Detecting a gas molecule moving through a small pinhole

The best way to do this is to detect the molecules you want in the presence of others that carry it along. You should not expect to detect every molecule of the type you want but just one of them at a time.

The most sensitive detection technique is mass spectrometry, more sensitive than fluorescence because if a molecule fluoresces in a direction away from your detector its photon is lost. All ions can in principle be collected in a mass spectrometer.

To get the molecules into the spectrometer mix them with a solvent, (if necessary) and heat this with a pulsed IR laser. This has to be sufficient to vaporise the mixture/substrate  and it has to be positioned close to the input of the spectrometer since this is done outside the spectrometer. The vapour is at atmospheric pressure and mixed with air but because of the pressure differential efficiently enters the spectrometer (it is easy to see this). Here by differential pumping and using skimmers before the entrance to the mass spectrometer part some of the target molecules will be present in the beam produced. The mass spectrometer should probably be of the time of flight/reflectron type to increase throughput and mass resolution. By pulsing the laser it is then possible to gate the detector to observe the molecules at just the right time and so reduce noise. Of course once you have detected the molecule it is destroyed on the detector.

If you do not want to destroy the molecules then fluorescence detection is the only possibility provided the molecule does fluoresce as not all do with a reasonable yield. A second laser is thus needed after the mass selection and then a photomultiplier is used to count single photons of fluorescence as they arrive (v. common technique now for 50 years or more). I'm not sure what you could do after that but again there is time as photons travel effectively instantaneously compared to the sped the molecule will be travelling at. As I said in a comment not cheap, not trivial.


When can we add multiple reactions to compute yield of products in an equilibrium?

When you combine the reactions and the equilibrium constants like you do, you're calculating the amount of reaction which goes all the way to completion (the fraction which reacts in both steps).

The complication that you're seeing is that the protons are not just coming from the $\ce{H_2A}$ which completely dissociates, but also a (rather substantial) portion of the molecule which undergoes only partial dissociation.

To see this, think about what happens when the hydrogens in the molecule are distinguishable: for example, the dissociation of $\ce{HDA}$ (hemi-deuterated compound), where the reactions are

$$\ce{HDA <=> DA^{-} + H^{+}} $$
$$\ce{DA^{-} <=> A^{2-} + D^{+}} $$

(in reality this wouldn't be the obligate order, but just assume it is for a thought experiment).

The $10^{-15}$ equilibrium constant is only applicable for the formation of $\ce{D^{+}}$ - you still have the rest of the $\ce{H^{+}}$ from the first step, which will also contribute to the pH.

So your assumption in your last step that $\ce{[H^{+}] = 2 [A^{2-}]}$ is incorrect, thus resulting in an incorrect result.

Further explanation

Let's actually work through the $\ce{HDA}$ case, assuming (incorrectly) that the dissociation has an obligate order. You start off with a certain amount of $\ce{HDA}$ ($T = 0.1M$). Now, imagine that at equilibrium we get $x$ total turnovers for the first reaction. A certain amount of $\ce{DA^-}$ (call it $y$) is then going to go and dissociate further. So in the end at equilibrium, we have the following concentrations:

$$\ce{[HDA]} = T-x$$
$$\ce{[H^+]} = x$$
$$\ce{[DA^-]} = x-y$$
$$\ce{[D^+]} = y$$
$$\ce{[A^{2-}]} = y$$

Now we can set up our equilibrium equations:

$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} = \frac{(x)(x-y)}{T-x} = 10^{-5} M$$
$$\frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HD^{-}}]} = \frac{(y)(y)}{x-y}= 10^{-10} M$$

We have a system of two equations with two unknowns. We can thus solve for $x$ and $y$ and get $x=0.995\cdot10^{-3} M$ and $y = 3.15\cdot10^{-7} M$.

Now, let's do the combined equation. First off, we can see where it comes from:

$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} \cdot \frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{DA^{-}}]} = \frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = 10^{-5} M \cdot 10^{-10} M = 10^{-15} M^2$$

So we set up the equation:

$$\frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = \frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$

Um ... we have two unknowns, but only a single equation. We can't solve this without additional information. We can't assume that $[\ce{H^{+}}] = [\ce{D^{+}}]$, or that all the $\ce{HDA}$ which disappeared went completely to $\ce{A^{2-}}$ - some of it can (has to) form $\ce{DA^-}$ which doesn't react further. The equation is valid (if you substitute $x=0.995\cdot10^{-3} M$ into the equation, as determined above, you'll get the correct answer for $y$), but only if we have some way of knowing how much of the $\ce{HDA}$ reacted and didn't go completely to completion.

