How can difference in neutron number cause a difference in ionisation enthalpies?

There are a handful of additional corrections needed to calculate the exact differences. But the biggest factor is just the mass of the nucleus. Not only is the electron attracted to the nucleus, the nucleus is attracted to the electron. So not only does the electron "move", so does the nucleus. A heavier nucleus moves less. This in turn alters the wave function of the electron.

The difference between protium, deuterium and tritium is reasonably explained by the Bohr model of the atom. The gist is that the electron does not travel around a nucleus of infinite mass, but rather that the electron and nucleus travel around their common center of mass.

For one electron traveling around the nucleus the energy levels of the Bohr system are:

$\text{E}_n = \dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} \cdot \dfrac{\text{Z}^2}{\text{n}^2}\cdot \text{E}_R$

$\text{E}_n$ Energy of the $n$th orbital of the one electron system
$\text{M}_e$ Mass of the electron = $5.489\cdot10^{−4}$ atomic mass units
$\text{M}$ Mass of the nucleus
$\text{Z}$ Charge on the nucleus
$n$ orbital = principal quantum number $n$ = 1, 2, 3, 4, ...
$\text{E}_R$ Rydberg energy = 13.605693 eV

For the $1s^1$ electron of protium, deuterium and tritium, Z=1 and $n$=1.

protium
$\text{M} = 1.007825 - 0.0005489 = 1.007276$ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{1.007276}} = 0.999455$

ionisation enthalpy = $0.999455 \cdot 13.605693 = 13.598$ eV

deuterium
$\text{M} = 2.01410178 - 0.0005489 = 2.0135529 $ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{2.0135529}} = 0.999727$

ionisation enthalpy = $0.999727 \cdot 13.605693 = 13.602$ eV

tritium
$\text{M} = 3.0160492 - 0.0005489 = 3.0155003$ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{3.0155003}} = 0.999818$

ionisation enthalpy = $0.999818 \cdot 13.605693 = 13.603$ eV

Ionization energies From NIST webbook
hydrogen = 13.59844 eV
deuterium = 13.603 eV
tritium = 13.603 eV

Therefore protium < deuterium < tritium

I'll add that when you're doing this sort of thing other factors creep in the calculations.

(1) For a hydrogen atom the electrons aren't relativistic. If you get to uranium though that wouldn't be true.

(2) The 13.6 eV binding energy also reduces the mass of the hydrogen atom. So subtracting the rest mass of the electron isn't absolutely correct, but good enough for the precision used.

13.6 ev $=\quad\quad 0.000 000 014 5$ amu
mass e used $= 0.000 548 9$ amu


Calculation of overall rate constant of a parallel reaction

The key to answering these kinds of riddles is to think about units.  Rates of reaction have units of $\mathrm{\frac{mol}{L\cdot s}}$, but that' isn't fully specified.  You need to know moles of what.

In your case, the confusion is stemming from confusion of what the "moles" in $\mathrm{\frac{mol}{L\cdot s}}$ refers to.

The answer can either be moles of "reaction", or moles of A.  The other key thing to realize is that you can write "the" reaction however you want.  The value of the "rate constant" $k$ depends on how you define the rate.  If for the overall reaction you write:

$$\ce{3A -> \varnothing}$$

...and if the "rate" you are calculating is the rate of reaction, then there are three moles of A lost per mole of reaction, so

$${dA \over dt}=-3k_{\alpha}\ce{[A]}$$

If instead you decide "the" reaction is

$$\ce{A -> \varnothing}$$

then you can rate the rate of disappearance of A as

$${dA \over dt}=-k_{\beta}\ce{[A]}$$

The choice is up to you and it is arbitrary.  You just have to keep track of the moles of what the rate laws you write refer to.

  Then to calculate the k where k is the rate constant for the overall reaction we write [[Equation (i)]]

You need to think carefully about what you meant by the words "the overall reaction".  When you write equation 1, you are adding together rates of loss of A, i.e. the units are $\mathrm{\frac{mol~of~A}{L\cdot s}}$.  The only reaction you can write that has the same rate is one where the coefficient of A is 1.

"The" half-life for radioactive elements usually refers to units where units are $\mathrm{\frac{mol~of~element}{s}}$.  The fact that the unit is $\mathrm{\frac{mol~of~element}{s}}$ and not $\mathrm{\frac{mol~of~reaction}{s}}$ means that if we wanted to write the corresponding chemical reaction, the stoichiometric coefficient had better be 1.$%edit$


Can aromatic rings be cyanated using cyanide as a nucleophile?

Chlorobenzene is generally inert to nucleophilic substitution. It was observed by Rosenmund and Braun that using an ethanolic solution of potassium cyanide and cuprous cyanide as catalyst at 200°C, chlorobenzene converts to benzoic acid. But if the ethanol-cyanide solvent is removed and temperature is raised, then the reaction proceeds to form cyanobenzene. Yield was further improved when ionic liquids like DMF was used. This is Rosenmund-Braun reaction.

Note: Working with cyanide is highly dangerous. Do not handle cyanide until and unless you are a professional chemist.

Reference

Wikipedia: Rosenmund-von Braun reaction
Will cyanide substitute bromine in bromobenzene?
Potassium ferrocyanide, $\ce{K4[Fe(CN)6]}$ can be used as a non-toxic source for cyanation.
There are loads of other catalyst for cyanation like $\ce{ZnCl2}$, palladium or nickel salts.[@Waylander]


How does heat energy interact between two bodies of different heat capacities in order to attain thermal equilibrium?

Say the initial temperature of $A$ is $T_1$ and the initial temperature of $B$ is $T_2$. To determine the temperature ($T_f$) of $A$ and $B$ at equilibrium, we start with the equation $q_A=-q_B$, where $q_A$ and $q_B$ are the heat gained by $A$ and $B$ respectively as they reach equilibrium. We have $q_A=mc_A\Delta T_A=mc_A(T_f-T_1)$ and $q_B=mc_B\Delta T_B=mc_B(T_f-T_2)$. Thus the equation becomes $$mc_A(T_f-T_1)=-mc_B(T_f-T_2)$$ From here, we can solve for $T_f$. We can also note that if $c_A=c_B$, $T_f=\frac{T_1+T_2}{2}$ , which in your example is $T+\frac{\Delta T}{2}$. Otherwise, $T_f$ will not be the midpoint between $T_1$ and $T_2$.


No, no it's not a resonance arrow my friend, it has to be an arrow for equilibrium because it is a reaction equation of $\ce{CH3COOH}$ in its aqueous solution. It will dissociate feebly into its constituent ions because its a really weak acid. Thus establishing equilibrium quite a lot first.

Moreover in a resonance arrow we describe the same compound with different structures (which are all correct) and that is why we establish a resonance hybrid to denote all such possible resonating diagrams in a single diagram fulfilling the properties of the compound.

Nevertheless a resonance arrow can't anyway turn into a dissociation reaction arrow.


What do different forms of ice look like?

As you know, ice can exist in at least 15 or so solid phases. Of these, the first phase, $Ice$ $I$ (ice-one) is the only phase found naturally on earth, mostly in the form of $Ice$ $I_h$ hexagonal crystal as seen in glaciers and snowflakes.

The visible appearance of ice (eg: transparency and clarity) depend also upon the level of purity of the crystals formed. For example, air can get trapped within inclusions of the ice, making it translucent or opaque white, as shown in the image below.

The cubic form $Ice$ $I_c$ is metastable so is rarely found, generally in high clouds in the upper atmosphere, where it is formed at ambient pressure by the condensation of water vapour at very low temperatures (-80°C).

Most other forms of crystalline ice (designated as $Ice$ $II$ to $XV$ are metastable, being produced in the laboratory by the application of high pressures (above 200MPa) and/or low temperatures (below 72K), resulting of denser packing of the water molecules.

For example, $Ice$ $II$ (see image below) is formed by compressing 'normal' $Ice$ $Ih$ to 200MPa at temperature of 198K.

The SEM image above is of $Ice$ $I$ decorated grain boundaries in fine-grained $Ice$ $II$. The $Ice$ $I$ stands in raised relief relative to the $Ice$ $II$ due to volumetric expansion to the lower-density phase.

Scientists believe this form of ice can be found on the interior of some large icy moons orbiting Jupiter.

The macroscopic appearance of the 'exotic' forms of ice depends upon many variables such as crystal grain size. The laboratory production of these 'novel' forms of ice to ensure proper formation of fine crytallite grains (less than 10 microns) is time consuming, requiring careful control of temperature and cycling of very high pressures.

For example, $Ice$ $IV$ is formed by heating high-density amorphous ice at a slow (0.4K/min) rate from 145K and at a constant pressure of 0.81GPa. It forms a rhombohedral crystal (see image (a) below).

Similarly, $Ice$ $V$ is formed from liquid water at 500MPa by lowering its temperature to 253K. It has a more complex lattice structure and forms monoclinic crystals (see image (b) above).

"Square snow" or $Ice$ $VI$ is formed from liquid water by compressing it at 1.1GPa (10,850atm) and lowering its temperature to 270K. Its unit cell forms tetragonal crystals with all water molecules hydrogen bonded to four others, resulting in the formation of dendritic crystals (approx 45 microns across, see image below).

For more images from the laboratory, see: http://web.mit.edu/wbdurham/www1/short%20htmls/more%20ice%20photos.html

References:

Water structure and science, Ice phases, known ices, Martin Chaplin.

Grain size-sensitive creep in Ice II, Kubo, Durham, Stern & Kirby, Science, Vol 311 (2006).

Visual observations of crystal morphologies and melting of Ice IV at pressures up to 1GPa, Chou and Haselton, Review of High Pressure Science Technology Vol 7 (1998).

Dynamic pressure-induced dendritic and shock growth of ice IV, Lee, Evans & Yoo, Proceedings of the National Academy of Sciences, Vol 104, No. 22.


How can there be an enthalpy of formation for gaseous water?

The only explanation that comes into my mind is that,

$$\ce{H2 (g) + \frac{1}{2} O2 (g) -> H2O (g) <-> H2O(l)}$$
 And
$$\Delta H_f^\circ(\ce{H2O (g)}) = \Delta H_f^\circ (\ce{H2O (l)}) + \Delta H_\mathrm{vap} (\pu{25^\circ C}) = -285.8 + 44.99 = \pu{-241.8 kJ/mol}$$ almost as tabulated depending the sources, here the calculation is perform with tabulated values found in the 87th Edition of the Handbook of Chemistry and Physics.

