text:
Why is manganese(II) coloured although the transition should be spin-forbidden?
In every basic coordination chemistry class, at some early point the crystal field theory and LFSE will be taught, explaining that there will be an energy difference between d-orbitals (typically named $\mathrm{t_{2g}}$ and $\mathrm{e_g}$) and that electrons can be excited from one to the other to give colour. The next statement is typically that transitions have to be spin-allowed (the higher-energy orbitals need to be able to accomodate for the lower electron without spin-flipping being required) and Laporte-allowed (in a nutshell: no d→d transition in octahedral complexes).
The syllabus then goes on to say that the Laporte rule is not that strict a rule (typically, all near-octahedral complexes of 3d-elements which have spin-allowed transitions have a well visible colour; e.g. iron(II), chromium(III), manganese(III), nickel(II), copper(II) etc.) since small random vibrations can destroy the symmetry making the electron transition allowed and an incoming photon can excite the electron as needed. They also remark that the colour is much less intense than for Laporte-allowed transitions — hence why the colour of the $\mathrm{d^7}$ ion $\ce{[CoCl4]^2-}$ is much more intense than that of $\ce{[Co(H2O)6]^2+}$.
A very recent question on the colour of manganese(II) sulphate then had me confused. Manganese(II) salts are typically of a very pale pink. The difference to white can be observed when they are put before a white background. But manganese(II) is a $\mathrm{d^5}$ ion, so the transition is spin-forbidden.
I think that being spin-forbidden explains why manganese(II) complexes are so weakly coloured, but what is the mechanism that relieves the spin constraint and how does it work in practice?

I have some bits and pieces about this written up in this answer (especially in the second part of it) although in a different context.

Selection rules
The intensity of the transition from a state $\mathrm{i}$ to a state $\mathrm{f}$ is governed by the transition dipole moment $\mu_{\mathrm{fi}}$ (strictly, it is proportional to $|\mu_{\mathrm{fi}}|^2$):
$$\iint \Psi_\mathrm{f}^*\hat{\mu}\Psi_\mathrm{i}\,\mathrm{d}\tau \,\mathrm{d}\omega \tag{1}$$
where $\mathrm{d}\tau$ is the usual volume element (representing integration over spatial coordinates) and $\omega$ is an additional spin coordinate (since the states $\Psi$ also specify the spin). For an electronic transition, $\hat{\mu}$ is the electric dipole moment operator, and only depends on the spatial coordinates $x$, $y$, and $z$. (I've also ignored the vibrational component i.e. the Franck-Condon factor since it's not really relevant for the purposes of this answer.)
Normally in determining selection rules, we are more interested in whether the integral in $(1)$ is zero or not. If the integral is zero, the transition is said to be forbidden. The usual way to determine whether it is zero is by using symmetry: if the integrand is antisymmetric under any symmetry operation, then the integral is zero. (It is exactly analogous to the integral of an odd function $f(x)$ over a symmetric region; for example, $\int_{-a}^a \sin x \,\mathrm{d}x = 0$.)
If we assume that both the initial and final states can be separated into spin and spatial components, then we get:
$$\iint (\psi_\mathrm{f}^\mathrm{spin}\psi_\mathrm{f}^\mathrm{space})\hat{\mu}(\psi_\mathrm{i}^\mathrm{spin}\psi_\mathrm{i}^\mathrm{space})\,\mathrm{d}\tau\,\mathrm{d}\omega$$
(I also dropped the complex conjugates - they aren't important.) Since $\hat{\mu}$ does not operate on the spin coordinate, we can separate this multiple integral:
$$\left(\int \psi_\mathrm{f}^\mathrm{spin} \psi_\mathrm{i}^\mathrm{spin} \,\mathrm{d}\omega\right) \left(\int \psi_\mathrm{f}^\mathrm{space} \hat{\mu} \psi_\mathrm{i}^\mathrm{space} \,\mathrm{d}\tau\right)$$
The first integral is nonzero if $\Delta S = 0$. The quantum number $S$ is related to the eigenvalue of a hermitian operator (the actual eigenvalue is $S(S+1)\hbar^2$), and eigenfunctions of a hermitian operator corresponding to different eigenvalues are necessarily orthogonal, making the integral identically zero if $\Delta S \neq 0$. This is the spin selection rule.
The second integral is nonzero if $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ have a different character under the inversion operation, i.e. one is gerade and another is ungerade. This is because $\hat{\mu}$ itself is ungerade, so for the integrand to be gerade, we need $\psi_\mathrm{f}^\mathrm{space}\psi_\mathrm{i}^\mathrm{space}$ to be ungerade ($\mathrm{u \otimes u = g}$). In turn this means that $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ must have different symmetry. This is the Laporte selection rule. It can be relaxed when the geometry is not perfectly octahedral, or tetrahedral for that matter, since inversion is no longer a symmetry operation. However, I digress.

Spin-orbit coupling
The problems arise because the total states are not separable into spin and spatial components. This occurs because both electron spin, as well as orbital angular momentum, are both angular momenta. They therefore interact with each other, and this is the spin-orbit coupling spoken of. Mathematically, it is treated as a perturbation to the Hamiltonian; the result is that the transition dipole moment cannot be separated into two integrals and this leads to the invalidation of the selection rules.
When considering spin-orbit coupling, the term symbols such as $^3\!F$, $^1\!D$, which imply a simultaneous eigenstate of the total spin operator $\hat{S}$ and the total orbital angular momentum operator $\hat{L}$ are no longer meaningful since both angular momenta are coupled to each other (by $\vec{j} = \vec{l} + \vec{s}$) and behave as one "combined" source of angular momentum. The quantities $S$ and $L$ are called "nearly good" quantum numbers: they are no longer strictly conserved, but the "contamination" of different quantum numbers is relatively small. Mathematically we could express this as an admixture of different spin states into the original wavefunction:
$$^3\!F_J = c(^3\!F) + c_1(^1\!F) + \cdots$$
where $^3\!F_J$ denotes a state with spin-orbit coupling (not an eigenstate of $\hat{S}$ and $\hat{L}$), and $(^3\!F)$ denotes the original $^3\!F$ state before spin-orbit coupling is considered (which is a true eigenstate of $\hat{S}$ and $\hat{L}$). If $c_i \neq 0$, then one obtains a non-zero transition dipole moment between "states" of nominally different $S$.
The fact that the selection rules are still a thing is solely because spin-orbit coupling is a small effect ($c_i$ close to 0), so the compounds behave "almost" as if they obey the selection rules. More properly, the transition dipole moment between states of different $S$ is small, and therefore the d-d bands in $\mathrm{d^5}$ spectra are extremely weak. The magnitude of spin-orbit coupling, however, does vary with $Z^4$ ($Z$ being the atomic number). Therefore, while spin-orbit coupling is rather small in the 3d series, it becomes an extremely important mechanism by which the spin selection rule in the the 4d and 5d congeners, as well as the f-block elements, is relaxed.

That was a very enlightening response. I wasn't sure exactly where those selection rules actually came from. Can you give a reference or perhaps explain where the $Z^4$ dependence comes out?

I was expecting some final closing remark that would refer back to the question stated in OP. For me this is still not a finished answer.

The only question asked was what is the mechanism that allows systems to circumvent our "selection rules". I think orthocresol answered that question quite well.

@jheindel This is actually a very good question which would do well on main site! I looked up a couple of sources, and the crux of the matter is that the magnetic field $\mathbf{B}$ generated by the proton (in the rest frame of the electron), given by the Biot-Savart law, is proportional to the charge $Ze$. Spin-orbit coupling arises from the interaction of the electron's magnetic moment with this field, which is given by $-\mathbf{\mu}\cdot \mathbf{B}$. This incorporates the previous $Z$-dependence and can be shown to be proportional to $Z/r^3$. [...]

[...] In a hydrogenic atom, the expectation value of $1/r^3$ itself has a dependence on $Z^3$, which gives rise to the combined dependence on $Z^4$. The exact form of this spin-orbit coupling constant only strictly applies to hydrogenic atoms but the general trend is widely applicable to all the elements. For a full proof (!) see Townsend's A Modern Approach to Quantum Mechanics, 2nd ed., p 400 onwards. Atkins's Molecular Quantum Mechanics, 6th ed., pp 215-7 also discusses it, but not quite as well.

@orthocresol Thanks for the references. I'll check those out! That makes sense though.

To continue with the answer from @orthocresol. The breaking of formally ‘ forbidden’ transitions is very common indeed, possibly the best example is in the observation of phosphorescence from almost any aromatic or dye molecule.
Absorption in these molecules is generally a spin allowed transition, (single to singlet), however, triplet states, from which phosphorescence is observed,  can only be produced by a spin forbidden process, (singlet to triplet inter system crossing) and just as in the case of phosphorescence this is a spin forbidden process which becomes allowed due to spin-orbit coupling. In the case of the manganese ions and in other transition metals the forbidden transition is in absorption.
The treatment is the same in all cases and starts with allowing the two states to interact via some  operator. In the case of photons (absorption / emission) it is an electric dipole operator, i.e. dipole moment operator. Labelling the operator as $ H_0$, the rate of the transition from one state to another is
$$W_{if} \approx \int \Psi_f H_0 \Psi_i d\tau = \int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau $$
where $\sigma$ represents the spin wavefunction and $\psi$ the spatial one for initial (i) and final (f) states. As the operator does not depend on spin the integral can be separated into two parts as described in other answers
$$\int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau = \int \psi_f H_0 \psi _i d\tau\int\sigma _f \sigma_i d\tau = 0$$
and the final result is zero because the spin wavefunctions are orthogonal making the last integral is zero between states with different spin, i.e. $\Delta S \ne\ 0$. This is what we expect from states between which a spin changing transition is attempted.
To progress we next suppose that it is possible to mix a little of the final state wavefunction into the initial one and thus generate a new state with wavefunction $\psi_m$ then $$\psi_m=\psi_f + b\psi_i$$ where b is some mixing constant we can determine later on.
Now replacing the wavefunction for the final state ($\psi _f$) with this one ($\psi_m$) and repeating the calculation above and separating out terms as before, as orthogonal spin terms are zero, also as before, this leads to
$$W_{if} \approx \int \psi_m H_0 \psi_i d\tau = b\int \psi_i \sigma_i H_0 \psi_i \sigma _i d\tau = b\int \psi_i  H_0 \psi_i  d\tau\int\sigma _i\sigma_id\tau $$
in which case neither of the integrals is zero but evaluates to a number.  This equation shows that there is a finite rate for the process with operator $H_0$, for example the electric dipole operator.
The constant b remains to be understood and it is derived using perturbation theory and turns out to have the form
$$b=\frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i}$$
where $\Omega$ is the spin-orbit operator. Thus the rate for the process finally becomes
$$ w \approx \frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i} $$
and we can use symmetry to decide if this integral is zero or is finite.
In the case in point, the spin-orbit coupling the operator depends on the mass of the atoms involved and the product of orbital and spin angular momentum. This is why spin-orbit coupling is also called a ‘heavy atom effect’ when applied to emission processes.  In general, however,  any effects are enhanced in atoms with larger atomic numbers compared to those earlier in the periodic table, simply because $\Omega $ is larger in value. It is clear also that if the two states are close in energy the effect is also increased relative to atoms /molecules where the states are further separated in energy.
The second mechanism by which a 'forbidden' transition can become somewhat allowed is by 'Herzberg-Teller' coupling also called 'intensity stealing'. The standard example is in benzene where the $S_0 \rightarrow S_1$ transition is forbidden but is observed as a weak absorbtion due to a vibrational normal mode altering the symmetry of the electronic excited state so that it now has the correct symmetry to couple with a second and higher allowed electronic transition, and thus 'steal' intensity. The formalism is essentially the same as described. The terms that are important are thus (a) the presence of a normal mode of the correct symmetry (b) a second and allowed transition, preferably close by in energy and (c) an electronic coupling term between the allowed transition and the forbidden transition whose symmetry is altered by the vibrational normal mode.

