text:
How can the ground state electron configuration for a d3 complex correspond to an F term symbol?
The ground-state term symbol from a $\mathrm{d^3}$ Tanabe-Sugano diagram is $\mathrm{^4F}$. My question is how the total orbital quantum number $\Lambda=3$, or $\mathrm{F}$ term, arises.

For a $\mathrm{d^3}$ metal, I'd expect the following ground state d-electron configuration:

where notably the $\mathrm{t_{2g}}$ orbitals correspond to $\mathrm d_{xy}$, $\mathrm d_{xz}$, and $\mathrm d_{yz}$ suborbitals.

From the spherical harmonics, $\mathrm d_{xy}$ results from a linear combination of:
$$\begin{align}
Y_2^{-2}(\theta,\varphi) &= \frac{1}{4}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{-2\mathrm{i}\varphi}\cdot \sin^2\theta \\
Y_2^{+2}(\theta,\varphi) &= \frac{1}{4}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{2\mathrm{i}\varphi}\cdot \sin^2\theta
\end{align}$$
i.e. $m_l=\pm2$ equations.

$\mathrm d_{xz}$ and $\mathrm d_{yz}$ result from:
$$\begin{align}
Y_2^{-1}(\theta,\varphi) &= \frac{1}{2}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{-\mathrm{i}\varphi} \cdot \sin\theta \cdot \cos\theta \\
Y_2^{1}(\theta,\varphi) &= -\frac{1}{2}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{\mathrm{i}\varphi} \cdot \sin\theta \cdot \cos\theta
\end{align}$$
i.e. $m_l=\pm1$ equations.

From this assumption–which I may very well have presented incorrectly – the $\mathrm{t_{2g}}$ orbitals would therefore correspond to $m_l$ values of $\pm1$ and either $+2$ or $-2$. This would mean the greatest $\Lambda$ from one electron in each $\mathrm{t_{2g}}$ orbital (a $\mathrm d^3$ ground-state electron configuration), would be $\mathrm d_{xy} + \mathrm d_{xz} + \mathrm d_{yz} = \pm2+1-1=\pm2$, or a $\mathrm{D}$ term symbol.
Would anyone be able to correct me where my logic went wrong? I have a feeling I'm not allowed to restrict the $\mathrm{t_{2g}}$ orbitals to those $m_l$ values, but why would that not be allowed, when those are the equations that derive the $\mathrm{t_{2g}}$ d-orbitals?
Thank you!

The ground state of the free ion is $^4F$ but is $^4A_2 (t_{2g}^3)$ in a cubic field such as in an octahedral complex with $O_h$ symmetry.  This term is shown on the abscissa of the Tanabe-Sugano plot. Thus, even though there is an energy difference between the free ion and when it is in an octahedral  field, this is not shown in the plot. The lines representing the higher energy states measure the increase in energy from the ground state
The way to calculate the term symbol for the free ion is explained in detail in many textbooks and in my answer to How does one find the ground-state term symbol for a configuration that is exactly half-filled? .
Why the ground state term symbol is $^4A_2$ in an octahedral complex does need some explanation, however.
In  $O_h$ (and $T_d$) point groups the highest dimension of an irreducible representation is three-fold; Mulliken symbol T. As a result the states with orbital degeneracies greater than this e.g. $D, F, G..$ etc. must split into new terms of degeneracy no greater than three.
The calculation for S, P, D, F, G etc.  terms is outlined below with an example of F terms.
The effect that the symmetry imposed by the ligands has on the d-orbitals means that these have to be rotated, inverted or reflected according to the operations of the point group. This does not change the energy as it corresponds only to a change in the direction of the axes. Operating this way leads to a reducible representation that is then analysed to obtain its makeup as irreducible representations (irreps).
In $O_h$ the symmetry operations are $E, C_3, C_2, C_4, i, S_4,S_6, \sigma_h, \sigma_d$. The equations to use for rotation are shown in the notes below.
Applying these operations produces the following reducible representation for an F term with orbital angular momentum $L=3$.
$$ \begin{array}{c|cccccccccc}
O_h & E & 8C_3 & 6C_2 & 6C_4 & 3C_2 & i & 6S_4 & 8S_6 & 3\sigma_h & 6\sigma_d \\
\hline
\chi = & 7 & 1 & -1 & -1 &-1 & 7 & -1 & 1 & -1 & -1 \
\end{array}$$
Using the tabular method (see my answer to this question Understanding group theory easily and quickly ) produces the irreps $A_{2g}+T_{1g}+T_{2g}$. Thus an $F$ state splits into a non-degenerate $A_{2g}$ ground state and two triply degenerate states of higher energy.  The splitting of other terms (S, D, G etc) is determined in a similar manner.
As the d orbitals are inherently ‘gerade’ or g this subscript is usually dropped from terms in the Tanabe-Sugano plots. Unless spin-orbit coupling is exceptionally strong the spin of the final states is the same as that of the free ion.
The next table shows some free ion and $O_h$ terms.
$$ \begin{array}{clcr}
\text{Free ion} ~~&  ~~ O_h\\
\hline
S & A_{1g}\\
P & T_{1g}\\
D & E_g + T_{2g}\\
F & A_{2g} + T_{1g} + T_{2g}\\
G & A_{1g} + E_g + T_{1g} + T_{2g}\\
H & E_g + 2T_{1g} + T_{2g}
\end{array}$$
Using spherical harmonics to produce the energy splitting, which means calculating the potential energy and wavefunctions, is significantly harder and is only sketched.(See Balhausen, 'Introduction to Ligand Field Theory' for all the gory details.)
We suppose that the potential is caused by 6 charges around the central ion, and choose to use  the sum of spherical harmonics $Y_l^m$ to form the potential since these are solutions to a problem of full spherical symmetry. The general potential for i electrons is thus $V=\sum_i\sum_l\sum_m Y_l^m(\theta_i\phi_i)R_{nl}(r_i)$ where the R is the radial function which can be dropped from now on as a common factor.  The specific potential must transform as the totally symmetric representation of the point group of the molecule, ($A_{1g}$ in $O_h$) because the Hamiltonian must  remain totally symmetric under all symmetry operations. It turns out that only terms in $l=0, 2, 4$ can contribute to the potential. The $l=0$ term is the largest but as it is spherically symmetric it has little effect on electronic properties as it only shifts energy levels.  The $l=2$ harmonics produce irreps only of $E_g$ and $T_{2g}$, so are not suitable as the totally symmetric representation is absent, but the $l=4$ harmonics produce irreps of $A_{1g},~ E_g,~ T_{1g}$ and $T_{2g}$ which means that there is a linear transform of $Y_4^m$ that transforms as $A_{ig}$.
If the $C_4$ axis is taken as the axis to be quantised then the potential $V_4$ of $A_{1g} $ symmetry (excluding that from $l=0$) is proportional to a linear combination of the harmonics $V_4 \approx Y_4^0+b( Y_4^{+4}+Y_4^{-4})$, b is a constant. (These are the only harmonics that satisfy $\hat C_4 V_4=V_4$.)
To find the wavefunctions we use the fact that the d orbitals transform as $E_g$ and $T_{2g}$ in $O_h$. These can be combined to produce the familiar ‘real’ d orbitals shown in textbooks, $ d_{z^2}, d_{x^2-y^2} $ etc, by quantising along the $C_4$ axis.
The energy splitting $e_g-t_{2g}$ of a single electron in a d orbital e.g. $\ce{Ti^{3+}}$, is conventionally set to $\Delta = 10Dq$ and is positive.  The energy of each level  calculated as
$E_{eg} = \epsilon_0 +\int \phi^*(e_g)V\phi(e_g)d\tau$
and
$E_{t2g}=\epsilon_0 +\int \phi^*(t_{2g})V\phi(t_{2g})d\tau=-4Dq$
 where $\epsilon_0$ is the spherically symmetric part of the potential.  The energy gap is then $10Dq=E_{eg}-E_{t2g}$ and when all energy levels are filled with 10 electrons (an S state) then $0=4E_{eg}+6E_{t2g}$ from which $E_{eg}= 6Dq$ and $E_{t2g}=-4Dq$.
As the electron density of the $e_g$ orbitals are directed towards the ligands these are of higher energy than the $t_{2g}$.
Notes:
For quantum number k these relationships can be used with any point group because all point groups are subgroups of a sphere’s symmetry. Remember that $C_n$, is rotation by $2\pi/n$ radians.
$$
\chi(E)= 2k+1\\
\chi(C(x)) = \frac{\sin((k+1/2)x)}{\sin(x/2)}\\
\chi(i) = \pm (2k+1)\\
\chi(S(x)) = \frac{\sin((k+1/2)(x+\pi))}{\sin((x+\pi)/2)}\\
\chi(\sigma)=\pm \sin((k+1/2)\pi)
$$
The + sign is used with gerade, - with ungerade.

Free ion
The ground-state term symbol is only $\mathrm{^4F}$ in the case of a free ion. If you take a closer look at the Tanabe-Sugano diagram, the $\mathrm{^4F}$ term only appears at the far left-hand side of the diagram, where $\Delta = 0$. $\Delta$ refers to the ligand-field splitting parameter, and $\Delta = 0$ indicates that there is no ligand field, i.e. a free ion.
The quantum number $L$ (total orbital angular momentum) of the ground state can be obtained by coupling the individual orbital angular momenta of the d electrons using a Clebsch-Gordan series. The way to do this is described in most physical chemistry textbooks under the Russell-Saunders coupling scheme. For example in Atkins 10th ed. it is in page 386 under the chapter on "Atomic structure and spectra".
(Note that the symbol $\Lambda$ is used for diatomic molecules, not atoms.)
$L$ is said to be a "good" quantum number, in that the operator $\hat{L}^2$ (almost - this neglects spin-orbit coupling) commutes with the Hamiltonian $\hat{H}$. Quantum mechanically this means that the $\hat{H}$ and $\hat{L}^2$ (almost) share a set of eigenstates, and therefore for every state of the Hamiltonian (which correspond to the electronic configurations we are familiar with) one can (almost) calculate the corresponding value of $L$.
$$\hat{L}^2|\psi\rangle = L(L+1)\hbar^2|\psi\rangle$$
Octahedral complex
The ground state term symbol of a $(\mathrm{t_{2g}})^3$ ion is $\mathrm{^4\!A_2}$, not $\mathrm{^4F}$!
The $\mathrm{t_{2g}}$ set comprises the $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ and $\mathrm{d}_{xy}$ orbitals. These three d orbitals are what we call "real" spherical harmonics, which are linear combinations of the complex spherical harmonics which you have quoted. As such it is not possible to assign $m_l$ values like you have to the $\mathrm{t_{2g}}$ orbitals.
It's not correct to say that $\mathrm{d}_{xy}$ can have "either $m_l = +2$ or $-2$". That would mean that $\mathrm{d}_{xy}$, at any one point in time, is either equal to $Y_2^{+2}$ or equal to $Y_2^{-2}$, which makes no sense. It is not flip-flopping between the two spherical harmonics, it is its own thing: a linear combination of the two spherical harmonics, or a superposition, if you prefer that word. Furthermore, the spherical harmonics only have significance under spherical symmetry, where they act as simultaneous eigenstates of $\hat{H}$, $\hat{L}^2$, and $\hat{L}_z$. In octahedral symmetry, the spherical harmonics aren't of any significance at all and trying to "resolve" the $\mathrm{t_{2g}}$ orbitals into their components as spherical harmonics is physically unmeaningful (it helps with the maths, but that's all).
Under octahedral symmetry, the total orbital angular momentum $L$ is no longer a good quantum number (i.e. $\hat{L}$ no longer commutes with the Hamiltonian) and therefore the term symbol doesn't say anything about it!


text:
Hamiltonian: accounting for entropy
I might have a bit of a conceptual misunderstanding here, but how do we account for the entropy of a system when constructing a Hamiltonian? All entropy should be accounted, because the Hamiltonian includes the total energy, including the entropic contribution. However, when considering e.g. $\ce{H2+}$, the fact that the dissociated state is more entropically favoured doesn't go from the Hamiltonian.
Any suggestion that would clean up my understanding of the relationship between entropy and energy in this and other cases would be very helpful. Thanks!