The picture is further complicated because we're not dealing with distinguishable $\ce{H^+}$ and $\ce{D^+}$ which dissociate in an obligate order, but a single species which is involved in both reactions. That is, instead of

$$\frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$

we would have

$$\frac{(x+y)(x+y)(y)}{(T-x)} = 10^{-15} M^2$$

Again, the combined equilibrium constant is perfectly correct, and accurately expresses the relationship between the concentrations at equilibrium. It's just that the additional assumptions about the relationships between the concentrations that you're making are not. Specifically, $[\ce{H^{+}}] \neq 2 \cdot [\ce{A^{2-}}]$. Note this isn't particularly specific to this sort of combined equilibrium. Any additional reaction you're not considering can change the concentrations and make your assumptions less than correct - not just formation of intermediates which don't go all the way to completion to side reactions that draw off substrates, reactants or products into other forms.

The equilibrium constant tells you the relationship between the species concentrations, but the concentrations as they actually are in solution, not what the reaction generates. The latter is only true when the reaction in question is the only one under consideration.


How can I remove water vapor from an airstream without a cold trap?

You won't easily be able to use some other liquid to pull the water vapor out of the air. I think the cheapest option is a hydrophobic membrane or filter that you feel comfortable using. A commercial desiccant system will be the highest performance next to cryogenic cooling but may not be cost effective (depends on the scale you are working at). You can also try salt in place of silica beads for absorption drying versus adsorption drying; I imagine the silica will be more efficient but salt will be cheaper. When using things like silica, salt, or commercial desiccants you may run into issues of too much pressure drop as the gas travels through the packed column. You will have to figure out how often the desiccant media needs to be regenerated/replaced to make sure your drying is adequate.


Why is the rate inversely proportional to the square root of temperature in Grahams Law of Diffusion?

It helps to rewrite
$$\text{rate} \propto \frac{PA}{\sqrt{TM}}$$
by assuming the ideal gas law holds, as follows:
$$\text{rate} \propto \frac{RA}{V_m}\sqrt{\frac{T}{M}}$$
where $V_m$ is the molar volume or inverse of molar particle density.

Written this way it is clear that if the particles occupy the same volume, increasing their temperature increases the rate at which they effuse through the apperture.


So, is baking soda actually a strong base?

There is nothing to discuss (unless you find it necessary to compare two known numbers by discussion). $\ce{NaHCO3}$ is a very weak base.

Really, do you have baking soda around? Throw some in the water and see whether $\ce{CO2}$ actually escapes. It doesn't. It will if you heat the solution to near-boiling temperatures; then you'll have $\ce{Na2CO3}$ which is a base of moderate strength. You can't convert it all to $\ce{NaOH}$ this way.

So it goes.


The Evasive Ether: Existence of diethynyl ether?

You can find a nice discussion of diethynyl ether and some of its derivatives here:

Ooi, I. H., & Smithers, R. H. (1989). Bis [(trimethylsilyl) ethynyl] ether: a moderately stable C4H2O derivative. The Journal of Organic Chemistry, 54(6), 1479-1480.

https://pubs.acs.org/doi/pdf/10.1021/jo00267a053

They write that "little or nothing is known about diethynyl ether", and they were only able to find theoretical (INDO) calculations in their literature search about the compound, as opposed to actual reports of its synthesis. They do mention that the following authors managed to synthesize a dimethyl derivative, $\ce{Me-C#C-O-C#C-Me}$, but reported that it explodes at room temperature (and I infer, from what Ooi and Smithers wrote, that diethynyl ether would be even more unstable):

Brandsma, L., & Arens, J. F. (1962). Chemistry of acetylenic ethers LVI Di‐(alk‐1‐ynyl) ethers and thioethers. Recueil des Travaux Chimiques des Pays‐Bas, 81(6), 510-516.

https://onlinelibrary.wiley.com/doi/abs/10.1002/recl.19620810610

They themselves (Ooi and Smithers) were able to synthesize a di-trimethylsilyl derivative, $\ce{Me_3Si-C#C-O-C#C-SiMe_3}$ which, while stable when dissolved in $\ce{CH_2Cl_2}$ and stored in a freezer, degrades overnight when stored neat at room temperature.

