text:
Composition and pH of water produced by air conditioners
I would like to know if there are specific studies or research regarding the condensed water that is generated by air conditioners.
Many people feel that this water is acid and corrosive but could not find anything that confirms or denies this belief.
If someone could send me a link or document with a reliable scientific research would help me.

Acid? Why? Condensed water is essentially distilled water.

I suspect some people might think this because all that water dripping out of the AC unit over the course of a few years will begin staining/eroding whatever is in its path. That's just water being water, though.

Distilled water will dissolve carbon dioxide from the atmosphere, which could reduce the pH below 7. Here's a technical document for  water stills

True, but I deliberately tried to avoid this topic. Anyway, this is not what people outside of chemistry mean by "acid".

Additionally, if it's central AC and the blower unit is shared with a high efficiency gas furnace, the condensate from the furnace is quite acidic due to sulfuric acid in the combustion gas. So they may be confusing that with the condensate that comes off during cooling since they usually come out of the same drain pipe.

Pure condensed water is neutral (pH 7). However, water in contact with air is slightly acidic (approximately pH 5.6) due to dissolved carbon dioxide $\ce{(CO2)}$. Remarkably, the same applies to rainwater.
Furthermore, water condensed from air is not pure. The condensing water collects small amounts of various airborne substances (soluble gases and vapours, dust, and even radioactive particles from radon decay). (Note the large volume of air that is passed through the air conditioner while a certain amount of water is condensed.) Usually, these substances do not significantly affect the pH of the water. However, when the air contains large amounts of pollutants, such as sulfur dioxide or nitrogen oxides (e.g. produced in industrial processes), the condensed water can become more acidic (like acid rain, which can have pH values of about 4).
By way of comparison, typical values for black coffee are about pH 5.0–5.1, and for orange juice about pH 3.3–4.2.

What about the unpleasant smell emanating from the shop floor after rinsing the floor with the water  produced by the air conditioner? What is the reason for this smell?


text:
Why is Hessian Matrix called 'force-constant' matrix?
Hessian Matrix is a square matrix containing the elements as the second-order partial derivatives of energy-function of a molecule; the derivative is done with respect to geometric coordinates of the molecule.
However, I couldn't get why Hessian matrix is referred to as force-constant matrix; as is done in the book I'm following viz. Computational Chemistry: Introduction to the Theory and Applications of Molecular and Quantum Mechanics by Errol G. Lewars and in many other sources.
If the energy function is a quadratic function which is mostly the case for a molecule following 1D PES; the function which a Hookean spring follows for small displacement, then the second derivative does indeed give the value of force constant, that is $$\begin{align}E- E_0&= k(q-q_0)^2\\ \implies \frac{\mathrm{d}^2 E}{\mathrm{d}q^2}&= 2k\;.\end{align}$$
However, for molecules following multidimensional PES, most of the time, the energy function is not simple quadratic; how can the second order derivative give the force constant? Second order derivatives vary from point to point unlike the simple case above.
So, can anyone tell why Hessian matrix is called force-constant matrix?

I didn't know what to tag it: since I got the issue related to computational chemistry, I tagged it so. I wanted to tag it as [tag:soft-question]; but it doesn't exist here:(

Excuse me if this comment is more of a question than a comment, my background is math rather than chem, and, in math, Hessian matrix has a far more abstract definition than it is for chem, guessing by your question. So for the outsiders, how is your Hessian matrix defined? (Link preferred above explanation).

@GyroGearloose http://www.bioinfo.rpi.edu/bystrc/courses/biol4550/lecture20.pdf (please see slide 20-29)

Thanks, so "your" Hessian is about the energy of a molecule by displacement of the atoms, right?

The Hessian is calculated in the equilibrium geometry when used for force constants and such. The fact that the reality is non quadratic, it does not influence the fact that we use quadratic models (or else it wouldn't be constant, would it be) - most approximations still use the harmonic approach for the force fields.

You are right that multidimensional PES is not simple quadratic, but close to the PES well minima it can be approximated as harmonic oscillator. Also if you really want to include the real PES you can include anharmonicity in your calculation. In that case you have to include an anharonic constant $x_e\;.$ Your PES can be written as
$$E=\left(n+\frac{1}{2}\right)h\nu-x_e\left(n+\frac{1}{2}\right)^2h\nu$$
Usually in computational chemistry the energy can be expanded into a Taylor series
$$E=E_0+\frac{\mathrm{d}E}{\mathrm{d}x}(x-x_0)+\frac{1}{2}\frac{\mathrm{d}^2E}{\mathrm{d}x^2}(x-x_0)^2+\textrm{higher order terms}\;.$$
And you can usually neglect those first order term and higher order terms for most of the practical problems. And then your PES will follow the trajectory of a harmonic oscillator.

Thanks for the answer.... could you please tell how you came up with the formula above the expansion?

Please see the attached link:
http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/06._One_Dimensional_Harmonic_Oscillator/Harmonic_Oscillator/Anharmonic_Oscillator

@mamun The Taylor expansion is missing a  1/2

yes, you are right. I am changing it.

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text:
What is the lowest mass measurable using a 4 decimal point analytical balance?
I think it is 1 mg, but I've heard people saying 0.1 mg. As 0.1 mg = 0.0001 g, which is the uncertain digit, I think such a measurement is not reliable at all.
Any opinions ? Thanks.

An odd point is that for significant figures one would assume that the true measurement of such a balance would be ± 0.00005 g.

Keep in mind that for many analytical instruments, you cannot assume that the uncertainty is ± the last digit.
It's usually necessary to look at the specifications in the manual or on a sticker to find the measured uncertainty.
For instance, one of our balances, a Sartorius CPA225D (manual), is readable to 0.01 mg (i.e. the display can display increments of 0.01 mg), but the specifications list the repeatability as only ±0.05 mg (±1 s.d.).
This means that repeated measurements of the same mass will be (ideally) normally distributed with a standard deviation of 0.05 mg and that even though the instrument has the resolution to report in 0.01 mg increments, the uncertainty in a measurement is higher.

This precisely!

I think that by 4-digit analytical balance; you mean 4 decimal places. in which case the error would be ±0.0001 g.
If you were correct in that it was actually a 4-digit balance then the uncertainty would be ±0.001 g. However, this would mean that the balance only weighs up to 9.999g and balances usually allow more than that by my understanding. Hence why I think it would be 4 decimcal places and correct to ±0.0001 g.

EDIT

The scales I use have a repeatability of ±0.0001 g. My understanding is that if I have 5.0006 ±0.0001 g then there is doubt about the last digit. For every reading we take there will always be a measure of doubt. In this case it is important to highlight how big that doubt is, so it cannot simply be ignored. we have an uncertainty of ±0.0001 g meaning the true value could read 5.0005-5.0007 g. If you thought the uncertainty was 0.001g of the balance reading then our uncertainty would become 5.000-5.001 g. But the balance is certain that it doesn't weigh 5.000 g.

Sorry about that. I meant 4 decimal points. I've corrected the question.

i'll edit my answer about the reliability of the measurement

@jan I have seen analytical balances with 7 decimal places in the pharmaceutical industry

@jan ah I think I understand you. are we agreeing that the last value is the uncertainty?

@LiamH If the balance readout contains 4 decimal points, doesn't that mean the 4th decimal to be the uncertain digit ?

@LiamH Please explain "if I have 5.0006 ±0.0001 g then the last digit has some reliability because I know i don't have 5.007 g".

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text:
How can collission theory of chemical reaction and Law of mass action be connected?
From Wikipedia,

Collision theory is a theory ... that qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions. The collision theory states that when suitable particles of the reactant hit each other, only a certain percentage of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions. The successful collisions must have enough energy, also known as activation energy, at the moment of impact to break the preexisting bonds and form all new bonds.

From Britannnica,

Law of mass action, law stating that the rate of any chemical reaction is proportional to the product of the masses of the reacting substances, with each mass raised to a power equal to the coefficient that occurs in the chemical equation.

How can both explain and imply each other?
My theory: Due to more availability of reacting species (due to more amount), rate of reaction of increases. But it does not explain the reason of exponentiation of stoichiometric coefficients.

For exponentiation, we can say that it is actually repeated multiplication. Comparing two cases:

$$\ce{A + B -> AB}$$
Here, the rate expression is $r=K[\ce{A}][\ce{B}]$. This is because one molecule of $\ce{A}$ and one molecule of $\ce{B}$ is needed for the product to form.
$$\ce{2A + B -> A2B}$$
First, write it in this way: $$\ce{A + A + B -> A2B}$$
Now, rate expression is $r=K\ce{[A][A][B]}$.
That's it - $r=K[\ce{A}]^2[\ce{B}]$

An intuitive starting point for me to understand why concentrations are multiplied is probably to connect it with multiplication theorem in probability (for independent events).

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text:
Diagonalization of Hessian H= PkP$^{-1}$: what do the elements of P matrix physically mean?
This is quoted from Computational Chemistry by Errol J. Lewars' chapter 2's Stationary Points & Normal-Mode
Vibrations: ZPE:

\begin{align}\mathbf H &=\begin{bmatrix}
    \dfrac{\partial ^2 E}{\partial q_1 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_1 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_1 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_1 \partial q_9} \\
    \dfrac{\partial ^2 E}{\partial q_2 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_2 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_2 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_2 \partial q_9}\\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    \dfrac{\partial ^2 E}{\partial q_9 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_9 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_9 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_9 \partial q_9}
\end{bmatrix}\\ \\ &=\underset{\mathbf P}{\begin{bmatrix}
    q_{11} & q_{12} & q_{13} & \dots  & q_{19} \\
    q_{21} & q_{22} & q_{23} & \dots  & q_{29} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    q_{91} & q_{92} & q_{93} & \dots  & q_{99}
\end{bmatrix}} \underset{\mathbf k}{\begin{bmatrix}
    k_{1} & 0 & 0 & \dots  & 0 \\
    0 & k_{2} & 0 & \dots  & 0 \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & 0 & \dots  & k_{9}
\end{bmatrix}}\mathbf P^{-1}\end{align}
The 9 × 9 Hessian for a tri-atomic molecule
  (three Cartesian coordinates for each atom) is decomposed by diagonalization into a P
  matrix whose columns are “direction vectors” for the vibrations whose force constants
  are given by the $\bf k$ matrix. Actually, columns 1, 2 and 3 of $\bf P$ and the corresponding $k_1,k_2,k_3$ of $\bf k$ refer to translational motion of the molecule (motion of the whole
  molecule from one place to another in space); these three “force constants” are nearly
  zero. Columns 4, 5 and 6 of $\bf P$ and the corresponding $k_4,k_5,k_6$ of $\bf k$ refer to
  rotational motion about the three principal axes of rotation, and are also nearly zero.
  Columns 7, 8 and 9 of $\bf P$ and the corresponding $k_7,k_8,k_9$ of $\bf k$ are the direction
  vectors and force constants, respectively, for the normal-mode vibrations: and
  refer to vibrational modes 1, 2 and 3, while the 7th, 8th, and 9th columns of $\bf P$ are
  composed of the x, y and z components of vectors for motion of the three atoms in mode
  1 (column 7), mode 2 (column 8), and mode 3 (column 9).

