You can justify the formation of Benzaldehyde from Toluene by oxidising with peroxide just by looking at the mechanism of the reaction and think of migratory aptitude of Phenyl group , hydride and methyl group.

Firstly, if we delve into the mechanism of the oxidation of Cumene, we'll find that there is a rearrangement step in the mechanism. The rearrangement occurs when one of the oxygen in hydroperoxide gets protonated and leaves as water. If you now think carefully, there are both the possibilities of migration of Phenyl and methyl group to the oxygen. But Phenyl has certainly more migratory aptitude than methyl, and also the carbocation formed after the rearrangement is stable, as it has two methyl group on the $\alpha$-position to stabilise by electron release and one oxygen atom to stabilise it via donation of lone pair. Thus the migratory aptitude and stability of the carbocation formed both drive the rearrangement to prefer the phenyl migration. Therfore acetophenone is not formed due to unfavoured methyl migration.

On the other hand, the situation with toluene is slightly different. Here you have both the possibilities of hydride as well as phenyl migration. But hydride has almost similar (slightly more) migratory aptitude as Phenyl group, and what is more interesting is that the migration of hydride will lead to more stable benzylic carbocation, which is not possible by phenyl migration. These two effects cooperatively facilitate the hydride migration and therefore leading to the product Benzaldehyde.

Thus the migratory aptitude and stability of the carbocation are the two effects which create the differences in the product formation in case of oxidation of Cumene and Toluene.

(Note : For clarification, see the migratory aptitude order here : https://en.wikipedia.org/wiki/Migratory_aptitude )


I think the answer lies with the fact that alkanes are basically non-polar, even if they have halogen substituents. Water, on the other hand, is polar, so the solvent and solute are incompatible.


What is the coordination number for BeCl2?

Coordination number is determined by compound. It is normal for an element to have several different CN. Preferable CN can change for an element depending on the atom's neighbors. For example, Fe has preferrable CN=6 for fluorine anions, but CN=4 for chloride anions. When heated or under extreme pressure preferable CN may change as well. For example, $SiO_2$ has high-pressure modification with 6-coordinated silicon and 3-coordinated oxygen, while at normal pressure they are 4- and 2-coordinated respectively.

In case of $Be$ it prefers, just like all elements of the 2nd row, to have CN=4, but in many cases is forced to live with smaller numbers. BTW, in gas phase, $BeCl_2$ exists in equilibrium between monomer with Be's CN=2 and dimer with Be's CN=3.

In your case CN for Be in cristalline $BeCl_2$ is 4, while in gas phase it is 3 and 2 for dimere and monomer respectively. There is no 'one CN for Be atom right for all Be compounds.'


Relative stability of carboxylic acid derivatives

Firstly "stabilised" is very ambiguous. "Stable" is always defined with respect to something. The proper phrasing here should probably be stable towards nucleophilic attack.

For the anhydride - yes the actual structure is a hybrid. That just means that both C=O have to "share" the central oxygen's lone pair - in a sense each of them get half resonance each - which means that both C=O's are not very stabilised.

I don't buy the "less likely" explanation. Your teacher is trying to say that there's a "50% chance" that at any one point in time a particular C=O is stabilised, but resonance isn't something that you flip a coin and decide which way it goes.

For the acyl chloride, it is really to do with the leaving group ability of Cl-. See this question for a discussion: Do acyl fluorides exhibit resonance donation or inductive withdrawal?

Essentially the idea is that since $\ce{Cl-}$ is a better leaving group than $\ce{OMe-}$, for the acyl chloride (X = Cl), resonance form 2 contributes more to the actual structure than it does in the ester (X = OR):

As you can see, resonance form 2 makes the C=O more susceptible to nucleophilic attack because of the positive charge associated with the carbonyl group. Therefore, any substituent X that favours resonance form 2 will make the carbonyl group less stable towards nucleophiles.


Find gas pressures of injection knowing mixture pressure without knowing volume

Of course it is a pipe and it has a infinite volume hypotactically,
To use ideal gas equations is a bit tricky with mass flow rate need to convert to molar flow rate of each gas thus need to divide by $\ce{M_w}$ molecular weight respective to each gas.
Consider $\ce{1}$ time unit where inside of pipe gets
$$\frac{Q_1}{M_{w_1}}$$
$$\frac{Q_2}{M_{w_2}}$$
moles of gas to one volume section of pipe
At one time unit according to the ideal gas equation
$$\ce{PV=nRT}$$
Gas 1 will consume $\ce{V1}$ amount of volume
$$V_1 = \frac{Q_1RT_1}{M_{w_1}p_1}$$
Gas 2 will consume $\ce{V2}$ amount of volume
$$V_2 = \frac{Q_2RT_2}{M_{w_2}p_2}$$
Total of $\ce{V}$ volume per each time unit
$$V = V_1 + V_2$$
But $\ce{V}$ is just a section of the pipe it just shifts to what ever direction the gas is flowing and keeps going for ever in the infinite pipe.
But as further the gases go away just like you said turbulence will stabilize and final temperature and pressure inside pipe will be almost equal to $\ce{T_{mix}}$ and $\ce{p_{mix}}$
When it is stabilized,
Gas 1 will consume $\ce{V'_1}$ amount of volume
$$V'_1 = \frac{Q_1RT_{mix}}{M_{w_1}p'_1}$$
Gas 2 will consume $\ce{V'_2}$ amount of volume
$$V'_2 = \frac{Q_2RT_{mix}}{M_{w_2}p'_2}$$
Total of $\ce{V'}$ volume per each time unit
$$V' = V'_1 + V'_2$$
The important factor here is that number moles of gas will shift to next stage will be the same thus molar fraction for each gas is a constant. Think as $\ce{V'}$ stage in pipe
Molar fraction can be easily calculated
$$\ce{X_1 = \frac{(\frac{Q1}{M_{w_1}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})}}$$
$$\ce{X_2 = \frac{(\frac{Q2}{M_{w_2}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})}}$$
Using the partial pressure equation
$$\ce{P_A = X_A P_{tot}}$$

what is the relation?

Refer here!
$$\ce{p'_1 = \frac{(\frac{Q1}{M_{w_1}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})} p_{mix}}$$
$$\ce{p'_2 = \frac{(\frac{Q2}{M_{w_2}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})} p_{mix}}$$
There you have it final answer but consider that its one time unit thus take only the magnitude with respective unit of mass flow rate * unit time $\ce{|Q1| |Q2|}$ symbol ` is there to show gases are stabilized.


Why does osmolarity take into account both penetrating and non-penetrating solutes?

Collected and elaborated info from the chat with the OP:
To summarize the osmolality and tonicity difference, elaborated in the below link:

Osmolality takes into account all solutes.
Tonicity excludes solutes able to pass the semipermeable membrane.

So isotonic solutions need not to have the same osmolality.
The osmotic pressure is caused by decreased water activity, what does not depend on if solute is or is not able to pass the membrane. By other words, osmotic pressure and related osmolality  is property of solvent, not solutes, similarly as kryoscopic and ebulioscopic effects, all belonging to the class of colligative properties( depending on solute molality, but not on solute nature )
Primary is osmolality and osmotic pressure difference it causes. The difference then causes water osmotic flow.
When the osmolality is getting about balanced, then there is progressively coming into the play the difference in tonicity.
The same osmolality but higher tonicity means there is lower osmolality and therefore indirectly concentration of permeable solutes. This means there is concentration gradient powering the diffusion flow toward more tonic side. It is taking water with it, keeping osmolality balanced.
As result, volume of more tonic side increases even with the same osmolality.
At certain circumstances, the residual osmotic pressure may be compensated by the permeable solute concentration gradient, keeping status quo.


When most people think of the term solvent, a liquid medium comes to mind; however, in the technical sense of the definition, this does not have to be the case. Here is the definition of a solution present in the IUPAC gold book:

  A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.

The definition of a solvent is incorporated into the definition of a solution. This definition states that the substance present in the larger quantity is the solvent while that present in the smaller quantity is the solute. Based on this definition, it is entirely possible to have a solvent that is in the solid phase. For example, brass is a mixture of 64% copper and 34% zinc. According to the IUPAC definition, a solution can be in the solid phase. Because copper is present in a larger amount, it is considered the solvent -- a perfect, everyday example of a solvent in the solid phase.


When using a "dilute hydrochloric acid" solution, how diluted should it be?

The exact concentration doesn't matter that much. The stock solution that I used to use was 1 M aqueous HCl (~3%). If your compound is acid-sensitive then you might want to try using even more dilute HCl (e.g. 0.01–0.1 M), or skip this step altogether and just purify it using a different method. The important thing is to make sure that you have enough HCl to protonate all the basic stuff you want to remove – a relatively simple stoichiometry exercise.

I would suggest not using concentrated HCl (12 M, or ~37%) as for a typical acid-base extraction it's simply not necessary. You should save that for other circumstances, such as when you want to adjust the pH of a solution without diluting it too much with water.


Spin state change during crystallisation for [NiCl2(PPh3)2]

I made both isomers as a PhD student many years ago. The red diamagnetic form crystallises out when a concentrated CH2Cl2 solution is cooled in a dry ice-acetone bath, and you can quickly filter off the red product. But on leaving the solid at room temperature, it reverts in a couple of days at room temperature to the blue, paramagnetic form. So it's a thermal equilibrium.


X-ray fluorescence: where do expelled electrons go?

There are a number of techniques that use excitation of inner shell electrons. Auger spectroscopy, and ESCA for example. But XRF is a "bulk" analysis method, not surface such as Auger. The excitation energy from the initial x-ray will be keV whereas the empty orbitals in the atoms of the sample will be a few eV. The excited electron rips trough the sample creating various ions until its energy is reduced enough so that the emitted electron can drop into some empty low energy orbital. (In general it would a "different" electron that drops into the vacancy and creates the fluorescent x-ray.)

In general the x-ray can penetrated deeper into the sample than an excited electron can escape. So the bulk of the excited electrons stay within the sample.


Certainly nuclear fusion and nuclear fission can convert elements into other elements. Indeed, this is how different elements form in the first place.

The one issue I have is with your "arbitrary chemical element." You can use fusion to put together lighter nuclei or elements into heavier ones - up to iron, or use fission to divide more massive nuclei into lighter ones.. up to about iron.

The limit here is the binding energy of the nucleus which maximizes with $\ce{^{56}Fe}$.

In general, though, your point is a good one. Given enough energy (e.g., in a supernova), you can create lots of different elements.


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), polycyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system as follows.

The main ring of a polycyclic hydrocarbon ring system is selected so as to include as many skeletal atoms of the structure as possible (here: 8 atoms).

The main bridge is the bridge that includes as many of the atoms as possible that are not included in the main ring (here: 0 atoms).

Bridges other than the main bridge are called secondary bridges:

  P-23.2.6.1 Naming polycyclic alicyclic hydrocarbons

  Rings not designated by the bicyclic system described above (…) are defined by citing the number of atoms in each secondary bridge as an arabic number. The locants of the two attachment points of each secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower first) separated by a comma. The numbers indicating independent secondary bridges (bridges that connect atoms of the bicyclic system) are cited
  in decreasing order. (…)

  P-23.2.6.1.1 The prefixes ‘tricyclo’, ‘tetracyclo’, etc., in place of ‘bicyclo’, indicate the number of rings in the polyalicyclic system. The number of rings is equal to the number of bond cuts necessary to transform the polycyclic system into an acyclic skeleton, unbranched or branched.

