Reaction speed of Na2CO3 with vinegar vs that produced by NaHCO3 with CH3COOH

Sodium carbonate, $\ce{Na2CO3}$, will not create gas when mixed with an equimolar  portion of acetic acid (i.e. $1\  mol\ \ce{Na2CO3}:1\ mol\ \ce{CH3COOH}$). The reaction would proceed as follows: $$\ce{Na2CO3} + \ce{CH3COOH} \ce{->} \ce{NaHCO3} + \ce{CH3COONa}.$$ Therefore, you would need another equimolar portion of acetic acid to protonate the bicarbonate ion, according to this reaction: $$ \ce{NaHCO3} + \ce{CH3COOH} \ce{->} \ce{CH3COONa} + \ce{H2O} + \ce{CO2_{(g)}}.$$ This should illustrate that using sodium carbonate simply causes you to use an extra portion of acetic acid to make the same amount of gas.

Thus, sodium bicarbonate and acetic acid will produce gas in less overall time than sodium carbonate and acetic acid. When starting from sodium carbonate, the reaction will proceed through both reactions above. When starting with sodium bicarbonate, only the second reaction must proceed. Technically, the rate of gas evolution will be the same once gas formation begins (assuming equal starting molarities). However, the sodium carbonate reaction wouldn't produce any gas until the first reaction had completely finished, taking more time overall.


Letter case in chemical names sometimes matters too! E.g. when o-methylphenol (o-cresol, ortho-cresol, 2-methylphenol, I) improperly capitalized to O-methylphenol (instead of o-Methylphenol), it can be interpreted as a different compound, phenol methylated at oxygen (anisole, methoxybenzene, II).


CH3 radical is Sp2 hybridised carrying the lone electron in a pure p orbital. This is so because Sp2 hybridised orbitals are lower in energy than Sp3 hybrid orbitals (The more the S character the lesser is the energy of the orbital). When there is a lone electron present, it is more efficient to keep the bond pair electrons in lesser energetic Sp2 hybrid orbitals and lone electron in higher energy pure P orbital than keeping all bonding pairs and the lone electron in Sp3 hybrid orbitals. So most of the radicals are Sp2 hybridised.

But in situations like •CF3 where the atoms/groups attached to C are of high electronegativity things may change. According to Bent's rule more electronegative atoms/groups will favour less electro negative orbitals i.e. orbital s with lesser s character(Sp3 has lesser S character compared to Sp2). This is quite obvious because the more the S character of the C center the more is the tendency of the C center to attract the bond pair towards itself and in cases like electronegative atom, it will also try to attract the bond pair towards itself. This makes some unhappy situation. This is reflected by the Bent's rule. That is why •CF3 radical favours Sp3 hybridisation.

It's not that only such radicals like •CF3 can acquire Sp3 hybridisation. In many other cases like bridge head of adamentum also acquire Sp3 hybridisation becauseof opposition by its rigid structure to acquire planarity on generation of radical though it's not that much easy process to form radical here.

In conclusion, a radical will favour that hybridisation which will be energetically favourable for it.


If atoms are mostly empty space then how are we able to interact with things?

When we think about the structure of atoms and molecules we need to use a range of simplifying analogies as the objects we are talking about are too small to see and behave according to quantum rules we are unfamiliar with in the macroscopic world. None of the analogies we use are perfect and may mislead unless they are brought back to reality by recognising their limitations an checking their implications again the mathematics we know that does describe their behaviour.

That is the problem here. On one analogy atoms are built from nuclei containing protons and neutrons surrounded by electrons orbiting a long way away. This analogy is useful for explaining what happens when, for example, alpha particles are fired at a gold film a few layers of atoms thick. This is the famous Rutherford scattering experiment first done in 1909. This experiment led to the "solar system" analogy for what atoms look like. But really it only explains what they look like from the point of view of fast alpha particles. The mass of atoms is concentrated in the small nucleus and it is the nucleus that alpha particles bounce off, occasionally.

But that is only one, imperfect, way to describe what atoms and molecules look like. It doesn't describe how individual atoms join to form molecules, for example. For that we have to look at how electrons behave. Here a different range of analogies or mathematical descriptions are needed. The solar-system view of electron behaviour is deeply inadequate to describe what electrons actually do. And it is a very poor description of where electrons are. It is better to think of electrons as occupying 'clouds' of probability than being at a single point in space. Those clouds have a range of shapes from simple spheres to multi-lobed, multi-layer things in atoms with many electrons. Many of these have some probability of the electron being anywhere inside the cloud even, in some cases, right into the nucleus. In this analogy atoms are not empty at all but are filled with some probabilistic amounts of electron. No more than two electrons can occupy any orbital which imposes some strong constraints on what can happen when the clouds of two atoms or molecules meet.

Almost all the interactions we experience in normal macroscopic life involve those clouds of electrons interacting with each other. There are a range of different ways they can do this, but at the heart of things they all consist of electrostatic interactions at the atomic or molecular level where electron clouds push (or pull) each other. The low density of electron clouds doesn't matter since the quantum mechanical rules determine what can happen when the clouds meet. Normal interactions in the real world are just the macroscopic impact of these interactions.


Interpreting straight lines in a graph of isotherms of carbon dioxide

I will expand on the key points made in the comments. There are two issues that have been brought up:

Pressure is constant along line segment BC.

This is most easily intuited from a direct mechanical example. Consider a cylinder in which one face is a piston at vapor-liquid coexistence. Line segment BC then represents isothermal compression or expansion of the piston. As we push on the piston, the system volume decreases, and vapor condenses into liquid. As we pull on the piston, the system volume increases, and liquid evaporates into vapor.

That these processes occur just enough to maintain line segment BC as an isobar can be understood more quantitatively from thermodynamics. Phase coexistence actually places significant constraints on the system, as is quantified by Gibbs' phase rule---a system of a single chemical species at two-phase coexistence is specified by a single parameter, here the temperature $T$, because we are considering an isotherm. This implies that the pressure is known given $T$; that is, we have a relationship $P = P(T)$, and of course this is just the liquid-vapor coexistence line on a phase diagram. This pressure is, in particular, independent of the number of moles of substance in each phase, and so we must have the same pressure at phase coexistence; i.e., over the line segment BC.

Liquids cannot be compressed. (False!)

Liquids cannot be compressed easily. They are effectively incompressible for small changes in pressure (hence the steepness of the curve WX), but can certainly be compressed when subject to large pressures.


Contradictions in acidity and solubility of phenol derivatives

In the case of o-nitrophenol, the acidic hydrogen is hydrogen-bonded to the nitro group's oxygen atom, making it less acidic. In the case of catechol, one acidic hydrogen is hydrogen-bonded to the adjacent OH group's oxygen atom, but the other, more acidic hydrogen is not hydrogen bonded. This hydrogen is more acidic than that of hydroquinone. The resulting anion is even more strongly hydrogen-bonded than neutral catechol, stabilizing it even further. It is the second acidic hydrogen of catechol that is analogous to the acidic hydrogen of o-nitrophenol.


Is it important to use arrows as bonds in writing coordination compounds?

Historically, coordination compounds have been depicted in various ways.

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), bonds representing coordination from one atom to a single other atom should be represented as normal plain single bonds. The use of dative bonds (i.e. arrows pointing to the central atom) to represent coordination is not acceptable.


Why does F have a higher ionization energy than O?

At first approximation, we need to consider the orbital we are extracting an electron from. Both oxygen and fluorine (and nitrogen, carbon, boron and neon) have a $\mathrm{2p}$ orbital as their highest occupied, so we expect to ionise from that. If you draw a trend of the ionisation energies, you observe that the energy rises from boron to nitrogen, is lower for oxygen and rises from oxygen to neon. Disregarding the slight decrease at oxygen at first approximation we can say that it becomes increasingly hard to ionise an electron from a $\mathrm{2p}$ orbital with higher element number.

This is because essentially the shape and size of the $\mathrm{2p}$ orbitals are (close enough to, at first approximation) equal. So the wavefunction of the electrons we are removing does not change significantly. However, in their vicinity we have a nucleus which changes charge. For boron, we have an effective nuclear charge of $+1$ or $+3$ (depending on whether you count the $\mathrm{2s}$ electrons as shielding or not). For every step we take and every move we make to the right, the effective nuclear charge ($n(\ce{protons}) - n(\ce{core~electrons})$) will increase by one. We need to counter that nuclear charge to remove an electron which, naturally, because increasingly harder the stronger the charge holding it back.

The slight drop in ionisation energy at oxygen warrants a separate explanation. If you observe closer, you realise that the first three p-electrons (boron to nitrogen) are added as α spins while oxygen is the first element to obtain a β spin electron. A single β electron at first approximation has three degenerate orbitals it can populate. This is an unfavourable situation — compare with the Jahn-Teller effect in coordination chemistry that exists to counteract this situation. Thus, the energy required to remove this seemingly extraneous electron is lower than it would be just looking at effective nuclear charge. On the other hand, three α spins are more stable since the system is no longer degenerate. Thus it is slightly harder to remove an electron from nitrogen than it should be.


Best consumption-safe liquid for caffeine solubility?

Safety Warning

  Caffeine is a toxic substance. When following the advice below, always keep in mind the factors that will help you kill yourself if you are not careful! For more details you can start out on Wikipedia.

Supersaturation

Your idea of preparing a supersaturated solution is not that bad, but it might be that you have some nucleation sites in the bottle where you want to store it. Furthermore, it is probably not that easy to transfer the supersaturated solution from the container where you prepared it to the vial.

Acidic Medium

I've come across this PDF that mentions something interesting:

  Solubility in water is increased by adding dilute acid (e.g. HCl or citric acid).

It would be very easy to prepare an acidic solution (I'd aim at a pH of about 3 or so, in the general vicinity of energy drinks and Coca Cola, beverages known to contain large amounts of caffeine) from some citric acid¹ and dissolve caffeine in there. Maybe you could even conduct a study about the solubility of caffeine at different pH values!

Water/Ethanol Mixture

There is one dissertation (PDF) that touches on the solubility of caffeine in a mixture of water and ethanol. I wouldn't trust it too much, since it was written using MS Word and contains several formatting errors (it says $K_{ow}$ for example, instead of $K_\text{ow}$). The graphic containing the data can be found on page 52, but for your convenience it is reproduced below:

(Seriously, the figure caption is useless, and the plot is so bad I'm tempted to redo it just for this answer.)

Never mind the linear relationship that is indicated by the line, the diamonds are the important part here. As you can see, there seems to be a solubility maximum at around $x=0.5$, where $x$ is the volume fraction of ethanol in the water-ethanol mixture. So that's a whopping 50 vol.-% of ethanol, about equivalent to a cask-strength whisky!

However, it is probably not a good idea to use this knowledge to its fullest, because that will be quite a lot of ethanol, and as already mentioned in the comments it is probably not very wise to combine the consumption of caffeine with ethanol.