One mechanism is spin-orbit coupling.  There is an interaction between the magnetic field created by the orbital motion of the electron with the spin of the electron.  This allows transitions where the spin of the electron  changes in tandem with its components of angular momentum, whereas just changing the spin by itself would be forbidden.  But the spin-orbit interaction is weak, so transitions involving it give weaker coloration than simple d-d orbital transitions.
Read more about spin-orbit coupling here.


text:
Calculation of equilibrium constant for dissolution of iron(II) fluoride in acid
I am reviewing for the AP Chemistry test, and the question is this
$$
\ce{FeF2(s)  <=> Fe^2+(aq) + 2F^- (aq) } \quad  K_1 = 2\times 10^{-6}
$$
$$
\ce{F- (aq) + H+ (aq) <=> HF(aq)} \quad K_2 = 1 \times 10^3
$$
$$
 \ce{FeF2 (s) + 2 H+ <=> Fe^2+ (aq) + 2 HF (aq)} \quad K_3 = ?
$$
I solved for $K_3$ by multiplying $K_1$ and $K_2$, and then dividing by 2. I got $K_3 = 1 \times 10^{-3}$.  I would think that this would be not thermodynamically favorable since $K_3 < 1$, but the answer was the opposite: $K_3$ is thermodynamically favorable, because $K_3 > 1$.  Why is this the answer?

I think (but I should have really gone to bed three hours ago) that you should have multiplied the value but squared $K_2$ (you need that equation twice) because thermodynamics. I’m not $100~%$ sure though. Oh yeah: Please don’t use MathJax in titles ;)

Ah, that might be the answer.  My reasoning for the division by two was using $K = \frac{[HF]}{[H][F]}$, and replacing each one with twice the amount.

Let’s approach the entire problem from the other direction. Consider your final reaction:
$$\ce{FeF2(s) + 2H+ (aq) <=> Fe^2+ (aq) + 2 HF (aq)}\\
K_\mathrm{tot} = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2}$$
We have our initial dissociation with the following equation:
$$\ce{FeF2(s) <=> Fe^2+ (aq) + 2 F- (aq)}\\
K_1 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]}$$
And of course we have the second equilibrium which equates to:
$$\ce{F- (aq) + H+ (aq) <=> HF (aq)}\\
K_2 = \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]}$$
Now we want to express $K_\mathrm{tot}$ in terms of $K_1$ and $K_2$. Since we are dealing with products all the way through, it will somehow be a multiplication. So let’s try $K_1 \times K_2$:
$$K_1 \times K_2 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]} \times \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]} = \frac{[\ce{Fe^2+}][\ce{F-}][\ce{HF}]}{[\ce{FeF2}][\ce{H+}]}$$
We’re not quite there yet. We still have an extraneous fluoride concentration in the numerator and we are missing an $\ce{HF}$ concentration and a proton concentration. Luckily, that is exactly what an additional $K_2$ will give us:
$$K_1 \times \left( K_2 \right )^2 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]} \times \left ( \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]} \right )^2 = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2} = K_\mathrm{tot}$$
This is the mathematical way of saying ‘we need the second equation twice’. Due to the way equilibrium constants work, you always have to multiply or divide, so ‘doing something twice’ is equivalent to squaring.
Numerically, this means the answer is:
$$K_\mathrm{tot} = K_1 \times \left ( K_2 \right )^2 = 2 \times 10^{-6} \times \left ( 1 \times 10^3 \right )^2 = 2 \times 10^{-6} \times 1 \times 10^6 = 2$$

Yes, $\ce{FeF2}$ is a solid, so it has an activity of $1$ so we could ignore it when calculating the equilibrium constant. But we don’t actually use its value, so its being there doesn’t hurt either.

Yes, Jan is right. You need to multiply $K_2$ squared because it takes two additions of reaction 2 to get from reaction 1 to reaction 3. You will get $K_3$ = 2.
As a side note, and you probably meant this anyway, but saying that a molecule is "thermodynamically favorable" is meaningless. You should say "the dissolution of [molecule] in [condition] is thermodynamically favorable." And keep in mind that thermodynamically favorable in standard conditions refers to just that: standard conditions--1 molar solutions. Thus, even if K was less than 1, you could still say that it is thermodynamically favorable for [a certain small amount] to dissolve."

Yes, that is what I meant and it was  the original wording of the question was.  I omitted the details of the question because the bottom left corner says "Unauthorized copy or reuse of any part of this page is illegal"....but hopefully my educational purpose won't put me in jail.

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text:
Could alloys be electroplated onto a substrate?
For example, if I want to electroplate stainless steel onto copper, could that work?
I understand that different ions would need different voltages, but would those lower voltage ions merely make the lower voltage ion be deposited faster, or completely prevent the higher voltage ion from depositing at all?

Some alloys probably could be electroplated with carefully chosen electrolyte and voltage, but steel... well, I don't think so. How would you electroplate carbon, to begin with?

So theoretically it would be possible (as long as all the elements have soluble ionic compounds, e.g., not carbon)?

Yes, it looks theoretically possible . Might turn out impractical, but that's another story.

If it were easy then it would be a common industrial procedure. There are plenty of times when plating substrates with stainless steel would be really useful. The fact that it isn't done suggests either that it is impossible or very very impractical.

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text:
Differentiating esters from glycosidic linkages
What exactly is the difference between an ester linkage and a glycosidic linkage? I know both release a molecule of water, but what is the difference? Is it simply which types of monomers are involved in the bond?

In carbohydrates you can find examples of glycosidic linkages; the following example is α-maltose and the image is taken from Wikipedia, where a full list of authors is available.

This following example shows an ester; I chose one of the most simple esters (ethyl acetate) due to boredom. Again, the image is from Wikipedia, where a full list of authors is available.

One side of the molecule — the right-hand side in both of my examples here — always features a typical hydroxy group, i.e. is an alcohol. The difference is on the other side, the coupling partner. Look closely at maltose, and you may spot that the first carbon of the left-hand fragment also contains a hydrogen; its oxidation state is that of an aldehyde. Compare it with ethyl acetate, where the left-hand fragment linking carbon is in the oxidation state of a carboxylic acid (i.e. double bonded to an oxygen).
Technically speaking, glycosidic linkage is just a special case of acetal or ketal linkages; that is the correct name for the functional group $\ce{R^1O-CR^2R^3-OR^4}$ — acetal if either $\ce{R^2}$ or $\ce{R^3}$ is a hydrogen, otherwise ketal if both are carbons. ($\ce{R^1}$ and $\ce{R^4}$ must be carbon-based or we are dealing with hemiacetals/hemiketals/hydrates.) Examples of ketals for glycosidic linkages exist.
(And technically, ortho-esters exist, too, in which the linking carbon is bound to three $\ce{OR}$ groups. But these are very rare and special cases, even though they even exist in natural products such as tetrodotoxin.)

Simply and briefly put, esters’ vs glycosides’ “monomers”, or rather components, are (carboxylic) acids vs carbohydrates, and alcohols:

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text:
Introductory chemistry book with calculus prerequisite
I understand that the policy here is that book requests are allowed, as long there is an effort to narrow it down. This question is similar to Can someone recommend me a book ( preferably written long ago) that includes all the details of high school chemistry?.
I am trying to find a book which tries to approach chemistry from the perspective of someone who has a higher-than-high-school-math background, i.e., they're proficient in calculus, but they aren't familiar with chemistry. Does any such book exist?
Reason why I'm asking for this is that mathematics undergraduate students, in my experience, get very little experience applying their problems outside of mathematical theory. Chemistry, in my years of tutoring and teaching, gets virtually no mention in calculus courses unless a math professor decides to discuss balancing equations using techniques in linear algebra. It has been years since I've seen high-school AP chemistry, but I recall rate laws being a consequence of integration, although this was never actually shown in AP chemistry. I made an effort, when I was a Calc. I TA, to show a class of mostly chemistry majors how the first-order rate law is derived from a differential equation.

Introductory first-year chemistry doesn't involve very much math.

General Chemistry by Linus Pauling or Chemical Principles by Peter Atkins.  You may also want to skip general chemistry and go straight to P-Chem, e.g. Quanta, Matter, and Change by Atkins.

I would like to recommend Chemical Principles by Atkins as well. Its discussion of thermodynamics and kinetics involved some calculus iirc.

Most introductory physical chemistry courses and books don't rely very heavily on prior chemistry knowledge, and there is quite some math to do there. I think you can start most subjects there independently from scratch (QM, thermodynamics, kinetics).
To better prepare us for quantum mechanics, my (chemistry) faculty offered a two-semester lecture for "freshmen" where the professor derived the Schroedinger equation with us from first principles. The lecture was basically self-contained, both mathematically (introducing analysis) and chemically (first-years do basic chemistry and qualitative inorganic analytics, there is hardly any overlap with QM).
With students that are better in math than we were, this could work very well, and faster. :-)
I'm afraid I have no english book for this. The regular physical chemistry books don't dive into the actual mathematical derivation so much. We used "Wedler, Lehrbuch der physikalische Chemie".
Btw.: Students that are good in math AND chemistry are very valuable. Be prepared for your colleagues from the chemistry department to be after them. ;-)

Wedler <3! I always wondered why nobody has ever attempted to translate it. But then again, it is a typeset nightmare.

Yea, it's a typeset lecture, and those are rarely useful without a good lecturer. An ambitious translator might give it quite a boost!


text:
"Negative ions are tidier"?
From Chemguide:

That means that the only elements to form positive ions with noble gas structures (apart from odd ones like scandium) are those in groups 1 and 2 of the Periodic Table and aluminium in group 3 (boron in group 3 doesn't form ions).
Negative ions are tidier! Those elements in Groups 5, 6 and 7 which form simple negative ions all have noble gas structures.

