text:
The behaviour of Ag3PO4 in solution

To a $\pu{1L}$ flask, we add $\pu{4.1900 g}$ ($\pu{0.01001 mol}$) of $\ce{Ag3PO4}$, then fill it up with water. Knowing the following constants, we are asked to find the concentration of all species in solution:
$K_\mathrm{s}(\ce{Ag3PO4})=2.7\cdot 10^{-18}$,
  $K_\mathrm{s}(\ce{AgOH})=4.0\cdot 10^{-16}$,
  $K_\mathrm{a1}=10^{-2}$,
  $K_\mathrm{a2}=10^{-7}$,
  $K_\mathrm{a3}=10^{-12}$
$\ce{Ag2O}$ does not form.

By analyzing the problem, we notice that the formation of $\ce{AgOH}$ is more favorable:
$$\ce{Ag3PO4 + 3H2O <=> 3AgOH + H3PO4}, \\
K=\frac{K_\mathrm{s}(\ce{Ag3PO4})\cdot K_\mathrm{w}^3}{K_\mathrm{s}(\ce{AgOH})^3\cdot K_\mathrm{a1}\cdot K_\mathrm{a2}\cdot K_\mathrm{a3}} = 4.2\cdot10^7$$
Therefore, we start with $[\ce{H3PO4}]_0 = \pu{0.01001 mol//dm^3}$.
The equilibria we work with are:
\begin{align}
\ce{AgOH &<=> Ag+ + HO-}\\
\ce{H3PO4 + 3HO- &-> PO4^{3-} + 3H2O}\\
\ce{PO4^{3-} + H2O &<=> HPO4^{2-} + HO-}
\end{align}
The $\ce{HO-}$ formed from the insoluble salt is consumed in its entirety towards the formation of $\ce{PO4^{3-}}$, from which a fraction dissociates in water to form $\ce{HPO4^{2-}}$ and $\ce{HO-}$. Therefore, the principal conservation of mass law is:
$$[\ce{Ag+}]=3\ce{[PO4^{3-}]}+3\ce{[HPO4^{2-}]}$$

How do I tackle this problem in a more elegant manner? The equation on the top is solvable; but by looking at their results, we can see they do not fit at all:
$[\ce{Ag+}]=6.5\cdot10^{-14}$, $[\ce{HO-}]=6.2\cdot10^{-3}$, $[\ce{PO4^{3-}}]=3.8\cdot10^{-3}$
I'm thinking that, considering such a small concentration of $\ce{Ag+}$ compared to $\ce{PO4^{3-}}$, we might still have some $\ce{Ag3PO4}$ undissolved.
But this doesn't seem to make sense since the formation of $\ce{AgOH}$ is a lot more favorable, as seen from the value of $K$.


text:
Why does LiAlH4 reduce an amide to an amine, but only reduce a ketone/aldehyde to an alcohol?
Here are two different reactions in which lithium aluminium hydride, $\ce{LiAlH4}$, reduces a carbonyl group:

Why does $\ce{LiAlH4}$ completely remove the oxygen from the amide, but only reduces the carbonyl group of a ketone/aldehyde to a hydroxyl group?  Is the difference due to the fact that $\ce{H2O}$ is added in the first reaction (i.e. if the aldehyde/ketone reduction had a second step where water was added, would the carbonyl group be reduced to a methylene group)?

You've got it correct in your comment with the $\ce{N}$ lone pair assisting. The reason why the $\ce{N}$ lone pair kicks oxygen out, and not the other way round, is because oxygen-based groups are better leaving groups than nitrogen-based ones under the reaction conditions. (Just as a rough guide, you could compare $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$ and $\mathrm{p}K_\mathrm{a}(\ce{NH3}) = 38$.)

(Don't worry about how I've drawn the metals coordinating to oxygen - the exact details aren't important and sometimes aren't even fully understood.)
In the aldehyde reduction, there is no lone pair to assist in the expulsion of oxygen. So the intermediate alkoxide just sits there, until you add water or acid:

The aqueous workup serves two purposes: firstly to reprotonate the product (for example, with the aldehyde reduction, your final product is an alkoxide ion and adding water gets you the alcohol), and secondly to destroy any unreacted $\ce{LiAlH4}$. In fact, the aqueous workup is also carried out in the reduction of the ketone or aldehyde - we often just omit it because it's so common and it's tiresome to have to write it on every reaction we do. So even though it's not there, it's still implied.
If you wanted to reduce the carbonyl group of a ketone/aldehyde to a methylene group, the three most well-known methods available are the Wolff-Kishner reaction, the Clemmensen reduction, and the Mozingo reduction.

Why then, does an ester not undergo assisted reduction with the help of the neighbouring O's lone pairs?

I don’t get the point. Esters are fully reduced to alcohols. @YUSUFHASAN

No, I mean..I did exactly what you did up there with amide, just in place of N, used the ester's oxygen to remove lithium oxide...and then added a hydride at the carbon moiety of the doubly bonded oxonium ion so formed..But I ended up with an ether..So why can't assisted removal occur here?

@YUSUFHASAN because OR is a better leaving group than O[M], whereas NR2 is worse so N kicks out O[M]

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text:
Reactions of phloroglucinol
Does phloroglucinol respond to the

iodoform test?
cerric ammonium nitrate test?
2,4-DNP test?

I couldn't find any good sources that answer any of these.
I thought of this:

Tautomerism can "change" phloroglucinol to a 1,3,5-triketone,  which should respond to the iodoform test.
It is a phenol, so it should respond to the cerric ammonium nitrate test.
Again, the keto form should respond positively to the dinitrophenylhydrazine test.

Tautomerism is an interesting concept, isn't it?

Speculation aside, neither resorcinol nor phloroglucinol give positive iodoform tests. Fuson, R.C. and  Tullock, C.W., J. Am. Chem. Soc., 1934, 56, 1638.

In aromatic compounds, the enol form is more stable than the keto form due to the aromatic character. In normal conditions, aromatic compounds exist in enol form but in basic conditions, it tautomerises to keto form.  Phloroglucinol predominantly exist in enol form under normal condition.

iodoform test

Compounds containing ketomethyl group($\ce{-COCH3}$) will give iodoform test. e.g.- propanone, acetophenone etc. Thus phloroglucinol will not give iodoform test because it does not contain ketomethyl group.

ceric ammonium nitrate

Generally, CAN test is done for alcohols which give pink or red colour. But for phenols and phenolic compounds, it gives brown or black test. So, this test helps to differentiate phenols from alcohols. See the detailed question on Chem.SE

2,4 DNP test

Phloroglucinol predominantly exist in enol form under normal condition and thus not give  2,4 DNP test. However, in basic condition (pH- 11-13) it forms 3,5-dihydroxy-2,5-cyclohexadienone dianion,  the deprotonated intermediate in the interconversion of the two forms, referred to as the enolate anion which might give the 2,4 -DNP test. See here
One final note
Don't take tautomerism into account while experimentally determining the functional group in organic compounds . If you consider tautomerism, there will be two functional group for the compound and you will get confused as of which functional group should be considered. For ex- propanone tautomerises to form prop-1-en-2-ol which contains an alcoholic group and then you will get confused which test should be used- Ester or CAN test(for alcohol) or 2,4 DNP test (for ketone).

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text:
Sand Under Pool Turned Gold in Color
I'm removing an above-ground chlorine pool installed years ago by a previous owner of my house. Looking underneath the pool, I found that the sand on which it's resting is a gold/bronze color with a metallic sheen to it. The pool itself is steel, with patches of rust. Does anyone have any ideas about how the sand might have turned metallic? I'm doubtful that it started off that way, as no above-ground pool retailers I've contacted have ever heard of such a thing. Thanks!

A good photo might be of some help.

Is the sand uniformly colored?

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text:
What makes ionic conduction fast or slow?
Sometimes in certain electric circuits (such as in voltaic cells) ions are moved around as well as electrons.
I have a basic understand of what materials (mostly metals) conduct electrons well but what materials conduct ions well? What makes ionic conduction fast or slow?
I have read that molten salts are good ionic conductors and that seems sensible to me because all particles in a molten salt are charge carriers and there is nothing to get in the way of carrying charge. But I don't know much more than that such as what sorts of molten salts would be the best ionic conductors.

Solids can also be ionic conductors - the famous example is α-AgI. The idea is that there are lots of holes and spaces for the silver ions to move around. https://en.wikipedia.org/wiki/Fast_ion_conductor In general, you would want to look at the exact 3D structure of the solid.

related http://chemistry.stackexchange.com/questions/26863/what-is-the-meaning-of-superionic

@orthocresol Then would certain aerogels be fast ion conductors as they have lots of room for stuff to move around in?

I don't really know. "Holes" has a specific meaning here, it does not refer to macroscopic pores that you can see, like those of a sponge... it refers to empty cation sites, or interstitial sites, in the solid state structure. http://www.science.uwaterloo.ca/~cchieh/cact/c123/tetrahed.html


text:
Deriving kinetic equations for a complex reaction system
I have a system of protein reactions that I've derived differential equations for the kinetics of. I'm using them to run simulations to fit the rate parameters, but the simulations at present are working badly. I'd like to start by checking the system of differential equations. I realise they may be to complex for anyone to write them out here, but you know of software that could be used to check my work, please post it here. (A lot of people face similar problems)
The three monomers are $L$, $A_1$, $A_2$; the chemical equations for the monomers forming dimers are:
$L + A_1 \rightleftarrows^{k_{a1}}_{k_{d1}} LA_1$,
$\qquad$
$L + A_2 \rightleftarrows^{k_{a2}}_{k_{d2}} LA_2$,
and the dimers forming trimers are:
$LA_1 + L\rightleftarrows^{k_{a1}}_{k_{d1}} LA_1L$,
$\qquad$
$LA_1 + A_2\rightleftarrows^{k_{a2}}_{k_{d2}} A_1LA_2$,
$LA_2 + L\rightleftarrows^{k_{a2}}_{k_{d2}} LA_2L$,
$\qquad$
$LA_2 + A_1\rightleftarrows^{k_{a1}}_{k_{d1}} A_1LA_2$,
trimers forming tetramers:
$LA_1L+A_2\rightleftarrows^{k_{a2}}_{k_{d2}} LA_1LA_2$,
$\qquad$
$LA_2L+A_1\rightleftarrows^{k_{a1}}_{k_{d1}} LA_2LA_1$,
$A_1LA_2 + L\rightleftarrows^{k_{a1}}_{k_{d1}} LA_1LA_2$,
$\qquad$
$A_1LA_2 + L\rightleftarrows^{k_{a2}}_{k_{d2}} LA_2LA_1$,
dimers forming tetramers
$LA_1+LA_2\rightleftarrows^{k_{a1}}_{k_{d1}} LA_1LA_2$,
$\qquad$
$LA_1+LA_2\rightleftarrows^{k_{a2}}_{k_{d2}} LA_2LA_1$,
tetramer molecules changing their configuration
$LA_1LA_2 \rightleftarrows^{k_{a4}}_{k_{d4}} B$,
$\qquad$
$LA_1LA_2 \rightleftarrows^{k_{a3}}_{k_{d3}} B$.
Could anyone either point me in the direction of software to derive the system of differential equations, or do the derivation?

I'm afraid you should know how to do this yourself. Also if you want to only integrate ODE's you shouldn't worry about CPU time.

I'd use a matrix approach. Least likely to make errors. Set up the stoichiometric matrix.

