Why is 2H2O2 -> 2H2O + O2 a first order reaction?

Summarizing from the excellent comments:

  Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants?

Yes, that is how you get the overall order of reaction. If you compare the rate with a given set of concentrations to the rate with all concentrations doubled, the rate goes up by two to the power of the overall order.

  I don't see why that would be first order?

It is not, but it is pseudo-first order. If you start the reaction using certain initial concentrations, you will find an exponential decay of the reaction rate over time (i.e. there is a constant half-life).

As JM97 comments, the iodide is a catalyst, so its concentration does not change over time. So while the concentration of hydrogen peroxide drops over time and causes the reaction to slow down, the iodide concentration in the rate law does not contribute to a slowing down (no matter what order the reaction is in iodide).

  [OP responding to Ivan...] Actually the textbook statement is "_This reaction is first order with respect to both H_2 O_2 and I^- "

As Ivan comments, that means the reaction rate is proportional to the first power of the hydrogen peroxide concentration (if the iodide concentration is unchanged). It is also proportional to the first power of the iodide concentration (if the hydrogen peroxide concentration is unchanged). If both are changed at the same time, you use the rate law to see what the effect is.


Surprising values of intermolecular forces of ethane, oxygen and xenon

Van der Waals forces are weak and depend on polarizability. There is a correspondence between the boiling points of oxygen (−183 °C), xenon (−108 °C), and ethane (−88.5 °C) and their constants (1.364, 4.137, 5.507). These gases are polarizable to varying extents, but not polar with charge separation. The size of the molecule and its mass will affect polarizability - perhaps a way to look at this is to say electron density in the outer regions of the molecule will determine how easily dipole-dipole interactions can evolve.

But ammonia stands out with a boiling point (−33 °C) which is too high to be explained by van der Waals forces alone. Of all the gases listed, ammonia is the only one with significant hydrogen bonding capability. Perhaps if $\ce{H2O}$ were included, it would be an even greater outlier.


A good question. The term 'shoulder' is really an informal description used to describe the appearance of parts of peaks in the NMR spectrum. When a peak tends to deviate from the standard Lorentzian, we might describe this as having a shoulder. Shoulders can come about from typically three origins:

Shimming anomaly. Fairly common, but the key to diagnosing this cause is that it will affect all peaks in the same way (within the natural linewidth of the peak). So, have a look at the TMS peak, or solvent peak, or that internal reference of ethyl acetate that your supervisor suggested was an impurity. They will all have exactly that same lineshape anomaly, with identical shoulders. Always check these small molecule peaks when you run your spectra to determine shim quality. As an aside on this spectrum, these 'shoulders' you are looking at aren't really shoulders - they are split peaks. If this lineshape was genuinely due to a shimming anomaly, that indicates a pretty serious z shim issue, and you've probably got some floating solid in the sample.
Overlapping peaks. NMR spectra are simply the superposition of all nuclei in the sample. If two singlet species have a non-identical chemical shift within their natural linewidths, then they will appear as a shouldered peak. Obviously, the relative populations will determine how they actually appear. This could account for this type of spectrum (but more than likely doesn't), and that is why you would do other experiments such as COSY or HSQC to verify this.
Splitting due to coupling Often it is difficult to fully resolve small couplings, especially those that come from aromatic systems or long range pathways, where the size of the splitting might be about 1Hz or so. This is where an understanding of the expected coupling pattern for the molecule is very helpful. So, take a deep breath in.....

For Salicylamide, as described in your paper, we would expect H6 to be a large doublet (ortho to H5, ~7-8Hz), but also show meta coupling to H4 (~1-3Hz), and so we expect this large doublet to exhibit small splittings. Indeed, that is what you are seeing. On a good shimming day, in better solvents than dmso, we might even expect to see para splitting (<1Hz - don't hold your breath though). For the H5, we expect, again, a large ortho doublet coupling to H6, but also a large ortho doublet coupling to H4 (again, ~7-8Hz). That gives us the reported triplet, which is in fact a dd. But, we also expect to see some meta coupling to H3 (again 1-3Hz). Again, we see this, and each of these of these should be split due to the small meta couplings. (And we'll leave alone the discussion about whether they have reported apparent splittings, as opposed to J couplings, otherwise J56=J65),

And breath out.
So what you are seeing is genuine splitting coming from small couplings. These are often not reported in experimental data, because it is quite difficult (or at least time consuming) to resolve all of these couplings. You can get a better understanding of these small splittings if you use resolution enhancement when you process your data. Use a window function such as a small negative line broadened exponential function, or a gaussian function, and you can often split these lines almost to the baseline.


Dichlorine hexoxide has different structures depending upon whether it is in the gas or solid phase. In the gas phase, the molecule adopts structure (a) (reference 1, reference 2).  It is a relatively unstable molecule having a half-life measured in minutes.

In the solid phase it exists as the dark-red ionic material as shown in (b) and is often referred to as chloryl perchlorate.  Both forms are very strong oxidizing agents and react explosively with organic compunds.


Apparent contradiction in the name of group 2 elements

"Insoluble (in water)" is a relative term, especially for ionic compounds, and so is "alkaline".  That relativeness gives us enough wiggle room for two of our most common bases to slip through.

Roughly speaking, it takes a thousand volumes of water to react with and then dissolve one volume of calcium oxide, so the solubility of lime in water would not be evident from casual observation.  Yet if you put an indicator into the water it reacts in a way that indicates the water has turned alkaline.  The solubility of lime in water is hard to see with the eye, but it overwhelms the tiny intrinsic autoionization of water, so lime gives evidence of being both an "earth" and "alkaline".  Ditto (despite the still lower solubility) for magnesia.

Lime and magnesia, then, are the quintessential alkaline earth's, far exceeding anything else with similar properties in their common occurrence.  They have become the namesake of the all the "alkaline earth metals" in Group 2.


Does sublimation happen only on the surface or throughout the body?

The situation is akin to boiling.  If a molecule at the surface frees itself of its attractions to its neighbors, it can move into the vapor phase easily.  Same thing with evaporation at the surface of a solid.

Can a molecule evaporate from the bulk liquid?  The answer is no.  Imagine such a thing happening.  The escaping molecule(s) form a bubble inside the liquid.  The pressure inside the bubble is the vapor pressure of the liquid at that temperature.  But that temperature is less than the boiling temperature so the vapor pressure is less than the pressure external to the liquid.  The result is that the bubble collapses.  The boiling temperature is the lowest temperature at which the bubble pressure is equal to the external pressure.

Now think of sublimation in the bulk of a solid.  Here the "bubble" has to have enough pressure to break apart the solid.  It generally won't develop that much pressure.  Indeed, one can watch "dry ice" sublime well above its freezing point and nothing happens inside the bulk of the solid.

So the general answer is no, sublimation does not take place in the bulk of a solid.


Although we know that the helium dimer has a tiny bond dissociation energy, we don't consider it a molecule, as the "bond" does not survive a molecular vibration, even rotation of the molecule is enough to cleave it.

From an answer by Lighthart on "Molecular orbital theory & predicting the stability of a molecule?" I took the liberty to borrow the MO schemes:

In the scheme you can clearly see how we arrive at the bond order of zero for $\ce{He2}$. Both, the bonding $\sigma$ and the anti-bonding combination $\sigma^{*}$ are doubly occupied. Dihelium is only held together by weak van der Waals forces.
In the case of the dihelium cation $\ce{He2+}$, one electron is removed from the anti-bonding orbital. There are now more electrons in bonding than anti-bonding orbitals. This effectively results in a net energy loss or bond energy gain, and the overall bond order increases to 0.5. In a more phenomenological context, three negative charges are now in a field of four positive charges, this gives rise to Coulomb attraction.

Reference: Wikipedia article on dihelium and references therein.


How to verify that ethanol is infinitely soluble in water?

The better term is miscible in all ratios. The idea behind the term is that you can have a mixture from 99.9999… % water and 0.0000…1 % ethanol to a mixture of 0.0000…1 % water and 99.9999…% ethanol. No infinity necessary.
The obvious opposite to miscible is immiscible (in some or all ratios). A common pair of household liquids that are immiscible are fatty oils and water. In the chemistry lab, hexane and water might be a better example.
To verify the miscibility of water and ethanol, set up a parallel experiment with water and hexane/a fatty oil such as sunflower oil. You will notice that a fully immiscible pair will never mix, there will always be two different bits of liquid visible. You can try a little oil/hexane and lots of water and vice-versa. On the other hand, a miscible pair will mix and you get a solution where you cannot tell one from the other.
For completeness, partially miscible: if you repeat the experiment with ether and water you will observe that the first couple of drops of water will dissolve in sufficient volumes of ether (and vice-versa) but if more is added you will again get two phases.


Half cell reactions for oxidation of water by acidified solution of potassium dichromate

Look closely at the oxidation state of all atoms in the equation.
$$\ce{Cr2O7^2- (aq) + 14 H+ (aq) + 6e- <=> 2Cr^3+ (aq) + 7 H2O (l)}$$

The oxidation state in chromium changes from +6 to +3. This is what the electrons in the equation are necessary for. Oxygen retains its oxidation state of -2 in both sides of the equation, as well as hydrogen which remains its +1 oxidation state.

In the other equation
$$\ce{O2 (g) + 4H+ (aq) +4e- <=> 2H2O (l)}$$

the oxygen gets reduced by changing its oxidation state form 0 to -2. Again for hydrogen, the oxidation state of +1 is retained.

The total equation for this process would therefore be
$$\ce{Cr2O7^2- (aq) + 14 H+ (aq) <=> 4Cr^3+ (aq) + 8H2O (l) + 3O2}.$$

From the standard potentials you can see, that this reaction may happen, it does not tell you anything about the conditions under which this reaction will happen. There are a lot more factors to consider.


How do chemical bonds increase molecular refractivity?

The polarisability is the ability of any atom to have its electron 'cloud' displaced by an electric field. Polarisability is defined as the ratio of the induced dipole, formed by electron displacement, divided by the electric filed strength. It has units proportional to volume usually; values are in units $4\pi\epsilon_0\times 10^{-30} \mathrm{m^3}$. Larger atoms/molecules  naturally have larger values, in general, than smaller ones, e.g. $\ce{CH4}\; 2.6,\; \ce{CCl4}\; 10.5$. In a general sense also the number of electrons and how tightly they are held affects the magnitude of the polarisability.

In the case of light it is its oscillating electric field that affects the atoms in the material, a solvent medium for example. In doing so the speed of light in the medium is reduced and the ratio of this to that in vacuum is the refractive index. The energy of the photon remains the same inside as it is outside the material (assuming no absorption or scattering ) so the frequency remains the same, but inside the material the wavelength changes because $c/\lambda =\nu$ is constant and c changes. In forming the refractive index, the tails of electronic absorption bands  are important and are usually far into the ultraviolet.

In terms of working out the refractive index of some arbitrary molecule usually a semi-empirical method is used with the very many measured values available; values for different bonds can be extracted and parameterised. From these values then the refractive index of unknown compounds can be calculated. I found the pages www.pirika.com which has some explanation of how this is done.

In bringing atoms together to form molecules the atom/bond polarisability is used to form the refractive index and  should be additive provided that in forming the bond a strong absorption band is not brought into the visible part of the spectrum.  This is just what happens when a double bond or aromatic is formed so in these cases some extra empirical adjustment is needed.


Is the relative natural abundance of isotopes of an element the same everywhere?

