Why do substances ionise in water to turn into charged species? Wouldn't they prefer to stay as uncharged molecules?

While a picture as in Bob’s answer is sometimes better than a thousand words, it may also help to add the thousand words[1] to explain what the picture is supposed to be showing or why it is relevant.

Water is a very polar solvent. The electronegativity difference between oxygen and hydrogen is rather high and the bent shape of the molecule additionally allows a higher dipole moment. Therefore, you can think of water molecules having a decidedly negative side — the oxygen atom — and a decidedly positive side — the space between the hydrogen atoms. In pure water, the molecules will arrange (at first approximation) in such a way that the positive sites are close to the negative ones and vice-versa.

At second approximation we need to introduce hydrogen bonds, which go one step further. Remember how each single hydrogen atom is strongly positively charged due to the electronegativity difference/the polar bond? Well, because hydrogen does not have any core electrons, this means it can act like a ‘slightly weakened proton’ and interact well with lone pairs on electronegative atoms such as oxygen. Thus, a better description of the intermolecular interactions in water molecules is that hydrogen bonds are formed from each hydrogen to one of another molecule’s oxygen lone pairs. These interactions are pretty strong for intermolecular interactions since they are caused by rather large charge differences (and need to stabilise these).

After having considered the water molecules by themselves, what happens if we submerge ions into them? In solid state, ionic compounds form extensive networks of cations/anions approximately alternating. Each cation will typically only have anions as closest neighbours and vice-versa; therefore each ion’s charge is stabilised by the charges of the opposite ions close by. If you wanted to dissolve this ionic compound into something like hexane, it would not work well. It is just much better for the ions to stick together in their solid-state form and be submerged in the hexane liquid phase — as your question implies.

In water, however, we have the positively charged and negatively charged areas not unlike those of the counterions. Taking a single ion out of the ionic crystal and surrounding it with the appropriate number and orientation of water molecules as shown in Bob’s answer, means that electrostatically almost nothing is gained or lost. True, the water molecules, not being charged particles, can stabilise a little bit less at first approximation but it is still comparable.

At second approximation, we can on the one hand reintroduce hydrogen bonds gonig from water molecules to anions such as chlorine. These may stabile the anions in solution, but note that having a hydrogen bond directed towards oxygen is typically better than one to a ‘typical halide’. The cation side is much more interesting, though: oxygen’s lone pairs can interact with the cationic centre to form a coordination complex. This lowers the energy of oxygen’s electrons at the expense of unoccupied cationic orbitals — a win-win situation for both partners. Thus, we can expect additional stabilisation due to this effect. (To be fully honest, one can and should also consider most ionic crystals as chained local coordination complexes. However, these are typically better stabilised if only one ligand sphere is involved, thereby preferring the solution.)

Thus we have:

A net average enthalpic no-effect concerning charge stabilisation (ionic crystal versus Bob’s picture
A slight net enthalpic gain due to the formation of better coordination complexes
A strong net entropic gain from the fact that a single, highly ordered crystal is broken up into gazillions of minute entities that can move freely in solution (I did not touch this point in this answer, but it is true for all solutions).

Considering all these, the conditions for ionic compounds to dissolve in aquaeous solutions are very favourable. And Bob’s your uncle.

[1]: Well all right. 645 words then. Not quite 1000 but close enough.


Formation of imines is usually a nucleophilic subsitution of oxigen with nitrogen. For this to happen, the oxigen usually must be bound beforehand as carbon-oxigen bond is quite strong. The most straightforward way is to use acidic catalysis.

Most reactions that you referenced use acidic catalysis, though sometimes it isn't immediately apparent because they use solid catalyst such as molecular sieves a.k.a. H-form zeolites. A few use Lewis acids, such as $\ce{Ti(OEt)4}$. They still count as acids.

The cases when no acid is used are not pure nucleophilic substitution. The reference you provided features various oxidative and other catalytic reactions, proceeding through different mechanism, that do not need acid, and a few reactions where reagent itself bears acidic function, like tosyl isocyanate, that is a strong electrophile and can bind with oxygen on its own.

TL;DR: the cases when acid apparently isn't used usually include it in non-obvious way. Other cases proceed through wild varying different mechanisms, usually involving oxidation.


The 2-fold axes are parallel to the "paper plane ", i.e. perpendicular to the three-fold. They are depicted as arrows, the half-arrows indicate screw-axes. The numbers next to it show the "height" of the axis, i.e. where the plane parallel to the paper-plane cuts the unit cell: 1/2 would be in the middle, 1/3 would be a third from the bottom/top, etc. (obviously one only gives relative numbers here, as the absolute unit cell length varies depending on the real-life crystal you are looking at).

Also have a look here for the symbols: http://img.chem.ucl.ac.uk/sgp/misc/guide.htm

Otherwise, the two links you provided are equivalent, as you can see from the pictogram in the lower lefthand corner - the first link is in the "tri-/hexagonal", the second link in the "rhombohedral" setting. The symmetry for both is obviously the same (otherwise they would not be the same space group), it is just a different way of depicting the spacegroup by changing the point of origin.

See here how they relate to each other: https://www.uwgb.edu/dutchs/symmetry/rhombo-lattice.htm

This one should also be quite enlightening - but I am not quite sure whether it might be behind a paywall: http://it.iucr.org/Ab/ch7o1v0001/sgtable7o1o155/
(Alternatively, have a look at the hardcopy of the International Tables of Crystallography, Volume A and look up spacegroup 155 - any decent crystallography lab or library should have it).


Does esterification require reactants to be present in the same plane?

The two commenters are collectively correct. But to clarify and augment their answers I would say that structure (d) cannot be correct because it is missing 2 carbon atoms relative to (c), and if you tried to cyclize (c) by forming esters, the result would be highly strained. Try making a molecular model of it and see how it looks.

Below is the product you would get if (c) cyclized. What is the difference between this product and (d)?


The Reaction Between Nickel and Hydrochloric Acid

Try 30% and heat it to 80C. All of the chloride should be exhausted after a day or less. You can then evaporate the solution to generate crystals of nickel chloride. Any work with nickel compounds should always be done in a well-ventilated hood.


Reduction and Oxidation - Alternate definitions

The "official" definition I've always used is that reduction is the gain of electrons. IUPAC seems to use the same definition. It looks like organic chemistry books often use the addition of oxygen to identify oxidation, and the addition of hydrogen or the removal of oxygen to identify reduction. This is in line with the historical definitions (at least according to wikipedia) that defined reduction in terms of the change in mass on heating (if oxygen was lost, the weight would be reduced).

When using these shorthand rules, I think it's important to distinguish between the purposes of the definitions. The more formal "official" (according to IUPAC) definition is intended to get as close to the underlying physics as possible. The less formal "shortcut" definition used here is intended to quickly allow chemists to identify electron transfer just by looking at the structural formulas. In fact, in my copy of Wade's Organic Chemistry (4th ed) it says

  Most organic chemists habitually use the following simple rules, based on the change in formula of the structure:

It then goes on to define redox reactions in terms of gain or loss of oxygen and hydrogen. This implies that the author doesn't intend it to be read as a formal definition - just a shortcut you can use in a pinch.

This is a case of "it works well enough" being used to make life easy - but it is important to remember that reduction isn't really being defined as gain of hydrogen in the formal scientific sense. It is still defined as a gain of electrons - it just so happens that in organic reactions, it is very difficult to add electrons to something without also adding hydrogen, and so the shortcut works (due to electronegativity differences between O, C, N, P, and H, as you pointed out).

Here's an example of a redox reaction where this rule won't work, but looking at electrons will:

$$
\ce{Li(s) + 1/2H2(g) -> LiH(s)}
$$

We are adding hydrogen to lithium, and the oxidation state of Li goes from 0 to +1 - it is oxidized, not reduced.

I have been looking for an example of an organic reduction that does not involve the addition of hydrogen, but haven't been able to find one - which is probably why this shortcut rule works.


Why are axial bonds are longer than equatorial bond in sp3d hybridized atoms?

You are asking for $\mathrm{sp^3d}$ hybridisation, but I do not know of a case where $\mathrm{sp^3d}$ hybridisation actually happens. Either it does not make sense to discuss hybridisation at all (the iron in pentacarbonyliron) or the hybridisation is actually not $\mathrm{sp^3d}$ but $\mathrm{sp^2 + p}$ (the phosphorus in $\ce{PCl5}$). I shall discuss the latter, because I believe it could be the one you are on about.

Phosphorus pentachloride is often drawn like in figure 1 below for simplicity. This depiction implies five identical covalent 2-electron-2-centre bonds. However, that does not agree with the octet rule.

Figure 1: Simplified structure of $\ce{PCl5}$.

Instead, one can draw a set of two mesomeric structures, each one conforming with the octet rule — see figure 2.

Figure 2: Mesomeric structures of $\ce{PCl5}$ conforming to the octet rule.

This already hints us towards the answer: Rather than assuming two bonds which are equal to the other three bonds we need to consider three ‘classical’ 2-electron-2-centre bonds to the equatorial chlorines and one 4-electron-3-centre bond to the two axial chlorines. This bond’s order is $0.5$ rather than $1$. Typically, the lower the bond order the weaker a bond and therefore the greater the bond length is — the experiment is in fine agreement with theory.

But no bonding discussion is truly complete without an orbital consideration. Check out the three collinear p-orbitals of phosphorus and the two chlorines, that together form three molecular orbitals labelled $\Psi_1$ to $\Psi_3$ in figure 3.

Figure 3: Representation of the three molecular orbitals that form the 4-electron-3-centre bond.

As you can see, the lowest MO $\Psi_1$ is bonding with respect to both $\ce{P-Cl}$ bonds. The highest MO $\Psi_3$ is antibonding with respect to both. And the middle one is bonding with respect to $\ce{Cl-Cl}$ but nonbonding if we add the phosphorus atom. We need to fill in four electrons into these three orbitals ($\ce{PCl3}$ has one lone pair, and we are effectively using the bond of a $\ce{Cl-Cl}$ molecule as our second electron pair). Thus, the bonding and the nonbonding orbitals are filled. Bond order can now be calculated by:

$$\text{bond order} = \frac{(\text{electrons in bonding orbitals}) - (\text{electrons in antibonding orbitals})}{\text{number of bonds}\times 2}\\= \frac{2-0}{2 \times 2} = 0.5$$

Again, a lower bond order typically correlates with a greater bond length.


Ensuring constant ionisation of transition metals using plasma ionisation for accurate mass spec

Interesting question. Some of the stuff can be justified. As an analytical chemist, we should keep in mind that no technique is perfect or universal, even if the instrument costs thousands or millions of dollars. Mass spectrometer cannot distinguish isobaric species (or ions), for instance, if two ions have identical mass due to a different location of the isotope within the same molecule, the mass spec won't be able to you. However, if there is a small difference in mass at fourth, or even fifth decimal place, modern day mass spectrometers can distinguish them (see e.g., Fourier transform ion cyclotron resonance mass spectrometers). The more money you add, the fancier the mass spec gets, with it comes better and better mass resolution.

Now come to the ion generating source - the plasma, which is at say 10,000 K. Although the name plasma sounds very fancy as the fourth state of matter, and given that its temperature is higher than the surface of the Sun- this temperature is still too low for atoms to ionize like crazy. It is pretty hard to strip off one, then the second and then the third electron from an atom as you are thinking of Al. Therefore in a typical ICP torch, multiple ionization on a single atom is not a major concern. To satisfy yourself, do a Boltzmann distribution calculation for sodium atom at 10,000 K. You will notice that a very small fraction of sodium atoms will be in the excited state.