I'm assuming that you want time profiles not just equilibrium values. In this case use a Master Equation approach to set up a matrix and solve for eigenvalue  values & eigenvectors numerically. Use these to reconstruct the time profiles etc. You can use Maple or Mathcad, both commercial (& expensive), or use Python which is free, but just as easy to use for numerical work.

In many cases, complex systems of even simple differential equations do not have simple and obvious solutions. Therefore, it could be too difficult and not reasonable to find fully analytical solutions.
Usually, numerical solutions can be easily calculated using iterative algorithms. However, such computer codes can be susceptible to calculation artefacts. Furthermore, the validation and verification of the model and the code can be a difficult issue.
I made a compromise and used an analytical approach when solving systems of differential equations, but I allowed the code to evaluate subexpressions involving constants numerically. Thus, I made numerical calculations to combine all constant parameter values as soon as possible in order to limit the number of parameters in the equations (using excessive digits in order to avoid accumulation of round-off errors).
I found that this approach can be easily and quickly implemented using the commercial software Maple. However, other similar mathematical software could be suitable, too.
I successfully used this approach for the evaluation of radioactive decay of the inventory of a nuclear reactor core in dependence of time after shutdown. The calculation of the radioactive decay of the relevant radionuclides involved more than 100 individual differential equations, which mainly describe chains of simple first-order reactions, e.g.:
$$\frac{\mathrm da_{\text{I-131}}(t)}{\mathrm dt}=-\lambda_{\text{I-131}}\cdot a_{\text{I-131}}(t)+0.778\cdot\lambda_{\text{I-131}}\cdot a_{\text{Te-131m}}(t)+1\cdot\lambda_{\text{I-131}}\cdot a_{\text{Te-131}}(t)$$
> de_a_I131:=diff(a_I131(t),t)=-l_I131*a_I131(t)+0.778*l_I131*a_Te131m(t)+1*l_I131*a_Te131(t);

A large number of further differential equations could be added in order to model transport phenomena during postulated accidents (e.g. loss of coolant).
Differential equations that were dependent on each other and the corresponding initial conditions were grouped into systems of differential equations, e.g.:
> sys_131:={de_a_Xe131m,de_a_I131,de_a_Te131,de_a_Te131m,a_Xe131m(t0)=core_Xe131m,a_I131(t0)=core_I131,a_Te131(t0)=core_Te131,a_Te131m(t0)=core_Te131m};

The systems of differential equations were solved analytically and the subexpressions involving constant parameter values were evaluated numerically.
> sol_131:=evalf(value(dsolve(sys_131,[a_Xe131m(t),a_I131(t),a_Te131(t),a_Te131m(t)])));

The results could be easily rearranged into a suitable form, e.g.
$$\begin{aligned}a_\text{I-131}(t) ={} & -1.6030637832147524320\cdot10^{17}\cdot\mathrm e^{-6.4180294496291232354\cdot10^{-6}\cdot t}\\&-7.9769363886868464215\cdot10^{15}\cdot\mathrm e^{-4.6209812037329687295\cdot10^{-4}\cdot t}\\&+4.7213833147101620896\cdot10^{18}\cdot\mathrm e^{-9.9782796169607015473\cdot10^{-7}\cdot t}\end{aligned}$$
The results could also be directly used for further calculations within the same worksheet.


text:
Historically, how was the connection between NMR peaks and hydrogen atoms made?
A large part of the basis of this question may be due to the student who didn’t pay attention in the history section of the NMR course (and not during the technical details bit either) but I find the question intriguing nonetheless.
My basic knowledge of the history of NMR spectroscopy is that in the very old, almost prehistoric days, Bloch and Purcell performed some experiments that applied radiofrequency to a sample and by chance got some radiofrequency signal returned. Then I have a big black box and arrive at modern spectrometers that tune themselves into hydrogen spectra, carbon spectra and see beautiful Fourier-transformed spectra on my PC screen (unless it’s carbon, in which case the baseline is, of course, fuzzy).
I imagine that there had to be a lot of research invested until it was realised that each signal in the Fourier-transformed spectrum corresponds to a hydrogen atom, that they are shifted around depending on how electron-rich or -poor they are, how couplings work and so on. I can imagine that most of this stemmed from trial and error: a certain combination of magnetic field and radiofrequency gives a signal, Fourier transformation seems to happen anywhere, and once you have rationalised that ‘signals’ correspond to ‘hydrogens’ in a certain type of spectrum (likely the one that gave most resonance anyway) you’re half done and the rest is a walk in the park.
But there is one step that seem very unintuitive:
Realising that ‘signals’ correspond to hydrogens.
There is another important step, namely figuring out that solutes can be analysed if hydrogen-free ($\ce{CCl4}$) or at least protium free (deuterated) solvents are used. Now either it was figured out first that hydrogen is the cause for the signals, in which case choosing the correct solvents is a non-issue, but capturing signals from the solute may be. Or it was figured out that using certain solvents gave ‘cleaner’ spectra, in which case it might also be interesting to know how the solvents were chosen.
I realise that there may be a good deal of speculation, retrofitting, inflationary use of the term serendipitously, anecdotical evidence, storytelling and bad memories involved, but is it known how the discoverers of NMR spectroscopy realised that a ‘signal’ is a hydrogen atom in the molecule observed?

You are seriously underestimating physics, and the capability of physicists plan experiments and  predict their outcome.

The Fourier Tranformation was a later addition to the NMR spectrometers, the earlier spectrometers used Continuous Wave NMR. Pulsed FT NMR is a much later development

The NMR experiment is a classic example of experimental results supporting theoretical prediction. The first reports of NMR signals of bulk matter (water or paraffin) were confidently reported as signals from the $\ce{^1H}$ nucleus because that was what was expected from theoretical prediction.
When Bloch, Hansen and Packard carried out their Nuclear Induction experiment published in 1946 (Phys. Rev. 1946, 70, 474), they knew the signal they were observing came from the $\ce{^1H}$ nucleus because of the basic equation for the observing the NMR phenomenon (Larmor equation):
$$\omega = \gamma B_0$$
That is, the resonance condition will be met when a superimposed oscillating magnetic field (what we call excitation Rf) is applied to matter in a magnetic field $B_0$, for a nucleus with gyromagnetic ratio $\gamma$.
$\gamma$, the ratio of nuclear magnetic moment to angular momentum, is characteristic of a particular type of nucleus, and measurements by Stern and Rabi had provided quite precise values for $\ce{^1H}$ by the time Bloch carried out his experiment. However, it should be pointed out that field homogeneity was so poor in initial experiments that the error in observed frequency (actually the magnetic field) actually still created some doubt that $\gamma$ was constant for a given nucleus. Improvements in field stability and homogeneity finally showed that this was true. Observation of chemical shift some years later lead to the notion of a $B_\mathrm{eff}$; an effective magnetic field attenuated by local shielding.
It should be noted that there were other attempts to observe Nuclear Magnetic Resonance prior to Bloch and Purcell, and these were done on other nuclei. Gorter although unnsuccessful and little known, made an attempt to observe  $\ce{^7Li}$ signals at very low temperatures. Rabi first observed Nuclear Magnetic Resonance on $\ce{LiCl}$.
Of course, as we dig deeper and remind ourselves that all great science stands on the shoulders of those that have gone before, this understanding of resonance frequencies and magnetic fields all stems from the famous Larmor equation, published way back in 1897, long before the notion of NMR was born and even while the structure of the atom was still being refined. The relationship, though, holds true for any charged particle with angular momentum, and originally $\gamma$ was known as Larmor's constant - a measurement of the particle charge:mass ratio. Knowing this, then, once the $\gamma$ for $\ce{^1H}$ had been calculated, scientists could safely assume that at a given magnetic field, the signal observed when matter was irradiated with the correct Rf field was indeed $\ce{^1H}$.
And of course FT NMR as is widely used today was a late-comer to the table also; introduced commercially in the 1960's. Prior to this, CW spectrometers operated and NMR was largely an absorbance technique rather than the emission detection technique it predominantly is now.

1 Bloch's theoretical paper adds the footnote: "The first purely magnetic experiment to find an effect
due to nuclear moments by measuring the susceptibility
of liquid hydrogen was published by B. G. Lasarew and
L. W. Schubnikow, Sov. Phys. ll, 445 (1937)"

TL;DR: Scientists knew that electrons had a spin, and thought, "wouldn't it be nice if atomic nuclei did too?" When they passed molecular hydrogen through a homogenous magnetic field (in a modified version of the Stern–Gerlach experiment) and subjected it radio waves, they found energy was absorbed at precise frequencies. This set the stage for the usefulness of $^1\ce{H}$ NMR in the distant future.
Taken from A Brief History of Nuclear Magnetic Resonance:

In the mid-1920s it became apparent that many features in atomic spectra could be accounted for only if certain atomic nuclei likewise possessed spin and a magnetic moment. Refinements of the Stern-Gerlach experiment verified this concept by 1933.
In 1939 Rabi et al. made a major improvement in beam techniques by sending a stream of hydrogen molecules through not only the inhomogeneous magnetic field required for deflection, but also through a homogeneous magnetic field, where they were subjected to radio frequency electromagnetic energy. Energy was absorbed by the molecules at a sharply defined frequency, and the absorption caused a small but measurable deflection of the beam. This was the first observation of NMR, and Rabi received the Nobel Prize in 1944. However, such studies were limited to nuclei in small molecules under very high vacuum in a molecular beam, the deflection of which served to detect the resonance.

NMR reflects the difference of energy levels upon absorbance of a wave length precessing close to the Larmor frequency (Zeeman interaction). You must remember that NMR also obeys exclusion rules as any other spectroscopic methods, that is that only certain changes in energy of the spin of nuclei are observed, in NMR +1 and -1, which sometimes we call single quantum coherences. Now with respect to why certain solvents e.g. deuterated are NMR invisible is not because was a trial and error, it was because of the foundation of NMR is based also on 1/2 spins like 1H that obeys the exclusion rules of NMR, other nuclei like 2H or deuterated proton do not have spin of 1/2. This led us to immediately take advantage of this nice property and make a variety of experiments to analyse a variety of samples. NMR unlike any other spectroscopy, the observer is the nucleus spin and not the experimentalist, as often it is the case with IR or UV spectroscopy.

Deuterated molecules are not NMR invisible; 2H has spin=1, and can be observed just as most other nuclei that do not have spin=0 can be. I don't understand what you mean by the foundation of NMR is based on 1/2 spins.  Gorter's first attempts to observe NMR signal were made on 7Li - spin 3/2. How do you explain how we observe nuclei like 2H, 6Li, 7Li, 11B, 14N, 17O,  51V.......?

You are right when you say that it is possible to observe spins bigger than 1/2. Deuterium is considered NMR friendly with spin of 1, but most of these are technically challenging and considered NMR silent. since their spin will have electric energy terms (Quadropolar terms) that depend on the orientation or internal structure of the nucleus. This means that spins>1/2 integers will interact the electric fild gradient


text:
Hybridization of nitrogen in a ring
Is there a well defined way to discern the hybridization of a nitrogen atom in ring, like pyrrole? How can you know whether the nitrogen's lone pair are in the conjugated system or not?

related: How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?

If an $\alpha$-$\beta$ unsaturated heteroatom isn't already participating in a $\pi$-bond and it has a lone pair of electrons, these electrons will be delocalized and in the $\mathrm{p}$-orbital.
Let's compare pyridine and pyrrole:
$\hspace{5cm}$
The nitrogen in pyridine is already participating in a double bond, so it is clearly $\mathrm{sp^2}$ hybridized. The nitrogen in pyrrole is $\alpha$-$\beta$ unsaturated, so its lone pair is delocalized. Both of these molecules are heterocyclic aromatic rings, but these are not the only cases in which a nitrogen atom's lone pair are delocalized. They are also delocalized in amides (which resemble the $\alpha$-$\beta$ unsaturation of pyrrole):
$\hspace{4.6cm}$
If you want to know more about criterion for aromaticity, read about Hückel's rule.

