Why exactly are standard potentials additive?

Following on Derek's great answer, it is very important to remind that the conventional way we use to add half-cell potentials is a consequence of the conservation of energy. Therefore, we should look at this from the perspective of Hess's Law.

How so?

Well, if we add two reactions, no matter which ones, there's one certainty we keep in our souls: the new free energy will equal the sum of the individual ones.

This is a critical issue when trying to figure out the potential of a new reaction that is not the sum of an oxidation and a reduction.

For instance, take a look at these reactions:
$$ \ce{Fe^3+ + e- -> Fe^2+} \qquad {E^\circ}_1 = 0.771\, \text{V}_{SHE}  $$
$$ \ce{Fe^2+ + 2e- -> Fe_{(s)}} \qquad {E^\circ}_2 = -0.447 \, \text{V}_{SHE}  $$
$$ \ce{Fe^3+ + 3e- -> Fe_{(s)}} \qquad {E^\circ}_3 =\, ? \, \text{V}_{SHE}  $$

It is tempting to simply add these potentials and get the third value, but we would be incurring on a major error. The proper way to proceed would be to calculate each reaction free energy and then figure out the standard potential.

$$ \ce{Fe^3+ + e- -> Fe^2+} \qquad {\Delta G^\circ}_1 = -(1)\mathrm{F}{E^\circ}_1  $$
$$ \ce{Fe^2+ + 2e- -> Fe_{(s)}} \qquad {\Delta G^\circ}_2 = -(2)\mathrm{F}{E^\circ}_2  $$
$$ \ce{Fe^3+ + 3e- -> Fe_{(s)}} \qquad {\Delta G^\circ}_3 = -(3)\mathrm{F}{E^\circ}_3  $$

Since we know that ${\Delta G^\circ}_3 = {\Delta G^\circ}_1 + {\Delta G^\circ}_2$, we have:
$${E^\circ}_3=\frac{{E^\circ}_1 +2{E^\circ}_2}{3} = -0.041\, \text{V}_{SHE}  $$

Why don't we go through all this way when working with cell potentials? Because you always work your equations so that the number of electrons transferred on each of them has the same value. That way, the sum of the potentials is an indirect way of summing the free energies, which are now normalized by a common value.

This trick is evidently very useful, but sometimes it may be very misleading.


What's the minimum amount of energy needed to remove CO2 from the atmosphere?

The OP is specifically interested in the efficiency of removing $\ce{CO_2}$ from ambient air, rather than from point sources like power plants.  The former is obviously less efficient because the $\ce{CO_2}$ is far less concentrated.
According to p. 50 of the the SAPEA* Evidence Review Report No. 2
(Schlögl, Robert, et al. Novel carbon capture and utilisation technologies: research and climate aspects. 2018.  Available at : https://www.sapea.info/wp-content/uploads/CCU-report-proof3-for-23-May.pdf),
"the thermodynamic minimum energy required to extract CO2 from ambient air is about 250 kWh/(ton $\ce{CO_2}$)."  This equals 0.99 MJ/(kg $\ce{CO_2}$).
The OP asks that we assume the $\ce{CO_2}$ came from a coal-fired plant that produces 24 MJ/(kg $\ce{CO_2}$).
Hence (assuming SAPEA's number, and the OP's, are correct), at at the theoretical maximum thermodynamic efficiency, the energetic cost of capturing $\ce{CO_2}$ from ambient air is 0.99/24 x 100 = 4% of the cost of the energy generated in producing that $\ce{CO_2}$ from a coal-fired plant.
Of course, this is not a lifecycle analysis.  One needs not only to determine the actual thermodynamic efficiency of the process, one also needs to account for the energetic costs of putting the infrastructure in place for the carbon capture, and for maintaining that infrastructure.
*For more information on SAPEA, see:
https://en.wikipedia.org/wiki/All_European_Academies


Bromination of an alkene in presence of water

While there is no formal carbocation intermediate, the cyclic bromonium intermediate has a few different resonance forms (reference):

As you can see on the right, in the structure representing the resonance hybrid, partial positive charges build up on the electrophilic carbons connected to the bromine. The carbon that is more substituted with electron donating groups has a greater partial positive charge (and therefore more electrophilic character) because of the stabilizing effect of the substituents. That's why the nucleophile attacks at the more substituted carbon.


Why do Asn-Ser and Gln-Thr have different H bonding patterns?

I don't have the textbook you've gotten this from, but unless there's a statement to the contrary, I suspect these are intended as examples of hydrogen bonding, rather than an exhaustive list of the only hydrogen bonding pattern which can exist.

If you start to look at real structures, you will see both "directions" of hydrogen bonding (hydroxyl donor/carbonyl acceptor and nitrogen donor/hydroxyl acceptor) in real protein structures.

Take, for example, the structure 1GKT. S297 is very clearly in a hydrogen bonding orientation with N303, with the hydrogen of the hydroxyl pointing toward the carboxylate oxygen of the asparagine. Also, Q22 is very likely in a sidechain-sidechain hydrogen bond with T29, where the hydrogen on the glutamate nitrogen is pointing toward the oxygen of the threonine.

If you spend some time looking at various experimental structures from the PDB website, you should be able to find examples of a wide range of hydrogen bonding patterns - basically, any acceptor and any donor can be paired, depending on the geometry of a particular protein.

One caveat I should mention here is that because of the way X-ray crystallography works, it's often difficult to tell which way the sidechains of the ASN and GLN residues are oriented. With electron density, the NH2 group and the =O group look very similar. Often it's a bit of guesswork on the crystallographer's part to tell which orientation they have, often relying on the layout of neighboring residues to determine the most likely hydrogen bonding network.

However, 1GKT was not determined by X-ray crystallography, but by neutron diffraction. In neutron diffraction it's very clear where the protons are located. (I picked a neutron diffraction structure as the example for this very purpose.) I also double checked that there wasn't any reason for flipping those particular ASN/GLN residues, both by manual visualization and with the MolProbity web server.


The dissolution of glucose in an aqueous solution is an endothermic process.  As stated in a comment, a reaction need not be exothermic to be spontaneous. Don't forget about the entropy term.

Simply put, the glucose dissolved endothermically in your saliva causing a heat transfer from your mouth to the glucose/saliva solution, resulting in a slight decrease in the temperature of your mouth, making it cooler than it was prior do dissolving the glucose.


If the molality is $1.44 \ce{m}$, it means that, there is 1.44 moles of methanol in $1 \ce{kg}$ of water.

To prepare this solution, we add the mass of  $1.44\times 32= 46.08\ce{g}$ of methanol to $1000\ce{g}$ of water.

The mass of the final solution is $1000+46.08= 1046.08\ce{g} $.

Here, you have 1 kg of this solution (water + methanol). So, it contains $\frac {46.08 \times 1000}{1046.08}=44.05\ce{g}$ of methanol.

The mass of water in this one kg-solution is $1000-44.05=955.95\ce{g}$

The objective of the problem is to prepare a solution of molality $1 \ce{m}$,

i.e, it is prepared by adding $32\ce{g}$ of methaonl to $1000\ce{g}$ water.

It can also be prepared by adding $44.05\ce{g}$ of methanol to $\frac {44.05 \times 1000}{32}=1376.56
\ce{g}$ of water.

Our solution already contains $955.95\ce{g}$ of water.

So, we need to add only $1376.56-955.95=420.61\ce{g}$ of water.


The "hazardousness" of artificially made sweteeners

From the NHS website in the UK:
The US National Cancer Institute conducted a study of nearly half a million people, comparing those who consumed drinks containing aspartame with those who did not. Results of the study found aspartame did not increase the risk of leukaemia, lymphoma or brain cancer."
Also:
In 2013 the European Food Safety Authority (EFSA) conducted a comprehensive review of the evidence (there is a pdf on the website) and concluded that aspartame was safe for human consumption, including pregnant women and children.
Cancer Research UK also have stated that artificial sweeteners are safe.
Rigorous testing is carried out before a sweetener can be approved for use in the EU.
So, no evidence of cancer in humans. Rats, maybe. Humans no.


electron donation affecting carbonyl bond strength

Let's take the above molecules, acetyl chloride and ethyl acetate, as examples of acyl compounds with electron-withdrawing and -donating groups, respectively.

We know that Cl has a higher electronegativity than O as a halogen, so the electron density from the carbonyl carbon should be pulled toward the chlorine in acetyl chloride and pushed toward the carbonyl oxygen in the case of the ester. This might not be immediately obvious, as the two oxygens should have the same electronegativity, so I like to relate this latter point to electric force laws:
$$ F=\frac{kq^2}{r^2} $$
Since C=O has a shorter bond distance than C-O, the C=O will exert a greater force on surrounding electrons than the C-O (this assumes electrons to be point charges, so it is only useful as a trend).

Therefore, we draw the following picture:

Meaning that there is more electron density on the carbonyl carbon for ethyl acetate than for acetyl chloride.

Now, that's somewhat of a problem for the carbonyl oxygen of ethyl acetate! It already has a lot of negative charge as an oxygen, so it will experience repulsive forces due to the increased electron density on its adjacent carbon. Remember that oxygen does not like having more than a charge of -2 formal charge, which is already satisfied by the double bond it has to the carbonyl carbon; a similar relation can be made for actual charge density.

These repulsive forces, well, repel the oxygen away from the carbon a bit, increasing the C=O bond length. And we also know from physics that increasing the bond length decreases the frequency of vibration. This should make intuitive sense: if the bond distance increases, then the atoms have more distance to travel to complete one cycle of vibration, thus decreasing the number of cycles that can be completed per unit time.

Now onto our example electron-withdrawing group:

Acyl chlorides pull electron density away from the carbonyl carbon, as mentioned earlier. Following the same logic, we can conclude that the C=O bond distance decreases(... to a point*). It follows that the C=O vibrational frequency should increase.

Well, that was a lot of words. But let's approach this from another, much simpler, way. In an acyl chloride, the chlorine acts as an electron-withdrawing group, destabilizing the already electrophilic carbonyl carbon. While this decreases the stability of the molecule, that does not mean that all bond strengths are diminished. In acyl chlorides, the C-Cl bond is the weak, labile bond, not the C=O bond. In fact, the C=O bond strength increases for the reasons outlined above. In our example ethyl acetate, the C-O donates electron density to the electrophilic carbon, thus stabilizing the compound. This does not mean that all bonds are strengthened. If anything, it might indicate a greater C-O bond strength than would otherwise be expected. The C=O bond however,  is a different bond that is being affected inductively, so it must be considered differently.