This is a hypothetical state but actually, vapour water is in equilibrium with liquid water even under the standard state. It's just that this equilibrium is very unlikely to go in direction of formation of the gas water. But if you find an ingenious way to vaporize water under these conditions then the heat of formation of gaseous water will be the same as I just calculated.

Also in the 87th Edition of the Handbook of Chemistry and Physics, it is said (5-4):

  The standard state of a pure gaseous substance is that of
  the substance as a (hypothetical) ideal gas at the standard
  state pressure.

Although I've never figured that out and I can't really tell you about the need for this value, it exists. Probably you'd like to have this kind of values when performing a calculation considering a particular state but as far as I used to do, I've never used this one.


Nomenclature of organic compounds containing complex side chains

Simple prefixes are arranged alphabetically disregarding any multiplicative prefixes. The multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘1,2-dibromo-’ is considered to begin with ‘b’.
However, the name of a compound substituent is considered to begin with the first letter of its complete name.
For example, ‘1,2-dibromobutyl-’ is considered to begin with ‘d’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.
P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Therefore, the correct alphanumerical order corresponds to the name 7-(1,2-dimethylpentyl)-5-ethyltridecane (not 5-ethyl-7-(1,2-dimethylpentyl)tridecane) since the compound substituent name ‘dimethylpentyl’ starts with ‘d’, and ‘d’ is earlier alphabetically than ‘e’.
(Nevertheless, the preferred IUPAC name is 5-ethyl-7-(3-methylhexan-2-yl)tridecane.)

In your second example, the correct alphanumerical order corresponds to the name 6-(1,1-dimethylbutyl)-5,5-dimethyl-6-pentyltridecane (not 5,5-dimethyl-6-(1,1-dimethylbutyl)-6-pentyltridecane) since the compound substituent name ‘dimethylbutyl’ starts with ‘d’, and ‘d’ is earlier alphabetically than ‘m’.
(Nevertheless, the preferred IUPAC name is 5,5-dimethyl-6-(2-methylpentan-2-yl)-6-pentyltridecane.)


Chaotic melting points of n-alkyl carboxylic acids

My guess without looking at any literature:

Higher acidity $\implies$ higher hydrogen bonding $\implies$ higher melting point.

Thus, changing from H and Me where there is virtually no +I effect from the chain, to Et, Pr, Bu where the +I effect of the chain decreases the melting point, as the hydrogen bonds get weaker, as the COOH groups are less acidic.

Further increasing the chain length after hexyl does not really lead to any additional +I effects on the COOH group that is now far away. Thus, the melting point increases as there are additional van der Waals interactions between the longer chains.

Then there seems to be a factor that makes crystals of odd chains more favorable.


Reaction of 1,3-dimethyl-2,5-dintrobenzene with NaOMe/MeOH

You are correct that the nitro group can be displaced by methoxide, this has been reported examples here.
My instinct is that the less hindered nitro group is the one that will be displaced.


What is the freezing point of sulfuric battery acid?

The general effect you're talking about is freezing point depression, one of the basic colligative properties. There are several flavors of lead-acid batteries, so there are different additives that can modulate the freezing point.  The actual composition of the electrolyte varies with the charge state of the battery, so that means the freezing point also varies with how charged it is. I think all of the lead-acid batteries use water as a diluent (it will always be present as it's a species in the electrochemical equilibrium.) From the Wikipedia article on "lead-acid battery":

  Due to the freezing-point depression of the electrolyte, as the battery discharges and the concentration of sulfuric acid decreases, the electrolyte is more likely to freeze during winter weather.


are half of "H2SO4 pH sample problem" webpages just wrong, or am I wrong?

...and you didn't even get into the fact that almost all explanations will ignore the autodissociation of water!

In some cases your online resources will be wrong, of course, and it would be odd to expect uniform correctness in something people donate for free on the Internet (including here BTW).  So that's worth keeping in mind.

However, it's also the case that it is not always important to treat a problem in all its subtlety.  For example, the difference you get in the calculated pH of 1M H2SO4 by treating or ignoring the second dissociation equilibrium is likely to be smaller than your measurement error in a typical undergraduate lab.  So why bother?  On the other hand, if you were running a very careful experiment with 0.01 M H2SO4 you might well want to take those other equilibria into account, and if you were doing something super sensitive with 10^-7 M H2SO4 then you'll need to worry about the autoionization of water, too.

Generally we use the least complex model that gets us the answer consistent with our measurement limitations.  That maximizes insight and reduces the obfuscation of complex calculation, which can easily make it so we fail to recognize an erroneous answer.   This runs all through science and engineering, and it's why simple models have such great staying power even when they are superseded by more sophisticated understanding.  Naturally a concomitant is that the good scientist must become very adept at identifying the least complex model he can use.  That takes deep insight into the limitations of each model, the probable sources of error, the limits of measurement, and the quality of data on hand.  Being good at this is a much more important skill in the senior scientist than mastering the details of the most complex and realistic model there is.


The first step of the Cannizzaro reaction is nucleophilic attack of hydroxide on an aldehyde (generically $\ce{RCHO}$) to form a tetrahedral intermediate.

$$\begin{aligned}
&\ \ \ \ \  \ce{O} && \ \ \ \ \ \ce{O-} \\
&\ \ \ \ \  \parallel && \ \ \ \ \ \mid \\
&\ce{R-C-H +OH- ->}  && \ce{R-C-H} \\
& \ && \ \ \ \ \ \mid \\
& \ && \ \ \ \ \ \ce{OH}
\end{aligned}$$

Formaldehyde $(\ce{R}=\ce{H})$ is more reactive that benzaldehyde $(\ce{R}=\ce{Ph})$ for both steric and electronic reasons:

Sterics: The Ph group is larger than hydrogen, so nucleophiles will have more trouble getting to the aldehyde in benzaldehyde
Electronics: The Ph group is electron rich (all those $\pi$ electrons), which repels the electron rich nucleophile.

If formaldehyde reacts faster than benzaldehyde with hydroxide anion, then more of formaldehyde's tetrahedral intermediate will form, which means there will be more benzaldehyde left for the second step - hydride transfer.


Is it true that iron (Fe 0+) cannot exist when there's water around?

Iron metal can exist in presence of water. As you stated, important point is the $\ce{Fe^0 -> Fe^3+}$ reaction. Water is not an oxidizing agent, therefore is not capable of inducing this process.

To create rust, you need also oxygen, or as you noted, electrochemical potential.


What is the minimum pressure for existence of a liquid solution?

The only things I can figure for $a$ and $b$ in $p=\dfrac{a}{27b^2}$ are the coefficients from the van der Waals equation for nonideal gases:

$$\left(p+\dfrac{n^2a}{V^2}\right)\left(V-nb\right)=nRT$$

These constants have appropriate units to give you pressure: $a\ \left( \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}\right)$ and $b \ \left(\dfrac{\text{L}}{\text{mol}} \right)$.

$$\dfrac{a}{27b^2}\implies\dfrac{\dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left(\dfrac{\text{L}}{\text{mol}}\right)^2}=\dfrac{\text{atm}}{27}\implies p$$

However, these constants are empirically derived, and thus different for every substance, and so the critical pressure would be different for every substance.

For example, for argon $a=1.355 \ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}$ and $b=0.0320\ \dfrac{\text{L}}{\text{mol}}$, so $p$ for argon is

$$p=\dfrac{1.355\ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left( 0.0320\ \dfrac{\text{L}}{\text{mol}} \right)^2}=49.01\ \text{atm}$$

Were you told which substance you should be using? For all substances, $b$ appears to be less than $1$, so we need to find the minimum $a$ and maximum $b$ combination to get the minimum $p$. However, by my calculation, all values of $p$ will be greater than $1$. The value for water is $220.5\text{ atm}$ (which does not make sense given that a vast number of liquid solutions are stable at normal atmosphere pressure).


What do X-ray diffraction and NMR tell about the Proteins?

Dynamics can mean many things, and not all proteins undergo extensive conformational changes during their lifetime. Some are fairly rigid in solution.

XRD or NMR data represents an average over the conformations of a protein construct sampled under experimental conditions. A protein construct is a variant of the protein, selected to allow expression and purification, and possibly also to ensure stability and crystallization (as necessary for XRD).

NMR or XRD data is processed to derive an ensemble of geometries that are consistent with the data and satisfy other criteria, usually imposed by a protein force field or prior statistics about the frequency of geometric features, which allows scoring the quality of a geometry. Before being deposited, the ensemble undergoes quality control by the PDB, amongst other things to verify that there are not too many or extreme violations of expected protein geometric parameters (bond lengths, angles, dihedrals, vdw contacts). Deposited NMR geometries are intended as representative, not exhaustive.

Whether experimental geometries are relevant depends both on the protein and what structural (or dynamical) questions are of concern. Approximating physiological conditions improves the likelihood that the data is relevant.

Proteins under XRD have less conformational freedom, they are more "rigid" (having looked up the definition in Meriam-Webster's dictionary) than those in solution. Different XRD samples (crystals) may still capture views of significantly different conformations, and complementary techniques such as NMR can be used to fill in gaps.

Yes, bigger proteins are more difficult generally to study, but progress has been made. Some of the limits are instrumental, others with sample prep. There is no single answer (too broad). Q4 is a standard problem. You get the structure and then worry whether there are inconsistencies with what is known/expected. You can try different media and see if they generate different structures.

Why is crystallizing proteins hard? Crystallization has to happen spontaneously! The protein must be kept stable (folded) as it crystallizes, which means being restricted to a narrow range of solvent and temperature conditions. There's a lot more that can go wrong versus what needs to go right.  Even as you want the protein to associate and form an ordered crystal, you don't want it to form the wrong kind of interactions resulting in a disorderly aggregate.


usually when metal is bonded to a non-metal, they are bonded together by ionic bond but this is not the only condition , the more important condition is the electronegativity that must be greater than 1.7 to be sure that the bond is ionic if not like in your example 1.2 then the bond will be polar covalent bond which means that the electronegativity difference is not enough for the electrons to be lost by iron and gained by chlorine but they are shared and shifted toward the chlorine atom as it has the greater electronegativity.


Since the molecular weight of water is $18\ \mathrm{g/mol}$, $100\ \mathrm{ml}$ (density = $1\ \mathrm{g/ml}$, so $100\ \mathrm{ml} \overset{\wedge}{=} 100\ \mathrm{g}$ of water) of water contains
$$\frac{100}{18} = 5.56$$
moles of water, or 5.56 moles of atomic oxygen ($\ce{O}$), or 5.56/2 = 2.78 moles of molecular oxygen ($\ce{O2}$).

At standard temperature and pressure, 1 mole of a gas occupies 22.4 liters.  Therefore, 2.78 moles of molecular oxygen will occupy 62.22 liters.