  This is further complicated if you consider that the elements rarely
  exist in solution at oxidation states of 0, but rather aqueous ions
  with varying oxidation states.

Once the ions from solution reach the plasma they get reduced to "zero" oxidation state. Consider the plasma as a rich source of free electrons, which are circulating like crazy following the radiofrequency field of the coil. The emission in ICP mainly comes from neutral atoms.


The Franck–Condon region is a region of the final potential energy surface which is accessed by the vertical transition from the initial state. Usually, the initial state is the ground vibrational state of the ground electronic state and the final is some vibrational state of an electronically excited state. For instance, the Franck–Condon region is shaded below on the Figure from Wikipedia article on the Franck–Condon principle.

  But when a transition takes place from S0 to S1, it is said that
  molecule is excited to the FC region of the S1 state - but there is no
  guarantee that going vertically from an S0 minimum will lead to a near
  S1 minimum. So something is probably incorrect with my current
  understanding.

Nope, what you are saying is correct. Indeed, according to the Franck–Condon principle vertical transitions are the most probable. The minimum of the upper potential energy surface might lie outside of the Franck–Condon region, so a vertical transition from the ground vibrational state of the ground electronic state to the excited electronic state might indeed result in one of the vibrationally excited states like it is shown in the picture above. Once a molecule is excited in such way it quickly relaxes to the lowest vibrational level of the electronically excited state from which it can decay then back to the electronic ground state. And again, on this way back to the ground state the most probable transition is the vertical one, so that photon emission can lead to a vibrationally excited state of the ground electronic state, not necessarily to the ground vibrational state of it.


Aqueous Chemistry: Ranking Ion Concentration

Let's look at the overall reaction:

$\ce{Na3PO4(aq) + 3AgNO3(aq) -> 3NaNO3(aq) + Ag3PO4(s)}$

As you can see $\ce{Na^+}$ and  $\ce{NO_3^-}$ are spectators ions, they don't participate in the main reaction. Their concentrations, taking into account the dilution, are:

$\ce{[Na^+]= \frac {3 \times 1.0 \times 100}{200}}= 1.5 \ce{M}$.

${[\ce{NO_3^-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

The net ionic equation is:

$\ce{PO4^{3-}(aq) + 3Ag^+(aq) ->  Ag3PO4(s)}$

The initial concentration of phosphate and silver ions taking into account the dilution are:

${[\ce{PO_4^3-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

${[\ce{Ag^+}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

According to the stoichiometry of the reaction: 3 moles of silver ions react with one mole of phosphate ion. This means that silver ion is a limiting reactant. It will completely react to form the precipitate of silver phosphate. On the other hand, the concentration of phosphate at the end of the reaction will be:
${[\ce{PO_4^3-}]= 0.5-\frac{0.5}{3}=\frac{2\times 0.5}{3}=0.33 \ce{M}}$.

In order to determine exactly the concentration of silver ion, we have to use the solubility product of silver phosphate:
${[\ce{Ag^+}]= \frac{K_{sp}}{[\ce{PO_4^3-}]^3}=\frac{8.89\times10^{-17}}{(0.5/3)^3}}=0.04\times10^{-17}\ce{M}$


Is there a chemical which fluoresces only after coming in contact with water?

Many fluorescent dyes are not fluorescent in their solid states. Fluorescein and similar dyes (rhodamines, eosines) for example. They need to be dissolved in order to be fluorescent (so just wetting them with a bit of water won't work either).


The answer to your question does not need a single calculation.

You have got your 1 molar phosphoric acid that will get deprotonized by the 1 molar sodium hydroxide solution that is added. As both solutions have got the same concentration, you will reach the first inflection point after 100 mL and the second inflection point after 200 mL.

Now what is between those two volumes, at your desired volume of 150 mL?
A simple buffer system that can be described using the Henderson–Hasselbalch equation:
$$\mathrm{pH}=\mathrm pK_\mathrm a-\log_{10}\frac{\ce{c(H2A- )}}{\ce{c(HA^2- )}}$$

As both concentrations $\ce{c(H2A- )}$ and $\ce{c(HA^2- )}$ are equivalent, the logarithmic expression gets zero and your final pH equals the second $\mathrm pK_\mathrm a$ value of phosphoric acid, which is 7.21 (according to the German Wikipedia).

$\hskip3.5cm$


Thionyl chloride first reacts with the alcohol to form an alkyl chloro sulfite,which gives various stereochemical products as shown here.

Extending this to the current question , alkyl chloro sulfite 2 is formed.${Cl^-}$ attacks in a ${S_N^2}$ mode to give 3.

Base deprotonates 3 to give 4.

Alkoxide and chloride in 4 are in correct trans configuration for a ${S_N^2}$.This results in 5(B in your scheme).

If we take C in your question , and follow the scheme above, 9  is formed.In 9 alkoxide and chloride are not in correct configuration for ${S_N^2}$.Therefore
C in your question does not undergo epoxidation.


Compare stability of the given carbocation

The question is pretty easy though the INDUCTIVE and MESOMARIC effects are same but there is a diffrence in the HYPERCONJUGATIVE Effect as in HYPERCONJUGATIVE Effect the bond between $C-H$ breaks to provide stability thus any change in bond energy of the bond to be broken changes the strength of effect.

Now as the bond strength of $C-T$ bond if highest hence the HYPERCONJUGATIVE Effect of $\ce{CT3}$ decreases and hence the carbocation is least stable.And $\ce{CH3}$ is most stable

You can read more about Hyperconjugation in the following Site:
https://chemistry.tutorvista.com/inorganic-chemistry/hyperconjugation.html


How does aluminium amalgam work as a reducing agent?

Could aluminium amalgam be used to deoxygenate a carboxylic acid or amino acid to the corresponding alcohol or alkane?

No. Aluminium amalgam has been used to selectively reduce other groups in molecules containing a carboxylic acid group, while preserving the carboxylic acid group.

See for example in Industrial Organic Nitrogen Compounds at page 139:

  Some pyridine-carboxylic acids have been reduced to the corresponding dihydropyridine-carboxylic acids with aluminium amalgam.

See also the patent Polyene carboxylic acids and esters and manufacture thereof where aluminium amalgam is suggested as a mild reducing agent which preserves the carboxylic acid or ester group.


Factorizing Slater determinant into product of spin-up and spin-down slater determinant

When exchange effects are not taken into consideration, the simplest possible description of a many-body wave function is simply the Hartree product, i.e. the product of individual-particle wave functions. This is reasonable for bosons, but electrons are fermions, which means that the exchange of any two must flip the sign of the wave function.
Slater determinants are used to antisymmetrize the wave function so that fermion antisymmetry (Pauli principle) is obeyed. Since this is only relevant for electrons of the same spin, it is sufficient to describe the overall wave function by giving each spin its own determinant and then combining the determinants by taking the product. It is like up- and down-spin wave functions occupy different subspaces and interact via the electron–electron correlation term in the Hamiltonian.


Why does an increase in alkyl group substitution cause downfield shift H'NMR?

The short answer to this question is that C is more electronegative than H (H 2.2, C 2.5), and therefore one would expect a trend in increasing chemical shift with increasing carbon substitution.

The longer answer....

As you point out, alkyl groups are electron donating, and you are on the right track in surmising that alkyl groups are less electron donating than protons. Remember that all shielding is relative. Looking at shielding effects at the α-substituent;

a bare proton nucleus (H+) would come at δ35,
H-H comes at δ7.4, and
H-CH3 comes at δ0.23.

A clear upfield shift due to the electron-donating alkyl group. So, we need to look deeper at the effect of β-substituents:

H-CH3 (δ0.23),
H-CH2CH3 (δ0.86),
H-CH(CH3)2 (δ1.33), and
H-C(CH3)3 (δ1.56).

This longer answer involves looking at the magnetic anisotropy around the observed nucleus, and questioning some age-old depictions of the C-C bond. Historically, the C-C magnetic anisotropy was depicted as below, with shielding at the side, and deshielding along the ends. Have a look in your NMR textbook, and you'll probably find this. Taken in isolation, this model conforms with the observed trend of increasing shift with increasing substitution with C.

However, this model contradicts somewhat the notion that alkyl groups are electron donating, and indeed what is observed with α-substitution. And with good reason. Recent studies by Baranac-Stojanovic (DOI: 10.1002/chem.201204267) show that this anistropy model should be reversed, and that there are shielding zones at the end of the C-C bond. Why then do we observe the increase in chemical shift with increased C substitution? Again, it is because the shielding anisotropy of the contributing C-C bond component is considerably less than contributions from the geminal C-H anisotropy contributions.


Why was a Plimsoll symbol chosen to indicate standard state?

Prigogine's Chemical Thermodynamics gives a bit more context as to why a plimsoll symbol is somewhat more preferable and is indeed used to denote non-zero nature of standard values [1, pp. xxiv, 86]:

  The properties of a pure substance are denoted by a superscript $^0$. The use of this symbol to denote "standard" thermodynamic quantities is then permissible only when these refer to the pure substance as standard state. Some other convention is needed to denote the more general "standard quantities" defined in equation (7.51), where the standard state may for example refer to an infinitely dilute solution. After much consideration the symbol $^⦵$ has been introduced. This symbol is based on the circle, and is thus closely related to the more common (but occasionally confusing) notation for standard quantities. It seems useful, however, to associate it with the plimsoll mark which, appropriately enough, refers to a reference state of loading of a ship; this use of an ideogram has, we believe, some value in keeping the essential nature of "standard states" in prominence.

  Quantities denoted by the plimsoll symbol are still, in general,
  functions of temperature and pressure; the idea of "standardization"
  of temperature and pressure is not stressed. When necessary we denote
  quantities at the standard pressure (which for convenience is nearly
  always chosen as the unit of pressure) by the superscript $^†$, quantities at a standard volume (which for convenience is always chosen as the unit of volume) are denoted by $^‡$. Quantities referring to a (usually hypothetical) ideal system are denoted by $^\mathrm{id}$.

  [...]

  $$P = P^⦵(T,p) + P^M(T,p,x_1\ldots x_c) \tag{7.51}$$

It's also interesting that Lewis and Randall, known for defining the standard state in the beginning of XX century as we know it, used an upper circle [2, p.623]:

  $G^\circ$ A property of a substance in its standard state.

However, they referred to it as to a standard reference state (distinguished from a standard state) throughout the textbook [2, p. 88, 99]:

  While we cannot give the absolute magnitude of $\bar{H}$ for a substance in solution, we may ascertain how much greater or less this is than the heat content of the same substance in some chosen state. Thus, at any temperature, if we are dealing with the partial molal heat content of water in some solution in which water is the solvent, we may choose pure liquid water as the reference state, and denote its molal heat content by $H_1^\circ$.

  [...]

  It is highly convenient, especially for purposes of concise tabulation, to know the heats of reaction when various substances are formed from their elements. It is therefore desirable to choose some one standard reference state for each element. For this purpose, we shall, at all temperatures, take the element at a pressure of one atmosphere, and in that form which is most stable or most common at room temperature. Thus liquid mercury, gaseous oxygen, solid iodine, rhombic
  sulfur and graphitic carbon, all under a pressure of one atmosphere, will be considered to be in their standard reference state.

References

Prigogine, I.; Defay, R. Chemical Thermodynamics; Treatise On Thermodynamics; Longmans, Green & Co Ltd: Glasgow, 1954; Vol. 1.
Lewis, G. N.; Randall, M. Thermodynamics and the Free Energy of Chemical Substances, First Edition later Printing edition.; McGraw-Hill Book Company, 1923.