The calculation only involves solving for total energy as a function of internuclear separation and so  bonding and antibonding potential energy curves are produced. If you want now to dissociate the molecule say from its lowest energy state then energy has to be added, say from a photon or collision and this is where the entropy enters. Similarly to get atoms from infinity to the lowest molecular energy level involves removing energy and changing entropy.

@porphyrin Thank you! (and sorry for a delay in my reply) But when we dissociate a molecule, we account for the electrostatic interaction between its two nuclei and the electrons, rather that for the probability of $H$ and $H^+$ being found close together (this is how I understand entropy). Does electrostatic interaction already include this?

I don't fully understand you last question; you seem confused so my answer is to define entropy and perhaps this will help. Entropy is best defined from statistical mechanics as $s=k\ln(\Omega)$ where k is the Boltzmann constant and $\Omega$ the number of different ways any energy levels can be occupied. In a molecule, there is entropy associated with vibrations and rotations. These are calculated via the partition function; see most phys. chem. texts for details.

@porphyrin thank you! I'm still a bit confused but I think it is clearer to me now!

One definition of entropy is that the number of microstates that makes up a macrostate influences the probability of observing the macrostate.  So your wavefunction needs to be multi-configurational and include both $\mathrm{H}_2^+$ and $\mathrm{H}^+ + \mathrm{H}$. The usual approach is to invoke the Born-Oppenheimer approximation, which makes the wavefunction depend on a single static set of nuclear coordinates, i.e. single-configurational. This approach will just give you the energy and you must compute reaction probability using statistical mechanics.
Multi-configurational SCF does include entropy in the following sense.  Let's say you have a molecule with a HOMO-LUMO gap that is sufficiently small to allow for significant mixing. You run a CASSCF(2,2) to determine the LUMO contribution and is 5%.  Now suppose you have the same system, same HOMO-LUMO gap, but the LUMO is doubly degenerate.  Your CASSCF(2,3) would now have a larger LUMO contribution, though I am not sure it would be 10%.

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text:
Vapor pressure lowering
I have got this problem in my text book,

Consider these three
  mercury manometers. One of them
  has 1 mL of water placed on top of the
  mercury, another has 1 mL of a 1 m
  urea solution placed on top of the
  mercury, and the third one has 1 mL of
  a 1 m NaCl solution placed on top of
  the mercury. Identify X, Y, and Z with
  these solutions.

I got the answer in answer sheet is X = ${H_2}O$ , Y = $NaCl$  and Z = ${(NH_2)}_2{CO}$ . Can any one please explain me? Thanks in advance.

Hint - https://en.wikipedia.org/wiki/Colligative_properties

@MaxW but i'm confused whether  Y = NaCl or not, i think  Z = NaCl is more preferable

NaCl splits into two ions thus has more colligative influence than urea which the molecule dissolves as one thing. Scale in drawing obviously greatly exaggerated.

@MaxW sorry i couldn't get it. Can you please write your answer with proper description ?

The answer hings on the colligative properties of solutes. In essence solutes lower the vapor pressure of the solvent. NaCl splits into two ions (i.e. two solute "particles") and lowers the vapor pressure of water more than urea for which the whole molecule dissolves as one solute "particle."
If there was no liquid the Hg level would be higher than any of the three tubes shown since there would be a pure vacuum above the Hg.
The scale of the relative differences in drawing is obviously greatly exaggerated.

Interesting!! The problem statement says urea and the answer does give empirical formula for glucose. I didn't catch that.

sorry edited..:) so now Z = urea ?

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text:
Neutralising sulfuric acid but not boric acid
I have another question here which refers to me asking how to separate boric acid from other compounds, and this is partially a follow-up.
Suppose I was to go down the route of making boric acid by the reaction of:
$$\ce{Na2B4O7·10H2O + H2SO4 → Na2SO4 + 4 H3BO3 + 5 H2O}$$
Which in actual fact, I just did 10 minutes ago and failed miserably.
The issue I'm having is that no matter what happens in a home-experiment, you're going to have contaminants, and in this case, they are the sodium sulfate and left-over sulfuric acid.
I added sodium bicarbonate in to remove the sulfuric acid which was stupid because I required so much but then realised that it will also neutralise the boric acid. So next I am going to attempt to neutralise the acid with sodium hydroxide.
But sodium hydroxide will also target the boric acid and just change it back into borax. Is there a way I can neutralise the sulfuric acid without touching the boric acid?

You've got to know that sulfuric acid is a strong acid and boric acid is a very weak acid. So, sulfuric acid can neutralize most of the bases where boric acid cannot. So, you have to take a base that can be neutralized by only sulfuric acid and not boric acid. So, you should probably take group 2 hydroxides and try to react with the unreacted sulfuric acid because sulfuric acid reacts with group 2 hydroxides whereas boric acid can't.
$$\ce{M(OH)2 +  H2SO4 -> M2SO4 + 2H2O}$$
$$\ce{M(OH)2 +  H3BO3 ->nothing}$$
where $\ce{M=Be,Mg,Ca,Sr,Ba}$.
So, which group 2 hydroxide to use? Group 2 hydroxide react with the sulfuric acid to form group 2 sulfates. Now, solubility of group 2 sulfates decreases down the group. So, you should use calcium hydroxide so the calcium sulfate that forms will not dissolve in the solution due to its low solubility and form a precipitate. You can separate that precipitate by filtration.
Formation of calcium borate is highly unlikely and so only calcium sulfate will only form.
Also, barium hydroxide is the best chemical for this process because barium sulfate is very insoluble. But barium hydroxide is costly if you buy it or making it is difficult due to the cost of other reagents. Instead, calcium hydroxide is your best bet.

In actual fact, its a good thing it forms a precipitate, because the water left over in solution will take in the boric acid, and therefore the salt can be filtered off. Should I use maybe calcium hydroxide?

Oh no, I'm just asking weather or not calcium hydroxide will cause something to precipitate out. I don't have any on hand at the moment and since I'm going to be doing a fair few experiments along these lines, I might need to buy or synthesize some.

Solubility of group 2 sulfate decrease down the group. So, if you use calcium hydroxide, calcium sulfate will be formed which is not soluble in water and so, you can separate that precipitate. I will edit the answer.

Great! Thanks alot. Now to go get some calcium hydroxide... :P

Sorry to be a pain, just doing some quick research, and you obviously know much more than me, is boric acid soluble in methanol? Half the MSDS's I find say only ethanol and the other half say yes. :/ Because after filtering out the calcium hydroxide, I'm thinking of just pouring methanol straight in, filtering the other stuff out again, and then just boiling it straight to dry boric acid.

Something's funky here.  Calcium and heavier G2 hydroxides are generally considered strong bases and I would argue that magnesium hydroxide is close to one.  I think they would neutralize the boric acid too.  Beryllium hydroxide might work in principle but beryllium compounds are horrendous from a safety/health point of view.

Sodium bicarbonate is actually a good choice. Bicarbonate ion is a stronger base than both sulfate and bisulfate ions, but a weaker base than borate ion (corresponding to boric acid being only a very weak acid). So the sodium bicarbonate will form sodium sulfate with the sulfuric acid leaving the boric acid unreacted.
Alkaline earth hydroxides, mentioned elsewhere, are not recommended here. Beryllium compounds are toxic and magnesium hydroxide, if dissolved by acid, may be too strong to avoid reaction with the boric acid. Heavier alkaline earth hydroxides would be stronger still.

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text:
Major product of a cyclopropanol with silver ions
I'm having trouble predicting the major product of the reaction of 1-(chloro(cyclohexyl)methyl)cyclopropan-1-ol and silver ions:

I thought that in presence of $\ce{Ag^+}$, the three-membered ring would open forming a ketone with chlorine atom in the $\alpha$-position. Then I thought that there might also be some migration of the ketonic group to the ring, but I'm not sure.

A common test for haloalkanes (save fluoroalkanes) is reaction with $\ce{AgNO3}$, which initially yields the corresponding silver halide and a carbocation intermediate. The formation of the insoluble salt is the driving force for the reaction:

On a primary haloalkane this carbocation would be quenched with water, but adjacent to a three-membered ring this carbocation is primed for ring expansion. In addition to reducing ring strain, the resultant carbocation is even further stabilized by the oxygen atom:

Deprotonation gives the final substituted cyclobutanone product.


text:
What is the difference between "atomic hydrogen" and "nascent hydrogen"?
My book (Comprehensive Chemistry by Dr. N . K. Verma, S. K. Khanna, Dr. B. Kapila) mentions two forms of hydrogen — "atomic" & "nascent".
It says that these two forms of hydrogen are more or less the same except that the former is produced by passing $\ce{H2}$ gas through an electric arc struck between two tungsten filaments and the latter is formed during chemical reactions in aqueous solutions. For example, $$\ce{Zn + H2SO4 -> ZnSO4 + 2[H]}$$
The book also said that the reactivity order of these hydrogens is
$$\ce{H2} \lt \ce{[H]} \lt \rm atomic~hydrogen$$
But I think  that the reactivity of $\ce{[H]}$  must be same as atomic hydrogen. So please explain the difference between them and also their reactivity order.

Is your book by chance very old? From the Wikipedia entry for "nascent hydrogen":

Nascent hydrogen is purported to consist of a chemically reactive form
  of hydrogen that is freshly generated, hence nascent. Molecular
  hydrogen ($\ce{H2}$), which is the normal form of this element, is unreactive
  toward organic compounds, so a special state of hydrogen was once
  invoked to explain certain kinds of hydrogenations. Mechanistic
  understanding of such reactions is now available, and the concept of
  nascent hydrogen is discounted, even ridiculed.

Then, by example of how this concept came to be:

Reductions of esters to give alcohols using a mixture of sodium and
  alcohols is called the Bouveault–Blanc reduction. It is an old
  reaction that has largely been superseded by alternative methods. At
  the time of popularity, the process caused much puzzlement because
  esters are unreactive toward hydrogen. It was also known that sodium
  reacts with alcohols to release $\ce{H2}$. it was concluded that some freshly
  generated ("nascent") hydrogen was responsible for this remarkable
  reaction. Subsequent studies have shown that this reaction proceeds
  via electron-transfer from metallic sodium to the ester substrate
  followed by protonation of the reduced intermediate. The evolution
  of hydrogen by the reaction of sodium and alcohol is purely a
  competitive reaction, the sole benefit being that in the presence of
  sufficient alkoxide, the sodium/alcohol reaction slows.

So in general, at least toward organic compounds, your relative reactivities of hydrogen species are correct if you take out "nascent" hydrogen so that:
$$\ce{H2} \lt \rm atomic~hydrogen$$

Then what is the reason behind more reactivity of the so-called nascent hydrogen?