Personally, I am a fan of the first explanation, as that provides an actual understanding of what's going on, whereas the second explanation leads to a more memorization-focused approach to chemistry. I hope this helps!

*Extra information--Let's consult the following graph (ignore the specific numbers on it):

What this shows is the distance between two atoms and the potential energy of their "bond" (this is one of those areas where it should be noted that bonds are really just constructs to help us make sense of molecules... but this is not important for now). To the right of the minimum, attractive forces (nucleus --> electrons) dominate. To the left, repulsive forces (nucleus <--> nucleus) dominate. As two positive charges get further away from each other, their energy goes to zero (this can similarly be rationalized by simple physics II-type equations). As their distance approaches 0 (where a distance of 0 indicates superposition, or direct overlap), the energy between the two atoms goes to infinity. This is all just to say that there's a limit on how close the atoms can get to each other. You can see that the slope to the left of the minimum is much steeper and increases at a much faster rate than the slope the the right. It follows that there should be less difference in the C=O bond distance as compared with some fair reference molecule such as acetaldehyde than between the ester and that reference. Part of why I'm bothering with this at all is because of how useful this graph is in just about everything I have studied in some shape or form, so it is very useful to familiarize yourself with it.


Safe coating to make a 3D printed respirator mask airtight

You need something flexible and airtight. Could neoprene fabric, like it's used for diving suits do the trick, esp. for prototyping? Otherwise, I thought of injection moulded silicone, as it's used in diving goggles.


Which method can be used to determine the pKa of an acid-base indicator like bromothymol blue?

Yes, HCl would work with a pH meter. If you know the mass of base (indicator) you use, you can as 0.5 equivalents of a strong acid such as HCl and determine the pH. At that point, since the weak base would be half protonated, the pH would be equal to the pKa, based on the Henderson-Hasselbalch equation. Or you could just so the whole titration and find the pKa halfway to the equivalence point.

In this case, you have to make sure you start with the basic form of the indicator, however. Not sure what form it is supplied in.


What does "n" stand for in n-akyldiols, n-alkanes, n-dicholoroalkanes?

The "n" stands for "normal", which in this context means "straight-chain".  It is not italisized, since it is not Latin, but English.  In contrast, the prefixes "i" or "a" for "iso" and "anteiso" are Latin and are properly italisized, but rarely are.

Your second choice, "n is like x", can be found in the subscripts of chemical formulae, where it does denote a variable amount.


You mean, why we write $\ce{4OH- -> O2 + 2H2O + 4e-}$ and not $\ce{4H2O -> O2 + 4H+ + 4e-}$?

Indeed, you may find both these reactions in the table of standard electrode potentials (at $E^0=+0.401$ and $+1.229$, correspondingly; also, note the reversed direction). In some sense, the second equation may even be considered more appropriate, since $\ce{H+}$ is a more abundant species in the solution than $\ce{OH-}$ (zinc salts would produce slightly acidic solutions due to hydrolysis).

But the first reaction was chosen not without reason. See, when we electrolyze something, it is anions that rush to the anode to get oxidized. Water molecules are neutral; electric field would not drag them in either direction. So this might be the reason why the equation is written this way, even if the amount of $\ce{OH-}$ is minuscule and the overall reaction looks more like the second one.


How to calculate the 15 geometrical isomers of an octahedral complex with all ligands different?

Assign "a" to one of the six equivalent positions.
Note that there are now 5 choices (b,c,d,e,f) for the opposing position.
You are left with four equivalent equatorial positions.  Once a ligand is assigned to one of these four, there are 3 ligands remaining to select for the opposing equatorial position.
Finally, there are 2 ways to assign the last two positions. The two possibilities are mirror images of each other.

So $5\times3\times2= 30$

30 isomers (15 pairs of enantiomers).


To first order, the power dissipated in an electrochemical cell is equal to that of an electrical circuit: $P=IV$. Thus, as the overpotential necessary to achieve a given current throughput increases, the power consumption increases also.

In the fluid flow analogy, think of an elevated overpotential as a throttling orifice inserted into one of the pipes.  In order to keep the water moving through the pipe at the same volume flow rate, you have to increase the pressure drop across the system.  A greater pressure drop to achieve the same flow rate requires a pump with a greater power rating.


Polymer Slurry used in boring foundation holes?

After reading the link provided in the question, and doing a little googling, of "polymer slurry" and "foundation", I found something useful.

Polymer slurry seems to be made of one or more hydrophilic superabsorbent polymers, such as polyethylene glycol, polyvinyl alcohol, or carboxymethylcellulose that produce a thick viscous slurry when mixed with water. It does not repel water so much as the opposite. It is so hydrophilic that it absorbs water before the water can reach the concrete or the steel. Then this slurry becomes thicker the more water it absorbs so that the rate of diffusion of water through the slurry decreases. Ultimately, this action creates a hydrogel, which may be quite strong, despite being mostly water, due to extensive hydrogen bond crosslinks.

Other applications of polymer slurry seem to be in the drilling industry to modulate the viscosity of drilling fluids.


Ideal gas and ideal fluid are not the same. Ideal fluid has molecular interaction that is even throughout different entities, and has no viscocity, and no compression.

Yes, ideal fluid is modeled so that it will go through phase change, but you would have to consider the gas from ideal fluid as a pseudo- real gas, since it has molecular interaction.

An ideal gas has no volume, has no interaction, will not go through any sort of phase change.

Hope this helps.


You are completely correct in your dimensional analysis and stoichiometry.

  36.8g is the correct answer.

Molar mass $\ce{Na2CO3 = 106g/mol }$

${25g \rightarrow 25/106 = 0.236}\text{ mol } \ce{Na2CO3}$

This will react with ${0.236*2 = 0.472} \text{ mol } \ce{Na2S}$

Molar mass $\ce{Na2S} = 78.0 \text{ g/mol}$

$0.472 \text{ mol } = 0.472*78.0 = 36.8 \text{ g } \ce{Na2S}$ will be needed.

I will now prove your book wrong

25.9g $\ce{Na2S}$ = 0.332mol $\ce{Na2S}$

0.332mol $\ce{Na2S}$ reacts with 0.166mol $\ce{Na2CO3}$

0.166mol $\ce{Na2CO3}$ = 17.5g of $\ce{Na2CO3}$, which is different from our starting quantity of 25.0g.


Why does a NOx container get cold when NOx is released? and What is the relationship of the vapor pressure curve with the liquid-vapor dome?

The vapor pressure curve describes the relationship between temperature and pressure for which the pure substance can exist as liquid and vapor, together in equilibrium.  So it is not necessarily all liquid or all vapor.  Both can be present in different proportions.  As far as specific volume is concerned, your tables should list both the vapor specific volume and the liquid  specific volume.  The specific volume of the combination of vapor and liquid is proportional to the mass fraction of each.  (So you need to know the mass fractions of vapor and liquid).


Could dimerization account for weak IR signals?

Intensity is (in general) proportional to sample concentration, so yes, if you are performing IR in solution, then an overly dilute sample would have less intense peaks. But solution IR is somewhat uncommon; I suspect you are either doing IR on your sample in neat form (using an ATR stage) or in a pressed KBr plate? If the latter, you may not have used enough sample.

Regarding your missing peak: sometimes peak intensities can change, depending on your sample prep. For example, if the peak in question corresponds to proton capable of hydrogen bonding (e.g., -COO-H, -NR2-H, -O-H, etc.) then the sharpness and intensity of the peak will depend on the extent of H-bonding your sample, which, in turn, can depend on things like sample concentration, the presence of water, etc. Sometimes these peaks get broadened to the point of being invisible if your sample is not very concentrated.

If your missing peak corresponds to a C-H, C-C, C-O, C-N, or some other non-labile bond, then there really isn't an explanation other than your compound not being what you think it is...


Calculate pressure equilibrium constant from concentration equilibrium constant

If you intend to get an answer in Pascal, which is an SI unit, you need to have your original values in SI units as well.

Note that:

$$K_\mathrm c=\pu{4.2M^-2}=4.2~\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}=4.2~\left(\frac{\text{moles}}{\pu{10^-3m^3}}\right)^{-2}$$

The conversion factor for litre (non-SI) to metre cubed (SI unit) is $10^{-3}$. Since you have $K_\mathrm c$ in units of $\pu{M^-2}$, the final answer would be $10^{-6}$ of the answer you had arrived at.

Elaboration:

Let's do dimensional analysis on the RHS factor-by-factor.

$K_\mathrm c$ has the units $\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}$ as shown above.
$RT$ has the units = $\pu{Jmol^-1T^-1}\cdot\pu{T}=\pu{J\cdot mol^-1}$
$(RT)^{-2}$ has the units = $\pu{J^-2\cdot mol^2}$.
Since one joule $=\pu{Pa\cdot m^3}$, hence $(RT)^{-2}$ has the units = $\pu{Pa^-2\cdot m^-6\cdot mol^2}$
Multiply $K_\mathrm c$ and $(RT)^{-2}$ we get
$$\text{Units of }K_\mathrm p=\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}\cdot\pu{Pa^-2\cdot m^-6\cdot mol^2}=\pu{Pa^-2\cdot m^-6\cdot litre^2}$$

Thus your final answer is not in the units of $\pu{Pa^-2}$ but rather $\pu{Pa^-2\cdot m^-6\cdot litre^2}$. Can you now see why we need the conversion factor?


Does an electrically conductive gas exist at room conditions?

It depends on what you are prepared to consider a gas and what you are prepared to consider room conditions.

The gas inside all discharge lamps (fluorescent lamps and neon lamps in shop signs, for example) conducts electricity and the lamps work under normal room conditions. However, the gases are often at low pressure and are, strictly speaking, plasmas not room temperature neutral gases. Some lamps, especially high intensity ones like the xenon discharge lamps in floodlights) exist with high pressure gas when working so low pressure isn't the key.

And air itself will conduct electricity given the right conditions. Lightning couldn't exist if it didn't. Nor could you get the (much safer) discharges from Tesla Coils or Van de Graaff generators. But, again, the effect relies on the creation of a plasma from the breakdown of the neutral molecules in the gas.