Does it have to be dissolved?

There are also caffeine tablets and pills that you can take. They come with the additional guarantee that some FDA-like agency has looked into the production of these products, so you can assume they are safe to ingest. Also, they come with instructions and more than enough warning labels so you can educate yourself further.

Conclusion

If you really want a caffeine solution, it is probably your safest and best bet to create an acidic solution where you can dissolve above-normal (i.e. pH=7) amounts of caffeine.

¹ Though please do not ask here how to calculate the amount you need, that would be considered a homework question and get deleted pretty quickly...


How does alcohol keep smell and taste flavors in wine?

First of all, ethanol has its own smell (which we perceive as taste), it also can stimulate pain sensors in the mouth. Therefore alcohol free wine will taste different.

The process of separating alcohol from wine is quite invasive: the water/alcohol is separated from flavour and colour compounds by reverse osmosis, then the water is distilled to remove alcohol, the water is added back, to the concentrated wine flavour. This process is surprisingly good at not removing anything but the alcohol, but it does change the flavour. Other small light volatile compounds are lost by this process. Tannins are also lost in this process, resulting in reds that lack body

Next the presence of alcohol helps some aroma compounds to evaporate. This means that the aroma of the wine changes. The scent chemicals can't get into the air without alcohol to carry them.

Finally, of course, the best wine is sold with alcohol. When you drink non-alcoholic wine, you are usually drinking a lower quality source wine.

source: http://winefolly.com/tutorial/non-alcoholic-wine/


What to do with (large) imaginary frequencies for constrained minimum structures?

The first I noticed is you said:

  The two gradient tolerances have been loosened slightly in order to converge the calculations. By using the default gradient tolerances, the calculations simply do not, I have found (especially the maximum gradient).

At first sight, the structure of your molecule should not leads to serious problems, but the above is a bad signal, and turns the results unreliable (at least doubtful). The constrains must be considered in the calculation of vibrational frequencies for they make sense.

Maybe no one can give you a foolproof steps series to follow. My suggestions are below.

What to do/check

If comments are right, and Orca does not care about fixing atoms for
the calculation of the vibrational frequencies, then you need to
change the software for this task (because in this case results are not meaningful) (This also answer the question "Is it "safe" to disregard the frequencies that mainly involved fixed atoms?"). The simpler way to check this is by looking at the number of freedom degrees reported (or deduced form the output). I was referring directly to vibrational f.d. because low frequency modes may be treated like rotations. I doubt that Orca lacks of this basic functionality. In such case I bet that this is described their User Manual.
The step you are doing are reasonable. My guess is that maybe you are far from a true minimum, in some unstable geometry, and that is why you have problems with accuracy changes. You said: "Should I converge my geometry tighter to come closer to the true minimum? ". Yes, it is always advisable to get good frequencies. So I would use not only tighter geometry convergence criteria but the TIGHT convergence option for the SCF part.
In the same lines, you can try using modern semi-empirical methods and then make the fine tuning with DFT. Even using a simpler DFT functional.
I agree with comments that you are not using the best basis set out there, but I think that this is not a problem here.
If you want to discard weird behaviors you can try with another functional, but I do not think that this is the problem (because it is known that B3LYP performs well in compounds with the elements involved here).

In brief. I would check if Orca compute correctly the frequencies, if not: change the software. If yes: discard those results, try to converge the structure using cheaper methods (if you are lack of computational resources) from different initial conditions, store those results and start new geometry optimizations and vibrational frequencies calculations starting from those previous converged states. The last step using the tight convergence criteria for SCF and normal convergence criteria for geometries, and using the target basis set (for energy and its derivatives).

Good luck

Note:

  Once I start computing reaction paths, I plan to try different
  functionals and basis sets to see if this give drastically different
  results.

Why do you expect that something like this would happen? You can save time by using a little large basis set and a better functional once you get the B3LYP/6-31g(d,p) results. It should not take too much time.

EDIT DUE TO COMMENTS

This answers generated some discussion about B3LYP or not B3LYP?

In that respect, my viewpoint is as follows:

B3LYP is not the most accurate functional (there should be no doubt about that).
It is reasonably good for geometries in systems like yours. Errors in geometries can be larger due to small basis set than due to B3LYP, or at least of the same order.
While there are functional with better results for geometries, you also must take care of how large basis sets those functional needs to be a better option. I mean, maybe using 5$\zeta$ basis sets they are better, but not using a small basis set.  (and maybe not)
The above discussion can be very long, but in any case, you are using approximations that (in my opinion) are much stronger than the errors due to  B3LYP. For example, the permittivity value, or the geometry of the constained parts, etc..
Also, even if you could perform a FullCI/CBS you shoud notice that the simulated system is different to the real system, so extremely exacts results in the simulated system may no improve your knowledge about reality.
So, if the functionals are accurate enough, what it is important here is obtaining the corresponding results for the functional/basis set. The functionales can be more or less problematic. B3LYP should not be problematic for your system. My guess is what I said in the answer.
Because of the above, I think that B3LYP is reasonable and enough for your case.


The staggered, conformation of (2R,3S)-butane-2,3-diol (1a) that you have shown does not have mirror planes but rather a center of symmetry, which is the reason that this conformation does not rotate plane polarized light. There are two other stable, staggered conformations of the diol, namely 1b and 1c. They are a racemic pair, each of which is chiral. But like any racemate, it is optically inactive. Each of these staggered conformations has also been represented as Newman projections with the S-carbon in the front (1a', 1b' and 1c''). Rotation of 1c'' by 180o about a vertical axis passing through the center of the C2, C3 bond leads to structure 1c', which is the mirror image of structure 1b'. A time tested device for ascertaining whether or not a compound is meso is to pair like groups in an eclipsed conformation  (1a''). While this technique serves its purpose and conformation 1a'' has a plane of symmetry, its population in solution is negligible as is the racemic pair of eclipsed conformations (not shown).


Difference between non-thermal and thermal ionized gases

A thermal plasma has its electrons and ions at the same temperature.  A raw plasma has hot electrons and cold ions from the vast difference in masses when microwave excited.  One way to thermalize is to run the plasma through a Helmholtz pair of coils with a strong uniform magnetic field parallel to plasma flow.  The low mass electrons tightly spiral around the parallel lines of force (that, of course, are not there), colliding with the nuclei to thermal equilibrium.

(That can sound a lot better than it works.)


Do conjugated double bonds take priority over a +1 formal charge?

Given the symmetry of this benzyl cation, the "direction" does not matter. I have put direction in quotes because there actually is no such thing. Electron pushing arrows are a helpful tool (at the very least for accounting purposes), but it is not what is happening in this molecule (or most other cases, either). What you have here is a resonance structure, meaning that the true situation is not well modeled by a single Lewis structure, thus we use several, and the true situation is somewhere in between the structures with varying contributions from the individual ones. So it does not matter how one arrives at these structures, it is important that they be reasonable and do contribute.

One could argue that the answer given is more extensive than your attempt, but I would consider both equal. The reactivity of the molecule is not dominated by the aromatic ring, thus placing extra emphasis on it seems superfluous.


Acid/Base dissociation constants relationship

The concentrations do change in the presence of acid or base and are no longer $10^{-7}$M.

Instead, [H3O+][OH-] $= 10^{-14}M^2$

This comes from:

$\ce{2H2O <=> H3O+ + OH-}$

$K = \frac{a(\ce{H3O+})a(\ce{OH-})}{(a(\ce{H2O}))^2}$

Where "a(x)" is activity of species "x".

Then, approximating activity of $\ce{H2O}$ as 1, and the activity of the ions as the concentration of the ions there is a constant $K_w$, such that:

[H3O+][OH-] $= 10^{-14}M^2 = K_w$


Writing a system of differential equations with the given arbitrary reactions

$\displaystyle\frac{\mathrm d}{\mathrm dt}\alpha(t) = k_1 - k_2\alpha(t)$ is correct.

Just to show you the way, here are the next equations:

$$
\begin{align}
\frac{\mathrm d}{\mathrm dt}\beta(t) &= k_2\alpha(t) - k_3\beta(t) \\
\frac{\mathrm d}{\mathrm dt}\gamma(t) &= k_2\alpha(t) - k_4\gamma(t) + k_5\delta(t)^2 \\
\frac{\mathrm d}{\mathrm dt}\delta(t) &= 2k_4\alpha(t) - k_6\delta(t) - k_5\delta(t)^2
\end{align}
$$


The reaction of $\ce{HBr}$ with olefins is a relatively fast reaction, in other words, the activation energy required is small.  Further, formation of the carbocation intermediate is the rate determining (slow) step in the overall process.  Considering both of these facts and applying the Hammond postulate we can say that the transition state leading from the reactants to the carbocation intermediate will be an early transition state, it will look more like the reactants than the carbocation.

This suggests that resonance and inductive effects need to be examined with regard to the reactants, rather than the transition state, in order to determine how the molecule will react with an electrophile (the proton).

As the above diagram illustrates, both resonance and inductive effects (in an early transition state that resembles the reactant) suggest that the beta carbon will be more nucleophilic than the alpha carbon.  Therefore addition of the proton (electrophile) will occur at the beta carbon.  This will be followed by rapid addition of $\ce{Br^{-}}$ at the carbon already bearing a bromine atom (the alpha carbon) to yield a geminal dibromide.


Ideal gasses reaction under const temp and pressure

Got it!

as $H=U+PV$ and $PV = nRT=const$ we ger $\Delta(PV)=0 => \Delta U= \Delta H$


What is the influence of an alkyl-hydroxy group on a benzene ring?

According to Hansch, Leo, and Hoekman (Exploring QSAR: Hydrophobic, Electronic and Steric Constants 1995) the Hammet sigma value for $-CH_2-OH$ is 0 (zero) in both either the meta or para position.  Positive values indicate deactivators or electron withdrawing (+.778 for nitro in para position) and negative values are activators or electron donating (-0.268 for methoxy group).  The Hammett relationship gives a way to compare electronic effects of substituents on a benzene ring.  Further info on Wikipedia.


There is no easy general pattern.

The dissolution can be formally divided into 2 steps:

breaking the ion lattice into free ions in vacuum
hydration these "naked" ions by water.

Both lattice forming and hydration enthalpies are negative and their absolute value increases with increasing ion charge and decreasing ion radius.

The relation of their values and the overall enthalpy sign  depends on the particular element chemistry, that modifies hydration and lattice enthalpy values.

There are some trends, but hard to generalize.

You can compare sodium and potassium salts of big anions with delocalized charge, like $\ce{NO3-, ClO3-, ClO4-}$. Dissolution of respective potassium salts is significantly more endothermic, compared to sodium ones. Sodium ion is smaller and is more intensely hydrated with lower enthalpy of hydration.