The relative natural abundance of isotopes is not the same everywhere.  Depending upon what you mean by "everywhere", there are two cases to consider.

Extraterrestial

Dust from before the sun was formed (stardust, presolar grains) has a very different elemental and isotopic composition than that found on earth.  Depending where a star is in its life cycle different elements and isotopes will be created and destroyed, so it is not surprising that dust from different stars has a very different composition than what is found on earth. As Wikipedia notes,

  In these materials, deviations from "natural abundance" are sometimes
  measured in factors of 100

Terrestial

When our solar system was forming, elements and isotopes were not uniformly distributed - close to uniform, but not exactly uniform.  This is due to processes like diffusion in which a mass-dependent fractionation occurs.  A heavier isotope won't travel as far as a lighter isotope in a given amount of time, so the resulting distribution of the two isotopes will not be exactly equal.  While these mass-dependent processes account for most of the isotopic variation in our solar system, mass-independent processes can play a smaller role.  It is also suggested that catastrophic events in nearby stars may have also influenced local elemental and isotopic homogeneity.  These "original" inhomogeneities will be further altered by a continuation of the mass-dependent (diffusion, bond-breaking and making, etc.) and -independent processes. Natural radioactive decay processes can also lead to changes in local isotope ratios.  For example, $\ce{^238U}$ ultimately decays to $\ce{^206Pb}$, so it wouldn't be surprising to find an altered lead isotope ratio (e.g. enriched in $\ce{^206Pb}$) around large uranium deposits.

This paper presents a study of the natural isotopic variation found on earth.  Here is a table from the paper that gives an idea of the range in variation.  As expected it is much smaller than the extraterrestial variation, but certainly not insignificant.  For boron and copper, the isotopic variation is on the order of parts per hundred, but more typically isotopic variation is on the order of parts per thousand.


You don't have enough liquid in the recipe. Double the lemon and water to 4 tablespoons each, and you should have better luck. Adding more liquid will bring the sugar/liquid ratio more in line with what is needed to make the paste pliable. Add more liquid if you still have issues.


What is the use to know that a compound optically active or not?

Yes, it is important.

For reasonably enantiopure chiral molecules, it is expected to record and publish a specific rotation when it is first described. For the person recording these, they are a little bit annoying, especially if you have a polarimeter as old as the one I had to work with during my PhD. Therefore, you want to cut time as much as possible. By immediately determining that a compound is achiral, you eliminate the need to record a specific rotation because theory predicts it is zero.

Conversely, if you were to simply record all specific rotations of your compounds and then just remove the ones that end up as zero you would likely miss the chiral molecules you synthesised as a racemate. If, however, you take a look at the molecule on paper, are able to determine it should be chiral—given the methods and chemicals employed—but still find a specific rotation of zero you have something that must be discussed in the final paper or thesis.


Preliminaries: I am using the wrong (but still common) notation of the dipole moment. Please see the question about the direction of the dipole moment.

The reason why carbon monoxide is often referred to a being practically nonpolar is because of its very small dipole moment.
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &&
\text{Direction:}~\longleftarrow\hspace{-1ex}+
\end{align}
Unfortunately the explanation of the bonding in carbon monoxide is far from simple, and it is perfectly understandable that you are confused. The dipole moment can be very well explained with molecular orbital theory, which I have already written about.
I will include a brief explanation at this point as I saw that the explanation on Wikipedia is from my point of view incomplete (not to say wrong).
Let's first state the obvious: Oxygen is more electro-negative than carbon. Therefore one can expect that all bonding orbitals are polarised towards the oxygen. That is of course only true when we are staying in the Lewis bonding picture. Based on this one would expect that the lone pairs cancel each other and since all orbitals of the triple bond are polarised to the oxygen, the dipole moment should also point in that direction.
This is obviously wrong, since the experiment tells us something different. There are ten valence electron in this molecule. They all occupy orbitals with bonding character, i.e. there is no nodal plane perpendicular to the bonding axis between the elements. The low lying orbitals are all polarised towards oxygen, because of electronegativity. The HOMO on the other hand is strongly polarised towards the carbon and is very large. In addition the nuclei contribute a dipole moment because of their asymmetry of charge.  With less positive charge on the carbon nucleus than on the oxygen nucleus, the dipole component of the nuclear charge contribution also has its negative end on carbon.  The combination of HOMO and nuclear dipoles in the reverse direction is larger than the combination of all the other occupied orbitals.

It is important to understand that it is often not possible to know about the properties of a compound simply by looking at a structure on paper. Very often the simple models we love and use break down easily for small molecules. A bonding situation has to be explained as is and not in terms of what would have been when we look at individual atoms and how does it change.


Burning a diamond produces how much energy?

Well it looks for me, the question was addressed by a non-Chemist.  Hence my attempt to answer this accordingly.

You have to know the reaction of the combustion, the starting materials, the product(s), and their relative quantities to balance the reaction equation.  Your case is described by $\ce{C} + \ce{O2} \rightarrow \ce{CO2}$, with "$\ce{C}$" for carbon (here: in form of diamond), $\ce{O2}$ for oxygen and $\ce{CO2}$ to yield the product, carbon dioxide. And the reaction equation is already balanced, i.e. the number of atoms per type on the left hand side (prior the arrow) equals the number of atoms per type written on the right, too.

There is a heat called standard enthalpy of formation, describing how much heat is necessary or liberated to form a compound if one (even theoretically) were to prepare this out of elements.  Normally, you look up these tabulated data for all your products and sum this up, and subtract the sum of all these enthalpies of formation of your starting materials.  In your example, by convention elementary gases (like oxygen, $\ce{O2}$) were attributed a zero value.  By means of thermodynamic, diamond is not the most stable form of carbon, and was attributed a value of ${+1.88}{\,\mbox{kJ/mol}}$ (same reference).  Carbon dioxide was attributed a value of ${-393.5}{\,\mbox{kJ/mol}}$ (reference) and the energy balance of the reaction equation equals to ${-391.62}{\,\mbox{kJ/mol}}$, i.e. if you would burn 1 mole of diamond.  (This is a quantity like dozen, used in chemistry.)

Now if we express the quantity of diamond actually burned (mass $m$ of 1 ct) in this above mentioned unity of mole, we need a proportional factor, the molar mass $M$, that happens to be $12{\,\mbox{g/mol}}$ for carbon (reference).
Now we have the pieces.  Hence we calculate the "chemical quantity" ($n$) of carbon deployed as

$n = \frac{m}{M} = \frac{0.2\,\mbox{g}}{12\,\mbox{g/mol}} = 0.017\,\mbox{mol}$
and than multiply this with the above calculated heat of the reaction, i.e
$\mbox{heat} = 0.017\,\mbox{mol} \times (-391.62\,\mbox{kJ/mol}) = -6.527\,\mbox{kJ}.$
So if you burn 1 ct of diamond in oxygen at 25 degree Celsius (about 77 Fahrenheit) and would collect the heat of the product by cooling the $\ce{CO2}$ to the same 25 degrees Celsius, this offers you 6527 Joules (or about 1559 calories).
Notes: The heat calculated here is the reaction enthalpy. The minus sign indicates energy is "liberated" in the course of this reaction. Actually, if all this heat were completely converted into electrical energy, you were able to switch on a light bulb of 60 W for less than 2 minutes (only 108 seconds).  In the beginning of modern chemistry, diamonds actually were burned by concentrated solar power, for example by Lavoisier (here) although to prove their composition.


Regioselectivity of an amination reaction of 2,6-dichloro-3-nitropyridine

I'd agree that nucleophilic substitution at C-6, leaving the nitro and piperazine substituents para to each other, would lead to the thermodynamic product.

The observed regioselectivity is therefore probably kinetically controlled. Although the nitro group activates both ortho (C-2) and para (C-6) positions towards nucleophilic substitution via resonance, it's worth bearing in mind that the nitro group is also extremely electron-withdrawing via the inductive effect in its own right.

This would make C-2 more electron-deficient and hence prone to nucleophilic attack.

Recall that in an SNAr reaction the nucleophile must attack the π* orbitals of the arene, which are above and below the plane of the ring. In this case, the nitro substituent is also (for the most part) in the plane of the ring, and doesn't lie along the trajectory of nucleophilic attack. So, the nitro substituent doesn't present all that much steric hindrance to the nucleophile.


How to find the temperature relationship between the isotherms in a compressibility factor (Z) vs pressure graph?

The compression factor $Z$ is defined as the ratio of the real molar volume with respect to the theoretical ideal

$$Z = \frac{V_m}{V_m^{ideal}}$$

where if $Z$ = 1, the gas behaves ideally.

If we want to get some sort of understanding for how this function behaves when you change the Pressures and Temperatures of your system, we need to substitute into this equation our variables of interest. To take the simplest approach, lets just use the ideal gas law as a substitute for $V_m$ (from a quantitative sense this doesn't make sense to do but qualitatively we can still derive some relationships therefrom)
$$Z = \frac{RTP^{-1}}{V_m^{ideal}}$$
What this shows us is that $Z \propto T$. Thus, an increase in $T$ demonstrates an increase in $Z$, which is the qualitative understanding we're looking for herein.

To note, using a more realistic model for $V_m$ gets complicated; for example if you wanted to use the Van der Waals equation
$$P = \frac{RT}{V_m-b} - \frac{a}{V_m^2}$$
instead of the ideal gas equation, the $V_m$ term doesn't reduce too nicely because of that $\frac{a}{V_m^2}$ term. Gets even worse for the Virial equations.


Inside nanobubbles would probably be "a high-pressure gas". Whether it be an inert gas or vaporized surrounding liquid depends on the temperature, pressure, components, and many other experimental variables.
Regarding theory on nanobubbles, and starting from a classical, macro-scale standpoint, the Laplace pressure describes the pressure inside a bubble due to surface tension on the curved vapor-liquid surface. The pressure will be necessarily higher inside the bubble in order that it doesn't collapse. The back-of-the-envelope calculation on that Wikipedia page indicates around 10 bar for a 300 nm bubble in water.
For smaller nanobubbles, things get more interesting. The surface tension is not constant with curvature and at the extreme curvatures necessary for nanobubbles this will change things. A first-order approximation to surface tension corrections is the Tolman length. Also, at the higher range of pressures involved gases would not be ideal, so some equation of state is likely necessary to describe the interior phase. As things get even smaller, then you get into the details where the continuum approximations like surface tension break down entirely.
Song et al. comes to my mind of a paper where they look at nanobubbles in an actual experiment. I'm sure there are many other experiments in this area.


Can heteroatoms with lone pairs be chiral centres?

Generally, amine nitrogens will not behave like a normal asymmetric carbon. Simple amines are roughly $\mathrm{sp^3}$ hybridiized and the molecules you use as examples do have 4 (we include the lone pair of electrons as a substituent) different substituents around the central nitrogen atom. So in principle me might consider it asymmetric or chiral.  But simple amines can undergo a process called nitrogen inversion, which essentially coverts one enantiomer into the other.

(image source)

However, if one can find a way to slow down or eliminate the nitrogen inversion process, then chiral amines can be isolated. One way to achieve this is to incorporate the amine nitrogen into a 3-membered ring (an aziridine). Now to achieve the planar state necessary for nitrogen inversion requires the bond angle in the 3-membered ring to open from 60° to 120°, an impossible task. Aziridines containing a chiral nitrogen atom have been isolated and characterized.