When is it necessary to check wavefunction stability in density functional theory?

The very short answer is: whenever you are in doubt.

The stability check is probably unnecessary when you are dealing with closed shell molecules, where you can be pretty sure that a restricted methodology is reasonable. For example, when you calculate water at the RHF level of theory, you won't really need to look for a more stable wave function, because it would not be physically sensible. An UHF approach in principle will still lead to a lower energy since it has more flexibility.

Whenever you are calculating molecules in which you would assume some open shell character you should be considering checking wave function stability. It usually doesn't hurt to perform one check too many. Discovering instabilities later can be a lot more frustrating. That does not necessary mean that you are gunning for the correct ground state, as you are still restricted on spin. I'll include a small, but well-known example below: $\ce{O2}$. I am using Gaussian 09 Rev. D.01 for the calculation, but it should work with anything more recent. (It won't work the same way with anything older.) Also note, that I am not conducting geometry optimisations, but use $\mathbf{d}(\ce{O-O})=\pu{120pm}$ to keep it simple.

First let's run an ordinary, restricted calculation. If you are not using the SCF(xqc) keyword, you should immediately notice, that something is wrong.

%chk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit             ! Use density fitting to spead up calculation
gfinput            ! Prints basis set in a form suitable for use as general basis set input
gfoldprint         ! Prints the basis set information in the Gaussian format
iop(6/7=3)         ! Switches on printing of all MO
symmetry(loose)    ! Use correct symmetry
scf(xqc,conver=10) ! Quadratic convergence after conventional (for tough cases)
INT(SuperFineGrid) ! Largest grid, most accurate

title card unused

0 1
O  0.6 0.0 0.0
O -0.6 0.0 0.0

You'll find a summary of the energies at the end, if you want to check and run your calculations.

Next up, we'll check and optimise the wave function.

%chk=bp86svp.rhf.sing.stab.opt.chk
%oldchk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(xqc,conver=10) INT(SuperFineGrid)
stable=opt

title card unused

0 1

You should find the following in your output:

 The wavefunction has an RHF -> UHF instability.

Followed by another SCF run. Pay close attention to the following:

 SCF Done:  E(UB-P86) =  -150.201660211     a.u. after    6 cycles
            Convg  =    0.3033D-11                    52 Fock formations.
              S**2 =  1.0016                  -V/T =  2.0049
 = 0.0000 = 0.0000 = 0.0000 = 1.0016 S= 0.6187
 = 0.000000000000E+00
 Annihilation of the first spin contaminant:
 S**2 before annihilation     1.0016,   after     0.0126

You are still finding a singlet state, but now as a UHF wave function. You should later find:

 The wavefunction is stable under the perturbations considered.

What you have found is a broken symmetry singlet. We do know that the ground state should be a triplet. Therefore we'll perform a restricted open shell calculation. We already know this one will have and RHF -> UHF instability, and since we cannot use the quadratic convergence, we cannot use stable either. But it'll provide us with a good starting point for the UHF calculation, or at least an energy to compare others to.

%chk=bp86svp.rohf.trip.chk
%oldchk=bp86svp.rhf.sing.chk
#p ROBP86/Def2SVP/W06
DenFit
geom=check guess=alpha
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(conver=10) ! Quadratic convergence not available for RO
INT(SuperFineGrid)

title card unused

0 3

The important part here is, that it generates a pure spin state wave sunction:

 = 0.0000 = 0.0000 = 1.0000 = 2.0000 S= 1.0000

Another calculation is using the the mixing of HOMO and LUMO to generate a broken symmetry ansatz wave function directly. You'll find the same result as with the approach above.

%chk=bp86svp.uhf.sing.chk
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
geom=check guess=(read,mix)
scf(xqc,conver=10)
INT(SuperFineGrid)
stable=opt

title card unused

0 1

You'll therefore find again:

 SCF Done:  E(UB-P86) =  -150.201660211     A.U. after   11 cycles
            NFock= 11  Conv=0.88D-12     -V/T= 2.0049
 = 0.0000 = 0.0000 = 0.0000 = 1.0016 S= 0.6187
 = 0.000000000000E+00
 KE= 1.494731832724D+02 PE=-4.114759744301D+02 EE= 8.357834648903D+01
 Annihilation of the first spin contaminant:
 S**2 before annihilation     1.0016,   after     0.0126

The wavefunction is stable under the perturbations considered.

Now for the last part, run a real UHF calculation:

%chk=bp86svp.uhf.trip.stab.opt.chk
%nproc=8
%mem=8000MB
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(xqc,conver=10)
INT(SuperFineGrid)
stable=opt

title card unused

0 3

You'll find as expected

 Annihilation of the first spin contaminant:
 S**2 before annihilation     2.0031,   after     2.0000

 The wavefunction is stable under the perturbations considered.
 [...]
 The wavefunction is already stable.

For giggles, why not consider a quintet?

Command file              Functional            Energy / Hartree ( cycles )
bp86svp.rhf.sing          E(RB-P86)       =       -150.155030787 (      4 )
bp86svp.rhf.sing.stab.opt E(UB-P86)       =       -150.201660211 (      6 )
bp86svp.rohf.trip         E(ROB-P86)      =       -150.213771982 (     10 )
bp86svp.uhf.sing          E(UB-P86)       =       -150.201660211 (     11 )
bp86svp.uhf.trip.stab.opt E(UB-P86)       =       -150.215679340 (      9 )
bp86svp.uhf.quin.stable   E(UB-P86)       =       -149.651916836 (      9 )

In conclusion, whenever you suspect some open shell character, check for stability. If you have well-behaved molecules (singlet), like most organics are, you can probably skip it.
Be aware that a stability optimisation will not necessarily give you the ground state.


Effect of phenyl and vinyl substituents on Acidity of carboxylic

When the cases of carboxilic acids are considered, we must note that the conjugation of $\ce{COO-}$ group is so strong in itself, that it's conjugation with the rest of the compound is rendered relatively weak. In such cases, the inductive effect is dominant.

Let us take the example of benzoic acid. The carbon to which $\ce{COOH}$ is attached shall be named as the key-carbon $\ce{C}$1. The conjugate base would be $\ce{PhCOO-}$, and the +I(Inductive effect) of $\ce{COO-}$ group will be dominant on $\ce{C}$1. Any group attached to this compound will further increase or decrease the acidic strength of the compound as a whole, depending upon the attached group's contribution towards electron density at $\ce{C}$1.

For example, in case of para-nitrobenzoic acid, electron density at $\ce{C}$1 decreases due to $\ce{NO2-}$'s strong -R (Mesomeric) Effect(partial positive charge is developed on $\ce{C}$1 which is stabilised by $\ce{COO-}$'s +I). Hence it turns out to be a strong acid.

The opposite case is that of para-methoxybenzoic acid. The electron density at $\ce{C}$1 increases due to +R of $\ce{OCH3}$ group and makes the conjugate base unstable, hence acid weaker. This can be cross checked by the $\mathrm{p}K_\mathrm{a}$ values of above mentioned two compounds as well.

Now coming to the compounds mentioned in the question, which are phenyl and vinyl, one must note that both of these would reduce the charge density on $\ce{C}$1 because of the $\mathrm{sp^2}$ hybridised Carbon atom which contributes to the overall electro-negativity of these two groups.

Long story short, the resonance effect between $\ce{COO-}$ and the rest of the compound is suppressed by a relatively stronger Inductive effect which works mutually between $\ce{COO-}$ and the rest of the compound.


How to figure out which two solutions will emit largest heat when they are mixed in same volumes?

You should just think about a fact: all chemical processes will correspond to enthalpy changes.

The basic process of acid-base reaction is
$$\ce{H+ + OH- -> H2O}$$

This process will have an enthalpy change of $\pu{-55.8 kJ mol-1}$ at $\pu{298 K}$. (I don’t know why, but in different books or articles this value will be slightly different.) However,

Precipitate formation is usually exothermic, since there are some ion bonds formed.
Weak acid and base usually need some energy, for they are not completely ionized, during the reaction, they will to ionize more since $\ce{H+}$ or $\ce{OH-}$ is consumed. Ionization will be endothermic.

Therefore the answer would be the second reaction, since $\ce{BaSO4}$ is precipitated, which is exothermic, while weak acid and base reactions are endothermic.

(However, there seems to be always some exception, for example, ionization of $\ce{HF}$ will be exothermic. But to discuss this may be out of topic, and to be frankly, I’m not quite sure what its mechanism is.)


How to make foggy and cloudy plastic fully transparent?

The tiny scratches on the plastic surface reflect and diffract light in different directions, so edges of objects behind the plastic become distorted, and the clear areas of the plastic are less clear.
There are several ways to make the plastic more transparent.

Use a finer grit sandpaper - really really fine: 1500 grit or finer, if you can get it. Plastic abrasive sponges are more gentle and can be used wet, so may create smaller, less deep scratches.

Use toothpaste to "sand" the plastic in the first place. You have to get the paste to be pasty thick, not watery thin. It will reduce scratches, but if you don't put deep scratches on in the first place, it will be easier to shine up a plastic surface. Toothpaste will reduce the shininess of a perfect surface, but usually will improve an ordinary surface. There are commercial pastes for cleaning and shining stovetops that are more abrasive, and you can make your own with sodium bicarbonate (no vinegar) - but you have to give it some energy, some elbow grease, and it takes a while. You could use a buffer wheel on a power drill.

Filling in the tiny scratches with a liquid helps even out the surface, but water evaporates and the effect does not last. An oil (e.g., silicone) or petroleum jelly (very thin layer) might last long enough for you - days, maybe weeks.

Depending on the plastic film, you might be able to heat it slightly to allow the scratches to deform slightly and become slightly wavy instead of sharp valleys and projections. This will require a lot of experimentation to get enough heating without distorting the plastic beyond your requirements. Hot air or hot smooth metal, or perhaps hot water or steam might work.

The least likely method to work is using a solvent to smooth out the surface. You might get some immediate effect that looks like an improvement, but any solvent that dissolves the very top surface will also diffuse into the plastic and when it evaporates, you will have a plastic that is disturbed even deeper, and will be foggier. Working this approach further suggests that you could use what would be better described as a plasticizer, a high molecular weight liquid that is compatible with the plastic, but does not migrate (out, usually, but in here), buff the surface and quickly clean the plastic to remove all the solvent/plasticizer. Now, since you probably don't have access to different kinds of plasticizers, you could try some vegetable oil plus a little of a more powerful solvent, like a few percent of alcohol, or mineral spirits, or even acetone, and you could add some sodium bicarbonate for a little abrasion, and use a buffer wheel.


Molecular term symbols for excited state oxygen

What are the possible term symbols for excited state oxygen with configuration $(1\pi_u)^3(1\pi^*_g)^3$ (all other occupied orbitals are closed shell)? Based on the possible values of $S$ and $\lambda$ it seems that $^1\Delta_u$, $^3\Delta_u$, $^1\Sigma_u$ and $^3\Sigma_u$ are possible.

Term symbols for $(1\pi_u)^3(1\pi_g^*)^3$ - i.e. 1 hole + 1 hole - is equivalent to finding terms for $(1\pi_u)^1(1\pi_g^*)^1$, i.e. 1 electron + 1 electron. I will continue answering the question for the latter case as this is easier to handle.