In summary, for $\ce{R-C#C-O-C#C-R}$, larger R's provide added stability:

R = H (diethynyl ether): If it were ever synthesized, it would probably have a very short half-life, except under specalized conditions.

R = Me: Has been synthesized, but explodes at room temperature

R = $\ce{Me_3Si}$: Low to moderate stability, as described above.   Ooi and Smithers write that the added  stability provided by the trimethylsilyl groups "is probably both steric and electronic."

See also:

Arens, J. F. (1960). Ethynyl ethers and thioethers as synthetic intermediates. Advances in organic chemistry, 2, 117.


According to the article on Birch Reduction on the Organic Chemistry portal:

The question of why the 1,3-diene is not formed, even though it would be more stable through conjugation, can be rationalized with a simple mnemonic. When viewed in valence bond terms, electron-electron repulsions in the radical anion will preferentially have the nonbonding electrons separated as much as possible, in a 1,4-relationship.

http://www.organic-chemistry.org/namedreactions/birch-reduction.shtm


Why there isn't any unimolecular addition reaction?

The problem I see with your question is that you mix up two completely different concepts. The first concept is the reaction order (i.e. order of the kinetics, for example unimolecular = first order) which is dependent on the number of components that take part in the rate determining step of the reaction. The other concept is the type of the reaction, e.g. addition, substitution, functional group interconversion.

Naturally, most addition reactions will be of the type $\mathrm{A + B \rightarrow C}$ or something like that, but that neither means that they will be of second order kinetically, nor that all addition reactions are of that type. Intramolecular additions would be one example for a unimolecular addition.

Please be careful though not to mix up the order of the reaction and the stoichiometry of the written down reaction.

Imagine a reaction were you have $\mathrm{A+B \rightarrow C}$ and $\mathrm{B}$ is used as the solvent, so that $\mathrm{B}$ is available in large excess. While there are clearly two components needed for the addition to proceed it will appear to be unimolecular (first order reaction).

Strictly speaking, the stoichiometry of the reaction has nothing to do with the reaction order.

However, most addition reactions can be conceptualized to be of the form $\mathrm{A+B \longrightarrow C}$ whether that means that $\mathrm{A}$ and $\mathrm{B}$ are part of the same molecule or different chemical entities or if the rate law will actually appear to be of second order as well then depends on the specific case.

In that regard, I don't really see the point in asking whether an addition reaction might be unimolecular, as the kinetics will depend on the specifics of the reaction while the stoichiometry can be inferred just by looking at the equation. I hope that clearifies it.


If I can't draw resonance structures for a pi system, does that mean it doesn't exist?

If there are adjacent, parallel, overlapping p-orbitals in a molecule, the electrons will be delocalized over all such atoms. While the negative charge isn't delocalized over every atom in the conjugated system, it's more than you have shown. What you are failing to consider is:

$\hspace{2.4cm}$

What's important to note about these resonances structures is that the net charge remains constant. These structures are consistent with both the reactivities of the enolate and $\alpha$,$\beta$-unsaturated carbonyl compounds. There negative charge is delocalized between the carbonyl oxygen and its $\alpha$-carbon, and the positive charge is delocalized between the carbonyl carbon (giving rise to 1,2-addition) and its $\beta$-carbon (giving rise to 1,4-addition).


Is the Standard hydrogen electrode potential at T other than 298 K also zero?

According to the definition used by IUPAC, the standard electrode potential $E^\circ$ of the standard hydrogen electrode is zero at all temperatures.

  For solutions in protic solvents, the universal reference electrode for which, under standard conditions, the standard electrode potential ($\ce{H+}/\ce{H2}$) is zero at all temperatures.

The absolute electrode potential $E^\circ(\mathrm{abs})$, however, depends on temperature. Its value can be calculated from thermodynamic quantities (e.g. $\Delta G^\circ$). The recommended value for $T=298.15\ \mathrm K$ is $E^\circ(\mathrm{abs})=4.44(2)\ \mathrm V$.


Is the given stereoisomer of 1,2-dimethylcyclobutane chiral?