If air is 21% molecular oxygen, then we would need 62.22/0.21, or 296.3 liters of air to provide that much molecular oxygen.

Since 1 cubic meter equals 1000 liters, this volume converts to 0.3 cubic meters of air.

Therefore, 0.3 cubic meters of air contains the same amount of oxygen as is present in 100 ml of water.


Calculation of equilibrium constant in a gaseous system A ⇌ B + C

Your analysis is correct, except that you don't account for how the the units cancel out in your calculation.

The initial reaction quotient, $Q_P$, is indeed 3, as you have calculated.  And since $K_P$ is 1, the reaction will shift to the left, increasing $P_A$.  Hence Statement-1 is nearly correct.  I say "nearly" because $K_P$ is dimensionless, so it should be 1, not 1 atm (each partial pressure should be divided by whatever you define as the standard pressure, 1 atm or 1 bar -- that's what you've left out of your calculation; you should try this yourself as an exercise).

Statement-2 is not quite correct, because it says the "equilibrium shifts in the backward direction."  Assuming you are not changing the temperature, the equilibrium does not shift (it's a constant: $K_P$ = 1).  Rather, the reaction  (and thus $Q_P$) shifts in the backwards direction until equilibrium is attained.


Why is the hypoiodite anion unstable in solution, whereas hypochlorite is relatively stable?

When iodine is added to water, the following reaction results:

$$\ce{I2(l) + H2O(l) -> OI^{-}(aq) + 2H+(aq) + I^{-}(aq)}$$

$\ce{I2}$ molecules and water molecules react to form hypoiodite ($\ce{OI-}$). The reaction can move both ways of the equilibrium, depending on the pH of the solution. Iodine may also occur as $\ce{I3-}$(aq), $\ce{HIO}$(aq), $\ce{HIO3}$(aq) . The compounds that are form behave differently when they come in contact with water. When iodine ends up in surface waters, it may escape as iodine gas.

$\ce{IO-}$ rapidly disproportionates to form $\ce{IO3-}$ and $\ce{I2}$
$$\ce{IO- -> IO3- + I2}$$.

See here for more details.

For the disproportionation reaction, consider the Latimer diagram for iodine:-

A species has a tendency to disproportionate into its two neighbor species if the potential on the right of species in a Latimer diagram is more positive than that on the left. In this case, right hand potential($\ce{+0.42}$) is greater than the left hand potential ($\ce{+0.15}$). So , it disproportionates.


Why does calcium oxide react with sulfur dioxide?

One way of thinking about it is that calcium is a metal and therefore forms a basic oxide, $\ce{CaO}$. On the other hand, sulfur is a non-metal and forms acidic oxides $\ce{SO2}$ (or $\ce{SO3}$). So, the reaction between $\ce{CaO}$ and $\ce{SO2}$ can be considered an acid-base reaction.

There is a somewhat obscure acid-base theory, the Lux–Flood theory, which is applicable to such reactions. In this context, bases are oxide donors (like $\ce{CaO}$) and acids are oxide acceptors (like $\ce{SO2}$).


What are some of the canonical texts for polymers? Is materials science more related to polymers than chemical engineering?

I don't know of any undergraduate level texts that would be suitable for a freshman, unless you have already taken physics, statistics, and chemistry up through organic chemistry.

I can tell you that Fundamentals of Polymer Science by Painter and Coleman is a good introductory text, but I used it in the final year of my Chemical Engineering program. There could be better resources as well - that's just the only one I am personally familiar with.

It also depends a lot on which specific aspects of polymers you are interested in. This actually matters a lot for your second question as well. The three main branches of polymers studies are:

Polymer chemistry
Polymer physics
Polymeric materials

There are sub-fields in each of these, and a lot of overlap, but essentially they focus on:

Polymer chemistry - How are polymers made? (this would include engineering and production of polymers)
Polymer physics - How does the behavior on a molecular level affect the macroscopic properties?
Polymeric materials - How can we use the unique material properties of polymers to make useful stuff?

At the undergraduate level, you won't get too specific on any one of these areas, no matter which degree you choose. If you majored in chemistry, you would learn a little more about polymerization reactions. If you majored in physics (and took electives related to polymers) you might learn more about the statistical mechanics of polymers. If you majored in materials science and/or engineering, you would learn more about how we use the polymers, and less about how they are made and why they act the way they do. Chemical engineers learn a little bit of everything, but they don't get too deep into any one particular branch.

The real differences show up if you go on to graduate school or work in a particular role in industry. There, you will definitely focus more, and your degree can make more of a difference. So I would say, for your second question, major in whatever aspect interests you more - if you want to know the details about how polymers are made in a lab, then maybe chemistry is a better option. If you really like math and physics and want to understand how they relate to the behavior of polymers, then physics is good. If you are more interested in how to characterize and use polymers, then materials science/engineering is a good bet. And if you like a little bit of everything, or are interested in industrial-scale production of polymers, then chemical engineering is a good choice.

Keep in mind as well: it is hard to predict now what you will really want to do in four or more years - so if you make a choice based on which aspects of polymer science you are interested in, and then later decide that polymers are actually not what you really like, you will still be studying the "right" stuff for you.


As you state, the ubiquitous polyethylene (or polythene) bag allows many chemicals to slowly seep through. Pollen and dust particles themselves are far to large to get through, but water vapor, for example does.

For that reason, many foods are sold in plastic (e.g., mylar and polypropylene) sacks layered with aluminum (ah, aluminium for you polythene-sayers) to block evaporation, oxygen (which leads to fats and oils becoming rancid) and loss of flavor.

For storing small amounts of food, reuse these aluminized bags -- which also are effective shields to block RF tags in passports and licenses. They're available commercially, e.g. this list.

These bags are comparatively expensive, though, and last millenia in landfills, and impractical for general use waste bin bags.

Non-chemical work-around: store food wastes and other potentially odorous items in paper or plastic bags that would have been discarded anyway, and that bag inside the bin liner, to delay odor migration.


The chemical disinfection (inactivation of bacteria) in fresh water can occur through a number of mechanisms, including oxidation of cell walls, inactivation of key enzymes and disruption of nucleic acids, thereby rendering them non-functional. The precise mechanism of inactivation depends upon the nature of micro-organism (bacteria, spores, viruses).

When chlorine was first used as a disinfectant in the USA in 1908 (slightly earlier in Europe), its germicidal power was commonly attributed to the liberation of 'nascent oxygen' from hypochlorous acid $(HOCl)$.

$HOCl\rightarrow HCl+\frac{1}{2}O_{2}$

Subsequent investigations (Chang, 1944, Green and Stumpf, 1946) debunked this theory and showed that chlorine reacts irreversibly with the enzymatic system of bacteria, thereby killing it.

Chang demonstrated by showing that hydrogen peroxide or potassium permanganate which release considerable amounts of nascent oxygen, nevertheless exhibit only weak germicidal activity. Chang was able to demonstrate that there is no liberation of oxygen involved in the inactivation of bacteria by chlorine.

In any case, oxygen gas $(O_{2})$ is a much weaker oxidizing agent than hypochlorous acid, as shown in the table below. Hypochlorous acid is likely to be effective as both an oxidizing agent as well as direct chlorination of the protoplasm and reaction with lipoproteins to form toxic chloro compounts that interfere with cell division (Chang, 1944).

Hypochlorous acid $(HOCl)$ dissociates in water to $H^{+}$ and hypochlorite ion $(OCl^{-})$.

$HOCl\leftrightarrow H^{+} + OCl^{-}$

Because hypochlorous acid $(HOCl)$ is uncharged, it is better able to penetrate cell walls than other chlorine species and is about 80 times more effective than hypochlorite $(OCl^{-})$ at chlorination disinfection.

The microbial disinfection due to total chlorine ($HOCl + OCl^{-}$) therefore depends upon pH and temperature. Whilst low pH and temperature favour disinfection by chlorine as $HOCl$, rate of diffusion through the cell membrane occurs faster at high temperature. The metabolic activity of the cell,rate of reaction of chlorine with enzymes and hence rate of inactivation of the cell is faster at higher temperature.

Disinfection by redox reaction using ozone gas $(O_{3})$ (a fast and powerful oxisizing agent) is sometimes used as an alternative to chlorination to reduces or eliminate the toxicity caused by residual chlorine, especially where there is concern over organic matter in the water which may produce trichloromethanes.

Sources:

Chang, S.L., Destruction of microorganisms, J. Am. Water Works Assoc., 36 (1944), pp. 1192–1206

Green, D.E. and Stumpf, P.K., The mode of action of chlorine, J. Am. Water Work Assoc., 38 (1946), pp. 1301–1305

Water Quality Control Handbook, Second Edition, by E. Roberts Alley (2007, McGraw Hill).

White's Handbook of Chlorination and Alternative Disinfectants, 5th Edition, by Black & Veatch Corp (2010, John Wiley & Sons)


Why is ionization energy of indium less than gallium?

The theoretical values of ionisation potential are pretty close to each other.in fact ionisation potential of gallium is almost equal to that of aluminium.Now due to MORE shielding in indium but lack of lanthanoid contraction the affect balances out.As the atomic radii and shielding affect simultaneously takes place from your viewpoint you can claim that the increase in the atomic radii predominates over the zeff.thus indium will have lesser ionisation energy.


The meaning of the sulfite anion attached to the benzene ring

If your wondering about the bond, the line going to the center means that the sulfonyl group is attached at any available position (2, 3, or 4) on the ring, and no particular position is given.


Why don't unstable odd electron species dimerize so that they become stable?

Odd electron species do dimerize.

  $\ce{NO}$

Nitric oxide dimerizes, but only at low temperature (and probably high pressure}.

  $\ce{NO2}$

Nitrogen dioxide does dimerize. In fact, this is a well known property of $\ce{NO2}$. $\ce{NO2}$ (orange-brown) is in equilibrium with $\ce{N2O4}$ (colorless).

$$\ce{2NO2 <=> N2O4}$$
The position of this equilibrium is dependent on temperature and pressure (like all gas equilibria), and so it makes a nice demonstration of gas properties.

  $\ce{O3}$

Ozone does not dimerize. Dimerization would lead to $\ce{O6}$, which likely is a 6-membered ring of all oxygen atoms. While this structure violates no rules, the oxygen-oxygen single bond is not stable. This instability of the peroxide bond makes it both useful and dangerous.

  So why does $\ce{NO2}$ dimerize readily, $\ce{NO}$ dimerize under duress, and $\ce{O3}$ not dimerize at all?