I think better insights into Hydrogen state can be found here starting about page 13:
Title: The Chemical Elements and Their Compounds, Volume I, Sidgwick, 1950
Text: https://archive.org/stream/in.ernet.dli.2015.8077/2015.8077.The-Chemical-Elementns-And-Their-Compounds-Vol-i-1950_djvu.txt
Image: https://archive.org/details/in.ernet.dli.2015.8077/page/n7/mode/2up
Hydrogen atoms recombine to molecules rapidly and with a large
evolution of heat in contact with certain solids, especially metals.
Bonhoeffer has shown [20] that the efficiency of different metals in
causing the recombination of hydrogen atoms is almost exactly in the
reverse order to their overvoltage values, as the following list
shows; the metals are in the order of their catalytic efficiency, and
the overvoltage stands below each:
 Pt > Pd  >  W  > Fe  > Cr  > Ag  > Cu > Pb > Hg(zero)

0.000 0.000 0.157 0.175 0.182 0.097 0.19 0.40  0.57
This seems to show that the overvoltage is due to the slowness of recom-
bination of the hydrogen atoms after they have neutralized their ionic
charges at the cathode, and suggests that the activity of ‘nascent hydro-
gen’ is caused by the presence of neutral hydrogen atoms in the liquid.
The heat of recombination of the atoms (51.7 kcals. per gramme) has
been ingeniously utilized by Langmuir in his ‘ atomic blowpipe ’ in which
a stream of highly atomized hydrogen is directed on the metal to be
heated ; this has the further advantage that the hot metal is in a reducing
--
Monatomic Hydrogen pg 17
atmosphere [21] ; according to v. Wartenberg [22] the temperature of the flame
near the electrode, as judged by the reversal of the spectral lines in com-
parison with those of the sun, is 4,600-4,800°. From the heats of reaction
(H + H = H2 + 103.4 kcals. ; 2H2 + 02 = 2H20 + 116 kcals.) it follows
that for equal volumes atomic hydrogen is 34 per cent, more efficient than
‘Knallgas’, and for equal weights it is 24 times as effective.
In the gas the rate of recombination is much slower. A collision of two
hydrogen atoms cannot of itself lead to the formation of a molecule,
because the resulting pair cannot get rid of the energy of reaction, and so
must separate again ; it is only fruitful when there is a three-body collision
2H + M -> H2M -> H2 + M,
so that the energy can be removed as kinetic energy of the H2 + M (where
M of course may be the wall) ; this difficulty arises in all reactions of
association or addition. Various determinations of the rate of recombina-
tion of hydrogen atoms [23] have confirmed this view, and have shown that
the rate is that required by the triple -collision theory ; the half-life of the
atoms is about 1 sec. at 0-2 mm. pressure.


text:
Why do spin isomers of hydrogen (ortho and para hydrogen) change their nuclear spin with temperature variance?
My book says that ordinary dihydrogen contains 75% ortho and 25% para forms of hydrogen, while at significantly lower temperatures (like 20K) ortho and para hydrogens are 0.18% and 99.82% respectively. It also says that pure para form can be obtained at very low temperatures.
So what is the relation between the nuclear spins and temperature, and also why do the nuclear spin changes with temperature variance?

The effect is due to the symmetry properties of the rotational energy levels and those of the nuclear spin. The change with temperature is, as usual, traced back to the Boltzmann distribution.
In homonuclear diatomic molecules such as $\ce{H2}$ it is necessary to consider the spin of the overall wavefunction to exchange of nuclei. If this is $\Psi = \psi_t\psi_v\psi_R\psi_n$. The translational wavefunction $\psi_t$ depends on the coordinates of centre of mass, the vibration only on separation of nuclei. Thus we need only consider the product of rotational and nuclear wavefunctions, $\Psi = \psi_R\psi_n$. As there are identical nuclei the degeneracy is $g=2s_n+1$ with nuclear spin $s_n$ ($=1/2$) for hydrogen atoms.
As the product wavefunction $\Psi$ is neither odd nor even, it is necessary to ‘symmetrise’ by making the symmetric form in terms of states $\psi_1$ and $\psi_2$ for particle $(1)$ and $(2)$, as these also satisfy Schrodinger’s equation.
$$\psi_S=\psi_{1}(1)\psi_{2}(2)+\psi_{1}(2)\psi_{2}(1)$$
and asymmetric form
$$\psi_A=\psi_{1}(1)\psi_{2}(2)-\psi_{1}(2)\psi_{n}(1)$$
The number of symmetric functions is $g_n(g_n+1)/2$ and asymmetric $g_n(g_n-1)/2$ making a total of $g_n^2$.
The rotational wavefunctions are symmetric for even J rotational states and asymmteric for odd rotational quantum numbers J. As the product $\psi_R\psi_n$ must be asymmetrical for a molecule with an odd mass number for even J only asymmetrical nuclear states are accessible and  and vice versa.
The partition function for odd mass number is
$$Z= \frac{g_n(g_n-1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n+1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) $$
and for $\ce{H2}$ the partition function is
$$Z= 1\sum_{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) + 3\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) $$
From the equilibrium distribution equation the ratio of number of ortho- to para- molecules is therefore
$$\frac{N_{ortho}}{N_{para}} = \frac{ 3\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT))  } {   \sum_{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT))} $$
As the temperature changes so does this ratio; as the temperature tends to infinity the ratio tends to a limit of 3 and to zero as the temperature tends to zero.
For any spin degeneracy the ratio in the limit of very high temperature is
$$\frac{N_{ortho}}{N_{para}}=\frac{g_n+1}{g_n-1}$$
and for $\ce{D2}$ with spin of $1$ and $g_n=3 $ the ratio is $2$.
The figure shows the ratio for hydrogen.


text:
Surface-dependency of kinetic in speed law?
we can comprehend the quantitative dependency of the kinetic of a chemical reaction through the speed law: $v(t) = k \cdot c(...)^k \cdot c(...)^n$
Here I used the concentration $c$ for    that take place in aqueous phases. For the gas phase, I can use $v(t) = k \cdot p(...)^k \cdot p(...)^n $ All is fine, moreover I can convert $p$ into $c$ through $pV=nRT$
But what is with solid-solid reaction? Here the "reaction surface" is the crucial key word. But how can I express this in the speed law which I written above for aqueous and gas reaction? Or is this impossible for solid-solid reaction in this way?
Thank for your answers!

The expression you want will depend on surface concentration, normally in m$^{-2}$.
In practice the processes of adsorption of the reactants and desorption of products are also important and those are dependent on the surface area of the catalyst and the bulk concentrations (in solvent/solid surface heterogeneous catalysis) or partial pressure (in gas/solid surface heterogeneous catalysis).


text:
What is the purpose of a rate law?
Other than determining whether reactants' concentrations can affect the rate, what purpose does it serve? Does it have any real world application in chemistry research? Also what is the point of knowing that a reaction is "second-order"?
Thanks!

If you want to produce something in large scale, you may benefit from knowing how the reaction is affected by temperature, pressure, etc..
Besides, knowing which other reactions can occur may help you achieve a certain degree of optimisation in your process.
So knowing kinetics may mean making more money in this case.

Suppose you want to design an industrial scale continuous flow chemical reactor to produce some product stream from a set of feed streams.  You need to determine how big to make the reactor so that enough product is produced, given the reactant flow rates that you are feeding.  To figure this out, it's pretty important to know the rates of the reactions occurring within the reactor.  It's also important to know the amount of heat you have to add or remove from the reactor to keep it operating properly.  This is also determined by the reaction rates.
Another example is in atmospheric science where you need to know the rates of chemical reactions (including photolytic) occurring in the atmosphere so that you can quantitatively evaluate problems such as fluorocarbon/ozone and global warming.


text:
Why does the reduction of naphthalene with nickel/hydrogen lead to trans-decalin?
Why does the reduction of naphthalene using $\ce{H2/Ni}$ lead to trans-decalin instead of cis-decalin?
I learned in high school that catalytic hydrogenations are always syn. Why is this inconsistent with what I learned?

Because what you learned in high school is an oversimplification.  This statement from Chem. Absts.: ”The reduced NiO-MoO3/Al2O3 catalyst at 350° and 35 bar produced the highest decalin yield, i.e., 40 wt.% trans- and 15 wt.% cis-decalin”.

A similar question about this diagram is on ChemSE. Look here

@user55119 Even with that simplification ('all' instead of 'some' trans-decalin) the question still remains why trans-decalin is generated at all. My initial assumption is that the bridge is not hydrogenated in a single step which leads to steric effects causing the hydrogens on either end to be installed in different directions but I haven't checked the literature.

This is because trans decalin is more stable than cis decalin. In cis decalin, 1,3-diaxial interactions (which are repulsive in case of decalin ) destabilize roughly by 8.4kJ/mol than trans decalin.

I think to understand heterogenious hydrogenation you need to understand homogenious hydrogenation of metals by Wilkinson's cataylst and hydroformylation of alkenes by a mixture of hydrogen and carbon monoxide using a cobalt carbonyl cataylst. If we consider for a moment the reaction of an acetylene with a complex such as RuHCl(CO)(PPh3)2. Then what will happen is that the acetylene will bond to the vaccent site on the ruthenium to form a pi complex.

Next a rearrangement known as an insertion reaction will occur which will convet the acetylene into a vinyl group. The metal and the hydrogen which was originally on the metal will be attached to the alkene such that they will be cis to each other. This is the basis of why the hydrogenation of an alkyne normally gives the cis isomer with things like partly poisoned palladium.
If we consider for a moment the reaction in the oxo process then we have a hydrogenation with a little bit added onto it. Here is the cycle for cobalt.

The hydrogenation cycle for Wilkinson's cataylst is very similar, you are likely to be thinking of the same sort of thing occuring on the nickel surface. Based on what we know about the chemistry of acetylenes with hydride complexes we can feel confident that an isolated alkene or acetylene will undergo cis hydrogenation. Please see the following picture. What will happen with your substrate is that the very unreactive benzene ring will hydrogenate once to form a very reactive 1,3-diene  which will hydrogenate very quickly.

A lot of the selectivity for hydroformylation (linear vs branched product) relates to the energies of the alkyl complexes which exist in part of the cycle.
There are three homogenous hydrogenation mechanisms which exist but the one for Wilkinson's cataylst is the most common and it is the closest to what happens on a nickel or palladium surface.
Now for the 1,3-diene I am sure that as it goes through the hydrogenation process that at some point it will pass through an alkyl complex. This is likely to be in the final stage of the hydrogenation.

The lower energy one will be the main one in the reaction mixture. We need to be careful as the platinum gave the other product. The problem is that when we change from a first row to third row transition metal is the bond energies change. It is possible for the rate determining step to be different even if the catatyltic cycle looks the same. Consider the Monsanto process for the carbonylation of methanol vs the iridium process for the same reaction (even the same substrates) which BP invented later. There the RDS is a different step in the cycle.
I hope that I am not too heavy going, you have just had the associate prof level of answer for the problem.

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


text:
Sudden pressure rising in a pipe of an old plant transporting 33% ammonia solution
We pump 33% ammonia solution along a single pipe line before we introduce it to the mix. We've had two occasions now where the ammonia pressure has rapidly risen in the pipe. It happens when the fluid is at rest, meaning all valve shut and no pumps running. The pressure just escalates. The pipe was rinsed with recycled water, but this is a weak solution of what the ammonia is mixed with anyway.
Is there any material/condition that the ammonia solution could react with to cause this sudden over pressure?
Please note:

The ammonia tanks are outside (+1–2 degrees Celsius).
The internal temperature is (+15–19 degrees Celsius).
There is no pressure release system.
The system is 8yrs old.