The fastest way is to use group theory. $O_2$ molecule belongs to $D_\infty^h$ point group. We look at the character table for this group (e.g. here) and perform a direct product of irreducible representations corresponding to our electrons, i.e.:

$\Pi_u \otimes \Pi_g = \Sigma_u^+ \oplus \Sigma_u^- \oplus \Delta_u$

Note this only pertains to spatial symmetry, i.e. orbital angular momentum. It tells us nothing about possible spin states. However, since the two electrons are in different orbitals, all spin combinations are allowed in all cases. The resulting terms are:

TLDR: $^3\Sigma_u^+,$$^3\Sigma_u^-,$$^3\Delta_u,$$^1\Sigma_u^+,$$^1\Sigma_u^-,$$^1\Delta_u$

Your answer is thus correct, the missing piece being the +/- signs of the $\Sigma$ terms. Note that in answering we relied heavily on Pauli exclusion principle not playing a role - since the two electrons occupy different orbitals. Were that not the case, some of the terms would disappear (terms of e.g. $(1\pi_u)^1(2\pi_u)^1$ are $^3\Sigma_g^+,$$^3\Sigma_g^-,$$^3\Delta_g,$$^1\Sigma_g^+,$$^1\Sigma_g^-,$$^1\Delta_g$, while terms of $(1\pi_u)^2$ are a non-trivial subset, specifically $^3\Sigma_g^-,$$^1\Sigma_g^+,$$^1\Delta_g$).

  However, I am not sure in this case how to determine the +/- labels for the $\Sigma$ term symbols.

The following derivations draw heavily from this pdf, read for further details.

We refer to definition of +/- signs as describing the symmetry with respect to reflection in plane that contains both atoms, let's use an arbitrary choice of xz-plane. Wavefunctions symmetric when reflected in this plane are "+" terms, anti-symmetric are "-" terms. However, imagining reflections is not practical. Therefore, we introduce operator notation. The "+" term corresponds to a wave function with eigenvalue of +1 (with respect to $\hat{\sigma}_{xz}$ operator); "-" term corresponds to eigenvalue of -1.

We note:

$$\hat{\sigma}_{xz} \pi_x = \pi_x$$
$$\hat{\sigma}_{xz} \pi_y = - \pi_y$$
$$\pi_1 = -N(\pi_x + i\pi_y)$$
$$\pi_{-1} = N(\pi_x - i\pi_y)$$

, where N is some boring normalization constant. Following up:
$$\hat{\sigma}_{xz} \pi_1 = - \pi_{-1}$$
$$\hat{\sigma}_{xz} \pi_{-1} = - \pi_1$$

We need combinations of configurations (Slater determinants = SD) that are eigenfunctions of $\hat{\sigma}_{xz}$.
E.g.:

$$^1\Sigma^+_u = N(|SD_1\rangle - |SD_2\rangle + |SD_3\rangle - |SD_4\rangle) =$$$$= N(|\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1}{\pi}_{g,-1}\rangle + |\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1}{\pi}_{g,1}\rangle) $$
Let's apply the $\hat{\sigma}_{xz}$ operator:
$$\hat{\sigma}_{xz}{ }^1\Sigma^+_u = \hat{\sigma}_{xz} (N(|\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1}{\pi}_{g,-1}\rangle + |\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1},{\pi}_{g,1}\rangle)) = $$
$$ = N(|-\pi_{u,-1},-\bar{\pi}_{g,1}\rangle - |-\bar{\pi}_{u,-1}{-\pi}_{g,1}\rangle + |-\pi_{u,1},-\bar{\pi}_{g,-1}\rangle - |-\bar{\pi}_{u,1},{-\pi}_{g,-1}\rangle) =$$
$$ = N(|\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1}{\pi}_{g,1}\rangle + |\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1},{\pi}_{g,-1}\rangle) = $$
$$= N(|SD_3\rangle - |SD_4\rangle + |SD_1\rangle - |SD_2\rangle) = { }^1\Sigma^+_u$$

In going from first to second line we apply the operator, i.e. flip "1" subscript to "-1" and vice versa; in addition we multiply each orbital by "-1". Since we have two orbitals this multiplication disappears: ($-1*-1=1$), which is what we do in going from second to third line. In the last line we see that $\hat{\sigma}_{xz}|SD_1\rangle = |SD_3\rangle$;$\hat{\sigma}_{xz}|SD_2\rangle = |SD_4\rangle$ etc.

We could have used a WF in the form e.g.: $WF = |SD_1\rangle + |SD_3\rangle$. Such a WF is indeed an eigenfunction of $\hat{\sigma}_{xz}$. But it is not an eigenfunction of $\hat{S}^2$.

Similarly, a corresponding "-" term is:
$$^1\Sigma^-_u = N(|SD_1\rangle - |SD_2\rangle - |SD_3\rangle + |SD_4\rangle) =$$
$$\hat{\sigma}_{xz}^1\Sigma^-_u = N(|SD_3\rangle - |SD_4\rangle - |SD_1\rangle + |SD_2\rangle) =$$
$$ = -1 * N(- |SD_3\rangle + |SD_4\rangle + |SD_1\rangle - |SD_2\rangle) = -1 * { }^1\Sigma^-_u$$

The four SDs we use here are all eigenfunctions of $\hat{L}_z$ and $\hat{S}_z$ with $M_l = 0$ (i.e. one $\pi_1$ + one $\pi_{-1}$ orbital) and $M_s = 0$ (i.e. one alpha and one beta orbital). There are two more combinations of these SD and these are part of the $^3\Sigma^-_u$ and $^3\Sigma^+_u$ states.

Lastly, non-$\Sigma$ terms have both a "+" and "-" component that flips to each other upon reflection, e.g.: $\hat{\sigma}_{xz} \Delta^+ = \Delta^-$. But since $\Delta^+$ and $\Delta^-$ are different SDs (different configurations), we don't use the label in these cases - they are reserved for $\Sigma$ terms.

The take-away message here is that "+/-" label relates to some kind of symmetry. Similarly to "g/u" label, it only is relevant in some special molecules ("g/u" in molecules with inversion center; "+/-" in diatomic molecules). We use group theory where possible or a very technical procedure when not.

  For ground state oxygen, triplet states pair with - and singlet states with + to give overall antisymmetric wavefunctions. However, in this case, two different orbitals are involved, so does this rule no longer apply?

There is some confusion here. As mentioned above, for ground state ($(1\pi_u)^2$) the terms are indeed $^3\Sigma_g^-$ and $^1\Sigma_g^+$. But there is also the state $^1\Delta_g$ that has no "+/-" label, yet it is still anti-symmetric. The "+/-" label has nothing to do with anti-symmetry. Anti-symmetry is ensured by use of Slater determinants. The "rule" you mentioned is a result applicable to $\pi^2$ (and other doubly degenerate 2 electron) configurations, but is not related to anti-symmetry and gives no predictions for other configurations.


Is it possible to find the ratio of isotopes only given the mean mass number?

You have two equations and 3 unknowns, so you can't solve it with just that. Say a, b, c are the fractions (as a decimal) of each isotope...

$$ a(x) + b(x+1) + c(x+2) = (x+\frac{1}{2}) $$

$$a + b + c = 1 $$

The 4:1:1 solution works. Another that works is 3:0:1. Another is 7:4:1. There are infinitely many solutions.


Does dissolving salt in water change the pH?

When you dissolve a salt in water, you have to distinguish 4 cases:

The salt results from the reaction of a strong acid with a strong base, the pH will be 7. This is the case, for example, of $\ce{NaCl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ and $\ce{Cl^-}$ are spectator ions and don't react with water. So, the pH doesn't change.
The salt results from the reaction of a strong acid with a weak base, the pH will be less than 7. This is the case , for example, of $\ce{NH4Cl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NH3}$. The ions $\ce{Cl^-}$ are spectator ions and don't react with water. While the ions $\ce{NH_4^+}$  do hydrolyze and give ions  $\ce{H^+}$ to water. So, the pH will decrease.
The salt results from the reaction of a weak acid with a strong base, the pH will be more than 7. This is the case , for example, of $\ce{CH3COONa}$ that results from the reaction of $\ce{CH3COOH}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ are spectator ions and don't react with water. While the ions $\ce{CH3COO^-}$  do hydrolyze and accept ions $\ce{H^+}$ from water. In this case, water act as an acid leaving a hydroxide ion. So, the pH will increase.
The salt results from the reaction of a weak acid with a weak base, the calculus of pH is a little bit complex. We need to take into consideration the $K_a$ and the $K_b$ of the weak acid and the weak base and whichever is the stronger acid or weak will be the dominate factor in determining whether the solution is acidic or basic. We can investigate three cases:

$K_a= K_b$, the aqueous solution is neutral.
This is the case, for example, of $\ce{NH4CH3COO}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{CH3COOH}$. The ions  $\ce{NH4^+}$ and $\ce{CH3COO^-}$ do hydrolyze. As the hydrolysis constant of ion acetate equals the hydrolysis constant of ion ammonium, the same concentarion of ion hydronium and ion hydroxide is produced, and the pH of the solution is neutral.
$K_a> K_b$
This is the case, for example, of $\ce{NH4F}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HF}$. The ions  $\ce{NH4^+}$ and $\ce{F^-}$ do hydrolyze  and as  ammonium ions hydrolyze to a greater extent than  fluoride ions, i.e more hydronium ion is produced, and the pH of the solution is acidic.
$K_a< K_b$
This is the case, for example, of $\ce{NH4CN}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HCN}$. The ions  $\ce{NH4^+}$ and $\ce{CN^-}$ do hydrolyze  and as  cyanide ions hydrolyze to a greater extent than  fluoride ions, i.e less hydronium ions are produced, and the pH of the solution is basic.


what kind of suspension liquid should be used with ferrofluid (so it does not stain the glass)

I had success with the following:

1) Glass bottle, ideally with flat sides.  If you have a round bottle you may not be able to see the ferrofluid due to diffraction.

2) A solution of ~25% distilled water (a.k.a. deionized water) and ~75% of store bought isopropyl alcohol at 91% concentration.  That means that ~70% of my solution was actually isopropyl alcohol.  I eyeballed the ratio between the water and the isopropyl alcohol, those numbers are likely to not be exactly correct.

Procedure that I followed:

Add some isopropyl alcohol to the glass container.  Then add some (or all) of your ferrofluid to the container.  Then add some water to bring your isopropyl alcohol concentration to about 70%.  Next, use a magnet to settle the ferrofluid to the bottom and then remove the magnet.  If your ferrofluid has little bits that are floating, then you need to add more water.  If your ferrofluid looks too stringy then I would add more alcohol.  When I was done the ferrofluid had zero staining, like you see in the videos online, except for in a few spots.  I think for better results I would need a higher quality glass container.

notes: Don't put too much ferrofluid in your container, otherwise you will have an uninteresting blob.  I recommend using gloves.  To clean the ferrofluid stuff off my hands I used warm water, laundry detergent, and lots of scrubbing with a towel.  If you mess up, you can use a magnet to hold your ferrofluid in place while you pour off some of your suspension fluid.  In addition, you can use a straw and your thumb to pipet some of the ferrofluid, however some will get stuck in the straw, so I wouldn't do that unless you have a bunch of ferrofluid.  Finally, I only finished like 30 minutes ago, so I hope it will all look good tomorrow.  If something bad happens I will make another post.

notes, part 2: If all you want is a cool desk toy, then I recommend just buying one of the pre-made ferrofluid toys online. (I saw some for ~$30).  You will get a lot more ferrofluid for your money if you buy just the ferrofluid, however making the toy is a bit of work.

GOOD LUCK!

edit:  no combinations of the following worked for me: acetic acid, hydrogen peroxide, pep plastic.