The extra electron in the nitrogen oxides is found in an antibonding orbital. Conversion to the dimer gets that electron into a bonding orbital. There are no antibonding electrons in $\ce{O3}$.

  $\ce{BH3}$

Borane dimerizes because the boron atom is electron deficient - it has an empty p orbital. This species is so electron deficient that it will form bonding interactions with nearly any electron pairs, including those already in sigma bonds. (More properly the B-H-B bonds are three-center-two-electron bonds). Aluminum compounds dimerize the same way.

  So why does $\ce{BH3}$ dimerize and $\ce{O3}$ does not?

Ozone has no electron deficient atoms. Count the electrons. All atoms have an octet, and while one has a positive formal charge, the compound as a whole is not electron deficient.


From a historical perspective, this was done to account for the commonalities between how the various elements behaved. In other words, the Alkali metals all exhibited a strong reaction with $H_2O$, the noble gasses all exhibited inertness in reactions, and so on.

This is explained in modern day science via the quantum description of electronic structure which indicates that elements within the same group tend to have the same number of valence electrons in the valence shell. Simply put, the reason why there are such gaps between $H$ and $He$; $Be$ and $B$, and otherwise is because there are simply different quantum arrangements of the electrons which are readily accessible to the element. That is, the ground state of $He$ has both accessible $l=0$ states occupied by electrons, which constitutes the valence shell and so is in an energetically favorable state to remain inert - becoming of the noble gasses. Hence, it’s moved far to the right from $H$ to the 8th group; consistent with the historical structure of the table.

That’s a crude description, but about what’s happening here. I agree however that the chemists may feel more at home discussing the periodic table. I’ll stick to the chart of nuclides.

Also, the Wikipedia article is fairly informative on this topic:
http://en.wikipedia.org/wiki/Periodic_table


In a galvanic cell, why does an electrode with a higher potential "pull" the electrons from the electrode with a lower potential?

There are not many problems where electronegativity helps you find the answer.  It is popular for some reason amongst newer chem students but it shouldn't be.

Here's what happens in the Daniell cell:

Metallic bonds in zinc are broken and zinc atoms each lose 2 electrons.  Both of these processes are endothermic.

Zinc ions are hydrated. This is exothermic.

Copper ions are dehydrated. This is endothermic.

Copper ions gain 2 electrons each and metallic bonds are formed. Both of these processes are exothermic.

Each of the above processes also involves an entropy change.

Only when all of these factors are considered can the direction of the electrochemical reaction be determined. Or, you can look at a table of standard reduction potentials.

By the way, silver's electronegativity is 1.93 but the reduction potential for $\ce{Ag+}$ is $\pu{0.8 V}$ while $\ce{Cu^2+}$ is $\pu{0.34 V}$.

In terms of charge, electrons flow from anode to cathode. There is no charge built up because of the salt bridge.


Chirality of symmetrically substituted biphenyl

No, the biphenyl above is not chiral.
2,2′,6,6′-tetrasubstituted biphenyls are nonplanar at room temperature in order to avoid steric clashes between the substituents. This can in general give rise to chirality depending on the substituents, such as in this example.

However, this chirality is only true under certain conditions, namely:

The substituent groups must be sufficiently bulky in order to prevent rotation about the C–C single bond that connects the two rings. If this rotation occurs freely, then the two enantiomers will be able to interconvert, and chirality will not be observed (under typical conditions)

Both pairs of substituents on the phenyl rings must be non-identical; otherwise, the molecule will possess a plane of symmetry, rendering it achiral even if rotation about the C–C bond is restricted.

Here, since the two pairs of substituents on both rings are the same (i.e. two methyl groups on one ring, and two nitro groups on the other), the second criterion is not fulfilled. As a result, the molecule is not chiral.


Why doesn't chemical equilibrium lead to uncertainties?

The reaction quotient Q approaches zero when at least one of the product concentration approaches zero (as the reaction goes to completion in the reverse direction and the extent of reaction x is at the lowest possible value), and it approaches infinity when at least one of the reactant concentrations approaches zero (as the reaction goes to completion in the forward direction and the extent of reaction x is at the highest possible value). Also, the reaction quotient increases monotonically with x.

No matter what K is, there is always a Q that matches K because Q assumes all values between zero and infinity.

The only real case where this argument does not work (real as in concentrations have to be non-negative real numbers) is when at least one reactant and one product concentration is zero. In that case, there will be no reaction at all, and Q is not defined.


The concept of temperature arises from the zeroth law of thermodynamics:

If two systems are in thermodynamic equilibrium with a third system, the two original systems are in thermal equilibrium with each other. Basically, if system A is in thermal equilibrium with system C and system B is also in thermal equilibrium with system C, system A and system B are in thermal equilibrium with each other.

From here, many hypothetical scales for temperatures are stablished.

It's said than one body has higher temperature than another when there is a heat exchange from the former to the latter.

But hot and cold are just sensations or feelings defined by people. Something is called hot when it's at higher temperature than our body (~37 °C), so a positive heat exchange will take place on contact. The opposite for cold (heat transference from our body to the cold object).

So, you are right, the higher the temperature of an object, the hotter it will be, but the concept hot is just to define the thermal perception.


It would be extremely dangerous to use a flammable solvent for a heat bath because any spill on the hot plate (or even worse, a Bunsen burner) could cause a fire.

If you want to distill off ether, it is by far better to just use warm water from a tap or a kettle, and replace your hot bath regularly.

Trust me, ether can spontaneously catch fire VERY easily ! I have been there!


Strictly speaking, a Lewis acid is something that forms a bond by accepting an electron pair form another molecule (Lewis base).  Water as such does not do that, rather it is a hydrogen ion from the water that does so.  When ammonia acts as a proton acceptor in water, the Lewis theory calls it a displacement reaction where the acid, $\ce{H+}$, is initially combined with one base ($\ce{OH-}$) and ends up combined with another base ($\ce{NH3}$).


Why when deriving the Clapeyron equation can you assume that dH = dq, as surely the pressure is not constant?

At any given temperature, the change from a liquid to a vapor occurs at constant pressure.  What we are trying to find with the Clapeyron equation is how this constant pressure changes as the given temperature changes.

At equilibrium at a given temperature and constant pressure, we have $$G_{vapor}=G_{liquid}$$ and $$S_{vapor}-S_{liquid}=\frac{(H_{vapor}-H_{liquid})}{T}$$

Now, if we want to find out how temperature and pressure change in tandem along the equilibrium line, we can write:$$dG_{vapor}=-S_{vapor}dT+V_{vapor}dP$$and
$$dG_{liquid}=-S_{liquid}dT+V_{liquid}dP$$Since, along the equilibrium line, we must have $dG_{vapor}=dG_{liquid}$, we can write:$$-S_{vapor}dT+V_{vapor}dP=-S_{liquid}dT+V_{liquid}dP$$or equivalently:$$\frac{dP}{dT}=\frac{(S_{vapor}-S_{liquid})}{(V_{vapor}-V_{liquid})}=\frac{(H_{vapor}-H_{liquid})}{T(V_{vapor}-V_{liquid})}$$This is the Clapeyron equation.


How to convert SPC files to ASCII without original spectroscopy software?

If you don't mind Python on the command line, there's an spc converter on Github.

While the trial version of the GRAMS Suite 9.2 is available with a 30 days license, it seemingly doesn't work with any data but the samples provided with it and therefore isn't suitable for your purpose :(


Is it possible for a substance to absorb a longer wavelength of EM wave and emit a shorter wavelength?

It is possible, but impractical. You can "upconvert" light. This effect, as Wikipedia relays, is due to the sequential absorption of two photons per emitted photon.

This is certainly possible to make in a fluorescent / phosphorescent light scenario, but the engineering challenges (or, more specifically - the cost) involved in such a solution would be steep.

There is a chemical model of such a compound in the two-photon-absorbance wiki site, but I have not the nomenclature skills to name it for you. I even tried using online search for the structure (and it took me a while. I hate nested benzene groups - probably why I chose inorganic chemistry)


Why does boron trifluoride form boron tetrafluoride ion instead of boron pentafluoride ion?

As you point out, the $\ce{B}$ in $\ce{BF_3}$ is sharing a total of 6 electrons, 3 from the boron and 3 from the three fluorines.  Now add a fluoride anion that brings 2 more electrons to share.  This now makes for a total of 8 electrons around the boron.


Justify assuming first-order kinetics for a complex

What would be the justification for thinking this is first-order?

The KISS principle :)

Before you start to model complex kinetics, begin with a simple model. If it fits, consider your job done.

Good enough is good enough!

(Imagine that you perform therapeutic drug monitoring. It's not your job to elucidate the exact metabolism of a drug here, you just want to be sure that the patients receive the right dose.)


Can I split water without constant refill of solution?

Salt is not needed. Salt merely allows the water to be somewhat more conductive.

The King of Random on YouTube has a great video on making one. All you need is electricity and water.
https://www.youtube.com/watch?v=cqjn3mup1So

He uses sodium hydroxide, but this not needed. It will just be less efficient.

If you want to be more efficient, you can add an electrolyte like salt. Whatever electrolyte you use, should not be used up. However, if the concentration of electrolyte gets to high (after you have used up a lot of water) it will split it too. This is bad with salt, as you can end up with chlorine gas and sodium hydroxide, which can be nasty to deal with.

So, as long as you replace the water fast enough and don't put in too much electrolyte, it should be fine and you will not need to replace it often.


Explanation for solubility of group 2 halides

To me it does not make too much sense to try to compare and explain the solubility (mass concentration) of anhydrous group 2 chlorides.

First of all, mass concentration does not seem to be the adequate physical quantity for comparison. Some of the chlorides have such high solubility that their saturated solution contains significant less water per volume than others. Molar fractions of solute and water would be a better choice.

Second, all of the chlorides precipitate as the hydrated salt from water. At normal temperature these are

$\ce{BeCl2\cdot 4 H2O}$
$\ce{MgCl2\cdot 6 H2O}$
$\ce{CaCl2\cdot 6 H2O}$
$\ce{SrCl2\cdot 6 H2O}$
$\ce{BaCl2\cdot 2 H2O}$
$\ce{RaCl2\cdot 2 H2O}$

Besides differences in structure their different stoichiometry makes a comparison difficult. And, most important, the formation of hydrates shows that we probably should not discuss the solubility of anhydrous chlorides at all.

Third, some anhydrous halides are more covalent than ionic compounds ($\ce{BeF2, BeCl2}$). This makes it difficult to argue with lattice energies and ionic radii.


Electrolysis used to cause a Chemical Reaction?