This difference is bigger that difference of their lattice enthalpies. It is probably due anion size,  packing of bigger potassium cation makes less difference than for chlorides.


Saturated is an old term in organic chemistry, dating back when the atomic structures and electrons did not exist. The meaning of saturated, even today is "chemically combined to the maximum extent, having no ability to combine further." The carbon atoms in cycloalkane are saturated in the sense that all the four valencies of each carbon atom are satisfied by four individual atoms.

The degree of unsaturation is a related concept: It tells us how many rings or double bonds are in the molecule. Don't mix these two terms.


Why do atoms get smaller moving left to right on a periodic table?

Atomic size across a period, in fact, has everything to do with electron shielding and the effective nuclear charge (charge felt by the electrons).

Across the period, electrons are added to the same outer level, meaning that shielding by inner electrons does not change, however, the numbers of proton increases. This results in the effective nuclear charge rising significantly (as no. protons increases while added electrons are shielding inefficiently), which of course, means that the atom has a smaller size.


Two chlorides of iron were prepared.

I do not read that as: $\ce{FeCl_{n}}$ was prepared by two different methods, but as:

Reacting iron with dry chlorine gas yields $\ce{FeCl_{n}}$
Reacting iron with hydrogen chloride yields $\ce{FeCl_{m}}$

Apparently, iron is oxidized in both cases.

What are the oxidants (what is reduced)?
Which of them is the stronger oxidant?
What is the stoichiometry (what are n and m)?

Could it be helpful to divide the amount of chloride produced by the amount of iron used? Would different numbers indicate which of the products has a higher chloride content?

  (i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.

Is it possible that they meant the same as indicated above, i.e.:

Calculate the amount of iron chloride that would have been produced from 2.80 g (instead of 4.50 g) iron in the first experiment and compare with the outcome of the second experiment.


Does evaporation occur while boiling as well?

Evaporation occurs while a random molecule from the bulk collides with the surface molecules and though the upper molecules are less tightly bounded, they gain the energy due to collision and escape from the liquid surface as vapours. But at low temperature this collision frequency is very less. In boiling, we provide heat to the liquid, due to which the collision frequency increases and upper surface molecules more fastly escape. So, boiling includes evaporation. Now for boiling, surface area doesn't matter as the heat which we externally supplied plays a dominant role compared to the surface area of liquid exposed to atmosphere in boiling. The surface area plays negligible role in boiling.


Isopropyl Alcohol leaves a white haze on plastic, is there anything I can use to clean it off?

Even the common alcohol ( ethanol, ethylalcohol) is known to corrupt surface of some plastics, and rubbing alcohol ( isopropanol, isopropylalcohol) is even better in that, as it contains bigger organic chain and is less polar.

These alcohols do not dissolve the plastics in large extent, like acetone often does, but they may corrupt surfaces, especially if they are shiny or polished.

The typical case is compact polystyren or polymethylmethacrylate, that are often clear and shiny, where the white milky haze is very well visible. The opposite case is filled black shiny plastic where the haze and lack of shining  is notable.

For similar reason, lukewarm water with detergents is preferred for plastics cleaning.

The only way I see to get rid of the haze is polishing, or applying a selfpolishing  coating, if applicable and worthy. Perhaps polishing of wax coating.


What is the proof of complete ionic dissociation in water?

One method is by making electrical conductivity measurements on aqueous solutions of salts. If we take four strong electrolytes:

Sodium chloride
Potassium chloride
Sodium perchlorate
Potassium perchlorate

We can measure the electrical conductivities at very low concentrations which allows us to obtain the limiting specific conductivties of the salts (extrapolation to zero concentration).

We will then find that the sum of the limiting conductivity of sodium chloride and potassium perchlorate will be the same as that of potassium chloride and sodium perchlorate.

This is because the limiting conductivity of a strong electrolyte salt is the sum of the contributions of the cations and anions. This allows us in an indirect way to show that some salts can totally dissociate when very dilute.


What I ended up doing was using jmol to produce the different phases of the MO and then I used TinkerCAD and made the molecule itself a "hole" structure. This created the space to actually put the molecule inside the orbital.


Are there any common (preferably somewhat complex) organic molecules that help combat ocean pollution?

Chemical surfactants or dispersants are useful for cleaning up oil spills. They break up the oil slick and allow the oil to become dispersed in the water, so that naturally occurring bacteria can break down and oxidize the hydrocarbons in the oil. Dispersants are typically designed to be water-soluble at one end and oily at the other end--basically soaps or detergents. See http://masgc.org/oilscience/oil-spill-science-dispersant-bkgrnd.pdf for more details.


Why are larger alicyclic rings more flexible?

It's geometry and you can get a petty good intuition by building simple models of the molecules

Saturated carbon-carbon bonds have two key properties of relevance. The angle between a pair of bonds on a carbon prefer to point to the corners of a perfect tetrahedron (~109.5 ° between each bond) if there are no spatial or geometric constraints. Also, C-C bonds can rotate freely unless something else in the molecule prevents that rotation.

In small cyclic aliphatic molecules (cyclopropane to cyclopentane) there is no easy way to satisfy the tetrahedral preference. No combination of bond rotations can give C-C-C bond angles of 109.5 °. If you build skeletal models of the compounds with tetrahedral carbon atoms this is easy to see: cyclopropane is rigid and flat; cyclobutane can twist slightly away from planarity but can't get those bond angles close to the unstrained angle, cyclopropane has slightly more freedom but is still pretty rigid.

But, from cyclohexane up there are plenty of extra freedoms. Cyclohexane can twist into a "boat" or "chair" shape and those give essentially unstrained bond angles at the carbons (the angles would be 120° in a flat hexagon, but when the bonds can also twist this does not constrain anything as enough combinations of twists are available to give an unstrained result in contrast to any of the smaller cyclic compounds). Plus, the twists cost little energy so they can interconvert easily given the amount of energy available at room temperature. Again, this is easy to see with physical models.

Higher cyclic compounds have even more freedom because there are more bonds to twist to give the ideal bond angle at carbon. So the increasing flexibility is a product of the fact that bond rotation is easy to do in the absence of geometric constraints. The more bonds you have to twist, the larger the number of conformations of the molecule available to give unstrained angles.


What is the meaning of n in n-octane and similar compounds?

In this case, n means "normal", i.e. straight chain. This is somewhat of a holdover from the olden days, but in the case of octane in particular, it makes sense to specify n-octane if that is what you have. For example, the "octane" used as a standard gasoline engine fuel against which gasolines are rated (by definition, 100 octane) is actually not n-octane, but rather 2,2,4-trimethylpentane - a branched isomer of octane - in contrast to the straight chain version.


Understanding molecule infrared absorption

All forms of  radiation match the difference in two energy levels, electronic, vibrational, rotational & nmr etc. For  IR transitions we refer to molecular vibrations. The molecule must have a permanent dipole or produce a transient dipole as a result of vibrating in one of its normal modes.

So homonuclear diatomics have no IR spectrum, they of course have energy levels, but the radiation cannot 'couple' to these and so be absorbed. Heteronuclear molecule have a dipole and so also have IR spectra. The classical idea is that the electrons oscillate as the nuclei move in a vibration, this oscillating charge can interact with electromagnetic radiation of the correct energy and cause a transition from a lower energy level to a higher one.

Molecules such as benzene vibrate in many ($3N-6$) normal modes, some such as asymmetric stretches produce transient dipoles and have IR transitions. Others do not and so have no transitions. What determines the are called selection rules.

The only secure way is to look at the symmetry of the molecules using a point group table, its very easy when you know how, but a bit daunting at first sight. Have a look at this answer Understanding group theory easily and quickly for more details.

(The molecule you mention belongs to the $\ce{C_{2v}}$ point group, see molecule-viewer.com for a 3D model with symmetry elements.)


Yes, you are correct - just flip the sign. As quoted from the venerable UC Davis ChemWiki:

  To determine oxidation electrodes, the reduction equation can simply be flipped and its potential changed from positive to negative (and vice versa).

They give the value for your (forward) reaction as +1.358 V, so for the reaction in question, you'll use -1.358 V.


How can scientists be certain that there are no undiscovered elements missing from the periodic table from hydrogen to lawrencium?

You are talking about the first 103 elements.  An element is by definition identified with the number of protons in its nucleus, the so-called atomic number.  Any atom with 6 protons is carbon, it cannot be anything else.  So there cannot be any missing elements out to 103, as all of these elements have been confirm discovered by accepted scientific methods.

For example, here is a link to a CHEM Study film showing how transuranium elements (i.e. atomic number > 92 and man-made) can be separated from one another and verified as distinct.

http://archive.org/details/transuranium_elements


There is a very wide variety of chemicals that have a good smell - and the flavor and fragrance industry has discovered and profited from thousands of them. To have a good smell, a compound first of all has to be low molecular weight and able to easily evaporate into air so it can be carried to your nose--i.e. a "volatile organic compound". Esters are a good class to look at for starters, and are responsible for many of the fruity odors. For example, amyl acetate is a simple compound, easily made in the lab, that smells like bananas. Another important class is terpenes: hydrocarbons made from 5-carbon units that include pine and cedar odors. Some odor components like those of chocolate are more complex, have many components, and are hard to reproduce exactly.

One way to think of good (and bad) smells is how over time animals and humans have evolved to appreciate the smells of some things (like fruit or cooking onions) as good and other things like rotting flesh as bad, since the former is good for you and the latter definitely bad. If you got it backwards, you would not survive for long. But flowers, for example, smell good but do not materially benefit us. Maybe we have some residual bee genes in us.


There is no difference in properties of salt hydrates and non-hydrates after they are dissolved in water, as they are hydrated to same state. So, yes, you can.

However, there is difference in reaction of salt hydrates and non-hydrates outside water solutions.


We use a convention for the sign of potentials. The standard reduction potentials are referenced wrt  hydrogen reaction $\ce{2H+ + 2e^- <=>H2_{(gas)}}$ which is given the value $0.0$ V. Oxidising agents have positive values, for example $\ce{Ag^+ + e^- <=> Ag_{solid}}$ $E^\mathrm{o}=+0.799$ V, or $\ce{Cl2_{(gas)} + 2e^- <=> 2Cl^-}$, $E^\mathrm{o}=+ 1.36$ V, and reducing agents negative values, e.g. $\ce{Cr^{2+} + 2e^- <=>  Cr_{solid}}$, $E^\mathrm{o}=-0.91$ V.

The oxidation potentials of these reactions are just of the same magnitude but of opposite sign. Thus for the same species  the oxidation potential + reduction potential add to zero. Normally we only deal with reduction potential so if you have an oxidation potential just reverse the sign and it becomes a reduction potential.

Note that the same species can be an oxidising agent or a reducing agent depending what it is being reacted with.

A reducing agent is an electron donor, and oxidising agent an electron acceptor. If a species is oxidised it looses electrons but gains electrons if reduced. Thus a reducing agent is itself oxidised;  confusing, yes  but its just how it is.