Does it mean that generally negative ions are luckier in achieving the noble gas form? Is there any trend behind this, or is it just a coincidence?
Why did he chose those particular three groups as an example? There are negative ions formed by elements in other groups too.

Can you give an example of what you mean by your last sentence?

@orthocresol - the oxygen anion, $\ce{O^2-}$ I wanted to take $\ce{Al(OH)_4}$ as an example, but recalled that it's a complex ion. I just did not understand clearly what the author meant.

"Groups 5, 6, and 7" actually refers to Groups 15, 16, and 17 according to the current nomenclature, i.e. the N, O, F groups.

@orthocresol - and simple ions are formed only by elements in those three groups?

No, but the simple ions that are formed will have noble gas configurations. For example you don't see $\ce{O-}$ or $\ce{N^2-}$ but you do see $\ce{O^2-}$ and $\ce{N^3-}$.


text:
How potassium ferrocyanide is coloured?
The colour of a complex compound is due to unpaired electrons. As per crystal field theory, $\ce{K4[Fe(CN)6]}$ has no unpaired electrons so it has to be colourless. But then why is it coloured?

Color is due to electron transitions. Whether or not the electrons were paired before the transition is less important.

There is no direct connection between having unpaired electrons and having a colour. In fact, many colourful compounds have no unpaired electrons and yet display a bright colour: potassium permanganate, potassium dichromate, azobenzene, sudan red, phenolphthalein, iodine and many, many more. Arguably, there are more coloured compounds known in the singlet state (which means no unpaired spins in the vast majority of cases) than in the dublet, triplet or any other state.
So what causes the colour? It is the possibility of electron transitions from a lower to a higher orbital. If that transition is equivalent to a photon with a visible wavelength, we observe a colour. Examples:

Most phenyl groups absorb light at c. $\pu{250 nm}$ — to short a wavelength to be visible; they are ‘ultravioletly coloured’.
The electronic transitions of most transition metal complexes (except $\mathrm{d^0}$, $\mathrm{d^5}$ and $\mathrm{d^{10}}$ complexes) are well within the visible range of $400$ to $\pu{700 nm}$.
In rare cases, extensive electronic systems may even allow the absorption maximum to be shifted outside of the visible range into the infrared range. Chlorophyll comes rather close to this, absorbing red light (the longest visible wavelengths).

For potassium hexacyanidoferrate(II), the colour we observe is yellow so the absorption must be the complementary blue/violet, indicating a relatively large energy difference.

Some technicalities: "no unpaired electrons" is not synonymous with "singlet state" (usually it is, but not always). For example: the $^1\Sigma_\mathrm{g}^+$ excited state of dioxygen is a singlet state but has two unpaired electrons. Nor is "one or more unpaired electrons" synonymous with the triplet state. Free sodium atoms have one unpaired electron, but the ground state is a doublet ($^2S_{1/2}$) state.

@orthocresol Thanks, I was aware that unpaired $\ne$ triplet, but the organic chemist in me pays more attention to triplet than any other unpaired state, so I just … forgot that ;) Is writing unpaired spins technically correct?

Hmm, I guess so.

Just wondering: Laporte and spin selection rules should have prevented the transitions, right?

@WilliamR.Ebenezer The Laporte rule indeed should prevent a $\mathrm{t_{2g}\rightarrow e_g^}$ transition, that is correct. However, that transition is not spin-forbidden as all $\mathrm{e_g^}$ orbitals are empty. Which transition is actually responsible for the colour is a good question because right now I am at a loss.

Actually, thinking again and looking at the salt, the colour of hexacyanidoferrate(II) is not intense so the Laporte-forbidden $\mathrm{d \rightarrow d}$ transition seems fine. Hexacyanidoferrate(III) should have another transition (possible LMCT) that explains its much more intense red colour.

@Jan, I don't understand why the spin selection rule does not apply. From what I have been taught, I understand that change in spin multiplicity is forbidden, which seems to be the case here. And regarding Laporte, yes, I agree that the color is pale and that does make sense. But again, I feel (intuitively) that LMCT might not be very significant, as I have usually seen it happening only at high metal center oxidation states. Thanks for your reply, BTW.

@WilliamR.Ebenezer Look at the rightmost diagram in the question and tell me which of the electrons in the $\mathrm{t_{2g}}$ orbitals are prevented from being excited to $\mathrm{e_g^}$ by the $\mathrm{e_g^}$ orbitals being completely occupied with an electron of the same spin. (I write $\mathrm{e_g^*}$ because they happen to be antibonding orbitals according to MO theory but describing them as simply $\mathrm{e_g}$ as in OP’s diagram is fine and leads to the same result.)

Of course, hexacyanidoferrate(II) looks unlikely to allow any LMCT excitations; the $\mathrm{t_{2g}}$ orbitals are fully populated so any LMCT must go directly to $\mathrm{e_g^*}$ which is a very high-energy difference. However, if you remove one of the $\mathrm{t_{2g}}$ electrons (oxidation to hexacyanidoferrate(III)), this opens up space for an excitation from a lower electron level into $\mathrm{t_{2g}}$. Thus, hexacyanidoferrate(III) looks like an LMCT transition is much more possible.


text:
How does aluminium react with bases to form aluminates?
An example of reaction:
$$\ce{2Al + 2KOH + 6H2O->2 K[Al(OH)4] + 3H2 ^}$$
Aluminium is not ionic, how then does it attract the $\ce{OH-}$ groups to bond with them into the complex ion $\ce{[Al(OH)4]-}$?
I started re-reading non-organic chemistry at a slightly deeper level recently after going through the basics of organic chemistry. This reaction stumped me. I've re-read some stuff about aluminium and complex ions, but can't seem to understand it yet.
There's no positive charge on $\ce{Al}$, yet it reacts with hydroxide groups floating around. Where will the electron from a non-ionic aluminium atom go?
I cannot 'imagine' the reaction, visualize its mechanism, and this makes it hard to remember it.

P.S. Aluminium hydroxide seems to be mysterious in its structure. I quote Chemguide:

The chemistry textbooks that I have to hand aren't too clear about the structure of aluminium hydroxide as far as the degree of covalent character is concerened, and a web search (until I got totally bored with it!) didn't throw up any reliable chemistry sites which discussed it. Several geology sites describe the mineral gibbsite (a naturally occurring form of aluminium hydroxide) in terms of ions, but whether it actually contains ions or whether this is just a simplification as a convenient way of talking about and drawing a complicated structure, I don't know.

P.S. I try to edit the text using the $$ enclosed notation, but my old PC literally goes into heart attack mode, to the extent that I have to type in a word and then wait a second before it appears on the screen.

P.P.S. I've just found one possible explanation (in Russian), but it's not completely clear either. Stashing it here to read it later.

To put things into perspective: you might have heard that many metals react with acids. (In most inorganic curriculums, this is a pretty heavily used example, and even the defining feature of acids, at least until you get to Brønsted–Lowry). All your questions are applicable to that reaction as well. Indeed, it is just the neutral metal with protons floating around, and all of a sudden they just up and react.

It trumped you?? What card did it play? Or has the presidential race given us a new definition for that word...

@foobarbecue - I'm not a native speaker of English. Did I use a wrong expression?

The rest of your post was so fluent that it did not even occur to me English might not be your first language. I'm guessing you meant to say "The reaction stumped me" rather than "The reaction trumped me." The world "trump" comes from card games and also happens to be the name of a particular presidential candidate...

@foobarbecue - I see! Thank you for the explanation! I confuse "stump" and "trump". Yes, that guy is famous even in our neck of the woods thanks to his stump speeches. (0:

There's no positive charge on Al, yet it reacts with hydroxide groups floating around.

Your observation is correct. In the beginning, aluminium in its elemental state has the oxidation state 0. Apparently, that changes in the course of the reaction:
$$\ce{Al <=> Al^3+ + 3e-}$$

Where will the electron from a non-ionic aluminium atom go?

Again, your observation is correct. If one species is oxidized, another one must get reduced. In this case, protons from the water are reduced to form hydrogen gas:
$$\ce{ 2H2O + 2e- <=> 2OH- + H2 ^}$$
If you combine both half reactions and balance the stoichiometry, you end up with the initial equation for the whole redox reaction.

I guess, the part of the question for this answer would be: why "Apparently, that changes in the course of the reaction"..

There are multiple layers here. Let’s initially substitute aluminium for something as reactive as lithium. Throw lithium into water and it will react forming hydrogen according to the following equation:
$$\ce{2Li + 2 H2O -> 2 LiOH + H2 ^}$$
Technically, this is nothing else than the typical metal dissolution in acidic media, which typically follows the equation:
$$\ce{2 M + 2$n$ H+ -> 2 M^{$n$+} + $n$ H2 ^}$$
Albeit here we have a less acidic media, so instead of adding $\ce{H+}$ to the reaction, we add water and remove hydroxide:
$$\ce{2 M + 2$n$ H2O -> 2 M^{$n$+} + 2$n$ OH- + $n$ H2 ^}$$
Basically, this is equivalent to taking water as a Brønsted acid.
Metals are divided into noble and non-noble metals depending on whether they should react with $1~\mathrm{M}\,\ce{HCl}$ solution, i.e. their standard potential. (Aluminium would love to react with acids because it is a very non-noble metal.) But that is relatively irrelevant, we’re in water (or base) so the standard potential shifts and not everybody knows (or cares) what it shifs to. Rather, it is important to note that aluminium is doing nothing else than what I wrote up there for a general metal, namely dissolving in water with water being the acidic species:
$$\ce{2Al + 6 H2O -> 2 Al^3+ + 6 OH- + 3 H2 ^}\\\ce{2Al^3+ + 6 OH- -> 2 Al(OH)3 v }$$
Or at least that is what should happen. Taking your random tin foil, the aluminium has created a substantial layer of aluminium oxide and hydroxyide on the top which is difficult to penetrate for water and creates a passivation. So any type of aluminium you put into a jar of water should really be considered aluminium coated with aluminium oxide and put into water.
This is where the hydroxide of $\ce{KOH}$ comes into play. It is able to coordinate the aluminium cations in the oxide/hydroxide, turning them into aluminates $\ce{[Al(OH)4]-}$ which can dissolve in water. This destroys the passivation layer, allows fresh aluminium to be reached, which can again react with the remaning water to form more aluminium oxide/hydroxide which is attacked … etc.
This reaction happens because aluminium is so non-noble that the oxidation to $\ce{Al^3+}$ is favourable even under basic conditions (whereas some other non-noble metals are stable under basic conditions because there is not enough potential to oxidise them).