Why don't water molecules react alongside aqueous reactions?

why don't some of the water molecules present in the aqueous solution of $\ce{CuSO4}$ also react with the sodium (Na) to form sodium hydroxide and hydrogen ($\ce{NaOH + H2}$)

I agree that sodium metal would react with the water.  Where/how did you come up with the equation implying otherwise?

I think you'll enjoy this video: https://www.youtube.com/watch?v=X5yLtqlopTM


Can a charge smaller than that of an electron exist? If so how?

Dipole moments have a distance times an implicit charge, but you could give the moment as unit charge separation times a distance which is not the bond length. This would eliminate fractional charges and suggest some distance, which suggests that an electron and the nucleus are point charges separated by something less than the bond length. This is difficult to visualize. The electron is not really a point charge. Dipole moment is not separable into fractional charges or tiny distances, it measures a distortion of charge.
And when you think about it, an undistorted charge distribution (like electron density in benzene) is still "distorted" in that electron density is between atoms and above/below the atomic plane.
Other fractional charges might be imagined in compounds like magnetite: $\ce{Fe3O4}$. Each iron has a charge of $\pu{2.666}$. But if you describe magnetite as $\ce{Fe2O3.FeO}$, then two thirds of the iron atoms have a charge of +3, and one third have a charge of +2. More satisfying.
And then there is cyclopentadienyl anion:

where each carbon has an extra 0.2 extra electronic charge, and there are other ions like cycloheptatrienyl cation, cyclopropenium cation, etc. that have a surplus or deficiency of electrons.
But there is no fractional electric charge, just fractional molecules, or combined valences. Although, I still wonder how quarks can have a partial charge!


Standard reduction potential of MnO4-/MnO2 couple

The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.

Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations

$$\begin{align}
\ce{MnO4- + 8H+ + 5e-} &\ce{-> Mn^2+ + 4H2O} & \Delta G_1^\circ &= -728.46~\mathrm{kJ~mol^{-1}} \tag{1} \\
\ce{MnO2 + 4H+ + 2e-} &\ce{-> Mn^2+ + 2H2O} & \Delta G_2^\circ &= -237.35~\mathrm{kJ~mol^{-1}} \tag{2}
\end{align}$$

The trick is to find the appropriate combination of equations $(1)$ and $(2)$ that give the equation $(3)$, which you are interested in:

$$\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O} \qquad \qquad \Delta G_3^\circ = \,\,? \qquad \qquad \tag{3}$$

If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:

$$\begin{align}
\Delta G_3^\circ &= \Delta G_1^\circ - \Delta G_2^\circ \\
&= -728.46~\mathrm{kJ~mol^{-1}} + 237.35~\mathrm{kJ~mol^{-1}} \\
&= 491.11~\mathrm{kJ~mol^{-1}} \\
E_3^\circ &= \frac{-\Delta G_3^\circ}{3F} \\
&= +1.697~\mathrm{V}
\end{align}$$

If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.


Calculate the number of Carbon atoms in .0400g C7H5NO3S

.0400g is a really non-chemical way to write that number.

I expect the text meant 0.400g, which is 10 times higher than the number actually printed, and this would explain why the book's answer is 10 times higher than the actual answer: they are working from the number they didn't print.


How to perform IR spectroscopy on silica particles at various relative humidities when they are passed through a tube using pressurized air?

Despite your question is a bit vague about the sample, I assume «when they are in motion» refers to particles passing a tube, are close to each other -- in contrast to particles floating in air.

If so, than you should consider to couple infrared spectroscopy (about the energy band) with a probe based on attenuated total reflection.  Roughly speaking, classical optics predicts you a threshold incident angle where light of a certain frequency no longer is able to leave an optical dense material (high refractive index, for IR for example Silicon, Germanium, or a small diamond) to enter a less dense material (for IR, for example air, most of organic materials).  Recognizing light equally has character of a wave however «allows» that light at this interface still interacts sufficiently enough that one may extract information to calculate an IR spectrum.

What you see in the lab as an ATR-FT-IR spectrometer typically is a gate-like construction

(source)

but there are ATR-IR probes which you may immerse into fluids to monitor the advancement of reactions (e.g., Mettler's reactir)

(source)

where the probe window is at the side or on the tip of the probe itself:

(source)

Potentially -- compared to classical transmission IR -- you truncate any sample preparation (no KBr pill, nor nujol) and may record data of not-too wet samples, too.  Don't forget an ATR correction, as the small penetration depth $d_p$ of light across the phase boundery between window material and sample depends both on $n_c$ (window / crystal material), $n_s$ (sample), the incident angle $\theta$ and wavelength $\lambda$ applied.

$$ d_p = \frac{\lambda}{2\pi n_c \sqrt{\sin^2 \theta - \left(\frac{n_s}{n_c}\right)^2 }}$$


Put some potassium nitrate $\ce{KNO3}$ plus powdered charcoal and sulfur in a mortar. Wet it carefully and grind the whole until you get a black homogeneous paste. Let it dry overnight in the mortar. The obtained black gun powder may be detached from the mortar with a wooden spoon. It is a safe rocket fuel. The proportions of the powders are defined by stoichiometry. They must allow the chemical reaction : $$\ce{2 KNO3 + 3 C + S ->  K2S + 3 CO2 + N2}$$
I have done it many times with my students. It's a good exercice of stoichiometry. Never had an accident ! And remember : Never ! Never grind dry powders ! Grind as wet powders as possible ! In case of doubt add more water ! Pastes are even a better choice for grinding purposes !


Why isn't cyclopropane considered unsaturated?

In short, there are two definitions with separate purpose.

IUPAC defines alkanes as

  "Acyclic branched or unbranched hydrocarbons having the general formula $\ce{C_nH_{2n+2}}$, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms."

and notes notes about cycloalkanes, contrasting e.g., to unsaturated cycloalkenes:

  "Saturated monocyclic hydrocarbons (with or without side chains), e.g. cyclobutane. Unsaturated monocyclic hydrocarbons having one endocyclic double or one triple bond are called cycloalkenes and cycloalkynes, respectively. Those having more than one such multiple bond are cycloalkadienes, cycloalkatrienes, etc. The inclusive terms for any cyclic hydrocarbons having any number of such multiple bonds are cyclic olefins or cyclic acetylenes."

Thus, as in the earlier comment by @MaxW, cyclopropane equally is a saturated alkane because each carbon atom is connected with the next carbon atom only by a $\ce{C-C}$ single bond.  Or, as MSU condenses,

  "Alkanes and cycloalkanes are termed saturated, because they incorporate the maximum number of hydrogens possible without breaking any carbon-carbon bonds."

This set of definitions is not the one serving to determine organic structures from spectra, e.g. in Field's training book, where

  "[...] the molecular formular is reduced to $\ce{C_nH_m}$ and the degree of unsaturation is given by $$\mbox{Degree of Unsaturation} = n - \frac{m}{2} + 1$$ The degree of unsaturation indicates the number of $\pi$ bonds or rings that the compound contains."

(4th edition, p. 4; emphasis added by mine.)


Why aren't the amine groups in the nucleic acid bases protonated?

There are 2 answers:

As f' answered, the amino (NH2) group in a nucleobase is conjugated
with the heterocycle. Therefore, its pKa is closer to the one of
aniline. In other words, it is much less basic than an alkylamine.
Any drawing of a molecule in chemistry or biology is only a mere representation (agreed on by scientists) of the reality. For example, amino-acids are most usually drawn as H2N-R-COOH. We all know that it does not reflect reality but we all use that representation because it is much more readable than +H3N-R-COO- (which itself is also wrong), and because the real form of the amino acid will greatly depend on the experiment conditions. So you must always consider in what medium the molecule is to determine, more or less, its major form.


Identify the structure of the compound (C3H5NO) given its NMR chemical shifts

Dhruba Charan Dash published the problem in his analytical chemistry textbook (ISBN 9788120353008). This is the picture that goes with the question:

Notice that the scale of the chemical shift increases with the strength of the magnetic field $B$. This means they are using the $\tau$ scale instead of the $\delta$ scale. The modern description of the signals would be $\delta = 10 - \tau$, where $\tau$ is the value given in the figure.
So we have a singlet at 4.30 ppm (B = broad?) and triplets at 3.83 ppm (typo in the figure, should be 6.17 instead of 5.17) and 2.63 ppm.
$$\delta\ 2.63 \text{ (triplet, 2 H)}$$
$$\delta\ 3.83 \text{ (triplet, 2 H)}$$
$$\delta\ 4.30 \text{ (broad, 1 H)}$$
For completeness, I will post the answer given by the author Dhruba Charan Dash:

PROBLEM 13.17 PMR spectrum of the compound $\ce{C3H5ON}$ in an acidic medium is shown below. Suggest its structure. Given that the peak 1 disappears in the presence of $\ce{D2O}$.
Solution Analysis of the spectrum:
(i) Since the peak 1 disappears in the presence of $\ce{D2O}$, it may represent a group containing active hydrogen. This group may be a hydroxyl group.
(ii) The peaks 2 and 3 must represent the hydrogen atoms attached to the methylene group.
(iii) The analysis of integral trace indicates the presence of 1, 2, and 2 protons within the peaks 1, 2 and 3 respectively.
The point (iii) suggests the presence of two methylene groups. These must be attached each other (point (ii)). The only structure consistent with the above facts is $$\ce{HOCH2CH2CN}$$

It is clear why the question posed by the OP (and by the teacher of the OP, as I learned in chat) was hard to solve. A single problem transcribing the question (replacing $\tau$ with $\delta$) obfuscated the intended chemical shift values.
For reference, here is the problem as presented to the OP:


Shouldn't this compound be called 2-methyl-5-(1,2-dimethylpropyl)nonane?

‘5-(1,2-Dimethylpropyl)-2-methylnonane’ is not the preferred IUPAC name. The preferred name according to current IUPAC recommendations is 2-methyl-5-(3-methylbutan-2-yl)nonane.
If this is a homework exercise, however, it might be based on obsolete IUPAC recommendations from 1979 or 1993.
You are partly correct; simple prefixes are arranged alphabetically disregarding any multiplicative prefixes. The multiplicative prefixes are inserted later and do not alter the alphabetical order. For example, ‘1,2-dibromo-’ is considered to begin with ‘b’.
However, the name of a compound substituent is considered to begin with the first letter of its complete name. For example, ‘1,2-dibromobutyl-’ is considered to begin with ‘d’.
Likewise, ‘1,2-dimethylpropyl-’ is considered to begin with ‘d’. Therefore, it is arranged before ‘methyl’.


Hydrophilic substances are not necessarily water soluble, and the two terms are not synonymous.

On a molecular scale, "hydrophilic" is defined by the IUPAC Gold Book$^1$ as:

  'Water loving'. The capacity of a molecular entity or of a substituent to interact with polar solvents, in particular with water, or with other polar groups.

Note that the term hydrophilic does not necessarily refer to an entire molecule, and that interacting strongly with a molecule or a part of a molecule does not imply that it must be soluble in water.

Also, entire macroscopic surfaces can be said to be hydrophilic, meaning that they interact strongly with water but certainly are not soluble in water. In this context, the degree of hydrophilicity is measured by the contact angle, which can be simply stated as the angle formed when a droplet of water is placed on a surface, where a small angle (as seen from inside the droplet) means that the droplet has spread out over the surface, and that the surface is therefore hydrophilic.

So, although small, hydrophilic molecules are frequently water soluble, solubility plays no part in the definition of "hydrophilic" on any scale or context.

1) IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8


Electrode potential for metal-metal ion and metal-metal insoluble salt-salt ion electrode

My question is that $E^\circ$ is just the potential at some given standard condition, i.e. 1 bar and 25 °C....

No, $E^\circ$ is the potential at the given standard condition 25 °C, 1 ATM and activities of involved compounds equal to 1.

This may be the source of your confusion.

Standard potentials are for the unit activities. In my example, once for silver ions, once for chloride ions.

The activity of $\ce{Ag+}$ in 1M $\ce{AgNO3}$ Is very different to activity of $\ce{Ag+}$ in 1 M $\ce{KCl}$ with the $\ce{AgCl}$ precipitate.

You cannot find pressure in the Nernst equation. The pressure plays only very marginal role here, like being big enough to avoid water boiling at low pressure.

Usage of each of the standard potentials implies usage of activity of different ions to compute the final potential, or concentations as approximations.

One is using the metal cation activity, the other the anion activity.

One is using the metal cation activity directly, the other indirectly via the salt solubility product.

$$\begin{align}
E_\mathrm{Ag/Ag+}&=E^{\circ}_\mathrm{Ag/Ag+}+0.059\log{\left(a_\mathrm{Ag+}\right)}\\
&=E^{\circ}_\mathrm{Ag/Ag+}+0.059\left(\log{\left(K_\mathrm{s,AgCl}\right)}
-\log{\left(a_\mathrm{Cl^-}\right)}\right)\\
&=E^{\circ}_\mathrm{Ag/AgCl/Cl^-}
-0.059\log{\left(a_\mathrm{Cl^-}\right)}\\
\end{align}$$


How much energy is stored when H+ is converted to H2?

I'm seeing that The energy of formation of H+H+ is 0 J/mol0 J/mol (it can be treated as a free proton)

No. Nope. Noway.
Formation of proton from hydrogen molecule is extremely endotermic process. It requires to 1) dissociate H-H bond, which is, btw, 432 kJ/mol, and then ionization of hydrogen atom, which is around 1300 kJ/mol.

However. In solution hydrogen ion is bound to water molecules (or other solvent), which results in serious energy discount, around 1100 kJ/mol for the case of water.

However again :) this is of no significance here, since the dive into small details is not required here.

The gross process is $\ce{ Me2CHOH \space =\space  Me2CO + H2 }$, and since entalphy of formation of hydrogen is roughly zero, only change from isopropanol to acetone is relevant. Hence, the difference is 68.8 kJ/mol.


Why any non polar compounds dissolve in nonpolar substance like Benzene what is the reason behind it?

The simple answer is entropy. The explanation not so much.  If you consider solvation (a specific type of dissolving) to really be a melting process you can consider the resulting solution a eutectic/peritectic mixture.

To explain Figure 1 is the phase diagram of tin-lead compounds.  You could consider tin to be soluble in lead since at $300^\circ C$ lead is liquid and tin is solid but when tin is added to lead, it dissolves.  Or you could consider the mixture of the two to be a eutectic that lowers the finally melting temperature of both substances to $183^\circ C$ (at 61.9 wt% lead) which is commonly the case.  In fact if you were to make a phase diagram for any solute and solvent you would see similar behaviour.

Now sometimes the solvent forms a solid complex such as $\ce{CaCl2 *6H2O}$ which casuses a discontinuity. in Figure 2, point A is a eutectic point but points B and C are peritectic which is a result of the solid phase having a different equillibrium complex ($\ce{CaCl2*6H2O}$ vs. $\ce{CaCl2*4H2O}$.

That was some concept now for some math.  The entropy of mixing can be determined for a solution by: $\Delta S_{mix} = -R[x_a ln(x_a) + x_b ln(x_b)] $ where $x_a$ and $x_b$ are the molar ratios of the components  ($x_a = 1 - x_b$). Since $\Delta S_{mix}$ is always positive, $\Delta G_{mix}$ will always be negative as long as the heat of mixing ($\Delta H_{mix}$) is not too positive.

Figure 1: Eutectic Mixture Example:

Figure 2: Peritectic Example:


Heme: The Iron molecule is bound to 4 Nitrogen atoms, but take a look at porphyrin. All the nitrogens are part of one actual structure. You can visualize the iron molecule in the center of the larger compound's structure, replacing the two Hydrogen atoms in N-H:
The large structure surrounding and coordinating with Fe is also known as a polydentate ligand since it bonds to Iron at more than one site. So there is just one ligand on Fe.

Cytochrome C: is a heme protein, so you may expect the same answer: A single polydentate ligand binds to Iron. This page has a PDB Model of it in a plugin.
As one ligand coordinates to four sites in this case, other ligands typically attach (e.g. oxygen binding in hemoglobin) to fill the octahedral group (up to 6 ligands attached to Fe the coordinate center).


Change in H+ ion concentration of water on boiling

It is an interesting question, however there is a misconception. You are giving too much attention to the word neutral. The word neutral is just a qualitative version. Numbers and equations are more accurate way to discuss science.

You have to see how water's $\pu{pH}$ at 100 °C is calculated. How do you get those numbers? Lets us do it algebraically, so that the word neutral does not matter.

Autoionization of water at temperature T:$\ce{ H2O <=> H+  + OH-}$, if the conc. of $\ce{H+}$ is x, the quantity of $\ce{OH-}$ is also x, if you have accepted the ionic product of water is

$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = x^2=
51.3\times{10^{-14}}$

Solve for $x = [\ce{H+}] => -log [\ce{H+}] = 6.14$

So the numbers prove that $\pu{pH}$ has changed. Water has become slightly acidic. If you associate the word neutral = 7.0 at 25 °C. However, if you use the word neutral with respect to water temperature, then $\pu{pH}$ of 6.14 is neutral at 100 oC. Ultimately, the numbers Kw and pH matter, nobody will quibble about the word neutral in serious research.

Water is the most amazing solvent in Nature. People are still exploring its properties and it is yet to be fully understood. For example, I just heard in a conference that someone was trying to understand what happens on the surface of micron sized water droplets. "Pure" water showed very odd chemical effects which have never been shown before.

Read this well written Wikipedia article: https://en.wikipedia.org/wiki/PH#cite_note-2


How to determine which of the given redox reactions will occur?

Both 3 and 4 are impossible. $\ce{CO2}$ does not react with the $\ce{KMnO4}$, because $\ce{KMnO4}$ is strongly oxidizing. $\ce{C}$ is in the highest oxidation state and can not be oxidized, hence, it is impossible to react.


Why is the melting point of p-dichlorobenzene higher than those of o-dichlorobenzene and m-dichlorobenzene?

Generally, the melting point of para-isomer is quite higher than that of ortho- or meta-isomers. This is due to the fact that it has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. As a result, intermolecular forces of attraction are stronger and therefore, greater energy is required to break its lattice and its melts at higher temperature. This question is most likely to be answered easily.
If you consider solubility. These are insoluble in water but soluble in organic solvents. Generally, para-isomers are highly soluble in organic solvents than ortho-isomers. This is interesting question.
Every compound (more specifically solids) does not dissolve in a given compound. In general, a liquid dissolves in a liquid if the intermolecular interactions are similar. This is in accordance with the rule "like dissolves like," ionic or polar compounds dissolve more readily in polar solvents. Non-polar compounds (covalent or organic) are soluble in non-polar solvents.
If you consider p-dichlorobenzene, it has zero dipole moment and thus is more non-polar than o-dichlorobenzene which has dipole moment of about $\pu{2.54 D}$. So, non-polar p-dichlorobenzene dissolves more readily than o-dichlorobenzene in non-polar (organic) solvents.


What is the rate expression for decomposition of $\ce{H2O2}$ in the presence of a catalyst?

For catalase, it is most likely Michaelis-Menten kinetics.
At typically low enzyme concentration, this would be first order in catalase. For hydrogen peroxide, it is first order at low concentrations (<< Km) and approaches zero order at high concentration (>> Km). This behavior is also called saturation kinetics.

I don't know what it is for the catalysts $\ce{MnO2}$ and $\ce{Fe2O3}$. Another question in my mind is that even if it's similar to that of catalase, since these are powdered and so do not have a concentration, how can I express it?

For catalysts in the solid phase, it is the surface area (compared to the volume of the solution) that matters. You might see similar saturation kinetics, indicating that at some concentration of reactant, all the surface area is used so that an increase in reactant concentration no longer translates into an increase in rate.


Do strong acids actually dissociate completely?

Yes, for sufficiently strong acids, in sufficiently dilute conditions, in sufficiently basic solvents. However, things are hazier than you might expect, and depending on your definition, "clear-cut" examples of complete dissociation only become possible when you start swatting flies with nuclear bombs.
The problem is that chemistry works on equilibria, and at some point it becomes difficult to measure tiny concentrations - in a haystack of one billion straws, how can you be sure there isn't at least one needle? And in one trillion? And what about in a septillion?

First, let us perform a qualitative investigation by considering the generic acid dissociation equilibrium in water, as follows:
$$\ce{HA(aq)->H+(aq) + A-(aq)} \quad \quad \quad \mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}}=10^{-pK_a}}$$
To simplify the following discussion, we will make the very crude approximation where we equate the thermodynamic activity of a species with its molar concentration. In other words:
$$\mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}} \approx \frac{[H^+(aq)][A^-(aq)]}{[HA(aq)]}}$$
Our objective now is to virtually eliminate the presence of undissociated $\ce{HA}$ molecules. For clarity, we rearrange the above equation:
$$\mathrm{[HA(aq)] \approx \frac{[H^+(aq)][A^-(aq)]}{K_a}}$$
Clearly, to make $\ce{[HA]}$ as small as possible, the numerator must be very small, and the denominator must be very large. Let us set a goal of making $\mathrm{[HA]<10^{-27}\ mol\ L^{-1}}$, which implies less than $\mathrm{0.001}$ molecules of undissociated $\ce{HA}$ in a litre of solution.
At this point, I should mention that aqueous solutions of acids in ambient conditions must always have $\mathrm{[H^+(aq)]>10^{-7}\ mol\ L^{-1}}$, thanks to the self-ionization of water. We take the best-case scenario (which physically corresponds to very dilute solutions):
$$\mathrm{10^{-27} \gtrsim \frac{10^{-7}\ [A^-(aq)]}{K_a} \quad \longrightarrow \quad 10^{-20} \gtrsim \frac{[A^-(aq)]}{K_a}}$$
We now take a typical strong acid, hydrogen chloride ($\ce{HCl}$) which has an estimated pKa of -5.9 in water. In other words:
$$\mathrm{10^{-20} \gtrsim \frac{[Cl^-(aq)]}{10^{5.9}}=\frac{[Cl^-(aq)]}{790\ 000} \quad \longrightarrow \quad [Cl^-(aq)] \lesssim 8\times 10^{-15}\ mol\ L^{-1}}$$
We see therefore that meeting the criterion of full dissociation requires preparing an aqueous $\ce{HCl}$ solution with femtomolar concentration. Essentially every solution used in a chemistry laboratory will have a higher concentration, so every $\ce{HCl}$ solution anyone has ever touched will have undissociated $\ce{HCl}$ molecules. Strictly speaking, you could say the dissociation is never complete. However, if you have a mind to make a fully dissociated solution of $\ce{HCl}$, it is physically possible.