How do we know cyclopentadienylmolybdenum tricarbonyl occurs as a dimer?

The suggested answers from the textbook are to look at the output of a mass spectrometer, or to observe the macroscopic crystal structure.

However, NMR spectroscopy is also a perfectly acceptable way to go about it.


How can three p orbitals per atom bond in tin (II) sulfide?

I think your conclusion of a dative bond between sulphide anions and tin cations (assuming pure ionic bonding, which is clearly not the case if you study the image closely) is correct.

Note that the charges you are observing are nothing but formal charges. In a macroscopic picture, tin is still less electronegative than sulphur and the electrons would be considered sulphur’s e.g. when determining oxidation states. The mere fact that an electron pair with correct symmetry is adjacent to an atom is enough to stabilise said atom’s empty orbital.


Why can you blow bubbles in milk but not hot cocoa, which is milk based?

Milk contains lecitihin, which acts like soap to keep fat globules in milk suspended (although it does slowly separate into cream and skimmed milk, which you could demonstrate with non-homogenized milk).

Lecithin and soap work by having long molecules, one end of which is attracted to water (hydrophilic) and the other is attracted to oil or air more than to water (hydrophobic). This helps the water form a double layer, with air inside and outside and the lecithin holding things together. C. V. Boys Soap-Bubble and the Forces Which Mould Them is worthwhile reading and lists numerous experiments with detergent and water that you might try with your daughter's assistance.

Ah... but you ask why cocoa interferes with bubble blowing! Here are two tentative explanations:

The cocoa particles don't dissolve, so they "poke holes" in the soap/water/soap membrane, bursting the bubbles.
The cocoa adheres more tightly to the hydrophobic end of the milk lecithin than does air, preventing formation of the bubble layers.

You and she could experiment to see if either are correct. Try putting something other than cocoa in the milk (you might substitute dish-washing detergent in water for milk, because it makes longer lasting, cleaner and cheaper bubble stuff). Finely shredded cellulose is hydrophilic, and powdered candle wax is hydrophobic, for example. You could try finer powders, too. If you do experiment, please post your results here - I would be interested in your findings!


Sodium bisulfite is one of several sulfur-based chemicals commonly known as ‘sulfites’. Others include potassium bisulfite, potassium metabisulfite, sodium metabisulfite, sodium sulfite, or sulfur dioxide, and also your calcium bisulfite. They have fungicide properties and stop the natural browning process of produce. Dried apricots stay orange, and golden raisins stay yellow. Sulfites stop the browning and further ripening of fruit by blocking the natural enzyme, polyphenol oxidase. Sulfites are also added to some wines.

Sulfites destroy vitamin B1 (thiamin) and so the FDA does not allow sulfites on food containing vitamin B1 (thiamin). They estimate that 1 in 100 people are sulfite sensitive, so it is illegal to use sulfites on foods sold raw, except potatoes. They get an exception because there is no alternative to keeping french fries from looking brown.

The FDA lists sulfites as generally recognized as safe, but sulfites are linked to neurological, gastrointestinal, and respiratory problems. Trace amounts of sulfites have been known to cause hives, vomiting, wheezing, and anaphylaxis. I know from experience that too many dried apricots can cause intestinal cramping.

Modified, but essentially reporting from  https://traditionalcookingschool.com/preserving/dehydrating-preserving/sulfites-and-food-preservation/


Is this textbook bonding picture of hybridization in ethene wrong?

You are not mistaken. Your book is wrong. If one desperately wants to involve the hybridisation concept (which makes sense for carbon), then the description should be in accordance with the molecular geometry of ethene:

So the carbons can be described as $\ce{sp_{x}p_{y}=sp^2}$ hybridised, hence there are four $\ce{H_{s}-C_{sp^2}}$ $\sigma$ bonds, one $\ce{C_{sp^2}-C_{sp^2}}$ $\sigma$ bond and one $\ce{C_{p_{z}}-C_{p_{z}}}$ $\pi$ bond.

However the concept of hybridisation is no law, but yet another tool to describe bonding. In case of methane, (ethane,) ethene and ethyne it is quite simple to come to certain $\ce{sp^{x}}$ hybrid orbitals where $\ce{x}\in\mathbb{N}$ - different molecule may lead to fractional $\ce{x}$. (Not a rule, just a guideline)

In response to Dissenters comment:

The coordinate system is chosen arbitrarily to indicate the molecular plane to be $xy$, as indicated in the picture taken from the german article of wikipedia:

As the π orbitals have to be perpendicular to this plane they only can be composed of $\ce{p_{z}}$ orbitals.

The molecular plane can be chosen different, e.g. as the $xz$ plane, the the $\ce{sp^2}$ orbitals would be formed from $\ce{sp_{x}p_{z}}$ and the $\pi$ bond would consist of $\ce{p_{y}}$ orbitals.


I drink city water, city has lead pipe, how is this safe?

In light of the recent events in Flint Michigan, this issue has come to the forefront (see this article from the American Chemical Society's Chemical  & Engineering News and see this article). To summarize:

Active Measures to prevent lead corrosion:

Maintain high pH. This makes lead carbonates less soluble in water.
Add orthophosphates which create relatively insoluble layer on the inside of the pipes.

Without these treatments, the addition of chloride disinfectants aggravates the corrosion problem by forming lead (II) chlorides (which is water soluble).

As mentioned by Jaroslav, the time that water sits in the system also plays a factor.


Fermentation of different types of sugar

Micro-organisms eat the sugars then excrete all sorts of compounds. Acetic acid and/or ethanol is commonly excreted by anaerobic organisms.  Anaerobic organisms can derive energy from multiple reaction paths, even within the same micro-organism (consider the case of facultative anaerobes.)

Since the same organism can use different paths to derive it is quite natural to assume what comes out is naturally affected by what goes in.


How to determine which gas would have the lowest pressure?

Not all of them are polar. The gases  $\rm N_2$, $\rm O_2$,  $\rm Cl_2$ are not polar.

The $\rm HC\equiv C-CH_3$ is polar, but $\rm H_2C=C=CH_2$ should not be and $\rm C_6H_4$ is also polar, if it is right chemical formula you wrote. But I'm not sure which is more polar, but I will say that it is $\rm C_6H_4$

.

In more polar compound there will more interaction between molecules and that is why pressure will be smaller, simply because they "want" to be liquid form - not in gas.


Assigning the IUPAC name of this compound

This is because alkene has higher preference than alkyl halides. So when counting clockwise position of ene is less than that when counting anticlockwise. Hence the answer is as stated.

Check out Rule 6.2 of basic principles on Wikipedia.


What fraction of the vapour in electronic cigarettes is water?

The short answer to this is that there is no single answer to this problem as the vapor composition that is produced by any given atomizer is dependant upon a large number of components. To name a few:

The initial composition of the liquid, which will vary flavor to flavor and manufacturer to manufacturer. You can find out a bit more about this with the following articles

Characterisation of mainstream and passive vapours emitted by selected electronic cigarettes DOI 10.1016/j.ijheh.2014.10.001
Electronic cigarettes: overview of chemical composition and exposure estimation PMCID: PMC4304610

     A note to consider with the above is that many toxic organic solvents such as Formaldehyde, Acetaldehyde, Acetone, Acrolein, Propanal, Ethylene glycol, and    Diethylene glycol are found in eLiquids, in various concentrations.

The temperature that the eLiquid is heated to internally (this affects the vapor pressures of the various components in the liquid)
The concentrations of all of the, solvents, flavorings and such which are adsorbed by the person who is using the atomizer

Given all of these factors, and the large number that I am sure that I am missing, there is simply too many confounding variables to be able to give a straight answer to your question without directly measuring the concentration - not that I would know how you would make such a measurement.


Why a specific temperature for value of 1 calorie?

why temperature such as 0 or 1 degree celsius is not chosen??

That has been chosen by some.
According to Physical Laboratory Experiments: Part III.--Heat, 4th edition (1912) :

The following different calories should not be confused:

ZERO DEGREE CALORIE:- The quantity of heat required to raise one gram of water from 0 [degrees] C to 1 [degree] C.

ORDINARY CALORIE: cal.-The quantity of heat required to raise one gram of water from 15 [degrees] C to 16 [degree] C, this being taken as the mean room temperature.

MEAN CALORIE OR ICE CALORIMETER CALORIE: The one-hundred part of heat required to raise one gram of water from 0 [degrees] C to 100 [degrees] C.

...

Heat capacity of liquid water is a continuously varying function of temperature itself, so a particular temperature or temperature interval must be specified.  Various intervals were chosen historically, and when a small interval near 15 degrees C is chosen it is because that was thought of as room temperature 100-200 years ago.


If I understand your question correctly, you want to relate a droplet's geometry to it's environment. The curvature of a liquid droplet is described by the Laplace equation of capillarity, also known as the Young-Laplace equation. This is valid for both a sessile (sitting on a surface) and a pendant (hanging) droplet. I am assuming a static experiment in which the droplet is axisymmetric, with $z$ defined as the axis of symmetry. Such methods of solving for interfacial tension and contact angles are called Axisymmetric Drop Shape Analysis (ADSA) methods. ADSA methods are based on fitting the shape of an experimental droplet to the Laplace equation. This differential equation has been numerically solved in 1883, but thanks to computers and camera's this method is easier and more accurately to apply.

The Laplace equation is given by:

$$\gamma\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\Delta P\label{eq:laplace}\tag{1}$$

in which $\gamma$ is the surface tension and $\Delta P$ is the pressure difference over the interface. After some work, this may be written as a system of differential equations.
$R_1$ and $R_2$ are the principal radii of curvature.
$R_1$ is called the meridional curvature, which turns in the plane of the screen you are reading from.
Perpendicular to this is the azimuthal curvature $R_2$.

The meridional curvature is defined as:

$$\frac{1}{R_1}=\frac{\partial \phi}{\partial s}$$
The azimuthal curvature is defined as:
$$\frac{1}{R_2}=\frac{\sin{\phi}}{x}$$

`In the absence of external forces, other than gravity, the pressure difference is a linear function of the elevation' (Del Rio and Neumann, 1996):

$$\Delta P= \Delta P_0+\left(\Delta\rho\right)gz\label{eq:rio}\tag{2}$$
$\Delta P_0$ is the pressure at the origin, at which the droplet is not perturbed and $R_1=R_2=R_0$.
Using \eqref{eq:laplace} this leads to the following expression:
$$\Delta P_0 =\gamma\left(\frac{2}{R_0}\right)\label{eq:p0}$$
Combining \eqref{eq:laplace} and \eqref{eq:rio} and filling in the expression for $\Delta P_0$ and the curvatures leads to:

\begin{align}
\gamma\left(\frac{\partial \phi}{\partial s}+\frac{\sin{\phi}}{x}\right)&=\frac{2\gamma}{R_0}+\left(\Delta\rho\right)gz\\
\frac{\partial\phi}{\partial s}&=\frac{2}{R_0}+\frac{\left(\Delta\rho\right)gz}{\gamma}-\frac{\sin{\phi}}{x}
\end{align}
Now the coordinates are transformed to dimensionless coordinates:
\begin{align}
\bar{x}=\frac{x}{R_0}, \bar{z}=\frac{z}{R_0}, \bar{s}=\frac{s}{R_0}
\end{align}

This changes the differential:
\begin{align}
\frac{\partial\phi}{\partial s}=\frac{\partial\phi}{\partial\left(R_0\bar{s}\right)}=\frac{\partial\phi}{R_0\partial\bar s}
\end{align}
Multiplying both sides with $R_0$ gives:
\begin{align}
\frac{\partial\phi}{\bar s}&=2+\frac{\left(\Delta\rho\right)gR_0^2}{\gamma}\bar z-\frac{\sin\phi}{\bar x}\label{eq:phi}
\end{align}

Furthermore the following geometric interpretation leads to the remaining differential equations.
The arc length $s$ follows the curvature of the interface of the drop. $\phi$ is the tangential contact angle, so a change in $x$ over a change in $s$ is equal to the cosine of $\phi$:

\begin{align}
\frac{\partial x}{\partial s} &= \frac{\partial \bar{x}}{\partial \bar{s}} = \cos(\phi)\label{eq:cos}
\end{align}
 Similarly, a change in $z$ over a change in $s$ is equal to the sine of $\phi$:
\begin{align}
\frac{\partial z}{\partial s} &= \frac{\partial \bar{z}}{\partial \bar{s}} =\sin(\phi)\label{eq:sin}
\end{align}
The same holds for the dimensionless coordinates, resulting in the given system of differential equations.