I wonder if the reaction $\ce{Al2O3 + Ca(OH)2}$ is possible, along the same lines as with $\ce{KOH}$.

Al2O3 + Ca(OH)2 → Ca(AlO2)2 + H2O . Indeed, when you put aluminium to water and  add a little slaked lime in it. The reaction will happen slowly, took many hours. And then Aluminium lost it protected coat Al2O3 will react directly with water to create H2 and Al(OH)3


text:
Why do alcohols and ethers have approximately the same solubility in water but different boiling points?
In Morrison & Boyd, I found this question:

Butan-1-ol (b.p. $118~\mathrm{^\circ C}$) has a much higher boiling point than its isomer diethyl ether (b.p. $35~\mathrm{^\circ C}$), yet both compounds show the same solubility ($8~\mathrm{g}$ per $100~\mathrm{g}$) in water.

How do you account for these facts?

Now, as far as I know, the boiling point depends on the inter molecular forces. Since in the alcohol, there is hydrogen bonding in the O-H group, it has a higher boiling point than the ether. But in an aqueous solution, the solubility is also related to hydrogen bonding. Since there is more hydrogen bonding in the alcohol, I would expect it to have a higher solubility in water, but that is not the case. So how do we explain it? I am completely confused by this question. What is the answer?

Hydrogen bonding needs two parties - roughly speaking, H and O (not necessarily so, but this would suffice for the question at hand). Diethyl ether has only one of those. Therefore it can't form hydrogen bonds by itself, but can do it fairly well when mixed with water.

Ok, great! Now I got it. Thank you very much @IvanNeretin

Please don't capitalise names of chemicals, they are not proper nouns and should not be capitalised..

Ivan's comment is spot on. It take two to tango (or hydrogen bond) - a donor $\ce{Z-H}$, where $\ce{Z}$ is some electronegative atom (usually $\ce{O},\ \ce{N},$ or $\ce{F}$) and an acceptor (something with just $\ce{Z}$). 1-Butanol can hydrogen bond with itself, since it has an $\ce{O-H}$ group capable of being both donor and acceptor. Deithyl ether cannot hydrogen bond with itself since it lacks an $\ce{O-H}$ group. However, both can form hydrogen bonds with water. Thus, 1-butanol has the higher boiling point, but they both are similarly soluble in water.
Actually, Morrison and Boyd oversimplify the solubility difference (and there is one). 1-Butanol is slightly more soluble in water than diethyl ether, perhaps because it can interact with water in more than one way, although the dipole moment of the alcohol group is certainky larger than that of the ether group as well.
The NIH PubChem entries for diethyl ether and 1-butanol list several solubilities in water, but I want to focus on two entries that share the same source for both compounds:

Solubilities in water at $25\ ^\circ \mathrm{C}$ from the NIH TOXNET Hazardous Substances Database

$$\begin{array}{c|c|c|c}
\mathrm{substance} & \mathrm{Solubility\ in\ water\ (g/100\ mL)} & \mathrm{density\ (g/mL)} &\mathrm{volume\ percent}\\ \hline
\text{1-butanol} & 6.8 & 0.8098 & 8.4\\
\text{diethyl ether} & 6.0 & 0.7134 & 8.4
\end{array}$$

Solubilities in water at $20\ ^\circ \mathrm{C}$ from the International Labor Organization's International Chemical Safety Cards

$$\begin{array}{c|c|c|c}
\mathrm{substance} & \mathrm{Solubility\ in\ water\ (g/100\ mL)} & \mathrm{density\ (g/mL)} &\mathrm{volume\ percent}\\ \hline
\text{1-butanol} & 7.7 & 0.8098 & 9.5\\
\text{diethyl ether} & 6.9 & 0.7134 & 9.7
\end{array}$$
Note that I have used the solubility data to calculate percent compositions by volume in both cases. It was and still is common to report low solubilities as percents composition. It also is and was common to see percents composition listed without specifying whether they were mass/mass (which should be the default), mass/volume (convenient for dilute solutions), or volume/volume (convenient for solutions, but unhelpful due to colligative properties).
At both temperatures, the data supports that 1-butanol and diethyl ether have nearly the same percent solubility by volume. Morrison and Boyd is an older generation of textbook, first published before the world wide web. If they found a source of solubility data for these compounds that listed percents (but did not specify volume/volume), then the authors made a reasonable assumption that those percents were mass/mass or mass/volume and reported the same solubility.
The slight solubility edge for the n-alkanols persists over a larger range of compounds. The table below lists the solubilities in grams per 100 mL of water of the n-alkanol and the symmetric ether, with data drawn from the NIH Hazardous Substance Database.
$$\begin{array}{c|c|c|c|c}
\text{sol. in water in g/100 mL} & \ce{C2} & \ce{C4} & \ce{C6}  & \ce{C8}\\ \hline
\text{n-alkanol} & \text{miscible}  & 6.8 & 5.9  & 0.54 \\
\text{symmetric ether} & 4.6  & 6.0 & 4.9 & 0.30
\end{array}$$

Conversely, the solubility of water in diethyl ether is 1-2% (depending on method and reference source)  and for 1-butanol around 20%.


text:
HSAB: alkyl halides & carbonyls LUMO energy flaw
Hard electrophiles have LUMO of higher energy than soft electrophiles. Alkyl halides are considered to be soft electrophiles while carbonyl compounds are considered to be hard electrophiles. The sigma antibonding C-X orbital of an alkyl halide is higher in energy than the pi antibonding C-O orbital of carbonyls which is in contradiction with the previous statement.
How is this paradox explained?


text:
Why does CO2 accept electrons?
CO2 is a lewis acid as it can accept electrons. But as the octets of all atoms of CO2 are complete why does it accept electrons? Shouldn't it donate electrons and be a lewis base?

Most of the stable compounds have complete octets on all atoms. This has nothing to do with their willingness to donate or accept electrons.

@IvanNeretin Then how do we determine it?

Consider that the carbon in carbon dioxide is bonded to two electronegative oxygen atoms. This will cause shift the electrons towards the oxygens and away from the carbon, making it electron deficient:
$\hspace{7.5cm}$
This makes the central carbon electrophilic, and explains the reactivity of the carbonyl moiety towards nucleophiles. Similarly, $\ce{CO2}$ undergoes addition reactions with Grignard reagents to form carboxylate anions.
$\hspace{4cm}$


text:
Why lithium gives flame coloration?
Theoretically speaking,  beryllium and magnesium does not give flame test because their atoms are comparatively smaller and the valance electrons are strongly attached to the nucleus. Therefore their ionisation energies are high, so they need large amount of energy for the excitation of their valence electrons to higher energy level which is usually not available in the Bunsen flame. So they do not impart any colour to the flame.
Now, comparing the atomic radius of lithium and magnesium, both have nearly same atomic radius due to diagonal relationship.($\ce{152}$ pm and $\ce{160}$ pm respectively). Since both of them nearly have same atomic radius, why is lithium imparting flame coloration(crimson red) while magnesium is not? Both of the metals has to be supplied equal amount of energy to excite the electron which the burner is not able to supply and thus flame test should fail in this case but this is not happening.

I don't understand despite the reasoning for beryllium not giving flame coloration is quite correct, wikipedia claims beryllium  to give flame coloration(white).

IE1(Li) = 5.320 eV; IE1(Mg) = 7.642 eV (Shriver & Atkins 6ed p836) For what it's worth I'm not sure how the ionisation energies actually affect the flame test. Aren't you dealing with $\ce{Li+}$ and $\ce{Mg^2+}$ anyway?

No, you actually need to reverse that reaction; $\ce{Li+ + e- -> Li}$, because the gaseous metal atom is the cause for flame colouring.

@Jan Now I understand. btw I have found a question in which you have answered it by taking this case only. Can you brief it here or i'll brief it up here?

@NilayGhosh The thing is, according to theory I have no idea why beryllium and magnesium do not give a flame colour. I could think it through or research it but I don’t want to spend that time on it now. So … I don’t feel good about answering. My other answer is on why the colouring actually happens and is only partially relevant here. The basis of an answer but not the actual answer.

I would love to find numbers, but alas, I have failed. If anyone can point me towards energy differences between the orbitals I am lacking, please feel free!
There is no a priori physical reason why lithium, sodium, potassium and calcium give flame colours but beryllium and magnesium would not. For all these elements (and hydrogen), the principal mechanism works in the same way: An electron is thermally excited into a higher orbital, and when it relaxes back to its ground state it releases a photon of exactly that wavelength. The flame colour is always due to gaseous atoms (not ions), so the electrons are always excited from an s-orbital.
The strongest lowest-energy excitation and relaxation should always be $n\mathrm{s} \rightarrow n\mathrm{p}$, i.e. from one shell’s s-subshell to that shell’s p-subshell. For sodium, this energy difference corresponds to $589~\mathrm{nm}$ or $2.10~\mathrm{eV}$, for lithium it is $671~\mathrm{nm}$ or $1.85~\mathrm{eV}$, for potassium $767~\mathrm{nm}$ and $1.61~\mathrm{eV}$ and for calcium $657~\mathrm{nm}$ and $1.89~\mathrm{eV}$.[1]
We can see that the energy difference rises from potassium to calcium, even though they are neighbours and the excitation is from the same orbital to the same different one. This is because calcium’s higher nuclear charge contracts the orbitals, stabilises them and thus lowers their energy — compare with the ionisation energy which also typically rises across a period. This effect is, of course, more pronounced when going across the second or third period when compared to the fourth; thus, I would expect an even greater energy difference when going from sodium to magnesium and even greater than that from lithium to beryllium.
The best explanation I have for the colourlessness of magnesium and beryllium is that their excitations are too close to ultraviolet to be properly observed due to strong contractions of the orbitals, but without numbers that is a weak explanation at best.

Notes:
[1]: I would be grateful for confirmation that these values I found are correct.

For magnesium, the transition corresponding to the calcium 657 nm line is at 457 nm.  (See Fig. 7-25 here: http://bcs.whfreeman.com/webpub/Ektron/Tipler%20Modern%20Physics%206e/More%20Sections/More_Chapter_7_1-Multielectron_Atoms.pdf) However, in both cases the transitions are forbidden singlet-triplet.  I believe Russell Saunders coupling is less correct further down the periodic table, and the transition is less forbidden for Ca.  https://physics.nist.gov/PhysRefData/Handbook/Tables/calciumtable3.htm vs https://physics.nist.gov/PhysRefData/Handbook/Tables/magnesiumtable3.htm

Based on empirically-useful ICP lines, it looks like your calculations for the alkali metals are spot-on.  Appears things are substantially more complicated for the alkaline earths, though, even those for which a useful atom-emitting line is listed.