Just diluting acid solutions into oblivion until every molecule dissociates is kind of boring. The more exciting prospect is, of course, to increase the strength of the acid. The acidity scale goes far, far beyond aqueous solutions of $\ce{HCl}$.
Interestingly, for sufficiently strong acids (typically $\mathrm{pK_a<-10}$, meaning acids at least 10 000 times stronger than aqueous $\ce{HCl}$), it is possible to isolate hydronium salts - when the strong acid is mixed with water in a 1:1 molecular ratio, a solid salt is formed, where the cation is $\ce{H3O+}$ and the anion is supplied by the conjugate base of the acid. The stoichiometry can be made exact. Surely this represents full dissociation of the acid?
Well, again, it depends on your strictness. If you actually look at the crystal structures of many of these salts, it's clear that there is a strong, close  interaction between the $\ce{H3O+}$ cation and the anion, and you could argue this still does not count as complete dissociation.
That said, you can go crazy and consider some of the strongest acids we can put in a bottle, such as carborane superacids. The corresponding hydronium salt of these superacids can have extremely weak interactions between the hydronium cation and the anion. For example, when the hydronium salt $\ce{(H3O+)(CHB11Cl11^-)}$ is prepared in a benzene solution, the solid that forms actually contains three molecules of benzene surrounding the hydronium cation, which is completely detached from the anion. If this doesn't imply "complete dissociation" of the carborane superacid in water, I don't know what would.
The acidity scale goes further still than even carborane superacids. There are a few known acids which will practically always fully dissociate, no matter what. The representatives of this class are rather exotic things such as $\ce{H_3^+}$, $\ce{HN2^+}$, $\ce{HNe^+}$, $\ce{HHe^+}$, and so forth. These acids can only be detected in the gas phase, as they fully dissociate by protonating virtually anything they come into contact with (including any anion, which is why they cannot be isolated as a neutral compound).

The last way to ensure full dissociation of an acid in a solution is to change the solvent in which it is dissolved. All things considered, water is a pretty weak base. If, for example, ammonia (am) were used as the solvent, the relevant equilibrium for $\ce{HCl}$ would be:
$$\ce{HCl(am)->H+(am) + Cl-(am)} \quad \quad \quad \mathrm{K_a(am)=\frac{a_{H^+(am)}a_{Cl^-(am)}}{a_{HCl(am)}}=10^{-pK_a(am)}}$$
If I were to give a crude guess, I'd imagine that the $\mathrm{pK_a}$ for $\ce{HCl}$ in liquid ammonia would very roughly be somewhere around -15; that is, dissociation of $\ce{HCl}$ in liquid ammonia is about a billion times easier compared to water. This is because ammonia is a stronger base than water.
You know how we were talking about hydronium salts as if they were some unusual thing? Well, the equivalent concept in ammonia would be the formation of ammonium salts. And indeed, $\ce{HCl}$ does easily form a salt with ammonia in a exact 1:1 molecular ratio, namely ammonium chloride, $\ce{NH4Cl}$. Not exactly an unusual compound.
There's nothing stopping you from using an extremely strong base as a solvent. For example, in pure liquid phosphazene superbases, even very weak acids may be fully dissociated in ambient conditions, even in the most rigorous sense.


Chemical Reaction for making Acetic Acid from Glycerol inside Fermenter

Short answer: under anaerobic conditions, it is impossible to produce acetate as the sole product from glycerol with wild-type E. coli.

Explanation: Under anaerobic conditions (and assuming no substrates for anaerobic respiration have been added), a significant concern is redox balancing. Since electrons are not transferred to oxygen (as they are under aerobic conditions), the number of electrons in the substrates must equal the number of electrons in the products.

Glycerol as a substrate presents problems in this regard, because it is slightly more reduced than typical hexose substrates (eg glucose). Furthermore, the desired product acetic acid is more oxidized than glycerol (or hexoses for that matter).

Looking at the core metabolic pathways of E. coli, we immediately see that 1 molecule of glycerol is converted to 1 molecule of acetate. E coli is unable to utilize all three carbon atoms of glycerol make 3 molecules of acetate from 2 molecules of glycerol. The third carbon atom of glycerol is lost as either formic acid or $\ce{CO2}$ (with concomitant production of hydrogen gas). So we can write an initial stoichiometry of

glycerol ($\ce{C3H8O3}$) + $\ce{H2O ->}$ acetic acid ($\ce{C2H4O2}$) + formic acid ($\ce{CH2O2}$) + 4 electrons + 4 $\ce{H+}$.

[Note that the formic acid can be converted to $\ce{H2}$ and $\ce{CO2}$. The exact ratio of hydrogen to formate will depend on conditions.]

Unfortunately, there is no pathway for disposing of the excess electrons. You would need to add a respiration substrate such as fumarate (converted to succinate) or nitrate or nitrite (both can be reduced to ammonia).

One alternative approach would be to ferment the glycerol to a mixture of ethanol and formic acid or hydrogen gas and then chemically oxidize the ethanol to acetic acid. The balanced stoichiometry of that fermentation is

glycerol ($\ce{C3H8O3}$) $\ce{->}$ ethanol ($\ce{C2H6O}$) + formic acid ($\ce{CH2O2}$)

or

glycerol ($\ce{C3H8O3}$) $\ce{->}$ ethanol ($\ce{C2H6O}$) + $\ce{H2 + CO2}$


Fun!  weak acid versus weak base!
Ammonium cyanide, $\ce{NH4CN}$, is a solid where the atoms are grouped into the same ions that are generated in solution: $\ce{NH4+}$ and $\ce{CN-}$.  Although it can be sublimed with very mild warming, it is fairly unstable. In particular it reacts with atmospheric water vapor, evolving ammonia and HCN.  Dangerous stuff!
The first part of the answer to 1. is the decomposition of $\ce{NH4CN}$ into ions in aqueous solution, which should be obvious; one subtlety is that you may want to add (s) and (aq) next to each species, especially if your professor makes a point of doing so.  So that's the dissociation. The data in the problem don't allow us to compute the equilibrium constant of the dissociation but it's safe to say that it is extremely high, that is, dissociation is complete: there is no $\ce{NH4CN}$ as such in solution.
The acid-base equilibrium equations can be the ones you wrote.  Again, you may want to add more information, like this:
$$\ce{NH4+ (aq) + H2O <=> NH3 (aq) + H3O+ (aq)}\quad K=5.5\times10^{-10}$$
Here the equilibrium constant $K$ is the acid dissociation constant of the acid $\ce{NH4+}$, the protonated form (aka conjugate acid) of ammonia.  I got its value by dividing $K_\mathrm w$, the dissociation constant of water ($1.0\times10^{-14}$), by the $K_\mathrm b$ of ammonia, given in the problem.
I encourage you to add similar information for the  equilibrium involving cyanide ion.  Hint: if you keep the reaction as you wrote it, the appropriate $K$ is not the $K_\mathrm a$ of $\ce{HCN}$ but the $K_\mathrm b$ of the deprotonated form (conjugate base), $\ce{CN-}$.
(But you can instead consider the alternate reaction $\ce{HCN + H2O <=> CN- + H3O+}$, and then the equilibrium constant is the $K_\mathrm a$ of $\ce{HCN}$.)
How about part 2?  The only correct answer to

Will the salt react as an acid or a base?

is "It can do either, depending on what it's reacting with".  But maybe what they mean is whether the solution of the salt is acidic or basic.  If so, one way to find out is to combine the two equilibrium equations and their $K$s.  I don't want to give problem away completely, but think of it this way:

What is the pH of a (say) 1 M solution of $\ce{HCN}$, based on the $K_\mathrm a$ of $\ce{HCN}$?
(I'm assuming the calculation of the pH of a solution of a pure weak acid or a pure weak base has been covered in the course before getting to the question you posted today.)

What is the pH of a 1 M solution of $\ce{NH3}$, based on the $K_\mathrm b$?

Is $\ce{HCN}$ a stronger acid than $\ce{NH3}$ is a base, or is $\ce{NH3}$ a stronger base than $\ce{HCN}$ is an acid?  Based on this, which would you expect to "win out" if you mix the two in equal proportions?

(The point here is that what you get by dissolving $\ce{NH4CN}$ is indistinguishable from what you'd get by mixing equal amounts of equimolar solutions of $\ce{HCN}$ and ammonia.  In other words, if you mix 1 L of a 2 M solution of $\ce{HCN}$ with 1 L of a 2 M solution of ammonia, the result is very nearly  2 L of a 1 M solution of $\ce{NH4CN}$.   Footnote: "very nearly" because volume is not always preserved exactly, unlike mass.)


To test if it's ruby at home check it hardness. If it's harder than 7 even if it's not ruby it still something precious. :-) You can also try to find refraction index however for this you need some equipment. If you really want 100% prove, ask on a local chemistry faculty if they could "as a teaching example" make single crystal diffraction experiment. Simple run with 30 frames would be enough to find a cell what is enough to prove it's ruby.

Fitting your ruby on diffractometer would be challenge as a new diffractometers are working on crystals with size of 10--300 micrometers. To find if it natural or synthetic you need to go to specialize jeweler. They look for imperfections (natural stone are usually less perfect than synthetic).


Why is it advised to put a Styrofoam plate under an Erlenmeyer flask?

You provided the additional information in the comments that your experiment was about measuring the change in temperature that occurs upon mixing. This should give you a hint as to whether the substances mix with water in an endo or exothermic manner.

When measuring temperature changes it is crucial not to influence the temperature of the reaction mixture in any way. Naturally, that would for example mean not to keep the flask in your hand (body temperature is typically more than room temperature). But some error sources are not as easily visible. In your case, the table top was an error source. Whatever the material, it likely conducts heat better than air — meaning that if your flask was on the table a significant portion of the heat generated (or used) would be conducted away (or onto site) by the table resulting in a lower change in temperature. That is why these experiments are typically performed under maximally insulating conditions. One of the best and widely available insulators is polystyrene because of its high air content (air is a good insulator). Hence, keeping polystyrene under a flask reduces the error caused by the table conducting heat.


How does repeatedly freezing and thawing antibiotics decrease their antimicrobial effectiveness?

As indicated in the comments, freeze-thaw cycles should not affect molecules of size of typical antibiotics. Freeeze-thawing may affect colloids by disrupting their structure. Also, the effect would depend on how fast a sample was frozen/thawed etc (slow freezing would cause rearrangement of molecules and crystallization of water whereas rapid freezing would result in vitrification of water.)

The popular lab practice, in my opinion, would have perhaps arisen because it is better to keep the stock stored at the appropriate temperature (frozen) for longer life. Repeated thawing of the stock would definitely increase its chance of degradation because of mishandling and even otherwise. Therefore, it is better to aliquot smaller volumes; even if you have forgotten to keep it back into the freezer you'll end up losing only a small amount.

Not related to your question but this is an interesting fact: freeze-thaw cycles promote molecular assembly of RNA into higher order structures (ref).


How does the solvent determine whether mono- or tribromination of phenol occurs?

In aqueous medium, phenol is deprotonated to a certain extent, forming the phenoxide ion in which the ortho and para positions are even more activated than in phenol itself. Hence, trisubstitution occurs here.

$$\ce{PhOH + H2O <<=> PhO- + H3O+}$$

In an organic solvent, phenol is not deprotonated and monosubstitution tends to occur.

ron also pointed out in a comment that:

  The concentration of $\ce{Br^+}$ is higher in water than in $\ce{CCl4}$ making the reaction faster in water.
  The dielectric constant of water is much greater than that of $\ce{CCl4}$, so the former solvent will better stabilize the charged intermediate, again accelerating the reaction.