Therefore, the correct systematic name is ‘pentacyclo[4.2.0.02,5.03,8.04,7]octane’. However, the retained name ‘cubane’ is used in general nomenclature and as preferred IUPAC name (PIN).


Equilibrium constant for the neutralization of weak acid by strong base

Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:

$$\ce{HOAc + OH- <=> OAc- + H2O}$$

$$K' = \frac{[\ce{OAc-}][\ce{H2O}]}{[\ce{HOAc}][\ce{OH-}]}$$

Since $[\ce{H2O}] = \text{const}$ (reaction medium), $K'[\ce{H2O}] = K = \text{const}$:

$$K = \frac{[\ce{OAc-}]}{[\ce{HOAc}][\ce{OH-}]}$$

By multiplying both numerator and denominator by $[\ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_\mathrm{a}$ and ionic product of water $K_\mathrm{w}$:

$$K = \frac{\color{red}{[\ce{OAc-}][\ce{H+}]}}{\color{red}{[\ce{HOAc}]}[\ce{OH-}][\ce{H+}]} = \frac{\color{red}{K_\mathrm{a}}}{K_\mathrm{w}}$$

For acetic acid:

$$K = \frac{\pu{1.8e-5}}{\pu{1e-14}} = \pu{1.8e9}$$


p-Nitroaniline reaction with sulfuric acid

I think the abstract of the reference did not give away what's going on the reaction. However, I assume that the full paper should have described what's happening to at least certain extend. Since I don't have access to this paper at the moment, only I can do is speculate. Yet, I'd direct you to reliable source for certain information. Accordingly, when you heat the mixture of p-notroaniline and conc. sulfuric acid:

The only gaseous product you would obtain is $\ce{H2O}$ (with some nitroaniline subliming at higher mole ratios) up to the temperature of $\pu{230 ^\circ C}$. At $\pu{230 ^\circ C}$, it is called pre-expansion stage. Sulfur dioxide has been absent from gaseous products (confirmed by iodine test). It was noted that dehydration and sulfonation are the main chemical processes taken place at this stage.

There are two more stages to follow before you see the carbon snake, after stage of which called oxidative decomposition of the residue. This UCSB article also provides some other interesting excerpts from the original article (Ref.1), and listed below for readers' benefit:

The mole ratio of nitroaniline to sulfuric acid, however, is not critical; voluminous foams have been obtained from mixtures over a mole ratio range of 0.3 to 2.0 if a proper heating schedule is maintained. Sulfur dioxide and water are the principal gaseous products evolved during these expansions.

The foams were good thermal insulators inert to chemicals, e.g., concentrated mineral acids, alkalies, organic and inorganic solvents, and oxidants.

The exact structural nature of the insoluble residue constituting the foam is unknown. All the sulfur in the residue is present as sulfonic acid groups attached to the benzene nuclei. Secondary amino or azinium groups resulting from deammoniative coupling must be present in order to account for ammonium ions and Nitrogen content less than 2 per benzene nucleus. A high nitrogen-to-oxygen ratio indicates that few if any nitro groups remain. Thus, both nitro arid sulfonic acid groups are involved in oxidative processes.

Also, followings are three relevant demonstrations of the reaction: here, here, and here. Also, here is a demonstration of the dehydration reaction of sugar by concentrated sulfuric acid forming a column of graphite for comparison.

It is worth noting that the carbon form created by the p-nitroaniline reaction with sulfuric acid is found to be useful in $\ce{CO2}$ adsorption (Ref.2).

References:

A. C. Poshkus, J. A. Parker, "Studies on nitroaniline–sulfuric acid compositions: Aphrogenic pyrostats," J. Appl. Polym. Sci. 1970, 14(8), 2049–2064 (doi: https://doi.org/10.1002/app.1970.070140813).
Enrico Andreoli, Andrew R. Barron, "$\ce{CO2}$ Adsorption by para-Nitroaniline Sulfuric Acid-Derived Porous Carbon Foam," J. Carbon Res. 2016, 2(4), 25 (9 pages) (doi: https://doi.org/10.3390/c2040025).


What are the practices to naming organic chemicals by abbreviations?

The only accepted practice without any further constraints is to only use element symbols and sum or constitution formulae. Taking your and a few other examples, that would result in:

$\ce{C2H5OH}$ or $\ce{H3CCH2OH}$
$\ce{C6H5COCH3}$
$\ce{H3C(C6H5)2COH}$
$\ce{H3CCOOC2H5}$
$\ce{(CH3)3COCH3}$

The more complicated these get the less clear everything is. Thus, many people make use of a set of common abbreviations. However, these must be defined in a list of abbreviations, footnotes or wherever.

$\ce{EtOH}$
$\ce{PhAc}$[1]
$\ce{MePh2COH}$
$\ce{EtOAc}$
$\ce{$t$BuOMe}$

Where the symbols $\ce{Et, Ph, Ac, Me, $t$Bu}$ would need to be defined to mean ethyl, phenyl, acetyl, methyl and tert-butyl, respectively.

Finally, of course, you can suggest other common or your own abbreviations and use them (consistently!). For example, in my group it is rather common to abbreviate $\ce{EtOAc}$ as $\ce{EE}$ (because of its retained German name Essigester). Or a common way to denote the last compound in that list is $\ce{MTBE}$ or $\ce{TBME}$ (methyl-tert-butylether or tert-butylmethylether). Again, these abbreviations must be defined in footnotes or a list of abbreviations in best practice.

Finally note that an abbrevation typically includes all the hydrogens. Thus, methyl is always $\ce{CH3}$; $\ce{MeCl}$ is permissable for chloromethane but $\ce{MeCl2}$ does not mean dichloromethane. Consequently, neither $\ce{EtPh2OH}$ nor $\ce{Ph2EtOH}$ are permissable for 1,1-diphenylethanol, since carbon $\ce{C{1}}$ has no remaining hydrogens (not 2 as $\ce{CH3CH2}$ — ethyl — would imply).

[1]: Note that $\ce{Ac}$ can also mean actinium. If your paper uses actinium anywhere, do not use $\ce{Ac}$ as an abbreviation for acetyl or it shall cause confusion. The same is true for $\ce{Pr}$ and praseodymium and a few even less common others.


In most reactions that follow markovnikov's rule, a carbocation is formed and rearranged such that it is stabilised to the maximum extent .

The stability usually comes from the inductive effect, that is the electron releasing methyl groups that lead to a better charge distributed system. More is the number of such methyl groups surrounding the carbocation, more will be the stability and more such intermediates will be formed. These stable intermediates will be attacked by the negative reagent. This justifies the Markonikov's rule (at least in cases where carbocations are formed).

Note: This only applies to reactions such as acid catalysed hydration, action of halogen acids on alkenes, etc. In some reactions such as oxymercuration-demercuration no carbocation is formed but it still follows Markonikov's rule and uses a different reaction mechanism.


Can we separate positive ions from negative ions in a solution?

While it is true that solutions are not always locally electroneutral, that non-neutrality is small, and does not extend to larger scale. Thus, it does not impact chemical concentrations: you cannot separate positive and negative charges in a solution at a macroscopic level. See for example this excerpt from Ionic Transport Processes:in Electrochemistry and Membrane Science:


What is the structure of iodide of Millon's base?

First, ignore any monomeric formula. The compound is definitely polymeric.

The literature contains hints for existence of two separate lines of compounds. The first one is salts of polyamidomercury: $\ce{[Hg(NH2)]_{n}A_{n}}${1} with mercury-amide chains in the structure, and $\ce{(Hg2N)_{n}A_{n}}$ with structure based on crystoballite {2}{3}

It seems thatn the lines are not distinguished in earlier literature, but recent researchers use the words "Millon's base salts" for $\ce{(Hg2N)_{n}A_{n}}$ only.


Hybridisation of $\ce{BrF5}$ is $\ce{sp^3d^2}$ (involving one 4s, three 4p and two 4d orbitals) giving rise to octahedral geometry. But one hybrid orbital is occupied by lone pairs. So the effective shape of molecule is square pyramidal.


What is the meaning of "equivalent weight" and its formula?

Definition of Equivalent Weight (also known as equivalent mass):

The simplest definition of equivalent mass I can think of is the following:

The mass of a compound which will combine with 8g of oxygen or 1g of hydrogen is said to be the equivalent weight of the compound.

Lets take an example,

$\ce{2Ca + O2 -> 2CaO}$

From basic stoichiometry, you know that 2 moles of calcium react with 1 mole of oxygen to produce 2 moles of calcium oxide. We can say the same statement in terms of masses as 80g of calcium reacts with 32g of oxygen to produce 112g of calcium oxide.

Since 80g of Calcium reacts with 32g of oxygen, 20g of calcium should react with 8g of oxygen.

Now by the definition of equivalent weight mentioned earlier, the equivalent weight of calcium is 20g.

$$E = \frac{M}{V}$$

The formula you were talking about gives the equivalent weight of an element (can be extended to be used with ions).

Equivalent weight = (Atomic mass)/(Valency)

This could be intuitive and obvious for some. Something with valency of 1 would combine with just 1 hydrogen atom, or in other words, a mole of compound with valency 1 would react with 1g of hydrogen. If the element has a valency of 2, then it would love to combine with 2 hydrogen atoms, or in other words, a mole of the element will combine with 2g of hydrogen. By definition, equivalent weight is the mass of the substance that would react with 1g of hydrogen. Therefore you must divide the mass by 2 to get equivalent weight. In fact, this formula based definition of equivalent weight of an element is identical with the definition mentioned earlier in this answer.

Previously, we determined that equivalent weight of calcium was 20g using the other definition. We should obtain the same equivalent weight for calcium by using the formula.

Equivalent weight = (40g)/2 = 20g

Definition of equivalents:
Another terminology you might come across is equivalents.

Equivalents is similar to moles. The number of moles is given by (mass of the compound)/(molecular mass). Similarly the number of equivalents is given by (mass of the compound)/(equivalent weight).

For example, 1 equivalent of calcium contains 20g of calcium. 2 equivalents of calcium contain 40g of calcium.

The utility of equivalent weights is amazing. In any reaction, $N$ equivalents of any compound on the reactants side reacts with N equivalents of another reaction to produce $N$ equivalents of the product(s).

If that sounds complicated, this example should make it clear.

20g of calcium reacts with 8g of oxygen to produce 28g of calcium oxide.

An equivalent statement is, 1 equivalent of calcium reacts with 1 equivalent of oxygen to produce 1 equivalent of calcium oxide.


Why does a proton prefer to attack a double bond instead of protonating an alcohol in a reaction of unsaturated alcohol with sulphuric acid?

Both the reaction mechanisms are equally correct. Reaction takes place at numerous sites in most of the organic reactions, many kinds of products are formed, but the thing which actually give the major product is one of the intermediate, which is relatively stable than its peers.

We know, that 3° carbocation is a more stable intermediate than an allyl carbocation (irrespective of the fact that the latter is going to give a conjugated product), so the former will be present in adequate amount in order to give the major product.