I couldn't conceive what actually $\bf P$ implies; it is, of course, the eigenvector matrix of the Hessian $\bf H $ but what does it represent?
Lewars wrote that columns of $\bf P$ are actually direction vectors; now what did he mean by direction vector?
I thought $q$ represents the geometric coordinate; I got $k$ is the force-constant; but what are those series of $q$ s in $\mathbf P\;?$
Are the $q$ s in $\bf P$ coordinates of the atoms in molecules? If so, aren't they always changing since they are vibrating, rotating...?
Can anyone please explain this to me what those $q$ s actually represent? What did Lewars want to mean by direction vectors?

After a long time, I got a downvote; thank you- I was missing it. But I would be happier if Mr. Downvoter attempts to waste a few minute to tell where I committed a sin so that I may try to undo that. Please : (

Are the $q$s in $\mathbf{P}$ coordinates of the atoms in molecules?

Close, but not quite. The $q_{i,j}$ represent the displacements from the equilibrium geometry of the appropriate atom/coordinate $i$, for each mode j. Thus, for example, $q_{5,9}$ corresponds to the displacement from equilibrium of atom $2$ along axis $2$ (typically the $y$-axis) when the molecule is undergoing vibration in mode $9$.

If so, aren't they always changing since they are vibrating, rotating...?

There are no changes due to translation or rotation. It is customary to "project out" the translational and rotational motion from the Hessian and the geometry.  Qualitatively, this means that we simply ignore the translation and rotation and consider the molecule as "fixed" in its equilibrium configuration (allowing for vibrational 'wiggling'), with that configuration always oriented in the same way relative to the coordinate system.
As for the vibrational motion, though: yes, the $q_{i,j}$ values do technically also 'wiggle' as the molecule vibrates. However, the magnitudes of the perturbations $\delta q_{i,j}$ are in many cases very small, and can be neglected. In those cases where they cannot, the harmonic oscillator approximation to the vibrational motion breaks down and is no longer a useful model for vibration along that mode.
See my answers here and here for some links to literature on situations where this occurs (emphasis on the 'internal rotation' topics, though some of the 'anharmonic' and 'Coriolis' references might be somewhat relevant). Here is a Chem.SE question about nitrogen inversion that might be of interest.  More generally, a Google search for "low-frequency vibrational modes" should turn up lots of interesting reading.

First one should understand what normal coordinates and (mass-weighted) cartesian coordinates are: They are both a means to specify where in space each single atom of a molecule is located. In cartesian coordinates you do this simply by specifying the x,y and z coordinates of each single atom. Normal coordinates are a little trickier to understand, since they include always several atoms.
A simple example might illustrate this: Suppose you have two masses which are connected by a spring and can move only in one direction (the direction of the spring). Also assume that their mass is $m=1$ so we don't have to distinguish between the usual and mass-weighted cartesian coordinates. Then there are two ways to specify the position of the two masses in space: You can specify the x coordinates of the two masses (cartesian coordinates) or you can specify the center of mass and the distance between the two masses (normal modes).
To be more concrete, the dynamical matrix (Hessian of the potential in mass-weighted coordinates) of the above problem is
\begin{equation}
\mathbf H = \begin{pmatrix}
k & -k \\
-k & k
\end{pmatrix}
\end{equation}
where k is the force constant of the spring.
Diagonalization then yields the force constants 0 and 2k and the matrix
\begin{equation}
\mathbf P = \frac{1}{2} \begin{pmatrix}
1 & -1 \\
1 & 1
\end{pmatrix} \end{equation}
The matrix $\mathbf P$ now tells us how to compute the values of the normal coordinates $Q_i$ from the values of the cartesian coordinates:
\begin{equation}
Q_i = \sum_j q_i P_{ij}
\end{equation}
From this we see that the first normal coordinate is proportional to the center of mass and the second one proportional to the distance of the two masses. Note that my example used usual cartesian coordinates as $q_i$ while in real problems you usually use displacement coordinates. So you can see that the matrix $\mathbf P$ just let's you express the values for one set of coordinates in terms of the values for the other.
You can of course also do it the other way round and ask: what is the position/displacement of some atom, when I know that the normal coordinates have the values $Q_i$. Then you can calculate your cartesian coordinates via
\begin{equation}
q_i = \sum_j (P^{-1})_{ji} Q_j = \sum_j P_{ij} Q_j
\end{equation}


text:
Fluorine Preparation:How can this be explained on basis of Lewis Acid-Base Reaction?
How can this be explained on basis of Lewis Acid-Base Reaction ?
$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF3 + F2↑}$
References:
https://en.wikipedia.org/wiki/Fluorine#Industrial
Google Books reference (click me)

$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF3 + F2↑}$
Occurs in two stages. First there is a (Lewis) acid-base reaction
$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF4}$
where the Lewis acid $\ce{SbF5}$ accepts a lone pair from a fluorine in the manganese complex, and the adduct then falls apart to leave $\ce{MnF4}$. This is thermally unstable at ambient temperatures (Housecroft and Sharpe) and falls apart in the redox reaction
$\ce{2 MnF4 → 2MnF3 +F2}$

This is actually a double Lewis acid-base reaction. One Lewis acid-base complex is broken down while another is formed.
The starting compound potassium hexafluoridomanganate(IV) or $\ce{K2[MnF6]}$, features a complex hexafluoridomanganate(IV) ion which is a Lewis acid-base adduct. Upon adding $\ce{[SbF5]}$, we suddenly have two Lewis acids — $\ce{[MnF4]}$ and $\ce{[SbF5]}$ — both competing for the Lewis bases $\ce{F-}$. It turns out that $\ce{[SbF5]}$ is the stronger Lewis acid, i.e. the acid-base equilibrium is shifted to the product side.
$$\ce{[MnF6]^2- + 2 [SbF5] <=>> [MnF4] + 2 [SbF6]-}\tag{1}$$
Thus, the more stable Lewis acid-base adduct has been formed. Manganese(IV) fluoride is not stable with respect to degradation, i.e. the redox decomposition to form manganese(III) fluoride and fluorine is favourable.
$$\ce{2 MnF4 -> 2 MnF3 + F2 ^}\tag{2}$$


text:
Equivalent of Szabo and Ostlund book for DFT
Szabo and Ostlund book Modern quantum chemistry [1] is extremely useful to understand Hartree-Fock and post-Hartree-Fock methods. Not only it explains the theory behind such methods, but it is also oriented towards a practical implementation. Following this book I was able to implement an Hartree-Fock code from scratch in order to perform Born-Oppenheimer molecular dynamics.
Since I already work with software packages based on density functional theory (Gaussian09 for calculations in vacuum and QuantumESPRESSO for solid state systems), I would like to implement a simple DFT code in order to better understand what is going on inside the "black box". However, I would like perform this task using a plane wave basis set: this is somewhat more involved than using Gaussian basis sets since it involves periodic boundary conditions and pseudopotentials. In addition, I will have to take care of fast Fourier transforms everywhere and of the approximation of the exchange-correlation functional.
Supposing that I will be happy with the local density approximation, there is a book out there similar to Szabo's book for DFT? Notice that I am not interested in the basics of DFT but I am searching for a good book explaining in details the practical implementation using PW.

Here some comments on the books I already considered:
Martins [2]: This book contains a lot of informations and is oriented to solid-state electronic structure, therefore it explains plane waves and pseudopotentials. However I think that it works well as a reference but not as first reading to understand in details the concepts.
Giustino [3]: Very nice book with the goal of teaching DFT to undergrad students. For this reason is too simplistic and does not go into technical details.

[1] A. Szabo and N. S. Ostrlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Doever Publications, 1996.
[2] R. M. Martins, Electronic Structure: Basic theory and practical methods, Cambridge University Press, 2004.
[3] F. Giustino, Materials modelling using density functional theory, Oxford University Press, 2014.

I'm not confident enough to post this as an actual answer (might be too basic), but the CPMD folks cite v2, part 5 of Handbook of Molecular Physics and Quantum Chemistry, "Density Functional Theory: Basics, New Trends and Applications." Unfortunately, it looks terrifically expensive, though it's available in a few European libraries.

Assuming LDA with no exact exchange and pseudopotentials the job is plain and simple search of minimum of a functional. Assuming plane-wave, the function in question is decomposed into a set of plane waves, so the job is minimaztion of a function over the decomposition coeffecients.

Have you had a look at this question? Maybe you could comment on the sources that are referred there, to make it easier to not give duplicate answers.

@Martin-マーチン I don't think the question is useful to me... In the comments you read " you are not even supposed to understand how DFT calculations are done step by step" or "everything you need to know about SCF is in Szabo & Ostlund". I would like to go a step further and implement a "simple" DFT code from scratch. S&O was extremely helpful to implement the HF method in a Gaussian basis set and I would like to find something similar for DFT. The books on which I studied DFT are extremely superficial and only explain the basic theory...

@Brian Thank for pointing this out. I will try to give a look to this book, provided that I can access a copy from my University.

Check out The ABC of DFT by Kieron Burke  here. It's free and comes in at 104 pages. Also, see Kieron's group literature page for other gems here.

I can recommend the book of Yang and Parr, although I am not sure it deals  with plane wave bases at all. In principle, a DFT code is not really different from a HF code. In fact, you can use exactly identical codes, just replacing the Coulomb and exchange integrals by numerical integration of some functional. E.g., if you choose $X\alpha$-LDA, than your exchange functional is
$$K = C_\alpha \left(\frac{3}{\pi}\right)^\frac{1}{3} \int \rho(\mathbf{r})^\frac{4}{3} \mathrm{d}\mathbf{r}$$
You can evaluate your density numerically, choose any numerical integration scheme and obtain the exchange energy. That works even, if you still calculate the Coulomb integrals the HF way. However, you wouldn't save much time since you still have the $\mathcal{O}(N^4)$ step in the Coulomb integrals, where $N$ is the number of basis functions. But once you integrate the Coulomb integrals numerically, you have basically a working DFT code.
There is a number of papers about choosing the correct grids and integration schemes for numerical integration. Once, I read one that compared the radial grids quite well, but sadly I don't remember the reference. But the buzz words would be e.g. Becke grid, (Treutler?-)Ahlrichs grid or simply Gauss quadrature for the radial integration (most of the grids are a form of Gauss quadrature, though). For the angular integration there is e.g. the Lebedev grid, but its calculation is a little more involved.
My suggestion, based on teaching experience, is to write first a HF code in a Gaussian basis. Then you pimp it with a numerical integration scheme and something like $X\alpha$. In the next step, you replace your Coulomb integrals. Lastly, you can switch from a Gaussian basis to the basis of plane waves. The last step only changes your matrix elements but not the algorithm (HF/DFT) itself. Since you first implemented a HF code for a Gaussian basis, you can always check your algorithm before going into uncharted waters.