So the answer is "yes" but only if you allow plasmas created from the medium to count as a normal gas.


Restricted open-shell versus unrestricted methods for open shell systems

To be more precise, it's not so much that restricted open-shell calculations are unaffected by spin-contamination as that you are forcing them to give back the multiplicity you want.

Broadly speaking, closed-shell methods, restricted open-shell methods and constrained unrestricted open-shell methods all give correct spin multiplicities, but that's because all of them assume from the beginning which is the correct multiplicity and force it into the calculation - there is always a restriction in place. For closed-shell methods, it is that the orbitals occupied by both spin populations are the same; for restricted open-shell methods, the same applies to all but the excess spin electrons from the larger spin population. But more recent developments point to restricting multiplicity after the fact - by adjusting orbital population to minimise spin contamination. There are a couple of methods to do that, but they start from an unrestricted calculation (much faster, as you mention), and then force the population in a way that minimises spin contamination - either by forcing orthogonality constraints between the $\alpha$ and $\beta$ determinants, or by parametrising spin contamination in a way that allows it to be fitted to the expected total multiplicity by constraining occupation. This retains the computational advantage and simpler mathematical implementation of unrestricted methods but makes sure that the multiplicity of the model system reflects the multiplicity of the modelled system with generally very small additional computational effort.

I'd suggest that this is going to be an active area of research that will see increased use in the next few years.


This is a little tricky. I would argue that only the first proton dissociates completely and should be considered "strong" - using the rules for writing net ionic equations, where we only dissociate strong electrolytes, this would meant that we would write:

$\ce{H2SO4 + 2KOH -> 2H2O + K2SO4}$

$\ce{H+ + HSO4- + 2K+ + 2OH- -> 2H2O + 2K+ + SO_4^{2-}}$ - here only the first proton dissociates completely

Canceling out spectator ions gives:

$\ce{H+ + HSO4- + 2OH- ->2H2O + SO_4^{2-}}$

Your teacher (or whoever is giving you that answer) is dissociating the second proton as well, to give:

$\ce{2H+ + SO_4^{2-} + 2OH- ->2H2O + SO_4^{2-}}$

Which becomes:

$\ce{H+ + OH- ->H2O}$

There are a few reasons why they might do this, but I would argue that the 1st approach (which only dissociates the first proton) is best because it more accurately reflects what is happening, in the sense that $\ce{HSO4-}$ is a weak electrolyte. This will become important later when you study acid-base equilibria.

In general, you dissociate protons from acids as if they were regular metal cations - just "pull them off" and don't forget the positive charge. For polyprotic acids, the first proton is strongly acidic, or dissociates completely, but the other protons are not. There might be exceptions to this rule but I can't think of any off the top of my head.


The radical monobromination of bicyclo[2.2.1] heptane could give $\ce{1}$ ,$\ce{2}$ and $\ce{3}$.

However , $\ce{2}$ and $\ce{3}$ , are unstable .On observing $\ce{2}$ , planar radical at bridgehead is unstable due to angular strain introduced by planarity of radical at bridgehead .

Similarly in $\ce{3}$ , due to angular strain introduced in to the neighboring bonds,this  radical also is unstable.

The situation leads to product from ${1}$ , that gives  stereoisomers as shown below.


IUPAC name of trisubstituted benzene ring

As -CN has the highest priority, root-word is benzonitrile. Lowest locant is 2 for -Br.
So, the name given is:
2-bromo-5-hydroxybenzonitrile

As the functional group is -CN, with highest priority, no need to look for priority of -OH and -Br; just look for lowest locant.


How do I arrive at correct furanose structures when starting from a Fischer projection?

A Fischer structure is one way of depicting structures such as sugars or amino acids to quickly draw them and have a definite way of seeing stereochemistry on every carbon. The rule behind Fischer structures is that the highest oxidised atom should be at the top or as close as possible to the top. The following convention is used to turn tetrahedral carbons into a square environment:

This allows you to assign absolute configurations to every carbon atom.

When instead drawing the same structure as a ring like in your lower images, you just need to take care that every carbon has the same configuration. (Note that your drawing of β-D-fructose is wrong; it should be drawn with a very wavy line like in my image below, because of the constraints of Fischer projections.)

Once you have gotten there, you are free to rotate the ring around as much as you wish while retaining the absolute configuration at every stereocentre. Note especially, that a rotation of $180^\circ$ of my structure of β-D-fructose will lead to you one on the left. If you don’t believe me, print it out, turn the paper around and then nudge the forwards-pointing part of the ring downwards below the tetrahydrofuryl-oxygen.


Naming compound containing a ring (cycloalkane) side chain?

In view of the release date of your book (2005), the used names for organic compounds are probably in accordance with the 1993 Guide to IUPAC Nomenclature of Organic Compounds. However, note that the current version of the recommendations can be found in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

According to the 2013 recommendations, two methods are possible for naming systems composed of rings and chains: either the ring or the chain can have seniority. To construct the preferred IUPAC name (PIN), the ring is always selected as the parent structure. Therefore, the PIN for the compound given in the question is (6-methylheptan-3-yl)cyclopentane since the cyclopentane ring has seniority over the heptane chain.

In general nomenclature, however, the context may favour the chain, so that the parent structure has the greater number of atoms in the principal chain. The corresponding systematic name for the compound given in the question is 5-cyclopentyl-2-methylheptane.

According to the 1979 recommendations, a hydrocarbon containing a small ring attached to a long chain was generally named as a derivative of the acyclic hydrocarbon. The 1993 recommendations do not explicitly mention this rule but refer to the 1979 recommendations. Hence, it is not surprising that your book still uses such a name, although the 1998 Guide to IUPAC Recommendations already explains that the ring is always selected to be the parent structure.

Now coming to your actual question:

According to the 2013 recommendations, the prefixes ‘cyclopentyl’ and ‘methyl’ in the name 5-cyclopentyl-2-methylheptane are arranged alphabetically.

  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, ‘cyclopentyl’ begins with ‘c’ and ‘methyl’ begins with ‘m’. Note that, for example, ‘2,3-dimethyl’ is considered to begin with ‘m’ since the multiplicative prefix ‘di’ does not alter the alphabetical order. Whereas, for example, the compound substituent name ‘(1,2-difluorobutyl)’ in 7-(1,2-difluorobutyl)-5-ethyltridecane is considered to begin with ‘d’.


How is o-Fluorophenol more acidic than phenol even after having Hydrogen bonding?

Perhaps the words in the picture are explained at greater length in the text, or perhaps actual numbers are given for the dissociation constants (https://www.quora.com/Why-is-o-flurophenol-is-more-acidic-than-p-flurophenol).
As it turns out, the dissociation constants are 8.7, 9.3. 9.9 and 10.0 for ortho, meta, para and unsubstituted phenol. Ortho-fluorophenol is the most acidic - perhaps in spite of the possible hydrogen bonding. The inductive effect of the fluoro group is surely greatest at the ortho position.
In the series of nitrophenols (Acidity order of nitrophenols), the para (p$K_a$ = 7.16) is slightly more acidic than the ortho (p$K_a$ = 7.2), so it could be said that the ortho substituted compound, which would be expected to be more acidic (if you assume that the resonance effects at ortho and para positions would be similar), but isn't, we have to look for some other explanation, and hydrogen bonding could stabilize the neutral molecule.
Well, even if the actual effect is stronger than 0.04 p$K_a$ units, the hydrogen bonding effect seems to be rather small, and is overcome in the fluorophenols by the $-I$ effect of fluorine. Just scratch out the words "H-bond decreases acidity". (Lightly.)


Proper explanation for depression of freezing point

A simple "intuitive" oversimplification: intrusive foreign molecules get caught between solvent molecules, getting in the way of crystallization. That, however, would contradict the similar boiling point elevation.

Another oversimplification: the solute molecules pull the solvent away from the incipient crystals.

Yet another simplistic answer: osmosis exerts a force on the solvent, pulling it back into solution, away from the pure crystal of solvent.

Perhaps the last explanation is closest to a statistical analysis of freezing point depression.

Hopefully, chemistry will not be a depressing subject.


Chemistry needs a devoting reader. You should try to love chemistry as much as you want! That'll be the ignition for lots of reading and stuff. My pointers:

Love chemistry.
Learn in an organized matter. (For example, do not jump from studying about covalent bonds' basics to reaction mechanism.
Take notes of the most important points of what you learn.
Look for open-source online tutorials and not only the books.
Some chemistry concepts are related to each other. Learn the first one's basics and then read the other one's. This will build up your understanding, literally. For example, first learn about electronegativity and then study the covalent bonds' basics. This is the exception for rule 2.
Try to differentiate between what's taught in schools to be true and what really is. For example, look for exceptions of the periodic trends that are taught in schools.
Be imaginatively practical. Don't just say enough for the lewis structure of water, but do it about $\ce{SF6}$.
Always try to go relatively higher than what your textbook has written. We have a proverb in my mother language meaning "If you have 100, you do have 90."
Ask what misunderstandings you have in here!
Never look to doing a research as a waste of time.
When approaching to solve a chemical problem, do it step by step.
When approaching to solve a problem, don't forget to write units of measurement.
When you reached an answer to the problem, ask yourself: "Is this possible?" If you've calculated 5 grams of water to be $\frac{18}{5}$ mol, and you know that a mole of water is 18 grams, shouldn't something be wrong here?
For any definitions, uses, notions, and practical experiments you encounter, ask yourself: "What is this for, either in real life, experimental sciences or theoretical chemistry?"
Look for online exercises, free of charge, but usually important.
Follow the latest news that are being discussed everywhere in the world of chemistry.
For specific concepts in chemistry, e.g.: Periodic table, doing a research about the history is very helpful.

This is as far as I can count. You can ask "why"s in comments.


Does the enthalpy of formation include translational and rotational energy?

To add to @Kexanone's answer above, the enthalpy of formation refers to how much heat is evolved or adsorbed from the environment in a reaction. So it's the total energy, and in an ideal gas it has the form: dft + translational + rotational + vibrational + PdV, where the PdV refers to the energy needed to make space for the gas (grand canonical ensemble; in the canonical ensemble (fixed volume) this does not show up). At room temperature the vibrational and PdV terms are typically small compared to DFT energy errors (order 1 millihartree), with the PdV term only coming into play at pressures within an order of magnitude or two of those found in the detonation of thermonuclear bombs.