You wouldn't need to supply electric energy to get oxygen and hydrogen to recombine into water, because the free energy change for that reaction is negative. In other words, the products have a lower potential energy than the reactants, and so the reaction will proceed "downhill" on its own. You would need to overcome the very small activation energy, which you could do with electricity or with a flame or spark. But once it got started, you would not need to supply additional energy, and could in fact generate electricity from the reaction.

This is the basis of how hydrogen fuel cells work - hydrogen and oxygen react in a device that is set up to extract the energy that is produced in the form of an electric current.

An easy way to think about it is that if you need to supply energy to make a reaction happen, then the reverse reaction will produce the same amount of energy (if we ignore losses due to inefficiency) or less (if we don't). The reverse is true as well.

On a side note - electrolysis specifically refers to splitting a compound. The "-lysis" suffix means "break apart," and "electro-" prefix means "with electricty." In your case, you are thinking of "electrochemistry" - which just means "using electricity with chemistry." Electrolysis is one thing you can do with electrochemistry.


Why does the Förster resonance energy transfer scale as 1/r⁶?

From this article by Förster himself:

  This coupling is strongest if the corresponding optical transitions in
  both molecules are allowed for electric dipole radiation. Then these
  transitions are coupled not only to the radiation field but also to
  each other. Naturally, the interaction energy is of a dipole-dipole
  nature, depending on an inverse proportionality to the third power of
  the molecular distance. The probability of energy transfer is then
  proportional to the square of this interaction energy and decreases,
  therefore, with the sixth power of the distance.

This paper does not give a detailed derivation though. But in his seminal paper from 1948 Förster goes through the derivation in detail (unfortunately it is not in English but in German and the condition of the text is not very good).


Change in current in an electrolytic cell over time

The electrolyte conductivity is dependent on the temperature of the electrolyte. Higher temperature increases ions mobility, thereby increasing electrolyte conductivity. The conductivity increases around 2% per degree (assuming linear increase of conductivity). More info here.

As Burak Ulgut mentioned, temperature increase is due to ohmic (Joule) heating. This can be estimated accordingly: $Q=V \cdot I \cdot t$


What causes soda to fizz and how can it be stopped?

The soda is supersaturated with dissolved carbon dioxide since in the sealed container the soda is under pressure using carbon dioxide. A number of things can cause the formation of nucleation sites which degas the solution by forming bubbles which float to the surface. The foam is due to the surface tension of the liquid solution. The lay understanding is that the soda is "flat" when the soda is no longer supersaturated with carbon dioxide.

It should be noted that even in a saturated solution most of the carbon dioxide is in solution as a dissolved gas and not carbonic acid. Since soda solutions are acidic anyways, this also favors the dissolved gas as opposed to carbonic acid.

To minimize fizzing:

Use a clean glass. (Bubbles emanating from some sort on the
glass indicate that the glass is dirty.)
Use wet ice at the freezing temperature of water (not ice from
fridge below 32 F).
Cool the soda before pouring gently onto the side of the glass.
In other words do not just dump it into the center of the glass onto the
ice. Imagine trying to pour water into the glass without creating
any bubbles.


What is the pKa of the second ionizable group?

Since you're given that the maximum $\mathrm{p}K_\mathrm{a}$ of the second group is $7$, you can assume that the amino group will fully protonate before the second group starts accepting electrons.  (Perhaps this is what you meant by "at that point there should be no free $\ce{H+}$"?)
Therefore, from Henderson-Hasselbalch$^\ddagger$, you can calculate the fraction of the amino group that is protonated based on the starting $\mathrm{pH}$ and the amino group $\mathrm{p}K_\mathrm{a}$. (Hint: Yes, the amino group should be mostly, but not entirely, protonated.)
It's straightforward to then calculate:

The total number of moles of acid added
The number of moles of not-yet-protonated amino groups before adding the acid

which is equal to

The number of moles of acid consumed to protonate the rest of the amino groups

and, therefore, you can find

The number of moles of acid left over to protonate a portion of the second ionizable group in the molecules.

Having calculated #4, you can then calculate what fraction of the second ionizable group is protonated by the leftover acid.  This feeds back into Henderson-Hasselbalch, from which you can calculate the second $\mathrm{p}K_\mathrm{a}$ value since you are told the value of the final $\mathrm{pH}$.
$^\ddagger\,\mathrm{pH} = \mathrm p K_\mathrm a +\log\left(\frac{\ce{[A-]}}{[\ce{HA}]}\right) = \mathrm pK_\mathrm a + \log \left(\frac{[\mathrm{A_{total}} ] - [\ce{HA}]}{\ce{[HA]}}\right) = \mathrm p K_\mathrm a + \log\left(\frac{1}{f_\mathrm{protonated}}-1\right)$


Esters of catechol - Fries rearrangement

This Wikipedia article provides a nice overview of the Fries rearrangement. They note that the mechanism of the rearrangement is not fully understood, but the following mechanism is generally accepted.

The key step in the reaction is thought to involve the interaction of a lewis acid with a phenyl ester to generate an acylium ion ($\ce{[R-C=O]^+}$).  Once the acylium ion is generated the reaction proceeds like any other electrophilic aromatic substituion. The phenolic oxygen is a strong ortho-para director, so both ortho- and para-products are produced by the same mechanism. The $\ce{OCOR}$ group does not play a role in "directing" the substitution pattern.

The ratio of ortho/para isomers in most aromatic electrophilic substitution reactions is very temperature dependent. This is because of kinetic vs. thermodynamic control. Typically, the para product is the kinetic product (first formed), while the ortho product is the thermodynamically more stable product and will form if the reaction is run at higher temperatures and allowed to equilibrate. For example, when phenol is mono-brominated at low temperature, the para-bromo product is formed in >95% yield (reference). So your specific reaction is likely being run under kinetically controlled conditions to produce the para product as the major product. As the Wikipedia link points out, the para product can equilibrate at higher temperatures to produce more of the ortho product.

Finally, just a word to address @bon 's comment.  Indeed, Fries rearrangement of catechol diesters does produce mono-acyl catechols (reference, p A-172) - apparently one of the acyl groups is lost in the process.


How are atoms encapsulated in buckminsterfullerene molecules?

There’s a great summary to your question in the introduction of this article:

  Levitt, Malcolm H. "Spectroscopy of light-molecule endofullerenes."
  Philosophical Transactions of the Royal Society of London A:
  Mathematical, Physical and Engineering Sciences 371.1998 (2013):
  20120429.

It’s well worth a read, but I’ll summarize the basic ideas for you since I think it’s behind a paywall.

There are several techniques to get atoms and molecules in to fullerenes (or buckyballs). Note that the nomenclature for endofullerenes is A@C$_n$, where A is the atom or molecule trapped inside the fullerene and $n$ is the number of carbons making up the cage.

Early techniques involve vaporizing carbon and a metal or noble gas, and hoping that a) a fullerene would form, and b) a metal or noble gas would get trapped inside. For this reason, early endofullerenes were metallo-endofullerenes, which were fairly easy to create in low yield by laser vaporizing a metal oxide and graphite composite. This is how La@C$_{82}$ was first synthesized and reported. Fullerenes themselves are often made by a high temperature vaporization of carbon (arc discharge). If you do this in the presence of a metal, you can get metallo-endofullerenes. Some noble gases can be trapped inside a fullerene this way. They’ve actually found He@C$_{60}$ trapped in ancient rocks, which are hypothesized to have been formed before the formation of the Solar System in a similar process. These extreme techniques are limited to metals and noble gases.

Another way to get noble atoms into fullerenes involves heating fullerenes and nobles gases at high pressure. It’s thought that temporary openings in the fullerenes arise, and this allows the noble gases to insert and become trapped. You are limited to noble gases here. Yet another way is to use ion beams to “blast” atoms into the fullerenes. You can get P@C$_{60}$ and N@C$_{60}$ this way.

The above techniques give you low yield, and you are limited to metals and noble gases.

The latest techniques are often called “molecular surgery” and involve chemically creating a hole, or orifice, in a fullerene. The orifice may be enlarged by chemical means, depending on the size of the guest molecule you want to insert. These open-cage endofullerenes, which are relatively stable under mild conditions, are introduced to the guest molecules. Elevated temperature and pressure often helps get the molecules into the cage. Finally the fullerene is closed using chemical methods. In this way, H$_2$O@C$_{60}$ and H$_2$@C$_{60}$ have been made. Just in the past year, HF@C$_{60}$ was made with similar techniques.


Inability of alkyl fluorides to be reduced to an alkane in the following reaction

Other answers may be more elegant, but it boils down to the carbon-fluorine bond being too strong, in fact stronger than carbon-carbon or carbon-hydrogen bonds.  The required bond breaking simply cannot be done under normal conditions.  Similarly Grignard reagents (formed with a much more powerful reducing agent) are much less common with fluorine than with heavier halogens, and the bad things chlorofluorocarbons do to our ozone layer come from breaking $\ce{C-Cl}$ rather rhan $\ce{C-F}$ bonds.


How to determine the pressure to output a gas to achieve a particular gas flow rate?

Well, since the pressure in the tank will decrease with usage, you can't ultimately do what you seem to want to do.

At a given backing pressure, there are equations to determine the orafice size to get a flow rate - google 'choked flow'. Downstream of the regulator, your orafice should be good for quite some time.

For an experimental setup, one would use either a mass flow controller (MFC) to actively control the flow rate, or (much cheaper) one of those little flow indicators with the ball floating in the He (air) stream. Note the second option here is basically an adjustable orafice with integrated flow indicator.


In the following below, please interpret formal charges as marked below the atom.
$$\ce{O=N=O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{O\equiv N-O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{O-N \equiv O} $$
$$\ \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \ \ +\ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\ \ +\ \ + $$

The triply charged species are valid resonance forms, but very likely are small contributors.


Negative delta S ($\Delta S <0$) is a decrease in entropy in regard to the system.

For physical processes the entropy of the universe still goes up but within the confines of the system being studied entropy decreases.

One example is a freezer with a cup of liquid water in it.  The freezer will utilize the electrical energy coming in to pump heat from the water until it becomes a solid (ice). At which point the entropy of the system (the contents of the freezer) decreases, however the electrical energy needed to be produced to power the freezer such as coal (burning a solid to a gas) and heat was wasted by the freezer in the process both of which create larger amounts of entropy than was reduced in the system by the freezer.

For chemical processes entropy can be a great driver of many reactions but it is not absolute. A system's favorability to release energy (enthalpy) competes with entropy. For example, an electron of hydrogen may have higher entropy if it drifts from the core proton but the electrostatic forces (and quantum mechanics) energetically keep it bound to the atom.  For isobaric processes, you much determine the change in Gibbs free energy for the reaction to know which way it is driven. For isochoric processes, you must determine the Helmholtz free energy to know which way a reaction is driven.