Two of the books that I found close to your needs (i.e. not a basic intro but a detailed implementation including programming and algorithm) are the following:

Ab initio molecular dynamics by Marx and Hutter (published by Cambridge Uni Press)   - This has some info on the implementation of algorithms in CPMD code. Some flow charts of the algorithms.
A chapter (Ch 6) in the book "Computational Physics" by Thijssen (also by Cambridge Uni Press) comes with a short computer program (downloadable from the companion site written by the author of the book).

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text:
What is the Purpose of an Electrolyte in a Galvanic Cell?
I understand how a galvanic cell works and the purpose of erverything such as the ssalt bridge and the electrodes, however I don't understand why an electrolyte is required. For example, consider this scenario.
I have a copper electrode in one beaker, connected by a wire with a voltmeter to a graphite electrode in another beaker. The copper electrode is immersed in water while the graphite electrode is immersed in lead ions. There is also a salt bridge connecting the two beakers. Now my question is, will there be a reading in the voltmeter?
I believe that there should be one since copper is more reactive than lead so copper ions will form while solid lead will form on the graphite electrode. However my school teacher said it won't work as there is no electrolyte in the first beaker. Why does it make a difference if there is an electrolyte or not?

Pure water without any electrolytes doesn't conduct electricity (or only does so extremely poorly), so how can you expect a current to flow?

@orthocresol why does the current need to flow through the water. Forgive me for sounding stupid, but doesn't the current only need to flow from the copper electrode to the lead solution?

You can't have a current going from A to B that doesn't go back to A. This is simple physics - you need a closed circuit for any electricity to flow

Just because you have a potential difference (i.e., a voltage reading) does not mean you'll get much current. Without supporting electrolyte, the water will be too resistive and (as @orthocresol said) very little will happen.

@orthocresol it is my understanding that the copper electrode continuously supplies electrons to the graphite electrode via the wire. Hence why is there a need for a close circuit for electrons to continue circle around if there is already a supply of electrons?

LOL this just hits my funny bone. " What is the Purpose of an Electrolyte in a Galvanic Cell? I understand how a galvanic cell works..." If you understand how a galvanic cell works, why are you asking about the purpose of the electrolyte?

Because if your electrons only move from point A to point B, there will very soon be a buildup of excess electrons and negative charge at B. Electrons don't want to move to a place where there's too much negative charge, so you need some way for negative charge (in a galvanic cell - this role is performed by anions in solution) to go back from point B to point A. If there are no charge carriers, then that doesn't happen.

I understand what you mean, but isn't that why there is a salt bridge? My teacher told our class that the purpose of the salt bridge is to prevent build up of excess charge in both beakers. This is achieved by dipping the salt bridge in $\ce{KNO3}$, thus the K ions go to the cathode while the NO3 ions go to the anode. If that this true, I don't see why an electrolyte is required.

Since copper has greater potential, won't it deposit on the copper electrode whereas lead would dissolve?

Two issues are at play here.
1) You will measure a potential without any supporting electrolyte. As soon as you have everything connected, a potential will develop. Because a self-respecting voltmeter doesn't require much current (ideally zero), the potential will be measured.
2) As soon as you try to pass any current, as comments have mentioned, that current has to be carried from the electrode through the solutions and the salt bridge. Without any electrolyte in the solution, the solution resistance will be huge(as Geoff mentioned) and the amount of current passing will be cell voltage/cell resistance. This will essentially limit the current passing through the solution to very low values.

Thanks for your explanation, however I still don't get why the electrons need to come back to the anode to form a complete circuit if the anode electrode is  continuously supplying electrons. Also as the reaction proceeds wouldn't the solution in beaker A become an electrolyte anyway as Cu ions are being formed?

Electrons don't need to come back to the electrode, but overall electroneutrality has to be sustained. Otherwise, charge will buildup, and the current will not flow.

@Ulgut But as I mentioned before, isn't that why we have the salt bridge. It is dipped in KNO3, and the K ions go to the cathode solution while the nitrate ions go the anode solution  to prevent charge build up (this is what my teacher told me).

But as told before, that would take a very long time, hence the very high resistance.


text:
Why isn't Graham's Law working here?
In graham's law of diffusion most numericals I have seen assumes the formula
$$\frac{r_1}{r_2}=\frac{P_1}{P_2}\sqrt{ \frac {M_2}{M_1}}$$
Where they say $P_1,P_2$  are the partial pressures of the two gases receptively inside the container.But my question is should'nt it be the difference in partial pressures inside and outside the orifice?
Because say if the pressure outside equals pressure inside then the rate of diffusion becomes $0$ .Then graham's law fails in such a case.
Please tell me if I am going wrong.

Well, yes, it should be the difference of pressures inside and outside; it's just that we assume the gas is escaping into a kind of void, so the outside pressure can be ignored.


text:
Qualititative inorganic analysis
Why is it that a specific order is followed while performing qualitative inorganic analysis? I mean, why are we told to first test for the anion in the given salt (and there is an order to be followed for which anion to test for first, which second and so on) and then for the cation?
Does it really matter which way we do it?
How does it affect the analysis if we don't follow the order of cations/anions?

It is just a matter of preference. If you wish to test for the cation first, you can do that and nothing is going to change. Generally the order is given to make the job easy for students, so they can remeber the tests (and also because the profs. know which salt is generally given and the given order is the order of probability. Like if there are 4 salts with sulfate ions, it is better to do the test for sulfates first).


text:
Lucas test for alcohols
The Lucas test for alcohols is based on the way primary, secondary and tertiary alcohols react with Lucas reagent. If on adding Lucas reagent to an alcohol, the mixture immediately becomes turbid, then it is a tertiary alcohol.
I am aware that alcohols get converted to alkyl chlorides on reaction  with concentrated HCl, and that it the alkyl chloride that is insoluble and hence responsible for the turbidity.
But why do tertiary alcohols react the fastest with HCl?

Because the rate of the reaction is directly proportional to the stability of carbocation

The tertiary alcohol has more R groups compared to primary & secondary alcohols, there by removing the C-O Bond from a tertiary alcohol is more easier. (the tertiary carbocation is more stable)

In Lucas test, the alcohol acts as a polar protic solvent and the -OH group can get protonated to form water which is a good leaving group. So SN1 is the most likely mechanism and the order of reactivity is 3°>2°>1°.


text:
Why does benzaldehyde not respond to Fehling's test?
All aldehydes are supposed to respond to Fehling's test. So why doesn't benzaldehyde respond to it?
Does it have something to do with the fact that benzaldehyde is an aromatic compound? If so, then how?

Related: Why does formic acid give positive Fehling's test?

The rate-limiting step of the Fehling’s test reaction with aldehydes is the formation of the corresponding enolate:

The subsequent reaction of the enolate with copper(II) proceeds through a single electron transfer mechanism.
Aldehydes that lack alpha hydrogens, such as benzaldehyde or pivalaldehyde (2,2-dimethylpropanal) cannot form an enolate and thus do not give a positive Fehling’s test result under usual conditions.

From where did you find the mechanism of this reaction?

Could you kindly share the source/citation? Thank you!

In benzaldehyde, the carbonyl group is an electron withdrawing group so the carbonyl group pulls the electron from the electron-rich benzene ring. Due to this, the polarity of the $\ce{C-H}$ bond in the carbonyl group is reduced because the $\ce{C-H}$ bond now has a higher electron density. Thus, the $\ce{C-H}$ bond becomes stronger (the lesser the polarity of a bond, the stronger the bond ). So Fehling's solution (comparatively a weaker oxidizing agent than Tollen's reagent) can't oxidize benzaldehyde (an aromatic aldehyde). However, Fehling's solution can oxidize an aliphatic aldehyde.


text:
Chemical test for ferrous ions
I have a reference sheet for inorganic analysis which says that the confirmatory test for ferrous ions is as follows:

Make a solution of the given salt (whose cation is to identified) and add potassium ferrocyanide [ potassium hexacyanidoferrate(II) ] solution to it.
If a dark blue precipitate is formed, then the presence of ferrous ions in the given salt is confirmed.

What is the chemical composition of this dark blue precipitate? Or is there a flaw in the test that I have specified? If it is flawed, then what is the appropriate test?

The test is better suited for $\ce{Fe^3+}$.
$$\ce{Fe^3+ + [Fe(CN)6]^4- -> Fe4[Fe(CN)6]3}$$
The compound $\ce{Fe4[Fe(CN)6]3}$ has a specific name. Its called Prussian Blue.
If you perform the test for $\ce{Fe^2+}$, the precipitate won't be that dark. First partial conversion of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ takes place. Then the above reaction takes place to give the prussian blue precipitate.
There are specific tests for $\ce{Fe^2+}$ as well. You can use Dipyridyl to get a red colouration which is due to the formation of $\ce{[Fe(dipy)]^2+}$.
Or, you can use Dimethylglyoxime to get red colouration again which is due to $\ce{[Fe(DMG)2]}$.
Reference
JD Lee, Inorganic Chemistry (IITJEE Edition), page 278.


text:
Is there a convention as to what order to write reactants and products in a reaction?
When doing homework, it can be time-consuming to check the answers in a redox reaction due to different order of the reactants and products.
For example, while $$\ce{2MnO4- + 5Fe^2+ + 16H+ -> 2Mn^2+ + 5Fe^3+ + 8H2O}$$ and $$\ce{5Fe^2+ + 16H+ + 2MnO4- -> 5Fe^3+ + 8H2O + 2Mn^2+ }$$ are equivalent, they are not visually equivalent and to mark them as correct/incorrect takes a lot of time.
Is there a convention of what order to write these reactants/products in? If so, what is this convention?

There is no convention.
There might be some cases where one choice simply makes more sense to people. For example I have yet to see anybody write
$$\ce{NaCl(s) <=>[H2O] Cl-(aq) + Na+(aq)},$$
but there is no convention dictating that you have to write $\ce{Na+}$ before $\ce{Cl-}$.


text:
What are some Internal Standards used for Ion Chromatography?
What are some internal standards (ISTDs) used for ion chromatography (IC)?
For experienced IC users, what ISTDs have you used for your sample analysis?

This might be related: http://chemistry.stackexchange.com/q/14085/4945

Thank you Martin for your kind advice! I gotta edit the question. :)


text:
Balancing disproportionation reactions
How to balance the following disproportionation reaction
 $$\ce {XeF2 + H2O → Xe + XeO3 + HF + O2} $$
 My attempt: I tried by making total number of reductions equal to total number of oxidation but I couldn't do because Xe is both oxidised and reduced also the solvent is water so what should I do to balance  protons and oxygen on both sides?.  Please help by solving it on the basis of same procedure that is making oxidation equal to reduction

since acid is in the product side balancing should be done as in acidic medium

It can't be balanced unambiguously, for this is not a single reaction, but a mechanical sum of two reactions, and these can be formally mixed in arbitrary proportion.
The first reaction is indeed a disproportionation. To balance it, use the common trick: consider these two xenons different elements for a while.
$$\ce{\underbrace{XeF2}_{for\;reduction} + \underbrace{XeF2}_{for\;oxidation} +H2O->Xe + XeO3 + HF}$$
The second is a typical redox:
$$\ce{XeF2 + H2O -> Xe + O2 + HF}$$

Nertin how do you know that it is a sum of two reactions? Could you please suggest any good textbook for it

This is a sum of two reactions because it has two independent oxidized products: $\ce{XeO3}$ and $\ce{O2}$. As for the book, I don't think there is one; the matter is too trivial to deserve a whole book on its own.