What exactly causes equilibrium to shift when pressure is decreased?

This happens because the equilibrium constant $K_p$ is independent of the pressure, although it does depend on temperature, which we will assume to be constant. The equilibrium constant can be written as

$$\displaystyle K_p =\frac{p_A^2}{p_Np_H^3}$$

where for simplicity I have abbreviated the chemical names. The partial pressure $p$ can be written in terms of the degree of dissociation and total pressure $P_T$, i.e. $p=aP_T$ where $a$ is a mole fraction term in the degree of dissociation. As partial pressure is proportional to the degree of dissociation and total pressure, for example, $p_A =a_A^2  P_T^2$ then the equilibrium constant

$$K_p=\frac{a_A^2}{a_Na_H^3} \frac{1}{P_T^2}$$

where the terms in $a$ we have not yet evaluated and contain the degree of dissociation $\alpha$. However, you can see that as $K_p$ is constant if $P_T$ decreases then $1/P_T^2$ increases and so  $\displaystyle \frac{a_A^2}{a_Na_H^3} $ must decrease. This means that the degree of dissociation decreases and so the equilibrium 'moves to the left'.

Working out the partial pressure for this reaction is a bit messy.

Let $\alpha $ be the degree of dissociation then $(1-\alpha)$ of nitrogen reacts , $3(1-\alpha)$ hydrogen and $2\alpha$ ammonia is produced. The total of these is $6-2\alpha$ and so the mole fraction of ammonia is, for example, $a_A=\displaystyle \frac{2\alpha}{6-2\alpha}$ and the partial pressure $p_A=\displaystyle \frac{2\alpha}{6-2\alpha}P_T$. You can make a similar expression for the other species and substitute to get the mole fraction term $\displaystyle \frac{a_A^2}{a_Na_H^3}=\frac{16\alpha^2(\alpha-3)^2}{27(\alpha-1)^4}$. This expression increases as $\alpha$ increases. As you see the expression is messy but the main point is that as the pressure decreases the mole fraction term must decrease which it does by reducing $\alpha$.


Shifting of axial and equatorial protons

The distinction between axis and equatorial hydrogens is not going to be visible in an NMR spectrum for cyclohexanol. The conformations of the molecule interconvert too quickly at room temperature to make a distinction in the proton NMR spectrum.

So, in practice, you will only see 4 distinct C-H bond types in the molecule and the splits that will be visible on your spectrum depend on the resolution (which is not particularly high on the spectrum you have). The splits cause by neighbouring hydrogens for the groups between shifts of 1 and 2 overlap even in high resolution spectra and it isn't clear whether what you are seeing this jumble of peaks as 4 separate environments because of the low resolution or whether one of the peaks is an impurity.

Some software packages for structure drawing also generate theoretical NMR plots which can help interpret such spectra.


You do not detect equivalence points. What you detect is the endpoint. This is a very critical distinction.

You have some feature you are explicitly measuring, be it pH or color or opacity.

Once this feature changes you have reached the endpoint. The endpoint is not the equivalence point because you will never have exactly the right amount. Frequently, you overshoot a bit.

Also, the reason that the endpoint is detectable is because the transition is very sharp and sudden. That is because near the equivalence point, the pH (and whatever pH derived feature you have) changes very rapidly. The derivative is definitely not zero. In fact, the derivative you've provided is never zero but monotonically increasing or decreasing depending on whether you're tritrating with acid or base. It has to. If you add more base, the pH is only going to increase in value.


Determining actual safety from a vague material safety data sheet

General common sense should always remind you not to drink the liquid stick your finger in it, rub it all over your skin or breathe its vapours for extended periods of time. Paracelsus is best known in Germany for this quote:

  All things are poison and nothing is without poison; only the dose allows that a thing is not a poison.

If you are now scared out of your wits, then please relax and consider that preface a general disclaimer. Because all the toxicologic data I can see points towards the liquid being entirely harmless. Here’s why:

  Acute Toxicity: LD 50/oral/rat: >20000 mg/kg

That’s more than $20~\mathrm{g/kg}$ — an incredible number for $\mathrm{LD_{50}}$ values. Usually, the highest $\mathrm{LD_{50}}$ values I see are $2~\mathrm{g/kg}$. And the greater than basically means ‘we couldn’t kill the rats with that oral dose so we gave up.’

  Low oral toxicity proprietary antifreeze formulation based on refined vegetable extracts and non-toxic corrosion and scale inhibitors. Classified as GRAS (Generally Regarded as Safe) as a food additive and not classified as harmful for handling or transport purposes. However we recommend good working practices are employed and personal protection gear is worn when handling any chemical.

(Emphases mine) As you can see, there are numerous references to ‘nope, this thing really won’t harm you.’ The GRAS classification actually seems to be concerning food additives. No danger here.

  Respiratory Protection: Respiratory protection if there is a risk of exposure to high vapour concentrations.

I’ll admit, this sounds a tad unsafe at first glance. But if you know what MSDS write for truly unsafe compounds, you’ll realise that this is saying ‘If we write “safe”, somebody will stick their nose in, suffocate and then sue us, so we have to put some kind of caveat here.’

The only thing that actually hints at a potentially harmful substance is the thing about high vapour concentrations. But remember that this is a cooling liquid; you don’t want a cooling liquid evaporating away, that would reduce the cooling efficacy. The minute amounts you are losing will dilute well in the air in your room. Open the window once a day or every two days and you’ll have removed anything that can potentially harm you. The liquid is as safe as it gets.

However, don’t sue me if you decide to turn off common sense and start drinking the liquid or stop venting your room or whatnot.


Why the products of combustion of Cu2S are Cu2O and SO2?

The reaction gives different products when combusted at different temperature.

$$\ce{8Cu2S + 15O2 ->[\Delta] 6Cu2O + 4CuSO4 + 4SO2}$$

Copper(I) sulfide react with oxygen to produce copper(I) oxide, copper sulfate and sulfur dioxide. This reaction takes place at a temperature of 500-600°C.Here

$$\ce{2Cu2S + 3O2 ->[\Delta] 2Cu2O + 2SO2}$$

Copper(I) sulfide react with oxygen to produce copper(I) oxide and sulfur dioxide. This reaction takes place at a temperature of 1200-1300°C.Here

Sulfur monoxide is very unstable and it is unlikely to exist in Earth's atmosphere but it exist in interstellar spaces. Sulfur trioxide will not form from sulfur dioxide as it requires an oxidising agent.


For many common purposes we say there's weak acids and strong acids, and define strong acids to be ones that dissociate completely in water. This is because if the acid dissociates completely in water, then every proton dissociated will just protonate water, and turn it into $\ce{H_3O^+}$ (hydronium ion). The hydronium ion is the strongest acid that can exist in water as far as I understand, and hence why strong acids are simply strong acids in some contexts.

What you're describing is a super acid, that is not used in aqueous solution Take HCl for example. This is a classic strong acid, that will protonate a wide range of structures. But if we take something that is a VERY, VERY poor base (something that really does not want a proton), such as an alkane, HCl can't protonate it. The proton will rather stay with chlorine than protonate the alkane. That is one of the things you can do with the acid you mentioned.

The strength of an acid depends on many different factors, but it all comes down to how strongly the $\ce{H^+}$ is bound to the acid structure. If it is very loosely bound, the structure is high in energy, and the proton would like to find a more stable structure to partake in. In this case protonating an alkane to form a carbocation and molecular hydrogen is low enough in energy, that the acid can protonate it.

But remember that protonation is an equilibrium. If we achieve a vastly lower energy state by having the acid protonate another structure, then the protonation will be complete. If it is the same energy, half of the acid will have protonated (assuming equal concentrations of acid and base).


Is this the NMR spectrum of 3-methyl-1-butanol?

Welcome to ChemistrySE.
In short: is the spectrum consistent with 3-methyl-1-butanol? No. The triplet splitting and integral of the signal at 0.86ppm is inconsistent with an isopropyl sidechain. This suggests a methyl group attached to a methylene, $\ce{-CH2-CH3}$.
The signal at $\pu{3.6 ppm}$ has a chemical shift symptomatic of an alcohol, as you suggest. It has diagnostic splitting of $\pu{4.2 Hz}$, also consistent with coupling to the alcohol peak. So, this peak is a doublet of triplets..... which makes it look a bit like a quartet. To see this coupling, the sample needs to very dry to eliminate exchange decoupling. So where is the alcohol peak? In the absence of any exchange, and in chloroform (most likely solvent used given the spectrum shown), alcohol peaks usually come in the region of about $1.2$-$\pu{1.5 ppm}$. So possibly under the multiplet. Which would then leave you with 6 protons to account for......or $3 \times \ce{CH2}$ groups. The peak at $\pu{1.5 ppm}$ has coupling that suggests it is attached to the $\ce{-CH2OH}$, giving a $\ce{-CH2CH2OH}$ group. There are 14 protons in all. You've got a $\ce{-CH2-CH3}$, a $\ce{-CH2CH2OH}$ and the rest are probably $\ce{-CH2\!-}$. How might you put them together?
In the real world, identification of complete unknowns are rarely done using $\mathrm{^1H \ NMR}$ alone. Mass Spec, $\mathrm{^{13}C}$, and a range of $\mathrm{2D}$ options are very useful.
And to answer a follow-up question below, the splitting at 1.54 is from the protons on the beta carbon, and is a triplet of triplets. When coupling constants are close (6.5 and 6.8) this will appear as a pseudo-pentet, as shown below:


If I've understood your question properly, you are asking - in the condition $E^\circ_{\ce{M^3+/M^2+}} =\pu{4 V}$, $E^\circ_{\ce{M^3+/M}} =\pu{1 V}$, $E^\circ_{\ce{Y+/Y}} =\pu{3 V}$ - whether $\ce{M^3+}$ is a stronger oxidising agent than $\ce{Y+}$ irrespective of it going to $\ce{M \text{or} M^2+}$.
This would not be true since $\ce{Y+}$ has a higher reducing potential than $\ce{M^3+/M}$ but not $\ce{M^3+/M^2+}$, and so $\ce{M^3+}$ would be a stronger oxidising agent than $\ce{Y+}$ where it converts into $\ce{M^2+}$ but not when $\ce{M^3+}$ converts into $\ce{M}$.