@Radi: Critical changes to original post! I will guide you through a reasonable approach. You have to realize that you have not specified the stereochemistry of 11 -- of which there are 16 possible stereoisomers (8 racemates) -- but I will ignore the issue. You will recognize the symmetry in 3,4,7,8-tetramethyldecane (11), so we will look for a coupling reaction of two 7-carbon entities. (E)-2-Buten-1-ol (crotyl alcohol) (4) and triethylorthopropionate 5 are commercially available. The reaction of 4 and 5 is a Johnson-type Claisen rearrangement (Google it!) which produces racemic esters 7 via transition state 6. Reduction of the double bond of unsaturated esters 7 leads to saturated esters 8. Reduction of the ester group to an alcohol followed by conversion to a bromide gives structure 9. The Wurtz reaction (9 --> 11) is rarely used but here is an opportunity.
Alternatively, esters 8 under the same conditions of the Wurtz reaction produces the α-hydroxyketones 10 (acyloin condensation). While the Clemmensen reduction (10 --> 11) of aliphatic ketones does not work as well as with aromatic ketones, under these conditions you would lose both oxygens. If the ketone is produced (structure 10 less the hydroxyl group), a Wolff-Kishner reduction would lead to the tetradecanes (plural) 11.Read about these name reactions which will be of more value to you than your original proposal. Good luck!


Final amount of Hg present after $\beta$ decay

$$\ce{^198Au -> ^198Hg + e^{-}  ~~~~~~~~ \beta~ decay}$$

260 hours corresponds to 4 half-lifes (4 x 65 = 260).  After each half-life, one-half of the $\ce{^198Au}$ will have been transformed into $\ce{^198Hg}$.  After 4 half-lifes ((1/2)^4 = 1/16) 1/16 of the gold will remain and 15/16 of the gold will have been transformed into mercury. Therefor, after 260 hours 15/16 gram atom of mercury will be present.  15/16 gram atom of mercury corresponds to (15/16 x 198 = 185.63 gm) 185.63 gm of mercury.

Using your equation we (of course) get the same result.
$$\ce{N_t=N_o * e^{-kt}~~(I)}$$ where k is the rate constant and t the elapsed time.  For radioactive decay (a first order reaction)
$$\ce{k=\frac{ln(2)}{t_{1/2}}~~(2)}$$ where $\ce{t_{1/2}}$ is the decay half-life.  Substituting into equation (I) we find
$$\ce{\frac{N_t}{N_o}= e^{-\frac{ln(2)}{t_{1/2}}*4}=1/16}$$


Is there a method to separate iron from mercury?

As iron is one of few metals not forming amalgams,

  Almost all metals can form amalgams with mercury, the notable exceptions being iron, platinum, tungsten, and tantalum.

..it should preferably contaminate the mercury surface.

I would try, perhaps repeated,  dropping mercury, e.g via punched filtration paper to sulphuric acid bath.

It can be combined with  agitating mercury - acid phase border.

Important can be the origin of impurity and the intended purpose of purified mercury. There is possible the only reasonable way would be distillation, mentioned in the other answer, respecting the warning.


Internal energy change during phase change

You are considering the $\Delta H$ and the $\Delta U$ between the following two thermodynamic equilibrium states:

State 1:  1 mole of ice at 0 C and 1 atm.

State 2:  1 mole of liquid water at 0 C and 1 atm.

The relationship between $\Delta H$ and $\Delta U$ at constant pressure is:
$$\Delta H=\Delta U + p\Delta V$$where V is molar volume.  What is the molar volume of ice?  What is the molar volume of liquid water?  What is $\Delta V$?


What do the prefixes meta, ortho, pyro mean in inorganic chemistry?

In the older literature, the prefixes “hypo-”, “per-”, “ortho-”, “meta-”, and “pyro-” were used in some cases for the distinction between different acids with the same characteristic element.
The prefix “hypo-” was used to denote a lower oxidation state, and the  prefix “per-” was used to designate a higher oxidation state. (The prefix “per-” is not be confused with the prefix “peroxo-”.)

$\ce{HClO}$ hypochlorous acid
$\ce{HClO2}$ chlorous acid
$\ce{HClO3}$ chloric acid
$\ce{HClO4}$ perchloric acid

The prefixes “ortho-” and “meta-” were used to distinguish acids differing in the formal content of water.

$\ce{H3BO3}$ orthoboric acid
$\ce{(HBO2)$_n$}$ metaboric acid
$\ce{H4SiO4}$ orthosilicic acid
$\ce{(H2SiO3)$_n$}$ metasilicic acid

The prefix “pyro-” was used to designate an acid that is formally formed by removing one molecule of water from two molecules of an ortho-acid.

$\ce{H4P2O7}$ pyrophosphoric acid $(\ce{2 H3PO4 -> H4P2O7 + H2O})$.

Various traditional names are retained for use in IUPAC nomenclature, though the number of retained names has been reduced with each succeeding edition of the IUPAC recommendations. The Second Edition (1970 Rules) of the IUPAC Nomenclature of Inorganic Chemistry (published in 1971) still included all of the above-mentioned prefixes; however, the approved use was limited to only some selected retained names. The further development and the current situation of the nomenclature are well explained in @Jan’s thorough answer.


Will ice be cold or not when formed using pressure?

Water can be formed by applied pressure, but only at certain temperatures. Looking at the phase diagram below, you can see that water must be at 0 C until extreme pressures are reached. This can create ice, but it would only remain solid under these pressure conditions at that temperature. The ice would have a cooling effect on surroundings that have a higher temperature, in fact the rapid change back to liquid would cool the surroundings, but the ice would not be cold. This ice would also quickly revert to another phase if not held under these pressure conditions.

So, if at the bottom of a deep, high gravity sea, the ice will form at temp above 0C. If it starts floating upward, the lower pressure will allow it to melt.

If you create warm ice in a hydraulic ram, it will melt as soon as pressure is relieved.


Why does cotton dissolve in a dilute solution (1/2) of fuming sulfuric and nitric acids?

To address you second question first, in neither case did the cotton dissolve.  In the diluted acid mixture the cotton was just wetted, possibly also having enough oxidation of the cotton by the nitric acid to give it a different texture.

Regarding the effect of dilution on the nitration of the cotton, the reaction was basically just too dilute to proceed.  Also, according to this Wikipedia article, the sulfuric acid is present as a catalyst, producing the nitronium ion, $\ce{NO2+}$. The net reaction is given as:

$$\ce{3HNO3 + C6H10O5 → C6H7(NO2)3O5 + 3H2O}$$

Since water is a product of the reaction, by adding water to the reaction via dillution we further inhibit the formation of the products.


In my homemade electrolysis setup, only the negative end bubbles?

Your setup, using table salt ($\ce{NaCl}$ plus additives) will initially electrolyse according to the following two reactions, separated by cathodic reduction and aniodic oxidation:

$$\ce{2 H+ + 2 e- -> H2 ^}\tag{CatRed1}$$

$$\ce{2 Cl- -> Cl2 + 2 e-}\tag{AnOx1}$$

These reactions assume no counterions. However, you don’t have dissolved protons in your solution because you aren’t electrolysing hydrochloric acid. Instead, the proton is taken from water molecules and if we include them (and sodium counterions) we get the following improved cathodic reduction equation:

$$\ce{2 Na+ + 2 H2O + 2 e- -> H2 ^ + 2 NaOH}\tag{CatRed2}$$

This still displays the hydrogen gas bubbling away at the cathode but chlorine gas at the anode should also bubble after some time (or at least evolve noticeably). However, with the increase of $\mathrm{pH}$ due to the generated hydroxide, the chlorine produced at the anode will react further:

$$\ce{Cl2 + 2 OH- -> Cl- + OCl- + H2O}\tag{AnOx2}$$

And both of these are soluble ions. The overall reaction of your electrolysis setup is:

$$\ce{NaCl + H2O -> H2 ^ + NaOCl}\tag{RedOx}$$

There will be hydrogen evolution at the cathode but no gas evolution at the anode. The browning will likely be due to a number of side reactions induced by the oxidant hypochlorite anion.

In case you’re wondering why water’s oxygen is not oxidised, this is due to its higher standard electrode potential.


Why is zinc blende not directly reduced in smelting furnaces?

The standard free energy of formation of most of the sulphides are larger than those of $\ce{CS2}$ and $\ce{H2S}$. Therefore, these sulphides are most stable than those with carbon or hydrogen. So, neither carbon not hydrogen can reduce metal sulphides to metals.

But standard free energies of formation of oxides are much lower than those of $\ce{SO2}$.  Therefore, oxidation of metal sulphides to metal oxides is thermodynamically favorable. Also, the oxides are easily reducible.


What elements (in the pure form) would harm a human, if eaten?

Is, for example, eating pure carbon bad for the organism?

No, quite the contrary.

Activated charcoal (in the form of powder or tablets) was/is traditionally used to treat diarrea. It is listed on the WHO Model List of Essential Medicines on page 4.


Why is boric acid written as B(OH)3 and not H3BO3?

This is mostly because boric acid is commonly believed to have different acidity mechanism. While common acids generally dissociates

$\ce{HNO3 <=> H+ + NO3-}$

boric acid is commonly believed to associate with water

$\ce{H2O + B(OH)3 <=> H+ + [B(OH)4]-}$

this type of acidity is rather common for metal hydroxides, hence the way of writing.

For that matter, not every hydrogen should be written at the beginning of a formula of an acid, but only actually acidic ones. So, $\ce{H2SO4}$, but $\ce{HPH2O2}$


Increasing the length of a carbon chain in an alkane

One method could be metathesis of alkanes. There is a nice paper about it published by J. M. Basset et al., SCIENCE 1997, 276, 99-102, "Metathesis of Alkanes Catalyzed by Silica-Supported Transition Metal Hydrides" and for sure more recent ones have been published since then.

  Abstract
  The silica-supported transition metal hydrides $\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$ and $\ce{(≡Si-O-)xM-H (M, chromium or tungsten)}$ catalyze the metathesis reaction of linear or branched alkanes into the next higher and lower alkanes at moderate temperature ($\pu{25^\circ}$ to $\pu{200^\circ C}$). With $\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$, ethane was transformed at room temperature into an equimolar mixture of propane and methane. Higher and lower homologs were obtained from propane, butane, and pentane as well as from branched alkanes such as isobutane and isopentane. The mechanism of the step leading to carbon-carbon bond cleavage and formation likely involves a four-centered transition state between a tantalum-alkyl intermediate and a carbon-carbon bond of a second molecule of alkane.


Reaction between lead dioxide and nitric acid

You need to consider the reaction conditions.

The Wikipedia Lead Dioxide article is saying that in dilute nitric acid the first reaction occurs, while in warm nitric acid the second reaction occurs.

So the first reaction would only proceed to the second reaction under harsher conditions, heat and more concentrated acid.


Rotational Constant for CO2 from an IR Vibrational-Rotational spectra

I'm assuming you are measuring the rotationally well resolved band at $\approx 2349 \pu{cm^{-1}}$ and you are correct.

The rotational band structure for the antisymmetric vibration centred at $\approx 2349 \pu{cm^{-1}}$ (also called parallel $v_3$ band) has alternate lines missing, which are those having odd J. $\ce{CO2}$ is a linear symmetric molecule ($D_{\infty h}$ point group) but nuclear spins on $^{12}\ce{C}$ and $^{16}\ce{O}$ are both zero and means that antisymmetric rotational lines are missing. The ground vibrational state is $\Sigma^{+}_g $ resulting in the odd levels being absent.