Nertin how did you figure out what reactants and products would be in each reaction?

Why, the OP specified them all in the original question. I just figured that the oxidized products are independent, so there would be only one of them in each reaction.

The original equation cannot be solved using mathematics because it gives the coefficient of $\ce{XeO3}$ zero. The reason for that is exact reason given by Ivan Neretin (see else where). Yet, I'd like to show a way of solving a balancing problem of true chemical equation by using mathematics. First, assume the compound is $\ce{XeF4}$, which undergoes disproportion in water. The chemical equation for that disproportion can be written as:
$$\ce{a XeF4 + b H2O  -> c Xe + d XeO3 + e HF + f O2  \tag 1}$$
So the atom balance of equation $(1)$ yields:
$$\text{Balancing } \ce{Xe}: \quad a = c + d     \tag 2$$
$$\text{Balancing } \ce{F}: \quad 4a = e     \tag 3$$
$$\text{Balancing } \ce{H}: \quad 2b = e     \tag 4$$
$$\text{Balancing } \ce{O}: \quad b = 3d + 2f     \tag 5$$
Now you have four equations relate to the six unknown coefficients $a$ through $f$, each coefficient is in general, an integer greater than zero. Yet, keep in mind that, a fraction of coefficients are allowed in chemistry if at least one coefficient is an integral. Since there are only 4 equations for 6 variables, we can give an arbitrary value for at least one variable. Since $\ce{H}$ and $\ce{F}$ are appeared in only one compound in either side, we can give our arbitrary value to either $b$ or $e$, respectively. I'd prefer to give $e = 4$ for the sake of argument. Hence, we can proceed as follows:
Assuming $e = 4$ gives $a = 1$ from the equation $(3)$ and $b = 2$ from the equation $(4)$. However this makes mathematics complicated because in right hand side of equation $(1)$, there are two $\ce{Xe}$ atoms and that makes $c$ and $d$ fractions (since $a = c + d$). Thus, we assume $e = 8$, doubling the original assumption. Then, from equation $(3)$, $a = \frac14e =\frac14 \times 8 = 2$. Similarly, from equation $(4)$, $b = \frac12e =\frac12 \times 8 = 4$.
Applying $a = 2$ and $b = 4$ in equations $(1)$ and $(5)$ respectively give:
$$2 = c + d     \tag 6$$
$$4 = 3d + 2f     \tag 7$$
By subtracting the equation $(7)$ from the equation $(6) \times 3$ gives you:
$$4 = 3c - 2f = 2    \tag 8$$
The smallest values to satisfy the equation $(8)$ are $c = 1$ and $f = \frac12$, which is allowed since other coefficients are integers. From the equation $(2)$, $d = a - c = 2 - 1 = 1$. From the equation $(4)$, $e = 2b = 8$. Hence, all numerical values of coefficients are: $a = 2$, $b = 4$, $c = 1$, $d = 1$, $e = 8$, and $f = \frac12$. Therefore, the equation would be:
$$\ce{2 XeF4 + 4 H2O  -> 1 Xe + 1 XeO3 + 8 HF + 1/2 O2  \tag 9}$$
If you multiply whole equation $(9)$ by 2, it becomes:
$$\ce{4 XeF4 + 8 H2O  -> 2 Xe + 2 XeO3  + 16 HF  + 1 O2}  \tag {10}$$
The equation $(10)$ is the correct balance equation, using mathematics.
The given equation in the question is not true redox equation. Yet, it is a combination of two redox reactions, one of which is a disproportionation reaction (of $\ce{XeF2}$) while the other is regular oxidation-reduction of two compounds $\ce{XeF2}$ and $\ce{H2O}$. This complex mixture can be solved by step-by-step analysis:
As Ivan Neretin correctly pointed out, the given reaction is sum of two reactions: The first reaction is redox reaction where one molecule of $\ce{XeF2}$ is oxidized while another $\ce{XeF2}$  molecule is reduced. Let's look at oxidation first:
$$\ce{XeF2 <=> XeO3 + 2 F-}$$
Balance oxygen by $\ce{H2O}$, hydrogen by $\ce{H+}$, and charges by $\ce{e-}$:
$$\ce{XeF2 + 3H2O <=> XeO3 + 2 F- + 6H+ + 4e-} \tag{11}$$
The reduction of $\ce{XeF2}$ is:
$$\ce{XeF2 <=> Xe + 2 F-}$$
Just balance charges by $\ce{e-}$:
$$\ce{XeF2 + 2 e- <=> Xe + 2 F- } \tag{12}$$
Now add the equations $(11)$ and $(12)$ in order to cancel $e^-$s ($(11) + (12) \times 2$):
$$\ce{3XeF2 + 3H2O -> XeO3 + 2 Xe + 6 F- + 6H+} \tag{13}$$
or
$$\ce{3XeF2 + 3H2O -> XeO3 + 2 Xe + 6 HF} \tag{14}$$
This is the first redox reaction, now balanced. The second redox reaction, which is correctly given by Ivan Neretin:
$$\ce{XeF2 + H2O -> Xe + O2 + HF} \tag{15}$$
To balance this equation, follow above methodology:
$$\text{[Ox]}: \quad \ce{2H2O <=> O2 + 4H+ + 4e-}\tag{16}$$
$$\text{[Red]}: \quad \ce{XeF2 + 2e- <=> Xe + 2F- }\tag{12}$$
Add the equations $(12)$ and $(16)$ in order to cancel $e^-$s ($(16) + (12) \times 2$):
$$\ce{2XeF2 + 2H2O -> 2Xe + O2 + 4H+ + 4F-} \tag{17}$$
or
$$\ce{2XeF2 + 2H2O -> 2Xe + O2 + 4HF} \tag{18}$$
Now, these two equations, $(14)$ and $(18)$ can be add together in any proportion to get the sought equation of OP's. For example, add one to one proportion gives:
$$\ce{5XeF2 + 5H2O -> XeO3 + 4 Xe + O2 + 10 HF} \tag{19}$$
This is one of balanced equation in the equation given in OP's question. There are many of them can be written by different combinations of  equations $(14)$ and $(18)$.

Excellent! The circle is now complete! ;)


text:
Why is dimethyloxonium cation stronger acid than hydronium?
$\ce{(CH_3)_2O^+ H}$ has $\rm pK_a \; -3.8$ while $\ce{H_3O^+}$ has $\rm pK_a \; 1.74\;.$ So, the former is stronger acid than latter hydronium ion.
But why is it so?
Is it due to the $\rm +I$-effect of $\ce{-(CH_3)}$? But it would rather decrease the attraction of $\ce O$ on the electron pair of $\ce H$ and that would make it indifferent to release $\ce {H^+}$ isn't it?
What is actually going that makes $\ce{(CH_3)_2O^+ H}$ stronger acid than $\ce{H_3O^+}$ ?

A powerful but often overlooked factor in determining a substance's acidity and basicity is the effect of the solvent (or lack thereof) in which it is dissolved.
You are not wrong; in the gas phase, $\ce{H_3O^{+}}$ is indeed a stronger acid than $\ce{(CH_3)_2OH^{+}}$, by a factor of about $5\times 10^{18}$. This can be seen by comparing the proton affinities of the conjugated bases, dimethyl ether ($\mathrm{804\ kJ\ mol^{-1}}$) and water ($\mathrm{697\ kJ\ mol^{-1}}$), respectively.
In other words, a mole of gaseous dimethyl ether reacting with a mole of gaseous protons releases $\mathrm{804\ kJ}$, while a mole of gaseous water releases only $\mathrm{697\ kJ}$. This means that in the gas phase, water is a weaker base than dimethyl ether, which implies that protonated water is a stronger acid than protonated dimethyl ether.
Why does the tendency invert in aqueous solution? Water can form hydrogen bonds, a very strong type of molecular interaction. $\ce{H_3O^{+}}$ has three protons available for hydrogen bond donation, whereas $\ce{(CH_3)_2OH^{+}}$ only has one. This means that the solvating water stabilises $\ce{H_3O^{+}}$ more than $\ce{(CH_3)_2OH^{+}}$. Higher stability means reduced acidity, and the difference happens to be enough that the relative order of acidity in water is the opposite of that in the gas phase.

So, it has no relation with inductive effect?

The inductive effect suggests that hydronium should be a stronger acid than dimethyloxonium, which indeed is observed in the gas phase. However, the solvent effect overpowers the inductive effect for determining the acidity in aqueous solutions.

Here (1, 2), are some older answers of mine which touch on the importance of solvation for acidity and basicity.

So, the solvation is more important?

In this case, yes. If you were to analyze the same substances in another solvent (e.g. DMSO, toluene, etc), induction could still prevail.

@NicolauSakerNeto would like to make a short comment to supplement your excellent answer, which the OP might find interesting (since the importance of solvents with regard to determination of acidity was mentioned). The upper limit to the strength of any acid is fixed by the solvent. In essence in a aqueous media, nothing is more acidic than the hydronium ion (final proton acceptor); thus, a weaker proton acceptor solvent is needed to discriminate between very strong acids.
read more here: https://en.wikipedia.org/wiki/Leveling_effect


text:
(P1,T1) -> to (P2,T2) then what will be the change in enthalpy for ideal gas?
I need to prove that it is equal to the integral of $C_pdT$ from $T_1$ to $T_2$.
But should not be the case only when the pressure is constant?

The enthalpy of an ideal gas does not depend on pressure.  It is a function only of temperature.  So $C_pdT$ is always equal to dH for an ideal gas of constant composition.  The subscript p on $C_p$ merely indicates that $C_p$ can be measured directly by determining the heat Q absorbed by the gas in a constant pressure experiment.