The cell potential is defined in your second definition as this uses the standard reduction potentials. The others are just variants by messing with reduction vs oxidation and changing signs as appropriate.

( Note that in some papers and books you may come across the redox potential values quoted wrt the saturated calomel electrode, not hydrogen, and this has a value $+0.246 $ V so its always worth a quick check. )


The first thing you need to take into account is that cyclobutane rings are not planar, as they are drawn on paper, but puckered. Thus, a three-dimensional structure of the two isomers would look something like this:

I expect the structure on the left to be rather stable as it is, in the way I drew it, minimising steric interactions. Any other puckering would can either be traced back to that one by rotation or mirroring or put the two methyl groups into axial position which is sterically undesirable. Thus, we can expect only one significant species in solution.

This species should have $C_\mathrm{2v}$ symmetry if I didn’t miss anything. Using the principal axis — the one through the centre of the cyclobutane ring — we can identify four sets of homotopic hydrogens. We therefore expect four signals in the proton NMR spectrum.

Coupling constants, especially those along the ring, should be clearly visible and identifyable (if not confounded due to signal overlap).

The species on the right is less rigid. It can interconvert easily between different puckered states and in each one there will be one methyl group in axial position and one in equatorial position. Still, all structures can be traced back to the one I’ve drawn through mirroring and rotation unless I missed something. So here again, we have only one set of signals at all temperatures.

This species has a lower symmetry: $C_\mathrm s$. There is no rotation symmetry axis, so none of the ring protons are actually homotopic. Transforming one onto another is only possible by mirroring, hence they are enantiotopic. Since the methyl groups and the tertiary hydrogens are also non-identical, I expect a total of six signals at low temperature. However, repuckering would transform the axial into the equatorial methyl group and vice-versa. Therefore, I expect only a set of three signals to be visible at room temperature due to rapid interconversion.

These three signals would be:

the methyl group
the tertiary proton
the set of secondary protons

Since the protons rapidly exchange positions, I would expect the coupling constants at room temperature to be averaged. At very low temperatures, after freezing the rearrangements, these should become distinct.

So yes, rigidness of a ring due to stereochemistry does affect the number and type of signals. cis-1,3-dimethylcyclobutane is very rigid and unlikely to transform from one arrangement to another. This leads to a clean, full set of signals. trans-1,3-dimethylcyclobutane, on the other hand, interconverts rapidly between different orientations, mapping the otherwise non-identical methyl groups and protons onto each other. It therefore shows less signals than it should at low temperature.

A similar effect can be observed in cyclohexane and derivatives, where cooling to low temperatures also freezes the ring flip and causes the signals of axial and equatorial protons to be distinct.


Formating chemical equations for proteins binding in hugely complex configurations

Generally speaking, for these sorts of complex reactions, you want to go to more than two dimensions. (At least for the initial bit.)

For each potential binding reaction you have a different dimension. That is, for $L$ binding to $A_1$ on site 1, you go horizontally. For $L$ binding to $A_2$ on site 1, you go vertically. For $L$ binding to $A_1$ on site 2, you go in the Z dimension, and then for $L$ binding to $A_2$ on site 2, you go in the 4th dimension.

Obviously, you can't put an actual 4D representation on paper. In fact, you can't put a 3D representation on paper. But if you're lucky, you can do a bit of topological rearrangement of the network and then do a projection of your network onto a 2D plane.

For example, if you have three potential binding interactions, each independent, then your system can be represented as the corners and edges of a cube. You can then represent this system by any of the standard projections of a 3D cube into 2D space.

For four potential binding interactions, it's the the corners and edges of a tesseract. Projecting a tesseract onto 2D space is much harder, in that there's no way to do it without lines crossing. But depending on what you want to represent, you might be able to adequately mark the crossings, or add "jumps" between two sub-graphs. (That is, have two more sub-graphs and have a systematic way of going from nodes on one graph to the node on the other graph with implicit reaction edges.)

For an arbitrary reaction graph, there's no guarantee that the crossing number of the graph is going to be zero -- that is, there's no guarantee that you'll be able to represent the network on a 2D page without crossings or jumps. Often the best you can do is attempt to be systematic about things, and then add crossings/jumps as appropriate.


To model early earth chemical reactions which types of reactions should be included?

For the reactions look up Stanley Miller's experiment. The experiment used water (H2O), methane (CH4), ammonia (NH3), and hydrogen (H2). I think you should start with that.


Why does having equivalent resonance structures give more stability?

Preamble
One important thing to know is that what we call "resonance structure" is a byproduct of our chemical notation which can't describe the structure of some compounds effectively using only one chemical structure. Personally I find the old term mesomeric structure more appropriate (meso- Greek mésos in the middle;  -merism from Gk. merismos "dividing, partition") because resonance is already used for a phenomena, a real phenomena, in physics that has nothing to do with resonance in chemistry. Furthermore the compound is not continuously changing it's structure between the possible resonance structures, like the term resonance seems to suggest, but is a "normal steady" compound that simply has an electron distribution that cannot be described with a single chemical structure.
Why does having equivalent resonating structures give more stability?
The resonance energy is due to electron delocalization, so more the electrons are delocalized in the molecule higher is the resonance energy ( this means a decrease in the overall molecule energy). This is clear in the acetate ion - we can draw a resonance hybrid  like this:

Here the electrons are equally distributed (delocalized) between the two oxygen atoms.
However for some compounds the real structure is not always a structure right in the middle between the different resonance structures; the first assumption is true only if the two structures are two equivalent structures. So it can occur that the electrons are not well delocalized between all the atoms hence the resonance energy is lower and so the overall stability.
In the case of phenolate ion I think the first structure in the figure [2] gives a greater contribution to the final structure of the phenolate ion so the electrons are not well delocalized, but if I'm not wrong are mainly in the oxygen atom. Unfortunately  I can't find the energy of the different resonance structures.

Finally, however, I think that you should study case by case if the number of the resonance structures or their contributions are determinant to the overall resonance energy taking into account valency and electronegativity of the atoms involved.


Why can NCl3 be hydrolyzed but CCl4 cannot?

There are a few (mis-) conceptions in the question that require addressing other than giving the simple (yet correct) answer that permeakra gave.

$\ce{NCl3}$

The hydrolysis of this compound is different from the hydrolysis of the other two. This is because the $\ce{N-Cl}$ bonds are polarised towards nitrogen, not towards chlorine as they would be for $\ce{SiCl4}$ and $\ce{CCl4}$. Since the oxygen in water usually attacks nucleophilicly, the attack happens on the chlorine; it can proceed nicely with an $\mathrm{S_N2}$ mechanism (the rear side of all three chlorines are pointing into the solution). Note that another water molecule or the same will likely protonate the nitrogen either before or concertedly, so the leaving group won’t be $\ce{NCl2-}$ but rather $\ce{NHCl2}$. The other reaction product is $\ce{HOCl}$.

Other concerted mechanisms including only one water molecule might be possible, but sound less likely.

$\ce{SiCl4}$

Silicon atoms are rather large compared to both carbon and nitrogen: They have an additional electron shell. (Layman’s simplification, but it works.) As such, even in crowded cases like tri-iso-propylsilyl chloride $\ce{((CH3)2CH)3-SiCl}$ with three bulky iso-propyl groups, the silicon atom can be rear-side attacked by water (or catalysts such as 4-(N,N-dimethylamino)pyridine) liberating a chloride atom in a traditional $\mathrm{S_N2}$ mechanism.

No d-orbital contribution is required to explain it. In fact, silicon does not have ‘low-lying’ or ‘accessable’ d-orbitals; that is an old, superceded explanation.

The hydrolysis products of $\ce{SiCl4}$ are $\ce{Si(OH)4}$ (which will likely condensate to some polysilicate or $\ce{SiO2}$) and $\ce{HCl}$.

$\ce{CCl4}$

Technically, this could also proceed via an $\mathrm{S_N2}$ mechanism. However, carbon is much smaller and the chlorine atoms much larger in relation, so the rear-side attack is disfavoured sterically.

Instead, this molecule has access to the $\mathrm{S_N1}$ pathway, which can be described as follows:

$$\ce{CCl4 -> CCl3+ + Cl-} \\
\ce{CCl3+ + H2O -> Cl3C-OH2+} \\
\ce{Cl3C-OH2+ + Cl- -> CCl3-OH + HCl}$$

And so on, displacing the remaining chlorine atoms, ultimately liberating $\ce{H2CO3}$. (After the second chlorine displacement, phosgene will form which can be attacked directly by water with the $\ce{C=O}$ bond acting as an electrophile.)

Note that here again we gain $\ce{HCl}$, an indirect proof of chlorine being the more electronegative element.

Conclusion

The three molecules we considered are so different in their steric and electronic properties, that we need to consider three different mechanisms that lead us to two different sets of products (depending on which side gains the hydrogens and which formally gains hydroxides). Nowhere do we need d-orbitals.


What chemicals are mixed together to cause chemoluminescence?

If you want different colours, you might want to give the peroxyoxalate reaction, known from glow sticks, a try.

Typically, you will need three components:

a diaryl oxalate
hydrogen peroxide (to oxidize the diaryl oxalate, probably to a 1,2-dioxetanedione)
a fluorescent dye, which determines the final colour.

Wilhelm J. Baader has examined the reaction in great detail, and you'll find a lot of references on his website.


Why doesn't propynal give cannizzaro reaction?

It is known that terminal alkynes reacts$1$ with $\ce{NaNH2}$ and produces salt, much like an acid-base reaction, hence it is a fast reaction.
Furthermore, the aldehyde group is an EWG, which makes it easy to lose a proton, hence drives the reaction!
Whereas, cannizzaro reaction happens through multiple reversible steps as seen in the following image,

Therefore, acid-base reaction in this case has much higher probability than Cannizzaro reaction.

References
$1$: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch09/ch9-2.html


Is there a treatment for distilled water to become drinkable?

The content of mineralic ions in tap water is usually rather small compared to the amount that you take in with your daily food. Especially Calcium content varies wildly between sources, and nobody recommends calcium supplements in cities with soft, low calcium content water.

The common DI water you have from a tap in a laboratory building or buy in the supermarket is not safe for drinking, firstly and mostly because it is not tested. But who would want to drink it anyway? It is more expensive, often has a taste that varies between lame and slighly foul, and you do usually have a regular drinking water tap to use instead.

The only way you could run into problems is when you have little food, drink a lot (hot environment), and have nothing to drink but distilled/rain water. On the other hand even well water likely has not enough sodium to balance what you loose through sweating.

To answer your question: A big problem with distilled water is its corrosiveness to common household tubing. Regular tap water coats everything in a layer of calcium carbonate, which makes even lead pipes relatively safe. In a standard copper pipe, DI water will first slowly dissolve the oxide layer, and then a lot of electrolytic corrosion will take place.