Ammonia is more soluble at lower temperatures so the temperature change between the outside of the factory and the inside could raise the pressure inside the pipes.
Gas dissolved in water are metastable systems most of the times. Nucleation points in many cases suffice to promote the release of the gas from the solution (have you ever tried to put some salt in your beer?). It is possible that somewhere in the old pipes there are some areas with higher roughness (e.g. rusty connections) where some nucleation points promote bubble formation and so the ammonia can be released.

This certainly could be the case depending on the actual ammonia concentration (in other words how much it actually varies from 33%) and temperature variations.  At 25 degrees C ammonia is soluble to 31% and at 0 degrees C it s 47%.  So if his conditions are closely maintained to what he stated, the solution should be slightly sub-saturated and not "metastable".  Still, my guess would be that the conditions vary sometimes and then what you've described is what takes place.


text:
What single, pure compound would best replace petrol?
Petrol for use as automotive fuel is produced by blending of different product streams of oil refining. It consists of hundreds of different compounds, and the exact composition can vary widely. Certainly, not all components are absolutely necessary. Is there a single pure compound that could be used instead and that is similar or even superior to typical petrol?
In order to limit the scope of this question, I only consider pure hydrocarbons ($\ce{C_xH_y}$), i.e. no alcohols, ethers, amines, nitro compounds, or other compounds containing hetero atoms. I am only asking about fuel for spark ignition (Otto) engines, i.e. not about diesel fuel.
The relevant criteria for the comparison shall be only the knock resistance and the energy content of the compounds. Other properties may be ignored.
For similar compounds, there may be a tendency for fuel with a higher knock resistance to have a lower energy content. The challenge is to find a single pure compound with a suitable balance of knock resistance and energy content, with better values than found for typical petrol.
The knock resistance shall be expressed as research octane number (RON). Currently, the most important petrol blends in most countries have an RON of 95. However, blends with an RON of up to 102 are widely available. Therefore, the candidate compound should have at least an RON of 102.
The energy content shall be expressed as specific enthalpy of combustion $\Delta_\mathrm ch^\circ$ at a temperature of $T=25\ \mathrm{^\circ C}$. If necessary, this value can be directly calculated from the molar enthalpy of formation $\Delta_\mathrm fH_\mathrm m^\circ$ or the molar enthalpy of combustion $\Delta_\mathrm cH_\mathrm m^\circ$. For this purpose, the generated water shall be assumed to be in liquid form (although this assumption is not representative for vehicle engines and may favour small saturated compounds). The energy content of petrol can vary. For example, typical average values for $\Delta_\mathrm ch$ given by Aral in Germany are

Regular (RON = 91): $-44.2\ \mathrm{MJ/kg}$
Super (RON = 95): $-43.5\ \mathrm{MJ/kg}$
Super Plus (RON = 98): $-42.7\ \mathrm{MJ/kg}$

Therefore, the candidate compound should have at least an energy content corresponding to about $\Delta_\mathrm ch^\circ=-44\ \mathrm{MJ/kg}$.
In order to limit the choice, the candidate compound shall be a liquid at room temperature. The boiling point should be below $210\ \mathrm{^\circ C}$, preferably between $70\ \mathrm{^\circ C}$ and $150\ \mathrm{^\circ C}$. The melting point should be well below $0\ \mathrm{^\circ C}$.
By way of comparison, the following table shows the relevant parameter values for the reference compounds that are used for the octane number (heptane and 2,2,4-trimethylpentane) as well as for one candidate compound (2,2,3-trimethylbutane) that I have already found.
$$
\small\begin{array}{llllllll}
\hline
\text{Compound}&\text{Formula}&\Delta_\mathrm fH_\mathrm m^\circ&\Delta_\mathrm cH_\mathrm m^\circ&\Delta_\mathrm ch^\circ&\mathrm{RON}&T_\mathrm{mp}&T_\mathrm{bp} \\
&&\text{in}\ \mathrm{kJ/mol}&\text{in}\ \mathrm{kJ/mol}&\text{in}\ \mathrm{MJ/kg}&&\text{in}\ \mathrm{^\circ C}&\text{in}\ \mathrm{^\circ C} \\
\hline
\text{Heptane}&\ce{C7H16}&-224.2&-4817&-48.1&\hphantom{00}0&\hphantom{1}{-90.55}&98.4
 \\
\text{2,2,4-Trimethylpentane}&\ce{C8H18}&-259.2&-5461&-47.8&100&-107.3&99.22\\
\text{2,2,3-Trimethylbutane}&\ce{C7H16}&-236.5&-4805&-48.0&112&\hphantom{1}{-24.6}&80.86 \\
\hline
\end{array}$$
I would be interested in answers that search through literature data and find better candidate compounds as well as in answers that use a theoretical approach to estimate where to find the best compounds.

Our city has a some vehicles running on natural gas.  Also if the engines were modified to handle it, hydrogen could be used to fill a cylinder and a spark plug would surely ignite it.

It should also be noted that there is a plethora of other considerations: relative toxicity and carcinogenicity, cost to synthesize, melting/boiling points (for storage), etc.  When looking at a candidate, consumer safety and convenience will also be important.

If this is about fuel surrogates (compositions used to provide a repeatable, specific substitute for a fuel that's variable and tends to degrade) then most of them are mixtures of a small number of hydrocarbons.

Higher $\mathrm{RON}$ seems possible. Yet the boiling point rises accordingly while heat of combustion remains roughly the same. Here are two compounds that fit all posed criteria.
\begin{array}{|c|c|c|c|c|}
\hline
\mathbf{Molecule} & \mathrm{mp\ \mathrm{(^\circ C)}}  & \mathrm{bp\ \mathrm{(^\circ C)}} & \Delta_\mathrm cH_\mathrm m^\circ\ (\mathrm{kJ/mol}) \ \text{@}25\ ^\circ \mathrm{C} & \mathrm{RON} & \mathrm{MON}\\
\hline
\ce{1,3,5-TMB^{[a]}} & -44.7^{[1]} & 164.65(15)^{[1]} & -5193.1(1.3)^{[2]} & 137^{[3]} & 124^{[3]} \\
\ce{2,2,3,3-TMH^{[b]}} & -54^{[4]} & 160^{[5]} & -5405^{[6]} & 112.8^{[3]} & 92.4^{[3]} \\
\hline
\end{array}

[a] 1,3,5-trimethylbenzene, [b] 2,2,3,3-tetramethylhexane

Theoretical predictions of $\mathrm{RON}$
Tareq A. Albahri$^{[7]}$ summarises in his paper the usage of statistics $\mathrm{RON}$ and $\mathrm{MON}$.

The research octane number $\mathrm{RON}$, which is representative of the fuel performance during low-speed city driving, is more often reported in the literature than the motor octane number $\mathrm{MON}$, which is representative of the fuel performance during high-speed highway driving.$^{[7]}$

Theoretical intercalculation between the two characteristics is possible.$^{[7]}$ This work was done by Jenkins$^{[8]}$. First predictions of $\mathrm{RON}$ itself were and are still largely empirical$^{[7]}$. These models use regression techniques to assign effective octane numbers to various organic groups. Albahri examplifies the procedure via Anderson, Sharkey, and Walsh$^{[7]\ [9]}$.

Anderson et al.,$^{[9]}$ for example, developed an empirical model for calculating the RON based on chromatographic analysis of gasoline.$^{[7]}$
In their model, the gasoline is divided into 31 hydrocarbon groups or pseudocomponents, all of which are assigned an “effective” octane number that is estimated by regression of experimental data. The octane number of gasoline is calculated by adding the contribution of octane number[s] from each group.$^{[7]}$

Many similar methods include those proposed by Van Leeuwen et al.$^{[10]}$, Sasano$^{[11]}$, and  Lugo  et  al$^{[12]}$, Ramadhan and Al-Hyali$^{[13]\ [14]}$, Nelson$^{[15]}$, Baird$^{[16]}$,  Twu and Coon$^{[17]\ [18]}$, Rusin et al.$^{[19]}$, Habib$^{[20]}$, Cotterman and Plunkee$^{[21]}.$ Some of these fail to differentiate between isomers, or are valid in specific mixtures.$^{[7]}$ Additionally,

Although  these  techniques [by Van Leeuwen et al$^{[10]}$, Sasano$^{[11]}$, and  Lugo  et  al$^{[12]}$]  give reasonably accurate results, they are usually too time consuming for planning studies and often the compositional data are not available.$^{[7]}$

Tareq A. Albahri's method of SGC (structural group contribution)

After considerable testing, Albahri found that the best equation to take into account contributions of different groups is$^{[7]}$
\begin{split}
  \mathrm{ON} &=
  f\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^{-1}+ a + b\left(\sum_{i}\mathrm{\left(ON\right)_i}\right) \\
  &\quad + c\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^2 + d\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^3 + e\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^4
\end{split}
where $\mathrm{ON}$ is either $\mathrm{RON}$ or $\mathrm{MON}$, and $\left(ON\right)_i$ is the contribution from the $i$th group. Note how the powers increase from left to right.
The coefficients still have experimental basis.$^{[7]}$ This equation is a modified version of what Albahri put forward a year earlier.$^{[22]}$ For many compounds the difference between the estimate and experimatal data is just $5$.$^{[7]}$ If the $\left(ON\right)_i$ are known, one only needs the knowledge of the structure of the molecule to apply Albahri's method.
I suggest reading both papers by Albahri since they include tables of data and an example calculation.
Here is a selection of $\left(ON\right)_i$ values.$^{[22]}$

QSPR Model for octane number prediction

In a research article by Al-Fahemi, Albis, Gad, a quantitative structure-property relationship (QSPR) is performed. This includes establishing a correlation relation between $\mathrm{ON}$ and various physical parameters.$^{[23]}$ Again the model is based upon regression rather than derivation from first principles.
\begin{align}
\mathrm{ON} =
    & - (193.53 \pm 319.19) + (1.47 \pm 1.01)M\\
    & - (53.06 \pm 31.47)E_H - (8.67 \pm 2.73)B_P\\
    & - (24.94 \pm 19.44)M_R - (50.52 \pm 26.92)\log P\\
    & + (4.33 \pm 3.09)C_P + (3.72 \pm 2.04)C_V + (5.17 \pm 2.08)C_T\\
\end{align}
where

$M$ is molecular mass,
$E_H$ is hydration energy,
$B_P$ is boing point,
$\log P$ $-$ octanol/water distribution coefficient,
$M_R$ is molar refractivity,
$C_P$, $C_V$, $C_T$ is the critical point (pressure, volume and temperature, respectively).$^{[23]}$

TL; DR
There is no model to date that estimates or derives a result for $\mathrm{ON}$ from first principles. Different models depend more or less on multiparameter regression analysis and experimental data. Nevertheless, the methods of Albahri and others have predictive capability. Often the difference beteen estimated and experimental octane numbers is less than $5$.
Or as Albahri mentions,

Octane number is one of the most difficult properties to estimate or correlate because of its complex dependency on the molecular structure of the compound.$^{[7]}$
An estimation technique of the octane rating of pure hydrocarbons, though  essential, is nonexistent.$^{[7]}$

So for your second question, I suggest applying Albahri's method because

widest applicability among hydrocarbons,
relatively easy to use,
few $\mathrm{RON}$s are known from experiments,
Al-Fahemi et al.'s procedure requires many parameters which will be a pain to track down (if known at all).