Solving the curvature of the droplet is an initial value problem, with the initial values being $x_0$, $z_0$ and $\phi_0$. You can feed this to your numerical differential equation solver, e.g. Matlab's ODE45 function. With this method you will be able to solve the contact angle and surface tension from a given curvature, or draw a curvature for a given contact angle and surface tension.

Suggested reading on which I have based my answer:

Y. Rotenberg, L. Boruvka, and A. W. Neumann, “Determination of surface tension and contact angle from the shapes of axisymmetric fluid interfaces,” Journal of Colloid and Interface Science, vol. 93, pp. 169–183, May 1983.

Rı́o, O. I. del, and A. W. Neumann. “Axisymmetric Drop Shape Analysis: Computational Methods for the Measurement of Interfacial Properties from the Shape and Dimensions of Pendant and Sessile Drops.” Journal of Colloid and Interface Science 196.2 (1997): 136–147. ScienceDirect. Web.


Why does the reaction between Na (s) and Cl2 (g) occur spontaneously if the entropy of the system decreases?

As you point out there are two things to consider, the heat released in the reaction and the change in entropy. In forming NaCl from chlorine gas and solid sodium the entropy decreases because the gas is converted into solid NaCl which has a far far smaller total volume. The energy released has the effect of heating the solid NaCl just formed and then the surroundings as energy flows away to eventually reach equilibrium again but now at a higher temperature than that at equilibrium before reaction occurred.

When a quantity of heat $dq$ from a body at $T_2$ transfers to one at $T_1$ the overall entropy change is $dS = dS_1 + dS_2 = dq/T_1 - dq/T_2$. As $T_2 \gt T_1$ the entropy change is positive.

In the reaction the entropy change overall depends on comparing that decreased due to volume decrease vs. that increased due to the energy released on reaction.

Entropy is a measure of the number of ways that the particles can be placed into their energy levels, this is often called the number of configurations $\Omega$, and Boltzmann found that $S=k\ln(\Omega)$ so that increasing the distribution of energy into the energy levels increases $\Omega$ and hence S. As the temperature increases more vibrational and rotational levels are populated and the kinetic energy distribution widens, all this means that there is an increased number of configurations and hence entropy just because energy can be placed into  energy levels and in more ways.

If there was just one level then all 'particles' would have to be in this level and so entropy would be zero, $\Omega=1$.  When there are two levels some can be in the first and some in the second level and now the entropy is increased ($\Omega \gt 1$), and so on for many levels.


Electrochemical cell polarity and electrodes

The electrons which are "given up" by the oxidized species at the anode flow from the anode through the electric circuit to the cathode, where they are "needed" for the reduction reaction. The oxidation creates an excess of electrons (negative charge) at the anode, which is compensated by the consumption of electrons at the electron-deficient cathode (positive charge) for the reduction reaction. In summary, the flow of electrons in the electrochemical cell is from anode (electron source) to cathode (electron drain).


Determining the most stable enolic form of 1-phenylbutane-1,3-dione?

In opposition to the line of thought presented, I argue that the 4-hydroxy isomer is more stable than the 3-hydroxyl form for that the conjugated $\pi$-electron system is larger, than in the former.
This is based on multiple assumptions.

For one, given a molecular framework which may have multiple $\ce{C=C}$ bonds, if other parameters are not altered, the system with double bonds in conjugation is more stable than the one with isolated double bonds. As an example, the computed free enthalpy of formation, $\Delta{}_RG$ about 1,5-hexadiene (two isolated $\ce{C=C}$ bonds) is $\pu{92.70 kJ/mol}$ (link to computation on MolCalc.org), the about 1,3-pentadiene (two $\ce{C=C}$ bonds in conjugation) $\pu{86.57 kJ/mol}$ (link to computation).  And the less endothermic a compound, the more thermodynamically stable.

For the other, I said the 3-hydroxyl isomer would contain a less extended conjugated $\pi$-system, than the 4HP isomer.  Elements to participate in this system are the benzene ring, the carbonyl group, and the simple $\ce{C=C}$ bond, present in both.  If you imagine a water wave running from one to the other other end of the $\pi$-system, $\ce{C=O}$ only is tangent in 3HP, while in 4HP the wave runs all across these bonds.  Thus, equally a reason why the 4HP isomer is more favourable, than 3HP.
For a raw simplification, you may consider the benzene ring as "super atom" and count the longest pathway accessible. 3HP: 4 atoms, 4HP: 5 atoms participate in the $\pi$ system; again energetically more favourable for 4HP, than for 3HP.

Sadly, the engines to run MolCalc.org's web site constrain the computations to molecules with 10 non-H atoms maximum.  Thus, contrasting to earlier posts by others (e.g., this by Ben Norris), I'm not able to support my arguments by a similar computation for your substrates (12 non-H atoms).


Why does the conversion of SO₂ to SO₃ require high pressure?

As Philipp stated in the comments it is Le Chatelier's principle in play.

The principle in layman's language states that

  If you try to bring about any physical change in a system the system will try to but not necessarily succeed in cancelling the change.

The reaction you have given is an equilibrium reaction where all the compounds are in gaseous state.

Now if you count the total number of moles on the left-hand side it will be $\ce{2SO2 + O2}$
i.e. $\ce{2 + 1}$ that is $\text{3}$.
And on the right-hand side there are only $\text{2}$ moles of $\ce{SO3}$.

So ultimately if you consider the reaction to proceed in forward direction 3 moles of gas are converting into 2 moles of gas and hence the volume decreases in forward direction.

If you think according to Gas Law(or logically) an increase in pressure generally corresponds to a decrease in volume.

Now use Le Chatelier's principle to say that if I increase the pressure on the the system ,in order the system
to counteract to the change will decrease it's own volume and it has no other way of achieving that but to proceed in the forward direction.

But if the query arises that high pressure should favour any and every reaction take a look at the example below:

$$\ce{PCl5 <=> PCl3 + Cl2}$$

Here the gaseous moles on the left-hand side is $\ce{1}$ and on the right-hand side $\ce{2}$. So increasing pressure makes the reaction proceed in the backwards direction as there are less number of gaseous moles on the left-hand side.

  In fact, in higher level this idea is what is called $\Delta\text{n}_\text{g}$ i.e the difference($\Delta$) of gaseous($_g$) moles($\ce{n}$).

  $$\Delta\text{n}_\text{g} \quad = \text{gaseous moles on right-hand side - gaseous moles on left-hand side}$$

If $\Delta\text{n}_\text{g}$ is positive, then on increasing pressure reaction proceeds backwards and vice-cersa


The molecular orbitals of lithium hydride

We may want to consider also the energy change in the lithium 1s orbital.  Formation of the molecular orbitals also changes the energy level of the core orbitals even though these do not participate appreciably in the bonding.

This is because what I call "reverse shielding" occurs.  In a lithium atom in the ground state, there are of course two 1s electrons and one 2s electron.  We commonly suppose that the 1s electrons, packed close to the nucleus, partially shield the 2s electron from the nuclear charge and raise the energy level of the latter.  But, because the probability distributions of the 1s and 2s clouds overlap, there is a small probability that the 2s electron is instead shielding one or both of the 1s electrons.  The effect on the 1s wavefunctions is small, due to the small probability, but the effect on lithium 1s energy can still register in electron volts because the 1s wavefunction is crowded into a region of strong Coulonb potential.

Then, when the hydrogen atom is added and the bonding orbital forms, what had been the 2s electrons are drawn away from the lithium atom and thus the reverse shielding of the lithium 1s electrons is diminished.  Therefore, the lithium 1s orbital is lowered in energy, possibly enabling a stable molecule even if the bonding molecular orbital is higher than the isolated hydrogen 1s orbital as the OP suggests.


How Can I Induce Corrosion Electrochemically?

Corrosion is a more or less involuntary/undesired reaction. If you apply a current, the corrosion is defined by the current you apply, so you are controlling the system, not just observing it. In general, electrochemical forcing puts your system into a highly unnatural regime, but can certainly provide interesting information.

Corrosion can be affected (in ways that you can determine and document) by temperature, by contact with other metals, by the corrosivity of the solution (concentration and specific ions), and by inhibitors.

Some time ago, we were interested in stress-corrosion cracking (failure) of stainless steels. Hydrogen, (slowly generated in a natural environment) would embrittle the steel. By increasing the corrosivity of the solution (chlorides) and adding an activator (sulfides), we were able to speed up the hydrogen evolution/absorption so that steel specimens under tensile stress would fail within days or weeks, rather than months or years in nature. This enabled choosing alloy compositions that were most resistant to stress-corrosion cracking.


Cahn–Ingold–Prelog priority of carboxylic group versus ortho carboxylic acid

When using the Cahn–Ingold–Prelog (CIP) priority system and the ‘Sequence Rules’, double bonds are split into two bonds and the doubly bonded atoms are duplicated.

Thus

is treated as

where (O) and (C) are duplicate representations of the atoms at the other end of the double bond;

whereas

is treated as

Therefore, the name of the structure that is given in the question is (2E)-3-bromo-3-iodo-2-(trihydroxymethyl)prop-2-enoic acid.


Why are H+ and OH- ions preferentially discharged in the electrolysis of NaCl(aq)?

Water does not need to be dissociated into ions in order to participate in electrochemical reactions. While it is common to refer to electrode reactions as "discharging" this or that species, there is no requirement that the reactant of an electrode reaction must always be an ion.

In neutral solution, it's quite common for the water electrolysis reactions to be written as:

$$
\begin{align}
\text{Cathode:} &\quad \ce{4H2O + 4e- -> 2H2 + 2OH-} \\
\text{Anode:} &\quad \ce{2H2O -> O2 + 4H+ + 4e-}
\end{align}
$$

It's absolutely the case that the electrochemistry written using the ions as reactants in many cases is likely to be more facile:

$$
\begin{align}
\text{Cathode:} &\quad \ce{4H+ + 4e- -> 2H2} \\
\text{Anode:} &\quad \ce{4OH- -> O2 + 2H2O + 4e-}
\end{align}
$$

But it's still far (far) more favorable for $\ce{H2O}$ to be reduced than $\ce{Na+}$, and in many cases more favorable for $\ce{H2O}$ to be oxidized than $\ce{Cl-}$. You're quite right that the concentrations of $\ce{H+}$ and $\ce{OH-}$ ions are extremely low in neutral salt solutions, but that fact is mostly irrelevant to the electrochemistry.


How to search substructures with explicit hydrogen in RDKit?