Also, the formation of conjugated alkene takes place through multiple intermediates, and therefore it becomes important to care for the stability of each intermediate. I especially doubt the 3rd step, where along with a 1° carbocation, alkene is present, which acts as an EWG, and hence lowering the stability of the intermediate.

So, taking account of these factors, I would likely go with the given solution.

Edit: As per Yusuf Hasan's points on the stability of THF in comments, the given product, being a THF derivative is likely going to solvate the other intermediates and it is itself stable. That's why, THF is an unreactive compound and hence it is used as a solvent in many organic reactions. (source, under the heading "Reactivity of cyclic ethers - Oxacyclopropanes")


4 compounds in equilibrium with one another - determine their equilibrium concentrations

In case anyone else travels this road...

Using $\mathrm{\Delta G~=~-RTlnK_{eq}}$ and the 3 $\mathrm{\Delta G's}$ supplied above, we find that
\begin{aligned}
\mathrm{K^{1/4}_{eq}}~&=~0.003\\
\mathrm{K^{3/1}_{eq}}~&=~40.749\\
\mathrm{K^{2/3}_{eq}}~&=~2.163\\
\end{aligned}
We also know that the concentrations of the 4 species must total to 1

$${[1] + [2] + [3] + [4] = 1}$$

so we have the following 4 equations and 4 unknowns
\begin{aligned}%
[1]~ &=~~~ 0.003 ~[4]\\
[3]~ &=~ 40.749 ~[1]\\
[2]~ &=~~~ 2.163 ~[3]\\
[1] + [2] + [3] + [4]~ &=~ 1\\
\end{aligned}
solving for the 4 concentrations yields
\begin{aligned}%
[1]~ &=~~~ 0.23\%\\
[2]~ &=~ 20.27\%\\
[3]~ &=~~~ 9.37\%\\
[4]~ &=~ 70.13\%\\
\end{aligned}


Not sure what kind of explanation you are looking for, but

Electrode Dynamics, A. C. Fisher, Oxford Chemistry Press, 1996, ISBN: 019855690X

is a great concise explanation of overpotentials among other things.


Keeping one component concentration constant and varying another in a solution

Add varying amount of water and a standard B solution on top of the constant amount of A you have added. Then you will finish with several solutions with different B concentration.


Effects of a non-inert electrode in a half cell reaction involving hydrogen gas as the electrolyte ion

If you started with a full cell with two inert electrodes immersed in sulfuric acid and applied a large enough voltage between them, a current would flow. At one electrode, the easiest to oxidize species would be oxidized (in this case, that would probably be 2 H2O -> O2 + 4 H+ + 4e-). At the same time, the easiest to reduce species would be reduced at the other electrode (probably 2H+ + 2e- -> H2). If you switched the cathode electrode from some inert metal to copper, the easiest reduction would still be 2H+ + 2e- -> H2 (Remember, there is no Cu2+ in the system to reduce). If, on the other hand, you changed the anode from inert metal to copper, the easiest oxidation would probably be Cu -> Cu2+ + 2e-, so the net reaction would change.


The most important simplified criteria for the numbering are:

lower locants for the group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given has no principal characteristic group that would be expressed as a suffix. Thus, Rule (c) is not relevant in this case.
No locants are assigned to the delocalized electrons of the benzene ring. Thus, also Rule (e) is not relevant.
Usually, Rule (f) is important for benzene rings with multiple substituents. In this case, however, all six substitutable positions are completely substituted; i.e. the locant set (all considered together in a series of increasing numerical order) is always ‘1,2,3,4,5,6’ – no matter how we turn it.
Finally, Rule (g) stipulates that the lowest locant is assigned to the substituent that is cited first as a prefix in the name. Therefore, the correct name for the compound that is given in the question is 1,3-dibromo-5-chloro-2-ethyl-4,6-dimethylbenzene.


Why do many organic halogen compounds smell or taste sweet?

The taste of halogenated organic compounds mainly depends on the halogen atoms involved. Study showed that halogenation of aliphatic series imparted sweetness and halogenation of aromatic compounds increased/decreased the intensity of sweetness/bitteness. This increase/decrease of intensity is due to the efficacy of halogen atoms which is directly related to electron withdrawing power (Hammett $\pu{\sigma}$ value). This is called AH-B theory of sweetness$\ce{^{[3]}}$. Effect of halogen substituent on aromatic compounds to impart bitterness was in opposite trend when compared to effect of halogen substituent on aliphatic compounds to impart sweetness. For example, halogenation of  aminonitrobenzene compounds impart sweetness$\ce{^{[1]}}$. Nitroaniline has a relative sweetness of 40 and substituting halogen at para position respect to nitro group increases sweetness (chlorine: $\pu{x 400}$, bromine:  $\pu{x 800}$, iodine:  $\pu{x 1250}$).

Commercially, chlorine played an important role in imparting sweetness in organic compounds. One instance is halogenation of saccharin. Substituting fluorine at para position imparts sweetness but a bitter aftertaste. Substituting chlorine imparts sweet but with accentuated bitterness. Substituting bromine enhances sweetness but it is extremely bitter. Substituting iodine loses is sweetness leaving a slight bitter aftertaste. Chlorination of other halogen compounds like sucrose and its derivatives, tryptophan, saccharides, lactose, maltose has also been studied where sweetness/bitterness increases by many folds$\ce{^{[2]}}$.
Notes and references

Taste by Lloyd M. Beidler
Taste Chemistry by R.S. Shallenberger
Robert Shallenberger and Terry Acree proposed the AH-B theory of
sweetness. Simply put, they proposed that to be sweet, a compound
must contain a hydrogen bond donor (AH) and a Lewis base (B)
separated by about 0.3 nanometres. According to this theory, the
AH-B unit of a sweetener binds with a corresponding AH-B unit on the
biological sweetness receptor to produce the sensation of sweetness. (Wikipedia)


Rearrangement of a bicyclic dienone under acidic conditions

Your analysis of the possible products is excellent. I would consider both of the products to be major. Waring and coworkers studied this problem with a slightly different substrate (an extra methyl group at the 1-position, next to the carbonyl on the 'bottom' of the molecule). Under aqueous acidic conditions (sulfuric acid), the corresponding products were obtained in ~25% (mechanism A) and ~40% yield (product B). Only when run with sulfuric acid in acetic anhydride did mechanism B truly dominate (~90% vs 5% mechanism A).

There is actually a third pathway, where the methyl group shifts from ring juncture to ring juncture, and then goes through a mechanism similar to B. For the dienone in the question, the product is indistinguishable from mechanism B, but for the dienone studied in the publication, this gave another product in ~30% yield. I suspect that for the dienone in the question, the total yield of product B would be ~70%. That makes it appear as if mechanism B is dominant, but it is really two similar pathways operating that happen to give the same product.


Barron states that 4.18×10⁸ joules equal 1 kcal, is this correct?

This should never have happened in the scientific community, but:

  The energy values of food are measured by the Calorie, with a capital C. A Calorie is equal to 1,000 calories or 1 kilocalorie (kcal).

Source: here

It is possible that Barron's got confused between Calorie and calorie. Even so, they still would be off by a factor of 100.

According to Wikipedia:

  The SI unit of energy is the joule, with symbol "J"; one small calorie is now defined as exactly 4.184 J, and one large calorie is 4184 J. However, the two units are still used occasionally in technical work, and the large calorie is still widely used in nutrition.

So you are correct in saying that 1 "small" kilocalorie is 4180 joules. 1 "large" kilocalorie is $\pu{4.18e6 J}$, and maybe Barron's is now suggesting an "extra large" calorie.

More likely it is just a typo.


Why is an energy cost associated with splitting the HOMO spins in singlet oxygen?

Short story: The comment by orthocresol is correct, based on Hund's rules. Given the same spin multiplicity (i.e., the two singlet states), the state with higher orbital angular momentum ($\mathrm{^1\Delta_g}$) wins.

Loosely speaking, one can imagine that comparing the two singlet states, the higher angular momentum implies that the electrons are a bit further out, and suffer less electron-electron repulsion.

Longer story...

The "unified atom" model can apply to simple diatomics - that is, many of the same considerations for the electronic structure can be used for multi-electron atoms. There's discussion in both Mulliken (Rev. Mod. Phys. 1932, 4, 1. and Herzberg's Spectra of Diatomic Molecules).

The vibrational-electronic spectra of $\ce{O2}$ was determined fairly readily:
 (Figure from Chem. Rev., 2003, 103 (5), pp 1685–1758, adapted from Herzberg).

We can assign the lines as discussed by Mulliken in 1932:

  The $\mathrm{^1\Sigma_g^+}$, is 1.62 volts above the $\mathrm{^3\Sigma_g^-}$ according to theoretical considerations,[118] the $\mathrm{^1\Delta_g}$, is probably about halfway between $\mathrm{^1\Sigma_g^+}$, and $\mathrm{^3\Sigma_g^-}$ (cf. Fig 48).

Reference 118 is to Hund and Hückel:

F. Hund, Zeits. f. Physik 63, 726 (1930).
E. Huckel, Zeits. f. Physik 60, 442-3 (1930).

Both Mulliken and Herzberg discuss the Hund's rules in regards to many diatomics - that the triplet will lie lower than the singlet because of higher S, but then the two singlet states will differ based on angular momentum (here L).


Under standard conditions (100kPa, 0°C), nitrogen and hydrogen do not react with each other in any measurable way. It follows logically that also ammonia cannot decompose (as it would to a small extent in an equillibrium reaction), because otherwise it would decompose completely with time, although it is stable.

In equillibrium, the forward and backward reaction rates must be identical and finite.


How to pool fractions after chromatography?

If all fractions contain the same protein, I'd combined all and concentrated the combined solutions to desired concentration using biochemical techniques such as dialysis. If they contained more than one protein, then I'd combined all fraction collected under the correct peak. However, it is better ask your supervisor (or adviser) before you combined them. The person who wrote the protocol knows which one to collect. Yet, you can look at following reference to get an idea how to concentrate the combined protein solutions:

Sarah M. Andrew, Julie A. Titus, Louis Zumstein, "Dialysis and Concentration of Protein Solutions," Current Protocols in Immunology 1997, 21(1), A.3H.1-A.3H.5 (https://doi.org/10.1002/0471142735.ima03hs21).


So I think what you heard is about the right idea. The flat region is your buffering region, and isn't super helpful to deduce the pKa. Adding base consumes the weak acid, and because it is weak, you know you are mostly consuming the [HA] form, and equilibrating back to around the pKa. This is why your pH doesn't change much around the pKa.

As an example, if you had 10 mmol of HA to start, then at the pKa point you would have 5 mmol of HA and 5 mmol of A-. Now, if you keep adding base, at some point you will essentially consume the remaining 5 mmol of HA. Then, there will be negligible amount left and it can't buffer anymore - i.e. your pH will change rapidly because you are adding strong base. Here you will have ~0 mmol of HA, and ~10 mmol of A-. Note that you have twice the amount of A- now. The pKa will have been at the point where you had half of this.