Hückel's rules actually cover only a tiny subset of aromatic compounds. It's a start, but it should not be used to attest aromaticity. The rules were derived from hydrocarbons and are strictly speaking only valid for those systems. Ergo pyridine and pyrrole are not really covered by them. Also, hybridisation is a model and not all the orbitals have to be hybridised in the same way - it is only depending on the geometry. In your last example I would doubt that the nitrogen is planar.

I think it's bizarre to suggest that Hückel's rule wouldn't apply to heterocyclic rings. There are many, many examples on the web of them perfectly and accurately fitting the criterion. If anything I would caution against applying it to polycyclic species, where it is known to break down. Also, DMF is planar in the solid and liquid phases, and slightly non-planar in the gas phase according to this paper.

Strictly speaking, it does not apply. It is theoretically not justifiable and when you say that it fits perfectly and accurately, you are just over simplifying a very complex problem. Even trying to applying it to polycyclic compound is flat out wrong, it had nothing to do with cautioning. Just because it is easy, it does not mean it is correct. Just because many people make the same mistake, it does not make it correct. I find it bizarre, that you use the rules without questioning the validity.


text:
How to read the used volume from a burette?
I conducted an acid-base titration with vinegar and sodium hydroxide today, where sodium hydroxide was used as the titrant (solution in the burette). I collected the data for the initial and final burette reading to work out the volume of sodium hydroxide used. Is the volume used just the final burette reading, as the measurements are counted backwards?

It depends ;-)
The scale on a burette typically is ascending from top to bottom. With other words: zero is close to where you fill your $\ce{NaOH}$ solution in, the largest value is close to the valve. The total amount added (to your vinegar) is the difference between the final and the initial reading.
But if your burette was initially filled up to the zero mark, this would just mean to subtract zero from the final reading (which is just the final reading).


text:
What is the difference between ΔG and ΔrG?
Consider the reaction
$$\ce{A -> B}$$
The reaction Gibbs free energy, $\Delta_\mathrm{r} G$ is given by the following equation
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT \ln Q$$
Now what is the difference between $\Delta G$ and $\Delta_\mathrm{r} G$ ?
Which one of the above represents $G_{\text{products}} - G_{\text{reactants}}$ ?
I read in my book $\Delta_\mathrm{r} G$ is the slope of $G$ plotted against the extent of reaction at different instants. And at equilibrium, it is zero. So can we say that it has value at a particular instant while $\Delta G$ is for a process ($\ce{A}$ to $\ce{B}$ in my question) i.e. is for a time interval?
Does $\Delta G$ for the above reaction represent change in Gibbs Energy when 1 mole of $\ce{A}$ reacts completely to form $1$ mole of $\ce{B}$ ?

Not for the faint-hearted: There is an excellent, but very mathsy, article here: J. Chem. Educ. 2014, 91, 386 describing the difference.
The Gibbs free energy change, $\Delta G$
You are quite right in saying that $\Delta G$ represents a change in a system over a time interval. The notation $\Delta G$ itself implies that it is a difference in $G$ between two things: an initial state and a final state. Let's use a real example in order to make things clearer. Consider the thermal decomposition of ammonium nitrate:
$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$
If someone were to ask you "what is $\Delta G$ for this reaction?", you should really, technically, be telling them: it is not well-defined. That is because of two things. Firstly, the reaction conditions, e.g. temperature and pressure, are not specified.
But more importantly, a balanced equation does not tell you exactly how much ammonium nitrate is reacting. In your beaker, you could have $1~\mathrm{mol}$ of $\ce{NH4NO3}$ reacting according to the above equation, but you could equally well have $2~\mathrm{mol}$, or you could have $5000~\mathrm{mol}$ (I must say that's one huge beaker though), or you could have $0.001~\mathrm{mmol}$.
The stoichiometric coefficients in the equation do not indicate the amounts of compounds reacting.
So, a more proper question would read like this:

$\pu{2 mol}$ of ammonium nitrate fully decompose according to the balanced equation $\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. Calculate $\Delta G$ for this process.

Great. Because the decomposition is complete, we now know that our final state is $2~\mathrm{mol}~\ce{N2O(g)} + 4~\mathrm{mol}~\ce{H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$, and our initial state is $2~\mathrm{mol}~\ce{NH4NO3(s)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. So,
$$\begin{align}
\Delta G &= [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot G_\mathrm{m}(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{NH4NO3(s)})]
\end{align}$$
We can't quite find the absolute molar Gibbs free energies, so the best we can do is to use Gibbs free energies of formation.
$$\begin{array}{c|c}
\text{Compound} & \Delta_\mathrm{f}G^\circ / \mathrm{kJ~mol^{-1}}\text{ (at }298~\mathrm{K}\text{)} \\ \hline
\ce{NH4NO3(s)} & -183.87 \\
\ce{N2O(g)} & +104.20 \\
\ce{H2O(g)} & -228.57
\end{array}$$
$$\scriptsize \text{(data from Atkins & de Paula, }\textit{Physical Chemistry}\text{ 10th ed., pp 975-7)}$$
(Note that the use of standard formation Gibbs free energies is only because of the conditions specified in the question, which conveniently corresponds to the standard state. If we specify different conditions, we can still find $\Delta G$, but we would have to use different data.) So:
$$\begin{align}
\Delta G &= [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{NH4NO3(s)})] \\
&= [(2~\mathrm{mol})(+104.20~\mathrm{kJ~mol^{-1}})] + [(4~\mathrm{mol})(-228.57~\mathrm{kJ~mol^{-1}})] - [(2~\mathrm{mol})(-183.87~\mathrm{kJ~mol^{-1}})] \\
&= -1073.62~\mathrm{kJ}
\end{align}$$
Note that we have units of kJ. Since $\Delta G$ is the difference between the Gibbs free energy of one state and another, $\Delta G$ has to have the same units as $G$, which is units of energy.
Now, this does not necessarily mean that $\Delta G = G_\text{products} - G_\text{reactants}$. For example, if I changed my question to be:

In the Haber process, 100 moles of $\ce{N2}$ and 300 moles of $\ce{H2}$ are reacted at $800~\mathrm{K}$ and $200~\mathrm{bar}$ according to the equation $\ce{N2 + 3H2 -> 2NH3}$. Only 10% of the starting materials are converted under these conditions. Calculate $\Delta G$ for the process. (These numbers are made up.)

then, your final state would not be the pure products. Your final state is not 200 moles of $\ce{NH3}$. Your final state is $90~\mathrm{mol}~\ce{N2}$, $270~\mathrm{mol}~\ce{H2}$, and $20~\mathrm{mol}~\ce{NH3}$.
In general, one could write, for a chemical reaction,
$$\Delta G = \sum_i (\Delta n_i) G_{\mathrm{m},i}$$
where $\Delta n_i$ is the change in the amount of compound $i$ (in moles), and $G_{\mathrm{m},i}$ is the molar Gibbs free energy of the pure compound $i$, under the $T$ and $p$ conditions specified. Going back to our ammonium nitrate example, we would have $\Delta n_{\ce{NH4NO3}} = -2~\mathrm{mol}$, $\Delta n_{\ce{H2O}} = +2~\mathrm{mol}$, and $\Delta n_{\ce{H2O}} = +4~\mathrm{mol}$.
The Gibbs free energy change of reaction, $\Delta_\mathrm{r} G$
As you have correctly stated, $\Delta_\mathrm{r}G$ is the slope of a graph of $G_\mathrm{syst}$ against the extent of reaction, commonly denoted $\xi$. This is the easiest way of interpreting $\Delta_\mathrm{r}G$. This question contains a slightly fuller derivation and explanation of what $\Delta_\mathrm{r}G$ means.
However, $\Delta_\mathrm{r}G$ does indeed, somewhat, refer to the instantaneous difference between the "molar Gibbs free energies" of the products and reactants. This is different from $\Delta G$ in three main ways.
First, $\Delta G$ is the difference between the Gibbs energy of the entire system at two points in time. Here, we are interpreting $\Delta_\mathrm{r}G$ as the difference between the "molar Gibbs free energies" of two components of the system: reactants and products. In other words, the system generally contains both reactants and products, and $\Delta_\mathrm{r}G$ may be thought of as the difference between the Gibbs energies of the product part of the system, and the reactant part of the system, even though you cannot separate them in the laboratory.
Secondly, it is the difference between the chemical potentials, not the molar Gibbs free energies. The molar Gibbs free energy is simply defined by $G_i/n_i$; the chemical potential is a partial derivative and is defined by $\mu_i = (\partial G/\partial n_i)_{T,p,n_j}$. In the case where there are no other species present (i.e. species $i$ is pure), then the chemical potential is identical to the molar Gibbs free energy.
Lastly, it is weighted by the stoichiometric coefficients $\nu_i$ instead of the change in the amount $\Delta n_i$. The stoichiometric coefficient is a dimensionless quantity, which is negative for reactants and positive for products. So, in the ammonium nitrate decomposition as written at the very top, we have
$$\nu_{\ce{NH4NO3}} = -1; \qquad \nu_{\ce{N2O}} = +1; \qquad \nu_{\ce{H2O}} = +2$$
and our expression for $\Delta_\mathrm{r} G$ is
$$\begin{align}
\Delta_\mathrm{r} G &= \sum_i \nu_i \mu_i \\
&= \nu_{\ce{N2O}}\mu_{\ce{N2O}} + \nu_{\ce{H2O}}\mu_{\ce{H2O}} - \nu_{\ce{NH4NO3}}\mu_{\ce{NH4NO3}} \\
&= \mu_{\ce{N2O}} + 2\mu_{\ce{H2O}} - \mu_{\ce{NH4NO3}}
\end{align}$$
Now, note the units again. The chemical potential is a partial derivative of the Gibbs free energy (units $\mathrm{kJ}$) with respect to the amount of $i$ (units $\mathrm{mol}$), and so it must have units $\mathrm{kJ~mol^{-1}}$. And from our above expression, $\Delta_\mathrm{r}G$ must also have units of $\mathbf{kJ~mol^{-1}}$, since the stoichiometric coefficients are dimensionless.
Does it matter if you start with $1~\mathrm{mol}$ or $2~\mathrm{mol}$ of ammonium nitrate? The answer is now, no. The amount of starting material does not affect $\mu_i$, nor does it affect $\nu_i$. Therefore, $\Delta_\mathrm{r}G$ is independent of the amount of starting material.
Why is $\Delta_\mathrm{r} G$ an instantaneous difference between the chemical potentials? Well, that is because the chemical potentials of the reactants and products, $\mu_i$, are changing continuously as the reaction occurs. Therefore, if we want to calculate $\Delta_\mathrm{r} G$, we have to take a "snapshot" of the reaction vessel: otherwise it makes absolutely no sense to speak of $\mu_i$ because we wouldn't know which value of $\mu_i$ to use.
Compare this with $\Delta G$ above: the quantities of $G_{\mathrm{m},i}$ are constants that do not vary depending on the extent of reaction. Therefore, we do not need to specify a particular extent of reaction to calculate $\Delta G$.
A diagram to sum up

Note that, to adequately define what $\Delta G$ is, you need a starting point and an ending point. I showed two possibilities for $\Delta G$; there are infinitely many more.
Let's say the reaction goes to completion. If you double your starting material and double your product, the difference between $G_\text{products}$ and $G_\text{reactants}$ will also be doubled. So, it's important to specify!
Likewise, to adequately define $\Delta_\mathrm{r} G$, you have to define the single specific point at which you intend to calculate $\Delta_\mathrm{r} G$.
If you double your starting material and double your product, the curve is stretched by a factor of 2 along the y-axis, but it is also stretched by a factor of 2 along the x-axis because $\xi$ is also doubled. (For a mathematical explanation see the definition of $\xi$ given in the linked question earlier.) So, while the difference $\Delta G$ is doubled, the gradient $\Delta_\mathrm{r} G$ remains unchanged.
The units should also be clear from this. Since $\Delta G$ is a difference between two values of $G$, it has to have units of $\mathrm{kJ}$. On the other hand, $\Delta_\mathrm{r}G$ is a gradient and therefore has units of $\mathrm{kJ/mol}$.