So, to use distilled water as tap water, you either have to make everything out of PE (doesn't really work if you want hot water), or add calcium carbonate. I believe this is what's done in areas where desalinated water is fed into the communal water supply. You just have to mix in some amount of local well water.


Unfortunately, the sense in which orbitals are orthogonal is more or less impossible to define rigorously without recourse to functions of some kind.  So, I'll give an explanation a shot using some simple, 1-D functions to illustrate the concept, followed by the pictorial orbital example you've asked for.

At a basic level, in order to have any two functions be orthogonal, you need to define a "mixing rule" for them. The most useful mixing rule for orbitals turns out to be integrating the product of the two functions$^\dagger$ over some defined domain.

So, consider two real 1-D functions, $f(x)$ and $g(x)$, defined over an interval $x=[a,b]$. The mixing rule $\Pi\,_a^b\!\left(f,g\right)$ I'll define as follows:

$$
\Pi\,_a^b\!\left(f,g\right) \rightarrow \int_a^b{f(x)g(x)\, dx}
$$

Simply put, two functions $f(x)$ and $g(x)$ are orthogonal under the mixing rule $\Pi\,_a^b$ (or any other mixing rule!) if and only if $\Pi\,_a^b\!\left(f,g\right) = 0$.

A pretty standard example of two orthogonal functions in this framework would be the following:

\begin{align*}
  f(x) &= \cos(2\pi x) \\
  g(x) &= \sin(2\pi x) \\
  a    &= 0 \\
  b    &= 1
\end{align*}

The integral to check from the above is:

$$
\Pi\,_0^1\!\left(f, g\right) \rightarrow \int_0^1{\cos(2\pi x)\, \sin(2\pi x)\, dx}
$$

Since this is Stack Exchange and not a math exam, I will now cheat and just link to Wolfram Alpha to demonstrate that the above integral is, indeed, zero. Thus, on the domain $x=[0,1]$, $\sin(2\pi x)$ and $\cos(2\pi x)$ are orthogonal.

On the other hand, to pick some random examples, $f(x)=-x$ and $g(x)=\ln(x)$ are not orthogonal on the interval $x=[0,1]$.

$$
\Pi\,_0^1\!\left(f,g\right) \rightarrow \int_0^1{\left(-x\right)\ln(x)\, dx} = 0.25
$$

Similarly, the original two functions are not orthogonal on the interval $x=[0.2,0.9]$:

$$
\Pi\,_{0.2}^{0.9}\!\left(f, g\right) \rightarrow \int_{0.2}^{0.9}{\cos(2\pi x)\, \sin(2\pi x)\, dx} = -0.0444852
$$

Ok, enough with the math lesson. How does this help define orthogonal orbitals?

Well, orbitals, both atomic and molecular, are (usually) defined just as as 3-D functions over all space.  If you have two orbitals $\phi\!\left(\vec r\right)$ and $\psi\!\left(\vec r\right)$ and want to know if they're orthogonal, you use a mixing rule somewhat similar to $\Pi$ above and calculate the appropriate integral. If the resulting value is zero, then the orbitals are orthogonal.

For example, consider a $1\mathrm s$ and a $2\mathrm p$ orbital (yes, these are atomic orbitals, but the principle holds for molecular orbitals, too):

Here, the colors of the two lobes of the $2\mathrm p$ orbital are important: one color, say orange, represents positive function values; the other, say blue, represents negative values.  For the $1\mathrm s$ orbital, though, it's all orange, or all positive).

What happens if, per $\Pi\!\left(\phi, \psi\right)$, you center these two orbitals on top of one another, multiply their values together, and integrate over all space? Well, the half of the $\mathrm p$ orbital that is positive will result in positive contributions to the integral of $\Pi$, and the half of the $\mathrm p$ orbital that is negative will provide negative contributions to the $\Pi$ integral.  And, further, due to the spherical symmetry of the $\mathrm s$ orbital and the reflection symmetry of the $\mathrm p$ orbital, those two positive and negative contributions will exactly cancel. Thus, the $1\mathrm s$ and $2\mathrm p$ atomic orbitals are orthogonal.

Epilogue: So, given all of this, why do we also say that vectors that are $90^\circ$ from one another are orthogonal?

Well, it just involves a different mixing rule (the domain is implicit, deriving from the dimensionality of the two vectors, which must be the same):

$$
\Gamma\!\left(\vec v_1, \vec v_2\right) \rightarrow \vec v_1\cdot\vec v_2
$$

Under the (extraordinarily common) mixing rule $\Gamma\!\left(\vec v_1, \vec v_2\right)$, the two vectors $\vec v_1$ and $\vec v_2$ are said to be orthogonal  when $\vec v_1\cdot\vec v_2=0$.

Note: In theory, I could have just linked to this question on Math.SE. However, none of the answers there were really couched in suitably readable terms.

$^\dagger$ Technically, integrating the product of one function with the complex conjugate of the other: $\int_a^b{f^*(x)g(x)\,dx}$


How to choose which reactants to have in excess?

Sometimes the excess is the cheap/unlimited reagent you use to get a high yield on your expensive/limited reagent, and sometimes you use the one that you don't mind/can easily extract from the products.

In this case, if you wanted to collect and record the lead(II) sulphate as a precipitate, using an excess of the lead(II) nitrate would be a waste.  If your intention was to produce a large quantity of sodium nitrate solution, and you don't mind having a load of the lead nitrate contaminating it, then you could use an excess of the lead nitrate.


Why are the two CH2 groups in allene perpendicular to each other?

No, VSEPR theory does not allow for a coplanar arrangement. Let us consider allene and its orbitals:

Source. Green indicates $\mathrm{p}$ orbitals; blue $\mathrm{sp^2}$; and red $\mathrm{s}$.

Hybridization

The terminal carbons are $\mathrm{sp^2}$ hybridized, and form three $\sigma$-bonds each. This means that each terminal carbon has one unhybridized $\mathrm{p}$-orbital.
The central carbon atom is $\mathrm{sp}$ hybridized, and forms two $\sigma$-bonds. This means it has two unhybridized $\mathrm{p}$-orbitals. For our sake, we will call these two orbitals $\mathrm{p_x}$ and $\mathrm{p_y}$. These orbitals are perpendicular to one another.

$\mathrm{\pi}$-Bonds

A $\pi$-bond is the result of the sharing of two $\mathrm{p}$-electrons, and it is particularly important for us to consider them as belonging to a particular region of space. In order for the central carbon's $\mathrm{p_x}$-electron to form a $\pi$-bond with one of the terminal carbons, it must be with another $\mathrm{p_x}$-electron. If the two did not share the same spacial orientation no bond could form. Since the $\mathrm{x}$-dimension is occupied by the electrons of a $\pi$-bond between the central carbon and one of the terminal carbons, the only way for the central carbon to form an additional $\pi$-bond is by sharing its $\mathrm{p_y}$-electron with the other terminal carbon. Hence, the two terminal carbons are necessarily perpendicular to one another.


Why don't beryllium and magnesium dissolve in ammonia?

Magnesium and beryllium are not really anomalous.  Their poor solubility or nonsolubility in ammonia can be correlated with the fact that their ions are more densely charged than those of the alkali and heavier alkaline-earth metals that do readily give solvated electrons in liquid ammonia.  However, at least for magnesium solvated-electron solutions can be formed in some other solvents or, in ammonia, with an electrolytic process.
Let's look more closely at magnesium, for which much more is known experimentally about solubility and solvated-electron generation than for beryllium.  Compared with lithium, calcium and the elements below these, the greater charge density of the magnesium ion (and by extension, the beryllium ion) inhibits solvated electron formation in at least two ways.

The metal gets passivated.  Solvated-electron solutions of metals in ammonia are not completely pure metal-ammonia solutions.  The highly electropositive metals will slowly react to form the amide and hydrogen, but the amide ion in solution stabilizes the solvated electrons and thus also slows the reduction reaction.  But with the doubly charged and compact magnesium ion, the hard-base amide ion form a salt with little solubility.  This inhibits metal dissolution rather than stabilizing solvated electrons.

What does dissolve, reacts faster.  With its compact size and double charge, magnesium ion polarizes the ammonia molecules and tends to displace protons.  These then react (as solvated ammonium ions) with the solvated electrons.

IUPAC has published a review by Lagowski [1] which includes a discussion of metal solubilities in ammonia solvent.  Magnesium is considered "soluble" in their table, but the text indicates that this is a borderline case and other solutes, such as ammono-bases, are needed to get a stable solvated-electron solution.  Magnesium is shown with a higher ion charge density than any of the clearly soluble metals, including the lanthanides europium and ytterbium.
Thus compared with metals that form those blue solutions, magnesium dissolves more slowly and reacts faster, and so few solvated electrons can accumulate in a proposed magnesium/ammonia solution.
Breaking through
Magnesium is certainly electropositive enough to generate solvated electrons.  Xu and Lerner [2] report that magnesium along with its heavier congeners calcium, strontium and barium generate solvated electrons in ethylenediamine, these electrons being captured by graphite to form blue or green intercalation compounds.  (Curiously, magnesium and barium appear to adopt a +1 oxidation state while the intervening metals go to +2.)  Significantly, magnesium requires more aggressive conditions of higher temperature and longer time to react, a difference the authors attribute to the effects of passivation (correlated with compact ion size) and compare with lithium versus other alkali metals.
More relevant to dissolution in ammonia is the electrolytic process described by Combellas et al. [3].  In this process, the magnesium is dissolved electrolytically into a magnesium salt solution, generating reasonably stable solvated-electron solutions.  Such solutions find use in reductive activating PTFE surfaces without undesirable carbonization that alkali metals tend to produce; and since the process is electrolytic it can be controlled by way of the electric current.  The electrolytic process breaks both barriers cited above.  With the polarization the magnesium dissolves much faster than it would as an isolated metal, and observations of solvated electron lifetimes made by the authors indicate that the dissolved salt is stabilizing the blue solution against hydrogen evolution.  (The authors do not give a solution-chemistry explanation for this, but the observations are consistent with the magnesium being incorporated into singly-charged ion pairs rather than remaining as independent +2 ions.)
References
1.
J.J.Lagowski, "Solution Phenomena in Liquid Ammonia", http://publications.iupac.org/pac/25/2/0429/pdf/index.html.
2.
W. Xu and M. M. Lerner, "A New and Facile Route Using Electride Solutions To Intercalate Alkaline Earth Ions into Graphite", Chemistry of Materials 2018 30 (19), 6930-6935
DOI: 10.1021/acs.chemmater.8b03421
3.
C. Combellas, F. Kanoufi, A. Thiébault, "Solutions of solvated electrons in liquid ammonia: Part 1. Chemical properties of magnesium solutions", Journal of Electroanalytical Chemistry ,
2001 499, 1, Pages 144-151.   ISSN 1572-6657,
DOI:  10.1016/S0022-0728(00)00504-0


Is there a goal here? If you wanted to react aluminium with concentrated aq ammonia, you would get hydrogen but the rate would be proportional to temperature.  However, this would be quite a poor way to create a reductive environment and as someone has already alluded, aluminium and sodium hydroxide or a strong mineral acid is better. From what I've seen, if you use alkali, a catalyst is needed.
Ethylamine can be more conveniently used in place of ammonia.
Calcium can also be used in reductions.I think I've also seen Indium, but that seems a bit expensive.