Experimentally, I know two simple ways. Using a pH meter and no indicator, you have to measure the sharp region more carefully (i.e., drop by drop), because you want to be able to find the exact point where the pH changes the fastest. By approximating the derivative, i.e.

$$ \frac{d}{d(V_{base})} pH(x_i) \approx \frac{pH(x_{i+1})-pH(x_{i-1})}{x_{i+1}-x_{i-1}}$$
you can find the point of greatest change, or the equivalence point.  Obviously, better data in that region will give you more clarity. I used to do one titration quickly so that I would know the approximate region where I would need to be and then slowly titrate the rest.

The other method, i.e. using an indicator solution and then just very carefully reaching the equivalence point essentially gives you the same information, and I know people who can very quickly do this. But in general, this would not be as accurate.


How pigments actually function as colorants?

Consider natural indigo, formerly used to color blue jeans. It is insoluble; reduction converted it into water-soluble "indigo white". Fabric was soaked in indigo white solution; on air oxidation, it became insoluble blue indigo again, with indigo molecules now presumably mechanically entangled in the fiber and held by van der Waals forces.

More often "binders" called mordants are used. They form an insoluble "lake" when they coordinate to the pigment; the lake can bind to fabric fibers. You can read more about the chemistry of the process here.


The answer is yes, ionic force does influence real pH signal. Because pH meter is sensitive to activities (instead of concentration of H+ only) of ions that modifying it junction potential you must take in account ionic force when it is strength enough to significantly impact activities. There is a lot of limitation with pH-meter: above pH>12 or bellow pH<2 they are not precise at all (see for acidic and basic errors). Or if they are used in different solvents than water. Organic chemists should tell you what pH could mean in hexane. Assessing pH is a complex task that must be addressed with the care it deserves.

Anyway, you should find the user-guide of your pH meter on internet. I am sure the manufacturer will provide limitation and use tips.

There is a detailed chapter on pHmetry in Skoog and West Analytical Chemistry. It may help you to deeper understand those phenomena.


This is a bit touchy.  The problem is that a bare proton, $\ce{H^+}$, won't exist in water.  It attracts electrons too strongly.  Once this was recognized, chemists started to write the bare proton as being bound to a water molecule, thus $\ce{H3O^+}$.  However, studies also showed that water molecules are typically linked together in clusters containing a fair number of water molecules held together by hydrogen bonds (Google for this if you are not familiar with it).

Thus today's version really is $\ce{H+_aq}$ but that looks awful, so it is as you note $\ce{H3O+_aq}$ is more often used.

The important point is that how you write it isn't really important.  It is the realization that the bare proton attaches itself to any nearby source of electrons.

How you should handle your homework depends on what your teacher does in class.  I'd follow that example.


Phenol preparation from Grignard reagent

As requested by the OP - the route that lab chemists use to form phenols from aryl halides (not aryl fluorides) is via formation of the aryl boronic acid or boronate. These may be formed by either metallation of the aryl halides to form the aryl lithium or Grignard and reaction with a trialkyl borate $\ce{B(OR)3}$, or by Pd catalysed reaction with pinacol borane (Suzuki-Miyaura reaction). These boronates or boronic acids may be isolated, but in practice an oxidative workup, most often with basic peroxide, yields the phenol in good to excellent yield.

(read more links: 1, 2, 3)

image from US7678158B2.


Why is the equilibrium constant defined that way intuitively?

It can be understood in simple terms with just few intuitive keys (of course it is not very rigorous).

Take in mind that $K$ is a measure of how displaced to reactive or equilibrium the reaction is. So it should be greater when products concentration is relatively high (that is why they go into the numerator) and small when reactive concentrations are relatively high.
In elemental reactions (technical term) is like a balance between the speed of the reaction in the direction of products and the reverse reaction.
A reaction only occurs when its reactives are in the same region of the space. So, if you have an elmental reaction of the form $a A + b B \rightarrow $, for the reaction takes place it is needed that $a$ molecules of $A$ and $b$ molecules of $B$ collide. This is the main point: How probably is this event? Well, if you only need one molecule of $A$ ($a=1$), the probability that an $A$ molecule be in this specific place (at some time) is proportional to the $A$ concentration. If $(a=2)$ we need another $A$ molecule, as the probability for a second molecule being in that space region (at some time, in particular at the same time) is independent, the probability of finding two $A$ molecules in the same space region (at the same time) is the product of the two individual probabilities (is like throwing a dice two times). In general, for the mixture the probability of the encounter is proportional to $[A]^a [B]^n$.
The reaction speed in that direction is proportional to the number of encounters.

That is why you need to make products. Of course equilibrium constant $could$ be defined in other ways, for example, suppose that we define them as $K_{new}=ln(K_{old})$. In such case for the logarithm properties you'll have other expressions that includes sums. But there is not advantages of defining $K$ in that way.


Cooling down and heating of solution unexpectedly during cation identification test

After the help by @paracetamol, I found that the process of dissolution of $\ce{NH4Cl}$ is an endothermic reaction, which results in the decrease in the temperature of the solution.

An experimental demonstration can be found here.

Greater information may be found here.

Important parts of the second reference:

  When discussing this question, one needs to think about all the processes involved (or at least those that account for most of what happens). In general, any process will occur spontaneously, if $\Delta G$ is negative. To determine $\Delta G$, equation (1) is used.

  $$\Delta G = \Delta H - T \Delta S\tag{1}$$

  There is an enthalpic term related to the heat a reaction produces or requires and an entropic term that is temperature-dependent. We cannot say a lot about the entropic term a priori, but we can about the enthalpic term — which is luckily what the question is about. For that, we first need to write a reaction equation (2):

  $$\ce{NH4Cl (s) + n H2O <=>> NH4+ (aq) + Cl- (aq) + m H2O}\tag{2}$$

  Dissolving of the ammonium chloride can also be thought as two separate processes, see equation $(2')$.

  $$\ce{NH4Cl (s) -> [NH4+ (g)] + [Cl- (g)] -> NH4+ (aq) + Cl- (aq)}\tag{2'}$$

  Meaning that first we break up the ammonium chloride crystal structure and second we dissolve the bare ions. If we want to write that in single enthalpy terms, we can do that as shown in equation (3).

  $$\Delta H_\mathrm{tot} = -\Delta H_\mathrm{lattice}(\ce{NH4Cl}) + \Delta H_\mathrm{solv}(\ce{NH4+}) + \Delta H_\mathrm{solv}(\ce{Cl-})\tag{3}$$

  All of these values can be looked up. I have given those as calculated by Jenkins and Morris in table.

  $$\textbf{Table 1:}\text{ Values of enthalpies used in this answer as}\\\text{quoted from Jenkins and Morris (reference 1).}\\\begin{array}{cccc}\hline
\text{compound} & \Delta H_\mathrm{lattice} [\mathrm{kJ/mol}] & \Delta H_\mathrm{solv} [\mathrm{kJ/mol}] & \Delta H_\mathrm{tot} [\mathrm{kJ/mol}] \\\hline
\ce{NH4Cl} & -709.1 & - 694.7 & 14.4 \\ \hline\end{array}$$

  So we need energy (a lot of it!) to break up the crystal lattice of $\ce{NH4Cl}$. We then re-gain energy by creating the hydrated, i.e. dissolved, ions in solution. If energy is required, it is typically (excluding photochemical reactions — not the case here) heat energy which is simply drawn from the surroundings. If energy is released, it is typically (same caveat) released as heat into the surroundings.

  Thus:

  The heat is absorbed by the solid ammonium chloride. It is used to break up the salt crystal according to equation $(2'.1)$. Part of it is re-released (unnoticed) by forming solvent–ion hydrogen bonds (the solvation mechanism of $\ce{NH4Cl}$ — equation $(2'.2)$).
  This heat is drained from the surroundings, which in this case is primarily the water in which you wanted to dissolve the ammonium chloride. (Of course, the solution will then further exchange heat with whatever is around it, meaning that either the air around a bottle/flask or an outer flask as in your described experimental setup will lose heat at the benefit of the $\ce{NH4Cl}$-solution.)


The variation from ionic to covalent to metallic bonding is not clear-cut, but a matter of degree. For example, the bond between hydrogen and oxygen in water is mostly covalent (sharing electrons), but at room temperature, about one in 10,000,000 bonds break, with the hydrogen losing an electron to oxygen -- the ionic bond you discussed.

However, in the case lithium and beryllium, there is not enough difference between them to favor one getting an electron instead of the other. However, something more interesting happens: both are willing to let electrons go, so there is a sea of electrons, unattached to a particular atom! This is called a metallic bond, and it enables many metals to mingle to form alloys, such as bronze, amalgams and stainless steel.


Boiling point comparison between cyclohexane derivatives

It's not the double bond in C that is significant here, it's the oxygen. The carbon-oxygen bond is polar and will make C more "electrically sticky" with its neighbors compared to A. Since the $\ce{O-H}$ bond in B is even more polar than the ketone in C, its boiling point will be even higher. Read more about it on this page.

(If you'd like to skip to the relevant section, on Google anyway, search "ketone vs alcohol boiling point".)


Mechanism of three-membered lactone hydrolysis in base

According to Organic Chemistry: Second Edition,[1, pg 934](from which I assume that the question was taken):

Enantiomerically pure (R)-2-bromopropanoic acid reacts with concentrated sodium hydroxide to give (S)-lactic acid. The reaction goes with inversion and is a typical SN2 reaction—and a good one too, since the reaction centre is adjacent to a carbonyl group. If, on the other hand, the reaction is run using $\ce{Ag2O}$ and a low concentration of sodium hydroxide, (R)-lactic acid is obtained — there is overall retention of stereochemistry.

Therefore this reaction that has been questioned takes place at a low concentration of $\ce{OH-}$ in the presence of $\ce{Ag2O}$.
How do we decide between competing mechanisms?
The next paragraph clearly outlines what we need to look for in order to proceed.

In the first, we have conditions suited to an SN2 reaction: a very good nucleophile ($\ce{OH-}$) and a good leaving group ($\ce{Br-}$). Improve the leaving group by adding $\ce{Ag+}$ ($\ce{Ag+}$ assists $\ce{Br-}$'s departure much as $\ce{H+}$ assists the departure of $\ce{OH-}$ by allowing it to leave as $\ce{H2O}$) and worsen the nucleophile ($\ce{H2O}$ instead of $\ce{OH-}$, of which there is now only a low concentration), and we have the sorts of conditions that would favour an SN1 reaction. The trouble is, without neighbouring group participation, the cation here would be rather unstable — right next to a carbonyl group. The carboxylate saves the day by participating in the departure of the $\ce{Br-}$ and forming the lactone. The key thing to remember is that a reaction always goes by the mechanism with the fastest rate.