Enthalpy is defined as $$H= E + PV$$
Its change is defined by the differential $$\mathrm dH= \mathrm d E +\mathrm d\, (PV) \;.$$
Using Thermodynamic Identity $$\mathrm d E=  - P\mathrm dV + T\mathrm dS  $$  for quasistatic process, we get $$\mathrm dE= \delta W + \delta Q$$
So, infinitesimal change in enthalpy becomes $$\begin{align}\mathrm dH &= \delta Q + \delta W +P\;\mathrm d V + V\;\mathrm d P\\ &= \delta Q + V\mathrm d P\;.\end{align}$$
Now, take again the thermodynamic identity
\begin{align}\mathrm d E& =   T\;\mathrm dS- P\;\mathrm dV\\ &=\mathrm d(TS)- S\;\mathrm dT -\mathrm d (PV)+ V\;\mathrm dP\\ \implies \mathrm d(E+ PV - TS)&=-S\;\mathrm dT + V\;\mathrm dP\\& = \mathrm dG  \;_;\end{align} this implied change in $G$ (which is Gibbs Free Energy) depends on the variation of $T$ and $P\;.$
Therefore we can write; $$\mathrm dG= \left(\frac{\partial G}{\partial T}\right)_P\, \mathrm dT +\left(\frac{\partial G}{\partial P}\right)_T\, \mathrm dP \,.$$ Comparing the above equations, we get $$\left(\frac{\partial G}{\partial T}\right)_P= -S\\\left(\frac{\partial G}{\partial P}\right)_T= V $$
Equating the cross derivatives, we get $$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P\;.$$
Also, $$\mathrm d S = \frac{\mathrm dE}{T} + \frac{P}{T}\, \mathrm d V\,$$ and this implies $S$ depends on $T, V\;.$
Now, the definition of $C_P$ is $$C_p = \left(\frac{\partial Q}{\partial T}\right)_P= T\,\left(\frac{\partial S}{\partial T}\right)_P\;.$$
Using all these, we can write the differential of $H$ as
\begin{align}\mathrm dH &= T\left[\left(\frac{\partial S}{\partial T}\right)_V \,\mathrm dT+ \left(\frac{\partial S}{\partial V}\right)_T \,\mathrm dV\right] + V\,\mathrm dP\\ &=C_P\,\mathrm dT - V\left[1 - \frac{T}{V}\,\left(\frac{\partial V}{\partial T}\right)_P\right]\,\mathrm dP\\ &=C_P\,\mathrm dT - V\left[1 - \frac{T}{V}\,\left(\frac{\partial \left(\frac{\nu R T}{P}\right)}{\partial T}\right)_P\right]\,\mathrm dP \\ &= C_P\,\mathrm dT - V\left[1 - \frac{\nu R T}{PV}\,\left(\frac{\mathrm d T}{\mathrm dT}\right)\right]\,\mathrm dP\\ &= C_P\,\mathrm dT\;.\end{align}
So, the total change in enthalpy $$\Delta H=  \int_{P_1,T_1}^{P_2,T_2}\; C_p\,\mathrm dT\;.$$

References:
$\bullet$ F Reif's Fundamentals of Statistical and Thermal Physics.
$\bullet$ E Fermi's Thermodynamics.

That equation for $\Delta H$ is incorrect.  It should be calculated from $$dH=C_pdT+\left(V-T\frac{\partial V}{\partial T}\right)dP$$For an ideal gas, this reduces to $\Delta H=C_pdT$.

The starting equation for dH should be dH=TdS+VdP.  One then writes $$dS=\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP$$.  Then, there is a Maxwell relation giving $$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$There is also the equation $$\left(\frac{\partial S}{\partial T}\right)_P=\frac{C_p}{T}$$If you combine all these equations, you obtain the equation that I wrote.

This is a well-known derivation that is found in most Thermodynamics textbooks.  Please just consider presenting the final result with a reference.


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How do Cryocoolers that use Helium as Coolant work?
I was unable to find elaborate information on helium based Cryocoolers. I like to know how they use helium to cool samples to cryogenic temperatures and why helium is the best option as a coolant.

A helium based cryocooler works just like an air conditioner, but helium is the gas rather than a halocarbon compound. The boiling point of helium is −268.928 °C, so such a cryocooler could get really cold.
Helium is a rare gas and relatively expensive. So to cool a small experimental apparatus it is much cheaper to buy one helium based cryocooler rather than dewars and dewars of liquid helium.

Could you tell me how it cools the samples?

http://inspirehep.net/record/852153/plots
The actual design depends on the experiment you want to perform, as you usually cool the sample for some reason, not just for fun.


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Free chemical database to search for physical properties?
I am looking for an Open-access tool that can search "backwards", eg. give me all the structures corresponding to a given heat of formation, critical pressure, etc. It would be even better if the database canbe automatically searched by API or by downloading it. Currently my best go is ChemExper, where I can search for some of the properties, but it is challenging to make it automatic. Is there anything similar in the field?

Related: http://chemistry.stackexchange.com/questions/22285/public-api-for-physical-properties

As I've mentioned on other related questions, this is very much an open topic. There aren't many open-access databases in chemistry, and far fewer that have APIs.

As suggested in the other answer, there are multiple ways to query / scrape Wikipedia.
ChemSpider offers some property search features.
Reaxys offers some property searching, including spectra features. It's not open access.

Indeed, what I see is open access chemistry databases have a long way to go. The reason I asked this question is exactly this: there might be some databases that slipped my knowledge. On an open forum like stackechange, there might be someone who knows something.

@Ezze I think it's a good question. It's just not that different from the last question

Yes, you are right. I didn't see the question. However, I am hoping for a better answer this time.

@Ezze - Nothing has changed since then. In a few months, we'll be rolling out API-based property search for PQR but these would be computed values.

You might search the German Wikipedia using SPARQL via DBpedia.
Go to http://de.dbpedia.org/sparql and run your query, e.g.
PREFIX dbo:<http://dbpedia.org/ontology/>

SELECT ?chem ?bp
{
  ?chem a dbo:ChemicalSubstance .
  ?chem dbo:boilingPoint ?bp .
  FILTER ((?bp >= 372) && (?bp <= 374))
}
ORDER BY ?bp

I've just discovered this website where you can find a tremendous amount of data, especially about inorganic chemistry with things like E-pH diagrams, phase-diagrams, things about alloys of different metals, ultra pure silicone, salts, elements, nuclear data and so on.
And if you prefer sugar derivatives here is a great database with online calculators about many sugar things, syrops etc.
Enjoy !!!


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How to approximate the pH of a box of water molecules with one proton?
I have a cubic box of length $L = 12.42\ \mathrm{\overset{\circ}A}$, containing $60\ \ce{H2O}$ molecules and $1\ \ce{H+}$ ion. How can I approximate the $\mathrm{pH}$ of the system?
Following the basic definition, $\mathrm{pH}=-\log\left([\ce{H+}]\right)$, I tried to calculate it like the following two ways:
\begin{align}
  \mathrm{pH} &= -\log\left(1/60\right) = 1.78 \tag1\\
  \mathrm{pH} &=
    -\log\left(
      \frac{1/6.022\times 10^{23}}{12.42^3\times 10^{-30}\times 10^3}
    \right)
    = 0.0621 \tag2
\end{align}
However, I am not sure whether these approximations are valid or not in this regime.

Chemistry is not about individual molecules at all. But if you insist, well, then your second way must be right.

For a rough, back of the envelope, sanity check sort of calculation, your second approach is more appropriate than the first. (Though check your value for Avogadro's number.) This is something that should jump out at you if you keep track of units in your calculation. The $\ce{[H+]}$ in the equation should be measured in moles of hydrogen ion per liter, so your calculation needs to result in an inserted value that's in moles of hydrogen ion per liter. You can also rule out the first equation by performing a thought experiment: if the size of the box doesn't factor into the equation, does it make sense that the pH wouldn't change if the box size changes? (No matter how much the size changes?)
However, the statement that pH is equal to the negative log of the hydrogen ion concentration is strictly speaking only true for dilute aqueous solutions near standard temperature and pressure. For concentrated acids/bases, solutions which are not primarily water, or for pressure and temperatures well outside standard, things get more contentious. There's a number of different interpretations of what "pH" means, and each have their advantages/disadvantages, depending on the situation.
One of the big considerations here is that "concentration" is not really really the thermodynamically relevant parameter when figuring out how chemical systems behave. Instead, it's the compound activity which is typically the more important quantity to keep track of. As it happens, the thermodynamic activity of a compound is normally proportional to the concentration in dilute solutions, so by convention we set unit activity to be equivalent to 1M solution.* This means the distinction between concentration and activity is normally ignored/glossed over in most introductory chemistry courses, but make no mistake, it's the activity of the compounds which should be used in things like the Henderson–Hasselbalch equation, not the concentration.
*(This normalization by 1 mol/L is the reason we can take the log and not run up against any strange results with dimensional analysis - it's technically a dimensionless value inside the log: the actual concentation, divided by the reference concentration of 1 M.)
The reason this matters for you is that you're dealing with a situation which isn't quite the same as a dilute aqueous solution. Nominally, it sounds like you're close. Your water density is around standard water density, and the amount of hydrogen ions are close to 1 mole per litre. But there are other considerations to take in to account. For example, what temperature and pressure are you using? Extreme values can change the hydrogen ion activity. Also, you didn't mention anything about the presence of a counter-ion. The unbalanced electrical charge might affect the predicted activity as well. The other issue is that just a single hydrogen ion will result in some pretty extensive fluctuations in the system as the system changes state. The "pH" (or hydrogen ion activity) of the system may actually exhibit time dependence, which might be important for the system you're modeling.
If you're truly interested in what the equivalent pH of the solution is, you should look into simulation techniques which attempt to actually gauge the chemical activity of the hydrogen ion under the conditions you're interested in. -- That is, figure out why you want to know the "pH" of the system, and look for more direct measures of that.

Thank you for the answer. To be more be specific, I have been trying to see the activity of 1 $H_3O^+$/2 $H_3O^+$ ions in presence of oxyanions, $XO_4^{2-}$. The systems have been modelled at NTP having overall density of about 1.0-1.1. Can you please help me with some more insights?

And thank you very much for pointing out the mistake in putting Avogadro's number, which now been corrected!

@SangkhaBorah I've actually never done such simulations myself, so I'm a little short on practical help. What I would highly recommend is an extensive literature search for people which are doing similar things, (looking at hydrogen ion activity in simulations) and attempting to replicate the approaches they used.


text:
How is dimerization done?
Nitrogen dioxide is dimerized to form dinitrogen tetroxide according to the equation
$\ce{2NO2 -> N2O4}$. How do two molecules of one compound fuse to form another?

I think you can answer your question easily. Draw the Lewis structures for NO2 and you'll see how the dimer is obvious solution to some problem.

The same way as any other reaction takes place?

The electronic structure of NO2 leaves an unpaired electron on the nitrogen. This is one of the reasons it is quite reactive. But it is also the reason why it can easily dimerise to N2O4 as the two unpaired electrons in two molecules of the gas can pair up creating a weak bond. The dimer is favoured at low temperatures when the thermal energy isn't large enough to shake the dimer apart (I'm simplifying a bit). This is why the colour of the gas changes depending on temperature as the dimer is colourless but the NO2 is brown.

This is very correct. However, when reading the questions it appears to me the person asking it (on a rather basic level) did not understand what the dimer would look like. There seems to be a misconception about the binding process, given the phrasing of "fusing". The crucial bit here is to understand that the atoms of the 2 nitrogen dioxide molecules are not broken apart and refused to form something new, but that the molecules stay intact and bind together. See also my answer.

Maybe the confusion arises because you're unclear about what a dimer is. The molecules don't "fuse". They form a bond.
It's better to think of the dimer as $(NO_2)_2$ rather than $N_2O_4$.