A caveat
Unfortunately, the notation $\Delta G$ is often loosely used and treated as being synonymous with $\Delta_\mathrm{r}G$. You will therefore see people give $\Delta G$ units of $\mathrm{kJ~mol^{-1}}$. For more information refer to Levine, Physical Chemistry 6th ed., p 343. I would personally recommend making a distinction between the two.

"To adequately define $\Delta_\mathrm{r} G$, you have to define the single specific point at which you intend to calculate $\Delta_\mathrm{r} G$".
Consider following definition.
"The standard Gibbs energy (change) $\Delta_\mathrm{r} G^\circ$ for a chemical reaction is the change in G for converting stoichiometric numbers of moles of the separated pure reactants, each in its standard state at T, into the separated pure products in their standard states at T."
As $\Delta_\mathrm{r} G^\circ$ is special case of $\Delta_\mathrm{r} G^\circ$ when $Q=1$, then why it is defined as above?

Please don't edit the question. I have rolled back the changes. Anyway, let me use an analogy. If you have a function $y = x^2$, $\mathrm{d}y/\mathrm{d}x = 2x$. The value of the derivative $\mathrm{d}y/\mathrm{d}x$ varies with $x$ and therefore we cannot assign it a specific numerical value, unless we specify one particular value of $x$ that we want to evaluate the derivative at. Likewise, we cannot find a numerical value for $\Delta_\mathrm{r}G$ unless we specify the point at which we want it to be evaluated.

For $\Delta_\mathrm{r}G^\circ$, the additional condition that the $\circ$ imposes is that the particular point we pick must correspond to $Q = 1$. For $\Delta_\mathrm{r} G$, we have to pick a point, but it does not have to be the point corresponding to $Q = 1$.

Brilliant Answer by the way. Exactly what i was looking for. Thank you very much, Sir.

I read in my book that
The first term on the right in the equality is
the standard reaction Gibbs energy, $\Delta_\mathrm{r} G^\circ$ :
$\Delta_\mathrm{r} G^\circ$= {cμC + dμD} − {aμA + bμ B }
Because the standard states refer to the pure materials,
the standard chemical potentials in this expression
are the standard molar Gibbs energies of the (pure)
species. Therefore, eqn is the same as
$\Delta_\mathrm{r} G^\circ$ = {cGm(C) + dGm(D)} − {aGm(A) + bGm(B)}
Is this why we can define $\Delta_\mathrm{r} G^\circ$ like difference in Standard Molar Gibbs Energies?

Yes, that is correct. Because $\mu_i^\circ = G_{\mathrm{m},i}$ you can replace the chemical potentials with molar Gibbs energies in the expression for $\Delta_\mathrm{r} G^\circ$.

Last question
So We can define $\Delta_\mathrm{r} G^\circ$ for a process ( converting stoichiometric numbers of moles of the separated pure reactants, each in its standard state at T, into the separated pure products in their standard states at T)
and it made sense But we cannot do so for $\Delta_\mathrm{r} G$ and it has a physical meaning only at a instant?

I would hesitate to say that that is a "process", because you are not converting 1 mol of something to another, you are converting 1 of something to another ($\nu_i$ being dimensionless). But I get what you mean. If you want to look at it that way, then $\Delta_\mathrm{r} G$ at a particular instant can also be thought of as a process of converting the reactants, at the particular state at that instant, into the products, at the particular state at that instant. This state needs to be defined by multiple parameters, including $T$, $p$, and all $n_i$'s of species present.

(...) So, when we add the standard state symbol in $\Delta_\mathrm{r} G^\circ$, all that changes is that "that instant" corresponds to the time where $Q = 1$, and the "particular states" correspond to standard states.

"Then ΔrG at a particular instant can also be thought of as a process of converting the reactants, at the particular state at that instant, into the products, at the particular state at that instant"
Can you give an example?

Using the Haber process example above, if the reaction proceeds to equilibrium, there will be 90 mol N2, 270 mol H2, and 20 mol NH3. The "state" of N2 is therefore completely specified by: temperature 800 K, pressure 200 bar, amount of N2 present 90 mol, amount of H2 present 270 mol, and amount of NH3 present 20 mol. Changing any one of these variables will affect the value of the chemical potential; the state is specified by the variables $T$, $p$, $n_i$, $n_j$, $n_k$, $\cdots$

Compare this to our "standard state" at 800 K (standard state does not specify temperature, so it must be additionally specified): temperature 800 K, pressure 1 bar, amount of N2 present does not matter, amount of H2 present 0 mol, and amount of NH3 present 0 mol (since N2 must be pure). The amount of N2 present does not matter because you divide by it anyway (whereas in the earlier case it matters because it is not a division, but a partial derivative).

Sir, I have understood everything. But this thing  that "ΔrG at a particular instant can also be thought of as a process of converting the reactants, at the particular state at that instant, into the products, at the particular state at that instant" is not getting into my mind. I tried reading books but it was not explained. I have read your other answer couple of time but i am still having this as doubt. Can you please add a detailed explanation to this in this answer? It would be really helpful. Why it can be thought so?

In my books they have written $\Delta_\mathrm{r} G^\circ$ as the energy which can be extracted from system but I think that should be $\Delta G$  and not $\Delta_\mathrm{r} G^\circ$. Because as you wrote $\Delta G$ represents change in Gibbs Free Energy for a process and not Reaction Gibbs Energy.
For example see this http://i.imgur.com/UEqrLYg.png
Why $\Delta_\mathrm{r} G^\circ$ is for a reaction and not for an instant?

@CLAPorSLAPII Come over here: http://chat.stackexchange.com/rooms/39235/thermodynamics

@orthocresol In the formula for calculating equilibrium constant $$K=e^{\frac{-\Delta G}{RT}}$$ what $\Delta G$ represents?

@adosar that is not correct: it should be $\Delta_\mathrm r G^\circ$. Standard Gibbs energy of reaction. see https://chemistry.stackexchange.com/q/41862/16683

@But if it is $\Delta_\mathrm r G^\circ$ then what it represents? A derivative with respect to the extent of reaction? I have always related the above equation to stability. So it must turn out that this stability factor is hidden in $\Delta_\mathrm r G^\circ$. I am completely lost. Does there exist a free book for chemical thermodynamics?

@adosar I don't have much left to say. What I wanted to say has already been said in this post and the one in my last comment. Yes it is a derivative with respect to $\xi$. When you want to get a derivative, you must also specify a point at which you want to evaluate it. The $^\circ$ means that you evaluate this at the point where each species is in their standard state. If this point is towards the left of the minimum (see graph above), then the slope $\Delta_\mathrm r G^\circ < 0$, $K > 0$, products are favoured, which makes sense because the minimum (equilibrium) lies to the right.

@adosar Of course it is about stability; equilibrium is all about stability, but simply saying that doesn't give us any insight as to the problem. It is a big word that has no meaning on its own. The crux of the matter is that there are two points on the graph that are of interest. (1) the standard-state where activities are equal to unity. (2) the equilibrium, which is the minimum, i.e. lowest $G$. If 1 lies to the left of 2, then $K > 0$ as discussed previously. If 1 lies to the right of 2, then $K < 0$. In both cases, 2 is more stable than 1!

@orthocresol Appreciated the answer. Thanks.

@orthocresol But how we can calculate ΔG for a sytem if it is not in equilibrium? All the thermodynamic variables must be measured at equilibrium. How can we measure arbitrarily two G energies at non-equlibrium states?

@CLAPorSLAPII "The standard Gibbs energy (change) ΔrG∘ for a chemical reaction is the change in G for converting stoichiometric numbers of moles of the separated pure reactants, each in its standard state at T, into the separated pure products in their standard states at T." How we can convert all reactants into products if the reaction will reach equilibrium?

@orthocresol What an excellent answer. As a new thermodynamics student, I'm having trouble due to lousy definitions. I really appreciate that you go into minute detail and explain things from the ground up. Thanks!

@orthocresol Just one doubt-in the expression for $ \Delta_r G$-can I take the chemical potential=$G_m$ of the reactant/product at that particular time if the reaction is at constant temperature and/or constant pressure?

@Boson I'm not sure. I'd probably be inclined to say no, because $\mu$ is like $dy/dx$ and $G_m$ is like $y/x$ -- if $y = f(x)$ then the value of $f(x)/x$ is not necessarily equal to the value of $f'(x)$. But I don't know, I'm not really very good at thermodynamics.

@orthocresol In an earlier comment to this you said "Yes, that is correct. Because $∘=m,i$, you can replace the chemical potentials with molar Gibbs energies in the expression for $Δr∘$ Could this be extended to non-standard state conditions but with constant T and P so it can be used for $Δr$?

@orthocresol Btw, I think you are actually very good at thermodynamics, at least for undergrads. I've read several of your other answers on thermodynamics and they always strike me as both elaborate and understandable, which is really helpful in thermodynamics which is quite confusing to me due to lousy definitions. On introspection, my question is actually for a rather niche case and not really important. Thanks again!

I think the fact that I contradicted myself suggests that I don't really understand things ;) Well, yes, chemical potentials aren't associated with standard states, they are defined for any system so that shouldn't be an issue.

@orthocresol 1 more question-if the reaction is at constant T and P, and we take $=$, then in the expression $Δ_r=∑_G{m,}$ would $Δ_r$ be constant throughout the reaction because $$ is obviously constant and $G{m,}$ is also constant since T and P is constant(all above for ideal gases)?

@orthocresol Also, is $ \Delta_r G$ analogous to $ \Delta_r H$? If yes, then why is there no mention of extent of reaction here ?

@Boson $G$ is constant at constant $T$ and $p$ if there is no reaction occurring. You are probably thinking of the equation $\mathrm{d}G = -S\mathrm{d}T + V\mathrm{d}p$ but this is only true for a system which is not undergoing chemical change. In general one has $\mathrm{d}G = \mu \mathrm{d}n - S\mathrm{d}T + V\mathrm{d}p$ (in which case, you can see that the no-reaction case simply has constant $n$, so reduces to the original). I'm not entirely sure about the link to enthalpy. Sorry, but I'm going to stop here; if you have further questions I gently urge you to post a new one.

@orthocresol Could we continue in chat?