If you work in a Jupyter Notebook you can visualize your substructure patterns and the search results, so you can see how the patterns work.

from rdkit import Chem
from rdkit.Chem import Draw

smiles = ['CC1=CC=CC=C1','CC1=C([H])C([H])=C([H])C([H])=C1[H]',
          'C1(C2=CC=CC=C2)=CC=CC=C1','C12=C(CC3=CC=CC=C3C2)C=CC=C1',
          'C12=CC=CC=C1C=C3C(C=CC=C3)=C2','C12=C(C(C(C=C3)=CC=C4)=C4C=C2)C3=CC=C1']

params = Chem.SmilesParserParams()
params.removeHs=False # draw and work with explicit Hs
mols = [Chem.MolFromSmiles(s, params) for s in smiles]
Draw.MolsToGridImage(mols, molsPerRow=3)

This will give you the compounds and especially the one with the explicit Hs.

Now we get the patterns and especially the one with the explicit Hs.

patt1 = Chem.MolFromSmiles('CC1=CC=CC=C1')
patt2 = Chem.MolFromSmiles('CC1=C([H])C([H])=C([H])C([H])=C1[H]', params) # search with explicit Hs
patt3 = Chem.MolFromSmiles('Cc1ccccc1')
patt4 = Chem.MolFromSmarts('Cc1ccccc1')
patt5 = Chem.MolFromSmarts('[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]')

patts = [patt1,patt2,patt3,patt4,patt5]
leg = ['CC1=CC=CC=C1','CC1=C([H])C([H])=C([H])C([H])=C1[H]','Cc1ccccc1 smiles',
       'Cc1ccccc1 smarts','[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]']
Draw.MolsToGridImage(patts, molsPerRow=3, legends=leg)

Search and highlight the result.

ms = []
patt = []
allsubs = []
for m in range(len(mols)):
    for p in range(len(patts)):
        sub = mols[m].GetSubstructMatches(patts[p])
        if len(sub) > 0:
            ms.append(mols[m])
            patt.append(leg[p])
            allsubs.append(sub[0]) # substructures could be find multiple time - just take the first

Draw.MolsToGridImage(ms, molsPerRow=3, legends=patt, highlightAtomLists=allsubs)

Compounds 5 and 6 have no nonaromatic bonds, so there is nothing to find.


Acid Catalyzed Hydration of an Alkene Synthesis Problem

You propose to use but-1-ene rather than but-2-ene as a starting material to synthesise butan-2-ol via addition of water to the double bond. There is no reason why it would not happen the way you drew.

However, using but-2-ene allows only one possible reaction pathway. With the relatively high symmetry of the molecule ($C_\mathrm{2h}$, assuming trans), attacking either side of the double bond will lead to an identical product: The molecule’s intrinsic $C_2$ axis will transform one into the other. Thus, it is only possible to generate butan-2-ol.

Starting from but-1-ene, you have two non-equivalent first steps, generating either the primary or the secondary carbocation. These obviously cannot be transformed into each other by symmetry (the alkene’s symmetry is now $C_\mathrm{s}$). Thus, you will get a side product: butan-1-ol. These products will be difficult to separate. If anything, careful distillation would do the job.

As RobChem already noted in the comments, (organic) chemists typically attempt to avoid side reactions and different products, hence why the book suggests using the more symmetric alkene.


Predicting pH of an acidic salt of an weak acid and weak base

The main reactions which take place are

1)Ionization of $\ce{HA-}$: $\ce{HA- -> H+ +A^2-}$

2)Hydrolysis of $\ce{HA-}$: $\ce{HA- + H+ -> H2A}$

3)Hydrolysis of cation: $\ce{B+ + H2O -> BOH + H+}$

for reaction $1$, $$k_{a2}=\frac{[H^+][A^{2-}]}{[HA^-]}$$

for reaction $2$, $$k_{a1}=\frac{[HA^-][H^+]}{[H_2A]}$$

for reaction $3$, $$k_h=\frac{k_w}{k_b}=\frac{[BOH][H^+]}{[B^+]}$$

At any moment, $$\ce{[H^+]=[BOH] +[A^{2-}]-[H2A]}$$
$$[H^+]=\frac{[NH^{4+}].k_w}{[H^+]k_b}+\frac{k_{a2}.[HCO_3^-]}{[H^+]}-\frac{[HCO_3^-][H^+]}{k_{a1}}$$
Rearranging the terms to get $\ce{[H+]}$,
$$[H^+]=\sqrt{\frac{\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]}{1+\frac{[HCO_3^-]}{k_{a1}}}}$$
$$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{k_{a1}+[HCO_3^-]}}$$

Now we will have to make 2 assumptions. First lets assume that $k_{a1}<<[HCO_3^-]$ so that $k_{a1}+[HCO_3^-]=[HCO_3^-]$

$$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{[HCO_3^-]}}$$

If you assume that neither hydrolysis nor the dissociation goes too far, you can assume the concentration of bicarbonate ion to be equal to the initial concentration (say C). Since Ammonium bicarbonate dissociates to give two products in 1:1 ratio, concentration of ammonium produced is equal to the concentration of bicarbonate ion produced.

  I saw this formula for calculation pH of such salts in my textbook but
  no derivation was given:
  $$[H^+]=\sqrt{k_{a1}\cdot\big(\frac{k_w}{k_b}+k_{a2}\big)}$$ The above formula can
  be obtained if you consider the last assumption. (Clarification
  needed)


What ion remains in the largest concentration at equilibrium: 0.1 M H₃PO₄ and 0.2 M KOH?

Actually, the ion with the largest concentration is $\ce {K+}$.

But that is not what they want. Since potassium hydroxide is a strong base, then the concentration of $\ce {OH-}$ starts with 0.2 M (roughly pH of 13.3). To neutralize this, it would take 2 hydrogens from $\ce {H3PO4}$ at the specified concentration to get both concentrations near $10^{-7}$ (pH ~7). So the major anion is $\ce {HPO4^{2-}}$.

You can find the exact concentrations if you need.

Addendum:
Since $K_w = 10^{-14}$, then $\ce {[H^+][OH^{-}]=10^{-14}}$
Adding 1 L of each solution would give 0.2 moles of $\ce {OH^-}$ to react with any liberated hydrogen from the phosphoric acid. Removing one hydrogen from all the phosphoric acid gives 0.1 mole of $\ce {H+}$, which reacts with the hydroxyl, leaving 0.1 mole of $\ce {OH^-}$ (pH of 13.0). Since this is smaller than ${K_{a2}}$, more hydrogen is released from $\ce {H_2PO_4^{-}}$.

Assuming 0.2 moles of hydrogen is released, then there is an equal amount of $\ce {H+}$ and $\ce {OH^-}$ in the solution. To have $K_w = 10^{-14}$, then each of the ions needs to be near $10^{-7}$M.

Calculations:
Find $\ce {[H_2PO_4^{-}]}$:
$$\ce {H+ + HPO4^{2-} <=> H_2PO_4^{-}}$$
$K_{a1}=\frac {[H_2PO_4^{-}]}{[H^+][HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$7.5\times 10^{-3}=\frac {[H_2PO_4^{-}]}{10^{-7}\times 0.05}$
$[H_2PO_4^{-}]={10^{-7}\times 0.05}\times {7.5\times 10^{-3}}$
$[H_2PO_4^{-}]=3.75\times 10^{-10}$M

Find $\ce {[PO_4^{3-}]}$:
$$\ce {HPO4^{2-} <=> H^+ + PO_4^{3-}}$$
$K_{a3}=\frac {[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$4.5\times 10^{-13}=\frac {10^{-7}\times [PO_4^{3-}]}{0.05}$
$[PO_4^{3-}]=\frac {0.05\times 4.5\times 10^{-13}}{10^{-7}}$
$[PO_4^{3-}]=2.25\times 10^{-7}$


What are some popular science books in chemistry?

1) Curiosity: How Science Became Interested in Everything
by Philip Ball
 (https://www.goodreads.com/book/show/13650233-curiosity)

2) Molecules: The Elements and the Architecture of Everything
by Theodore Gray, Nick Mann
(https://www.goodreads.com/book/show/20729580-molecules)


What is change in internal energy of a system in which combustion occurs at constant temperature?

The change in internal energy of a reaction mixture is a function not only of the temperature change but also of the changes in the amounts of reactants and products in the mixture.  This is because energy is consumed and released when we break- and make chemical bonds.  The equation you wrote for the internal energy change applies only to a pure species or to a mixture of constant chemical composition.  The occurrence of chemical reactions causes a change in the internal energy even at constant temperature.


Stability of Gauche and Anti form in conformational Isomerism

This problem should be referred to concept called stereoelectronic effect. It is stabilized by the donation of $\sigma_{C-H}$ bond to $\sigma_{C-F}$ bond, leading to decrease energy of the whole system.


Didn't you consider the way through sulfanilic acid, the para-position will be protected. In this case both substituents will direct the electrophilic substitution in ortho- position with respect to amino group. Then remove sulfonic acid group by desulfonation and you will obtain your ortho-bromoaniline.


Can toluene be oxidized to benzoic acid using Pyridinium Chlorochromate(PCC)?

PCC is generally regarded as a milder oxidizing agent, used to effect selective oxidation of alcohols to aldehydes (from $1^{\circ}$ alcohols) or ketones (from $2^{\circ}$ alcohols). While over-oxidation to a carboxylic acid from a primary alcohol is definitely possible, PCC should not react with a simple unfunctionalized benzylic carbon.

However, oxidation at the benzylic position is easily achieved with more aggressive oxidizing agents. In particular, chromic acid (generated in situ from chromium salts in aqueous sulfuric acid, as in the Jones oxidation) as well as aqueous potassium permanganate will fully oxidize virtually any benzylic carbon all the way to the carboxylic acid, providing the benzylic carbon has at least one hydrogen. Of course, the rest of the chain is completely cleaved in the process.

Edit: Out of curiosity, I consulted my copy of March's Advanced Organic Chemistry to determine if any other common reagents were able to effect the same benzylic oxidation. Nitric acid is listed along with the two methods I described above. Naturally, a survey of various more exotic methods is also given. Intriguing was the use of sodium hypochlorite in acetonitrile, or NBS in aqueous sodium hydroxide under photochemical conditions, for the specific complete oxidation of aryl methyl groups.


As it's been stated in the comments, both formulas are equivalent, however, I'd like to clarify the terminology and use cases for both.

$\ce{CaF2 * 3Ca3(PO4)2}$ is a formula of a formal addition compound (often used for clathrates, crystallohydrates and multiple salts). Widely used several decades ago when crystal structure of a given compound hasn't been elucidated enough to speak of exact atomic positions, hence it's often been attempted to present it as a ratio of simpler compounds packed in a crystal lattice.

$\ce{Ca5(PO4)3F}$ is a formula unit. Constituents are ordered from the most electropositive to the most electronegative groups; the ratio between them is denoted with numerical indices. By multiplying the formula unit $\ce{Ca5(PO4)3F}$ by $Z = 2$ ($Z$ is the number of chemical formula units per unit cell), you get $\ce{Ca10(PO4)6F2}$ — the content of the entire unit cell. $\ce{Ca10(PO4)6F2}$ is equal to $\ce{CaF2 * 3Ca3(PO4)2}$ in terms of stoichiometry, but the use of latter middle-dot-notation is deprecated for multiple salts as it doesn't carry additional structural information and often leads to confusion.

I think it's important to mention that not only $\ce{Ca5(PO4)3F}$, but $\ce{CaF2}$ and $\ce{Ca3(PO4)2}$ are also formula units, and not molecules!


Why can't I dry/concentrate ethanol using calcium chloride ... or can I?