Expanding a bit on the last paragraph of Jan's answer:

The best explanation I have for the colourlessness of magnesium and beryllium is that their excitations are too close to ultraviolet to be properly observed due to strong contractions of the orbitals, but without numbers that is a weak explanation at best.

While the physics and chemistry of inductively-coupled plasmas as used in ICP-AES/ICP-OES are rather different from flame tests, given the orders-of-magnitude difference in energies involved, the usable emission lines for these various elements indirectly support Jan's supposition.  The following data are drawn directly from the fantastic Interactive Periodic Table maintained by Inorganic Ventures, which lists the top few most-usable emission lines for each element; the wavelengths below are ordered so as to best highlight trends as a function of period. Lines marked with an asterisk derive from the ion as the emitting species, and thus are probably irrelevant to flame test behavior; unmarked lines are for the atom as emitting species.  Bear in mind that the visible/ultraviolet transition occurs around 400 nm (per Wikipedia).
$$
\require{begingroup}
\begingroup
\newcommand{\nm}[0]{\ \mathrm{nm}}
\begin{array}{cccc}
\hline
\ce{Li} & 670.784\nm & 460.286\nm & 323.261\nm \\
\ce{Na} & 589.595\nm & 588.995\nm & 330.237\nm \\
\ce{K}  & 766.490\nm & 771.531\nm & 404.721\nm \\
\ce{Rb} &    -       &    -       & 420.185\nm \\
\ce{Cs} &    -       &    -       & 455.531\nm \\
\hline
\ce{Be} & 313.042\nm^* & 313.107\nm^* & 234.861\nm   \\
\ce{Mg} & 279.553\nm^* & 280.270\nm^* & 285.213\nm   \\
\ce{Ca} & 393.366\nm^* & 396.847\nm^* & 422.673\nm   \\
\ce{Sr} & 407.771\nm^* & 421.552\nm^* & 460.733\nm   \\
\ce{Ba} & 230.424\nm^* & 233.527\nm^* & 455.403\nm^* \\
\hline
\end{array}
\endgroup
$$
Unfortunately, I don't know much about the details of why certain lines are strongly emitting and some aren't--the constraints of forbidden transitions mentioned by DavePhD likely play a significant role, though, as well as complex chemical and physical phenomena occurring at the high temperatures of the plasma. Regardless, some interesting, if highly approximate, trends are present in the data.
Of relevance to this specific question, though: the alkali metals all have atomic emission lines in the visible spectrum, whereas it's only the heavier alkaline earths that have useful atomic lines juuuuust over the threshold between the near-UV and visible.  While the uncertain connection between emission lines at ICP versus flame conditions must be re-emphasized, this at least provides weak corroborating support for Jan's argument.


text:
Isn't the reaction of chloral with sodium hydroxide a disproportionation reaction?
Chloral undergoes a haloform reaction with sodium hydroxide. Now the carbon atoms in chloral are in the $\mathrm{+III}$ and $\mathrm{+I}$ oxidation state. Chloroform’s carbon is in the $\mathrm{+II}$ oxidation state, as is sodium formiate’s. Now, carbon has both been reduced and oxidised. Then why is it not a disproportionation reaction ?

Who said it wasn't? o.o Anyway, it's not always useful to classify organic reactions into classes like disproportionation etc. But if you really want to, yes you could probably still call it a disproportionation.

OK! It's given in my text that it isn't a disproportionation reaction.

I’m not sure if that was intended or a slip of the pen: If anything, the reaction would be called a comproportionation.

Oh, true true, it's a comproportionation not a dis... My bad

related http://chemistry.stackexchange.com/questions/45172/why-does-trihaloacetaldehyde-formed-during-the-haloform-reaction-not-undergo-a-c

You should not call the haloform reaction of trichloroacetaldehyde with sodium hydroxide a comproportionation. (And even less a disproportionation.)
Disproportionations are defined such that you start at two species of the same type and oxidation state (e.g. two formal $\ce{Cr^4+}$ species) and end up with two different species (e.g. a formal $\ce{Cr^3+}$ and a formal $\ce{Cr^6+}$ species). This is not the case here, the oxidation states start out differently.
Comproportionations are defined oppositely: You start at two different species and create two of the same. But in no way can chloroform and sodium formiate be called ‘the same’, even though their oxidation states are. Carbon monoxide (also $\mathrm{+II}$) is not the same as either, either.


text:
Can aluminium hydroxide be split into Al 3+ and OH-?
When writing a net ionic equation containing aluminum hydroxide, can the aluminum hydroxide be split into ions of $\ce{Al^3+}$ and $\ce{OH-}$, or does it stay as $\ce{Al(OH)3}$?

When you deal with soluble compounds, you may write them either way as you see fit. But when a compound is insoluble, leave it as is.

Aluminum hydroxide has poor solubility in water (solubility product $3 \times 10^{-34}$).
However, $\ce{Al(OH)3}$ is amphoteric and soluble in acids and bases. An excess of $\ce{OH-}$ will form tetrahydroxidoaluminate:
$\ce{Al(OH)3 + OH- <=> [Al(OH)4]-}$


text:
Equilibrium Problem Involving Solids: Faulty?
I am convinced that this is a faulty problem:

$\ce{2 Cu + S -> Cu2S}$
For the reaction above, $\Delta H$, $\Delta S$, and $\Delta G$ are all negative. Which of the substances would predominate in an equilibrium mixture of copper, sulfur, and copper(I) sulfide at $298~\mathrm{K}$?  Explain how you drew your conclusions...

The problem as written does not include state symbols, but they are obviously all in the solid states. However, solids don't achieve equilibrium, at least as far as I know. Thus, how can one talk of an "equilibrium mixture" of solids?
Update:
How can I explain the fact that using the ubiquitous equation $\Delta G^{\circ} = -RT\ln K$, I do indeed get an equilibrium constant?

I think it’s a trick question. At $298~\mathrm{K}$ it should take longer than the age of the universe for $\ce{2Cu + S <=>> Cu2S}$ to reach equilibrium …

Why, solids do achieve equilibrium, it's just that it happens to be shifted all the way to one side.


text:
Making a glow in the dark substance glow only when heat is applied
I want to make a glow in the dark substance that glows only if it gets warm.
How do I obtain this substance, or what chemicals to I need to mix to make it?
It's going to go in a sealed glass jar with some magnetic materials, and I'll use a magnet to move around the magnetic part (iron filings?) to heat up the substance and make it glow.
Ideally, the substance should only glow if it goes above 110 degrees Fahrenheit.
It can't be a gas, because it's going in a sealed glass jar. It also can't produce toxic fumes, but it can be toxic. Also, I don't want the magnets to be the only things to make the substance glow. I want to put it next to an oven, in the bathroom while I take a hot shower, next to the heater, or near a light-bulb as a temporary light source for when the power goes out.
I don't care if it's a solid or a liquid, or uses a state change to achieve the glow effect. However, it absolutely CANNOT be a gas, because if the glass breaks, I don't want to have all of the fire alarms going off at the same time while everyone has passed out from lack of oxygen.
An idea for a use for this:
because it's going to be warm when it's glowing, It could act as a good solid-state heater, and you could reheat it when the glow starts to fade. Kids will love it on a cold winter's night, because it's a night light that helps to warm their cold room (or their feet).

A half joke:

Now I need to make it glow yellow when it's warm and turn cold gray
  when it's cold, and then I can say that I'm an accomplished alchemist,
  and that I can turn lead into gold!

On a serious note,
I have a budget of around $100 to make this work, but the cheapest way of doing this would be best.
However, any way of doing this will work because it will be helpful to lots of people even if I can't afford it.

After a lot of searching, I have found this video but I don't know it would help it or not. It works the  opposite.  It glows yellow when it is cold and green at room temperature. You need liquid nitrogen to cool it and UV light to  see it. Otherwise it fulfils other specification. It also comes well within 100$ if you see the price for the chemicals in sigma-aldrich or so.

Hmmm... I'm not sure that is quite what I'm looking for... You can't just carry around a bottle of liquid nitrogen all the time! That is interesting though... even though it needs need the opposite of that. My substance would be a nice thing to take on a camping trip. It would light up next to a campfire and stay lit for a while.


text:
Can gamma radiation cause transmutation?
In irradiation of food for sterilisation, is gamma radiation absorbed by the food? If so, can it theoretically cause production of radioisotopes? Or does it ionise atoms in the food?

While photo-induced fission is possible, it generally speaking would require elements in your food that should not be there in the first place unless you live near Chernobyl...

@JonCuster Gamma sterilization of food, i.e. irradiation with gamma radiation from external sources (e.g. Co-60), is permitted by several countries. The applied doses are very high (several Gy or even some kGy).

Gamma radiation (unlike alpha or neutron irradiation) has no effect on most nuclei so can't cause transmutation. The effect is mostly chemical or biological where complex chemicals in biological systems are disrupted preventing reproduction.

@Loong - Photofission, which is prompt fission initiated by gamma radiation, certainly exists and is well studied in the nuclear physics community. However, the nuclei that show significant cross sections for this reaction, not surprisingly, tend to be things like uranium that are unlikely to be in food grade products. Perhaps my comment was not clear. I would think that a fission reaction would be considered a transmutation event.

@JonCuster What about photodisintegration of deuterium by gamma rays, http://adsabs.harvard.edu/abs/1934Natur.134..237C ? or beryllium http://journals.aps.org/pr/pdf/10.1103/PhysRev.61.19

@DavePhD - disintegration of deuterium needs (per the cited paper) 1.8MeV gammas, which Co-60 does not provide. And then you have to hope the free neutron gets absorbed somewhere else to do something nasty. As for beryllium, I hope you aren't eating too much of it. Yes, lower Z elements can be photofissioned, the question was on production of radioisotopes which is much more likely from high Z fission events.

@JonCuster Magnesium is definitely present in food.  https://inis.iaea.org/search/search.aspx?orig_q=RN:14740554

@DavePhD - at GeV gamma energies, yes. Basically, if you hit the nucleus hard enough, 'bad things' will happen. The question is, how much will happen in a standard Co-60 sterilizer, and the answer for Mg is nothing. I'm sure you can find photofission cross section for many, if not all elements. Keep an eye on the gamma energy required.

@JonCuster Co-60 is just one source used for food irradiation.  Linear accelerators are also used.  See "several energies can be selected at: 5, 7.5, 10 MeV"  http://ccr.ucdavis.edu/irr/how_food_irr.shtml So that is definitely enough to photodisintegrate deuterium.  Magnesium would take slightly over 10MeV, but not GeV.

@JonCuster Also, that reference explains that C0-60 does produce 2.159 MeV and 2.505 MeV photons at low intensity.

@DavePhD - the branching ratio is about 0.013% into that channel. No convolute that with the photofission cross section and you get, well, not very much impact.