The effect is due to the symmetry properties of the wavefunction, and  what happens on exchange of coordinates. For fermions ( multiples of spin 1/2 , eg. electrons, protons,) the wavefunction must be antisymmetric but symmetric for Bosons (integer spin 0,1,2 etc) which is the case for $\ce{CO2}$. Look for 'statistical weight', 'linear molecules', and 'nuclear spin' in a good spectroscopy book, Herzberg 'Infra Red and Raman Spectra' for example.


What's the difference between FFKM and PTFE?

Teflon (PTFE) and Kalrez (FFKM) are different polymers.

PTFE is made from tetrafluoroethylene, and contains only carbon and fluor.
The material is stiff.

FFKM is made from tetrafluoroethylene, vinylidene fluoride and perfluoromethylvinylether, so besides carbon and fluor, it also contains hydrogen and oxygen.
These additional monomers make the polymer an elastomer.


Wave function for particle in a infinite well located at -L and +L

It is very simple, there is no need to solve it again since you already know the solution of infinite square well. The idea is very simple, the trap looks the same, but wider now $2L$. So first you should extend it by a factor of two and then move it back to the center, it is given by the rescaling and translation:
$$L \mapsto 2L \tag{1}$$
$$x \mapsto x-L \tag{2}$$

So the solution is:
$$\psi_{n}=\sqrt{\frac{1}{L}}\sin\left(\frac{n\pi x}{2L}-\frac{n\pi}{2}\right) \tag{3}$$
$$E_{n}=\frac{n^2\pi^2 \hbar^2}{8mL^2}$$
As expected. Note that $\sin(\pi/2-x)=\cos(x)$ so that you can split it into odd and even series, but the Eq (3) given above is more succinct.


How to check if azobisisobutyronitrile (AIBN) is still active?

If azobisisobutyronitrile (AIBN) underwent homolytic cleavage over time, do you expect the resulting cyanoprop-2-yl radicals would survive as free radicals for a long period of time? If not, what would be the decomposed products?

Alkali metals reduce benzophenone in THF to form blue diphenylketyl radical anions. I'm not sure azobisisobutyronitrile (AIBN) or cyanoprop-2-yl radical can do that.

Maybe you can test the structural integrity of AIBN by 1H-NMR, IR or Raman spectroscopy?
The -N=N- stretching vibration of a symmetrical trans-azo compound like AIBN is "IR-forbidden" but absorbs in 1576cm-1 region of the Raman spectrum (1). If it decomposes, perhaps you can see the decomposed products in the mass spectrum.
Maybe you can try to activate the subject AIBN as a radical initiator?

a. Perhaps one can observe a minute amount of nitrogen gas upon homolytic cleavage?

b. AIBN can be used as radical initiator in radical polymerization. Once activated as cyanoprop-2-yl radical, it readily attacks olefinic double bond of vinyl monomers. Perhaps you can test such reaction and see if some polymerization occurs using mass spectroscopy?

Reference:

R.M. Silverstein and F.X. Webster: Spectrometric Identification of Organic Compounds (6th ed.)


Impact of detergent on internal living tissue?

Caustic Ingestion Injury
Triton X-100 has a pH of 9.7 at 100% concentration (according to Wikipedia) which can cause alkaline caustic injury, however it is less likely to cause severe full thickness injury, as those injuries tend to happen with ingested products of pH <2 or >12.5 or with other corrosive properties that Triton X-100 does not have. Ingestions at pH 6.0-8.0 are unlikely to cause injury unless there is long term exposure (as in gastro-oesophageal reflux disease or bile reflux); note that saliva itself often has a pH of 7.8!
Alkaline caustic injury has a large spectrum. The alkali can cause liquefactive necrosis via saponification and protein dissolution. In the best case scenario, there is only superficial injury to mucous membranes, causing pain and exudates. In the worst case scenario there is full thickness involvement, usually involving the oesophagus, where you get a perforation. One of the worst cases I have had any involvement in was a young lady who drank a large quantity of Drano (concentrated NaOH with pH 11.5-13.5) and had oesophageal, gastric and duodenal perforation with extensive chemical damage to her pancreas, liver and fortunately to a much lesser extent, mediastinum.
For reference, Zargar's grading system for caustic ingestion (usually based on findings on gastro-oesophagoscopy +/ pharyngolaryngoscopy) is as follows:

Grade 0: Normal examination
Grade 1: Oedema and hypermia of the mucosa
Grade 2a: Superficial ulceration, erosions, friability, blisters, exudates, hemorrhages, whitish membranes
Grade 2b: Grade 2a plus deep discrete or circumferential ulcerations
Grade 3a: Small scattered areas of multiple ulceration and areas of necrosis with brown-black or greyish discoloration
Grade 3b: Extensive necrosis

Triton X-100 would most likely cause Grade 1-2b injury. Grade 3 injury is unlikely but I suppose it depends on dose as well.
Ingestion at lower concentrations and effects not related to caustic ingestion:
This link contains information on the general toxicity of non-ionic surfactants such as Triton X-100.
In summary:

They cause mild irritation to mucous membranes
This is primarily due to foaming
Foaming and irritation can cause vomiting and diarrhoea
Foam can also be aspirated causing chemical pneumonitis
Alcohol ethoxylates (as Triton X-100 appears to be) are readily absorbed into the bloodstream via mucous membranes and is likely to have toxic effects in this way

The rest of the evidence I could find also states:

They can cause haemolysis- makes sense, as surfactants in the circulation would come into contact with red cell membranes
Triton X-100 causes direct cytotoxicity due to cell membrane disruption
It also inhibits NAD(P)H, which is a coenzyme involved in cellular respiration
There is dose-dependent destruction of the cytoskeleton
There is also some associated double-stranded DNA breakage
It is possible for cells to recover from this poisoning

Thus, the likely scenario from swallowing dilute Triton X-100 is this:

First you get foaming, oral & oesophageal pain and abdominal pain
Next you get vomiting and diarrhoea, at which point you may inhale some of the foam
If you've inhaled any foam, you develop respiratory distress
You develop dose-dependent haemolysis which presents as tiredness (and/or shortness of breath, chest pain, palpitations), red or dark brown urine, jaundice
In severe cases potential multi-organ failure as a result of direct cytotoxicity, haemolysis, anaemia and/or chemical pneumonitis

It seems it'd be a huge spectrum all the way from general foaminess and some pain all the way to multi-organ failure and death. I'm not entirely sure what doses would be toxic as it appears no-one has forced animals or humans to drink the stuff!
Regarding whether there are more toxic and corrosive but chemically similar compounds out there
yes.


Why is carbon monoxide a "weak" lewis base?

$\ce{CO}$ is usually considered a Lewis base, as the lone pair on carbon readily donates it's electrons, for example forming transition metal coordination complexes. Although oxygen is more electronegative than carbon, the carbon is more rich in electron density in the $\ce{CO}$ molecule as it basically has a full formal negative charge. Additionally, products formed by the carbon bonding to Lewis acids are much more favorable that products that would be formed by bonding to the oxygen atom.

It is also possible for CO to act as a Lewis acid however. When acting as a ligand with certain low-valent transition metals, the metal can redistribute electron density back to the $\ce{CO}$ ligand in a process called back-donating. In this case the $\ce{CO}$ molecule accepts electrons via it's antibonding $\mathrm{\pi}$ orbitals, making it act as a Lewis acid.


What happened to this stainless steel tool when exposed to vinegar?

What grade of stainless steel was the clipper made of? Was it labeled stainless steel, or is this an assumption because it looked shiny or similar to some other corrosion-resistant object (which may itself be stainless steel or not)?
Test with a magnet: if they are non-magnetic, the clippers could be stainless steel. Without knowing more about the source of the clippers, I could guarantee you that they are not stainless steel - after, they were not stain-less even in vinegar.
It is difficult to imagine any kind of heat treatment or surface treatment (short of plating with a more active metal) that could result in this kind of generalized corrosion.


How does increase in volume change the speed of reaction in production of NO2?

I'm not an expert but this is how I would do it.

Determine the reaction rate formula from the reaction equation

$$r = K[\ce{NO}]^2[\ce{O2}]$$
Changing the volume by a factor $2$ means the concentration of $\ce{NO}$ and $\ce{O2}$ will become half

$$r' = K\cdot\frac{[\ce{NO}]}{2}^2\cdot\frac{[\ce{O2}]}{2}$$
Compare the reaction rates

$$r' = K[\ce{NO}]^2\cdot\frac{1}{4}\cdot [\ce{O2}]\cdot\frac{1}{2}$$

$$\frac{r}{r'} = \frac{K[\ce{NO}]^2[\ce{O2}]}{K[\ce{NO}]^2[\ce{O2}]\cdot\frac{1}{8}} = 8$$

Conclusion: increasing the volume by $2$ will decrease the concentration (or pressure) of $\ce{NO}$ and $\ce{O2}.$ This causes the reaction rate to decrease by a factor of $8$.


In general just by looking at these equations there is no surefire way to know which is endo or exothermic. However you can look at the bond strength or the ∆H of formation for each molecule. In short you have to know something.

The energy stored in the molecules(specifically in the bonds) is lower before the reaction in an endothermic reaction and higher in an exothermic reaction. This means if you know relatively how strong the bonds or how much energy it takes to form each of the molecules then you can know whether the reaction is endo or exothermic.

If you have a few ∆H of formation values you can use Hess's Law


There's a bit of a misconception on what actually happens in the video linked in your question:

The reactions between alkali metals, namely lithium ($\ce{Li}$) and sodium  ($\ce{Na}$) and water, diluted acid and concentrated sulfuric acid are not neutralisations, but redox reactions, that is transfer of an electron from one reaction partner to the other.

\begin{align}
\ce{2H2O &<=> 2H+ + 2OH-\\
2Na &-> 2Na+ + 2e-\\
2H+ + 2e- &-> H2 ^}
\end{align}

If you take this three reactions as a (mathematical) equation system, with which you're probably more familiar with, and sum it up, the total reaction is
\[\ce{2Na + 2H2O -> 2NaOH + 2 H2 ^} \]
To put this into words:

The metal is dissolved and forms metal hydroxide.
The resilting solution is alkaline (basic), hence the name alkali metals.
The reaction releases a lot of heat!
Hydrogen gas is formed.

Since the reaction releases a lot of heat, the hydrogen gas is burnt with the oxygen from the air, yielding water, This is the explosion that you noticed.

Taken these vigorous reactions into account, it is not astonishing that in nature, sodium and and lithium do not exist as metals, but in the form of various salts. It wasn't until 1807 that sodium metal was actually made in the laboratory, followed by lithium in 1818.

Turning back to your question on neutralisations, let's define what it actually means.

(I apologize if I'm getting to colloquial here!)

The term neutralisation is typically used to describe that two effects don't sum up to result in something stronger but rather incapacitate each other. In the case of acid-base reactions, this translates to:

\[\ce{effective\ acid + effective\ base -> something\ ineffective + water}\]

In water, a typical example would be the reaction of of a strong acid (such as hyrochloric acid) with equimolar amounts of strong base (such as sodium hydroxide):

\[\ce{HCl + NaOH -> NaCl + H2O}\]

Adding the right amount of a strong and corrosive acid to a solution of also corrosive and harmful caustic soda a harmless solution of sodium chloride (table salt) in water.
The only side effect of the reaction is that the solution can get pretty hot, but an explosion is not to be expected here.

Typically, usable explosions do not result from neutralisation reactions but from fast combustion reactions in with large volumes of gases are formed, possibly (but not necessarily) accompanied by the release of heat.