The apparent answer is yes, to quote from a source (composed from extracts of old chemistry journal):

  Calcium chloride forms addition compounds with the alcohols. On evaporating a solution in ethyl alcohol at a low temperature rectangular plates of $\ce{2CaCl2.7C2H5OH}$ are deposited. The compounds $\ce{CaCl2.3C2H5OH}$ and $\ce{CaCl2.CH3OH}$ have also been separated, as well as a mono- and a di-acetone compound, and compounds with isobutyl and amyl alcohols.

So, if you are working with a cold solution of $\ce{C2H5OH}$ with $\ce{CaCl2}$ exposed to air undergoing evaporation, it is possible to form an addition compound (taking the form of rectangular plates) with the formula, $\ce{2CaCl2.7C2H5OH}$ or $\ce{CaCl2.3C2H5OH}$.


$\newcommand{\conj}[1]{\overline{#1}{}}
\newcommand{\braket}[2]{\langle{#1}\,|\,{#2}\rangle}
\newcommand{\bracket}[3]{\langle{#1}\,|\,{#2}\,|\,{#3}\rangle}
\newcommand{\mat}[1]{\mathbf{#1}}
\newcommand{\rel}{\vec{r}_{\mathrm{e}}}
\newcommand{\rnuc}{\vec{r}_{\mathrm{n}}}
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\Tnuc}{\linop{T}_{\mathrm{n}}}
\newcommand{\Tel}{\linop{T}_{\mathrm{e}}}
\newcommand{\Velel}{\linop{V}_{\mathrm{ee}}}
\newcommand{\Velnuc}{\linop{V}_{\mathrm{en}}}
\newcommand{\Vnucnuc}{\linop{V}_{\mathrm{nn}}}
\newcommand{\Hel}{\linop{H}_{\mathrm{e}}}
\newcommand{\psiel}{\psi_{\mathrm{e}}}
\newcommand{\psieli}{\psi_{\mathrm{e},i}}
\newcommand{\psielj}{\psi_{\mathrm{e},j}}
\newcommand{\Eel}{E_{\mathrm{e}}}
\newcommand{\Eeli}{E_{\mathrm{e},i}}
\newcommand{\Eelj}{E_{\mathrm{e},j}}
\newcommand{\psinuc}{\psi_{\mathrm{n}}}
\newcommand{\psinuci}{\psi_{\mathrm{n},i}}
\newcommand{\psinucj}{\psi_{\mathrm{n},j}}$

Setup

In short, our goal is to separate the coordinates of all electrons $\rel = \{ \vec{r}_{i} \}_{i=1}^{n}$ from that of the nuclei $\rnuc = \{ \vec{r}_{\alpha} \}_{\alpha=1}^{\nu}$. Thus, we write the time-independent Schrödinger equation in the following form
\begin{equation}
    \linop{H} \psi(\rel, \rnuc) = E \psi(\rel, \rnuc) \, , \tag{1}
\end{equation}
where the Hamiltonian operator is given as follows
\begin{equation}
    \linop{H} = \Tnuc + \Tel + \Velnuc + \Vnucnuc + \Velel \, ,
\end{equation}
or, explicitly in atomic units,
\begin{equation}
    \linop{H}
    = - \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^2
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^2
    - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{\alpha=1}^{\nu} \sum\limits_{\beta > \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
\end{equation}
where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $\beta$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.
Here for simplicity we ignore magnetic interactions and use the non-relativistic treatment.

Exact separation

We know that nuclei are much heavier as compared to electrons, $m_{\mathrm{n}} \gg m_{\mathrm{e}}$, and assuming that the momentum of electrons and nuclei is of the same order of magnitude, $p_{\mathrm{n}} \sim p_{\mathrm{e}}$, we conclude that kinetic energy of nuclei is far smaller than kinetic energy of electrons $p_{\mathrm{n}}^2/(2 m_{\mathrm{n}}) \ll p_{\mathrm{e}}^2/(2 m_{\mathrm{e}})$, or, talking in terms of classical mechanics, nuclei are moving considerably slower than electrons.
For a coupled motion of two subsystems with significantly different speeds it is often assumed that one could treat the motion of a relatively fast moving subsystem (electrons) as if it depends only on the static configuration of a relatively slow moving one (nuclei).
This assumption results in a procedure with two major steps, described below.

First step

First, we solve an equation describing the motion of electrons for a number of static configurations of the nuclei.
The equation is known as the electronic Schrödinger equation and it can be written as follows
\begin{equation}
    \Hel \psieli(\rel, [\rnuc]) = \Eeli([\rnuc]) \psieli(\rel, [\rnuc]) \quad \text{for} \quad i = 1, 2, \dotsc \, .
\end{equation}
where $\Hel = \Tel + \Velnuc + \Velel$ is the electronic Hamiltonian, $\psieli$ is the electronic wave function describing $i$-th electronic state, and $\Eeli$ is the electronic energy of the $i$-th electronic state --- the total energy of a molecule with nuclei being fixed at particular coordinates less the nuclear repulsion potential energy $\Eel = U - V_{\mathrm{nn}}$.

Note that the electronic wave function is a function of electronic coordinates only, though parametrically it still depends on nuclear coordinates. The parametric dependence of $\psieli$ and $\Eeli$ on $\rnuc$ means that for different sets of fixed nuclear coordinates $\rnuc$ we have different electronic Schrödinger equations with different solutions $\psieli$ and $\Eeli$, or, in other words, that the electronic Schrödinger equation written above is a family of equations parametrized by $\rnuc$, rather then one equation.

Second step

Once solutions of the electronic Schrödinger equation $\{ \psieli(\rel) \}_{i=1}^{\infty}$ and $\{ \Eeli \}_{i=1}^{\infty}$ are found for a number of different fixed nuclear configurations $\rnuc$, we took the second step: the task is to solve an equation of motion for nuclei which is obtained as follows.
Note first that since $\Hel$ is self-adjoint, its eigenfunctions $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ form a complete orthonormal set.
Now we use the assertion that a function of two independent variables $\psi(\rel, \rnuc)$ can be expanded over a complete set of functions of one variable $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ in the following way
\begin{equation}
    \psi(\rel, \rnuc) = \sum\limits_{i=1}^{\infty} \psieli(\rel, [\rnuc]) \psinuci(\rnuc) \, . \tag{2}
\end{equation}
Substitution of (2) in the Schrödinger equation (1), multiplication of both sides by $\conj{\psi}_{\mathrm{e},j}$, and integration over electronic coordinates $\rel$ only, leads to
\begin{equation}
    \Big( \Tnuc + \Eelj + \Vnucnuc - E \Big) \psinucj
    -
    \sum\limits_{i=1}^{\infty} \linop{\Lambda}_{ji} \psinuci
    = 0 \quad \text{for } j = 1, 2, \dotsc \, , \tag{3}
\end{equation}
where operator $\linop{\Lambda}_{ji}$ is defined as follows
\begin{equation}
    \linop{\Lambda}_{ji} =
    \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}}
    \big(
        2 \bracket{ \psielj }{ \nabla_{\alpha} }{ \psieli } \nabla_{\alpha}
        + \bracket{ \psielj }{ \nabla_{\alpha}^2 }{ \psieli }
    \big) \, .
\end{equation}
and arguments $\rel$ and $\rnuc$ were dropped for brevity.
The last equation describes a system of differential equations (one equation for each value of $j$), and once the system is solved and a set of functions $\psinuci$ is obtained, one could write down the exact solution of the Schrödinger equation (1) in the form we introduced above (2).

Approximations

The task to solve the system of equations (3) is extremely difficult and the reason is that equations in the system are coupled in a sense that the solution of any $i$-th equation, $\psinuci(\rnuc)$, enters all other equations as the $\linop{\Lambda}_{ji} \psinuci$ term. One obvious way to decouple equations in the system (3) is by assuming that $\linop{\Lambda}_{ji} = 0$ either for all $i \neq j$, or for all $i$ without any exception. Operators $\linop{\Lambda}_{ji}$ may be regarded as the elements of a square matrix $\mat{\Lambda}$ and of course the fact that we equate these elements with zero is an approximation: in the first case, when we equate with zero all off-diagonal elements $\linop{\Lambda}_{ji}$, we introduce the adiabatic approximation, while in the second case, when we equate with zero all elements $\linop{\Lambda}_{ji}$, we introduce the Born–Oppenheimer approximation.

Both these approximations lead to a system of the so-called nuclear Schrödinger equations
\begin{equation}
    \linop{H}_{\mathrm{n}} \psinucj(\rnuc) = E \psinucj(\rnuc) \quad \text{for } j = 1, 2, \dotsc \, ,
\end{equation}
where $\psinucj(\rnuc)$ is known as the nuclear wave function and $\linop{H}_{\mathrm{n}}$ as the nuclear Hamiltonian, with the only difference being that in the first case the nuclear Hamiltonian takes the following form
\begin{equation}
    \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj} \, ,
\end{equation}
while in the second case the last term $\linop{\Lambda}_{jj}$ is absent
\begin{equation}
    \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc \, .
\end{equation}

The first thing to notice is that in both the above mentioned approximations the only terms in the nuclear Schrödinger equation for $j$-th nuclear state $\psinucj$ that depend on electronic states, $\Eelj$ and also $\linop{\Lambda}_{jj}$, are related to the same and the only electronic state $\psielj$.
This means that in both these approximations each nuclear state $\psinucj$ depends only on one electronic state $\psielj$, while in the accurate physical picture (3), electronic states are coupled through off-diagonal $\linop{\Lambda}_{ji}$ terms, known as non-adiabatic coupling terms (NACT).

Note also that the first term in the nuclear Hamiltonian in both adiabatic and Born–Oppenheimer approximations represents the kinetic energy of nuclear motion, and consequently, the remaining terms collectively should stand for the potential energy of nuclear motion.
It is said thus that nuclei move on the so-called potential energy surface (PES) defined by the sum of these remaining terms: $\Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj}$ in the adiabatic approximation and just $\Eelj([\rnuc]) + \Vnucnuc$ in the Born–Oppenheimer one.

Non-adiabatic coupling terms usually (but not always) can indeed be neglected, especially around the equilibrium nuclear configuration.
Besides, often the errors due to neglecting these terms are less than the errors resulting from some further approximation introduced when solving electronic and nuclear equations.
However, when two potential energy surfaces come close to each other or even cross, non-adiabatic coupling terms becomes large and are no longer negligible.

Answers

  What adiabatic means?

The term "adiabatic" from Ancient Greek αδιαβατος (α - "not", δια - "through", βατος - "passable") literally means the situation when something "is not passing through" something else.
In thermodynamics adiabatic process is a process occurring without exchange of heat of a system with its environment, i.e. a process in which heat is not passing through system enclosure.
In quantum mechanics adiabatic refers to a process in which no abrupt transition from one state to another with respect to continuous changes of some parameters happens, and thus, energy of a system changes continuously with respect to continuous changes of these parameters.

And the reason why we call the above mentioned approximation adiabatic lies in the fact that in this approximation when solving the electronic Schrödinger equation for different fixed nuclear configurations by continuously varying the coordinates of the nuclei we indeed assume that electronic state of the system continues to be the same. In other words, we assume that we are "staying" at the same PES. This could be the case, of course, only when potential energy surfaces for different electronic states are well separated, so, in the regions of nuclear coordinates where potential energy surfaces come close to or even cross each other adiabatic approximation is inadequate.

  What would be the case of a non adiabatic appox. to Schrodinger eq.?

Non-adiabatic solution refers to the solution of the system of coupled equations (3).