The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$
\begin{array} {rlllc} \require{cancel}
\ce{N^y+ + ye- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x)\\
\ce{ M^x+ +xe- &-> M}  &E^\circ = E_m^\circ &\quad|\times(y)  \\  \hline
\ce{yM^x+ + xN  +\cancel{(xye-)}  &-> yM + xN^y+ + \cancel{(xye-)}} &E^\circ = E_{\text{cell}}^\circ  \\
\end{array}
So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$
Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.
For just one mole of M being reduced , we get
$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$
For just one mole of N, we get
$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$
Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to
\begin{align}
\Delta G_{\text{cell}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\
&= xyF(E^\circ_m -E^\circ_n) \tag{3}
\end{align}
Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:
$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$
Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get
\begin{align}
\cancel{(xyF)}E^\circ_\text{cell} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\
\implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n
\end{align}
For a case where you have the same compound disproportionate, this question and its answers should be enough.


How many valence electrons does the azide ion have?

Nitrogen (the element) has 7 electrons. But two of those electrons are in the $1s$ orbital, and so are not considered valence.  Therefore, a single elemental nitrogen has five valence electrons.

There are three nitrogens, and so $5 \times 3 = 15$ electrons for $\ce{N_3}$.

Finally, azide is $\ce{N_3^{-}}$, there is an additional electron beyond the neutral form.  So, the total number of valence electrons is $15 + 1 = 16$.


What is the conversion rate in a chemical reaction?

The paper you linked to pertains to birth-death processes. An example of a birth-death process in the physical chemistry context that you're interested in  would be a set of molecules where each molecule is in one of two states, $X$ or $Y$, which are isomers of each other. The model assumes that due to the molecules' environment, each isomer can turn into the other isomer at times that are modeled as being random.

In the paper you provided a link to, $w_1$ is the probability per unit time that within a short period of time $\Delta t \ll 1/w_1$, a given molecule that's in state $X$ will transition to state $Y$. I.e., if a given molecule is in state $X$ at time $t$, then there is a probability of $w_1 \Delta t$ that the molecule will have randomly transitioned to state $Y$ by time $t+\Delta t$.


At Last! Success!
I found this research paper, which describes a "Self-assembling nanocapsule", which is held together by plenty of hydrogen bonding, and can assemble around a linear molecule of all-anti tetradecane, $\ce{(C14H20)}$ using the energy afforded by the hydrogen bonding to squeeze/twist the alkane into a helical all-gauche conformation. Have a look:

Cool, right?
Sources:
Scarso A, Trembleau L, Rebek J Jr. Helical folding of alkanes in a
self-assembled, cylindrical capsule. J Am Chem Soc. 2004 Oct 20;126(41):13512-8.
PubMed PMID: 15479107.
Note:
Other answers are still greatly appreciated!


Why 4‐methylidenespiro[2.4]heptane cannot be converted to 1‐chlorobicyclo[3.2.0]heptane?

Here the concept of angular strain comes into the picture. The ideal tetrahedral and is around 109° 29'. As the angle between the C−C bond increases, it is more away from the ideal $\mathrm{sp^3}$ hybridization and hence will become more unstable.

For a cyclohexane molecule, the angle between the C−C bonds is 120°, therefore the deviation from the ideal angle is 10° 31', whereas for a cyclopentane the angle between C−C bonds is 108°, therefore the deviation from the ideal tetrahedral angle is around 1.5°. This is a very low deviation.

Since the transition is from cyclopentane which has a relatively lower deviation from the ideal angle to cyclohexane which has a relatively higher deviation from the ideal angle, the above transition does not take place.

Plus the number of carbons are not balanced in your reaction.


What is more acidic: D3O+ in D2O or H3O+ in H2O and why?

For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined.

The entire point of the above reference is that

$\ce{H3O+ + H2O <=> H2O + H3O+}$

(which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same.

$\ce{D3O+ + D2O <=> D2O + D3O+}$

would also correspond to an equilibrium constant of 1

So when Clayden and the OP write

  $\ce{D3O+}$ in $\ce{D2O}$ is stronger than $\ce{H3O+}$ in $\ce{H2O}$

it is wrong for the above reason.

Two genuine thermodynamic equilibriums are

$\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$

Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).

But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$.  $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.

As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species.  See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.

Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles.  According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$."  My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.

Also the equilibrium constant for

$\ce{H2O + H2DO+ <=> HDO + H3O+}$

can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751

Explanation of Normal/Inverse Solvent Isotope Effect

For a kinetic normal/inverse solvent isotope effect
there will be a reactant and transition state.  If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect".  Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.

Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.


How to reproduce the Gaussian 09 SDD basis set in Gamess-US?

SDD is a basis set description, which uses Stuttgart/Dresden ECPs. They are readily available for download at the BSE, but also via Univ. Köln. And they are pretty easy to pick out. See the Pseudo keyword for what is Gaussian 16 is using currently, which should be the same for 09. At least in the BSE there hasn't been an update to the S/D ECPs since their initial upload. See the related notes included at BSE and pick the appropriate size. The more problematic part is finding the BS for the orbitals that are actually described. I believe this is the correct choice, i.e. Stuttgart 1997 RSC. Please review the literature specified on the Gaussian homepage about the implementation of SDD (should be about 20 to 30 papers), and then you can cross-reference them to the references saved in the BSE library.

The work is from the earlier days of computational chemistry, so it is quite fragmented. I believe the ECPs are still popular and used in the correlation consistent Dunning basis sets. The orbital parts I wouldn't be too sure about. There are plenty of other choices out there and it would be wise to review the literature to see what is the state of the art for strontium.

Why not use the move away from Gaussian also as a move to a less fragmented, different level of theory? If you want to compare the data to previous calculations, you have to recalculate them anyway, why not choose a method which is a bit more consistent than the SDD basis. And depending on what you are calculating, you might also want to look at different software suits, e.g. Orca, Psi 4, Turbomole. Gamess certainly has its strengths, basis set implementation and handling is unfortunately not among those.


Relationship between the ideal gas constant and entropy

The gas constant is equal to Avogadro's constant times Boltzmann's constant, the latter serving as a proportionality constant between the average thermal (kinetic) energy of the particles in an ideal gas and the temperature:

$$\left(\frac{\partial \bar U}{\partial T}\right)_p=\frac{3}{2}k_\mathrm{B}$$

The entropy can be regarded as a proportionality constant between the change in free energy $G$ with change in temperature of a system at constant pressure, since

$$\left(\frac{\partial G}{\partial T}\right)_p=-S$$

Note also that the average entropy of a particle can be written as

$$S=k_\mathrm{B} \log\Omega$$

So both $R$ and $S$ can be regarded as proportionality constants between energy and temperature.

The Wikipedia page on the equipartition theorem may provide some enlightenment on the origin of these proportionalities.


White granular substance from restaurant with interesting properties - what is it?

You kept some of the unknwon as evidence, didn't you?  And a witness, too? Because if you get harmed, document it, and seek both medical as well legal counsel.

Turning to the chemistry: Likely, while dissolving X in water, you felt some heat, too?  Was the colorless solution obtained, when diluted further with additional water and then applied on skin a bit "soap like", felt slippery?

Vinegar is acidic, because of the acetic acid present.  Hence, mixing / dissolving the unknown proceeds with evolution of heat is indicative for the unknown acting as base that is neutralized. Such reactions generate heat. Similar token with lemon juice, equally reacting acidic, because of the presence of citric acid (even a more acidic than acetic acid): heat evolution possibly because of the neutralization of X (as a base) with it.

An additional indication that X is something basic were the action on diluted neutral juice of red cabbage.  If basic, the addition of X would yield a blue/greenish stain, in contrast to acids turning it red.

Speculation: it could be remaining NaOH.

Because sodium hydroxide actually (NaOH) is a base, that dissolves with ease in water, it reacts with the emanation of heat while dissolved in acids.  (Even if I am surprised the lemon juice experiment heated up enough to shatter the container.)  Even while preparing a dilute solution of NaOH in water, say 1 M as commonly used in the chem lab, the heat generated is sensible across the glass wall of the container used.
Because sodium hydroxide actually is used as a cleaning agent, to remove oily residues, to remove fat and grease (in other instances, to prepare lye soap) from pans (provided they are not made of aluminum).
Because NaOH is hygroscopic, in other words, the little white grains of it quickly look wet and stick to each other when exposed to humid air.
Because in pure form, sodium hydroxide is corrosive; and aqueous solutions of it are hazardous, too (pointing to your bleeding sensation).

Thankfully, such accidents in restaurants occure not often; yet there are examples ref1, ref2.  Bars ref3 use caustic soda to rinse their pipes of beer & other beverages, too, connecting the counter with the stock, often out of sight, in the basement.


From the reaction

$$\ce{2A -> B}$$

we have $a = 2$, which leads to the differential equation:

$$\frac{d[A]}{dt} = - k [A]^2$$

which then leads to the integrated rate law you've cited.  The $2$ is nowhere as a coefficient, it is related to the order of the equation, and determines which integral/form should be used.  Similar conventions also apply for 1st and zeroth order kinetics.  I disagree with your instructor.

It should be stressed that these rate laws apply to elementary reactions, not complex reactions.  Experimental measurement may differ from that predicted using an elementary approach, which then leads to the conclusion there are other steps involved.  In the case of OP's comment regarding oxidation of carbon monoxide with NO2, elementary reactions would predict a rate proportional to $$k [\ce{NO2}][\ce{CO}]$$ if the reaction is the result of a simple bimolecular collision. The fact that a different rate is observed indicates this process is more elaborate.


How can I calculate the equilibrium constant for this reaction?

You might want to check your units.

The energy should be in J/mol, according to the values you put in, and not in kJ/mol.

$e^{-22200/(8.314*298)} = 0.000128397 \approx 1.3*10^{-4}$


How does electrolytic conduction depends on the size of ions & their solvation?