One thing you should always remember is that you should always consider the actual three-dimensional structure when determining chirality. Sometimes there are elements of chirality that are not visible on paper but clear if the full structure is considered. For cyclobutane, this means you should consider the puckered structure of the core structure. However, this often rapidly equilibrates between different practically identical configurations and does not add a layer of complexity to this molecule. Thus, we are able to analyse with the paper plane only.

If you put a plane of symmetry through the centre of the $\ce{CMe-CMe}$ bond, at first glance it may look like it’s a perfect fit: all four carbon atoms of the ring are put on top of another and the wedge and hash simply switch positions.

However, note that wedge and hash mean something entirely different. In one case, the methyl group is above the paper plane (wedge), in the other, it is below it (hash). The corresponding opposite position is occupied by a hydrogen atom that completes the four valences of the cyclobutane’s carbon. Thus, this plane of symmetry you propose transforms hydrogen onto methyl and vice-versa. It is not a symmetric transformation.

You can maybe visualise this by taking a sheet of paper, marking the bottom left corner with a filled dot (the wedge-methyl group) and the bottom right with a circle (the hash-methyl group). Turn it around and put a circle behind the dot and a dot behind the circle, so that you may identify it from both sides. You may already notice that in both cases the dot is bottom-left while the circle is bottom-right — this already indicates that we have a $C_2$ axis, i.e. we can rotate around an axis that is perpendicular to the $\ce{CMe-CMe}$ bond and in the ring plane.

Prepare a second piece of paper but with the reverse orientation: this is the molecule you created by mirroring along the plane. Now attempt to orient the two pieces in such a way that dot aligns with dot and circle aligns with circle. Spoilers: you will fail. anti-1,2-dimethylcyclobutane is chiral.

You can retry the experiment with the syn-1,2-dimethylcyclobutane, i.e. two dots at the bottom of the sheet of paper (and two circles on the back side). But you should immediately see that another piece of paper with identical dots/circles can immediately be aligned perfectly with the original. Hence, syn-1,2-dimethylcyclobutane is achiral.

This can be explained well with point groups. The anti isomer has the point group $C_2$ which does not feature a plane or centre of symmery. The syn isomer has the point group $C_\mathrm{s}$ whose only element of symmetry is a plane, hence it is achiral. You can read more about point groups in this question.


Options for long term storage of helium?

In my experience, for small quantities in laboratories helium usually comes compressed in cylinders. Except possibly in the case that you require extremely high purity, I don't see why helium couldn't be stored for long periods in this form, it would be relatively safe and would avoid the need for refrigeration.

I don't know what sort of quantities you are thinking about, but it seems that you can get big cylinders. For example the largest cyclinder size available from BOC gases, a British company, is 194.7 $m^3$ filled to a pressure of 300 $bar$ (source: BOC website).


Lowercase "a" in organic chemistry nomenclature?

'4a' and '8a' are locants which refer to a specific atom. They are fundamentally no different from the numbers '1', '2', '3', '4'... which are also locants referring to specific atoms.

The locants in geosmin come from the locants in naphthalene, which are assigned based on the following excerpt from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

  P-25.3.3.1.1 The numbering of peripheral items in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. Each fusion carbon atom is given the same number as the immediately preceding nonfusion skeletal atom, modified by a Roman letter 'a', 'b', 'c', 'd', etc.


Is there a relatively simple rule to solve (and understand) such
  relatively simple molecules?

The most electropositive atom is usually at the center of a molecule because

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


Is the product of Lindlar hydrogenation of (2Z,4S)-4-methylhept-2-en-5-yne optically active?

The two cis-configured pro-1-en-1-yl groups present on the central carbon remove the asymmetry. You can put a plane of symmetry through the $\ce{Me-C-H}$ plane of the central carbon to transform the molecule onto itself. You can confirm that by rotating the $\ce{(C=)C-C(HMeR)}$ bond.

Thus, this hydrogenation removes chirality.


Linking an electrochemical cell to an electrolytic cell

For an electrolytic cell, a galvanic cell is nothing else but a power source.
For a galvanic cell, an electrolytic cell is nothing else but a powered electronic circuit.
The galvanic cell anode/cathode does not form with the electrolytic cell cathode/anode a respective separate cell, as there is no ionic flow between them. Similarly for a simpler case, 2 half cells do not form a cell if there is no ionic connection between them.
The reason for it is simple, based on the electrostatic neutrality requirement and enormous forces between unbalanced charges. Imagine you managed to push the current 1 A for 20 s through the wire, without being balanced by ionic motion. The comparable unbalanced charge causes in atmosphere a lightning several kilometres long.