Answer to a More General Form of Your Question
The answer to the question

Can gamma radiation make non-radioactive stuff radioactive?

is yes, but only if the gamma ray has enough energy. "Enough" is different for all atoms but for most of the lighter ones it's over 2 MeV, and what basically happens if the gamma ray has so much momentum that when it hits a proton or neutron in the nucleus of an atom it can knock that proton or neutron right out of it.
Answer to Your Specific Question
The answer to the question

In irradiation of food for sterilization, is gamma radiation absorbed by the food? If so, can it theoretically cause production of radioisotopes? Or does it ionize atoms in the food?

is no in the grand majority of cases. Either way it ionizes the atoms in the food (That is how it sterilizes after all), but the gamma rays used in most food irradiation are produced by 60Co and therefore have a max energy of ~1.3 MeV. If you recall the answer to the more general question above, these gamma rays are not high enough in energy to cause photodisintegration in any of the light atoms of food.
In some cases however, the radiation is produced in a high energy accelerator. If these energies are high enough, gamma rays can start to actually do things. These high energy accelerators are not really used for food sterilization that much as far as I know, so most of the time the bolded answer above holds true.

C0balt-60 does produce 2.159 MeV and 2.505 MeV photons at low intensity. http://www.nucleide.org/DDEP_WG/Nuclides/Co-60_tables.pdf  Also, here is an example of a food irradiation facility that operates up to 10 MeV: https://nucleus.iaea.org/fitf/FacilityDisplay.aspx?ID=228

@DavePhD Thanks for the feedback! I'm aware of those things, but those high-energy gammas make up almost none of the total dose and won't cause much to happen. I knew some places use accelerators - hence the words "that much" - but I admit that I probably could have worded that better.

Yes.
See Photo-Fission in Heavy Elements (1947) Phys. Rev. 71, pages 3-10 :

fission should be possible for all heavy nuclei which lie well
  beyond the minimum of the packing fraction
  curve, provided sufficient excitation is provided
  to produce the necessary deformation of nuclear
  fluid which precedes division of the nucleus.
  Such excitation can be provided by particle capture or by absorption of a $ \gamma $-ray into the
  nucleus.

"Packing fraction curve" is an older term for the nuclear binding energy curve, so the reference is saying that gamma ray induced fission is possible for nuclei that are substantially larger than nickel.
On the low end of the binding curve, there is a different process called photodisintegration.

For example, magnesium-25, upon absorbing a photon of sufficient energy, emits a proton and becomes sodium-24.

Even the most stable nuclei such as iron and nickel can be disintegrated by energetic gamma rays.
See Photoneutron Thresholds Phys. Rev. (1951) vol. 84, pages 387-394 for quantitative information concerning how much energy a photon must have to photodissociate a given nucleus.

Still, I doubt energy of rays used in sterilization is high enough.

@Mithoron The energy range 5-10 MeV for food sterilization by beam source is discussed here: http://ccr.ucdavis.edu/irr/how_food_irr.shtml  That is well over the energy needed for deuterium photodisintegration (2.225 MeV).

Generation of radioisotopes outside of those already in your food cannot be done by irradiating it with gamma radiation. Transforming a non-radioactive isotope into a radioactive one involves changing the neutron number, and gamma rays are not neutrons and cannot induce a change in the number of neutrons.

You can have two substances with the same neutron number, one radioactive, one non-radioactive. Even if everything you said was 100% correct, it wouldn't answer the question because it doesn't address the issue of creating a radioisotope from an atom of a different element.


text:
How could we figure out the recipe of Coca Cola?
I know that the ingredients and the manufacturing process are both important, as discussed in this question. What would be a modern approach to figuring out the ingredients?
The ingredients listed on the website are: "Carbonated Water, High Fructose Corn Syrup, Caramel Color, Phosphoric Acid, Natural Flavors, Caffeine." My ideal answer would provide information about how to get more detail about the specific ingredients (not just 'natural flavors') and their proportions.

Aside from reading the label?

@ringo Edited, clarifying question.

I'd say some combination of a GCMS / LCMS. The chromatography separates the components & can often be relied upon to ppm level identification.
Then compare the MS peaks for hits against one of the standard databases e.g. NIST.
I suppose that strategy would identify most of the peaks. The left over peaks, which don't give a good match against a GCMS database, would be more challenging to identify. One would need to screen potential raw materials one suspects may have gone into the blend to see if one generates a hit.
Of course, this identifies individual chemical species. Any practical recipe on the other hand is unlikely to be based on pure species amounts but would use available ingredients which are each complex mixtures of pure species.
The GCMS  / LCMS approach won't tell you what pseudo-components (e.g. lemon oil  ) went in & in what proportions. So you figured out the chemical composition but not really the recipe.
Sometimes, rarely, two compounds can have very similar MS peak patterns. In those cases (or to reinforce a GCMS analysis) one could also compute the Kovats Index (KI) of each peak on the GC based on retention times alone (and the injection of a standard Hydrocarbon mix). There's a good amount of info. and databases on the KI flavor / natural product ingredients.
That would be another strategy to confirm peak identity.
On the molecular biology side of things I'm sure there are techniques using PCR etc. that could amplify and characterize traces of DNA segments that came via ingredients of plant origin. e.g. One might be able to say a certain plant oil extract was used. Although at high dilutions & after extensive processing it would be a challenge to get enough starting matter for these techniques to work.


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Quantitative Analysis of Urea in aq. solution
What would be a good way to assay Urea in aq. solution? The test doesn't need to be super accurate. 1% accuracy is fine.
But it needs to be relatively fast & robust to carry out.
NaHCO3 is a co-dissolved impurity so the test needs to tolerate NaHCO3 interference.

In clinical chemistry, urea is quantitativly determined as follows:

Hydrolyze urea with urease to ammonia and carbon dioxide
react the ammonia with 2-oxo-glutarate in the presence of glutamate dehydrogenase and $\ce{NADH}$: $\ce{NADH}$ is converted to $\ce{NAD+}$
monitor the decay of the UV absorption of $\ce{NADH}$ at $\lambda$= 340 and 410 nm

Thanks! Is there any titration based method you know of? There may not be a spectrophotometer available.

@curious_cat I'm not aware of any, but that's doesn't necessarily mean that there isn't any. There's another method that involves the reaction of urea with diacetyl monoxime. But again, this requires a photometer to monitor the absorption at $\lambda$= 525 nm.


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Is it possible for one of reactants to deplete before achieving equilibrium
Consider the following reaction where $\ce{A}$ & $\ce{B}$ reacts to produce $\ce{C}$ & $\ce{D}$.
$$\ce{ A(g) + B(g) <=> C(g) + D(g)}$$
Is it possible for $\ce{A}$ or/and $\ce{B}$ to run out before equilibrium at some given starting moles of $\ce{A}$,$\ce{B}$,$\ce{C}$ & $\ce{D}$.

When one of the reactants depletes, this is the equilibrium. It can't happen before achieving equilibrium, for pretty much the same reason that a man can't die before own death.

No, that is not possible.
Consider the definition of equilibrium from the IUPAC gold book:

chemical equilibrium
  Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy, $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:
  \begin{align}
\Delta G_\mathrm{r} &= \Delta G_\mathrm{r}^\circ + R T \ln K &&= 0\\
\Delta G_\mathrm{r}^\circ &= − R T \ln K\\
\end{align}
  The equilibrium constant, $K$, is given by the mass-law effect.

The important thing is that the reaction proceeds in both directions. It means, that as soon as some of the product is formed, it will react back towards the reactants. The equilibrium is reached when the rates of both of the directions are identical.
From the formula in the definition, you can also see, that the location of the equilibrium, i.e. the concentration of all species or $K$, is only determined by the difference in the Gibbs energy between products and reactants.
In other words, a system that forms an equilibrium will always form an equilibrium, independent of the concentration of the starting materials.

No, this is not possible, because given a certain set of conditions (temperature, pressure), there will be an equilibrium constant $K_\rm{eq}$. Given enough time, an equilibrium will be reached, and this constant tells you the ratio of products to reactants that will exist at this equilibrium. In case you are unfamiliar, equilibrium is the state a reaction reaches when the rates of the forward and reverse reactions are equal.
If you deplete your reactants in a chemically reversible reaction, then:
$$Q=\frac{[\text{products}]}{[\text{reactants}]}=\frac{[x]}{[0]}=\text{undefined} \ne K_\rm{eq}$$
Whether you are considering the acidity of $\ce{HI}$ or $\ce{CH4}$, there is an equilibrium constant. To put this in perspective, there are not enough molecules of methane in the entire universe to protonate one molecule of water, and yet the equilibrium constant is not $0$.


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Is it safe to inhale O2 from electrolysis of a pH 13.7 KOH solution?
I'm conducting electrolysis of water with a 0.25M KOH solution in water (about 14 on my pH meter) using graphite electrodes, and I was wondering if the O2 would be safe to inhale, as in an oxygen bar?

No. Untreated and unpurified, the gas coming directly out of your reaction vessel would be unsafe to inhale. It is likely that water vapour containing traces of KOH would be present, and this poses a significant safety risk to the person breathing this vapour. You could test this yourself by measuring the pH of the headspace above your reaction vessel. Hold a piece of indicator paper here for a period, and it will slowly show a change.

Technically, it's rather tiny droplets than vapor that poses as issue.

@IvanNeretin I can confirm.

@IvanNeretin Point taken - you are absolutely right.

So how can I make the oxygen safe to inhale? I guess drying the O2 would work, maybe passing it through a fine filter.  Would a carbon fiber filter work?  Are there other electrolytes I could use that wouldn't require filtering?

You have a couple of new questions that should be asked separately. Without knowing your experience handling chemicals, I wouldn't advise inhaling any gas generated from this setup, regardless of post-filtration.

The only thing in the gas phase at the anode will be pure oxygen, but when these small oxygen bubbles break the surface they create a fine mist of the electrolyte. This can be eliminated as an issue by running the resulting gas through a wad of cotton wool or a wash in water. There is, however, a different reason to be cautious.
Oxygen toxicity.
At high enough partial pressures, oxygen becomes toxic to humans (really all life, but to different degrees). This becomes a major problem for deep scuba dives, divers have to reduce their oxygen to inert gas ratio beyond a certain depth.
For short exposure times, O2 partial pressures well above one atmosphere (~101.3 kPa) can be tolerated by humans, but on longer exposure times even O2 partial pressures as low as 30 kPa (about 30% O2 at sea-level pressure) have been known to cause issues for some individuals.
Because of this, I strongly advise against breathing pure O2 from any source for an extended period of time.

+1 for the scrubber.  No points for the scaremongering.  There are lots of people who are on extra oxygen for long periods.  An O2 bar does not imply  inhalation of 100% O2 indefinitely.  I would get a used O2 concentrator for as they are designed for the purpose.