I think it would be even clearer if you were to include the Lewis structure.


text:
pka from molecular dynamics
How can one calculate pKa values from first principle calculations, static or dynamics?
I have encountered some works using CPMD where people try to find pKa values from -(i) constrained Molecular Dynamics (MD) and (ii) metadynamics. These are somewhat more intricate ways, I believe. Is there any other simpler ways (theoretical/computational)? Can not we do it from simple MD?
Please suggest some references as well, if possible.
Thanks in advance.

Initially I tried to use the basic formula for pKa, given in wikipedia..                      $$pK_a=-\ln \left[ \frac{[A^-][H_3O^+]}{[HA][H_2O]}\right].$$For my case, I used this formula as follows, $$pK_a=-\ln \left[ \frac{69\times 2}{31\times 58}\right]=1.12$$ against experimental values of about 2.0. Here the numbers that I used represent not the molar concentration, but rather the proper number of molecule/ion. How well do you think the result is reliable?

The calculation you suggest would be valid if the given concentrations were at the system's equilibrium. If you don't intend on directly simulating a grand-canonical ensemble, feel free to accept the (only) answer below.

Have you considered a free-energy calculation?
That is, calculate the free-energy of the dissociated species and of the neutral one, and calculate the pKa via the formula for activity.
There was this GROMACS tutorial, but it seems the link is broken.

That's what I'm trying to do now, as you have suggested! I would not be able to use gromacs as my calculations are based on AIMD simulations. For free energy calculations, as statistical physics says, the formula is $F=-kT \ln Z$, where Z be the partition function. How can I calculate Z from MD trajectory?

If you can't be bothered to look up a 'smart' way, you can do this directly by binning position and momentum space and build a histogram from real-space trajectories (gotten from a MD simulation, of course).


text:
If for reversible process, ΔS= 0 for universe (system+surroundings), why would there be non-zero change in gibbs energy?
As Second Law says, the entropy must increase for universe
$$\Delta S_\textrm{universe}\ge 0$$
Now, we know, $$\Delta G= -T\Delta S_\textrm{universe}\le 0^\dagger$$
For reversible process, $$\Delta S_\textrm{reversible, universe} = 0$$
That would mean $$\Delta G_\textrm{reversible}= 0$$
But does it happen so?
Peter Atkins, in his book writes:

At constant temperature and pressure, for a reversible process: $\mathrm dG = \mathrm dw'_\rm{rev}\;.$

Now, isn't it contradicting that $\mathrm dG\ne 0$ for reversible process? If $\Delta G= -T\Delta S_\textrm{universe}$, then wouldn't $\mathrm dG= 0$ for $\mathrm dS_\textrm{universe}= 0$ during reversible process?

\begin{align}^\dagger \Delta S_\textrm{universe}&=\Delta S_\textrm{system}+ \Delta S_\textrm{surroundings} \\ &= \Delta S_\text{system}+ \frac{-\Delta H_\text{system}}{T} \\ \implies   -T\Delta S_\text{universe} &= \Delta H_\text{system} - T\Delta S_\text{system} \\&=\Delta G  \end{align}

Forgot about enthalpy H?

@Mithoron: Huh?

G=H-TS so your equations look wrong...

@Mithoron: Edited.

Think more about it and you'll see why you're wrong.

@Mithoron: Periodic Table?

The equation in Atkins' book refers to the system only, and not to the universe, and, moreover, in this equation, w' specifically refers to non-PV work, like electrical.

G at constant temperature and pressure doesn't talk about the entropy of the universe?

I thought dG would be zero as $\mathrm dG= -T\mathrm dS_\rm{universe}$ and $\mathrm dS_\textrm{universe}= 0$ for reversible process?

Doesn't Atkins' book specify these constraints when it talks about the change in G for a system experiencing a process at constant temperature and pressure?  What exactly does the book say leading up to your highlighted statement?  How exactly does it define w' (with the prime)?  You seem to have omitted important details in your question.

He actually doesn't elaborate anything; only he does mention the term maximum non-expansion work at const. temp. and pressure

I guess he should have been more precise.  I don't have a copy of Atkins, but I can tell you the Denbigh (Principles of Chemical Equilibrium) has a development that spells this out with more precision.

https://books.google.co.in/books?id=BV6cAQAAQBAJ&pg=PA117&lpg=PA117&dq=non+expansion+work+gibbs+free+energy+peter+atkins+physical&source=bl&ots=N4rC6dkgPO&sig=kUlgR5ig24YI3D8V4-dbYtqBfd8&hl=en&sa=X&ved=0ahUKEwjs5_GA047LAhUQC44KHQfVBuUQ6AEIHDAA#v=onepage&q&f=false

Two Thermodynamics books I highly recommend are Fundamentals of Engineering Thermodynamics by Moran et al, and Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness.

Thanks for the recommendation; but I can't immediately buy either of the two now ;/

Don't you think Gibbs energy is related to non-expansion work?

In the online version you provided, the Justification 3.3 is similar to the development in Denbigh, although Denbigh is more precise. I do think that, for a reversible change at constant temperature and pressure, the change in the Gibbs energy is equal to the non-expansion work.

Oh! But reversible process means the change in entropy of the universe is 0, isn't it?

So?  What is your point?

As $\Delta S_\rm{universe}= 0$, the equation (valid at const. temp and pressure) $\Delta G= -T\Delta S_\textrm{universe}$ yields 0  but also the derivation that Atkins shows gives $\Delta G= w'$ anon-zero quantity. So, isn't it contradictory?

It's not contradictory.  The equation we have been discussing applies to a specific closed system, and not necessarily to the universe.  If you want to apply the equation to the universe, then, for the universe, w' = 0, because the universe is not doing any reversible non-PV work on anything.  The reversible non-PV work that a closed system does on the surroundings is exactly equal in magnitude and opposite in sign to the reversible non-PV work that the surroundings does on the system.  So, if you include both system and surroundings, the reversible non-PV work is zero.

Are you saying $\Delta G= -T\Delta S_\rm{universe}$ (at const. temp. and pressure) is wrong?

I'm saying that, for a reversible process at constant temperature and pressure, the non-PV work done by the universe is zero, and $\Delta G$ of the universe is zero.  But please understand that, the real applicability of the equation in Atkins is for a closed system (of lesser extent than the entire universe) experiencing a reversible change at constant temperature and pressure.  I've run out of ideas on how to explain this more clearly.  Maybe someone else can help out.  I need to go to bed now.

Good night ;) So, at reversible process, the non-expansion work of the universe is zero?

I think it is a very bad idea to think of this in terms of the universe.  The focus should be on a finite closed system.  If you include both the system and surroundings, then this combination can't do work on anything (either reversible or irreversible).  The work the system does on the surroundings is just the negative of the work that the surroundings does on the system, so the sum of the two (the net work) is zero.  Try to picture in your mind the universe doing work on anything, given that work only occurs at the boundary of a system.

Let us continue this discussion in chat.

I don't think so.  I'm pretty much out of ideas on how to explain this.

@user36790 In the future, it's better to formulate some of these into new questions rather than having a back and forth in chat between just two people.

@Chester Miller : Have you visited the chat-room & read my comment?

At constant temperature and volume, \begin{align}W_T & \le (U_1- U_2)- T(S_1-S_2)\ \implies W_T & \le (F_1 -F_2) \end{align} Now, if reversibility exists then $$W_T= F_1- F_2$$ though $dS_\textrm{universe} = 0$

Generally it is written that $$\mathrm dS_\textrm{surr}= \frac{-\mathrm dH_\textrm{sys}}{T}$$ but it is valid only when the non-expansion work is zero. Generally $$\mathrm dH = \underbrace{w'}_{\textrm{non-expansion work}} + \delta q$$

So, when one uses $\mathrm dH $ to denote heat transferred (or absorbed) in the surroundings, then he is definitely taking $w' = 0,,$ isn't it?

The proper equation is
$$\mathrm dG=\mathrm dH-\mathrm d(TS)$$
Since $H=U+pV$, we rewrite the above equation as
$$\mathrm dG=\mathrm dU+(p\mathrm dV+V\mathrm dp)-T\mathrm dS-S\mathrm dT$$
Furthermore, we know from the first law that
$$\mathrm dU=\delta Q + \delta W$$
and
$$\Delta S=\int{\frac{\delta Q_\text{rev}}{T}}$$
which is equivalent to writing
$$T\mathrm dS=\delta Q_\text{rev}$$
so, putting it all together and remembering that temperature is constant, we have
$$\mathrm dG=T\mathrm dS+\delta W_\text{rev}-T\mathrm dS$$
Then,
$$\mathrm dG=\delta W_\text{rev}$$
where the pressure-volume work went to zero because, as you point out in one of your comments, all of this is only valid for zero expansion work.
I hope this kind of answers your question, but you seem to have a pretty good grasp on this already.

Also, it probably shouldn't be written that
$$\mathrm dG=\mathrm dW_\text{rev}$$
because this seems to imply that integrating this expression will have you take an integral with respect to work, but this cannot be done because work is not a state function. You would figure this out, however, when trying to figure out what values of work to use as the boundaries in that integral.

$\delta Q\ne 0,,\textrm{but},,=  T\mathrm dS$

For a reversible process the heat transfer does equal zero.

No, no; this does happen only for adiabatic process.

Ya you're right about that. I think there shouldn't be heat transfer in this case because I believe this is how Atkins gets to the result you're talking about.

@jheindel It is not correct to say that "for a reversible process, there must be no heat loss from the system."  What about an isothermal reversible process?

@jheindel Also, your expressions for dG are incorrect.  dG=dH-d(TS), which leads to dG=VdP-SdT

Oh I see where it's going wrong. I'll edit it later. And ya you're right. At the time I couldn't figured out what was thing wrong so I reasoned very poorly.


text:
Reactivities: interhalogen compounds versus halogens
My textbook says that interhalogen compounds are more reactive than halogens, because the $\ce{X-X}$ bond in halogens (for example, dichlorine or $\ce{Cl2}$) is stronger than the $\ce{X-X'}$ bond in interhalogens (for example, chlorine monofluoride or $\ce{ClF}$).
(In the symbols that I have used, $\ce{X}$ and $\ce{X'}$ have been used to denote dissimilar halogen atoms.)
So my question is: why is the bond between dissimilar halogens weaker than the bond between atoms of the same halogen?
I am aware that the bond between two identical atoms will be a completely covalent bond due to no difference in the electronegativities of the participant atoms. So, the $\ce{X-X}$, bond as in $\ce{Cl2}$ will have no ionic character.
In case of an interhalogen compound, however, the difference in the elecronegativities of the (dissimilar) participant atoms will cause the bond to become polar and thereby acquire ionic character.
So is it the ionic character of the interhalogen compounds that causes them to be more reactive than halogens? But then, this means that ionic bonds are weaker than covalent bonds.
But, the answers here as well as here say that an ionic bond is stronger.
But this link says the very opposite. Okay, the people writing the answers there have written it to suit an eight grader's understanding, but even so, they have stated that covalent bonds are stronger than ionic bonds.
So which alternative am I to believe? I know that most bonds are neither purely ionic not purely covalent, but intermediate between these two extreme characters, but this only serves to increase my bewilderment.
Can somebody please resolve this, or tell me where I have gone wrong in my reasoning as regards the reactivities of the interhalogen compounds in comparison with halogens?
And finally, please tell me the reason why interhalogens are more reactive. Thanks!