What is the molecular weight of bilirubin?

The discrepancy is simply due to round-off error.  The atomic masses of these elements are usually given as follows.  These are known as the standard conventional form, or standard formal short form, since they are given without an interval indicating the uncertainty in their values (see  https://en.wikipedia.org/wiki/Standard_atomic_weight):
C = 12.011
H = 1.008
N = 14.007
O = 15.999
Hence 33 C + 36 H + 4 N + 6 O = 584.673
However, if someone were to round off all the masses to four significant figures, but then report the result to five significant figures, we would have:
C = 12.01
H = 1.008
N = 14.01
O = 16.00
And now 33 C + 36 H + 4 N + 6 O = 584.658 = 584.66


Gaussian09 vs home made Hartree-Fock implementation

Diagonalization of the core Hamiltonian provides usually not the best guess for the SCF procedure to say the least, and thus, by default Gaussian uses a more sophisticated guess obtained by diagonalizing the Harris functional (Guess=Harris). With this default guess one get the same energy as OP reported:

 SCF Done:  E(RHF) =  -107.495842181     A.U. after    5 cycles

Now, Pop=None was a terrible idea, of course. It is always advisable to look at the final orbitals. Everything looks fine for the case of the default guess

 Alpha  occ. eigenvalues --  -15.51806 -15.51612  -1.44283  -0.72249  -0.57311
 Alpha  occ. eigenvalues --   -0.57311  -0.53949
 Alpha virt. eigenvalues --    0.28131   0.28131   1.12344

but results with the core Hamiltonian guess look suspicios

 Alpha  occ. eigenvalues --  -15.44896 -15.44853  -1.34440  -0.64250  -0.60670
 Alpha  occ. eigenvalues --   -0.52980  -0.16921
 Alpha virt. eigenvalues --   -0.07243   0.31047   1.20826

The energy of the first virtual orbital is negative which is odd. It is difficult to say what went wrong with the core Hamiltonian guess, but some clue can be provided if turning the symmetry back on (this is another very sensible default turning which off was a bad idea, I think).

The default guess yields the electronic state with a particular symmetry ($\sigma_{\mathrm{g}}$)

 Initial guess orbital symmetries:
       Occupied  (SGU) (SGG) (SGG) (SGU) (PIU) (PIU) (SGG)
       Virtual   (PIG) (PIG) (SGU)
 The electronic state of the initial guess is 1-SGG.

and converges within this symmetry

 Orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (SGG)
       Virtual   (PIG) (PIG) (SGU)
 The electronic state is 1-SGG.

But the core Hamiltonian guess yields the electronic state with no symmetry

 Initial guess orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (PIG)
       Virtual   (PIG) (SGG) (SGU)

and converges within this state with no symmetry

 Orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (PIG)
       Virtual   (SGG) (PIG) (SGU)
 Unable to determine electronic state:  partially filled degenerate orbitals.

which is again suspicious.

So, there is something wrong with the core Hamiltonian guess for this system (I suspect the basis is too small for the core Hamiltonian guess to yield sensible results), thus, SCF spits out some "strange" numbers. Strictly, speaking there is nothing strange with the numbers. Remember that SCF usually converges to a local minimum, and which particular depends on which part of the wave function space the initial guess placed the system in. And besides, you're not even guaranteed to have a minimum, SCF might converge to a stationary point of any kind.


AlCl3 + RCl → AlCl4(-) indicating that AlCl3 is a Lewis Acid
but
SiCl4 + RCl → SiCl5(-) ??????  So, this tells me that AlCl3 > SiCl4 in terms of electron accepting behavior.
Now, don't get me wrong: pentacoordinated Si is well known. One of the failures of the Lewis Acid concept is that, as far as I know, there is no unambiguous way to measure it. It is much more qualitative than quantitative, despite years of attempts to do so. You may know that Phosphorous is typically pointed to as the first example of the breakdown of the octet rule. (see Hypervalence) After Period 3, the concept has little use, imho. Since Lewis Acidity is tangled up with Lewis structures, perhaps it is not too surprising that the concept has some serious problems. Usually, we condsider both electronic effects as well as steric effects to 'explain' Lewis Acid/Base strength. One popular alternative is Hard.Soft.Acid.Base (HSAB) "theory". Anyway, comparing a tri-coordinated species with a tetracoordinated species and predicting which will form a bond with an electron donor is really difficult. Did you know that AlCl3 doesn't really 'exist' in the solid or liquid state? And even in the gaseous state, the dimer is common. The structure of the (solid) dimer (Al2Cl6) has two Al-Cl-Al bonds. Anyway, as far as I know AlCl3 is the stronger Lewis Acid in most apple-to-apple comparisons. To be honest, I don't think that if the Lewis Acid concept weren't a historically established meme, that it wouldn't be invented today. Personally, scales should actually be founded on measurement. The Lewis Acid strength scale is just sorta kinda correlated to what we can actually measure.


Are all solutions of weak acid/bases buffers?

I totally agree with @DavePhD and I want to show you how the pH dependence of the degree of acid-base ionisation provides the background of the buffering mechanism. Titration curves effectively illustrate the influence of the pH on the ionisation of weak acids and bases. The Figure below shows the change in the pH of a 0.1 M acetic acid solution on its titration with a strong base. If you just dissolve acetic acid in water, yes it is true that there will be a small fraction that ionizes to yield the conjugate base, acetate. However if you add the slightest amount of a base, for example, it will result into a steep increase in the pH (look at the beginning of the titration curve) and the solution is not a buffer.

That initial steep increase in the pH slows down quickly, and a pH plateau appears in the range of 4.0-5.8. As in this range $[\ce{CH3COOH}]$ is comparable to $[\ce{CH3COO-}]$ (i.e. their ratio is within a tenfold range), we will have a two-component system containing both a weak acid and a weak base—i.e. a buffer. The centre of the plateau is at 50 % titration where $[\ce{CH3COOH}]=[\ce{CH3COO-}]$. Therefore, according to the equation $\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log{\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$, the $\mathrm{pH}$ will just equal the $\mathrm{p}K_\mathrm{a}$. Following the plateau (upon the exhaustion of the buffer capacity), the pH will again increase steeply, and the top of the titration step will be reached, i.e. the point of neutralization at which the degree of titration of the acetic acid will be 100 %

From the figure above, the region marked inside the rectangle is a buffer. Because it can provide the most efficient pH stabilization against both acidic and basic shifts. In this region, the ratio of the two components (acid and conjugated base) is one or close to one. Accordingly, a buffer is effective in the $\mathrm{pH}$ range of its $\mathrm{p}K_\mathrm{a} \pm 1$ $\mathrm{pH}$ unit


How to calculate the rate of consumption of aluminium with hydrochloric acid?

Since one of the products is a gas $\ce{H2}$, the decrease in system mass must reflect the escape of hydrogen. Thus, after thirty seconds, the system has lost $\pu{0.030 g}$ of hydrogen, which is $\pu{0.01488 mol}$ of $\ce{H2}$.

Thus, the rate $r$ of hydrogen production is:

$$
r (\ce{H2})
= \frac{\Delta n(\ce{H2})}{\Delta t}
= \frac{\pu{0.01488 mol}}{\pu{30 s}}
= \pu{4.96E-4 mol//s}$$

Now, how can you use this rate and the stoichiometry of the reaction to determine the rate of aluminum consumption?

   $\ce{2 Al (s) + 6 HCl (aq) -> 3 H2 (g) + 2 AlCl3 (aq)}$
 For every three moles of hydrogen produced two moles of aluminium are consumed, therefore $r(\ce{Al}) = -\frac{2}{3}r(\ce{H2}) = \pu{3.31E-4 mol//s}$.


Electrolyse of water is a redox reaction, meaning that there is a transfer of electrons between atoms. Now, every redox reaction can be written as a combination of 2 half-equations. One half-equation shows one species getting oxidised (losing electrons) and the other half-equation shows the other species getting reduced (gaining electrons). When you combine these two half-equations you get the overall reaction, which is what you are asking for. Here is guide that should help you find the overall equation of any redox equation, not just this one example.

Step 1: Determine what species gets oxidised and species gets reduced

Now, what I think the reaction you are trying to describe is the electrolyse of water. Correct me if I am wrong. If that is the case, there is no need for $\ce{NaOH}$ in the equation as it appears on both side, meaning that they cancel each other out (also $\ce{NaOH}$ has nothing to do with electrolyse of water). Secondly, it is $\ce{H2O}$ that gets oxidised and reduced, not $\ce{H+}$ and $\ce{OH-}$.

Step 2: Determine  half-equations

These can be easily found by using your table of standard reduction potentials. Oxidation half-equation: $$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Reduction half-equation: $$\ce{2H2O + 2e- -> H2 +2OH-}$$

Step 3: Balance each equation so that there is the same number of electrons

We can't immediately add the half-equations as the electrons won't cancel each other out. This means that we need to multiply the reduction half-equation by 2:$$\ce{4H2O + 4e- -> 2H2 +4OH-}$$

Step 4: Add the half-equations

When you add each half-equation, you get: $$\ce{6H2O   -> 2H2 + O2 + 4H+ + 4OH-}$$ The $\ce{H+}$ and $\ce{OH-}$ combine to form water which then cancel out with the water the other side. Therefore the overall reaction is: $$\ce{2H2O  -> 2H2 + O2}$$


Calculation for buffer of pH 5 in sodium fluoride solution

Acetic acid is a liquid. It has a $\mathrm{p}K_\mathrm{a} = 4.75$. A mixture of acetic acid and sodium acetate form a buffer solution which means that the solution resists $\mathrm{pH}$ changes. If the acetic acid and sodium acetate are present in equal molar quantities, then the $\mathrm{pH}$ of the buffer solution will be $4.76$. Since $\mathrm{pH}$ of buffer was $5$, there are more moles of acetate ($\ce{OAc^-}$) than acetic acid ($\ce{HOAc}$).

$$\mathrm{p}K_\mathrm{a} = 4.76$$
$$K_\mathrm{a} = 1.74 \times 10^{-5} = \frac{\ce{[H^+][OAc^-]}}{\ce{[HOAc]}}$$
$$\mathrm{pH} = - \log{\ce{[H^+]}}$$


C-4 oxidation of octanoic acid to 4-oxooctanoic acid

If I was tasked with making 4-oxooctanoic acid I would not start from octanoic acid! There is no easy way of making 4-oxooctanoic acid directly from octanoic acid. It is however a relatively trivial exercise from other starting materials.