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From a Quantum Chemistry Point of View how can I break Bonds without Heating?
Is it possible to dissolve/break bonds using strong alternating magnetic fields at high frequency? (Perhaps from tesla coils?) Could you use resonance to cause the bond to weaken over time?  If so how would I calculate or approximate what frequencies those may be?
Basically, how can I liquefy iron without heating it? Could you also apply that to $\ce{SiO2}$?

Only some things are solid because of chemical bonds (I'm excluding inter-molecular forces). Most melting is caused by putting in enough heat to break the bonds between molecules not the bonds inside molecules. And, even if you could specifically excite some bonds, the energy would usually redistribute across other bonds and other vibrations/rotations in the substance ending up as heat.

You can (sort of), but there is a catch. As soon as a bond is broken, the very same atoms will throw themselves into each other's arms and form that bond again, releasing the extra energy in the form of light and heat. So your method of breaking bonds without heating, whatever it might be, is going to produce lots of heat. By applying more power, you may break bonds faster than they are formed and thus liquefy the material, but this would have the side effect of producing even more heat. This is no different from ordinary melting.
Alternating magnetic fields, when intense enough, would indeed melt iron and other metals. With $\ce{SiO2}$ we'll have to invent something else.

So, what do you think the Hutchison effect is doing then ? It seems to have allowed wood to sink inside the metal without burning it.

David Copperfield once seemingly made the Statue of Liberty disappear. Hutchison effect is pretty much the same, only on a smaller scale.


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Rate of Evaporation of solution
Why does rate of evaporation of solution decreases on adding a non volatile component?
Also what happens if we add a volatile solute?

The rate of evaporation decreases because the activity of the solvent in the liquid phase decreases, while the activity in the vapor phase (or solid, for that matter) stays the same. This is also the reason for the increase in boiling point and the decrease in melting point of the solvent.
If we add a volatile solute, it will evaporate too, thus changing the solvent activity in the vapor phase as well. The resulting behavior can't be characterized universally; you'll have to consult the binary phase diagram of boiling for the particular system, and these can be pretty diverse.


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Strictly speaking, is air homogeneous?
Why would air be considered homogeneous in its gaseous state (by Wikipedia and several other sources) even though its concentration keeps varying with time?
A homogeneous fluid as i think of it is as follows: If I take a small part of the fluid the number of particles of each kind is always the same.

Homogeneous only means that all the components are in the same phase. Since they are all gaseous this is not an issue.

Is a mixture of water and oil homogeneous then?

You are confusing phase and state of matter.

The air is not a secluded, isolated volume, instead there is a constant exchange of matter and energy with other packets of air, the sun, open space, the solid ground, etc. It  makes no sense to ask "is air homogeneous". Some properties of the air are certainly, others are on certain time and lenght scales. But everything else is a generalisation and not useful at all.

The issue depends on both the timescale and the volume to be considered. varying over time doesn't imply a lack of homogeneity, it just implies that the contents change over time. On a small scale, air is homogeneous; on larger scales it has, for example, varying quantities of water vapour. It might be better to ask whether the atmosphere is homogeneous.

Strictly speaking you might have a point there. But it is probably more difficult to judge.
I was unable to unearth an official recommendation by the IUPAC, so I unfortunately have resort to Wikipedia ([1] Kristi Lew (2009). Acids and Bases. Essential Chemistry. New York: Chelsea House Publishing.):

Homogeneous is a term in physical chemistry and material science that refers to substances and mixtures which are in a single phase. This is in contrast to a substance that is heterogeneous.[1] The definition of homogeneous strongly depends on the context used. Generally it refers to smooth variation of properties with no discontinuities or jumps.

If we talk about all the air around the world, it is certainly not uniform. There are concentration differences across certain areas. When we consider a vector into space, the total concentration will decrease along this vector. As we know, we will pass certain areas with different chemical compositions, for example the ozone layer.
But all these changes will be gradually; there are no discontinuities. In this sense one would call air homogeneous, even though it does not fit in the definition of phase.
Additionally, if you take an arbitrary (volume) unit of air, all at the same temperature and pressure, and you allow it to establish an equilibrated state, then yes, this will not only be homogeneous, it will also form one phase.
The above linked article further considers air as a homogeneous mixture. It rambles on with considering it a gaseous solution of oxygen, carbon dioxide, etc. in nitrogen. This is technically incorrect, since the gaseous state of matter is specifically excluded from the IUPAC definition of solution. Also a solution can never be at the same time a mixture, as the former can be separated by mechanical means, while the latter cannot.
The problem here is obviously the lack of a unique and rigorous definition for homogeneity. For whatever purpose when it really matters, one should probably use a more robust definition like phase, which is essentially what your second statement describes.

Depending on the context you might even obtain a homogeneous mixture of water and oil - the question is then more or less how stable such a suspension is. After a while it will certainly separate in two phases.

Air is heterogeneous if you think practically. It is a mixture of nitrogen, oxygen, $\ce{CO2}$ etc. But the concentration of oxygen and $\ce{CO2}$ is not uniform everywhere. Polluted regions have more $\ce{CO2}$ while $\ce{H2O}$(steam) concentration is more in tropical areas or sea sides. Unpolluted air, which is rare to find, however is homogeneous.
Most text books don't mention the difference between polluted and unpolluted air....so there this is might be a good explanation!

The concentration of CO2 and oxygen in the atmosphere very very little outdoors across the world over short time scales (CO2 is increasing slowly but over long timescales). Polluted regions don't have notably more CO2 or notably less oxygen. Pollution is mostly about traces of other things (NOx, ozone, hydrocarbons...).


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Is it safe to drink reboiled water?
There have been some claims saying that drinking reboiled water is dangerous because it produces arsenic, nitrates, etc. Is this true?

The claims are not true, they are just rubbish pseudoscience misinformation. Google "reboiled water" and skip the websites that claim to be "natural" "healthy" "chemical-free" etc. Those people don't know what they're talking about but just like to fear monger.

A different concern in reboiling water is that the water has been degassed. Such water can "bump" rather violently when it starts to boil.

Drinking boiling water is a bad idea in general - you'll burn your mouth!  Wait for it to cool first and you should be fine.

Boiling water produces arsenic, nitrates, etc?   Can it produce gold?

Some interesting comments here

@Paparazzi: sure, about 1 microgram of gold per kilogram of seawater apparently.

I've found an additional leg of the claim, namely that "High temperatures transform nitrates to nitrosamines" (http://www.whydontyoutrythis.com/2015/06/warning-why-you-should-never-reboil-water-again.html). Which is technically true, but this is a potential issue when cooking food, and especially at frying temperatures. Not so much when briefly boiling water that contains only trace amines to start with. If you cook your food then (re-)boiling water is nothing by comparison.

@SteveJessop But I would say the gold was not produced.  The gold was there - just increased the concentration by boiling off the volatile components.

See http://skeptics.stackexchange.com/q/11118/8755

It is safe but if you have the option to use fresh water that will taste better. I empty my kettle each morning and refresh the water. The more the same pot is boiled, the less aerated it becomes, which impacts the brewing and taste, but not the safety. You saw that video of the plant being watered with (chilled) boiled water, I'm guessing.

Reboiling water will do precisely three things:

Some volatile contaminants in the water that survived the first boiling might be driven out by the second boiling.
Some additional evaporation will occur, causing anything dissolved in the water (salts, contaminants) to be slightly more concentrated.
Some additional material from the surface of the vessel you're boiling the water in might dissolve into the water.

It's virtually impossible to imagine any scenario in which this would matter, unless you were taking a huge amount of barely drinkable water and boiling away most of it or using a vessel whose lining could produce toxins in your water. Needless to say, you really shouldn't be drinking barely drinkable water at all, nor should you be boiling water even once in a vessel whose lining can produce toxins. If you're doing these kinds of things, the additional risk from reboiling would be miniscule.
It will definitely not summon aresenic or lead from nowhere.

I used to go camping a lot, and often times we boiled water to make it safe to drink. In many cases, we would "re-boil" the water because it was easier to do that then keep track of (or separate) "good" water from "bad" water.  #2 is your "largest risk" and to be honest, you would have to boil down hundreds of gallons of "bad" water to a few ml worth to be at any kind of risk. That being said, boiling water is usually done to remove bacteria, it won't remove dissolved things (like lead). So don't go drinking a contaminated water, regardless of the number of boilings.

@coteyr and if you were that worried about it, get a distilling apparatus! You might lose some minerals, but it doesn't get much safer!

@Tanenthor There are also micro filters that do not require boiling and remove almost all contaminants that are not dissolved.

I am glad you asked this question. There is an element of truth here. If the water already contains contaminants such as arsenic and nitrates, and/or if the pot that the water is being heated in is contaminated, then boiling (or equally well, reboiling) the water could result in higher levels of these contaminants.
On the other hand, if you are starting with pure (i.e. distilled) water in an inert vessel, no amount of boiling is going to cause these contaminants to magically appear.

"Inert vessel" - so no Philosopher's Stoneware?

The very same thing could be same about anything you're putting into any container. There's nothing specific to water, boiled water, or reboiled water.

And if the pot is contaminated, it's contaminating each new batch of water, potentially more than any added contamination to first-boiled water.

"If the water already contains contaminants such as arsenic and nitrates, [...] then boiling (or equally well, reboiling) the water could result in higher levels of these contaminants. "
How? If the water already contains them, boiling won't magically add more contaminants!

@Josef your answer is actually in that [...] which you didn't post ;)

@Josef: Only because some of the water comes off as steam, so if we assume that the steam contains a lower concentration of contaminants than the water, then the water that doesn't boil away will be left with a higher concentration. This is essentially the bit left after distilling water; the process works because the contaminants are generally left in the water. If the original water is comparatively clean compared to safety limits, then the boiled water will be too. It will likely only be a problem if you reboil many times with water that is already contaminated.

To put it another way, in terms of these toxins drinking reboiled water is as dangerous as drinking more water. How much more water depends on how long it was boiled for. In practice you aren't going to reduce the volume of water by even 10% just by bringing it to the boil twice under a lid, but you could simmer it. In theory, yeah, you could boil millions of gallons of water down to a nasty dose of arsenic if you could also figure out a way to separate it from the fistfuls of non-toxic salts you're precipitating (limescale).

Does the concentration even matter if I plan to drink the whole amount of water anyway?

@AndreKR exactly, I hardly doubt it matters. The amount of whatever poisonous stuff in there doesn't change. In fact this whole discussion is entirely moot because there's not enough poisonous stuff in there to begin with (unless you're in some kind of slum with no clean water, but I think reboiling water is the least of their concerns).

I might suggest being careful with the word "amount." Reboiling water doesn't create more contaminant, which is what (on the surface level) the word "amount" implies.