Let's see how this plays out for the case of a batch reactor in which a reversible reaction of an ideal gas is occurring at constant temperature and total pressure:  aA + bB --> cC + dD.  Let the initial concentrations of the four species in the reactor be $n_A,\ n_B,\ n_C,\ and\ n_D$. At time t, the concentrations of these 4 species are:
$$n_A=n_{A0}-a\xi $$
$$n_A=n_{B0}-b\xi $$
$$n_A=n_{C0}+c\xi $$
$$n_A=n_{A0}+d\xi $$where $\xi$ is the "extent of reaction" at time t. If we are willing to make the assumption that the free energy of the system can be precisely determined at any time during the process (even though the process is irreversible/spontaneous), we can express the free energy at time t during the reaction by:$$G=n_A\mu_A+n_B\mu_B+n_C\mu_C+n_D\mu_D$$where $\mu$'s are the chemical potentials of the four species in the reactor mixture.  For an ideal gas mixture, the chemical potential of the 4 species in the mixture are given by:
$$\mu_A=\mu_A^0(T)+RT\ln{p_A}=\mu_A^0(T)+RT\ln{(Px_A)}$$
$$\mu_B=\mu_B^0(T)+RT\ln{p_B}=\mu_B^0(T)+RT\ln{(Px_B)}$$
$$\mu_C=\mu_C^0(T)+RT\ln{p_C}=\mu_C^0(T)+RT\ln{(Px_C)}$$
$$\mu_D=\mu_D^0(T)+RT\ln{p_D}=\mu_D^0(T)+RT\ln{(Px_D)}$$where the  $\mu^0$'s are the molar free energies of the pure components at temperature T and a pressure of 1 Bar, the p's are the partial pressures of the various species in the reaction mixture at time t, P is the total pressure in the reactor (constant), and the x's are the mole fractions at time t.  For example:
$$x_A=\frac{n_A}{n_A+n_B+n_C+n_D}$$
If we focus on the change in free energy G between time t and time t + dt,
we have:$$dG=(dn_A\mu_A+dn_B\mu_B+dn_C\mu_C+dn_D\mu_D)+(n_Ad\mu_A+n_Bd\mu_B+n_Cd\mu_C+n_Dd\mu_D)$$
If we make use of the various relationships presented above to evaluate the two terms in parenthesis in the above equation, we obtain the following results:$$(n_Ad\mu_A+n_Bd\mu_B+n_Cd\mu_C+n_Dd\mu_D)=0$$
$$(dn_A\mu_A+dn_B\mu_B+dn_C\mu_C+dn_D\mu_D)=(c\mu_C^0+d\mu_D^0-a\mu_A^0-b\mu_B^0+RT\ln{Q})d\xi$$where $$Q=\frac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b}$$
So, we are left with :$$dG=(c\mu_C^0+d\mu_D^0-a\mu_A^0-b\mu_B^0+RT\ln{}Q)d\xi$$
The first four terms on the right hand side are usually exprfessed as $$c\mu_C^0+d\mu_D^0-a\mu_A^0-b\mu_B^0=\Delta G^0$$where $\Delta G^0$ can be seen to represent physically the change in free energy in going between the following two thermodynamic equilibrium states;
State 1:  a moles of pure A and b moles of pure B, each in its own separate container at temperature T and pressure 1 bar
State 2:  c moles of pure C and d moles of pure D, each in its own separate container at temperature T and pressure 1 bar
$\Delta G^0$ is typically referred to as the "standard free energy change for the reaction."  So, in terms of $\Delta G^0$, we end up with:$$\frac{dG}{d\xi}=\Delta G^0+RT\ln{Q}$$
This equation shows that, on a graph of G versus $\xi$, the right hand side represents the slope. The slope becomes equal to zero, and equilibrium prevails, when the reaction has proceeded to the extent that Q = K, where K is the equilibrium constant.

In this context $\Delta G$ and $\Delta_\mathrm{r} G$ are different notations for two similar things, both are changes in Gibbs Free Energy. The "r" means, that you can only interpret it for a very specific equation.


text:
Inert Pair Effect
My notes say that inert pair effect is the tendency of electron in the outermost atomic s orbital to remain unionised or unshared in compounds of post-transition metals. (Eg. Sncl2 is formed preferentially over SnCl4)
I've watched a video lecture on it from youtube but they just explained that electrons in s orbitals are ionised before d orbitals and hence the change from example +2 to +4 in Pb.
But I don't understand why s block tends to remain unionised in these post-transition metals. Could someone explain this for me?
This is for a general chemistry first year module! So I don't think the explanation is very difficult.
Thanks so much!

Related: http://chemistry.stackexchange.com/questions/8717/what-is-inert-pair-effect/13873


text:
In a galvanic cell where the two electrodes are in the same electrolyte solution, why do reduction and oxidation occur separately?
In a certain book, I was presented with the following solution:
An iron nail is attached by a piece of wire to a magnesium ribbon, and the iron and magnesium are placed into the same container, with dilute sulfuric acid.
My understanding is that due to the difference in reduction potential of the iron and magnesium, the magnesium ribbon will oxidise in preference to the iron, so the Mg metal will oxidise and lose its electrons to form Mg2+ ions, and the electrons will travel to the iron electrode and react with the H+ ions in the electrolyte solution, and H2 gas will be evolved at the iron electrode.
My question, is why must oxidation and reduction occur at the two different electrodes? If the magnesium isn't connected to the iron, it will just react with the H+ ions there, and the same overall redox reaction will occur. Why is it, that if there is a wire attached from the magnesium to the other electrode, the electrons from the oxidation of magnesium would prefer to follow the path of the wire, rather than just react with the H+ ions already present at the magnesium strip?

Magnesium blocks are connected to iron structures to serve as cathodic protectors. A key element is that the magnesium should react slowly enough (i.e., selectively) to last for a long time (years), while providing enough electron flow to keep the iron negative in the midst of all its electron acceptors (oxidizers). If the iron were in a severely corrosive situation and you needed more current, and you think of using sodium as a sacrificial anode to provide a faster reaction, it wouldn't be practical for many reasons.
The sodium situation is where the electrons wouldn't bother traveling to the iron cathode; things are happening too fast because of the energetics of sodium reacting with water (forget the acid!). This situation does not permit the selectivity to form separate anodes and cathodes.
Dry cells are constructed of zinc anodes and carbon cathodes; impurities in the zinc surface can short-circuit the cell: the zinc becomes both anode and cathode, and is consumed inefficiently. This condition was ameliorated by amalgamating the zinc surface with a tiny bit of mercury. Mercury has a high hydrogen overpotential (i.e., it takes a higher potential to force enough protons to hydrogen atoms and then to molecules and then to bubbles which rise up.) The mercury prevented the initiation of local cathodes, and did not prevent the migration of zinc to its surface and thru it. So the selectivity was improved by the mercury amalgamation, allowing the zinc to be a purer anode. (The next step was to improve it further so as to eliminate the mercury!)
A single iron nail can have electrochemical corrosion all over its surface with anodes and cathodes that switch around and/or are so small that you can't distinguish them. You see only a rusty nail. One way to separate the anodes and cathodes and to make them visible is to put the nail in an agar gel with a little NaCl, phenolphthalein and potassium ferrocyanide. The nail begins to corrode at a random spot. This first spot will lose Fe++ ions to the gel, where they will react with ferrocyanide ion to turn blue. Electrons from the reaction will flood the iron nail; somewhere else, O2 will be reduced to hydroxide ions, which turn the phenolphthalein red. Typically, the head and point of the nail will become anodes (because they are more highly stressed, therefore a little more anodic) while the shaft will be cathodic. Pieces of rebar were more random in their reactivities.
So, oxidation and reduction can occur at the same electrode; it's just more illuminating (and useful) to see them separated instead of all mixed up.


text:
Why do acids donate protons?
I would have thought the answer to this question would be simple, but after searching this site I was unable to find any answers.
I know Brønsted-Lowry acids are defined in terms of proton donation, but what is it that makes the acids donate protons?
I guess the same would apply to Lewis acids, but in terms of electron pair acceptance.

Have you tried wikipedia? https://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction

Technically, acids donate protons the same way I donate my wallet to a mugger. The protons are relatively easy pickings for a base to commit its robbery.

Using the following generic reaction, where an acid donates $\ce{H+}$, i.e. a proton:
$$\ce{HX + B <=> X- + HB+}$$
If a lone pair of electrons on $\ce{B}$ has a higher energy than than than the resultant lone pair of electrons on $\ce{X-}$, then the $\ce{H+}$ on $\ce{HX}$ will be transferred to $\ce{B}$ to form $\ce{HB+}$ in order to stabilize the high energy lone pair on $\ce{B}$ in the form of a bond.

So, there are two things that influence whether an acid will transfer a proton:

The energy of the lone pair on the base
The energy of the lone pair on the resultant the conjugate base

If the conjugate base of the acid is more able to stablize a lone pair than the basic starting material is, an acid-base reaction will occur, and a majority of the basic reactant will end up being protonated.

Perhaps, given the question, it is worth pointing out that 'donating a proton' is the same as 'donating a hydrogen ion'.

Fixed. Good point.


text:
Does the dispertion medium affects Quantum Dots fluorescence efficiency?
I would like to use CdTe Quantum Dots to make a fluorescent coating on a transparent support. There's a link to the material here: http://www.plasmachem.com/shop/en/62--cdte-quantum-dots-powder-hydrophilic-water-soluble
I observed that the fluorescence of these QDs is much higher when they are dispersed in water than when they are dispersed in a dry layer of acrylic.
Wondering why this happens, and being a very bad chemist/physicist myself, I was wondering if this loss of fluorescence was due to some physico-chemical interaction between the media (acrylic) and the QDs.
Does the interaction between the media and the surface of QDs affects in some way the luminescence mechanism of QDs? Or is it more a problem of structure, e.g. the QDs are "clustered" in the thin layer of dry acrylic while they are distant from each other in water?
Another simpler way of saying all that would be: what is the best media to make a fluorescent QDs coating? (I was thinking of acrylic or epoxy resin...)
(I also posted this question on the physics StackExchange)


text:
Hund's rules for Li, contradiction?
Hund's rules basically state that we sould maxiumise $S$ and for a given value of $S$ maximise $L$. So let us take the case of Li. Li has 3 electrons, clearly the first 2 are in the $n=1$ state, with $S=0$ and $L=0$. For our remaining electron it must go into one of the $n=2$ states. No matter which state we put it in $S=1$, so we now use Hund's rule to maximise the value of $L$, which would indicate that we put it into the $p$ state giving $L=1$ rather then $s$ state giving $L=0$. But it does go into the $s$ state, which appears to violate Hund's rule. Why does it do this?
(p.s. my notation may be slightly out)

The $2\mathrm{s}$ orbital is lower in energy than the $2\mathrm{p}$ orbital. Their energy difference is larger than the energy you might gain by maximising $L$. Thus $2\mathrm{s}$ gets populated.

Now is a good time to learn that many chemistry 'rules' are, at best, general advice with the caveat that you have to pay attention to all the 'outliers'...