Qualitative considerations$^{[3]\ [7]}$

shorter alkane chain $\ce{->}$ higher $\mathrm{ON}$ (but higher volatility)
branching of alkane $\ce{->}$ higher $\mathrm{ON}$
aromaticity in hydrocarbon $\ce{->}$ higher $\mathrm{ON}$ (but tends to burn sooty with carcinogenic byproducts)
effects on $\mathrm{RON}$ and $\mathrm{MON}$ can be very different

Arithmetic average of $\mathrm{RON}$ and $\mathrm{MON}$ $-$ anti-knock index or $\mathrm{AKI}$ $-$ is also used.
$$\mathrm{AKI} = \frac{\mathrm{RON} + \mathrm{MON}}{2}$$

$[1]$
T.F. Ardyukova, I.K. Korobeinicheva, A.I. Rezvukhin. Atlas of
Spectra of Aromatic and Heterocyclic Substances. $(1973)$, 7.

$[2]$
W. H. Johnson, E.J. Prosen, F.D. Rossini. 'Heats of combustion and
isomerization of the eight $\ce{C9H12}$ alkylbenzenes', Journal of
Research of NIST, $(1945)$, 35, pp 141$-$146.
$[3]$ A.
Demirbas, M. A. Balubaid, A. M. Basahel, W. Ahmad, M. H. Sheikh.
'Octane Rating of Gasoline and Octane Booster Additives'. Petroleum
Science and Technology, $(2015)$, 33(11), pp 1190$-$1197. DOI:
10.1080/10916466.2015.1050506.

$[4]$
Jean-Claude Bradley Open Melting Point Dataset. (January 29, 2017)

$[5]$
R. Stenutz. 2,2,3,3-tetramethylhexane. A collection of tables for
chemistry. (January 29, 2017)

$[6]$
Pure Components Data. (Excel file.) Petroleum Engineering, Texas A&M
University. (January 29, 2017)
$[7]$ Tareq A. Albahri. 'Structural Group Contribution Method for Predicting the Octane Number of Pure Hydrocarbon Liquids'. Industrial & Engineering Chemistry Research $2003$, 43(24). DOI: 10.1021/ie020306+.
 $[8]$ G. I. Jenkins. 'Calculation of the motor octane number
from the research octane number'. Journal of the Institute of
Petroleum, $1968$, 54, 14.

$[9]$
Anderson, P. C.; Sharkey, J. M.; Walsh, R. P. 'Calculation of  the
Research  Octane  Number  of  Motor  Gasoline  from  Gas
Chromatographic Data and a New Approach to Motor Gasoline Control'.
Journal of the Institute of Petroleum, $1972$, 58, 83.

$[10]$
J. A. Van Leeuwen, R. J. Jonker, R. Gill. 'Octane Number Prediction
Based on Gas Chromatographic Analysis and Nonlinear Regression
Techniques'. Chemometrics and Intelligent Laboratory Systems,
$1994$, 25(2), pp 325$-$340. DOI: 10.1016/0169-7439(94)85051-8.

 $[11]$  Y. Sasano. 'Measuring Research Octane Number of
Gasoline by Gas Chromatograph'. JP Patent 09138613, $1997$.
 $[12]$ H. J. Lugo, G. Ragone, J. Zambrano. 'Correlations
between Octane Numbers and Catalytic Cracking Naphtha Composition'.
Industrial & Engineering Chemistry Research, $1999$, 38, 2171.
 $[13]$ O. M. Ramadhan, E. A. S. Al-Hyali. 'New Experimental and
Theoretical Relation to Determine the Research Octane Number (RON) of
Authentic Aromatic Hydrocarbons that Could be Present in the Gasoline
Fraction'. Petroleum Science and Technology, $1999$, 17, 623.

 $[14]$ O. M. Ramadhan, E. A. S. Al-Hyali. 'Aromatic
Hydrocarbons in Some Iraqi Gasoline and Their Influence in the Value
of the Research Octane Number'. Petroleum Science and Technology,
$1999$, 17, 607.
 $[15]$ W. L. Nelson. 'Octane Numbers of Naphthas'. Oil and Gas
Journal, $1969$, 67, 122.
 $[16]$ C. T. Ch Baird. Address: De  la  Haute-Belotte  6,  1222
Vezenaz, Geneva, Switzerland. Unpublished results.
 $[17]$ C. H. Twu, J. E. Coon. 'Predict Octane Numbers Using a
Generalized Interaction Method'. Hydrocarbon Processing, $1996$, 71,

$[18]$
C. H. Twu, J. E. Coon. 'Estimate Octane Numbers Using an Enhanced
Method'. Hydrocarbon Processing, $1997$, 76, 65.
 $[19]$ M. H. Rusin, H. S. Chung, J. F. Marshall. 'A
Transformation  Method  for  Calculating  the  Research  and  Motor
Octane Numbers of Gasoline Blends'. Industrial and Engineering
Chemistry Research. $1981$, 20, 195.
 $[20]$ E. T. Habib. 'Effect of Catalyst, Feedstock, and
Operating Conditions  on  the  Composition  and  Octane  Number  of
FCC Gasoline'.  ACS  Symposium  Division  of  Petroleum  Chemistry,
Miami, FL, Sept $1989$.
 $[21]$ R. L. Cotterman, K. W. Plunkee. 'Effects  of  Gasoline
Composition  on  Octane  Number'.  ACS  Symposium  Division  of
Petroleum Chemistry, Miami, FL, Sept $1989$.
$[22]$ Tareq A. Albahri. 'Structural Group Contribution Method for Predicting the Octane Number of Pure Hydrocarbons and Their Mixtures'. Fuel Chemistry Division Preprints, $2002$, 47(2), 532
$[23]$ Jabir H.
Al-Fahemi, Nahla A. Albis, Elshafie A. M. Gad. 'QSPR Models for
Octane Number Prediction'. Journal of Theoretical Chemistry,
$2014$, vol 2014, article ID: 520652, 6 pages. DOI: 10.1155/2014/520652.

Even longer branched hydrocarbon chains or benzene derivatives will probably have higher $\mathrm{RON}$s. Finding data gets harder by each carbon atom though.

To reiterate, I think Tareq A. Albahri's method is the best we have got so far.


text:
Does the Lewis acid have to match the halogen in electrophilic halogenation of benzene
$\ce{FeBr3}$ catalyzes bromination and $\ce{AlCl3}$ chlorination. I'd like to know whether bromination and chlorination can be done in sequence without changing the catalyst? So - will using the $\ce{FeBr3}$ for chlorination and $\ce{AlCl3}$ for bromination work? Is it less efficient or slower or is there another reason?

The reason you use the respective halides for halogenation is that the halogen and Lewis acid can inter-react to form a chemical equilibrium if you don't.
$$\ce{FeBr3 + Cl2 <=> Cl+ + [FeBr3Cl]- <=> Br+ + [FeBr2Cl2]- }\\
\ce{AlCl3 + Br2 <=> Br+ + [AlBrCl3]- <=> Cl+ + [AlBr2Cl2]- }$$
Given these equilibrium reactions you can see that now you have two electrophile species which means your halogenation is not as controlled. It will work and halogenation will still occur, but now and you will have bromination and chlorination occurring in competition with each other when you only wanted one or the other.


text:
Is Elemental Sulfur a Lewis Acid or Lewis Base?
Is pure sulfur a Lewis base or a Lewis acid?
I am aware of the general concept behind what a Lewis base is (electron donor) and Lewis acid (electron acceptor).
This await on confuses me because I thought elemental sulfur could be either depending on the context in which it is found( I.e. What specific molecule it's found in)
I'm not sure if it is a Lewis acid or base when in elemental form.

Heat sulfur under hydrogen to form $\ce{H2S}$, and it's a Lewis base.
Heat sulfur under oxygen to form $\ce{SO2}$, and it's a Lewis acid.
In other words, its amphoteric.  It can act as either a Lewis acid or base depending on its environment.

I don't think these are good examples.

@Mithoron , appreciate the input, but could you elaborate?

These are good examples of S acting as oxidant and reluctant. Action as Lewis acid/base is debatable.


text:
What happened to mesoporous materials for drug delivery?
A review on the topic can be found here: Wang, S. Ordered mesoporous materials for drug delivery. Microporous Mesoporous Mater. 2009, 117 (1–2), 1–9. DOI: 10.1016/j.micromeso.2008.07.002.
So the 21st century started with great hopes for these mesoporous materials - particularly for drug delivery. After initial success it seems the research in this field has fizzled out.  I'm just curious to why there hasn't been more progress?  Thanks so much!

I think that research of mesoporous materials for several uses is alive and well.  The link referenced within your question is for an article written in 2008.  Wikipedia has a good, though brief, article on its current research status without respect to medical use in particular:

A procedure for producing mesoporous materials (silica) was patented
  around 1970, and methods based on the Stöber process from 1968 were
  still in use in 2015. It went almost unnoticed[8] and was reproduced
  in 1997. Mesoporous silica nanoparticles (MSNs) were independently
  synthesized in 1990 by researchers in Japan. They were later produced
  also at Mobil Corporation laboratories and named Mobil Crystalline
  Materials, or MCM-41.

This is an abstract available from PubMed published in 2007:

Research on mesoporous materials for biomedical purposes has
  experienced an outstanding increase during recent years. Since 2001,
  when MCM-41 was first proposed as drug-delivery system, silica-based
  materials, such as SBA-15 or MCM-48, and some metal-organic frameworks
  have been discussed as drug carriers and controlled-release systems.
  Mesoporous materials are intended for both systemic-delivery systems
  and implantable local-delivery devices. The latter application
  provides very promising possibilities in the field of bone-tissue
  repair because of the excellent behavior of these materials as
  bioceramics. This Minireview deals with the advances in this field by
  the control of the textural parameters, surface functionalization, and
  the synthesis of sophisticated stimuli-response systems.

One possible competitor for sustained drug delivery is transdermal drug delivery.  It seems that this method, which has been used for centuries, is well established and thoroughly approved by agencies like the FDA in the US.  In the US anyway, approval of new medicine or medical devices happens very slowly, and mesoporous materials seem to still be more in the research phase and may not have the level of agency approval that competing systems have attained.


text:
Significance of phase of atomic orbitals
I am learning about orbitals and bonding and antibonding MOs. So far, I know that when you combine s orbitals, you form both bonding and antibonding MOs- bonding when the wavefunctions are added in phase and antibonding when added out of phase. I have had several questions along the way:

What is the significance, if any, of the phase of the wavefunction for electrons in atoms when the atom is not interacting? Suppose you had a free hydrogen atom in space. Would it make a difference if you switched the phase of its 1s orbital? What about if you had a multi-electron atom? Would there be a consequence of switching the phase of the 1s orbital but not of the other orbitals? My guess is that, if this is at all possible, then yes this would make a difference due to the penetration of higher orbitals- i.e. the 2s orbital has some penetration into the 1s orbital. Thinking about it in this way, I think atoms are most stable when all the s orbitals are in the same phase and the p orbital phases are aligned. Otherwise you would get cancellation of the wavefucntion in regions closer to the nucleus so the electrons would be found further out, and therefore would have a higher potential energy. There would also be more localisation of electrons further from the nucleus so they would have a higher kinetic energy too. I think it therefore would be significant when we change the phase of orbitals relative to other orbitals (whether in the same atom or in interacting atoms), but if you changed the phase of all wavefunctions in the universe right now then it wouldn't matter.

My second question:

How do we form the bonding and antibonding MOs? Is it becaue each orbital is actually in a superposition of being 'phase 0' and 'phase pi' which is why you get both bonding and antibonding MOs? I'm just not sure if superposition of states is at all important here, or if I have misunderstood this.

Finally, when I was considering p orbitals, I noticed that th phase of the two lobes in a single p orbital is opposite.

So here we have two opposite phases existing together. Is there a significance of this? Or is it perhaps again only important in the interaction of p orbitals?

You have quite a few questions indeed!