@AndreKR: if you planned that then it wouldn't matter. But you don't plan that: you don't pour slightly less water into your tea when it's been boiled twice than you do when it's been boiled only once. You aim to fill the cup either way. This is not to say re-boiled water is dangerous (it isn't), but that the reason it isn't dangerous is because the "concentrated form" is still safe (having been only slightly concentrated from a very low starting point), not because concentration doesn't affect your dose at all.

There has been some claims that drinking reboiling water is dangerous claiming it produced arsenic, nitrates etc. it this true?

It doesn't produce anything.  Re-boiling water doesn't change the water into anything dangerous.
However, if the water is already contaminated, then lengthy boiling will evaporate some of the water, and the concentration of the chemicals already in the water will increase due to the lower concentration of water.
If you follow the standard recommendation and boil the water for a full minute, then depending on heat input and environmental conditions you could be evaporating as much as a cup of water a minute.  Let's assume the worst case - dry air, very hot stove, lots of surface area on pot, and only two cups of water to start with.  If you boil it for a full minute, you might lose half your volume, increasing the concentration of whatever is in the water twice.
So how can you avoid this without capturing the water from the steam?
Use a lid - a lot of steam collect on there and drip back into the pan.  Boil a lot of water at once, rather than boiling two cups for a minute, boil 4, 8, or sixteen cups.  Boiling a small amount requires little heat, but often people will turn the burner all the way up and leave it there, boiling it off far faster.  By boiling a larger amount, it's matched better to the stove and you'll only be adding the heat needed to get it to a rolling boil, rather than adding more heat than necessary.
Once the pot gets to a boil, turn the heat down - it needs less heat to stay boiling than it does to start boiling.
Lastly, once the minute is up, place the pot on a cool surface to cool the water down more quickly.  It can take 30 minutes to cool to drinking temperature depending on the environment, and during that time it's evaporating faster than normal.  Cooling it quickly will also save water.

Is it safe to drink reboiled water?

It's no safer than drinking boiled water, but each time it's boiled the contaminants become more concentrated.  Limit boiling to only when necessary - if the water has been contaminated since the last boiling.  Generally you won't need to re-boil water, but if you're unsure then you're simply going to have to weigh the risks of biological contamination against the risks of inorganic contamination.

There is a very unusual case in which reboiling water makes water (marginally) more dangerous to drink.
Suppose that you have a toxin or pathogen in the water that is neither volatilized nor denatured at 100 degrees Celsius. By boiling the fluid, you reduce the amount of $\ce{H2O}$, while the amount of the dangerous substance remains constant. Then, a fixed amount of boiled fluid will have a higher concentration of the dangerous substance than the same amount of unboiled fluid.
This may seem like a bizarre hypothetical, but this exact situation arose during the 2014 Toledo water crisis, which stemmed from a Microcystis bloom in Lake Erie. Microcystis, a variety of cyanobacterium, produces substances called microcystins, which are not destroyed or removed by boiling. The relevant authorities issued instructions not to boil the water during the crisis. (Of course, it is not the case that boiling converts potable water into unpotable water. Rather, it converts unpotable water into very unpotable water.)
That said, the other answers are correct in stating that reboiling water is almost always a safe practice. Even in water that contains a dangerous substance that is not removed or denatured by boiling, boiling the water is unlikely to make it significantly more dangerous to drink, since in practice, one does not typically remove a large fraction of a fluid's $\ce{H2O}$ content when boiling it.

First need to separate produce from concentrate.
If two or more molecules reacts to form different molecule(s) then you have produced.  Or a single molecule could split.  Or a single could could be changed like trans to a cis double bond.  Cannot produce an element (e.g. O) - it is present or not.  Not a lot of this going on at the boiling temperature of water.  The non bacterial components of water are pretty stable.
Boiling water removes volatile components. Arsenic is a chemical element - it cannot be produced. Arsenic is not volatile alone or as part of another molecule.  What is going to boil off is H20, N2, CO2, CO .....   The non-volatile components will be concentrated.
Nitrate is NO3 so I guess it is possible to produce it as N and O are present in water but that is highly unlikely.  And even if so boiling a second time would not change that.  I don't think a nitrate or anything it is a part of is volatile so it would not boil off.  It would just concentrate.
You cannot produce arsenic as it is chemical element.  It is in the water or not.  It would just be concentrated.
What does happen at the boiling temperature of water is the bacteria is killed.  If you killed the bacteria the first time then it should not grow back.  There is no purpose to boiling a second time if boiling did what it was supposed to do the first time (kill bacteria).
Yes potentially bad stuff could be concentrated but boiling a second time is not a factor.  Boil once for 10 minutes is equivalent to boiling twice for 5 minutes in regards to concentration.

You can produce arsenic from compounds which contain it. Either way it still isn't dangerous.

@bon  What?   If it is present then yes it is dangerous.   Arsenic is not just dangerous in element form - in fact that is not a common form.  Did you vote me down?  Regardless you are not going to break down mineral containing arsenic at the boiling temperature of water.

@bon Either way it still isn't dangerous?  Not according the medical community and  wiki.   https://en.wikipedia.org/wiki/Arsenic_poisoning  They even made up symptoms, diagnosis, and treatment?


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Mechanism of Friedel-Crafts acylation with succinic anhydride

Here is my proposed mechanism, something tells me that the intramolecular deprotonation happens after, and that this is not concerted. Is my mechanism correct?

Don’t draw it concerted. It is a stepwise process; the attack of the aromatic electrons onto the electrophile first with a delocalised carbocation as intermediate followed by deprotonation and rearomatising.

The reactivity of acid chlorides and acid anhydrides are actually quite similar. It may be helpful for you to think of an acid anhydride like this:

When we do, the acylation reaction is almost identical to that of an acid chloride, where an acylium cation $\ce{R-C#O+}$ is generated:

The subsequent proton exchange restores aromaticity:


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Electrode Potential at Equilibrium

At each electrode-electrolyte interface there is a tendency of metal ions from the solution to deposit on the metal electrode trying to make it positively charged. At the same time, metal atoms of the electrode have a tendency to go into the solution as ions and leave behind the electrons at the electrode trying to make it negatively charged. At equilibrium, there is a separation of charges and depending on the tendencies of the two opposing reactions, the electrode may be positively or negatively charged with respect to the solution. A potential difference develops between the electrode and the electrolyte which is called electrode potential. When the concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential.

Above is an excerpt from my book.
It says that at equilibrium the potential difference develops between electrode and eletrolyte and it is electrode potential.
But at equlibrium, should not the electrode potential should be zero as at equlibirum there should not be any tendency of reaction to occur?

From paragraph of textbook.
At the electrode-electrolyte interface the metal ions in electrolyte leak out, leaving a negative charge on electrode side of interface. While metal ions migrate at the other side of the interface from the electrolyte solution.
This eventually comes to a stop when the equilibrium is reached. That is the rate of negative charge build-up is equal to the positive charge. Like you say at this point there is no more drive for the reaction to go one way or the other.
However, now because of the charge separation at the interface, there is a potential difference between both sides of this interface which leads to the reduction potential we read out which is not zero. This is where the Nernst equation comes in handy.


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Relationship between pKa and pKb
The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $38$. Does it stand for the following equilibrium:
$$\ce{NH4+ <=> NH3 + H+} \tag{1}$$ With $K_\mathrm{a} = 10^{-38}$ for reaction $(1)$, indicating that the conjugate acid of $\ce{NH3}$ is quite weak?
The $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$. Does it stand for the following equilibrium:
$$\ce{NH3 + H2O <=> NH4+ + OH-} \tag{2}$$ With $K_\mathrm{b} = 10^{-4.74}$ for reaction $(1)$, indicating that $\ce{NH3}$ is a weak base?
Also: $$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 42.74 \neq 14$$

Regarding the last equation, it is only applicable to the sum of the pKa of a species HX and the pKb of its conjugate base, X-. Since both the pK values above are referring to NH3 itself, the formula is not valid. However, if you were to add pKa(NH4+) and pKb(NH3), you would indeed get 14.

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $38$.

This means, the reaction this value is describing is:
$$\ce{NH3 <=> NH2- + H+}$$

The $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$.

This means, the reaction this value is describing is:
$$\ce{NH3 + H+ <=> NH4+}$$

There is a lot of misuse going on with the term $\mathrm{p}K_\mathrm{a}$ since many people are too lazy to say

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$’s conjugate acid is $9.26$
$$\ce{NH4+ <=> NH3 + H+}$$

rather than

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $9.26$.
$$\ce{NH3 <=> NH2- + H+}$$

but that is really what they should be doing. Thankfully, when looking up values in a table the person compiling the table (hopefully) knew what they were doing and used the correct terminology.
Only for a $\mathrm{p}K_\mathrm{a}$ and a $\mathrm{p}K_\mathrm{b}$ that describe the same reaction (from opposite viewpoints) is it true that $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$.

Since the pKa of ammonia is 38, does that mean the pKb of the azanide ion is -24?

@LamGyro Yes. One bloody hell of a strong base.


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Name this amine
Consider the following two amines:

I've been given the name for the second amine as N,N-dimethylamino-2-propane.
Can this same approach be used to name the first amine? If so, I think its name would be N-methylmethyl-2-amino-2-propane. Is this correct?

Related: Nomenclature of Amines

The Short Answer
The name you list does correspond to the first compound shown under the naming scheme you were given for the second compound, however last I checked the IUPAC scheme recommends "N,N-dimethyl-2-propanamine" not "N,N-dimethylamine-2-propane" if you want to be really strict about it.
A More Realistic Look at Naming
Putting all that aside, you'll not often see simple substituted amines referred to this way by chemists. IUPAC naming in general is useful for unambiguously naming large, tricky compounds but for smaller garden-variety ones it's often seen as unnecessarily cumbersome.
The IUPAC scheme names amines in reference to the longest carbon chain (e.g. "N,N-dimethyl-2-propanamine"). The more common scheme for amines with relatively simple substituents is to name them in reference to the amine group instead (e.g. "dimethyl isopropyl amine").
This is similar to the common-name schemes for most other rotationally symmetric functional groups. In general, they're formatted:
[substituents, alphabetical if you're a pedant] followed by [type of compound]
For example, the compound consisting of a methyl group and an ethyl group attached to an ether, ketone, or sulfide would be called "methyl ethyl ether", "methyl ethyl ketone", or "methyl ethyl sulfide" respectively.
Finally returning to the example you've given, the first compound would most commonly be referred to as "methyl tert-butyl amine" and the second compound would be called "dimethyl isopropyl amine".

“… however last I checked the IUPAC scheme recommends …” – Apparently that was before 1993 since you are still using IUPAC nomenclature from 1979. The current recommendations are from 2013.