Hund's rules do not apply to different configurations that arise from the occupancy of orbitals with different energies.
For example, with a $\mathrm{2p^2}$ configuration (e.g. carbon), the allowed term symbols are $^1D$, $^3P$, and $^1S$. From Hund's first rule, you can predict that the most stable one out of these three is the $^3P$ term.
You cannot compare these three terms with a term arising from a different electronic configuration, e.g. a $\mathrm{2p3p}$ configuration, which could give rise to a $^3D$ term which would be "more stable" on the basis of Hund's second rule.
The aufbau principle is much more important here. A simplified diagram would look something like this (don't judge me, it's not to scale and I have no time to calculate the exact splitting):

Term symbols are associated with specific electronic configurations and while we often write term symbols on their own, we should always remember that a term symbol without an electronic configuration makes no sense.
Just to show you why it doesn't make sense: with your lithium example, if you really wanted to, then you might as well promote the 2s electron to a 3d orbital, since that would give you a $^2D$ state. Higher $L$, wonderful! Why not go further and promote it to a 4f orbital, or a 5g orbital? We might as well ionise it...


text:
Why is the CASSCF method multi-configurational, while the CI method is not?
The CASSCF method is perhaps the most commonly used theoretical method for studying difficult chemical systems exhibiting multi-reference character or non-dynamical/static/strong correlation. CASSCF is quite similar to CI, in the sense that a full CI wavefunction is generated within the active space used in the CASSCF method.
However, the CI method is a so-called ground state method, while the CASSCF method is multi-configurational. My question is simply "Why?". I feel it may be related to that during the CASSCF optimization, both the orbitals and CSF expansion coefficients are optimized, while in CI only the expansion coefficients are optimized. But I fail to understand how optimizing the orbitals gives the CASSCF method its multi-configurational nature.

I think it is just a convention.. If we put HF and MCSCF into equal footing. They both serve a common starting point to include dynamical correlation. HF->CI; MCSCF->MRCI. We name HF: single configurational, CI: single reference method; MCSCF: multi configurational; MRCI: multi reference method.

When you say multiconfigurational I think you mean that the ground state wavefunction does not have one single dominant MO configuration.
CI can be multiconfigurational depending on the type of excitation included.  For example, Full CI is of course multiconfigurational since it is the exact solution for the basis set used.  However, CIS (single excitations) or CISD (single + double excitations) calculations are the only two CI methods that are computationally efficient enough to be used in practice.  So that’s probably what you mean when you say CI. You also probably mean CIS or CISD based on an RHF wavefunction, to distinguish it from multi-reference CI.
The lowest energy solution of CIS based on RHF is the RHF energy, so you can think of the SCF procedure as a form of CIS.
So a CASSCF(2,2) calculation (e.g. where the active space is HOMO, LUMO) is a bit like a CISDT calculation.  For example, there will be contributions to the wavefunction where the LUMO is doubly occupied and another MO outside the active space is “singly excited”.  Thus, the LUMO is allowed to change when electrons are put into it, which becomes more and more important as the HOMO-LUMO gap decreases.  CISD cannot account for this, so in that sense it is not multiconfigurational.

CASSCF(2,2) can never have a triple excitation, because it will never have a change in orbitals that are not in the active space. It would not make sense to define an active space and then construct configurations outside of it.

The orbitals outside the active space are also SCF optimised. They are optimised by mixing in virtual orbitals, which has the same net effect as a single excitation.

I did, it does not make any sense to me. In a CASSCF(2,2) you have four configurations, which only differ in the occupation within the active space. There will only be doubly occupied and unoccupied orbitals outside the active space. Or I simply too stupid to understand the words you use.

@JanJensen Perhaps I should not have used multiconfigurational, but instead multi-reference. Or, like this: truncated CI is said to be a "ground state" method, right? While CASSCF is not.

@AndersMB yes, multi-reference is more accurate.  I knew what you meant though.  Bottom line: whether CI can handle multi-reference problems depends on the number of dominant MO configurations in the ground state and the number of type of excitations in the CI.  A minimum of CISDT is needed and that's already too expensive for most molecules


text:
Crusty white armpit stain that can't be removed by washing

Some of my t-shirts (the ones that are 100% cotton) started developing this strange armpit stain:

it makes the fabric feel brittle and crusty around it;
it can't be washed off with water and detergent;
it can't be scrapped off easily.

Now here's some information about the environment my clothes live in:

our water is moderately hard (60-120 ppm);
I use a deodorant that has the following ingredients:

I sweat a lot and our climate is quite dry.

I was thinking I'm dealing with a boring calcium carbonate deposit, but I don't really want to start drowning all my stained shirts in vinegar without first knowing what I'm dealing with. What do you guys think it is and how would you get rid of it?

I don't suspect it would be calcium carbonate in your armpits if it isn't everywhere else. Salt would come out, but that would have been my first guess. Maybe it's aluminum salts (or something else) from the antiperspirant. I've always noticed it to be pretty hydrophobic. Does it make your fingers feel strange if they're wet? I don't know how to describe the feeling. Kind of dry but covered in sort of a pasty substance.

@sendersreagent if feels pretty much the same, maybe less crusty. Do you know anything that dissolves aluminium salts so we can test this out?


text:
How can I tell if a certain carbon/hydrogen molecule can actually exist?
My 11 year old daughter asked me if the following compound exists in real life, and if so, what is it called. Not only do I not know, I have no idea how to find the answer to such questions. Any ideas?

It doesn't exist, although it looks very cute. You might want to have a look at the resources provided by the Royal Society of Chemsitry and the American Chemical Society for different age groups, including curious and creative girls at the age of 11.

Oh, and by the way: Welcome to Chemistry.SE!

ChemDraw tells me that "A name could not be generated for this structure". :( I am a little bit disappointed

Your compound is called 1,2,3-butatriene and belongs to the group of cumulenes :)

@KlausWarzecha, I'm not sure why you say this doesn't exist. I was even able to find this paper describing the $\ce{^{13}C}$ NMR signals of cumulene (butatriene), though it was at very low temperatures. It's certainly true that it doesn't exist very long at ambient temperatures, but I saw plenty of retro-Diels-Alder reactions leading to its formation.

@SendersReagent Please have a look at the edit history of the question. Initially, H and C were interchanged ;-)

@KlausWarzecha Ahhh, that makes a drastic difference.

Have a look at the Wikipedia page for butatriene or cumulene and https://pubchem.ncbi.nlm.nih.gov/compound/1_2_3-Butatriene#section=Top for other information about this molecule.

When drawing organic structures Carbon should have 4 "lines" coming from/to it while Hydrogen can have only 1 covalent bond. So if you could draw your molecule following these rules, it probably can exist. Here is an example:
H       H
 \     /
  C=C=C
 /     \
H       H

Alternatively there could be this structure with the same number of "lines" coming from/to Carbon:
  H   H
   \ /
    C     H
   / \   /
  H   C=C
     /   \
    H     H

As for the naming.. It's more complicated since it would depend on whether there is only one = bond or more than that. You could watch CrashCourse Organic Chemistry intro that talks about some naming conventions. But if you're stumbled, you could count the number of Carbons and Hydrogens and google for it: C3H4 or C3H6.

You've missed a carbon in the first structure (assuming we're just swapping C/H around). Also I'm not clear how you get the second one - are you saying it has the same valence sum or something?

Of course, it's not particularly relevant, but I would have thought that cyclobutane was 'closer' to the original structure in terms of overall bonds (although that is C4H8, or course).

Answer is more or less obsoleted since the edit to the question.

Hm.. I don't see how it becomes obsolete. This answer wasn't based on the picture drawn by Rolandus, so even after the edit it's up-to-date.

There are sites you can use like scifinder, chemistry journals, and even just going to a chemical supplier's website, most of them have places where you can draw a chemical and they will link you to it.  If they have it then it exist obviously, but if they do not sell it, it may be too dangerous or unstable to ship, thus you would have to go to the journals.
This chemical does exist, 1,2,3-butatriene.  I'm not saying it is common, but it can be made.

As a chemist, when asking myself if something exists, I open SciFinder, draw the structure, select exact search and single component and hit find. The caveat is that using SciFinder costs quite a large amount of money. (Yes, a really large amount of money, although I don’t know the actual figure.) If it has been published or proposed, SciFinder will tell you the paper.
Another test is checking out whether Sigma Aldrich sell it. Their catalogue is in no way complete because it only contains shipable items, but it is one of the most comprehensive to the best of my knowledge. You probably don’t want to check the prices, though. The caveat here is that you need a name first.
I am unaware of any website or freeware program that lets you draw almost anything and search for it or get a name. The problem is that knowing whether something exists (i.e. was published) required indexing all the journals in which publications happen, and all the way back to the mid 19th century. You cannot simply machineread everything; it will be in different languages sometimes with antiquated spelling, compound names will be all over the place, structures proposed and revoked. For every structure, you would need to render it in a machine-parseable format. The Chemical Abstracts Service (CAS) of the American Chemical Society (ACS) was founded in 1907 and has since systematically scanned journals to build up a repository of structures including a unique CAS number for every compound known (or proposed). This made it possible for them to digitalise that information and launch the SciFinder. That is 109 years of work steadily growing. It cost so much money that I just don’t see a freeware solution approaching.

You can use REAXYS https://www.reaxys.com/reaxys/secured/search.do similar to SciFinder to know actually this compound exist with out getting registered as paid member. I often use this site before conducting any reaction to get some idea about the reaction condition extra..
I hope this would be helpful...

There doesn't appear to be a free offering there anymore.


text:
Why is oxygen paramagnetic?
Paramagnetic molecules are molecules that have single electrons. When I draw the lewis structure of $\ce{O2}$, it appears to be a diamagnetic structure. What makes it paramagnetic?

Perhaps you could show us the structure you are drawing? I don't think that Lewis structures are going to help you here because they hold no information about the spin of electrons.

You are right about the single electrons. But note the plural. What if you have two unpaired electrons and your molecule is a biradical? ;-)

Is there even an answer on chem.SE explaining why O2 is paramagnetic? Everything I can find uses it as an example, but there's nothing explaining it... Edit: http://chemistry.stackexchange.com/a/39218/16683 The question isn't even about O2, sigh

You might look at these links:-http://www.physlink.com/Education/AskExperts/ae493.cfm .........     http://chemistrynotmystery.blogspot.in/2014/09/why-is-o2-paramagnetic-while-n2.html

To understand the paramagnetic nature of $\ce{O2}$, we must first understand how atomic orbitals mix together to form molecular orbitals. In the diatomic molecules of the elements in the second period, a phenomenon known as $\mathrm{s}$-$\mathrm{p}$ mixing  results in an increase in the energy of the $\sigma_\mathrm{2p_z}$ molecular orbital, and a decrease in the energy of the $\sigma_\mathrm{2s}$ orbital. This is also observed in the $\sigma^{*}_\mathrm{2p_z}$ and $\sigma^{*}_\mathrm{2s}$ orbitals.
The degree of $\mathrm{s}$-$\mathrm{p}$ mixing is determined by the energy gap between the $\mathrm{s}$ and $\mathrm{p}$ orbitals. The higher the energy gap, the less the orbitals mix. Because the $\mathrm{s}$-$\mathrm{p}$ gap increases across a period, $\mathrm{s}$-$\mathrm{p}$ mixing in $\ce{O2}$ and $\ce{F2}$ is actually so low that the $\sigma^{*}_\mathrm{2p_z}$ orbital is lower in energy than the $\pi_\mathrm{2p_{x,y}}$ orbital, unlike the preceding elements.
This can be seen here:

[source]
By constructing the molecular orbital diagram for $\ce{O2}$ and filling each  orbital according to Hund's rule, it becomes evident that $\ce{O2}$ is a diradical, with two unpaired electrons of the same spin. This is what gives oxygen its paramagnetism.

This answer does not address the question's statement that the Lewis structure has no unpaired electrons. It would be nice if it clarified if and why Lewis is wrong in this case.


text:
Are there any suitable catalysts for ferric oxide and aluminum reaction?
As an exercise, I am trying to figure out how to catalyze the reaction of ferric oxide and aluminum
Fe2O3 + 2Al -> 2Fe + Al2O3
I've been trying to find a substance A such that the reaction could be broken down into the steps:
Fe2O3 + A -> B + 2Fe
B + 2Al -> Al2O3 + A
And ΔGrxn for the first step is much less than that for the original reaction. I've had quite a bit of trouble finding online resources to assist me in my search. In general, are there any better methods for determining a suitable catalyst for such redox reactions? (education level high school junior, AP chem student).