For a moment I would suggest you to forget about electrons,their flow ,batteries and stuff. Assuming that you have the basic intuitions regarding the principles of Newtonian mechanics and work done and all I would like to proceed in a different way. Suppose you are standing now on a ground and you have a ball in front of you lying on the ground. It doesn't lift itself up in the air, which is a common observation we all have made by now. But when you wish to lift it up to some height,be it h, then you expend your ATPs and chemical energies in your muscle to lift the ball slowly to that height h , and in doing so you did a work exactly equalling mgh ( neglecting air drag) . Now you take a tube of some arbitrary shape ( just to give a proper intuition) such that the ball is at its one end and the other end meets the ground. Then you leave the ball and it cleverly curves through the tube and gets back to ground level. This is exactly what we find in a battery system. It pumps electrons, against an opposing conservative field and then allows them to fall back. Here ,there's no man power but the chemicals do the trick by getting indulged into some redox reactions. Now ,if you think of those questions from this perspective hope you will be able to answer them efficiently.


If Beer's Law appears to apply to a high concentration (>0.01M) of a solute, is it valid to use for concentration calculation?

Internet consensus seems to define "high concentration" for Beer's Law
  as >0.01M

Keep in mind that Beer's law is an approximation. Look at the more advanced version. Beer's law-advanced version. The statement that a concentration of 0.01 M is the upper limit of Beer's law is incorrect due to several reasons. It is the absorbance value which we should worry about rather than the concentration. One can have a solution which is 0.01 M yet its extinction coefficient is not that large at that particular wavelength, we would still be fine.

This trick is used in atomic absorption spectroscopy, where you choose wavelength, which is not a resonance line of that particular atom. As a result, one can use relatively high concentration without getting a non-linear curve.

Secondly, chemists tend to infatuate little bit with linear calibrations. It is okay to have a non-linear curve. Quadratic equations can be used to fit the empirical curve and one can get accurate analytical results.


Why do polysaccharides consisting of alpha-glucose have helical structures?

Note: This is an off the top of my head answer (I might be horribly wrong), but it has prompted me to do some research. I will add more details, and/or make corrections once I am done. In the meantime, I would love to hear what others have to say on this.

Pictured below is the 2D structure of amylose

This primarily composed of $\alpha$-(1-to-4)-D-glucose units. Although the $\alpha$-(1-to-4) links are capable of relatively free rotation about their torsional angles, hydrogen bonding between the $\ce{O}$3 and $\ce{O}$2 encourages helicogenesis.

Single helical amylose has hydrogen-bonding between $\ce{O}$2 and $\ce{O}$6 atoms on outside surface of the helix and the ring oxygen pointing inwards.

Contrat this with cellulose which is a linear polymer of $\beta$-(1-to-4)-D-glucopyranose units.

The fully equatorial conformation of $\beta$-linked glucopyranose residues stabilises the chair structure, minimising its flexibility.

Pictured below, the 2D structure of cellulose and displaying the intrastrand hydrogen bonding

Interactive 3D visualisation for amylose and cellulose.


Why does the Grignard reagent of CH3I can only be prepared using diethyl ether as solvent but there is no reaction when using THF?

MeMgI is usually prepared in ether, but it can also be prepared in THF. However, magnesium iodide is sparingly soluble in THF, so RMgI reagent in THF should not be very concentrated, so that no MgI2 precipitate forms.


Can class D fire extinguishers be used for other fire types than metal fires?

Unfortunately, as far as I can tell, specific class D fire extinguishers aren't even suitable for all types of class D fire, much less other types of fire.

The copper powder variety in particular has limited effectiveness against fires that aren't lithium-based, and the thermoplastic additives in the sodium chloride-based extinguishers may also catch fire.

Dry sodium carbonate-based class D extinguishers should be generally effective but suboptimal against many types of fire, but I'd say in general, even if you have to go a little further, you should go for the proper sort of extinguisher to minimise accidental horrible disasters.