What is the significance, if any, of the phase of the wavefunction for electrons in atoms when the atom is not interacting? Suppose you had a free hydrogen atom in space. Would it make a difference if you switched the phase of its 1s orbital?

No, see this question: Physical intuition behind negative values for wave function? and the answers within.
The phase is not related to any of the physical properties of the atomic orbital. For example the probability density of the electron in the orbital is given by $|\psi|^2$. If we write $\psi_\mathrm{ps} = \mathrm{e}^{\mathrm ix}\psi(r,\theta,\phi)$ ("ps" for phase shifted) it is easy to see that the probability density is independent of the phase chosen $x$:
$$|\psi_\mathrm{ps}|^2 = \psi_\mathrm{ps}^*\psi_\mathrm{ps} = \mathrm{e}^{-\mathrm ix}\psi^*(r,\theta,\phi)\mathrm{e}^{\mathrm ix}\psi(r,\theta,\phi) = |\psi^2|$$
so the phase shift has not done anything. Likewise, the energies etc. are independent of any constant factor that is placed in front of the wavefunction. This is easy to show using
$$E = \frac{\langle \psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}.$$

What about if you had a multi-electron atom? Would there be a consequence of switching the phase of the 1s orbital but not of the other orbitals? My guess is that, if this is at all possible, then yes this would make a difference due to the penetration of higher orbitals- i.e. the 2s orbital has some penetration into the 1s orbital. Thinking about it in this way, I think atoms are most stable when all the s orbitals are in the same phase and the p orbital phases are aligned. Otherwise you would get cancellation of the wavefunction.

In hydrogen the 1s and 2s orbitals are orthogonal to each other (in general all AOs are orthogonal to all other AOs). This is because the 2s orbital has an inner radial node and two regions which are phased differently (conventionally, positive + negative; however you can draw it negative + positive and the overlap would still be zero).

How do we form the bonding and antibonding MOs? Is it becaue each orbital is actually in a superposition of being 'phase 0' and 'phase pi' which is why you get both bonding and antibonding MOs?

The absolute phase of the orbital has nothing to do with it, but rather the relative phase. See the linked question above.

So here we have two opposite phases existing together. Is there a significance of this? Or is it perhaps again only important in the interaction of p orbitals?

I'd be inclined to say that there's no real significance, except that that's how the maths works out. More fundamentally the presence of a node (in this case an angular node) means that there must be regions of different phases. The same applies to the 2s orbital which has a radial node instead of an angular node. This is analogous to the solutions to the particle in a box: the ground state solution has no nodes and is therefore positive throughout, but the first excited state has one node and therefore has one positive and one negative region.


text:
Change in standard Gibbs energy is always zero?
So I have this statement from our professor :

We have $\Delta G^\circ = -RT\ln K$ and under standard conditions the activities of all substances in the reaction is $1$ ($a=1$), therefore $\Delta G^\circ > 0$.

I dont understand it.If every $a=1$, then $K = 1$ (as the division of activities) and thus $\ln K =0$ which would make $\Delta G^\circ = 0$ as well instead of $>0$,right?

Welcome to chemistry.SE!. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. If that is really the statement given, then it is wrong indeed. The main thing is that $^\circ$ points to standard state which does not include a rule that activities must all be one. See this.

No. $K$ is not a ratio of instantaneous activities! Let's say you hypothetically prepare a system where all components have unit activity. In that case you would have the reaction quotient
$$Q = \prod_i a_i^{\nu_i} = 1$$
but $K$ is the ratio of activities at equilibrium. If that hypothetical system is not at equilibrium, then it will go towards equilibrium. When it reaches equilibrium, the activities are no longer going to be equal to unity, and $K$ will not be equal to 1.

The complete equation is $$\Delta G=\Delta G^\circ+RT\ln{Q}$$If $Q=1$, then $$\Delta G=\Delta G^\circ$$Maybe your professor meant to conclude that $\Delta G \neq 0$ if all the activities are equal to 1.  It's hard to know what he meant to say.


text:
What is the nature of interstitial hydrides?
In my Chemistry book(NCERT Chemistry,Part 2, textbook for class 11, Pg 281). I found the mention of interstitial hydrides:

Unlike saline hydrides these compounds are nonstoichiometric, for
  example $\ce{LaH_n, TiH_n}$ and $\ce{PdH_n}$, where the chemical composition is variable. Typical formulae are $\ce{LaH_{2.87}, YbH_{2.55} , TiH_{1.8}, ZrH_{1.9}, VH_{1.6}, NbH_{0.7}}$ and $\ce{PdH_{0.7}}$.

The book also said:

Earlier it was thought that in these hydrides hydrogen occupies the
  interstices in the metal lattice producing distortion without any
  change in it's type. However recent studies have shown that except for
  hydrides of $\ce{Ni, Pd, Ce,  Ac}$ other hydrides of this class have
  lattice different from the lattice of the parent metal.

What does this mean? Please explain the nature of these compounds and type of bond present in them.

Nonstoichiometric compounds are simply compounds that cannot be represented with integer number of atoms. What the book probably is trying to say is that the hydrogen atoms do not simply fill the empty space between the atoms in the crystal. In other words, the hydrides will cause a change in the structure of the lattice. In some compounds, the atoms that are being inserted in the free space in the crystal will not cause any noticeable change in structure. The crystal may become slightly distorted. But it will retain its original shape before insertion of the new atoms. What the book is trying to say is that this not true for hydrides. Scientists used to think that this was the case, but it is not true.


text:
How many grams of CuSO4*5H2O yield 10 grams of H2O?
If I want there to be 10.00 g of water in the copper (II) sulphate pentahydrate sample that I'm measuring, what mass of the substance will I need in order to have 10.00 g of water?
I solved this question by...
$$\ce{CuSO4.5H2O -> CuSO4 + 5H2O}$$
Therefore, since we get five waters for each copper (II) sulphate pentahydrate,
we would only need 2 copper sulphate pentahydrates.
Thus, i think I'll need 2 g of the substance but is it correct?

You are violating conservation of mass!  Take a close look at the chemical equation. Is it units or moles that you should be considering?

You need to think about moles first and molecular weights second in order to solve your problem. Although the decomposition reaction you wrote is correct, you dont' need it to work this problem.
(1) The molecular weight of $\ce{CuSO4\cdot 5H2O}$ is 250 $\mathrm{g\cdot mol^{-1}}$.
(2) Each mole of $\ce{CuSO4\cdot 5H2O}$ yields 5 moles of $\ce{H2O}$, and each mole of $\ce{H2O}$ has a molecular weight of 18 $\mathrm{g\cdot mol^{-1}}$.
(3) For every mole of $\ce{CuSO4\cdot 5H2O}$ you have, there are $5\,\mathrm{mol}\times 18\,\mathrm{g\cdot mol^{-1}} = 90\,\mathrm{g}$ of $\ce{H2O}$ present (the $\mathrm{mol}$ terms cancel here).
(4) You want $10\,\mathrm{g}$ of $\ce{H2O}$, or $10/90 = 1/9$ of the amount you get from 1 mole of $\ce{CuSO4\cdot 5H2O}$, which is the same as saying that $1/9$ $\mathrm{mol}$ $\ce{CuSO4\cdot 5H2O}$ gives you $10\,\mathrm{g}\,\ce{H2O}$.
(5) Finally, you need $250/9 \approx 27.8\,\mathrm{g}$ of $\ce{CuSO4\cdot 5H2O}$ to yield $10\,\mathrm{g}$ of $\ce{H2O}$.


text:
About dative covalent bonding and expanded octet rule
I'm beginning my studies of chemistry at high school (I'm not a native English speaker, though) and I have some basic questions:
1)  Consider the compound xenon tetroxide. In this molecule, does the xenon have its valence layer expanded? In another words, the oxygen atoms only have double covalent bonds with the Xenon?
I've tried to look for this answer but I'm still in doubt whether xenon does four dative covalent bonds (with each oxygen) or four double covalent bonds. In the first case, does not occur the expanded octet rule for xenon, right?
2) Another question; is it possible for a dative covalent bond to occur with just one electron? I mean, the atom which has more electrons in its valence layer "donates" just one electron instead of two? If the answer is "yes", can you provide me an example, please?
I'm sorry for the English mistakes. I started to learn this "chemistry vocabulary" just now.

First of all, welcome to CE.Chemistry!  Don't worry too much about your English, it doesn't look too bad to me. There are some capitalization a couple grammar issues that I think I should just fix for you to look at rather than turning this comment into an essay on linguistics ;)  Also, have you done any online searches for "dative covalent bond"?  You might try that and discuss briefly your thoughts on it within your question.

Please do not be discouraged by how many edits I made to your question.  It was pretty understandable as it was, but I made more edits than I normally would have just because you stated that you are studying English and I thought that would be helpful.  If I inadvertently changed your meaning anywhere please accept my apologies and just change it back.

@airhuff I've done some google research, but I havent found a conclusive answer. For instance, in the first question, Ive found two structural formulas for xenon tetroxide, one of them was wih dative covalent bonds, and another with double covalent bonds.. So, I dont know which is the answer... (or maybe this compound can occur with two different bonds...)

For a description of the bonding in $\ce{XeO4}$ see this earlier answer,

To answer your first question: the concept of octet expansion for main group metals is no longer generally considered correct and only upheld by the more traditional organic chemists that don’t deal with inorganic bonding situations on a day-to-day basis — or anybody else who didn’t need to recheck what they were taught decades ago. Nowadays, these seemingly octet-expanding structures such as $\ce{XeO4}$ or $\ce{XeF2}$ are explained by dative bonds and/or four-electron-three-centre bonds; both have the beauty of not requiring d orbitals (whose energy would be comparable to the next shell’s s orbital; remember the aufbau principle!).
To answer your second question: no. A dative bond always requires two electrons that are given to one atom upon supply of the bond dissociation energy. It doesn’t matter how a certain bond is created, it only matters how it dissociates; and if it dissociates and was a bond, it will have to have had an electron pair.

So, if I understand you correctly ( my studies of chemistry started  last week), the xenon, in XeO4, does four dative bonds? I only know dative bonds and ''normal'' covalent bonds...

@gustavoreche Dative and ‘normal’ covalent bonds are two sides of the same picture. But yes, there’s a single bond between the two and charge separation ($\ce{\overset{+}{Xe}-\overset{-}{O}}$). These bonds can be called dative bonds, if you wish.


text:
What is the name of the CO2Na group?
I'm trying to identify the name of every functional group sticking off the sides of tartrazine (FD&C Yellow No. 5), and the one I can't find anywhere is this $\ce{COONa}$ group. Does anyone know how to describe it? Is it really two groups put together?

This is simply the salt of a carboxylic acid and sodium hydroxide, with the general reaction looking something like this:

The names of carboxylate salts can be derived from that of the original carboxylic acid and the base used, "-oic acid" becoming "[metal] -oate" (e.g. benzoic acid reacting with sodium hydroxide to become sodium benzoate). If part of a more complex molecule, the $\ce{COO-}$ group is treated like any other substituent and given the name "carboxylate".
This can be seen in the name of tartrazine:
trisodium [...]-3-carboxylate, as well as the name of the sodium salt of citric acid:
trisodium 2-hydroxypropane-1,2,3-tricarboxylate.


text:
How to test for lead and nitrates in water
My son is doing a science fair project and I need to know what chemicals I can use to test water contaminates. What chemicals can test for lead and nitrates/nitrite?

Welcome to SE.Chemistry!  Have you done any online searching for your answer?  You may find your answer pretty quickly that way.  Also, if you can include ideas that you have found yourself, that shows a degree of effort on your part. This will make it more likely that people here will invest there time and efforts to help you further.