From what I have understood the question lies in the step after the lactone is formed. What reacts, $\ce{C-1}$ or $\ce{C-2}$? Let's assume that both are possible and proceed with the two possible cases.
Firstly, the $\ce{OH-}$ attacks the carboxylic carbon($\ce{C-1}$). We get the possible reaction mechanism to be:

In the second case, let's assume that $\ce{OH-}$ attacks $\ce{C-2}$. The possible reaction mechanism would be:

So, now we are faced with the choice of which reaction could take place, attack on
$\ce{C-1}$ or attack on $\ce{C-2}$.
From the second quotation, we can say that if we are able to decide which is faster, that would be the reaction that takes place and so we reach our next question.
Which reaction is faster?
In pathway 1, we are talking about the base hydrolysis of esters. However, esters are not very reactive and the concentration of base in this case is low as well.
In pathway 2, we are talking about an SN2 reaction taking place on the $\ce{C-2}$ carbon which removes the ring strain that is seen in the lactone. Therefore, this would be faster than base hydrolysis that takes place at $\ce{C-1}$
Another way of checking the speed of the reaction would be to use the principle of least motion, (as pointed out by Yusuf Hassan).
According to Advances in Physical Organic Chemistry, the
principle of least motion is defined as:

The principle of least motion, states that those elementary reactions will be favored that involve the least change in atomic position and electronic configuration.

Using this, we can compare the relative rates of the two reactions.
During the base hydrolysis of esters, the hybridisation of the $\ce{C-1}$ carbon changes from $\mathrm{sp^2}$ to $\ce{sp^3}$. This leads to a big change in the atomic arrangements since you go from a trigonal planar configuration to a tetrahedral arrangement.
In case of the SN2 reaction on the $\ce{C-2}$ carbon, the configuration as a whole doesn't change. It remains $\mathrm{sp^3}$ throughout the reaction, so there is less rearrangement taking place.
This implies that according to the principle of least motion, the SN2 reaction is favoured over the the basic hydrolysis of the lactone formed via the anchimeric effect.
Hence, we can say that the reaction that would take place would be the attack on $\ce{C-2}$ and not the basic hydrolysis of the lactone.
Reference:

Organic Chemistry: Second Edition; Jonathan Clayden, Nick Greeves, Stuart Warren, 2014; ISBN 978-0198728719


why are simple organic molecules usually white?

Most substances are white in powder form if they have no strong absorption of visible light. They whiteness is a consequence of light scattering by small particles and the fact that there is no absorption within the visible range.

The lack of visible absorption is another issue. Light is absorbed by molecules when the energy difference between a filled orbital and an unfilled one in the molecule matches the energy of a photon of light. For light photons the energy is the equivalent of wavelength and therefore colour.

In most small, simple organic molecules, the lowest unfilled orbitals are a long way from the highest filled ones. So they need a lot of energy to cause them to jump to an unfilled orbital. So they absorb only high energy photons (UV light is higher energy than visible light so this implies the absorption bands will be in the UV and we will not notice them as our eyes can't see UV).

The characteristics that create colour in molecules are often long runs of unsaturated bonds where the electrons can spread along the chain of unsaturation. In these delocalised systems the gap between occupied and unoccupied orbitals is smaller than in simpler molecules and often falls into the visible spectrum. These molecules are perceived as coloured.

The molecule below is a simple example of a dye (most are bigger and more complex than this):

The electrons in this molecule can delocalise over all three rings via the NN bridge and this creates orbitals with the right energy gap to absorb blue light.

In simpler molecules with little delocalisation the energy gap is far into the UV. Even in polycyclic aromatic systems the energy gap is often still in the UV.

Of course, this is a simplification, but the essence of what sort of molecules you need to cause visible colour is a relatively large unsaturated system and by definition this doesn't happen in very simple organic molecules. Wikipedia lists many more examples of dyes so you can get a feel for what sorts of structures make strongly coloured compounds.


How to calculate atomic number of an atom when wavelength is known?

Wikipedia (Bohr Model > Rydberg formula) tells us:

  The energy of a photon emitted by a hydrogen atom is given by the difference of two hydrogen energy levels:
  $$E=E_i-E_f=R_\mathrm{E} \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right)$$
  where $n_f$ is the final energy level, and $n_i$ is the initial energy level.

  Since the energy of a photon is
  $$E=\frac{hc}{\lambda}$$
  the wavelength of the photon given off is given by
  $$\frac{1}{\lambda}=R_\infty \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right).$$

  This is known as the Rydberg formula, and the Rydberg constant $R$ is $R_\mathrm{E}/hc$, or $R_\mathrm{E}/2\pi$ in natural units.

Wikipedia also tells that it can be extended with $Z^2$ for hydrogen-like elements:

  The formula above can be extended for use with any hydrogen-like chemical elements with
  $$\frac{1}{\lambda_{\mathrm{vac}}} = R_\infty Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
  ... where $\lambda_{\mathrm{vac}}$ is the wavelength of the light emitted in vacuum, $R_\infty$ is the Rydberg constant for this element, $Z$ is the atomic number, i.e. the number of protons in the atomic nucleus of this element and, $n_1$ and $n_2$ are integers such that $n_1 < n_2$, corresponding to the principal quantum numbers of the orbitals occupied before and after.

Now you can see, that your first equation can be derived from your second equation. You also can see, that you a) forgot the squares in your first equation and b) made a sign error.

$\lambda = \mathrm{64.1 \cdot 10^{-10}\ m}$
$R_\infty = \mathrm{10973731.568508\ m^{−1}} \left(= \frac{R_E}{h\ c} = \frac{\mathrm{2.18\ 10^{-18}\ J}}{h\ c} \right)$

With all this in mind, you should be able to solve the equation to get $Z$.

   $$Z^2 = \frac{n_f^2\ n_i^2}{\lambda\ R_\infty\ (n_f^2 - n_i^2)} \xrightarrow{Z>0} Z \approx 4$$


IUPAC nomenclature for an interesting looking organic compound

It indeed can be named quite nicely using the recent "phane" nomenclature from the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, here is the example*

By abstracting to the bicyclo phane composed of nine benzene nodes ("superatoms") of three kinds, we get

1,5(1,3,2),2,4,6,8,9(1,4),3,7(1,2)-nonabenzenabicyclo[3.3.1]nonaphane

Note that the "superatom" (more precise "amplificant") locant ordering matters. If the numbers in the first parentheses were sorted, it would mean a different structure, that is little bit hard to draw in 2D:

1,5(1,2,3),2,4,6,8,9(1,4),3,7(1,2)-nonabenzenabicyclo[3.3.1]nonaphane

Publicly accessible references:

Phane Nomenclature, Part I: Phane Parent Names
Phane Nomenclature, Part II: Modification of the Degree of Hydrogenation and Substitution Derivatives of Phane Parent Hydrides

*) Please note there is a mistake in the example from the book, the pyridine (4) has wrong internal ring numbering - 4 should be replaced with 5. The name is correct, though. (It is mentioned in errata, for page 310).


Is 1,2- Hydride shift possible in ketonic cations?

The carbocation doesn't rearrange since oxygen donates it's lone pair and the resulting carbocation is quite stable since the octet of all the atoms is complete.


Hydrogen fuel cells which produce electricity from hydrogen and oxygen are over a hundred years old
https://www.thoughto.com/hydrogen-fuel-cells1991799

The recently introduced Toyota Mirai runs off a hydrogen fuel cell.
www.toyota.co.uk/mirai  as does the Hyundai ix35 and Honda Clarity

There aren't very many hydrogen filling stations in the UK which rather limits its appeal


What is the functional difference between a radical anion and a nucleophile?

Radical Anions are hundreds if not thousands of times more reactive
  than nucleophiles (right?).

Radical anions are very reactive, but some nucleophiles (such as $\ce{CH3^{-}}$) are also very reactive.

A radical anion is a molecule that has had an electron added to it.  For example, treating naphthalene with sodium in liquid ammonia transfers an electron from sodium to naphthalene forming the naphthalene radical anion, the electron is unpaired.

A radical arises from the homolytic breaking of a bond.  This process leaves an unpaired electron with each of the two molecular fragments.

  If an acidic organic compound is deprotonated to form a nucleophile by
  removing a proton, then is there not an unpaired electron availible?

Removal of a proton $\ce{H^+}$ (just a proton, no electrons) would leave 2 paired electrons (and a negative charge) on the remainder of the original molecule.

Nucleophiles come in a lot of flavors.  They can be neutral, like the nitrogen in an amine, or they can be negatively charged.  If negatively charged, the nucleophile could be a radical anion or a carbanion.  So radical anions just belong to a sub-class of nucleophiles.  It's probably fair to say that carbanions and radical anions are usually better nucleophiles than a neutral nucleophile.  Comparing the nucleophilicity of a carbanion and a radical anion is harder to generalize since they are both so reactive, but the carbanion with 2 electrons would probably be more nucleophilic than the radical anion with one electron..


Thionyl Chloride and enols, carboxylic acids, alcohols

Obviously it's not true that thionyl chloride only works with sp3 substrates. As you say, a carboxylic acid, an sp2 hybridized substrate, can be chlorinated with thionyl chloride.

Ultimately, the difference is that the carbon of an enol is not readily attacked by a nucleophile, while the carbons in the other substrates can be. Each of the reactions go through the same first step(s): addition of the hydroxyl to sulfur with chloride leaving sulfur (likely through addition/elimination, not shown below). However the next steps are different depending on the functional group of the substrate. For an alcohol, chloride attacks the activated alcohol (looks very much like a sulfonate) in an SN2 reaction to give the alkyl chloride. For a carboxylic acid, chloride attacks the carbonyl carbon, and after addition/elimination (not shown in detail), the acid chloride is produced. Starting from an enol, reaction with thionyl chloride gives an enol chloro sulfate. Unlike the sp2-carboxylic acid case, the pi bond doesn't have an effective electron sink (such as oxygen), which means that the reaction would have to be a direct SN2 displacement. As I'm sure you learned, SN2 at sp2 carbons is very slow.

A SciFinder search shows that other reagents, notably phosphorus pentachloride can accomplish the transformation.


Thermodynamics adding salt to water changes the temperature

The keyword is that the salt dissolves. Dissolution entails at least two steps:

1) Overcoming solvent-solvent interactions and bonds. An extreme example: your dinner plate doesn't dissolve in your kitchen table. One reason is that there is an extremely high energy barrier to overcoming the hypothetical solvent's (in this case, the kitchen table's) intermolecular forces. You'd have to take an ax to the table to overcome the strong intermolecular forces that hold the molecules of your dinner table together.

On the other hand breaking these solvent-solvent "bonds" or intermolecular forces in water is easy; you can jump into a swimming pool just fine. I wouldn't suggest jumping into a table.

Also, as this step's suggests, this is an energy intensive process or an endothermic process. Energy must be consumed to break these solvent-solvent bonds or interactions.

2) Breaking solute-solute interactions and bonds. This however isn't required. For example, some unionized $\ce{NaCl}$ may be solvated by water. You'll find this more true with some of the less soluble salts. But most of the sodium chloride you toss in water will have dissolved (we'll get to why later), so what is solvated generally isn't $\ce{NaCl_{(s)}}$ or $\ce{NaCl_{(aq)}}$ (this is somewhat misleading terminology) but rather $\ce{Na^+}$ and $\ce{Cl^-}$.

This again is an endothermic process as bonds are being broken.

3) Solute-solvent stabilization (solvation). This steps entails the formation of solute-solvent intermolecular forces. For example, the sodium ion, with its positive charge, may form a hydration shell of water molecules.

This is an exothermic process, as "bonds" or more accurately, stabilizing intermolecular forces are formed.

Couple all three of these processes together and you have the actual dissolution process of $\ce{NaCl}$.