See this:https://in.answers.yahoo.com/question/index?qid=20090728202251AA0E7oV

This question makes less than little sense. Some covalent bonds are stronger than some ionic bonds. And some ionic bonds are stronger than some covalent bonds.

Here is the answer for which bond (ionic or covalent) is stronger.
There is no direct answer to this question as it depends on atoms that take part in bond formation.For example you might be knowing that a Fluorine- Fluorine bond is weaker than a chlorine-chlorine bond due to high inter electronic repulsion.In another example iodine-iodine bond is weaker than chlorine-chlorine bond.Similarly there is certainly this sort of variation in ionic bond as well.
So if you compare the weakest ionic bond and strongest covalent bond then covalent bond would be stronger.And if you compare stongest ionic bond and weakest covalent bond, then of course ionic bond would be stronger.Also in many substances its not only the bond strength that matters but also the crystal structure.
An example where covalent bond is stronger:Carbon-Carbon bond is stronger than NaCl bond.
Whereas NaCl ionic bond is more stronger than Br-Br covalent bond.
So it depends on the atoms that participate in bond formation.
visit https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf for bond enthalpy.


text:
Do bare protons exist, even transiently, in aqueous solution?
Generally $\ce{H^+}$ ions in water remain in the form of hydronium ions ($\ce{H3O+}$), and not as "free" aqueous protons, as far as I've been taught.
My question: is this always the case, even when considering dynamics at extremely short time scales? By this I am not meaning interconversion with the Zundel ($\ce{H5O2+}$) or other states. I mean to say, does bare $\ce{H+}(\mathrm{aq})$ ever exist  on, say, a femtosecond time scale?

I think this is one of those things where there probably is a single proton floating around somewhere in a solution, but the formation of this is so unlikely. It means that you need to have just a hydrogen atom somewhere in solution and some other molecule which will take that electron and form a radical, which is certainly not stabilizing. Then, the proton needs to somehow not react with anything for a femtosecond. My bet is that this just doesn't happen...

This is probably more of a philosophical question - when is a H bonded to something,  and when is it not?  That's quite tricky to answer,  particularly in the liquid state...

Relevant: https://chemistry.stackexchange.com/questions/32558/why-are-hydrogen-ions-always-associated-with-another-molecule/32559#32559

According to Myths about the proton. The nature of H+ in condensed media Accounts of Chemical Research, vol. 46, pp 2567–2575:

But to put this into
  perspective, an electron-free proton has an estimated acidity $10^{56}$(!) times greater than 100%
  H2SO4, the threshold of superacidity.$^2$ Adding a bare proton to a molecule is a real event in the
  gas phase but only a solvated hydrogen ion can be transferred in condensed phases.

No.
The proton is always within some electron density, even when transitioning between the hydronium/water and whatever anion it was bonded with.

To expand on this answer:  Say a proton moves from one water molecule to another.  It does not jump off and become a bare proton, even for a moment.  Rather the second water molecule begins to form a bond with the proton as the first water molecule begins to loosen, leading to a transition state where the proton is partially bonded to both water molecules.


text:
Using Dalton's law of partial pressure to find percentages of composition

A mixture of helium and neon gases is collected over water at $28.0 \rm~^\circ C$ and $745\rm~ mmHg$. If the partial pressure of helium is $368\rm~ mmHg$, what is the partial pressure of neon? (Vapor pressure of water at $28\rm~^\circ C =28.3~mmHg$). Calculate the weight percentage of He and Ne in the original mixture.

The question is based on Dalton's law of partial pressure. I got the first part right.In the second part, since we have to deal only with the original mixture, total pressure should be $$745-28.7=\pu{716.7 mmHg}$$ So we calculate the total moles in terms of "v" (total volume) and then moles of neon and helium in terms of "v" from which we calculate the weight of neon and helium and calculate the weight percentage. "v" will eventually cancel out and we get weight percentage of neon as $34.54\%$ and that of helium as $65.45\%$ but that is not the right answer.
In the solution to this question, they took the total pressure while dealing with the second part as $\pu{748 mmHg}$ only but why would we do that. We are only talking about the original mixture. Where am I wrong?
Answer in my textbook : Neon weight% = $82.58 \%$, helium weight% : $17.43\%$

Sorry, I wrote hydrogen in place of neon. The question does not mention any other detail besides these, and I don't think we need any more details. Rest of the variables can be assumed as they will eventually cancel out.

OK.  What do you get for the mole fractions of helium and neon in the original mixture (excluding the water)?

Weight % = 100*(weight of substance/ weight of SOLUTION),so whats your doubt?

@SujithSizon The question is not about simply applying the weight % formula. All I want to know is why should we take total pressure as 748 mm Hg when we are just concerned about the original mixture. The vapour pressure of water was just given to solve the first part, it should not have any relation with original mixture.

@ChesterMiller I already mentioned it; weight percentage of Neon=34.54% and Helium=65.45%

@bluebellae by 748 you mean 745 ?

@SujithSizon Oh sorry, that was a mistake. I mean 745.

@SujithSizon We will need the w/w relation but that is not the integral part of the question. The question, according to me, is not framed to test your knowledge about this formula.

@bluebellae the w/w formula is based on the "solution" of the mixture so shouldn't we take the total pressure of the solution (=745)? (Also while doing salt analysis you write O.S. instead of X solution right? )

@SujithSizon Even if I had not known this formula, to calculate the weight of a substance in a compound, I would have used : (weight of that substance /total weight)* 100 using some simple maths. Why you are so fascinated by that formula? Also, we will need the total pressure, but that will be the one in the original mixture because that is what is asked.

@bluebellae its a solution you have to include partial pressure of water

@SujithSizon Who said it is a solution? I said it is a mixture.

Let us continue this discussion in chat.

First find the mole fraction of each gas in the original mixture using the formula, P(gas) = X * P(tot). For He:
$$X = \frac{368}{716} = 0.514$$
For Ne:
$$X = 1 - X(\ce{He}) = 0.486$$
Say you have 1 mol of mixture then you have 0.514 mol He and 0.486 mol Ne.
Converting these to weights using periodic table:
$$\rm 0.514~mol~He = 2.06~g~He \\\
0.486~mol~Ne = 9.81~g~Ne$$
The total mass of your mixture is the sum of these two masses, 11.9 g. Divide the individual masses by the total mass to get mass percent.
$$\rm\% w. He = 2.06/11.9 = 17.3 \%$$
Subtracting this from one for Ne (or you could apply the formula again):
$$\rm\% w. Ne = 1 - 17.3\% = 82.7\%$$
Note: I didn't carefully carry sig figs but you should!


text:
Nuclear zero-point energy and quantum delocalization effects of water molecules
What do we mean by "Nuclear zero-point energy" and "quantum delocalization effects" of water molecules ? It's said that without these effects "water" becomes somewhat unbounded, which is not very well understandable to me. Can you please help me to clear my doubts.

First, I think I know what "nuclear zero-point energy" is, but I have no idea what "quantum delocalization effects" are. Secondly, usually when something is said somewhere there should be an explanation there as well. So, I want to have a look at the actual source of this knowledge you shared with us. Personally I don't like question of the "I've heard that" form here.

Chemical Physics Letters 460 (2008) 423–431

So, what was actually said in the paper1 with respect to AIMD simulations of water is the following:

Nuclear zero-point energy and quantum delocalization effects are neglected making classical water overbound.

To understand the meaning of this statement we first need to know how AIMD works. Ab Initio Molecular Dynamics (AIMD) is a molecular dynamics simulation in which potential energy surfaces for nuclear motion are obtained from first principles (ab initio) rather than represented by a force field. The quantum chemical calculation of potential energy surfaces are done in the Born–Oppenheimer approximation in which nuclei are considered to be stationary classical particles.
As its very name signifies, it is, of course, just an approximation. It ignores that nuclei move and aren't perfectly localized in space. The nuclear motion can be approximately separated into vibrational and rotational the first of which energetically dominates the second. Taking the vibrational motion on the nuclei into account results in some additional so-called zero-point vibrational energy that rises the total energy. Other quantum effects (tunneling, delocalization) can be also taking into account for the nuclei which results in some further changes in energy.
Altogether, the more and more quantum effects are taking into account, the closer are the results of the simulations to the physical reality. AIMD simulations ignore few quantum effects, and thus, do not quite match the reality. For instance, water tends to be overbound in such simulations comparing to what we have in the real world. That is, essentially, the meaning of the statement quoted above.

1) A.D. Hammerich, V. Buch, F. Mohamed, Ab initio simulations of sulfuric acid solutions, Chemical Physics Letters, 2008, 460, 423-431, http://dx.doi.org/10.1016/j.cplett.2008.06.053.


text:
hydroxide ion in water -- How does the dynamic change?
It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions. Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion. If I am correct at this point, how would the dynamics be different in these cases?

It certainly wouldn't be $\ce{H_3O_2^{3-}}$. If anything, then $\ce{H_3O_2^{-}}$.

It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions.

This is not quite right. Because of the hydrogen bonding a wide variety of $\ce{H3O+(H2O)$_n$}$ ions are in dynamic equilibrium, where $n$ can take values as large as few tens. It has been shown that $\ce{H3O+}$ can donate three hydrogen bonds, so that it's first solvation shell of 3 water molecules, and the hydrogen bonds between $\ce{H3O+}$ and $\ce{H2O}$ are quite strong. This means that the $\ce{H3O+}$ cation can actually be considered as $\ce{H9O4+}$ in solution. For more details, see here.

Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion.

Reasoning by analogy is a very dangerous road. As mentioned by sheogorath there is no possibility for a coordinate covalent bond as in the case of $\ce{H3O+}$. But the hydrogen bonding is possible and the hydroxide ion is indeed strongly hydrated as well as $\ce{H3O+}$. The details of the hydration are not well known yet, but recent studies indicate $\ce{OH- (H2O)4}$ as the prominent complex in an aqueous solution. For details, see here.

$\ce{H3O}$+ is formed by
$\ce{H2O  +  H+  = H3O+}$             where $\ce{O}$ share $2$ electrons to $\ce{H+}$ with no electron to form coordinate covalent bond .There is no possibility for oxygen to form such bond with $\ce{H2O}$ and $\ce{OH-}$


text:
Does the addition or multiplication rule have precedence deciding the significant figures in 5×5.364?

How many significant figures should be present in the answer of
  $5\times 5.364$?

Addition rule: The result cannot have more digits to the right of the decimal point than either of the original numbers.
Multiplication rule: The result must be reported with no more significant figures as there are in the measurement with the few significant figures.

Using rule 1,
$$5\times 5.364 = 5.364 + 5.364 + 5.364 + 5.364 + 5.364 = 26.82,$$
one would use 4 significant figures.
Using rule 2,
$$5 \times 5.364 = 26.82 = 30,$$
one would use 1 significant figure.
Which one is correct?