Searching a prominent online store for "lead test kits" I found products for a little as $12 US that test drinking water for lead, nitrites/nitrates and more.  This would be a semi-quantitative solution.  Your question doesn't state what your specific goals and requirements are; is something like this what you are looking for?

Since this is a science fair experiment, you will need only common chemicals to test the contaminants in water.

Lead

You can use hydrogen sulfide, $\ce{H2S}$ to test lead. (Warning: it has rotten egg smell.) It is a common reagent and can be found in laboratory. You can also use sodium sulfide instead but it too has rotten egg smell.
$$\ce{Pb^2+ + H2S -> PbS + 2H+}$$
Due to the insolubility of lead sulfide in water ($\ce{4.9 \times 10^{-11}g l^{-1}}$), hydrogen sulfide test is such a sensitive test for the detection of lead and also can be detected in filtrate from separation of sparingly soluble lead chloride and other salts and hydrochloric acid.

Nitrate

You can perform the brown ring test because this test is very sensitive to nitrates in solution. You need conc. sulfuric acid and ferrous sulfate. A brown ring is formed at the junction of  two layers probably due to formation of $\ce{[Fe(NO)^2+]}$.
$$\ce{2NO3- + 4H2SO4 + 6Fe^2+ -> 6Fe^3+ + 2NO^ + 4SO4^2- + 4H2O}$$
$$\ce{Fe^2+ + NO^ -> [Fe(NO)^2+]}$$
Sensitivity: $\ce{2.5 \mu g}$ ;
Concentration limit: 1 in 25000

Nitrite

Similar to brown ring test but uses dilute sulfuric acid.

Extra info.
The regulatory standard method for testing for lead in water uses an Atomic Absorption (AA) spectrophotometer or XRF machines(cost $30,000)(source)
For testing nitrate/nitrite in water, see here.
Other very specific and expensive test used to detect contaminants are Gallocyanin test and diphenylthiocarbazone test for lead, diphenylamine test and nitron test for nitrate and sulfanilic test and indole test for nitrite but I won't elaborate them since it is a science project. :)

Thank you so much! How much of each chemical do I need to be able to test?

The complex is $\ce{[Fe(H2O)5(NO)]^2+}$

If you're looking for characteristic reations, two come to mind (unfortunately I do not know what chemicals you have access to).
Nitrates can be tested for with the "brown ring test", sulfuric acid and iron(II) sulfate (see here).
For lead you can test with the "golden rain" reaction using any iodine salt that is soluble in water (see here, they're using lead(II) nitrate in this example).
Depending on what else is in your water solution, you might need to divide it into several solutions of ions from the 3 distinct analytical groups that are generally discussed in literature. You can do this using distinct group reagents for each of the groups. If that's what you're looking for, I suggest reading up on it on the internet, it's a big topic but there's plenty of resources covering it quite well.

Drinking water contaminates are in the parts per million range. I highly doubt that either of these tests has enough sensitivity.

If your son has to perform this experiment in the science fair, I'm sorry to disappoint you but it will not be possible (only if his school has adequate instrumentation such as an Atomic Absorption Spectrophotometer or a Mass Chromatograph. Contamination of water by lead is in the order of parts per billion (1 part of lead per 1 billion parts of water), and although it is sometimes enough to cause health problems, it is indeed a very low amount of lead to be detected by simple experiments such as the reaction with potassium iodide. And I wouldn't recommend leaving a teenager with a big amount of lead in hands because of its extreme toxicity.
To test for nitrates, the brown ring experiment cited by Diego Zamora can possibly provide a visual result in water, since the nitrate concentration in it is not that small. However, test it before with your son so he don't get frustrated if the experiment goes wrong during the fair. Concentrated sulphuric acid can be found as car battery liquid in some specialized stores (is not very difficult to find). Ferrous sulphate can be found in pool products stores. Handle it with extreme care since sulphuric acid is corrosive and a powerful dehydrating compound that can cause severe burns if in contact with skin.


text:
Flame color vs electrical discharge color
I don't know the right terms to frame the question but here it goes:
Neon and argon give their characteristic colors when a current is passed through them at low pressure.  Balmer series etc, I understand what's going on there.
Other compounds like boron will give a nice green in a flame test.  Boron in alcohol for example.  But in an electrical discharge, nothing happens.
I'm not understanding where the difference is in the manner of exciting the boron atoms.
How can I find a compound that will give a strong green color with an electrical discharge, at low pressure (1 torr)?

I don't know if you could produce atomic oxygen in your system, but this article may (or probably may not) give you some ideas along those lines: http://www.ann-geophys.net/30/695/2012/angeo-30-695-2012.pdf

It would be a sealed tube like a neon light.   I thought of Oxygen, but I don't think it would be a good idea, my only way to get the low pressure is with a vacuum pump.  Oil, oxygen. :(  Seems risky.

There are certainly pumps available that would be safe, but you know your equipment and yea, don't go there if that could be an issue, :( indeed.

Boron is not a gas. Related http://chemistry.stackexchange.com/questions/53281/compound-with-a-bright-green-emission-when-electrically-excited

Boron compounds can be.

Here is an interesting interactive site that gives the emission spectra for eight different elements.  None of them would give you a pure green without filtering other emission wavelengths, say with a coating on your tube.  Neon has good emission in the green range for example but the red is so intense I doubt you could filter it out and have any significant green left.  Anyway, here's the site: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/atspect2.html#c1

@user103218 That's why trimethyl borate colors the flame green. It brings boron atoms to the gas phase where they can be excited and emit green light. However, this is impractical in a gas discharge tube.

Ok, why is trimethyl borate impractical?   I haven't gotten it to work yet.  I don't think it matters how the electrons get excited, and slamming high voltage current through them seems like it ought to be working.

I think trimethyl borate will be unstable under the harsh conditions in a gas discharge tube.

Ok, so how do I go about finding a compound that is a gas at around 1 torr, which will give me a bright green emission line without much else, when electrically excited?


text:
When we say "the energy in a system" (in chemistry) do we mean the energy stored in the chemical bonds in the compounds present in that system?
Is it the energy stored in the chemical bonds of the compounds present in a system, that we refer to as the energy in/of a system ?

This is a duplicate of a recent question, which I cannot locate right now. The correct answer was given by @Ivan in the comments thread. The net of it is: draw a box/cube around your "system" and add it all up. That's the energy. The rest is the surroundings/

@ToddMinehardt, I just came across it here.

The answer to this question depends on the nature of the system we have defined.
Firstly, to define the term system according to this Wikipedia article:

“...a physical system is a portion of the physical universe chosen for
  analysis. Everything outside the system is known as the environment.
  The environment is ignored except for its effects on the system. The
  split between system and environment is the analyst's choice,
  generally made to simplify the analysis. For example, the water in a
  lake, the water in half of a lake, or an individual molecule of water
  in the lake can each be considered a physical system.”

Every system has some relationship to it’s surroundings.  As another way of stating the above, we can describe the surroundings, or environment, as all parts of the universe that is not part of the system. We can then generally categorize the system as being isolated, closed or open:
Isolated system: Neither matter nor energy may pass into or out of the system.
Closed system: The exchange of energy is allowed but matter may not pass into or out of the system.
Open system: Both energy and matter can pass into and out of the system.
In the context of your question (as I think I understand it), we are talking about isolated systems.  It wouldn't make much sense to discuss the energy of chemical bonds without constraining the system, even if hypothetically, in this way. To use the lake analogy, we talk about the chemical bonds within a single water molecule in a manner that excludes the matter or energy in the rest of the lake (and the rest of the universe).  If we are talking about hydrogen bonding between the water molecules for example, then we are considering there to be some hypothetical container within the lake that excludes all matter and energy from outside the container.  The contents, matter and energy, of the container, comprise the system.


text:
How the crystal structure varying for the same compound?
i have synthesized the crystal Cyclohexylammonium acetate. Structure of this compound was already solved and it s under centro symmetric space group I2/a. But when i took single crystal analysis it comes under monoclinic C2/c space group. What is the reason for getting this kind of results?

This may pretty well be the same group in different settings.

a) Your structure analysis is wrong
or
b) the publication is wrong (your looked up the wrong substance, typo in the original work, ...)
or
c) You crystallised from another solvent, at different temperature, (pressure?) and got another modification.


text:
Are there pressure driven reactions such that the product remains stable even after it is depressurised?
High pressure chemistry tend to return compounds with very interesting and unusual bonding structures, such as $\ce{NaCl7}$, metallic hydrogen and $\ce{Na2He}$
However, for most of these, the resulting product is only at best metastable (e.g. diamond) when returning to ambient conditions, and does not survive when pressure returns to ambient conditions (r.t.p.).
For the following requirements, a compound that has unusual bonding similar to those 3 reported above, that is things like n centre m electron bonds is desired

Are there exceptions to this rule, where the high pressure is mainly needed to overcome the energy barrier and the resulting product is more stable than the reactants such that the product stays stable (instead of metastable) even when it is decompressed (That is, a spontaneous but kinetically slow reaction where pressure is the major factor to activate the reaction)?
If there are no known exceptions, what are possible theoretical reasons that products formed from high pressure are usually metastable and cannot survive at low pressures. Put it simply, why are most reactions reversible under the change of pressure?

"Usually" is a wrong generalisation. Completely wrong. Heard of Le Chatelier? Phase diagrams? Enthalpy, entropy? Activation energy? The full answer to your question is several months of chemistry lectures.

For substances with phase diagrams, a given phase can only exists at the given pressure and temperature (e.g. liquid CO2). While my undergrad physical chemistry course have went through all the concepts you mentioned above, there are few examples of pressure activated reactions (except when gases equlibria are involved). Most reactions we went through are heat, light activated. Graphite to diamond would have been the answer had diamond being not metastable near ambient conditions

To overcome an activation energy only by increasing pressure is nonsense. At high pressure, the activation energy can however be smaller, the kinetics can be different, etc. But why do complicated, dangerous pressure experiments, when very often a rather small temperature increase does the trick?

"But why do complicated, dangerous pressure experiments, when very often a rather small temperature increase does the trick?" The edited questions shows how compounds with unusual bonding tend to form under pressure driven reactions. In particular, metallic hydrogen requires cooling in addition to high pressure to form, which probably due to more complicated reasons then energy barrier arguments

What do you want, a STP stable product or some arcane high pressure phase? Metallic hydrogen still does not exist, that  recent article was humbug. Any speculation what is "complicated" about it just that, speculation. Unless you are some highly talented theoretical chemist. :-/

An arcane high pressure phase product that, once formed, can survive back in RTP without decomposing due to thermodynamic reasons

What for? Some context? At the moment this is like asking for black-and-white striped zebra elephants. Likely not against natures laws, but why and where do you want to look for them?


text:
Gas solubility in water
Good morning,
I know that Henry's law describes the gas solubility in water with partial pressure of a specific gas.
For example the oxygen concentration in water, under atmospheric pressure, is
$\mathrm{C = P_{O_2} * k}$, where $\mathrm{P_{O_2}}$ is $\mathrm{0.21}$ bar.
But if I have a steam-water mixure, how I can calculate the solubility (for example in a deaerator)?
What data I need to calculate that solubility?
Thanks

Why do you feel that you can't use Henry's law for this?

Well if you are constantly deaerating, then then you are not at equilibrium with air (oxygen = 0.21 bar).  The equilibrium would be with the small amount of oxygen in the deaerator, but presumably your goal is to approach "zero" oxygen.

If that's the case, I don't think we have enough data to solve for such a dynamical system.