What's interesting about $\ce{NaCl}$ is that even with the third, exothermic step, the dissolution (which comprises all three steps) is still a tiny bit endothermic. So that's what the question is premised on - the addition of sodium chloride to water kicks off a spontaneous process which is endothermic. So one might ask why might this endothermic process still occur. The answer lies with entropy; the dissolution of sodium chloride is not enthalpically favorable (dissolution consumes energy - specifically - from the water) - yet the dissolution is entropically favorable.

Remember that all processes tend toward greater disorder over time, and the dissolution of sodium chloride is a perfect way to increase disorder; we go from unionized sodium chloride to two ions! The two ions are definitely going to have a lot more degrees of freedom than a single, unified molecule.

$$\ce{NaCl_{(s)} + H_2O ->Na^+_{(aq)} + Cl^{-}_{(aq)}}$$


It suffices to consider only the reaction

$$\ce{CO3^{2-} + H2O <=> HCO3- + OH-}$$

with $k_{B_1} = 2.1 \cdot 10^{-4}$, since the reaction

$$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$

with $k_{B_2} = 2.3 \cdot 10^{-8}$ will only contribute a small amount of $\ce{OH-}$.

$$k_{B_1} = \frac{[\mathrm{{HCO_{3}}^{-}}] [\mathrm{OH^-}]}{[\mathrm{{CO_{3}}^{2-}}]} = \frac{x \cdot x}{c - x} \approx \frac{x \cdot x}{c}$$

$$x = [\mathrm{OH^-}] = \sqrt{k_{B_1} \cdot c} = 6.455 \cdot10^{-3}$$

which gives a pH of 11.8.


How does monoethanolamine (MEA) increase pH?

It is the latter case where "the whole MEA molecule is taking a hydrogen from $\ce{H2O}$ leaving the $\ce{OH-}$". Alcohols very seldom acty as arrhenius bases adding $\ce{OH-}$ to solution. One exception is triphenylmethanol but I cannot think of a second that isn't a variation of the given example. Instead the unbonded electron pair of the amine will abstract a hydrogen from water to form an ammonium ion leaving a hydroxide counter ion and thus raising the pH.


Identification of unknowns with varying solubilities and states of matter

The solutions which you have already found look good so far. For solid A, we need to find a compound which thermally decomposes into $\ce{N2}$ and an element which is only soluble in oxidizing acids like $\ce{HNO3}$, and whose chloride is insoluble in water or acids. Silver azide could possibly be a good fit, as it can be cleaved into $\ce{Ag}$ and $\ce{N2}$ by heating, and silver chloride has a very low solubility in water. Care would have to be taken with the temperature when actually performing the decomposition of $\ce{AgN3}$, as it is explosive.

$$\ce{2AgN3 -> 3N2\uparrow + 2Ag}$$

Dissolving silver (solid B) in nitric acid gives a silver nitrate solution D and $\ce{NO}$ as colorless gas E.

$$\ce{3Ag + 4HNO3 -> 3AgNO3 + 2H2O + NO\uparrow}$$

$\ce{NO}$ turns red on contact with air when it is oxidized to nitrogen dioxide by atmospheric oxygen.

$$\ce{2NO + O2 <=> 2NO2}$$

From solution D, insoluble $\ce{AgCl}$ (F) precipitates when $\ce{HCl}$ is added.

$$\ce{Ag+ + Cl- -> AgCl\downarrow}$$


How to derive composite acidity constant for H2CO3*?

[OP] Because if I make summation of (19) and (20), it will be: [...]

They meant they combine the mathematical expressions, not the chemical equations.

To express

$$K_\ce{H2CO3^\Huge{*}} = \frac{K_\ce{H2CO3}}{1 + K}$$

in terms of concentrations, you substitute equations (1) and (2) into the expression:

$$K = \dfrac{\ce{[CO2(aq)]}}{\ce{[H2CO3]}}\tag{1}$$
$$K_\ce{H2CO3} = \dfrac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}\tag{2}$$

So

$$\frac{K_\ce{H2CO3}}{1 + K} = \frac{\dfrac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}}{1 + \dfrac{\ce{[CO2(aq)]}}{\ce{[H2CO3]}}}$$

Multiplying top and bottom of the fraction by $\ce{[H2CO3]}$ gives:

$$\frac{\ce{[H2CO3]}}{1 + K} = \frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]} + \ce{[CO2(aq)]}}$$

If you want to make the intention clearer, you define a total concentration of carbon dioxide as

$$[\ce{H2CO3^\huge{*}}] = \ce{[H2CO3]} + \ce{[CO2(aq)]}$$

and then write

$$K_\ce{H2CO3^\Huge{*}} = \frac{K_\ce{H2CO3}}{1 + K} = \frac{\ce{[H+][HCO3-]}}{[\ce{H2CO3^\huge{*}}]}$$


You're asked to find the mole fraction of $\ce{N2}$ after the reaction. That is defined by:

$$x_{\ce{N_2},\text{f}} = \frac{n_{\ce{N_2},\text{f}}}{n_{\text{tot},\text{f}}}$$

where $x$ indicates a mole fraction, $n$ the number of moles and the subscript $\text{f}$ means "final". Furthermore, since the container is sealed, no gases can escape the container. This means that:

$$\begin{align}
n_{\ce{N_2},\text{f}} &= n_{\ce{N_2},\text{i}} + \Delta n_{\ce{N_2}} \\
n_{\text{tot},\text{f}} &= n_{\text{tot},\text{i}} + \Delta n_{\text{tot}}
\end{align}$$

where the symbol $\text{i}$ indicates "initial" and $\Delta n$ indicates the change in the number of moles caused by the decomposition of $\ce{NH4NO3}$. The four quantities on the right-hand side of these two equations are the quantities that you will have to find.

$n_{\text{tot},\text{i}}$ - This can be found by using the ideal gas law, $pV = nRT$ with the data given before the explosion. Substituting $p = 56.01\mathrm{~kPa}$, $V = 1.18\mathrm{~L}$, and $T = 286{~K}$, you will find that $n_{\text{tot},\text{i}} = 0.0276\mathrm{~mol}$.
$n_{\ce{N_2},\text{i}}$ - Since 79% of pre-explosion air is $\ce{N_2}$, $n_{\ce{N_2},\text{i}} = (0.79)(0.0276\mathrm{~mol}) = 0.0218\mathrm{~mol}$.
$\Delta n_{\ce{N_2}}$ - This is related stoichiometrically to the number of moles of $\ce{NH4NO3}$. In fact, it is numerically equal, as the coefficients of $\ce{NH4NO3}$ and $\ce{N2}$ are the same. The molar mass of $\ce{NH4NO3}$ is $80.04\mathrm{~g~mol^{-1}}$, so $\Delta n_{\ce{N_2}} = (8\mathrm{~g})/(80.04\mathrm{~g~mol^{-1}}) = 0.100\mathrm{~mol}$.
$\Delta n_{\text{tot}}$ - This is also related stoichiometrically to the number of moles of $\ce{NH4NO3}$. Each mole of $\ce{NH4NO3}$ leads to the formation of $(2+4+1)/2 = 3.5\mathrm{~mol}$ of gaseous products (observe that water will be gaseous after the explosion due to the rise in temperature). Therefore, $\Delta n_{\text{tot}} = (3.5)(0.100\mathrm{~mol}) = 0.35\mathrm{~mol}$.

Putting it all together, we have:

$$x_{\ce{N2},\text{f}} = \frac{0.0218 + 0.100}{0.0276 + 0.35} = 0.323$$

Note that the mole fraction is dimensionless. The exact final temperature is also of no use; its only purpose is to tell you that water will exist as a gas (and therefore contribute to the total number of moles). It could be any other number above $100^\circ\mathrm{C}$ and the final answer would be the same, as the mole fraction is independent of temperature.

I hope I didn't mess up the arithmetic.


Is it correct to say that ethenol is an alcohol?

The golden book states that ethenol is an alcohol:

  Enols

  Alkenols; the term refers specifically to vinylic alcohols, which have the structure $\ce{HOCR'=CR2}$.

However, the modifier "vinylic" is rather important, so I would never omit it.


A cited a pH value below 3 or above 9 is in line with the usually citation for the stability of the Al2O3 protective layer. Per a source on the effect of pH, to quote:

  It is well known that aluminum resistance is related to the thin and compact layer of naturally formed oxide on aluminium surface, but this oxide layer is stable only in pH range 4-8. Lower or higher pH values caused prominent destroying of protective layer and thus the significant metal dissolution [15].

So, outside the pH range, your nano SiO2 coating formulation may be subject to attack. The presence of active Aluminum may serve to introduce some new chemistry.

Most likely, in my opinion, the unprotective Al can react even with water to form hydrogen and aluminum hydroxide. The latter reacts with so-called reactive silica per Science Direct:

  Silica forms complex precipitates with iron, aluminum, and magnesium hydroxides.

which disrupts the SiO2.

Another but unlikely path starts with some atomic hydrogen ingressed into the Aluminum metal (see, for example, this Science Direct discussion and also a 2008 thesis "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari discussing surface atomic hydrogen formation on Aluminum). Suggested reaction mechanics could then be if such a reaction is favorable (unlikely):

$\ce{.H ⇌ H+ + e-}$

$\ce{4 .H + SiO2 (nano) -(?)-> Si + 2 H2O}$

potentially directly destroying the nano SiO2. Here relatedly to the reaction above is a hydrometallurgy source discussing a commercial leaching application with $\ce{.H}$ on Al metal acting on $\ce{PbS}$ at the bottom of Page 218. However, the heat of formation of PbS is 10 times less than SiO2, hence this reaction path is not likely.

Proposed Solution: Test Copper vessels for suitability (see this reference).


Any popular experiment in chemistry that digital signal processing played a crucial role in?

Many spectroscopic experiments record analog data, which is subsequently Fourier transformed to give a frequency-domain spectrum. For example, in a nuclear magnetic resonance experiment, the signal emitted by an excited nucleus is picked up as a voltage in an RF coil for 1-2 sec. This time-domain signal is processed then converted to the frequency domain by FT.

As for 'popular', Richard Ernst won the 1991 Nobel in Chemistry for his contributions to this field of research.


Can the enrichment of Uranium be done via manipulating the boiling/melting point difference between differing isotopes of the same element?

Of course isotopes may be separated by distillation, but it's useful, AFAIR, only for hydrogen isotopes.
For uranium, gas centrifuge is proven much cheaper, to some orders of magnitude: https://ru.wikipedia.org/wiki/Газовая_центрифуга
(translate this page, it contains more information than English Wikipedia article).
The essence of this technology is known for decades, and included in basic chemistry textbooks. Only engineering/technological know-hows are sort of classified.
But in the politics, mass media, and perhaps on stackexchange, the whole topic is still considered taboo and is surrounded by fear.


How to derive the rate laws for three consecutive reactions?

Your first set of four equations look correct. Next you need to integrate the first equation to get [A] vs. time i.e. $\ce{[A]}= \ce{A_0}\exp(-k_1t)$ where $\ce{[A_0]}$ is the value at $t=0$, and substitute [A] into the second equation. Integrate this equation (integrating factor method) and substitute it into the third and so on. It gets messy!  You can obtain $\ce{[P]}$ easily as this is the sum of all other values less the amount you start with.