Molecular weight determination from freezing point depression

Molar freezing point depression
Benzene      5.12 °C/m
Cyclohexane  20.0 °C/m

Molar boiling point elevation
Benzene      2.53 °C/m
Cyclohexane  2.79 °C/m

I would base my argument on the constants, instead of waving hands.


The significance of the dissociation constant of an indicator

1. It all depends on the $\text{p}K_{\text{in}}$ of the indicator.

If we look at the Henderson-Hasselbalch equation:

$$\text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{[\ce{A-}]}{[\ce{HA}]} \right) $$

You can see that when $\text{pH}=\text{p}K_{\text{a}}$, then logarithm part must equal 0. This occurs when we have equal parts of the protonated and unprotonated state, because when $[\ce{A-}]=[\ce{HA}]$, the ratio between them will be 1, and $\log 1=0$.

Every time you increase $\text{pH}$ by 1, to balance the equation, the logarithm must also be 1 higher, and this happens when the thing you take the logarithm of becomes 10 times larger. This means that for a small change in pH, we have a large change in the ratio between protonated and deprotonated forms.

Typically, with indicators, you say that the colour change happens when $\text{pH} = \text{p}K_{\text{a}} \pm 1$, because outside of that interval, you will quickly have almost purely $\ce{A-}$ or $\ce{HA}$.

  If you follow this far, let's try an example with an indicator with a high $\text{p}K_{\text{in}}$

You put an indicator with $\text{p}K_{\text{in}}=10$ into pure water, which has a $\text{p}K_{\text{a}}=7$. From these values, we can see that water is a stronger acid than the indicator. This means that $[\ce{H3O+}]$ just in pure water will be high enough, that the indicator is largely protonated:

$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$

$$7=10+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$
$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=-3$$

Which means $[\ce{HIn}] = 1000*[\ce{In-}]$ - try plugging in 1/1000, 1/100, 1/1, 100/1 etc. into the logarithm on your calculator to see why this is so!

Anyway, we see that when the indicator has a $\text{p}K_{\text{in}}$ significantly higher than the $\text{p}K_{\text{a}}$ of the solution, it will mostly be protonated. Why? We will see this in part 2 of the answer.

  Another example. Let's see what happens if we have a low $\text{p}K_{\text{in}}$ indicator

You put an indicator with $\text{p}K_{\text{in}}=3$ into pure water ( $\text{p}K_{\text{a}}=7$). This time the indicator is a stronger acid. This means that $[\ce{H3O+}]$ is not high enough to keep the indicator largely protonated:

$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$

$$7=3+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$
$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=4$$

Which means $10000*[\ce{HIn}] = [\ce{In-}]$ - Statistically, every time you find a protonated indicator molecule, you will have found 10000 deprotonated indicator molecules!

  Titration scenario: acetic acid titrated with NaOH, using an indicator with $\text{p}K_{\text{a}}=10$

Acetic acid has a $\text{p}K_{\text{a}}=4.7$. The acetic acid is a stronger acid than water, so it will have increased the concentration of $[\ce{H3O+}]$. To raise the $\text{pH}$, we need to lower this concentration. As we add $\ce{OH-}$ to the solution, the following reaction occurs:

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

We are removing $\ce{H3O+}$ from the solution. But the acetic acid was only partially deprotonated, and what happens if the remove $\ce{H3O+}$ from the following reaction?

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

Well, by the principle of Le Chatelier, acetic acid and water reacts further, to replenish the $\ce{H3O+}$. This goes on until the acetic acid is deprotonated to a large degree, and then the $\ce{OH-}$ very quickly raises the $\text{pH}$ of the water, since there is no acid to react with anymore.

When the $\text{pH}$ gets into the area of the $\text{p}K_{\text{in}}$ of the indicator, the colour change starts to happen. It happens swiftly, because we are not using very much of the indicator. This means that it doesn't take a lot of $\ce{OH-}$ to do to the indicator what we did to the acetic acid.

2. Indicators have different equilibrium constants because they have different acid/base energetics
Indicators have different values for $pK_{in}$ for exactly the same reasons as acids have different $\text{p}K_\text{a}$. It depends on the stability of the bond to the hydrogen that can be lost as a proton. If the bond is very unstable (as in HCl), the proton will lose its share of the electrons in the bond, and become free very easily, and the compound will have a low $\text{p}K$.