They don't need a catalyst. Once ignited, they react all by themselves, and quite fiercely.

@IvanNeretin It still doesn't proceed very quickly though. Experimentally, a lot of the heat energy would be absorbed in subsequent reactions, not to mention by the iron itself. Is there any other way to lower the reaction's activation energy?


text:
Why will CuSO4 solution be acidic?
An end-chapter problem posed in an 11-grade texbook asks to analyse whether a water solution of CuSO4 will be acidic.
The standard answer is yes, because "$\ce{H2SO4}$ is a strong acid, hence $\ce{SO_4^2-}$ is a weak base. As a weak base, it will not be too hot about grabbing $\ce{H^+}$ from the environment. On its part, $\ce{Cu^2+}$ will strongly attract $\ce{OH^-}$ groups, and overall protons will prevail over hydroxides."
But I was told in the chat that the real answer is more interesting and complicated, so I post this question in the hope that someone will explain the process in-depth.

There is nothing complicated about it. A salt formed by a strong acid and a weak base will make the solution acidic due to hydrolysis.

@IvanNeretin $\ce{HSO4-}$ isn't a strong acid; $\mathrm pK_\mathrm a \approx 2$.

Still, that's strong enough to make the overall result acidic.

@IvanNeretin Do you know the magnitude of the $\ce{Cu^{2+}}\mathrm{-}\ce{OH-}$ stability constants offhand? In general I agree, it's likely to be acidic, but I wonder if it's a closer thing than it first seems.

The question does indeed boil down to the $\mathrm{p}K_\mathrm{a}$ value of $\ce{[Cu(H2O)6]^2+}$, which I don’t know off the top of my head but should be greater than seven. The remainder is basic maths.

@Jan Wouldn't $\ce{Cu(OH)2}$ precipitation and higher hydrolysis forms potentially also need to be considered? (They might prove negligible, yes, but...)

@Brian I’m pretty sure that they can be neglected at first approximation (and possibly even at second) because of the significantly higher $\mathrm{p}K_\mathrm{a}$ value of $\ce{[Cu(OH)(H2O)5]+}$. But as I learnt from aluminium complexes there might actually be some fascinating chemistry in there.

@Jan Exactly why I'm pressing the issue. It could be just as boring as you and Ivan have laid out. But, it might not be. I have some literature on which to fiddle -- I will probably be posting an answer either way sometime soon.

@Jan You're right, it's pretty boring.  The various hydroxyl complexes  only come into play if you add base.  DavePhD's answer is pretty much all there is to it.

Because the second pKa of sulfuric acid is weak, sulfate withdraws protons from water.  Therefore, the contribution of sulfate alone is to make the solution slightly more basic than pH 7.
However, as quantified in CE4501 Environmental Engineering Chemical Processes, Cu will form species such as CuOH+, which is the reason the solution will be acidic overall.  This could be approximated by thinking of $\ce{Cu(H2O)6^{2+}}$ as having a pKa of ~6.3.
Because  $\ce{HSO4-}$ is a stronger acid than CuOH+ is a base, the solution is acidic.


text:
AAS background correction with Zeeman effect
I've found here the explanation to how Zeeman effect can be used for background correction. I understand everything up to this part:
"When light is polarized parallel to the field, both atomic and background absorption occurs, and when light is polarized perpendicular to the field, only background absorption occurs."
I understand that light can be polarized, and that we are trying to somehow exclude either atomic or background absorption from one of our measures. I don't see, however, why this would be valid. Why is the perpendicularly polarized light not experiencing also the atomic absorption?

The very short answer is that properly applied magnetic fields, via the Zeeman effect, can cause atomic absorption bands to split into multiple bands, and these bands have different polarization properties. This is the simulation model I use to demonstrate the spliting and polarization behavior:

This model demonstrates the transverse and longitudinal Zeeman effects.  The theory of Zeeman splitting is covered elsewhere, but the basic idea is this: in a magnetic field, atomic transitions may be “split”, because the levels associated with the atomic transitions are no longer degenerate (due to those pesky magnetic quantum numbers and the associated selection rules for allowed changes in them).  Hence, the transition splits into several transitions.  The simplest case, that of ordinary Zeeman splitting, involves a splitting into three bands: 1 $\pi$ band and 2 $\sigma$ bands.  In the transverse field case, where the magnetic field is transverse to the light propagation direction, the $\pi$ band is linearly polarized parallel to the magnetic field direction.  It does not spectrally shift as a consequence of the application of the magnetic field.  The two $\sigma$ bands are spectrally shifted, by an amount proportional to the strength of the applied magnetic field, to each side of the $\pi$ band.  The $\sigma$ bands are linearly polarized orthogonally to both the light propagation direction and to the magnetic field direction.  The light propagation is the “z” direction.  In the transverse magnetic field orientation, the magnetic field axis is along the “x” axis.  Therefore, the $\pi$ component absorbs light that is plane polarized along the x axis.  The two $\sigma$ components absorb light that is plane polarized along the “y” axis. The next figure shows the unshifted $\pi$ band in the center and the 2 $\sigma$ bands shifted out of harms way:

So the light source is linearly polarized and the $\pi$ band cannot absorb it: it could only absorb the orthogonal polarization, which is not present. The 2 $\sigma$ bands would be able to absorb the light source's polarized light, except that the magnetic field has spectrally shifted them out of harm's way. Thus, when the magnetic field is on, only the broad background absorbance absorbs light. When the magnetic field is off, the background still absorbs light and the atoms do as well.
In the anomalous splitting situation, it is possible to have rather complex splittings, even neglecting hyperfine splitting complications, isotope effects, etc.  In the simplest anomalous splitting, there are two $\sigma$ bands and two $\pi$ bands.  The $\pi$ bands are spectrally shifted, but are still polarized parallel with the magnetic field.  Their spectral shift is less than that of the $\sigma$ components.  In the longitudinal Zeeman configuration, the magnetic field is along the light propagation axis, so there is no $\pi$ component to the absorbance because light cannot be polarized along its propagation direction.  The two $\sigma$ components are circularly polarized, with opposite handednesses, and are spectrally shifted by an amount proportional to the strength of the applied magnetic field. The next figure shows the splitting for calcium, in the longitudinal Zeeman configuration:

Both of these Zeeman configurations are used for background correction in atomic absorption spectroscopy.  The basic idea behind all background correction schemes, in atomic absorption spectroscopy, is this: perform two absorbance measurements and take their difference.  Both measurements should be designed to do an excellent job of measuring the interfering background absorbance, but only one of the measurements should do an excellent job of measuring the desired atomic absorbance.  The other measurement should do a poor job of measuring the desired atomic absorbance.  Differencing then discriminates against the interfering background absorbance, because it is common mode, and it yields the desired atomic absorbance, because it is differential mode.  Just how well the scheme works depends on a variety of factors, so, as it happens, there are several viable background correction schemes in use.
The Zeeman effects are useful for background correction because, by suitable use of polarizers and magnetic fields, the desired atomic absorbance can be effectively “turned off”.  This makes it possible to perform a measurement where the absorbance measured is almost exclusively due to just the interfering background absorbance.  As noted above, this measurement is one of the two needed for the differencing involved in background correction. Hope this answers your question.

Welcome to StackExchange Chemistry! Very cool to have you here.

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text:
Calculating the value of equilibrium constant in terms of pressure

A mixture of $\pu{1 g}$ of $\ce{H2(g)}$ and $\pu{1.06 g}$ $\ce{H2S(g)}$ in a $\pu{0.5 L}$ flask come to equilibrium at $\pu{1670 K}.$ At equilibrium, there are $\pu{8E-6 mol}$ of $\ce{S2(g)}$ present. Determine $K_p.$

I set up an ICE table:
$$
\begin{array}{ccc}
\ce{& &2 H2 &+ &S2 &<=> &2 H2S} \\
\text{Initial} & &\pu{1 M} && 0 && \pu{0.062 M} \\
\text{Change} & &2x && x && 2x \\
\text{Equilibrium / Final} & &?  && \pu{1.6E-5 M} && ?
\end{array}
$$
From the $\ce{S2}$ column we see that $x = \pu{1.6E-5 M}$, so I calculated the equilibrium constant as follows:
$$K_p = \frac{(0.062 + 2\times\pu{1.6E-5})^2}{(\pu{1.6E-5})(1-2\times\pu{1.6E-5})^2} = 240$$
What is wrong with my solution?

The problem lies in that our information of the system is in terms of concentration, so you actually calculated $K_c$.
$$\ce{2H2 +S2<=>2H2S}$$
$$\begin{array}{|l|l|l|}
\ce{H2 & S2 & H2S} \\ \hline
0.992\ \mathrm{M}& 0\ \mathrm{M}& 0.062\ \mathrm{M} \\
2(1.6\times 10^{-5}) & (1.6\times 10^{-5}) & -2(1.6\times 10^{-5}) \\
0.992+2(1.6\times 10^{-5})  &  1.6\times10^{-5}\ \mathrm{M}   &  0.062-2(1.6\times 10^{-5})\end{array}$$
$$K_c=\frac{(0.062\ \mathrm{M}-3.2\times 10^{-5}\ \mathrm{M})^2}{(1.6\times10^{-5}\ \mathrm{M})\cdot(0.992\ \mathrm{M}+3.2\times 10^{-5}\ \mathrm{M})^2}=245\ \mathrm{M^{-1}}$$
which is numerically the same as the value you obtained, except that I used a more precise molecular mass for $\ce{H2}$.
Using the relation $$K_p=K_c(RT)^{\Delta n},$$
where $$\Delta n=(\sum \text{coefficient of gaseous products})-(\sum\text{coefficient of gaseous reactants}),$$ we can calculate $K_p$:
$$K_p=(245\ \mathrm{M^{-1}})\cdot(8.3144598\ \mathrm{J\ K^{-1}\ mol^{-1}}\cdot 1670\ \mathrm{K})^{-1}=0.0176\ \mathrm{kPa^{-1}}$$


text:
Testing for nitrate ions with NaNO3 in Devarda's test
When we use Devarda's test on $\ce{NaNO3}$ solution (we add $\ce{NaOH}$ and $\ce{Al}$ powder, heat the mixture, and produce ammonia), what is the equation of the process?
I understand that $\ce{Al}$ reduces $\ce{NO3-}$ to $\ce{NH4+}$ then $\ce{NH4+}$ and $\ce{OH-}$ have a reversible reaction to produce $\ce{NH3}$, and water, but I'm not sure how to summarize this in an equation. In particular, what happens to the aluminium?

Aluminium oxidizes, as you may guess. To find its final form, keep in mind that $\ce{Al}$ is amphoteric.

Quoting from Svelha G., Vogel's Qualitative Inorganic Analysis,

A solution of nitrate ion is boiled with zinc dust/aluminium powder
  and sodium hydroxide solution to get positive result. Excellent result
  is obtained by the use of Devarda's alloy (45% Al, 5% Zn, 50% Cu) but
  the ammonium ions must be removed by boiling the solution with sodium
  hydroxide solution or almost evaporating it to dryness before addition
  to metal.
The overall reaction is:
$$\ce{3NO3- + 8Al + 5OH- + 18H2O ->3NH3 ^ + 8[Al(OH)4]-}$$


text:
Is the reasoning of spin contamination correct?
The common explanation of spin contamination is like the following:
The non-relativistic electronic Hamiltonian commutes with $S^2$ and $S_z$. Hence, the exact eigenfunction should be an eigenfunction for $S^2$ and $S_z$ operators.
However, the non-relativistic Hamiltonian has no spin variables. Any wavefunction contains spin, the spin part will commute with the Hamiltonian anyway.
For instance, in addition of angular momentum,
http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect15.pdf
One can form $\{J_1^2, J_2^2, J_{1z}, J_{2z}\}$ set of variables. They commute with Hamiltonian, but no longer eigenfunction of $S^2$, i.e. $\{J^2, J_z, J_{1}^2, J_{2}^2\}$.
If my argument is correct, what would be the actual justification for spin adaptation?