You need to know the oxygen content (partial pressure) of the steam-water mixture with which you are purging the system.
The water being purged will come to equilibrium with the purge gas based on Henry's law, just as in your example of atmospheric conditions, except that the partial pressure of oxygen in the steam should be much lower.


text:
Thermodynamics - Number Of Component

$\ce{CaCO3(s) -> CaO(s) + CO2(aq)}$

It is said that the number of constituents for this reaction is three .
I think number of constituents refer to the number of different elements in the system . As there are 3 different species so it is three.
But for the number of components it is said that as $\ce{CaCO3 and CaO are both in solid state they form only one component}$ and the other component is $\ce{CO2 which is in aqueous form}$
So there are 2 components .
I couldn't completely understand what is number of component ?
And is the above definition of number of constituents is correct ?

Is seems like there is some mixed up terminology here, particularly in the second part of you question:"$\ce{CaCO3}$ and $\ce{CaO}$ are both in solid state they form only one component...".  Is this possibly a translation of some text where "component" should really be "phase"?  Regarding the first part of your question, I would think the constituents of the reaction are $\ce{CaCO3}$, $\ce{CaO}$ and $\ce{CO2}$.


text:
Why is phenol soluble in sodium carbonate but not in sodium bicarbonate?
The original question was basically to predict which of sodium carbonate or sodium bicarbonate can be used to distinguish between phenol and benzene.
I was sure that phenol neither reacts nor dissolves in $\ce{NaHCO3}$ but I was not sure about the $\ce{Na2CO3}$.  By searching on Google I found that phenol is soluble in sodium carbonate. I am curious to know why this is, even though phenol doesn't liberate $\ce{CO2}$.

$\ce{NaHCO3}$ is less basic than $\ce{Na2CO3}$. Phenol is weakly acidic.

The liberation of $\ce{CO2}$ from $\ce{Na2CO3}$ requires two protonation steps.
$$\begin{align}
\ce{CO3^2- + HA &<=> HCO3- + A-}\\
\ce{HCO3- + HA &<=> CO2 ^ + H2O + A-}
\end{align}$$
Neglecting the concentrations, the position of these equilibria will depend on the acid strength of $\ce{HA}$. If $\ce{HA}$ is a stronger acid than $\ce{HCO3-}$, for example, the first equilibrium will favour the products.
This is the case for phenol, which has a $\mathrm{p}K_\mathrm{a}$ of $9.95$, whereas $\ce{HCO3-}$ has a $\mathrm{p}K_\mathrm{a}$ of $10.32$. Therefore, as long as you don't have too much phenol, it will exist in the soluble phenolate form:
$$\ce{PhOH + CO3^2- <=>> PhO- + HCO3-}$$
The solubility of phenol does not necessitate that it releases carbon dioxide.

Now, the second protonation. If you want to quantitatively determine whether $\ce{CO2}$ will actually be liberated, you need more data than the $\mathrm{p}K_\mathrm{a}$ values, since quite a significant amount of $\ce{CO2}$ can exist in the aqueous form. However, one thing is certain: if you want to produce $\ce{CO2 (g)}$ you first have to protonate $\ce{HCO3-}$ to form $\ce{H2CO3}$ (which dissociates to give $\ce{CO2 (aq)}$).
This is not possible with phenol, since $\ce{H2CO3}$ has a $\mathrm{p}K_\mathrm{a}$ of $6.3$ (even after correcting for dissolved $\ce{CO2}$).


text:
Extract ammonium sulfate from water
What is the best method to extract ammoniumsulfate from water?
The assignment is as follows:

Manure is being processed and the product left is a mixture between water and ammonium sulfate, transporting the mixture costs a lot of money. So we only want the ammonium sulfate.

What is the best method to do this on a big scale? Evaporation isn't a good option since it costs too much energy to heat the big amounts of water.

Depends on if you want the sulfate or ammonium or both.

You would probably have to add a lot of grain alcohol to precipitate the ammonium sulfate out (ammonium sulfate is insoluble in ethanol), and recovering the alcohol for re-use wouldn't be free either.  Does this route look too costly?

Evaporation isn't a good option since it costs too much energy to heat the big amounts of water.

Is it? I've seen prices around USD 130-150 per ton (minimum order of 25 tons) for ammonium sulfate with a purity around 99%. You might want to calculate the costs to operate a production plant using natural gas (or other fuels) for evaporation yourself.
There is however a different energy source that you might want to take into consideration: the sun.
Evaporation of water from solutions of sodium chloride in open ponds by solar heating alone has a long tradition in the production of sodium chloride in the form of sea salt.

Update
Ammonium sulfate will not evaporate from the open ponds. At temperatures between 235-280 °C, it will start to decompose.
However, if surface temperatures reach that level under irradiation with unconcentrated sunlight, the fate of the ammonium sulfate is the least of your worries ;)

Thanks for your quick answer, I have a question, as soon as the evaporation in an open pond starts, won't the ammonium sulfate start to evaporate too?

@KarimLeurink I've updated my answer.


text:
Would an amalgam filling under or next to a gold crown cause any electrochemical reaction?
Would a mercury amalgam filling deteriorate faster if it was in contact with a gold crown?

"Mercury amalgam"? An amalgam is an alloy that already has mercury in it. It'd help if you could mention what the other metal is: Zinc? Silver? As for the mercury component, that wouldn't really cause a problem (mercury and gold are located pretty close to each other when I looked up the standard electrode potential chart on Wikipedia); they'd react, but only under more "intense" conditions than your mouth could possibly provide ;)

There's no reason for two elemental metals, both of which are already in their most reduced form, to react with each other in a redox reaction.

Mercury amalgam: I mean standard dental amalgam which is not elemental mercury as I understand it.

Typical components of an amalgam filling are $\ce{Ag, Hg, Cu, Sn}$, smaller amounts of $\ce{Zn, In, Pd}$, according to Colgate( http://www.colgate.com/en/us/oc/oral-health/procedures/fillings/article/dental-amalgam-a-health-risk) website.

Yes, it would. Gold and amalgam can form a local cell with increased corrosion. This is why in some countries dentists are advised not to use amalgam together with gold fillings.

The answer is basically yes, if the dental procedures are done improperly.  Here is a good answer from a DDS:

Galvanic current is a term that has been used in dentistry for over
  100 years. It is a condition created by the presence of dissimilar
  metals in the oral cavity of the teeth and gums, with saliva serving
  as the electrolyte. Have you ever felt a “shock” to your teeth caused
  by a piece of tin foil or a spoon that touches a sliver or mercury
  filling in your mouth? If you answer yes to this question, you’ve
  experienced a galvanic event.
There are several different types of galvanism: 1) A silver/mercury
  filling is placed in opposition or adjacent to a tooth restored with
  gold. These dissimilar metals in conjunction with saliva and body
  fluids constitute an electric cell. When brought into contact, the
  circuit is shorted, the flow of electrical current passes through the
  pulp, and the patient experiences pain. 2) Dissimilar metals coming
  into contact when the upper and lower teeth come together and touch
  each other. 3) Two adjacent teeth are restored with dissimilar metals.
  The current flows from metal to metal through the dentine, bone and
  tissue fluids of both teeth resulting in discomfort and tooth
  sensitivity.

Additionally, from the same site:

Metals will not necessarily always cause the galvanic effect in the
  mouth. It depends on the specific metal and alloy being used along
  with quantity and placement in the mouth.

In other words the amalgam filling and the gold cap may form a weak galvanic cell.  The proximity of the amalgam to tooth nerves creates a situation in which only a very weak cell current is required in order for the resulting sensation be perceived.  The rate of degradation could thus be very small, or if the filling and capping procedures are done correctly there will be no galvanic affect at all.

Very funny @Martin-マーチン ;) In this context, DDS = Doctor of Dental Surgery, aka a dentist.

See the work of Dr. Andrew Cutler, PhD Chemist.  Gold next to amalgam MOST CERTAINLY forms a battery and causes the mercury in the amalgam filling to outgass at an accelerated rate.
I had a gold crown installed in late 1998 that had a large chunk of amalgam under it as a build-up material.  In early 1999, I saw my first Psychiatrist about medication for depression.  My health degraded over the years until I found Dr. Cutler's work in 2016.  I had the crown replaced as well as my other mercury fillings and am painstakingly trying to put my health back together.  Only recently I am able to think clearly again.


text:
Computing reaction mechanisms using solvation effects
I am about to compute reaction steps for a series of compounds using DFT.
A key point in our discussion is the dependency of the experimental rate of reaction w.r.t. solvent used, so it is very important that we incorporate solvation effects in our computational models.
I am aware of the best practices in choosing functionals, dispersion corrections, basis sets, etc..
By reading a little I decided on using the SMD model for implicit solvation.
I may use either ORCA or NWChem.
My questions are:

a. Should geometries be optimised in solvent, in the gas phase or both?
b. In case I use only gas phase geometries, should I calculate frequencies using implicit solvation (I doubt) or can I use the calculated free energy of solvation (done in a single point calculation) to correct the gas phase free energy?
a. In case I decide on using (one or two) explicit solvent molecules to better model the system, are there best practices on using them together with an implicit solvation model (in terms of calculating free energy of solvation and in deciding when it has converged/is adequate)?
b. What about the points addressed in 1. (gas phase versus condensed optimisation and frequency calculations) for this case?

References from literature are greatly appreciated.
I am looking forward to hear from you!

These questions also can depend on the system size. If you are pushing the limits of a quantum calculation you might be a bit more limited. If the system is somewhat smaller you can explore all these options, or perhaps using you could use a smaller model system to get a feel for how the system is behaving. This can help build your intuition as well for determining what is appropriate.

I have written about it here: http://dx.doi.org/10.1039/C5CP00628G

@JanJensen thanks for sharing, this has truly clarified my doubts.
Even though all answers where very useful, I wish I could accept your comment as the definite one.

1) Yes, geometries must be calculated both in the gas phase and in the solvent. To analize your results in the solvent you need something to compare that's why the gas phase calculation is always needed and often is considered "the only source of truth".
Frequencies must be calculated both in the solvent and in the gas phase as well, on the respective converged geometries.
2) I never worked with explicit solvent but I feel that the use cases of two methods are not overlapping. When performing calculations on a multi step reaction usually you don't have enough computational resources to calculate all the solvent molecules needed. This is why people use implicit solvent calculations. If you want to benchmark your solvent calculations, compare it with the gas phase ones and evaluate wether there is some improvement in the replication of experimental data.

Thanks for your answer!
A few points: a. how bad are going to be the results if I optimise geometries only in the gas phase and do single point calculations using implicit solvation? b. could you provide a reference to an article using the strategy you described?

I'll be honest, very little will change in the final reaction profile. However it is advised to reoptimize the geometry in the solvent starting from the already converged structure in the gas phase, since it will not be a long computation. I don't have any specific reference for this, but you can search a supporting information of a relevant article and check their procedure.

For explicit solvent interactions I would recommend for example FDE (Frozen-Density-Embedding) or embedding techniques in general.
In FDE the subsystems (solvent/solute molecules) "see" only the electron density of the other subsystems.
One problem you should consider when calculation with explicit solvent interactions is that molecules move all the time.
So to really get accurate measurements for that case you should consider doing an MD simulation to get a couple of snapshots for your QM-Level energy calculations.
Since this will get extremely expensive for your case, I wouldn't recommend using explicit interactions.
If you omit explicit interactions, consider using a Continuum Solvation model like COSMO/PCM or force field methods.
Optimize your geometries in gas phase (so you have a converged geometry already), but for calculations, optimize them again with solvent environments.
The change might not be big, but can be significant nonetheless.