If this is the question in its entirety, then you should not try to 'guess' what the origin, and therefore the accuracy, of the measurement 5 is. You should take it on face value, and complete the calculation using the multiplication rule, as the equation is requiring a multiplication operation. In this case, you should have one significant figure only.
The only ambiguity you need to consider in calculating your significant figures is whether the '5' is an exact number or not. Exact numbers come about either by definition (1 hour = 3600 seconds) or by counting (5 measurements were made) and have an infinite number of significant figures. Hence, they do not affect the accuracy of your measurements. Exact numbers are usually integers (speed of light and number of centimetres in an inch are a couple of exceptions). My guess is that your question is simply a case of going through significant figures calculations, using the information provided. To avoid ambiguity in the case of a single significant figure measurement, one could consider writing 5 as 5E0, or 5x100.
If you know 5 to be an exact number (ie the question gives that information), then you would still complete the calculation using the multiplication rule, but assume infinite number of significant figures for 5. In this case you would have 4 significant figures.

Using the addition rule doesn't make any sense because then you would always use the addition rule for multiplication.
You would use the multiplication rule, and if this question was given as-is, 30 would be correct. If you're told the 5 is exactly 5 (like in many/most conversions), then you would consider it infinite significant figures and answer with 4 significant digits.

If the measured width of a rectangle is 5 meters and the length is 5.364 meters, then the area would only have one significant figure, so 30 square meters by the multiplication rule.
If you measure the width of 5 bricks and the measurement of each brick is 5.364 cm, then the number "5" is not a measured value, it is an exact value.  Four significant digits would be appropriate, consistent with the multiplication rule.
Contrary to the OP, literal application of the addition rule "cannot have more digits to the right of the decimal point than either of the original numbers", would mean 5 significant digits not 4.
However, when measurements are being added, the uncertainty in the square root of the sum of the squares of the uncertainties in each measurement, so the addition rule increasingly fails as more values are summed.

It really depends on what you're trying to signify, but I'd say that neither of the answers that you have given is entirely satisfactory. The "rules" you cite are good for sort of rule of thumb, but don't really work well when you're working with a single digit. 9 +/- 0.5 is about a 6% error, but 1 +/- 0.5 is a 50% error.
If you were working with a number which has 3 significant figures by a number which has 4 significant figures then the rules you cite work well enough.
The quirk here is that you can assume that 5 is +/- 0.5 so that there is an error of 10%. So 27 +/- 3 would really be a better answer.
Another thing to consider is that 30 has two significant figures. If you want to denote just one significant figure then you'd write the answer in scientific notation as 3E1 to clearly denote just one significant figure.
Of course if the 5 was an integer value then it has infinite precision. Say you're working with oxygen in the following chemical equation:
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$

Perhaps I misunderstand you, but 30 has 1 significant figure.

@long - It depends on if you know it is 30 +/- 10 (which has 1 significant figure), or if you think it is 30 +/- 0.5 (which has two significant figures). I pass a speed limit sign that says 30 mph. I don't think the law means +/- 10 mph.


text:
Is there an algorithm that counts Joback groups in a structure?
So I was wondering, there are a lot of cheminformatics tools (eg Babel) out there, to analyze and convert representations. I also saw some codes which counts certain groups, but ignores the carbon chain, which is obviously not suited for a Joback script. My google research only turned me a "homemade" matlab code found at Reddit, but it is no way general, as the first step of it is to exclude most of the rings.
So is there any open cheminformatic tool for this problem?

It might be helpful to add some details to your question about what a 'Joback group' is - it seems like https://en.wikipedia.org/wiki/Joback_method is relevant? In any case, you are probably looking for a /descriptor/ which is simply a value calculated for a structure.

Yes, by Joback grousps I mean the groups used in Joback's method, but any other group contribution method would do. As to your comment, I am not looking for a calculated descriptor, but for a list of the groups that the molecule contains.

Right, great - maybe your question could be "is there an algorithm for calculating group contributions" or something. In any case, I would expect the first step for any Jobeck descriptor would be to find the groups. Looks like some kind of simple substructure search would do it.

So it looks like 'Joback groups' are some set of functional groups that you want to find in molecules.
Sounds a lot like substructure search, for which there are many cheminformatics tools (like SMSD, for example) however, since you mention OpenBabel:
http://open-babel.readthedocs.org/en/latest/WritePlugins/AddNewDescriptor.html
talks specifically about 'group contribution methods' and gives a simple example of how to get the counts for a set of functional groups.


text:
How would Dalton's law be affected when there are two ideal gases in a container at different temperatures?

How would Dalton's law be affected when there are two ideal gases in a
  container at different temperatures?

Let the gas with higher temperature be gas A and the gas with lower temperature be gas B. Then heat will be transferred from gas A to gas B due to which kinetic energy of the molecules of gas A will decrease and kinetic energy of molecules of gas B will increase. Hence, molecules of gas B should make more collisions to the walls of the container because of the increased kinetic energy and hence the total pressure exerted by gas B should increase and using the same argument, pressure exerted by gas A should decrease.
Now, we can no longer use P(A) + P(B)= P(T)
; P(A) is the partial pressure of gas A,
P(B) is the partial pressure of gas B
and P(T) is the total pressure exerted by the mixture of gas A and gas B
Am I correct with this logic?

A glass bulb of volume 400 ml is connected to another bulb of volume
  200 mL by means of a tube of negligible volume. The bulbs contain dry
  air and are both at a common temperature and pressure of 293 K and
  1.00 atm. The larger bulb is immersed in steam at 373 K ; the smaller, in melting ice at 273 K. Find the final common pressure.

In the above problem, we cannot simply use PV= nRT to calculate final pressure because we don't know the final common temperature. So, we have to individually calculate the partial pressure of both the gases and then sum them up to get the total pressure. But according to me, as stated above, that should not be correct.
Is my reasoning correct or I am missing out on something? Also, how should I solve using my logic and please tell if there are other ways to go about solving this problem.

According to the problem description, the two bulbs do not reach a common final temperature.  The temperatures in the two bulbs are at 273 and 373 in the final state.  Apparently, the bulbs are isolated from one another.  So even though the pressures of the gas in the two bulbs are equal, the temperatures are not.

When the stopcock is opened, moles of one gas will interact with moles of other gas, and because there is temperature difference, heat will flow and hence kinetic energy of the molecules of those gases will be affected. How will I take into account the pressure and temperature change at the same time?

In the end, when the system equilibrates, all the gas in one bulb will be at 273 and all the gas in the other bulb will be at 373.  We are not concerned with what happens in between the initial state and the final state (in this example).  Our focus is only on the final state.  There will be negligible heat flowing through the tube of "negligible volume" at the end.  You know the total number of moles and you know the two volumes and temperatures.  This is enough information to find the final pressure and the distribution of gas moles between the two bulbs.

The final state is when the stopcock is opened and the gases in individual bulbs interact, and we are required to find the final pressure corresponding to that state. According to you, what is the final temperature when the gases interact.

As I already said, they want you to assume that the final temperatures in the two bulbs are different.  I don't know how to say it any other way.  The temperatures do not equilibrate.

What makes you think when the gases interact they won't change temperature. When two entities come in contact and they are at different temperatures, heat will flow and they will come to equilibrium.  What makes you think they will not?

This is what makes me think that:  "The larger bulb is immersed in steam at 373 K ; the smaller, in melting ice at 273 K. "  The problem statement didn't say anything about next taking the system out of contact with these reservoirs, and then allowing it to thermally re-equilibrate.  So this was my interpretation of the problem statement.  Was my answer of 1.14 atm for the final pressure correct?

See the term : final common pressure? Why would the question maker want  you to find the individual partial pressures of the gases and then sum them up. It doesn't make sense to have connected the bulbs at all with this reasoning. Yes, your answer is correct. And even I got this answer using the same procedure as yours but it is not practically convincing to me.

I think that we have both concluded that the problem statement was confusing and not very well specified.  But, I didn't see anything in the problem statement about partial pressures.

Incidentally, we are dealing here with a single component system (if we treat air as a single component, with MW 29).  In over 50 years of experience in ChE, I have never heard of the term "partial pressure" being used in connection with a single component system.

Well, when two ideal gases are in the same bulb (they will be when the stopcock has been opened) we will talk of partial pressures, at least that's what I have been taught. YunfeiMa correctly pointed out that all the collisions are elastic and that answers the first part of my question. Maybe "we" have got the answer the wrong way because you treated both the gases to be non interacting and a mixture of gases do behave that way in  container. ALl this while you been assuming that both the gases are never in contact, maybe that's your interpretation but I don't agree with you.

Bottomline: I am done with this question. !!

You cannot have two IDEAL gasses in a homogeneous mixture at different temperatures. As the collisions are stipulated to be non-elastic, all the kinetic energy of the particles (temperature) will be instantly transferred and spread out so that all the gas particles follow the Boltzmann distribution.

Maybe not instant exchange of energy but it will be very fast, maybe you should consider the example of a beam of fusion neutrons (14 MeV) in a large tank of high pressure deutrium gas, or maybe C4D10 gas (perdeuterobutane). The neutrons can be considered to be gas (like noble gas atoms) which then hit the other atoms (which are parts of molecules). I am sure that if we have an infinite tank of C4D10 that the neutrons will soon be at thermal energies.

Determine total number of moles of air in bulbs:
$$n=\frac{PV}{RT}=\frac{(1)(0.4+0.2)}{(0.082)(293)}=0.025$$
Consider final state of system.  Let $P_F$ be the final equilibrium pressure, $n_1$ be the final number of moles in the larger bulb, and $n_2$ be the final number of moles in the smaller bulb:
$$n_1=\frac{(P_F)(0.4)}{(0.082)(373)}=0.0131P_F$$
$$n_2=\frac{(P_F)(0.2)}{(0.082)(273)}=0.0089P_F$$
$$n=n_1+n_2=0.022P_F=0.025$$
$$P_F=1.14\tag{atmospheres}$$

We can have two "gases" at different tempertures, neutrons can be considered to be gases.
If we consider a light water reactor then the neutrons formed by fission in the fuel pins are fast neutrons with high energies, they can be considered to be a very hot gas.
If they enter a medium where the temperture is much lower then when they strike atoms in the medium (such as water, graphite or even high pressure hydrogen gas) then exchange of energy occurs between the neutrons and the atoms which make up the medium.
If we choose a medium such as deutrium gas where the chance of capture is low (rather than something like boron trifluoride) then the neutrons will bang around from collision to collision and after a short while the translational energy of the neutrons will be lowered to the same level as gas molecules of at the same temperture.
Based on this observation from the operation of thermal nuclear reactors I strongly hold the view that as soon as you mix a hot gas with a cold gas that exchange of energy between the two gases will occur which results in the cold gas warming and the hot gas cooling to the same temperture.
While it is possible in some reactors to observe a change in the neutron energy spectrum as you change position in the reactor I can not think of an experiment which would allow you to do this with "normal" gases.
I hate to have to say it, but when a chemist spends 20 years of their life thinking about radiaoctivity and nuclear issues they start to see the world through a special lens.

How does this answer the question?