Small and/or highly charged ions tend to be highly solvated in water. These solvated ions may still be smaller than the solvated ions of larger ions.
However, water molecules are not as free as the molecules of a gas. Water has a much higher viscosity, for instance. So think of water as a loose gel. A large (unsolvated, or poorly solvated) ion has an easier time slipping between gaps in this gel. A smaller, more highly solvated ion is, by definition, more tightly bound to the gel network, and moves more sluggishly, as if it has to carry more of the network with it. This approach is more explanatory than predictive (see figure).

There are lots of numbers, some agree, some disagree with any simple explanation. But a clean example is sodium vs potassium: sodium bare ion is smaller than potassium bare ion, but their hydrated ions reverse that (hydrated potassium ion is smaller than hydrated sodium ion - and check out lithium!). So potassium has a greater conductivity, for the same number of ions.


How is this Steampunk patina being applied?

I haven't found much in literature about this process, we can try to find the solution reasoning, but some experiments should helpful.

What is being deposited?

If he connects the positive terminal to the object it means that the object is the anode. So something is reduced on the anode.  Copper(II) oxide quite common oxide used for balckening. How can we obtain this by reduction? I would say that it could be the reduction of copper acetate to copper(II) oxide.

So where the copper acetate come from?
In fact, to the cathode there is also a copper wire wrapped around the steel, it could be that the copper ions are produced from that wire on the cathode:

$$\ce{Cu + 2H^{+} -> Cu++ + H2 }$$

Vinegard contains acetic acid and reacts with copper:
$$\ce{2CH3COO- +Cu++ -> Cu(CH3COO- )2}$$

Or directly from the acid attack of acetic acid:
$$\ce{2CH3COO- + 2H+ + Cu ->Cu(CH3COO- )2 +H2}$$

What could happen is the following (decribed here):
$$\ce{Cu(CH3COO- )2 +H2 -> (CH3COO- )2CuH2}$$
$$\ce{(CH3COO- )2CuH2 + Cu(CH3COO- )2 +H2O -> Cu2O +4CH3COOH}$$

Is the deposit related directly to the cathode material?

If this is true there is no relationship with the steel used at the cathode.

What happens if I change the cathode to say normal steel /copper wire?

The cathode may oxidize. If the copper wrapped around the steel has a part on the process, the process could not work if you don't use copper.

Do we not understand the role vinegar and some salt plays?

Salt is probably used to increase the conductivity of the solution.
The vinegard could be used for different reasons, but as  I said before it is probably used for producing copper acetate.

Why is this simple process open to debate and conjecture?

This is an excerpt from an article written at the New Zeland Institute of chemistry:

  Electroplating is both an art and a science. Although based on several
  technologies and sciences, including chemistry, physics, chemical and
  electrical engineering, metallurgy, and perhaps others, it retains in
  some ways the aspects of an art, in which experience is the only
  teacher. In fact, of course, all the sciences have elements of art
  which can be learned only by experience; all the reading of textbooks
  on chemistry will not produce a chemist. No text on electroplating
  will produce an expert electroplater; there is no substitute for
  experience and what is somewhat inelegantly termed know-how.

Is not easy to predict the role of each component if the experiment is poorly described, there are many variables.


Why does Sulfur dioxide incompletely freeze when entrained in a helium stream inside tubing dipped in liquid nitrogen?

Phase diagrams are about thermodynamics, while you have to consider kinetics as well.
The stream may have too high linear speed at 30+ mL/min to reach equilibrium while still in the loop.
Another  option is dragging  particles of/containing SO2 by the gas stream by mechanical way, not allowing it to settle. Like wind taking snowflakes, not allowing them to land and stay.


Thermodynamics of chemical reactions "at constant pressure"

You have correctly stated the pressure of the system is $$p = \frac {(n_1 + n_2)RT}{V}$$
But at the end of your question you incorrectly imply the volume is constant and the pressure changes. That is not true. Constant pressure means the external constant pressure ( like the atmospheric one ) keeps the system pressure constant. Imagine there is a massless, frictionless piston, ensuring $p_\mathrm{ext} = p_\mathrm{sys}$. It is similar as a thermostatic water bath keeps the system temperature constant.
So less confusing is to rewrite the ideal gas state equation :
$$V = \frac {(n_1 + n_2)RT}{p}$$
There are 4 variables: $p, V, T, n_\mathrm{tot}=n_1 + n_2$.
But there are just 3 degress of freedom = 3 independent variables.
Either $V, n_\mathrm{tot}, T$ are given, and $p=f(V, n_\mathrm{tot},T)$,
either $p, V, T$  are given, and $n_\mathrm{tot}=f(p,V,T)$,
either $p, n_\mathrm{tot}, T$  are given, and $V=f(p,n_\mathrm{tot},T)$,
either $p, n_\mathrm{tot}, V$  are given, and $T=f(p,n_\mathrm{tot},V)$.
We have already given 2 variables: $p, T$, with $n_\mathrm{tot},V$ remaining free.
But there is the equilibrium relation
$$K_p = \frac {p_{\ce{N2O4}}} {({p_{\ce{NO2}} )}^2}=\frac {p \cdot ( 1 - x_{\ce{NO2}}) } {({p \cdot x_{\ce{NO2}} )}^2}$$
$$K_p \cdot p \cdot {({x_{\ce{NO2}} )}^2} + x_{\ce{NO2}} - 1 = 0$$
Solving the quadratic equation we would finally get $n_\mathrm{tot}$, so all 3 degrees of freedom are saturated with given values for $p, T, n_\mathrm{tot}$.
So in our case, the following applies:
$p, n_\mathrm{tot}, T$  are given, and $V=f(p,n_\mathrm{tot},T)=\dfrac {n_\mathrm{tot}RT}{p}$


Why is the E2 protein important for protein degradation?

Pretty hard question, "why", as we've not been watching for the millions of years it took to build that system... I think the better way to look at it is to name the advantages this system has and hypothesize on it's selection. BTW in biology a good answer to why is always because.

I would guess though that given the function of ligases, it has been selected against due most probably to lack of selectivity. E1s are very small and already need to do both activation of Ub consumming atp and doing a somewhat major conformation shift to accept ligation of the activated Ub. The amount of Protein Protein surface usually required to maintain the exquisite selectivity that is required for such a system can be assumed to be too great for the size of the E1s.

Then if you ask why it did not evolve in some kind of huge multidomain protein; I would probably argue to the point that you need to produce a very large amount of it all the time, and that it must be somewhat difficult to target that protein for proteosomal degradation / make sure it's folded properly. Having a smaller footprint might help in keeping the domains ordered and reducing entropy hence making sure the protein stays functionnal longer.

Then why not the E2 doing it directly, and why do you need the E3 in between... Well again, I would assume this is also a question of selectivity. If you need to regulate thousands of cytosolic proteins and control their regulators at the same time, you need some modularity, you need the ablity to create multi protein complexes that bring different layers of regulation and cannot assume that one protein can do that. The term E3 is also some horrible bin where people put anything. E3 are very often large and sometime huge multiprotein complexes, think of the APC complex, or in the most well documented cases, the Cullin complexes.

The way E3s evolved is pretty cool actually, to recruit E2-Ub a protein just needs to get a domain through some mutation (RING, U-Box, HECT...) then it can target whatever it used to bind to a degradation complex. That is some pretty neat and quick trick to create diversity rapidly if a system actually does something that brings an evolutionary advantage.

I see you left out the E4 which may play a role in Ub chain propagation but as they pretty much function as E3 anyways, there is debate as to why we would need to use that terminology.

This is by no mean some definite answer since I could be wrong on most of these things. I hope I helped you to further your reflection on the subject.


I'm not sure whether the term "half-life" means the same as "lifetime" but I found a very interesting website about hydrogen bonding in water (with a lot of references) where it is stated that the lifetime of a hydrogen bond is in the range of 1 - 20 ps (it gives the following reference for this: F. N. Keutsch and R. J. Saykally, Water clusters: Untangling the mysteries of the liquid, one molecule at a time, Proc. Natl. Acad. Sci. U.S.A. 98 (2001) 10533-10540). That is quite a big difference to the 10 seconds from your reference. Thinking of the very dynamic nature of a hydrogen bond network and of the typical time scale for molecular processes in general I would tend to think, that the breakage and formation of hydrogen bonds on a molecular level should take much less time than 10 seconds. So if you are looking for the lifetime of a hydrogen bond on a molecular level a few picoseconds sound reasonable.

Edit: Here is another website, which also states a hydrogen bond lifetime of about a few picoseconds.


What is the logical way to understand Iso, Sec, Tert and Neo prefixes?

The prefixes, iso-, sec-, tert-, neo- are all traditional names when structural organic chemistry was still developing. You have already rationalized all except neo-. Neo- is Greek term for new. I looked up the OED etymological dictionary for older meanings, and the earliest use is traced to 1876. Now you can see in no 2, that neo- was a good option.

Used in the names of chemical compounds, minerals, etc., which are closely related to, or new forms of, known substances.
neocatechu  n. (Obsolete) a kind of catechu.
Used in the names of compounds and radicals in which one carbon atom is linked to four others.
Not part of standard chemical nomenclature although well established in neohexane and neopentane (and related terms).
(a)
Categories »

† neo-acid  n. Obsolete
1876   W. Odling in London, Edinb., & Dublin Philos. Mag. 5th Ser. 1 216   The names which call to mind that the normal, iso-, and neo-acids are mono-, di-, and trimethyl & etc. derivatives would seem to be most advantageous.

Neo- even it is used today represents a terminal tert-butyl group, just like the prefix iso- represents a iso-propyl group at the end.

IUPAC names have superseded all.


Does the salinity in seawater affect the process of alkaline hydrolysis?

I believe that desalination would be necessary as the salinity of the seawater would interfere with the earth metal salts already utilized to perform the process of alkaline hydrolysis.


Changing solubility of metal complex by adding ethanol

Would adding some organic phase like petrol ether or a non-soluble solid such as ceramic pellets do the same?

No. But adding other organic solvents which are miscible with water would.

The idea behind the "trick" is to reduce the polarity of the aqueous phase. With other words, you're making water more ethanol-like.


Alkali and alkali earth metal carbonates and bicarbonates decomposition

From your question, the heating temperature is unclear. See these links for sodium and lithium carbonates, the hydrocarbonate or barium.

They are all also mentioned on Wikipedia.

In other words, they all seem to decompose at some point, so what is meant by heating?