The general solution for a chain of n reactions in the special case that only the first species A is present at $t=0$ has the form

$$N_n = C_1\exp(-k_1t)~+ C_2\exp(-k_2t) ~+ ...+ C_n\exp(-k_nt) $$

$$C_1= \frac{k_1k_2...k_{n-1}}{(k_2-k_1)(k_3-k_1)...(k_n-k_1)}\ce{[A_0]} $$

$$C_2= \frac{k_1k_2...k_{n-1}}{(k_1-k_2)(k_3-k_2)...(k_n-k_2)}\ce{[A_0]}$$

$$ C_3 = .....$$
etc,  for as many terms as necessary.

good luck!


Why does the phosphate ion have resonance structures?

A much better representation of phosphate is the one given below (You need to add three lone pairs to each oxygen atom which I was too lazy to do):

It does not expand phosphorous’ octet, which calculations have shown is unlikely.
It does not require phosphorous’ 3d-orbitals which are energetically too far removed to participate in hybridisation in a noteworthy manner.
It gives you four equivalent $\ce{P-O}$ bonds a priori without needing to resort to mesomery.
It gives all oxygen atoms the same charge a priori and also gives phosphorous a charge in better agreement with experiment/calculation.
The experimental bond lengths are better explained as single bonds rather than having a bond order of $\frac{5}{4}$.

This can all be summed up as ‘phosphorous obeys the octet rule much like every other main group atom (doublet rule for hydrogen/helium).’

But you probably want answers to your questions on the decidedly minor mesomeric representations your teacher drew. Well:

  Why should one oxygen be different than any of the others?

Those four mesomeric structures are entirely equal, just like in benzene. To get a ‘picture closer to the truth’ you would need to put them all on top of each other and look through them. You would see that every $\ce{P-O}$ bond has the same (minute) chance of being a double bond and thus no oxygen is different from the others.

  What diagramming rules imply that $\ce{PO4^3-}$ should be depicted as my teacher did?

Well: none, really. If you are going to stick with the double bond representation, the order of the four structures does not matter as long as they are all there. Also, you can go from every single one to every single other one in a single step by appropriate electron shifts. So, as MaxW commented, you really should add the two missing diagonal arrows to be correct.


Why is edta in its calcium complex still a tetraanion at pH 10?

The $\mathrm{pH}$ discussions only make sense in the way you are using them for free $\ce{edta}$. Free $\ce{edta^4-}$ can indeed accept up to six protons depending on the acidity of the solvent; the most common forms are probably the neutral compound $\ce{H4edta}$ and the disodium salt $\ce{Na2H2edta}$. For these species, a $\mathrm{pH}$ equilibrium exists as you know it for polyprotic acids and only for these species is the discussion you’re beginning your post with correct.

Once $\ce{edta}$ forms a coordination compound, e.g. with calcium, things change. Six atoms of $\ce{edta}$ — four oxygens and to nitrogens — are now octahedrally surrounding the central metal ion. The complex will look something like this:

The calcium ion is of course calcium(II), the $\ce{edta}$ ligand carries four negative charges resulting in the complex anion being a dianion $\ce{[Ca(edta)]^2-}$. Technically, this species can still accept up to four protons, but only the carbonyl oxygens can now be protonated to some extent. If you do, you will encounter a very acidic oxygen-centred cation which will want to get rid of the extraneous proton again. Note that the two sites which I intuitively expect to be the most basic, the two nitrogen atoms, cannot accept any more protons since they are already tetracoordinated.

Thus, once you form a complex from free $\ce{edta}$, the equilibrium situation changes completely. We now no longer may consider $\ce{edta}$ by itself but must consider the complex as a whole; the remaining acid-base equilibrium is

$$\ce{[Ca(edta)]^2- + H+ <<=> [Ca(Hedta)]-}\tag{1}$$

As I mentioned, protonating the ligand will lead to a very acidic species (akin to protonating a protonated carboxylic acid) and therefore the equilibrium is shifted to the left.

So even though the $\ce{pH}$ values may tell you that you are dealing with a tri- or tetraanion of free $\ce{edta}$ in solution, the complex has quite different acidity constants.


How does a hydrophobic environment help in bond formation?

The major reason of the ‘hydrophobic environment’ is to block the enzyme-complex from accessing the solvent (water), which might otherwise enter the active site and hydrolyse ATP.

The reduction of polarity also helps to speed up the subsequent nucleophilic substitution.

Note: Hexokinase is a soluble, cytosolic protein

The reasoning behind this statement comes from the X-ray structural studies of yeast hexokinase (HK) by Thomas Steitz.

  Comparison of the X-ray structures of HK and the glucose–HK complex
  indicates that glucose induces a large conformational change in HK.
  The two lobes that form its active site cleft swing together by up to
  11.5 Å so as to engulf the glucose in a manner that suggests the closing of jaws.

This movement places the ATP in close proximity to the group of $\ce{-C6H2OH}$ glucose such that water is sequestered from the active site.
Water is certainly small enough to fit into the phosphoryl acceptor group’s enzymatic binding site. Moreover, phosphoryl transfer from ATP to water is more exergonic than it is to glucose.  If the catalytic and reacting groups were in the proper position for reaction while the enzyme was in the open position, ATP hydrolysis (to yield ADP + Pi) would almost certainly be the dominant reaction.

However, the large conformational change of the enzyme induced by glucose causing the exclusion of water from the active site making the phosphorus atom more accessible for the nucleophilic attack of the C6-OH group of glucose ensures that HK catalyses the phosphoryl transfer to glucose rather than ATP hydrolysis and also reduces the polarity of the active site, which help accelerate the nucleophilic reaction process.

This substrate-induced conformational change in HK is responsible for the enzyme’s specificity.

References

Lehninger Principles of Biochemistry
Voet and Voet Biochemistry


Your synthesis is unlikely to become a huge commercial success ;)

Methylchloride is a technical product, but usually made by reaction of methanol with hydrogen chloride:

$$\ce{CH3OH + HCl -> CH3Cl + H2O}$$

This works nicely without side products, because

$\ce{CH3Cl}$ has a low solubility in water (around $\mathrm{5\,g\cdot L^{-1}}$)
under standard conditions, $\ce{CH3Cl}$ is a gas (bp around $\mathrm{-28\,°C}$)

The radical halogenation of methane will usually lead to various halogenation products and thus is not a good idea. The same is true for the similar reaction of ethane.

In addition, I fail to see how the reaction of ethanol and carbon dioxide will lead to glycerol.

You should further take into account that there's plenty of glycerol available as a side product in the production of biodiesel.

From a commercial and ecological point of view, methane therefore is better converted to cabon monoxide and hydrogen by steam reformation. Further hydrogen may be obtained from a subsequent shift reaction and the resulting carbon dioxide may be reacted with gycerol to yield 1,2-glycerol carbonate.


Application of the Woodward-Hoffman rules to a [14+2] cycloaddition

General rules for cycloadditions

Before getting into the actual Woodward Hoffman 'analysis', one can quickly check any proposed cycloaddition reaction against the following table of general results.

The table is essentially an overview of the Woodward Hoffman rules that doesn't require any consideration of orbitals/orbital symmetry. It can often be useful to check against a table like this to ensure you've done the real Woodward Hoffman analysis correctly.

\begin{array}{ccc} \hline
{\text{Number of }\\\text{Electron Pairs}} &
  \text{Photochemical} &
  \text{Thermal}\\[2ex]\hline
\text{even}~(4n)    &
  \text{suprafacial-suprafacial}  &
  \text{suprafacial-antarafacial} \\[1ex]
\text{odd}~(4n + 2) &
  \text{suprafacial-antarafacial} &
  \text{suprafacial-suprafacial}  \\ \hline
\end{array}

From your example, we can count $14 + 2$ electrons in the π system, giving a total of $16$ which is a $4n$ integer (where $n = 3$). We also know from the reaction scheme given that the cycloaddition is taking place under photochemical conditions. We can, therefore, conclude that the reaction is allowed photochemically if both components react suprafacially.

The actual Woodward Hoffman analysis

Note: I'm doing this for the thermal $[14 + 2]$, firstly because its a little more interesting in terms of orbital overlap, but secondly to give you the opportunity to do the analysis properly for your problem. The approach taken here is the one taken by Ian Fleming in his Molecular Orbitals and Organic Chemical Reactions book.

First: Draw a diagram of the reaction showing the orbitals involved. In this case we are interested in the π system of the heptafulvarene and the π system of the tetracyanoethene.

When drawing a diagram, no orbital phases are needed- this is already taken into account by the generalised Woodward Hoffman rule. (Also note that its not important how you define the components, the 14π system could be considered as a series of seven 2π systems).

Second: Consider how each component is reacting. In a cycloaddition, each component may either react suprafacially (the reaction takes place on the same side of the π system) or antarafacially (the reaction takes place from opposite sides of the π system) (also note that although this is theoretically possible, it is geometrically impossible in some cases due to the inability to get into the correct orientation for orbital overlap).

Third: Apply the Woodward Hoffman rule, as given below.

  A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is odd.

In the example above, the 14π component is taken to be reacting antarafacially (denoted π14a), and the 2π component suprafacially (denoted π2s). When the Woodward-Hoffman rule is applied, this gives an odd number of components, making the reaction thermally allowed.

We could, however, also have drawn the reaction as having the 14π component reacting suprafacially (denoted π14s), with side on overlap of the 2π component reacting antarafacially (denoted π2a). Both are equally valid.

From that,you should now be able to consider your photochemical reaction.  One slight difference exists in the formulation of the Woodward-Hoffman rules for photochemical reactions, taking account of the excited state that a component exists in.  Namely, the total is now even, not odd, for an allowed reaction.

  A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is even.

Further reading

All of the diagrams were taken from Fleming's Molecular Orbitals and Organic Chemical Reactions, student edition, 2009. The book contains many more examples and a slightly more rigorous explanation of the Woodward Hoffman rules.


Is Dalton's Atomic Theory a theory or a postulate?

It is most definitely a theory. The 4 points you mention could be construed as hypothesis by themselves. But Dalton went further, he put them to the test.
The results of his exeriments formed yet another component of his atomic theory called the Law of Multiple Proportions. This law served as proof that the other components of his theory were true. I used to spend most of a class period explaining this law to students, it's best you search the web for a complete explaination but I'l give you a brief summary via example.

Say an experiment shows it takes exactly 16 g of oxygen to combine with 12 g of carbon to make carbon monoxide. Now suppose we make carbon dioxide. We would find that it now takes 32 g of oxygen to combine with that same amount, 12 g, of carbon. So, for a fixed amount of carbon, the amount of oxygen required to make the two different compounds can be expressed as a small, whole number ratio. What would explain this better than to surmise that elements come as particles and thes particles combine in small, whole number ratios to form compounds.

When a hypothesis have been proven by repeatable experimentation, we call it a theory.


Why do most drugs (eg: oxycodone) have seemingly arbitrary names?