Taking into account that strong acids had to be made by men, I'd be surprised to find that these materials were wasted in simple neutralisation reactions.

You will however find historical accounts on the use of the acids or the bases. Nitric acid was used to separate silver from gold (Scheidewasser). Calcium oxide (burnt lime), the precursor of calcium hydroxide e.g. is used to make cement.


How to create a buffer using CH3COOH and CH3COONa that has a pH of exactly 3.75?

In my understanding, there is considered $$\mathrm{p}K_\mathrm{a}=4.75$$ and therefore $\mathrm{pH}=3.75$ was chosen for easy calculation, as:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log \frac {c_{\ce{A-}}}{c_{\ce{HA}}}$$

and therefore
$\frac {c_{\ce{A-}}}{c_{\ce{HA}}}=0.1$

For the buffer change to $\mathrm{pH}=4.75$, at least $\pu{25 ml 0.1 M NaOH}$ has to be spent. ("withstand")

$$\frac {c_{ws,\ce{A-}}}{c_{ws,\ce{HA}}}=1$$

Calculating molar amounts to be equal for both acetic acid and acetate:
$$0.1\cdot c_{\ce{HA}}\cdot V_\mathrm{buf}+c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}\\=c_{\ce{HA}}\cdot V_\mathrm{buf}-c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}$$

From that, we calculate minimal concentration of acetic acid in buffer:

$$2\cdot c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}=0.9\cdot c_{\ce{HA}}\cdot V_\mathrm{buf}$$

$$c_{\ce{HA}}=\frac{2\cdot c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}}{0.9\cdot  V_\mathrm{buf}}=\frac{2\cdot 0.1\cdot 0.025}{0.9\cdot  0.05}=0.005/0.045=1/9 \mathrm{~ mol/l}$$

$$c_{\ce{HA},min}=\frac19 \mathrm{~ mol/l}$$
$$c_{\ce{A-},min}=0.1 \cdot \frac19=\frac1{90} \mathrm{~ mol/l}$$

All acetic acid in $\pu{50 mL}$ of the buffer must react with not more than  $35~\text{ml}$ of $0.1~\mathrm{M~}\ce{NaOH}$.

$$\begin{align}
n_{\ce{NaOH}}&\lt 3.5\mathrm{~mmol}\\
n_{\ce{CH3COOH}}&\lt 3.5\mathrm{~mmol}\\
n_{\ce{CH3COONa}}&\lt 0.35\mathrm{~mmol}\\
\end{align}$$

As the latter 2 amounts have to be present in $\pu{50 mL}$ of the buffer:

$$\begin{align}
c_{\ce{CH3COOH}}&\lt \pu{70 mmol/L}\\
c_{\ce{CH3COONa}}&\lt \pu{7 mmol/L}\\
\end{align}$$

Summary:
Concentration of acetic acid in the buffer must be at least $\frac19=0.1111\pu{ mol/L}$ and not more then $\pu{0.07 mol/L}$.

Such requirements is impossible to meet and such a buffer is not possible to prepare.

By other words, the buffer cannot have at the same time so big buffer ("withstand") capacity and so low "break" capacity.

If usage of acetate buffer was not mandatory, the goal would be achievable with buffers with pKa closer to 3.75 than the acetic acid has.  In such a case, the buffer capacity would be decreasing and not increasing between these pH values and the final break would be faster.


Which is more electronegative: hydrogen or astatine?

An interesting way to determine which of $\ce{A}$ or $\ce{B}$ is more electronegative is to take their binary compound $\ce{A_xB_y}$ and observe its dissociation. Note that it can be $$\ce{A_xB_y -> xA^y+ + yB^x-} \tag{1}\label{a}$$ or $$\ce{A_xB_y -> yB^x+ + xA^y-} \tag{2}\label{b}$$

Reaction $\ref{a}$ indicates that $\ce{A}$ is less electronegative than $\ce{B}$, because in the dissociation of $\ce{A_xB_y}$, $\ce{A}$ carries the positive charge, while $\ce{B}$ carries the negative charge. Reaction $\ref{b}$ instead indicates that, for similar reasons, $\ce{A}$ is more electronegative than $\ce{B}$ 1

Going by the same logic, we may tend to observe a hydrohalo acid in this case, specifically, the hydroastatic acid/hydrogen astatide. Quoting from its Wikipedia page,

  This chemical compound can dissolve in water to form hydroastatic acid, which exhibits properties very similar to the other four binary acids, and is in fact the strongest among them. However, it is limited in use due to its ready decomposition into elemental hydrogen and astatine (my emphasis) Because the atoms have a nearly equal electronegativity, and as the $\ce{At+}$ ion has been observed, dissociation could easily result in the hydrogen carrying the negative charge. (my emphasis) Thus, a hydrogen astatide sample can undergo the following reaction:
  $$\ce{2HAt -> H+ + At- + H- + At+ -> H2 + At2}$$

which also explains the spontaneous decomposition into elemental hydrogen gas and the astatine precipitate.

If you also have a look at their electronegativities, both have a Pauling electronegativity $2.2$, further complicating the comparison.

Therefore, based on these evidences, I can say, without the least hesitation, that theoretical comparison of the electronegativities of hydrogen and astatine is not possible.

[1]: a shrewd observer would notice that the apt way to write the second reaction's reactant would be $\ce{B_yA_x}$ instead, since $\ce{A}$ is more electronegative


Why do we need the rms, mean, and most probable velocities?

Alright, here's my best attempt at it.

Root Mean Squared $v_\mathrm{rms}$

The root-mean-square speed (r.m.s. speed) of the molecules. This quantity is defined as the square root of the mean value of the squares of the speeds,$v$, of the molecules. That is, for a sample consisting of $N$ molecules with speeds $v_{1}, v_2,...,v_N$, we square each speed, add the squares together, divide by the total number of molecules (to get the mean, denoted by angle brackets $\langle\rangle$), and finally take the square root of the result:

$$\begin{align}
v_{rms} &= \langle v^2\rangle^{1/2}
\\&= \left(\frac{v_1^2+v_2^2+...+v_N^2}{N}\right)^{1/2}
\end{align}
$$

Mean Speed $\bar{v}$

The r.m.s speed might at first sight seem to be rather peculiar measure of the mean speeds of the molecules, but its significance becomes clear when we make use of the fact that the kinetic energy of a molecule of mass $m$ travelling at a speed $v$ is $E_\mathrm{k} = \frac{1}{2}mv^2$, which implies that the mean kinetic energy of a collection of molecules, $\langle E_\mathrm{k} \rangle$, is the average of this quantity, or $\frac{1}{2}mv_{rms}^2$. It follows from the relation $\frac{1}{2}mv_\mathrm{rms}^2= \langle E_\mathrm{k}\rangle$, that

$$v_\mathrm{rms} = \left(\frac{2\langle E_\mathrm{k}\rangle }{m}\right)^{1/2}$$

Therefore, wherever $v_\mathrm{rms}$ appears, we can think of it as a measure of the mean kinetic energy of the molecules of the gas. The r.m.s speed is quite close in value to another and more readily visualized measure of molecular speed, the mean speed, $\bar{v}$, of the molecules:

$$\bar{v} = \frac{v_1+v_2+...+v_N}{N}$$

For samples consisting of large numbers of molecules, the mean speed is slightly smaller than the r.m.s. speed. The precise relation is

$$\bar{v} = \left(\frac{8}{3 \pi}\right)^{1/2}v_\mathrm{rms} = 0.921v_\mathrm{rms}$$

For elementary purposes, and for qualitative arguements, we don't need to distinguish between the two measures of average speed, but for precise work the distinction is important.

Most Probable $v_\mathrm{mp}$

$v_\mathrm{mp}$ is the most probable speed of the molecules, and is the maximum value for the velocity distribution curve. When putting all of of these equations together we see the relation

$$v_\mathrm{rms} > \bar{v} > v_\mathrm{mp}$$

That $v_\mathrm{mp}$ is the smallest of the three speeds is due to the asymmetry of the curve.

References

Atkins, P.; de Paula, J. Elements of Physical Chemistry 6th ed. W.H. Freeman and Company. New York, NY. 2013.

Chang, R. Physical Chemistry for the Chemical and Biological Sciences. University Science Books. Sausalito, CA. 2000.


Find amount of acid and base required to make buffer

Usually when you are preparing a buffer, you can about the pH and the strength (total concentration). Let's say you want to combine acid $\ce{A}$ and its conjugate base $\ce{B}$ to make a buffer with total concentration $C=[\ce{A}]+[\ce{B}]$ and a set $\text{pH}$.

The Henderson-Hasselbalch equation provides a relationship between the ratio of the two concentrations to the pH:
$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{[\ce{B}]}{[\ce{A}]}\right)$$

Because $[\ce{B}]=C-[\ce{A}]$, we can write, and solve for $[\ce{A}]$ (and then for $[\ce{B}]$).

$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)$$
$$\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)= \ce{pH}-\ce{p}K_a$$
$$\dfrac{C-[\ce{A}]}{[\ce{A}]}=10^{\ce{pH}-\ce{p}K_a}$$
$$C-[\ce{A}]=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]$$
$$C=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]+[\ce{A}]=[\ce{A}]\left(1+10^{\ce{pH}-\ce{p}K_a}\right)$$
$$[\ce{A}]=\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}$$
$$[\ce{B}]=C-\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}=\dfrac{C\left(10^{\ce{pH}-\ce{p}K_a}\right)}{1+10^{\ce{pH}-\ce{p}K_a}}$$

So... if you have a total concentration $C$ and a $\ce{pH}$ in mind for an acid-base pair with a certain $\ce{p}K_a$, just plug them in.


How does the calorimeter affect heat released by the system (q)?

Let's take a numerical example. $0.2432$ $g$ magnesium metal ($0.01$ mol) is introduced into a calorimeter containing $50~ g$ HCl $1$ M. The chemical reaction will be $$\ce{Mg + 2 H^+ ->  Mg^{2+} + H2}$$ The composition of the solution is changing between the beginning and the end of the reaction. There is an excess of HCl.  But let's admit in first approximation that the specific heat of the solution does not change and remains as in water ($C_w = 4.185~ J g^{-1} K^{-1}$). The tabulated enthalpy change of the reaction is - $462~ kJ/mol$. The total amount of heat $Q$ produced by the reaction of $0.01 mol$ Mg is : $$Q\ce{= 0.01 mol·462 kJ/mol = 4.620 kJ = 4620 J}$$
If the calorimeter was ideal, without any weight, and with the already mentioned approximation, the change of temperature should be : $$\Delta T = \frac{Q}{C_w·m} =  \frac{4620~ J}{4.185~J g^{-1} K^{-1}·50~ g} = \frac{4620~ J}{209.25~ J K^{-1}} = 22.07 K$$
Now suppose that the calorimeter is made of copper and weighs $m_{Cu} = 20~ g$. The heat capacity of copper is $C_{Cu} = 0.385~ J g^{-1} K^{-1}$. The change of temperature will be reduced and becomes $$\Delta T = \frac{Q}{C_w·m + C_{Cu}·m_{Cu}} =  \frac{4620~ J}{(209.25 ~J K^{-1})~ + (0.385 ~J g^{-1} K^{-1}·20~ g)} = 21.3 K$$
And now if the calorimeter is not well insulated, the final temperature change will be smaller than $21.3 K$.
Sorry for my poor management of the italicized letters. Any help would be appreciated.