My suggestion is to react the commercially available ethyl succinyl chloride (ethyl 4-chloro-4-oxobutyrate) with n-butylmagnesium bromide (also commercially available) modified with manganese(II)chloride/lithium chloride as described in Tetrahedron Lett. 1992, 33 (31), 4439–4442. This will give the ethyl ester of the required 4-oxooctanoic acid. A simple base hydrolysis gives 4-oxooctanoic acid. You could probably use the widely available n-butyllithium in place of the Grignard.


"Carbon snakes" - what physico-chemical process causes the exothermic response?

$\ce{C_mH_{2n}O_n -> m C + n H2O}$

The hydration reaction of sulfuric acid acts as the driving force because it is highly exothermic (95.33 kJ/mol H2SO4).


How to name the tautomers of hexane-2,4-dione correctly according to IUPAC rules?

You numbered all locants together in a series and you found the lowest set of locants according to Subsection P-14.3.5 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) since the locant set ‘2,2,4’ is lower than ‘3,4,5’.

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

This approach would be correct if all locants refer to detachable prefixes (like ‘2-hydroxy’). However, in the given compounds, the locants refer to prefixes (‘hydroxy’), endings (‘ene’), and suffixes (‘one’). The assignment of low locants to such different features is stipulated in Subsection P-14.4 as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Therefore, the resulting name is (4Z)-5-hydroxyhex-4-en-3-one (not (2Z)-2-hydroxyhex-2-en-4-one) since the locant ‘3-one’ for the principal characteristic groups is lower than ‘4-one’.


Why do noble gases like Xe and Kr form compounds, despite offending the octet rule?

When we say that noble gases are stable it means that they are more stable than the rest. Yet it is to be considered that even these have ionisation potential ;though high. if we make a noble gases react with such ions that help in their bonding then we can have a compound of noble gas. Search this on Google.

  The first published report, in June 1962, of a noble gas compound was by Neil Bartlett, who noticed that the highly oxidising compound platinum hexafluoride ionised O2 to O+
  2. As the ionisation energy of O2 to O+
  2 (1165 kJ mol−1) is nearly equal to the ionisation energy of Xe to Xe+ (1170 kJ mol−1), he tried the reaction of Xe with PtF6. This yielded a crystalline product, xenon hexafluoroplatinate, whose formula was proposed to be Xe+
  [PtF
  6]−
  . It was later shown that the compound is actually more complex, containing both XeFPtF5 and XeFPt2F11. - Wikipedia, noble gas compounds

This was the first real compound of any noble gas.


You could buy the chlorine test strips from a pool supply store to test for hypochlorite. You may need to neutralize the acid before testing, so ask if the test is accurate in a strong acidic solution.


Work done in irreversible isothermal expansion against vacuum

A system does its maximum amount of expansion work when the external pressure is equal to the pressure of the system all the way through the expansion process.

At constant pressure work done is given by:
$$w=-p_\mathrm{ex}\Delta V$$
Where $p_\mathrm{ex}$ is the external pressure. Therefore the minimum work done is when the external pressure is $0$, which is the case under vacuum. This would mean that the work done under vacuum is $0$.

This is called free expansion. The greater the opposing force then the greater the amount of work done is (as work is equal to force applied over a distance). Therefore when the external pressure is $0$ there is nothing for the piston to push against, because there is no opposing force, no work is done by the system.

If you are interested to see where $w=-p_\mathrm{ex} \Delta V$ comes from then note that
$$w=Fd$$

Because pressure can be given by $P=F/A$ and volume can be given by $V=Ad$ we can see that $$w=\frac{F}{A} Ad = p\Delta V$$

This is work done on the gas. For work done by the gas, we put a negative sign in front of the pressure term.

*Note that this is at constant pressure.


What is the safest way to hook up car battery jump cables and why?

Many recommended procedures such as this one call for connecting (+) to (+), then (-) on the "good" battery to a metal component in the "dead" car engine.  The negative terminal of the battery is grounded to the metal components, so we may think of the last connection as (-) on the dead car.  If a spark were to form when the circuit is established on the last connection, it would occur at the metal component and not at the terminal where it could interact with electrolytically generated hydrogen or oxygen gas.


Quantitative relationship between corrosiveness of acid and other properties

In addition to the acid dissociation constant, some other factors that add to the corrosiveness of an acid are its ability to act as an oxidizing agent and its affinity for water (hence, ability to dehydrate material). Concentrated sulfuric acid has both these abilities in addition to simply being a strong acid. I do not know if these factors have been quantified, but they could be.

Mixing chromic oxide ($\ce{CrO3}$) with sulfuric acid makes it an even better oxidizing agent, especially when hot - so-called - "cleaning solution" is great in the lab for removing residues of tar from your glassware, but it is rather dangerously corrosive and has to be handled very carefully.


Reason for decreased boiling point of hexafluoroisoproponal?

Since fluorine is so much more electronegative than hydrogen, its electrons are considerably less polarizable. This decreases the extent to which induced dipoles are able to form (London dispersion forces), which in turn lowers the energy necessary to break the intramolecular forces. This is what gives fluorocarbons lower boiling points than their hydrocarbon counterparts.


What does the "D" group represent on a chiral center?

I assume it's deuterium, the isotope of hydrogen that has a single neutron as well as a proton in the nucleus. As such, it has the same atomic number as hydrogen, which means you break the tie on atomic mass --  deuterium is higher priority than hydrogen but lower than everything else.


Colour due to transmission and reflection

Since your experimental observation with copper salts negates your original hypothesis, it implies that the way you are trying to explain it is wrong. The main culprit and the source of all these problems is the color wheel which is taught in schools. Misconceptions persist for long. When you are looking at the copper solutions in two different settings, it is misleading to think that copper solution is reflecting blue light back to you. It is not.
In each case, copper(II) solution is absorbing a small portion of the red light from the visible spectrum and it appears to you like a pure blue solution.
Indeed it is our brain which has been created in such a way that when the visible spectrum has a certain red portion missing, it perceives the remaining spectrum as "blue".
Hint: Water in the ocean also appears blue? Water also very very weakly absorbs the red portion of the visible spectrum. You just need tons of water to perceive this effect.
There is a beautiful book by the name of The Physics and Chemistry of Color. The same author wrote an article "The fifteen causes of color: The physics and chemistry of color." It is certainly worth consulting. Article-behind paywall

The fifteen causes of color derived from a variety of physical and chemical mechanisms are summarized in five Croups in this article. Vibrations and simple excitations explain the colors of incandescence (e.g., flames), gas excitations (neon tube, aurora), and vibrations and rotations (blue ice and water). Ligand‐field‐effect colors are seen in transition‐metal compounds (turquoise, chrome‐oxide green) and impurities (ruby, emerald). Molecular orbitals explain the colors of organic compounds (indigo, chlorophyll) and charge‐transfer compounds (blue sapphire, lapis lazuli). Energy bands are involved in the colors of metals and alloys (gold, brass), of semiconductors (cadmium yellow, vermillion), doped semiconductors (blue and yellow diamond), and color centers (amethyst, topaz). Geometrical and physical optics are involved in the colors derived from dispersive refraction (rainbow, green flash), scattering(blue sky, blue eyes, red sunset), interference (soap bubbles, iridescent beetles), and diffraction (the corona aureole, opal).

Now you can imagine that all is not that simple.


One of the major components of exhaust is carbon dioxide, produced by the combustion of hydrocarbons - if in complete combustion, follows the general reaction:

$$\ce{C_{x}H_{y} + O2 ->CO2 + H2O}$$

Other possible emissions include $\ce{NO_{x}}$, particulate matter, carbon monoxide, VOCs etc. The amounts of these emissions vary from car to car (due to a lot of factors, e.g. age, wear-and-tear etc). According to a self study by Audi, in the document Motor Vehicle Exhaust Emissions, on page 6, they determined that asides from $\ce{N2}$, the main emissions were $\ce{CO2}$ and $\ce{H2O}$.

For this answer, the focus will be on the possibility of recycling the $\ce{CO2}$ emissions (and linking to the $\ce{H2O}$ emissions).

This very question is the subject of the Scientific American article Reverse Combustion: Can $\ce{CO2}$ Be Turned Back into Fuel?, the articles details that there are several research groups looking at this problem.

A couple of examples of ideas and experiments that have been put forward include:

  In the 1990s a graduate student named Lin Chao at Princeton University decided to bubble carbon dioxide into an electrochemical cell. Using cathodes made from the element palladium and a catalyst known as pyridinium—a garden variety organic chemical that is a by-product of oil refining—he discovered that applying an electric current would assemble methanol from the $\ce{CO2}$.

Inspired by the natural process that removes $\ce{CO2}$ from the atmosphere - photosynthesis, continued research has sought to emulate this - essentially, the current research is effectively looking at a possible solution by taking

  $\ce{CO2}$, water, sunlight and an appropriate catalyst and generate an alcoholic fuel

The research reported in the Scientific American article and the National Geographic article Carbon Recycling: Mining the Air for Fuel go along the lines of extracting the excess $\ce{CO2}$ from the air rather than directly from the car's engine or exhaust. The reason for this is the current practicality, with the National Geographic reporting that

  The Sandia prototype’s solar collector has an area of about 20 square meters (215 square feet) for a reactor the size of a beer keg, Stechel says. About 300,000 acres (121,400 hectares) of mirrors would be required to collect enough sunshine to make the equivalent of 1 million barrels of oil per day, she says. (The world currently consumes about 86 million barrels per day of petroleum and other liquid fuels, including biofuels.)

The research shows that, yes it is theoretically possible and practically possible on a larger scale to recycle exhaust gases.


Finding $c$ with Dulong-Petit law on non-molecular substance

It may be useful to start with the specific molar heat capacity $c'$ (per mole, instead of per kilogram), which is somewhat more fundamental. Here $c' = 3R$.

This is true even for your crystalline solid, though we have to decide what "moles" refers to in this case. From a physical perspective, the heat capacity is a result of vibrational motion in three dimensions, and each atom will contribute to this motion. Our "moles" should therefore be "moles of atoms"---for the purposes of this calculation, we consider the atom to be the fundamental unit.

The molar mass should then be $\text{mass of solid/moles of atoms}$, which can be interpreted as a weighted molar mass. You can see that your formula $M/n$ calculates exactly this: the mass per mole of a formula unit, divided by the number of atoms in a formula unit.

Hence $$c = \frac{3R}{M/n} = \frac{3R}{M}n.$$


When expressing a quantity, do the non-significant figures determine whether you round up the last significant figure?