One example is the oxidation of iron in air.  When the oxygen is in the gas state it has higher entropy but the energy of bonding with iron is so great that at normal pressures oxygen goes from the gas phase and the iron rusts ("enthalpy wins") as delta G is negative.

Now we must consider statistical thermodynamics, this process is pressure dependent. At normal atmospheric pressure, the forward rate of oxygen entering the gas phase is the same as the reverse process. If the iron oxide were held in a sufficiently high vacuum the reverse process would occur and the iron oxide would reduce back to iron like many asteroids ("entropy wins"; note: thermodynamics is equilibrium after an infinite time). Accounting for pressure modifies the Gibbs free energy equation to: $$\Delta G = \Delta G^\circ -RT \ln(P) = \Delta H^\circ -T\Delta S^\circ -RT \ln(P)$$

One thing to note is that for chemical reactions the entropy and enthalpy values are for a standard temperature (such as $298\ \text{K}$). For a spontaneous system with $\Delta S^\circ < 0$ enthalpy must be negative, this heat in reality is absorbed by the system or the environment and produces entropy according to:  $$ \int \mathrm dS \equiv \int \frac{C_v}{T}\,\mathrm dT \equiv \int \frac{\mathrm dQ}{T} $$

This in and of itself produces some entropy in the universe though may not net above zero as the bonding energy is still the major driving force.


How to explain the selectivity between methanol and methanethiol in an SN1 reaction with an halogenated hydrocarbon?

Obviously saying "the rate is same but the activation barrier is different" is a contradiction, as both are directly related via the arrhenius equation. The rate is predominantly determined by the rate determining step. If there are several rate determining steps (aka several steps with similar activation energies), the overall rate will be a product of all steps.

So what does that mean for your question?

You stated that methanethiol is a better nucleophile than methanol. This would suggest that the activation barrier in the last step is slightly lower in the case of methanethiol. However, the rate determining first step is unaffected. Hence, the overall rate will be only slighter smaller for methanol.

The ratio of both products will be determined by the last step.


Your book is glossing over some details.

There are three things occurring when you dissolve non-metal oxides in water (e.g., $\ce{SO2}$, $\ce{CO2}$, $\ce{NO2}$, etc.):

The gas simply dissolves in the water. You are correct that this step is a physical change.  How much dissolves depends on the partial pressure above the water and the Henry's law constant.
Once dissolved in water, some of the non-metal oxide will react with water to form an acid (sulfuric/sulfurous acid, carbonic acid, nitric/nitrous acid, etc.) and this is the chemical change that your book is referring to. There will be an equilibrium constant that will dictate the relative concentration of dissolved gas to the corresponding acid that is formed.
The acid (usually weak) will dissociate a little bit.  Again, the exact amount of dissociation will depend on the equilibrium constant, $K_\mathrm{a}$.

Using your example, we would have:

\begin{align}
\ce{SO2(g) &<=> SO2(aq)}\\
\ce{SO2(aq) + H2O(l) &<=> H2SO3(aq)}\\
\ce{H2SO3(aq) &<=> HSO3^- (aq) + H+(aq)}
\end{align}

Similar things happen when metal salts dissolve in water (although not $\ce{Na^+}$ or the other group I and II metals). The metal cations will react with water and form a weak acid.


Are there compounds more conductive than Silver (Ag)?

Graphene supposedly has higher conductance.

The [...] resistivity of graphene sheets would be 10−6 Ω⋅cm. This is less than the resistivity of silver, the lowest otherwise known at room temperature.

but it is not a compound - it is a homonuclear molecule.


Why is o-toluic acid (2-methylbenzoic acid) more acidic than benzoic acid?

Having one methyl group ortho to the carboxylate group will not prevent the carboxylate group from becoming reasonably planar with the aromatic ring.  If you had a second methyl group in the other ortho position, then planarity would be a problem.

A methyl group attached to a benzene ring is electron releasing in both an inductive and a resonance sense.

Inductively, the electrons in an $\ce{sp^3}$ hybridized methyl group prefer to shift towards the lower energy (more s-character in the orbital) $\ce{sp^2}$ aromatic carbon.
The methyl group is also electron releasing in a resonance sense due to carbon-hydrogen hyperconjugation (see figure below).

The resonance structure pictured below on the left is one of many that we can draw for the o-toluic carboxylate anion.  Note that it places a positive charge on the methyl bearing ortho carbon. Both inductively and through resonance (see the hyperconjugated resonance structure below on the right) the methyl group will stabilize this resonance structure.  It is this stabilization provided by the methyl group that makes o-toluic acid a stronger acid (pKa=3.91) than benzoic acid (pKa=4.19).

Note: As pointed out by both @Mithoron and @GaurangTandon, this answer does not explain why p-toluic acid is a weaker acid than o-toluic acid or benzoic acid. The answers in this earlier question better answer this question.


Why are non-spontaneous reactions driven in the presence of energy?

You are incorrect in your assumption that the addition of a power source makes both the forward and the reverse redox reactions more spontaneous. An electrochemical cell operates on the same principle as a battery. On the anode of the cell, you will find a compound that is being oxidized, or losing electrons. These electrons then flow through some sort of a wire to the cathode, where the electrons are being used to reduce another compound. The electricity will flow in the "spontaneous" direction, in other words, from the compound that prefers to be oxidized to the compound that prefers to be reduced. The "voltage" across an electrochemical cell is determined by adding up how badly the reducing agent wants to get oxidized ($E°_{ox} = -E°_{red}$) and how badly the oxidizing agent want to be reduced ($E°_{red}$), and is a measure of how much the reaction wants to occur.

Now, lets try adding a battery in between the anode and the cathode. If the battery is added such that its anode points towards the cell's cathode, the battery will add to the voltage of the cell, causing the effective voltage for the reaction to increase. On the other hand, if the battery is placed such that its anode points towards the cell's anode, it will be creating a voltage in the other direction. When this happens, it becomes a question of whichever voltage is stronger. If the battery has a stronger voltage than the cell, electrons will flow in the opposite direction. This is what occurs in electrolysis, causing the originally non spontaneous process to become spontaneous. This results in charging the electrochemical cell. On the other hand, if the battery has a smaller voltage than the cell, the electrons will flow in the original direction - charging the battery as they pass through it.

While a power source can be used to either make a spontaneous process more spontaneous or to reverse the flow of electrons, it cannot do both at the same time.


For oxidation of aromatic side chains, you must have at least one Benzylic hydrogen. tert-butyl Benzene (which is actually 2-methyl,2-phenyl propane) doesn't have any benzylic hydrogen.

-By @Soumik Das


Methods to quantify lead(II) concentration in water on the ppm/ppb level

Russian interstate standard GOST 18293-72 "Drinking water. Methods for determination of lead, zinc and silver content" (PDF in Russian) suggests to use sulfarsazene (plumbone), which forms orange-yellow complex with lead(II).
Reported method sensitivity (spectrophotometry): $\pu{0.5 μg L-1}.$

A brief review in English [1, p. 315]:

  To determine lead in solid particles of industrial facilities, natural waters, biological objects and salts, sulfarsazene is used (4''-nitrobenzene-1'',4-diazo-amino-1,1-azobenzene-2''-arsono-4-sulfoacids).
  While it is less sensitive than dithizone , it has a number of advantages: it is water-soluble, easy to handle and needs no cyanide as masking agent.
  But, it has insufficient reaction selectivity.
  Manganese, nickel, copper, zinc, mercury, cadmium, iron, zirconium, and other elements interfere with the determination. The methods developed on the basis of sulfarsazene require a preliminary ion separation; for instance, lead is separated by extracting with dithizone solution in carbon tetrachloride …

  Study of the reaction of sulfarsazene with lead ions points to simultaneous participation of the lead cation with triazene and arson groups in complex formation.
  Polarographic investigation confirmed the existence of lead bonds with the o-arsonotriazene group of the reagent, and cleared up the azo-group contribution to the formation of the complex which may be due to coordinative bonding with lead atoms.
  The role of the nitrogroup as an electrophilic substituent that improves the mobility of the hydrogen atom of the aminogroup was experimentally confirmed.
  The structure of the resultant compound is as follows:

GOST 18293-72 also includes polarographic method for lead detection using the dropping mercury electrode.
Reported method sensitivity: $\pu{10 μg L-1}.$

References

Roeva, N. N.; Savvin, S. B. Spectrophotometric Determination of Minute Lead Contents of Natural and Industrial Objects. Reviews in Analytical Chemistry 1986, 8 (4). DOI: 10.1515/REVAC.1986.8.4.313.


There are 32 isomers of C6H12O6, of which you have named 6. The most common one is D-glucose (dextrose). I don't think all of the 16 L-saccharides are found in nature.

α and β refer to the connection between rings, whether the ketal (or aldehyde) oxygen is axial or equatorial.

  Structures made with ChemDraw.


I think this link should really help you out in understanding how the equilibrium constant depends on temperature.

Now, considering the case of water dissociation:
$$\ce{H2O <=> H+ + HO-}$$

We can calculate the free energy of the reaction ($\Delta_\text{r}G$), using the Gibbs energy for the reactant and the products (in aqueous solution):

$$\Delta G(\ce{H2O}) = -237.13\ \mathrm{kJ/mol}$$
$$\Delta G(\ce{H+}) = 0\ \mathrm{kJ/mol}$$
$$\Delta G(\ce{HO-}) = -157.24\ \mathrm{kJ/mol}$$

$$\Delta_\text{r}G = \Delta G(\ce{H+}) + \Delta G(\ce{HO-}) - \Delta G(\ce{H2O}) = -157.24\ \mathrm{kJ/mol} + 237.13\ \mathrm{kJ/mol} = 79.89\ \mathrm{kJ/mol}$$
The values of the Gibbs energies were taken from the 8th edition of Atkins' Physical Chemistry.
Now, applying the formula below (which can be obtained from $\Delta_\text{r}G = -RT\ln K_\text{eq}$) for $T = 298\ \mathrm{K}$ (or $25\ \mathrm{^\circ C}$):
$$K_\text{eq} = \operatorname{e}^{-\frac{\Delta_\text{r}G}{RT}}$$
we get something like this :
$$K_\text{eq} = 2.71^{-\frac{79890\ \mathrm{J\ mol^{-1}}}{8.3145\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm{K}}} = 9.93 \times 10^{-15}$$
This value can be approximated to ${10^{-14}}$.