The reaction for sodium carbonate decompositon at melting point should be:

$$\ce{Na2CO3 -> Na2O + CO2}$$


Judging from the location and colour, it is most likely residue from liquid polish that was not properly rubbed/polished off. Quite a few types of liquid polish for brass turn white when they dry, and any folds and creases in the design is where they would be most likely to accumulate if the person doing the polishing is not very thorough.
Depending on how long its been there, you may be able to shift it with just warm water and a soft brush or cloth, leaving the patina mostly in place.  Otherwise you may need to use brass polish and polish the whole thing.
You could try asking advice from some of the better antique dealers or ideally someone who specialises in conserving antique brasses if the piece is a valuable antique.


Entropy Changes in Electrolytic/Galvanic Cell?

[OP] One of the most fundamental equations in chemical thermodynamics states: $$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$

This is just the definition of the Gibbs energy. By itself, it does not give you any insight into chemical reactions.

[OP] If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.

This is not the usual interpretation. Instead $ \Delta_rG_m^⦵ $ is the maximal work you can get out of the reaction (or, if the reaction moves away from equilibrium as written, the minimal work you have to put in to run it in that direction). $ \Delta_rH_m^⦵ $ is the sum of heat and non-PV work exchanged with the system. It could be negative or positive irrespective of whether this is an electrolytic or galvanic cell. Finally, the entropy of reaction is the same as for any reaction - roughly how much the dispersion of energy and particles changes within the system when the reaction proceeds.

[OP] In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.

You can't make any generalizations about the heat transfer when there is also work exchanged between system and surrounding. In many thermodynamic arguments, you talk about processes that happen in the absence of non-PV work, and in a reversible manner (i.e. the entropy of the universe is near-constant, which is the same as saying it is a near-equilibrium process).

[OP] What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.

Without a specific link, it is hard to understand what statement on the hyperphysics site you are referring to, and it what context it was made. In general, the entropy change of the system reflects changes in the system.
EDIT: The remainder was written after OP provided the link.

[Hyperphysics] Since the electrolysis process results in an increase in entropy, the environment "helps" the process by contributing the amount TΔS. The utility of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.

If there is no work and the reaction is run near equilibrium, the environment shows a change in entropy of $-\frac{\Delta H}{T}$. TΔS is not directly related to the change of entropy in the surrounding. The reason that $\Delta H$ is related to the entropy change of the surrounding is that for a boring surrounding (i.e. one that only shows changes in temperature, and very little at that) the change in entropy is proportional to the heat transferred.

I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

That is correct. Entropy is a state function, so if the entropy of the system changes, it is because the system changes, no matter what happens in the surrounding (you can't break the second law, though). The hyperphysics quote does not say that the entropy change of the system is associated with heat exchange. Their (implicit) argument is the second law:
$$T \Delta S_\mathrm{system} + T \Delta S_\mathrm{surrounding} > 0 $$
So if we have a positive change in entropy of the system like in the example, we can get away with cooling down the surrounding a bit (negative change in entropy of the surrounding). Typically, though, we don't want a super-slow reaction, so you would apply a higher voltage than necessary, and the heat transfer would go in the other direction (as you can tell by the heat typically dissipated from a fast battery charger).
Applying this to the first law, some of the energy input for the reaction can be in the form of heat transfered from the surrounding, it does not have to be all work in this case. (In other cases you have to put in more work, not for the first law energy balance but to satisfy the second law.) For a graphic illustration of the different cases, see https://chemistry.stackexchange.com/a/112958/72973.


It's not necessarily a strict definition of a single intermingling of phases, however the closest thing I can think of corresponding to this in reality would be interstitial metal hydrides.

These materials are being investigated as hydrogen storage materials, so essentially the hydrogen molecules interact with the metal to form hydrides, "$\ce{H-}$", which recombine under different conditions to reform $\ce{H2}$ as required. A review of this technique is given here.

It would be interesting, to answer your question more directly, if liquid hydrogen could be used instead of hydrogen gas (this requires temperatures of <33K). I imagine this process would give exactly the same result, albeit would happen more slowly than the gas phase.


In the gold foil experiment why didn't thallium form?

The Rutherford formula, as derived, assumes purely elastic scattering from the Coulomb force. No formation of a compound nucleus is considered. Generally, for most of the initial experiments, the available alpha particle energies from various decays (in the range of a few MeV) were not high enough for large deviations from pure Rutherford scattering. Of course, folks rapidly wanted both other energies (or controlled energies) or other charged particles (such as the protons used by Cockroft and Walton for their first accelerator measurements).
These days, much higher energy particles are available, and are commonly used in nuclear physics. A nice recent paper on the topic is in Physics Review C, "Cross section of $\alpha$-induced reactions on $^{197}$Au at sub-Coulomb energies". Here, the authors measure the reaction cross-sections for $^{197}$Au($\alpha$,$\gamma$)$^{201}$Tl, $^{197}$Au($\alpha$,n)$^{200}$Tl, and $^{197}$Au($\alpha$,2n)$^{199}$Tl. Although mainly of interest to astrophysical heavy-element nucleosynthesis, the results bear on this question.
The bottom line is that high (> 13MeV) $\alpha$ energies are needed to drive these reactions. At the lowest energy used (~ 13.7 MeV), the cross-section for the $^{197}$Au($\alpha$,n)$^{200}$Tl reaction is 0.67 $\mu$b (micro barns). At 20 MeV it has increased to 37 mb (milli barns).
Anyway, it is quite clear that these reactions do not occur to any measurable extent in the few MeV regime.


Memory foam chemicals from pillow spoiled bed and rendered it useless

I would suggest that perhaps it wasn't the pillow but another issue altogether.

If in two days your pillow outgassed enough to saturate the wall and the plywood, a couple days should also be sufficient for those to also release any trapped gasses. As I doubt they have any properties better at trapping gases than the foam in your former pillow.

Just a couple thoughts:

Do you have a carbon monoxide monitor in your home? A faulty furnace or water heater could be causing some of your problems.

Have you opened a window to provide better airflow?

Have you febreezed the area? The active ingredient in febreeze, cyclodextrin, does trap unwanted chemicals and gasses.

Or another chemical trap, is activated charcoal.


Why do E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane?

When the $\ce{C-X}$ and $\ce{C-H}$ $\ce{sp^3}$bonds break, the orbitals left behind on the two carbons will rehybidize into p orbitals and form the pi bond in the forming double bond.  For these two p orbitals to form a pi bond they must be in the same plane (or very close).

This is illustrated in the following figure.  This is called the anti-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~180° with respect to each other) and is the preferred geometry for E2 elimination.

image source

Elimination can also occur from the syn-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~0° with respect to each other) as shown below, but this conformation is slightly less favored.

This is why in molecules where the $\ce{C-X}$ and $\ce{C-H}$ bonds are locked in a conformation (perhaps by an inflexible ring system) where they are oriented 90° apart, E2 elimination becomes very difficult, if not impossible.


Stability Comparison between the Tropylium and Tricyclopropylcarbinyl Carbocation

It seems likely that the tricyclopropylcarbinyl carbocation would be more stable than the tropylium carbocation for the reasons I'll outline below, but if you have a reference proving this point, it would be nice to add it to your question.

Background

The cyclopropyl group is similar to an olefinic double bond in that it is very effective at stabilizing adjacent positive charge. The reason for this is that the $\ce{C-C}$ bonds in a cyclopropane ring are close to being $\mathrm{sp^4}$ hybridized (see this earlier answer for further elaboration of this point), said differently, there is a lot of p-character in the bond. In fact, the p-character is high enough that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the adjacent carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.

Hart and Sandri [J. Am. Chem. Soc. 1959, 81 (2), 320–326] studied the solvolysis of the following 3 compounds (Ar = p-nitrophenyl) and found that the relative rates of solvolysis were 1 : 246 : 23,500.

The corresponding tricyclopropyl p-nitroester could not be synthesized, so to complete the series Hart and Law [J. Am. Chem. Soc. 1964, 86 (10), 1957–1959] prepared the analogous phenyl esters and found the relative rates of solvolysis to be 1 : 1,080.

Clearly, every time a cycloproyl group is added significant additional stabilization of the carbocation center occurs.

Answer

The number of substituents stabilizing an adjacent positive charge is critical in determining the stability of a carbocation. Just as the trityl carbocation is much more stable than the benzyl carbocation, we see above that the  tricyclopropylcarbinyl carbocation is much more stable than the cyclopropylcarbinyl carbocation. Three stabilizing substituents are better than two substituents which is better than one.

In the tropylium carbocation there are 2 double bonds and a hydrogen stabilizing the carbocation center.  In the tricyclopropylcarbinyl carbocation there are 3 cyclopropyl groups stabilizing the carbocation center. Three groups provide  better stabilization than two groups, hence we might expect the tricyclopropylcarbinyl carbocation to be more stable than the tropylium carbocation.

A Word about Resonance Structures

The number of significant resonance structures for the tropylium cation is 7, the positive charge can be delocalized to each of the seven carbons as shown below. However 7 resonance structures can also be drawn for the tricyclopropylcarbinyl carbocation. So actually both carbocations have the same number of resonance structures.


Splitting across a double bond in ¹H NMR

The hydrogens are not 'isolated' - they are still vicinal protons (3-bonds away) and so the J-coupling between them is strong. The strength of a splitting interaction is primarily determined by the number of bonds in between the hydrogens involved.

The angle between them also plays a role but to a lesser extent. The Karplus equation describes the three-bond J-coupling between two protons as a function of their dihedral angle.


What research methods can I use in identifying Aluminum in a solid sample?

If you want a simple one, then you might want to just perform qualitative analysis (QA) using aqueous sodium hydroxide and aqueous ammonia. Firstly, dissolve a portion of the cube using a small volume hydrochloric acid. Then, add a small volume of sodium hydroxide to a small volume the metal chloride solution. To another small volume of the metal chloride solution, add a small volume of aqueous ammonia.

For the first test, aluminium ions would first give a white precipitate but will dissolve in excess. For the second test, aluminium ions would first give a white precipitate but will not dissolve in excess.

Lastly to confirm, add aqueous potassium iodide to the sample. If no precipitation occurs, the sample is indeed aluminium chloride solution and the cube does indeed contain aluminium.