The international standard ISO 80000 Quantities and units – Part 1: General includes the normative Annex B Rounding of numbers, which stipulates:

Rounding in more than one stage (…) may lead to errors. Therefore, the rounding shall always be carried out in only one step.
Example: 12.254 should be rounded to 12.3 and not first to 12.25 and then to 12.2.

(Note that, in the given example, the ISO standard follows the round half to even rule in which the even multiple is selected as the rounded number. This is the default rounding mode for floating-point computation. Anyway, the rule “rounding shall always be carried out in only one step” is independent of the used tie-breaking rule.)


Are alkenes or alkynes reduced first with diimide?

Your given answer to the textbook question is seemingly correct. I said seemingly, because this reaction is not well understood to date, even though its first introduction was given at time as far as 1929 (Ref.1). The reaction was mostly hidden without getting attention in literature until to the middle of 20th century. This is mostly because the reactive species of the reaction, the unstable molecule, diimide ($\ce{HN=NH}$) was not well understood.

By 1960's, the unstable $\ce{HN=NH}$ was generated in a variety of ways such as oxydation of hydrazene (by $\ce{H2O2}$ or atmospheric $\ce{O2}$), or decarboxylation of azodecarboxylic acid, or thermal decomposition of p-toluenesulfonylhydrazene. It reacts with alkynes (e.g., diagram A below; Ref.2) and unhindered alkenes (e.g., diagram B below; Ref.3) to transfer two hydrogens in a cyclic process (diagram C below; Ref.4), resulting in a strictly cis reduction. This is the mildest way of performing such reductions. As shown in the diagram A below, one generally useful application of this reduction by diimide is it did not cleave the weak $\ce{C-I}$ bond during the reduction process. Most other reducing agents will cause competing or complete cleavage of the $\ce{C-I}$. The most important advantage in this case is in the conversion of iodoalkyne to vinyl iodide, it gives predominantly the cis-isomer. In addition, it is note worthy that this method of reduction with diimide does not lead to the cleavage of other sensitive functions such as $\ce{O-O}$ and $\ce{N-O}$ bonds.

To my understanding, the type of functions reduced by diimide are depend on the way it was generated. Besides, the rate determining step is always the rate of diimide generation. In general, trans-alkenes react more rapidly than cis-alkenes. During the mechanistic study of reaction, it was reported that diimide reduction of symmetrical multiple bonds proceeds readily. Those bonds included $\ce{C=C}$, $\ce{C#C}$,$\ce{N=N}$, and $\ce{O=O}$. On the other hand, more polar functions such as $\ce{C=O}$, $\ce{C#N}$,$\ce{C=N}$, and $\ce{O-N=O}$ are either very slowly reduced or not reduced at all (Ref.4).

The table 1 has given some uniques alkene substrates used in this process (Ref.5). In this research, 2-nitrobenzenesulfonylhydrazene has been used to generate diimide in situ.

  Abstract of Ref.5: A one-pot protocol for the formation of 2-nitrobenzenesulfonylhydrazide (NBSH) from commercial reagents and subsequent alkene reduction is presented. The transformation is operationally simple and generally efficient for effecting diimide alkene reductions. A range of 16 substrates have been reduced, highlighting the unique chemoselectivity of diimide as a reduction system.

A recent publication has shown that diimide can be generated by refluxing haydrazene in air (with solvent). The reduction of diimide generated by this method gives somewhat different results. For example, a norbornene derivative gives 100% conversion in Table 1. However, similar substrate used in Ref.6 has given only 42% yield (Entry 15; Ref.6). Further, a styrene derivative (cis-1-phenylpropene) gives 61% conversion in Table 1. Yet, styrene substrate used in Ref.6 (should be better since it is a terminal alkene) has given only 16% yield (Entry 17; Ref.6). Therefore, it is very convincing that the outcome is highly depend on the metheod of generating diimide,

Based on all, the result of the reaction in the question is uncertain.

References:

J. Hanus, J. Voříšek, "Action de L'hydrazine sur Quelques Acides Incomplets $\ce{C_nH_{2n-2}O_2, \ C_nH_{2n-4}O_2,\ C_nH_{2n-6}O_2}$," Collect. Czech. Chem. Commun. 1929, 1, 223-227 (https://doi.org/10.1135/cccc19290223).
Scott E. Denmark, Shyh-Ming Yang, “Total Synthesis of (+)-Brasilenyne. Application of an Intramolecular Silicon-Assisted Cross-Coupling Reaction,”  J. Am. Chem. Soc. 2004, 126(39), 12432-12440 (https://doi.org/10.1021/ja0466863).
James D. White, Rich G. Carter, Kurt F. Sundermann, Markus Wartmann, “Total Synthesis of Epothilone B, Epothilone D, and cis- and trans-9,10-Dehydroepothilone D,” : J. Am. Chem. Soc. 2001, 123(23), 5407-5413 (https://doi.org/10.1021/jo9710137).
E. E. van Tamelen, R. S. Dewey, M. F. Lease, W. H. Pirkle, “Selectivity and Mechanism of Diimide Reductions,” J. Am. Chem. Soc. 1961, 83(20), 4302-4302 (https://doi.org/10.1021/ja01481a057).
Barrie J. Marsh, David R. Carbery, “One-Pot o-Nitrobenzenesulfonylhydrazide (NBSH) Formation−Diimide Alkene Reduction Protocol,” J. Org. Chem. 2009, 74(8), 3186-3188 (https://doi.org/10.1021/jo900237y).
Hao Chen, Jianmin Wang, Xuechuan Hong, Hai-Bing Zhou, Chune Dong, "A simple and straightforward approach toward selective C=C bond reduction by hydrazine," Canadian Journal of Chemistry 2012, 90(9), 758-761 (https://doi.org/10.1139/v2012-057). Here is the available PDF of the article through Science Gate.


Good catch - you are absolutely correct. This is a mistake in the Wikipedia page, as a quick check with the reference listed there confirms. The values are the wrong way around.

Changing it now...

Edited to add: this of course raises the question of why it should be this way around. A somewhat hand-wavy answer is that, just like in benzene, the $p_z$ orbitals overlap within each graphene sheet so that the electron contributed by each carbon is delocalised in this plane (the other three are bound up in the three $\sigma$ bonds). So in a sense this is conjugation, but over a whole 2D plane rather than around a ring or along a single chain of atoms. On the other hand, the graphene sheets are rather further apart from one another than the covalently bonded atoms within a sheet (as you might expect since the sheets are only weakly bound through van der Waals interactions), and so the overlap is greatly reduced and conduction is much more difficult.

A more complete answer would rely on a study of graphite's electronic band structure, which is an active research area and beyond the scope of this response!


Where do the electrons come from to make the carbon stable during bombardment of alpha particles on beryllium

When beryllium is bombarded with alpha particles, where do the electrons come from to make the carbon stable?

Nowhere.  This is a nuclear reaction, not a chemical one.  Charge is conserved and no extra electrons were supplied.

The carbon nucleus is stable with respect to nuclear decay.

Chemical stability relates only to the behavior of electrons with respect to atoms.  It has no place in this nuclear  reaction.

Eventually the carbon nucleus may attract additional electrons from some external source, but that's nothing to do with stability of the nucleus.  The nucleus does not require any electrons to be stable.


The last statement

  …sodium sulfate and sodium sulfite are both basic and would turn pH indicator the same color

is true only for $\ce{Na2SO4}$ as it's formed by both strong base and strong acid and won't noticeably affect pH.
$\ce{Na2SO3}$, on the other hand, undergoes hydrolysis:

$$
\begin{align}
\ce{Na2SO3 + H2O &<=> NaOH + NaHSO3} \\
\ce{2Na+ + SO3^2- + H2O &<=> Na+ + OH- + Na+ + HSO3-} \\
\ce{SO3^2- + H2O &<=> HSO3- + OH-}
\end{align}
$$

resulting in elevated pH (basic solution) up to 9 (according to Wikipedia).


Preservation and bleaching of sliced mushrooms with hydrogen peroxide

Hydrogen peroxide is quite a strong oxidizer. First, let's address the bleaching effect: It bleaches by oxidizing colored compounds, making them colorless. From Wikipedia:

  An oxidizing bleach works by breaking the chemical bonds that make up the chromophore. This changes the molecule into a different substance that either does not contain a chromophore, or contains a chromophore that does not absorb visible light. This is the mechanism of bleaches based on chlorine.

This is also the mechanism of hydrogen peroxide as a bleach, since both chlorine-based bleach and hydrogen peroxide are oxidizers.

As for the antibacterial effect: hydrogen peroxide in this particular concentration (5%) may or may not be effective against bacteria (and viruses, fungi, etc.) due to the presence of the enzyme catalase in some bacteria and fungi. In those organisms which it is effective against, hydrogen peroxide would probably work by causing oxidative stress in the organism.


Your goal is to reduce the surface tension of the water so that it does not support the formation of large bubbles or inhibit the whetting of the dirt particles.

When the bubbles are small, the dust inside will be more likely to make contact with the liquid surface of the bubble within the lifetime of the bubble. Also, when it does make contact, it will be much more likely to irreversibly adhere to the water if you have reduced the surface tension.

The word "surfactant" is a blend of "surface active agent"$^1$, and this is what you need to interfere with the surface tension of your water.  I believe that you may have written off detergents too hastily. I think the best solution will be to add a small amount of laundry detergent, or possibly non-foaming carpet cleaner. Each of these products are designed to foam to a lesser degree than, say, dishwashing detergent or most hand soaps.

I suspect that you will have to experiment a bit, but that something on the order of a single cc per liter may be sufficient to increase the effectiveness of dust removal without causing excessive suds. Your goal should be to see how much detergent you can add without adversely affecting your vacuum, and decide whether this has resulted in a decreased throughput of dust particles.

As to your title question, "How do I lower the surface tension of water without using a detergent?", what I'm really saying is 1) that most anything that does what you want could probably be deemed a detergent and 2) that a detergent is likely the most practical solution to your problem.

1) Surfactants and Interfacial Phenomena, Fourth Edition, Rosen and Kunjappu, John Wiley and Sons, Inc., 2012


t-butanol + allyltrimethylsilane + Lewis acid

This is called the Hosoni-Sakurai Reaction (see J. Org. Chem. 2006 71 8516)

Can only find an image for the more common addition to carbonyl compounds but there are examples of additions to alcohols (see the JoC reference)


For energy levels in nuclei (and their gamma emission) the first place to look is at the Evaluated Nuclear Structure Data Files. I usually use the Brookhaven site, but there may be a closer mirror to you. Enter 24 into the Nuclide or Mass box and hit Enter. Check the 'Adopted Levels, Gamma' box by $^{24}$Mg, and request the pdf version (back up at the top).
On the last page you will find the lower levels like this (although rotated by 90 degrees):

So, the 4122.889 keV level has a decay time of 22 fs - this is the level that the first gamma decays from. The 1368.672 keV level has a decay time of 1.33 ps - this is the second gamma in your sequence.