How to correctly write "7 apples" according to the international system of units (SI)

In accordance with the International System of Units (SI) [Brochure in English, 8th edition, 2006; updated in 2014] and the corresponding International System of Quantities (ISQ) [ISO/IEC 80000 Quantities and units (14 parts)], you can define a suitable new quantity, for example with the quantity name “number of apples” and the quantity symbol “$N_\text{apples}$”.

The number of apples $N_\text{apples}$ is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):

$$\dim N_\text{apples} = 1$$

A quantity of dimension one acquires the unit one, (symbol: $1$); i.e. the coherent SI unit for the number of apples is the unit one.

Generally, the unit one is an SI derived unit; for example, the derived SI unit for friction factor is newton per newton equal to one, (symbol: $N/N = 1$). However, the unit one for counting numbers, e.g. number of protons in an atom or number of apples, is considered as a base quantity because it cannot be expressed in terms of any other base quantities. Hence, in this case, the unit one is usually considered as a base unit, although the CGPM has not yet adopted it as an SI base unit.

The name and symbol of the measurement unit one are generally not indicated. Therefore, you may write: “The number of apples is $N_\text{apples}=7$.”

The unit one or its symbol $1$ may not be combined with SI prefixes. For example, if you have 2000 apples, you must not write “$N_\text{apples}=2\ \mathrm k$” for $N_\text{apples}=2000$. (And by the way, when you see something like “10K reputation” mentioned on any stackexchange site, you are looking at at least three nonconformities at the same time.)

Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted. Expressions for units shall contain nothing else than unit symbols and mathematical symbols. Therefore, write

“the maximum electric potential difference is $U_\text{max}=1000\ \mathrm V$”, not “$U=1000\ \mathrm V_\text{max}$”
“the gauge pressure is $p_\mathrm e=0.5\ \text{bar}$”, not “$p=0.5\ \text{bar(g)}$”
“the electric power is $P_\text{el}=1300\ \mathrm{MW}$”, not “$P=1300\ \mathrm{MW_{el}}$”
“the water content is $170\ \mathrm{g/l}$”, not “$170\ \mathrm{g\ \ce{H2O}/l}$”

and also

“the number of apples is $N_\text{apples}=7$”, not “$N=7\ \text{apples}$”


What is the correct way to illustrate a bond to something irrelevant in a structural formula?

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), in the structure diagram of the parent structure, variable substituents should be indicated by an unambiguous label.

  This label should be chosen so that it cannot easily be confused with other valid element symbols or fragment abbreviations. Traditionally, the letter “R” is widely used, followed by a superscripted number (R1, R2, etc.) or prime(s) (R', R", etc.) if necessary to distinguish different groups. It is not acceptable to use a letter followed by a subscripted number, since subscripted numbers are more commonly used to indicate repeat counts, and confusion between the two is inevitable. Other common symbols include X, Z, and G. (…)

The label may be used with or without an explicit list of allowed substituents (e.g. in tabular form).

The wavy line is preferred in structure diagrams of the variable substituents, in cases where the external connection points must be indicated for the variable substituents.

Therefore, if you consider nucleobases with variable substituents, you may use labels such as “R”. However, if you consider the nucleobases to be the variable substituents of another (larger) parent structure, you may use a wavy line.


What is the highest possible expanded octet?

I took an interest in this question because it's something I recently wondered myself. First of all, I should clarify that while you mention hypervalency, what you seem interested in is hypercoordination, or even more generally, just compounds with high coordination numbers (hypercoordination is used specifically when the number of ligands in a compound is larger than "normal"). Hypercoordination and high coordination numbers are entirely independent of hypervalency or VSEPR theory altogether. Regardless of the precise electronic structure in a compound, coordination numbers can often be determined far more directly, especially when the compounds can create crystal structures for x-ray crystallography or neutron diffraction.

Usually there is little focus given to compounds with more than six ligands, as the vast majority of compounds will have atoms surrounded by six or less ligands. However, there are some representatives for higher coordination numbers. Some books will mention iodine heptafluoride, $\ce{IF7}$, as a seven-coordinate compound, with its pentagonal bipyramidal structure. It is still possible to add a fluoride and obtain an example of coordination number 8 in the octafluoroiodate(VII) anion, $\ce{IF8^{-}}$, which has an interesting square antiprismatic geometry (take a cube and twist a face by 45°).

For coordination number 9, representatives can be found in the transition metal hydrides, such as the nonahydridorhenate anion, $\ce{ReH_9^{2-}}$, and the lowest energy configuration in this case is a curious tricapped triagonal prism.

Is it possible to go higher than coordination number 9 in coordination compounds? While there are examples, at the present it seems that none of them contain solely monodentate ligands, that is, individual ligands which only bind to the centre once. This is because either there would have to be a lot of crowding over the central atom, creating repulsions between the ligands, or because to allow enough space the ligands would have to stay relatively far from the central atom, making their bonds weak.

However, if a single ligand is allowed to bond to the centre through more than atom simultaneously, then the coordination number can keep increasing without requiring the presence of too many ligands. Actinides have a very rich coordination chemistry and are capable of generating several impressive compounds, such as uranocene with coordination number 16, but it seems the current record is held by actinide elements surrounded by four cyclopentadienyl rings, each with five carbon atoms, reaching an amazing coordination number of 20 in tetrakis(cyclopentadienyl)thorium(IV) ($\ce{Th}\mathrm{(\eta ^5-}\ce{C5H5}\mathrm{)_4}$ or its uranium analogue.

To finalize, it's interesting to note that though there are no examples yet of compounds with coordination number 10 or above containing only monodentate ligands, their expected geometries can be calculated even for much higher coordination numbers (under certain assumptions). For very high coordination numbers, calculations and physical reality will likely diverge, but perhaps a few more coordination numbers with the expected geometry will be unlocked by the study of ultraheavy element chemistry. Of course, the calculations linked here don't help much to study the geometry of coordination compounds with polydentate ligands, as they rely significantly on the geometry of the ligand itself

Edit: This very relevant article raises some interesting points. For example, endohedral fullerene compounds could be thought of as a central atom surrounded by a single ligand in the shape of a cage, so one could possibly make a case for structures with coordination numbers of 60, 70, 80 or even more. The article also calculates the possible existence of a compound containing 15 monodentate ligands, the cation $\ce{PbHe_{15}^{2+}}$, though it would be very weakly bound (as you might expect from a helium compound) and possibly restricted to the gas phase.


Complexity in computing the integrals involving higher angular momentum

It appears that the integrals that are difficult are the two-electron repulsion integrals. There exists a recursive relationship that can turn integrals with high-angular momentum functions into integrals with only s-type functions. Refer to this lecture for details.

See the following articles for details concerning these recurrence relations:

M. Head Gordon and J. Pople, J. Chem. Phys. 89, 5777 (1988)

S. Obara and A. Saika, J. Chem. Phys. 84, 3963 (1986)


Should compounds in a generic reaction be italicised (typography)?

Going by the basic rule of a variable (that can hold multiple values) being italic, I would go for the $r_nR_n$ style (4 times italics), but then I found the following in the IUPAC Green Book (printed page 53):

  $$0 = \sum_j \nu_j \mathrm{B}_j \quad \text{where }\mathrm{B}_j\text{ denotes a species in the reaction ...}$$

Here, the B is upright although it is definitely a placeholder. I don't understand, why.

For the subindices, I think (hope) there is no doubt that the $n$ and $m$ should be italic, because they clearly are variables standing for numbers (either any number as in the sum, or for an unknown maximum number $1, 2, 3, \ldots, n$ as in the generic reaction). Both form are well established in mathematics (and I hope chemists don't deviate).

The IUPAC Gold Book, R05156 uses the same convention.

  For the general chemical reaction :
  $$a\,\mathrm{A} + b\,\mathrm{B} \rightarrow p\,\mathrm{P} + q\,\mathrm{Q} + \dots% not using \ce to have proper formating regardless of mhchem version$$

And they do not mean Boron and Phosphorus.

(Spotted via this answer).


What is the pressure belonging to in Boyle's Law?

Your derivation of Dalton's Law of partial pressure is fine.

I think you are confused about how pressure is "applied" to a gas, and what that does to the internal pressure of a gas.

Imagine that you have a "sphere" of gas surrounded by a "force field" (like in Star Trek) in empty space, then the force field containing the gas is dropped. Since there is nothing to confine the gas it just expands into infinite space.

A lot of ideal gas models, like the Wikipedia article on Boyle's Law, show the gas as being contained in a cylinder with a piston pushing on the gas to create the volume and pressure of the system. The external force on the piston is counterbalanced by the pressure of the gas times the surface area of the piston. The pressure of the gas is the same on all surfaces of the container. The gas doesn't just "press back" on the piston.

Pressure and volume are inversely related for the same system. So at the same temperature,

$PV = k$

Thus if the pressure increases, then for the same amount of gas the volume must decrease to keep $k$ a constant.


Has anyone actually prepared Zn(HCO3)2 absent the presence of pressurized aqueous CO2?

We have to distinguish between the bicarbonate in solution and the bicarbonate as a solid.  While many metal cations are compatible with bicarbonate anions in dilute aqueous solution, only in a few cases can a solid bicarbonate be expected as a precipitate.  In most cases, such as with lithium or with hard-water forming metals such as calcium and magnesium, the carbonate is formed instead.  The only solid metal bicarbonates I know of are those of the heavier alkali metals than lithium.


Why does hand sanitizer leave your skin feeling cool?

As user137 noted, this is from evaporation of the alcohol. Evaporation of a liquid takes the enthalpy of vaporization away with it, making you cooler.  Just like how sweating helps keep you cool in the summer, and how swamp coolers (evaporative coolers) keep houses cool in the southwest.


Water becomes cold on mixing energy drink

Well, the solution enthalpy of sugars is positive.
I found these numbers on the internet

$\ce{C12H22O11}$ (sugar(sucrose)) : 5.4 kJ/mol
$\ce{C6H12O6}$ (glucose) : 11 kJ/mol
$\ce{C6H12O6·H2O}$ (glucose monohydrate) : 19 kJ/mol

So if your "energy drink" is a dry powder (and not a readymade drink in an aluminum can), this could explain your observation.

You should however put a thermometer into your experiment, and get us some numbers. As is, my above is just another piece of guesswork.


Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions.

If the epoxide is asymmetric, the structure of the product will vary according to which mechanism dominates.

When an asymmetric epoxide undergoes solvolysis in basic methanol, ring-opening occurs by an SN2 mechanism, and the less substituted carbon is the site of nucleophilic attack.