The reason why the generic name of most drugs have seemingly little semblance to their chemistry is in the interests of practicality; a doctor would find it easier to write sildenafil (generic name) or Viagra (brand name) in a prescription than 5-{2-Ethoxy-5-[(4-methylpiperazin-1-yl)sulfonyl]phenyl}-1-methyl-3-propyl-1$\ce{H}$,6$\ce{H}$,7$\ce{H}$-pyrazolo[4,3-d]pyrimidin-7-one (chemical name)  unless he had a weird fetish for IUPAC nomenclature .
As for how manufacturers (and the relevant regulatory bodies) arrive at those generic and brand names,  msdsmanuals.com has a take on this,

When a drug is under patent protection, the company markets it under its brand name. When the drug is off-patent (no longer protected by patent), the company may market its product under either the generic name or brand name. Other companies that file for approval to market the off-patent drug must use the same generic name but can create their own brand name. As a result, the same generic drug may be sold under either the generic name (for example, ibuprofen) or one of many brand names (such as Advil or Motrin).

Furthermore, they add,

Generic names are usually more complicated and harder to remember than brand names. Many generic names are a shorthand version of the drug’s chemical name, structure, or formula. In contrast, brand names are usually catchy, often related to the drug’s intended use, and relatively easy to remember, so that doctors will prescribe the drug and consumers will look for it by name. Brand names often suggest a characteristic of the drug. For example, Lopressor lowers blood pressure, Glucotrol controls high blood sugar (glucose) levels, and Skelaxin relaxes skeletal muscles. Sometimes, the brand name is simply a shortened version of the drug’s generic name—for example, Minocin for minocycline.

I found equally useful information on popsci.com,

As for commercial names, pharmaceutical companies are reluctant to share their strategies. These are hugely important brands, even after the patents expire--nobody knows what Sildenafil citrate is, but everyone knows its brand name of Viagra. And the odds are you know someone who insists on brand-name Tylenol instead of discount-store acetaminophen.
Though pharmaceutical companies contacted by PopSci wouldn't share their marketing secrets, you can still derive a drug's molecular and popular heritage just by studying its name. Like, for example, Tylenol. Its chemical name is $\ce{N}$-acetyl-p-aminophenol--see the -tyl and -ol in there? Then there's Viagra. It doesn't sound like sildenafil, but it certainly suggests vigor and virility--exactly what Pfizer wants you to think. Xeljanz, which was approved earlier this year after almost 20 years of research, targets a protein called Janus kinase. It works in a different cellular pathway from other arthritis drugs. That's where you get the -jan part of that name, which, to someone with a drug decoder, connotes its uniqueness.

As for oxycodone specifically, it's anyone's guess, since that information can't be found (online) easily enough.

 Note: I've italicized all generic and brand names because, well... they're generic and brand names. Italicizing actual chemical/IUPAC names is against convention, so don't get the wrong idea ;-)


Aluminum, just like zinc, is already widely used to coat iron surfaces to prevent corrosion (in the elemental form, or in combination with other metals, like zinc). You can deposit aluminum on practically any surface (including plastics). On metals it is done by electrochemical deposition.

There are several companies working with this type of coating nowadays, including AlumiPlate (http://www.alumiplate.com/).


How to prepare elemental gallium from a mixture of gallium and aluminum nitrate?

If you have a mixture of aluminum nitrate and gallium nitrate, then you have to separate the aluminum nitrate from the mixture. This patent describe how to separate aluminum nitrate from a mixture of other nitrate and hence, this process can be applicable for separating aluminum nitrate from gallium nitrate.

Then, when you have gallium nitrate, just heat to produce gallium oxide.

$$\ce{4Ga(NO3)3 -> 2Ga2O3 + 12NO2 + 3O2}$$

  The thermal decomposition of nitrate gallium(III) to produce
  gallium(III) oxide, nitrogen dioxide and oxygen. This reaction takes
  place at a temperature of 110-200°C.(chemiday 1)

Then reduce gallium oxide to gallium using some reducing agent like carbon or hydrogen.

$$\ce{Ga2O3 + 3H2 -> 2Ga + 3H2O}$$

  Gallium(III) oxide react with hydrogen to produce gallium and water.
  This reaction takes place at a temperature near 700°C.(chemiday 2)

$$\ce{Ga2O3 + 3C -> 2Ga + 3CO}$$

  Gallium(III) oxide react with carbon to produce gallium and carbon
  monoxide. This reaction takes place at a temperature of 850-950°C.(chemiday 3)

You will find this youtube video of help. It shows how to synthesize gallium metal.


Will I get bromine by electrolysis of the brine from the Dead Sea?

The short answer is probably a bit, depending on the potential you apply.

There are at least a few factors involved:

How easily are $\ce{Cl-}$ and $\ce{Br-}$ oxidized?

$$\begin{align*}\ce{2Cl^{-}_{(aq)} -> Cl_{2(g)} +2e-}\; &E°_{\mathrm{ox}}=-1.36\ V\\\ce{2Br^{-}_{(aq)} -> Br_{2(aq)} +2e-}\; &E°_{\mathrm{ox}}=-1.09\ V\end{align*}$$
At standard activities, the two ions are pretty similar in oxidation potentials, so there is probably a range of potentials where both reactions occur to a similar degree or it might be possible to achieve some partial selectivity for bromine formation. The Nernst equation can be used to calculate equilibrium concentrations for a given potential, but so long as the potential applied is well over the standard oxidation potential, the exact values don't really matter.
What difference do the concentrations make?

In most cases, what determines how much of either reaction occurs is mass transport. For a redox reaction to occur, the reacting molecule must be very close to the electrode surface (<10 nm). For a planar electrode at a single potential in a standing solution, mass transport occurs by diffusion and the current that flows should be directly proportional to concentration in the bulk. In principle, the Cl to Br ratio should be the same as their activities in solution, though in some situations, other factors like the speed of the electron transfer can be limiting, so this may not be the case, especially since there is so much chloride compared to bromide.
What about the electrodes?

As electrolysis is a surface phenomenon, the electrode and its interactions with the reactants can be important. Assuming you pick an electrode which doesn't itself react with the chloride or bromide, the exact material can affect the behaviour of the cell. Even though the Nernst equation may predict a certain oxidation potential, interactions of the reactants and the electrode may result in varying amounts of additional potential (overpotential) needed to cause the reaction.
What happens to the bromine?

A certain amount of bromine can dissolve in water, but depending on the pH, it can rapidly disproportionate back to bromate and hypobromate: $\ce{Br2 + 2OH- → Br- + OBr- + H2O}$ Also, the chlorine gas can react with the bromide to make bromine and bromine is also known to react with metals (use carbon electrodes?). In short, you can probably create an electrolytic cell that produces some bromine, but it's difficult to say in what ratio with chlorine and what its exact fate would be.


Why is Sulfur trioxide a neutral electrophile?

Sulfur is bonded to three oxygens, twice each. The sulfur in the middle technically has a formal charge of zero, but the oxygens it is bonded to are extremely electronegative (they're electron hogs), and so the sulfur atom in the middle has a partial plus charge on it. Ergo, it will be willing to accept electrons to try to compensate.


Conversion of a value between g/gmol and lb/lbmol

A gram-mole ($\pu{gmol}$),often just called a mole ($\pu{mol}$), is the amount of a material whose mass in grams equal to the numerical value of the atomic or formula mass. For example, a gram-mole of water has a mass of 18.017 grams, this number being the formula mass for water.

A pound-mole ($\pu{lbmol}$) is the amount of a substance whose mass in pounds-mass equals the atomic or formula mass. For water, one pound mole has a mass of $18.017$ lb-mass.

The pound-mole is bigger than the gram-mole by the same factor as a pound is bigger than a gram - so pounds per pound-mole equals grams per gram-mole.  You do not divide by anything.


Covalent bonds are the bonds between the atoms within the same water molecule. Hydrogen bonds are the bonds between two water molecules. All molecules have covalent bonds, but only some molecules have hydrogen bonds. As an example, water has hydrogen bonds, but carbon dioxide does not.

One of the requirements for hydrogen bonding is that the molecule must be polar. Water molecules are polar because of two effects. The first is that oxygen is more electronegative than hydrogen, so each of the OH bonds will be polar. The electrons shared in each of those bonds will spend more time near the oxygen than near the hydrogen. The other effect is the geometry of the molecule: the oxygen atom has two free electron pairs in addition to the two bonds, so it assumes a tetrahedral arrangement. The $\ce{O-H}$ bonds are two corners of that tetrahedron, and the free electron pairs are the other two corners. This arrangement means that, if you only look at the atoms, the molecule has a bent geometry. Because of that, the polarity of the two bonds partially adds together. By contrast, $\ce{CO2}$ has a linear geometry, and the two polar bonds cancel each other out, leaving the molecule non-polar.

So the water molecule is polar, with a lot of excess positive charge around the hydrogen atoms, and a lot of negative charge on the side of the oxygen atom away from the hydrogen atoms. Hydrogen bond is the name given to the electrostatic interaction between the positive charge on a hydrogen atom and the negative charge on the oxygen atom of a neighboring molecule. The covalent bond is the electrostatic interaction between two atoms in the same molecule. Covalent bonds are much stronger than hydrogen bonds: the $\ce{O-H}$ has a strength of 467 kJ/mol, while the hydrogen bond is usually between 4 to 40 kJ/mol.


You may also think of hydrogen bonding here. See that in case of y, intramolecular hydrogen bonding is possible and hence greater percentage of enol content is of y than z. You may note that this is a general factor of esters (in this case a cyclic ester) in determining enol content percentage and that this factor may not always be the 'winning' one (as you can compare between x and y).


Reaction of concentrated nitric acid react with metallic lead?

Well, you were going along the right track.... just one correction. The products will be Pb(NO₃)₂, NO₂ and H₂O. Nitric acid is a very powerful oxidizing agent. It oxidized H₂ to H₂O.


What is the approximate lifetime of an SN2 transition state complex?

26-60 femtoseconds depending upon impact angle for $\ce{F- + CH3Cl}$ according to Mugnai et al. (Table IV.)


How to determine the amount of barium in a sample of contaminated water?

Because the client is ultimately interested in removing the barium by precipitation, gravimetric analysis is likely a suitable solution. This can be done in a few hours in a lab with basic equipment like a filter flask, a drying oven, an analytical balance and the appropriate reagents.

You can take advantage of the fact that barium sulfate is very insoluble in water by adding some saturated solution of sodium sulfate to a known volume of a sample of the water. Then just filter out the barium sulfate onto some pre-weighed filter paper, dry, weigh, and calculate the mass of barium (from the measured mass and the molecular weight of barium sulfate) per volume of water. You should repeat the analysis to be sure you added enough sodium sulfate solution to precipitate all of the barium.

If there is little or no precipitate found, you have a couple options. Firstly, you can optimize the gravimetric analysis by using large volumes of sample with a small filter and a good analytical balance. Also be sure to pre-filter the sample prior to adding the sodium sulfate in case there are suspended particulates, which would get measured just as if they were barium.

If a gravimetric analysis does not provide the detection limit required by the client, the next best options are atomic absorption spectroscopy (AAS) and inductively coupled plasma - atomic emission spectroscopy (ICP-AES). As these analyses can be done very cheaply (~\$10-$30 per sample) you may actually want to do this to begin with. The down side of outsourcing is that the turnaround time could be weeks, or the price could quadruple if you need results fast.