G = -nFE$$
Transfer of $\ce{e-}$ cannot be negative.

This answer has been flagged as low quality, probably because (1) you didn't make the effort to write full sentences, including punctuation or (2) you did not provide a clear answer. Please make changes or the answer risks being deleted.


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Why is the ratio between Silicon and Oxygen 1:3 in single chain silicates?
I thought that the ratio of silicon to oxygen in a silica tetrahedron was 1:4, so if a single chain is just many of these linked together, why does the ratio become 1:3?

Silicon has 2 complete oxygen atoms ($2\times1=2$) and 2 equally shared oxygen atoms ($2\times1/2=1$), total 3.


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Does Wurtz reaction follow an ionic mechanism or a free radical mechanism?

In a Wurtz reaction, is the intermediate formed a carbanion or a free radical?
And if we can assume both mechanisms to be true, how can we decide the reactivity of a tertiary halide?

I think your second question is fine as is. But, in the context of your first question, I just want to verify for myself that you do mean overall reactivity of a tertiary halide via the Wurtz reaction, irrespective of reaction pathway, and that you are not asking whether the reactivity will be dominated by carbanion-forming v.s. free-radical forming pathways.  Wow, ancient question ;) I hope we can lead you to an acceptable answer!

I think both mechanisms are valid. As far as I know, both are supported by some experimental evidence. Byproducts, like alkanes (other than the expected one) and alkenes, can be explained by the free radical mechanism. On the other hand, the formation of Organometallic compounds can be verified by the ionic mechanism.
The essential thing to note is that the mechanism involves a $\mathrm{S_N2}$ step. And because of that sterically hindered alkyl halides, like tertiary alkyl halides, instead of giving a substitution product yield an elimination product and you get an alkene.

A mechanism is only a hypothesis which can explain a particular reaction. Wurtz reaction can be explained by both ionic and free radical mechanism.

Well I personally think it is better explained by free radical mechanism, as using this mechanism, you can easily analyse all the products formed, whether it be symmetrical compound formed from same alkanes, or an unsymmetrical compound.