In the case of acid catalysed ring opening, the mechanism is not exactly SN1 or SN2. It is a hybrid of both.

In the first step, the oxygen gets protonated.

Second step the C-O bond starts to break ( the C with more alkyl substituents is preferred for cleavage of C-O bond. In your case the tertiary carbon atom is preferred to the secondary C atom)

Third step the nucleophile ( or alcohol in this case) attacks the electrophilic carbon (the C with which C-O bond is still breaking). Unlike in SN1 mechanism the reaction does not wait for the carbocation to be formed completely. The incoming nucleophile attacks the carbon from a direction opposite to that of direction of C-O bond because C and O are still bonded to some extent.

So hence the mechanism proceeds similar to what you have shown.
In acid catalysed ring opening the nucleophile attacks the more substituted carbon because this carbon has more fractional positive charge then the secondary one.


Nucleophilic Addition of Hydrohalic Acids

As you noted, hydrogen cyanide will react with carbonyl compounds to form cyanohydrin and alcohols to form ketals/acetals. The list can be expanded to include aminals after attack of a nitrogen nucleophile, or thioacetals after sulphur attacks.

So why does it not happen with hydrogen halides? Well, in the case of the above mentioned, the reverse reaction is usually not favoured. In fact, ketals are used as protective groups for ketones (usually the reaction with ethan-1,2-diol or propan-1,3-diol); and propan-1,3-dithiol is similarly used for umpolung reactions. Likewise, the cyanohydrin has a new carbon-carbon bond formed, and these are relatively stable. To remove the ketal protective group, a rather strong Brønsted acid is often required along with an excess of water.

But in the case of a halide, that can just easily diffuse away again as a happy little halide; carbonyl double bond will be reformed. In fact, this reaction can be (and is!) used in exactly the opposite direction: liberate a hydroxide on the same carbon as a halide, and the halide will be eliminated to create the double bond.


I am just wondering how it comes that 60-100% methanol fractions from MCI gel CHP20P eluted very early on C8 column?!

The gel you are using seems to be a porous divinyl benzene / styrene resin.  The aromatic groups of the resin may interact more strongly with aromatic side chains on your molecule (e.g. Y residues) than does the aliphatic C8 groups of the reversed-phase HPLC column you are using.

"Hydrophobic" vs. "hydrophilic" is a very useful axis for understanding solute interactions with stationary phases, but unfortunately not everything can be collapsed into this single-variable scale.

Another possibility is that residual solvents or very large amounts of co-solutes are interfering with the RP-HPLC method you are using, causing premature elution.  As an extreme example of what I mean, imagine if you injected a few μL of chloroform along with your sample in the RP-HPLC method.  The chloroform would prevent column binding and cause premature elution.  Now, it's unlikely your samples have chloroform, but they could have large amounts of methanol or other "solvent-like" impurities.

These are just general points; unfortunately I can't be sure that either is the precise source of your particular problem, without more (and more clearly presented) information on your former method from five years ago, your method today, and the differences in observed vs. expected results.


Titration of acid with base or base with acid: can I always use the same indicator?

When phenolphthalein is used as an indicator, NaOH is taken in a burette because then the colour change is from colourless to pink. And this time, it will be easier to observe.

On the other hand, if HCl is taken in a burette, colour change will be from pink to colourless which can't be observed in a single drop.

When NaOH is titrated with HCl using methyl orange as an indicator, HCl is taken in a burette because then the colour at the end point changes from yellow to red which is easier to observe rather than that from red to yellow.


According to the Oxford English Dictionary entry on "ethylene", the first English usage was in 1849 in the Chemical Gazette (alternate name Journal of Practical Chemistry), published in London. The specific citation is Vol. 7 p. 116 (Mar. 15, 1849). The sentence in which it is used is in an article by "A. Wurtz" titled "On a Series of Organic Alkalies Homologous with Ammonia.", and the relevant sentence reads "By adding to ammonia the elements of ethylene, $\ce{C4H4}$, ethylic ammonia, $\ce{C4H7N}$, is obtained."
The full text of the journal is available digitally via Google Books.


Electrostatic repulsion between electrons is the key to understanding $\ce{H-}$. The electrons might be delocalized, but they are also correlated - they avoid each other.

Both electrons of the hydrogen ion have the same energy level and they don't shield the charge of the core from one another.

The Aufbau and Pauli exclusion principles (and the assumption of hydrogenic orbitals) are insufficient to explain the structure and stability of $\ce{H-}$. In particular, it helps to drop the assumption that the electrons occupy the same spatial orbital and differ only according to their spin quantum number [1]:

As  pointed  out  in  the  introduction  Bethe  [5]  and Hylleraas [6] were the first authors to prove the stability of the negative hydrogen.  But it was Chandrasekhar [1]who first introduced a clever wave function to describe the H− system which leads to a beautiful physical picture. The key concept introduced by Chandrasekhar was to break the symmetry between the two electrons, which is a way to introduce implicitly the electron correlation.  [...]
We conclude that the negative hydrogen ion is a special  atomic  system  whose  stability  depends  completely on the electron correlation. [...] mean field theories and other approximations do not take into account correlation, preventing the possibility of predicting stability for the negative hydrogen ion.

Reference
[1]. Ignacio  Lopez  de  Arbina  de  Frutos. Stability of the Negative Hydrogen Ion:  Variational Approach with Electron Correlation.


Why is a double bond given priority over substituents during IUPAC nomenclature?

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic 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;

  (…)

You have correctly identified the principal characteristic group and assigned the lowest locant to the ketone in accordance with Rule (c).

Next, a low locant is assgined to the double bond according to Rule (e). Therefore, the name of the structure without further substituents is cyclohex-2-en-1-one rather than cyclohex-5-en-1-one since the locant 2 for the double bond is lower than 5.

Thus, the correct name is 6-ethyl-5-methylcyclohex-2-en-1-one.


Significant figures in a calculation of atomic mass from isotopic abundance

First, for reasons I'll point out later this is a poorly made-up problem which was only designed to test significant figures, not to teach anything real about chemistry.
Second there are no special rules for "word problems" using significant figures.
The gist is that significant figures are a simple method to do error propagation in calculations. It is essentially based on the notion that the error is +/- 1/2 of the last significant digit, as if the value had been rounded to that digit. So 151 means that the value should be between 150.5 and 151.5.
The whole point of significant figures is to prevent you from dividing 151 by 3.00 and getting some ridiculous precision like 50.333333333333333333333333333333.
Now on to the problem. 151 and 153 are mass numbers for the europium  isotopes. Mass numbers are poor values to use for the isotope  masses. If fact the known masses are 150.9198502(26)  and 152.9212303(26). The numbers in parenthesis give the standard deviation of the measurement. So for 150.9198502 the standard deviation is 0.0000026.
Likewise according to the Wikipedia page, $\ce{^{151}Eu}$ has an abundance of 0.4781(6) and $\ce{^{153}Eu}$ has an abundance of 0.5219(6). The point is that the isotopic variation in samples taken from all over the earth limits the precision of the atomic mass for any element, not the mass of the isotopes which are know with much greater precision.
The same Wikipedia page linked earlier gives the atomic mass of europium as 151.964(1) which is inconsistent with the given percentages. I do not know the how/why there are so many more significant figures in the atomic mass (and I am too lazy to turn this answer into a research project to figure that part out...).


Would osmosis occur between two solutions of equal moles of water but unequal moles of a nonpolar molecule?

Your example with the no-polar substance wouldn't qualify in a discussion of osmosis. There needs to be a solution on at least one side of the membrane. Since a typical non-polar substance would be insoluble in water, osmosis would not apply.

However, in your diagram, water would move from left to right due to a concept called pressure potential. You will eventually learn about this in AP Bio (I taught it for 24 years.) Pressure potential and osmotic potential work together, in a concept called water potential, to calculate which way water will move across membranes. You will probably perform a lab utilizing this concept this year. Pressure potential, in your example, is just a fancy term stating that because both the atmosphere and the non-polar substance on top of the left side are greater than just the atmosphere pushing down on the right, water will move to the right.

One of the things you should be careful about is that it is not the number of moles that are important, it is the concentration of substances. Get rid of your non-polar substance. Move the membrane over to the left so there is much more water on the right. Imagine the water levels are equal on both sides so we don't have any pressure potential to deal with. In this scenario, there would be no net movement of water even though there's far more total water on the right. This is because the concentration of water on both sides is the same, 100%


Major Product of Alkene in Bromine Water

@orthocresol mentioned the main points in his comment but I would just like to expand a little more. The main electrophile in bromine water is $\ce{Br2}$ and this is what does the initial attack on the double bond, forming a cyclic bromonium ion. The bromonium ion is a strong electrophile and so it will react with any nucleophiles that are around. Since the water molecules will far outnumber bromide ions in the solution then the major nucleophile is water, even though it is not as good a nucleophile as bromide. The nucleophile attacks in an $\mathrm{S_N2}$ reaction, leading to a halohydrin product, after loss of a proton. This mechanism also explains the trans stereochemistry observed in the product.

As @orthocresol mentioned, the equilibrium constant for the formation of $\ce{BrOH}$ will be very small and so $\ce{BrOH}$ will be negligible as an electrophile. This is explained in a bit more detail in @Jan's answer here.


Regioselectivity in Claisen condensation and aldol reaction

Claisen and aldol condensations are thermodynamically controlled,* so it is not a question of which enolate forms, it is a question of which product forms.

In your first example, product A can be deprotonated by ethoxide as there is still an acidic hydrogen between the two carbonyls. Product B, on the other hand, has a quaternary carbon between the two carbonyls and cannot be deprotonated.

Because the pKa of a 1,3-dicarbonyl (~10) is significantly smaller than the pKa of ethanol (~15),** this deprotonation of A is almost complete and it therefore pulls the equilibrium position over to one which favours product A, simply by le Chatelier's principle. In fact if you somehow make product B via a different route and then treat it with ethoxide, you should exclusively get product A out of your flask, via retro-Claisen and Claisen condensation.

For the second example I would expect product D (tetrasubstituted alkene) to be preferentially formed over C (trisubstituted).

* Unless you use kinetic deprotonation conditions such as LDA. In this case you will then have to look at the stability of the transition states leading to enolate formation, but that's a story for another day.

** Very rough values.


Why does the E1cb mechanism proceed in the specific way it does?

-OH is a poor leaving group and so E1 and E2 are not favourable. In the case of E1, the formation of a primary or secondary carbocation is not stable either.

For E1cb, the intermediate enolate is more stable and since the removal of the -OH is not in the rate determining step, it is more favourable than E1.