Alternatively, this paper here also provides some insights on verifying if something is indeed aluminium:

http://www.iaea.org/inis/collection/NCLCollectionStore/_Public/08/346/8346019.pdf

Unfortunately, the method I have mentioned and the methods in the paper are destructive as they require you to chip of a piece of metal from the cube for analysis. So... if you can't afford to do that then they cannot be used.


As you noted, the link you provided was to a reaction run under rather exotic conditions (low-temperature plasma). Under more everyday laboratory conditions, for example, in solution, activated aromatic rings can react with $\ce{NO^+}$ as shown below.

However if there is a ring-activating amino substituent present, then the $\ce{NO^+}$ will react with the nitrogen to produce the synthetically important diazonium ion.


HBr reacts with ROH?

Exactly. It's just a nucleophilic substitution.


This is actually a trickier question than first inspection.

Your statement about primary alcohol substitution is more or less correct. Unless you have a very good reason to form the primary cation, you won't.

The problem in this example is that cyclobutylmethyl cation is not a primary cation even though if you were to draw it out, you might naively think that. Cyclobutylmethyl cation belongs to a special class of cations that are considered non-classical in the sense that the positive charge is not heavily localized to one atom, for example like methyl cation or tert-butyl cation. This essentially means that the positive charge is spread out over several atoms, and this increases the relative stability of the cation, making it more likely to form.

The more famous example is a 2-norbonyl cation, which is covered in many introductory classes.

For cyclobutylmethyl, the literature I found (hard to link) seems to suggest that the contribution from cyclopentyl cation is much larger than the cyclobutylmethyl cation, so we might consider this to be more neighboring group participation (from the C-C bond) than a raw non-classical ion. This is likely due to the fact that the ring expanded version is so much better in terms of strain.

Either way, a dissociated cation is stabilized, so I would think that this mechanism is more $S_{\mathrm{N}}1$-like.


Background

It seems that you understand the mechanism of the Canizzarro and Crossed Cannizzaro reaction, but let me supply a link to an earlier answer on the Cannizzaro reaction itself for others who might want to review the reaction. Also, here is the mechanism for the Cannizzaro reaction from the same link (but note, as we'll discuss below, the first step is really an equilibrium).

In the Cannizzaro reaction an aldehyde disproportionates into the corresponding  alcohol and acid. The maximum yield of the alcohol or acid is therefore only 50%.  If you've worked in the lab for 2 weeks synthesizing 50 mg of the aldehyde, and if your next step is the Cannizzaro reaction to prepare the alcohol from it, then a 50% yield is fairly distressing!

Here is where the Crossed Cannizzaro reaction can come in handy.  If it's the alcohol that your after, then using formaldehyde as a sacrificial reagent allows you to achieve a much higher yield of the desired alcohol.  In the Crossed Cannizzaro the formaldehyde is always oxidized to the acid and the other aldehyde is always reduced to the alcohol - you could get a 100% yield of alcohol!

Your Question - Why is that?

The first step in either the Cannizzaro or Crossed Cannizzaro reaction is an equilibrium between the aldehyde(s) and hydroxide ion to form a tetrahedral intermediate as pictured below. In the case of Crossed Cannizzaro with formaldehyde, the formaldehyde carbonyl is extremely reactive and the equilibrium will lie strongly on the side of the formaldehyde tetrahedral intermediate, and consequently most of the other aldehyde will remain present as the aldehyde.

You are correct to note that the rate determining step is the second step where hydride is transferred from the tetrahedral intermediate to any remaining carbonyl compound. But look at the 2 possible transition states for the Crossed Cannizzaro with formaldehyde (pictured above) - they are virtually the same, just some minor differences.  Hence the 2 possible rate determining steps should have similar activation energies.  This means that the energetics of the rate determining step will not control the reaction outcome (e.g product distribution), rather the initial equilibrium will be controlling.

If most of the formaldehyde is converted to the tetrahedral intermediate, then most of the other aldehyde remains as the aldehyde; and if the 2 possible rate determining steps from the tetrahedral intermediates have similar rates (they do as explained above), then the most likely reaction will involve the formaldehyde tetrahedral intermediate (major tetrahedral intermediate present at equilibrium) and the other aldehyde (major remaining carbonyl compound present at equilibrium).

Hence, in a Crossed Cannizzaro the formaldehyde is always oxidized to formic acid and the other aldehyde is always reduced to the corresponding alcohol.


Equilibrium constant. Can it be reached?

Dave's mathematical answer is perfectly valid, but I feel like another, perhaps more chemical point of view is based on the the law of large numbers, i.e., the "thermodynamic" limit.

"Equilibrium" is a statistical concept that only applies to large numbers of molecules.  Another way to say it is that equilibrium is a thermodynamic concept and thus only applies in the thermodynamic limit.  What is "large" depends on the particular system of interest.  In particular it depends on the equilibrium constant.  For your reaction of interest, $K_{eq} = 10$.  Thus the amount of $A$ and/or $B$ at equilibrium will be smaller than the amount of $C$.  For the equilibrium concept to apply to a system of $A$, $B$, and $C$ molecules, the calculated equilibrium amount of $A$ or $B$ molecules must be very large, say tens of thousands or so for precise work, or perhaps as low as hundreds for the equilibrium concept to be "approximately" correct.

If the equilibrium constant were enormously large, say $K_{eq} = 10^{100}$, then by the same logic, the system would need to contain on the order of $10^{102}$ total molecules for the equilibrium to be meaningful.  (And by the way that is larger than the number of particles in the universe!)

Speaking loosely, if $K_{eq}$ is $10^{100}$ and there aren't close to that many molecules, the reaction is irreversible.

A probabilistic view of equilibrium for small numbers of particles, such as that mentioned in DavePhD's answer, is occasionally useful but not always valid.  For example, consider if a second reaction consumed $A$ irreversibly, converting it to $D$ at some rate.  If you take DavePhD's probabilistic view of the $\ce{A + B <=> C}$ equilibrium, you would assume that the probability distribution of persistence times for $A$ is exactly the same as in the thermodynamic limit.  However, that is not necessarily true with small systems.  Statistical fluctuations in the $\ce{A + B <=> C}$ system will mean that some of the time, $A$ persists far longer than in the thermodynamic case because due to a fluctuation, more than the "expected" amount of the molecules were in the $C$ form.  Some other times, $A$ will resist degradation to $D$ for far shorter than the thermodynamic equilibrium calculation would suggest.  To treat these cases, more complex equations such as the chemical master equation are required.


Why are limes alkaline, but lemons are acidic? They taste about the same

The word "lime" has two meanings and they have essentially no relationship between them.

The "lime” that is inorganic oxide quicklime, CaO, reacts with water to give a relatively strong-base hydroxide::
$$ \ce{ CaO + H2O -> Ca(OH)2 }$$
$$ \ce{ Ca(OH)2 <=> Ca^2+ + 2OH-}$$
Due to the presence of these hydroxyl ions, the solution is alkaline.

"Limes" that are the green, hybrid citrus fruits are acidic with citric acid, just like lemons.  Both have juice with a pH in the range of 2-3, for limes as in fruit a typical pH level is about 2.8.  Hence the tart flavor for which both fruits are known.  Like all citrus fruits, lemons and limes are a rich source of ascorbic acid, vitamin C.


I'll try and simplify the answer.

Looking at the image above, imagine the products sitting above the letter D and the reactants sitting below the letter B.  Now, imagine the particles involved in the reaction moving since they have kinetic energy.  It is much easier for the reactants to climb up the top of the hill labeled C then it is for the products to climb up the hill.  Once a reactant climbs the hill and slides down to products it will not be able to go back to reactants by climbing up the hill unless we give them more energy by raising temperature.  Raising temperature allows the products to climb up the hill and go to reactants.

The forward reaction in the diagram is exothermic and the reverse is endothermic.  So, the exo is favored at low T and the endo at high T.

You can also look at the equation delta G = delta H - Tdelta S.

Make up some numbers for delta H and delta S.  You will see that endo reactions becomes more spontaneous as you raise T and exo reactions become more spontaneous as you lower T regardless of delta S.


DIY spectrometers limits as a water quality sensor

In general, with this setup, nothing. You would need a UV detector instead of a webcam. And, assuming you had a UV light source and the right detector, it would depend on the UV source spectrum. You could use a blacklight for mid 300 nm or a Steripen for 254. A broad spectrum lamp would be best. Water doesn't start to absorb UV strongly until you get below approx. 200 nm, so compounds that absorb above that will show up. This would be on a commercial unit like a Hach DR6000. On a DIY kit, I doubt you'll be able to detect much of anything unless it is either at high concentrations (very high) or your path length is long (ie Beers law). With all that being said, UV spec isn't a very good "fingerprint" analysis like ftir or mass spec. For example DNA and many organics all absorb at 254 nm, so you won't be able to differentiate them. Probably the easiest water quality parameter you could measure is called UV254, but that's really only useful for determining organic content and disinfection byproduct formation potential.

Okay, will all that being said, you might have a better option using visible light. Hach sells chemical kits for measuring certain constituents - they are coloromertric so they convert the constituent to a colored compounds that is detectable at a certain visible light wavelength. For example, you could test for nitrate or chloride. A used portable Hach meter will run you around 500, these are basically a fancy visible light spec. They work decent, but not great and even that would be worlds better than any of these DIY kits if that tells you anything.


Why are least electronegative elements usually in center of lewis
  structure?

If there is a "central atom", then there are multiple (more than 1) bonds to the central atom.  If the central atom is relatively electropositive, then it will be better able to share its electrons and form bonds with other atoms, at least more so than an electronegative central atom would.  Also, electronegative atoms tend to carry multiple lone pairs of electrons.  If this electronegative atom and all of its lone pairs were at the center of the molecular structure, then we would have many more destabilizing (lone-pair - bonding pair) electron-electron repulsions, then if all of these lone pairs were on the periphery of the molecule.

  why are many molecules symmetric. Is there a reason why symmetry in
  nature and in chemistry is preferred over non-symmetry?

The symmetric arrangements (linear, tetrahedral, octahedral, etc.) minimize destabilizing bonding electron - bonding electron repulsions.  Remember methane is a perfect tetrahedron - that is the lowest energy arrangement because all of the bonding electrons are as far apart from one another as possible in this geometry.