OK. Finally I have some time to continue the discussion. As it appears now, you question has very little to do with spin contamination, rather it is about the thing you called "spin adaptation". It is about why do we want a trial electronic wave function to be eigenfunction of total spin operators, not how do we achieve this goal in the unrestricted calculations. Am I right?

No. The concept spin-adaption and contamination are connected. If we want to have a spin-adapted wavefunction, it will not have contamination. But, why? I can build a spin wavefunction as uncoupled basis, they also commute with Hamiltonian. From your answer, it is just convention.

I don't understand where do you see a contradiction out there. Yes, you can require the electronic wf to be an eigenfunction of operators of the individual sources of spin, rather then of the total spin, i.e. you can use the uncoupled picture instead of the coupled one. But when working in the uncoupled picture you don't know the total spin of the system which is often desirable. So what is your problem with this?

The problem is, I do not think it is just a convention.

Commuting operators indeed admit a set of simultaneous eigenfunctions, and since the non-relativistic electronic Hamiltonian commutes with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, the exact non-relativistic electronic wave function (which is an eigenfunction of the electronic Hamiltonian) is also an eigenfunction of the total spin operators. And this requirement can be also imposed on an approximate non-relativistic electronic wave function.

This is certainly true that the non-relativistic electronic Hamiltonian commutes not only with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, but with operators of the individual sources of spin $\hat{S}_{iz}$ and $\hat{S}_{i}^{2}$ as well, or to be more precise, with their trivial extensions on the state space of a many particle system. Thus, this is indeed true that one could adopt the uncoupled picture as well, rather then the usual coupled one, but what is the usefulness of such picture? In the absolute majority of cases the most important characteristic of a system is a property which describes it as a whole, rather that its components individually. For this particular case, recall, for instance, that the total spin of a molecular system is a very important characteristic of it. For instance, it is heavily used in electronic spectroscopy: important terms like term symbols, spin-allowed and spin-forbidden transitions, fluorescence, phosphorescence, etc. are all directly formulated in terms of total spin angular momentum of electronic states.
Now if you decide to use the uncoupled picture in which the total spin is unspecified you basically loose all the body of knowledge of electronic spectroscopy. But what do you gain instead? What is the point of using the uncoupled picture? You can specify the spin of each electron, yes, but but so what?

Thanks. I have revised the second sentence. But you have not answered the core problem: I can write other spin operators than $S^2$, they will commute with the Hamiltonian. I can have other wavefunction which will be the eigenfunction of such spin operator, e.g. $S_1^2$ and $S_2^2$. But no longer eigenfunction of $S^2$. The aspect emphasising $S^2$ will let it has a more privilege position than other choice of spin operators, e.g. $S_1^2$ and $S_2^2$.

I don't quite understand what do you mean by "other spin operators"? You mean operators for individual sources of spin (spin of individual electrons)?

for instance, uncoupled basis in additional of angular momentum.

@Rodriguez, then you're wrong. The total angular momentum squared $\hat{J}^{2}$ commutes both with $\hat{J}{1}^{2}$ and $\hat{J}{2}^{2}$, thus, an eigenfunction of $\hat{J}^{2}$ is an eigenfunction of $\hat{J}{1}^{2}$ and $\hat{J}{2}^{2}$ as well.

The problem is, the coupled and uncoupled angular momentum are two sets of basis. $|j_1 j_2 m_1 m_2\rangle$ and $|j m j_1 j_2 \rangle$. Once you choose the uncoupled basis, it will be eigenfunction of corresponding spin operator $j_1^2, j_2^2, j_{1z}, j_{2z}$ and Hamiltonian, but not $j^2$.

@Rodriguez, in the uncoupled picture the total angular momentum is unspecified. And as I said, the total spin angular momentum is so important in chemistry, so that the coupled picture is used exclusively.

Why do I need to specify the total angular momentum? The question is not about the importance in quantum chemistry. If it is just a customary, there is unlikely to be a reason to emphasis spin adaptation.


text:
Why aren't soldering tips coated with platinum?
The tips in a soldering iron tend to oxidize quite quickly (therefore becoming unuseable) if left heated and unused for more than a few seconds.
Why aren't they coated with platinum or some other material that sticks to solder and doesn't dissolve in it? The cost of the coating should pay for itself fairly quickly.

I'd guess that if the molten solder would wet a metal then it would dissolve at least small amounts of that metal too. Platings would be thin and hence dissolve rather quickly.

It doesn't take much effort to keep the tip in, well, tip-top shape. And I'm not going to pay some multiple of the price of a new tip to get a platinum-coated one. In fact, I'm not sure that I have ever had to replace a solder iron tip, aside from wanting several different sizes/shapes for different work. Take better care of your soldering iron!

Sigh - as pointed out by @user83678 some 3+ years after first writing this, I answered for Pd, not Pt. That will teach me to pay attention. But, the general answer remains the same. There are enough interactions of platinum with either lead, tin, or silver (a common component in lead-free solders) such that a platinum tip does not make metallurgical sense.
Lets look first at Pt-Pb, with the CALPHAD parameters from Z.H. Long et al., J. Phase Equil. Diff 30 318-322 (2009):

While Pt does not have a high solubility in Pt, the presence of multiple intermetallics indicates problems: either Pb reacts with the Pt time, or Pt dissolves into the solder melt and forms an intermetallic there. Neither is desired.
For Pt-Sn, one gets (with CALPHAD parameters from V. Grolier and R. Schmid-Fetzer, J. Alloys and Compounds 450 264-271 (2008)):

Here the Pt happily sucks up lots of Sn, and then there are all those intermetallics.
For Pt-Ag (from I. Karakaya and W.T. Thompson, Bull. Alloys Phase Diagrams 8(4) 334-340 (1987)) one gets:

The fcc Ag and Pt have large solubilities for each other (but not complete solid solubility, somewhat surprisingly). Again, you get lots of interactions between a Pt tip and a lead-free solder containing Ag.

Aside from my comment that it is not hard to keep a soldering tip in tip-top shape, I will point out several materials issues.
For this purpose, I will stick to the interactions of Pd with classic Pb-Sn solder. A peek at the Pb-Pd phase diagram (ASM International, or G. Ghosh, Met. Trans. 30A, 5-18 (1999)) one readily sees that palladium has a large solubility of lead, more than 10% at room temperature. Since the solubility rises with temperature, cooling a Pb-saturated Pd tip from soldering temperature to room temperature would result in Pb precipitates forming, likely disrupting the mechanical stability of the Pd tip.
A glance at the Pd-Sn phase diagram (same paper referenced above) shows a fairly similar situation (not so surprising since Sn and Pb are in the same column, but that isn't a given in phase diagrams). Again, even at room temperature the solubility of Sn in Pd is in excess of 10%.
For both systems there is a binary phase, Pd$_{3}$Sn or Pd$_{3}$Pb, very close to the fcc palladium phase field, as well as a range of other phases stretching to higher Pb or Sn contents. The ternary diagrams included in the Ghosh paper show many phases present. This means that reactions between molten Pb-Sn solder and a platinum tip won't stop at just saturating the fcc Pd with Pb or Sn - the reaction will continue, making compounds across the phase diagram and rapidly ruining your fancy Pd tip.
For lead-free solders, note that Sn remains a primary component in many of them, so the same things apply.
I have included a (calculated from the CALPHAD parameters in Ghosh's paper) the Pb-Pd phase diagram.

He asked about platinum not palladium.

@user83678 - interesting that it took more than 3 years to notice that. Sigh. Fair enough, the Pt-Pb Pt-Sn and Pt-Ag phase diagrams can be added - the story is basically the same...

Excellent answer(+1) but do you want to keep the information regarding palladium? Do you think palladium is relevant to the question as the OP wants information regarding platinum? It is totally upto you if you want to keep the information but I am just asking just to make sure :D


text:
Hybridization of carbon in CO2
I know that if there exists a carbon atom with single bonds only (with other atoms),then it is $\mathrm{sp^3}$ hybridized. And if at least one of the bond is double bond then its $\mathrm{sp^2}$ hybridized (not talking about two double bond on same carbon atom). And if there is at least one triple bond then its $\mathrm{sp}$ hybridized. But then carbon in carbon dioxide should be $\mathrm{sp^2}$ hybridized. But if I am right our teacher told it was $\mathrm{sp}$ hybridized. How is that possible if the carbon atom bears no triple bond?

Carbon uses one of its non-hybridized p-atoms to bond to one oxygen, the other to bond to another oxygen. These bonds are pi-bonds, they lie in the form of a cross, at right angles to each other. Besides this, carbon is bonded to each of the oxygens via sigma-bonds, using its two sp-hybridized atoms.

Related: http://chemistry.stackexchange.com/q/10976/4945

Hybridization is determined by molecular geometry. Carbons participating in triple bonds like those in acetylene have two regions of electron density. Carbon dioxide is a linear molecule, and also has two regions of electron density. The fact that both have two regions of electron density are what make these carbons $\mathrm{sp}$ hybridized.
To reenforce hybridization is determined by molecular geometry, consider the benzyne intermediate. It has a triple bond, but is $\mathrm{sp^2}$ hybridized because it is too strained to become linear and only have two regions of electron density.

How is it that we can compare two regions electron density in acetylene with CO2?

The carbon atoms in both $\ce{CO2}$ and $\ce{HCCH}$ are participating in two $\sigma$ bonds (hence the two regions of electron density), and two $\pi$ bonds. The only difference between them is that in $\ce{CO2}$, the two $\pi$ bonds are not to the same atom.

It is because $\ce{CO2}$ is linear in nature and hence the $sp$ hybridization.
In ground state, the electron configuration of Carbon is $1s^2\ 2s^2 \ 2p^2$. You can consider one of the $2s$ electrons to be excited to fill the other empty $2p$ orbital to give a configuration of $1s^2\ 2s^1 \ 2p^3$.
Each $2p$ orbital, $2p_x\ 2p_y$ and $2p_z$ now holds one electron. The $2s$ and one of the $2p$ orbitals, say the $2p_y$ can hybridize and form 2 sp hybrid orbitals.
Oxygen has the ground state electron configuration $1s^2\ 2s^2 \ 2p^4$. Two of the $2p$ orbitals, say the $2p_x$ and the $2p_z$ only hold one electron. The $2p_x$ can now overlap with one of the sp hybrids from the carbon to form a σ bond. The $2p_z$ can now overlap with the unhybridized $2p_z$ on the carbon to form a $\pi$ bond.
The same thing can happen on the other side of the carbon forming another $\sigma$ bond with Oxygen’s $2p_z$ and a $\pi$ bond with the $2p_y$ orbitals from each atom.

A triple bond takes two $p$ orbitals out of the sigma-bonding network and into pi-bonds, leaving just one $p$ orbital to combine with the $s$ orbital in the sigma bonding.  Two double bonds do the same thing and thus also leave the atom, in this case carbon, $sp$ hybridized.