For a compound such as acetic acid, protons will become free, but protons from $\ce{H3O+}$ can also transfer back to the acetic acid. The more $\ce{H3O+}$ that exists in the environment, the larger the chance, that a deprotonated acetic acid molecule will collide with it and receive the proton.

For the indicator with a high $\text{p}K_{\text{in}}$, if there is a lot of $\ce{H3O+}$, then the indicator molecules will be re-protonated as fast as they can lose their protons. Thus, they overall stay protonated.

The equilibrium constants $K_{\text{in}}$, $K_\text{a}$, etc. basically tell you how likely the molecule is to shed its proton.


How is pumping "Air" into a bicycle make dW (delta work )negative , when in fact there is a increase in pressure and a slight increase in volume?

Depending on context and domain dependent convention, the work done is assumed in 2 different, opposite ways, with opposite signs.
Work done on a system $\Delta U = Q + W$, preferred by chemistry
versus
work done by a system $\Delta U = Q - W$, historically preferred by physics. It is used by the former way as well recently.

Speculatively said, physics way was defined during creation of theoretical background for thermal machines. Physicists may had been focusing on work done by machines.
Later chemists came, who invented their way, focusing on energy of systems.

As a prevention, it is always good to explicitly mention which way is used, to prevent confusion.
If applied, your pump is doing positive "chemical" work on the system=tyre, but the system=tyre does negative "physical" work on the pump. The textbook takes the convention of physics.

In an important sign convention, preferred in chemistry, work that adds to the internal energy of the system is counted as positive. On the other hand, for historical reasons, an oft-encountered sign convention, preferred in physics, is to consider work done by the system on its surroundings as positive.

Wikipedia - Work_(thermodynamics)


Atomic emission spectrum assumption n_f = 3

As the wavelengths are in the infrared region hence the transition occurs at $n > 2.$

Now, if you may have seen the data, these values are all common values of Paschen series. But even if you haven't, you can do a simple test by finding the shortest wavelength of all the series after Balmer. You will find only the shortest wavelengths of Paschen and Brackett are suitable for having values lying in this range as the shortest wavelength for Pfund is $\pu{2279 nm}.$

Then also as the shortest wavelength for Brackett is $\pu{1458 nm},$ you can be sure that the second and third are definitely lying in the Paschen series and check for the remaining wavelength which would be a fairly easy job now.

Directly assuming that the final energy level is $n = 3$ isn't correct, I think, but some analysis can help us to draw the above-mentioned conclusions.


Can I claim that a hot dip Aluminum coating in steel provides cathodic protection to a substrate since Al can act as sacrificial anode?

In practice, no -- at least, not at ambient temperature.  Aluminum passivates and this passive film breaks the potential electrochemical circuit.  The passivated aluminum can't react to release electrons to the steel.  For cathodic protection you need something that will corrode instead of the steel, meaning it can't passivate.

Usually, cathodic protection is offered by zinc which has less driving force, but it does not passivate and thus the electrochemical circuit remains intact.

Addendum:

Aluminizing does offer cathodic protection at high temperature where the passivation is less effective.


Why do acids mix in water faster than bases?

You cannot compare the rate of first mixing acid with water with the rate of mixing the base with the diluted acid. You should compare the rate of mixing the acid with water with the rate of mixing the base with water instead.

The reason is that adding the base to the diluted acid will neutralize portions of the base during the mixing process resulting in lower concentration gradients that make the mixing slower.

A simple gedankenexperiment will show this. Consider a cylindrical tank with water of volume 1 with a mixing rod along the axis. Now put an amount of 1 acid into the tank along its axis. When the acid has made half the way to the wall of the tank the concentration in the inner volume is 4. The gradient is 4.

Now think of adding an amount of 1 base to the completely mixed diluted acid. When the base has made half the way to the wall of the tank, 1/4 of the base has already been neutralized. The concentration in the inner volume is 3. The gradient is only 3 compared to 4 before.


How can I formulate a rate equation using a solid catalyst?

You will perform the reaction inside a container or solution of some particular volume. This means that you can change the catalyst concentration inside that volume by adding more or less catalyst in different experiments. So you don't increase the volume of the solid (i.e. change its density), you change the mass of solid inside the reaction volume.


Coupling constant in 1H NMR with phosphorus

Yes, this is the 2JP-H coupling. You could verify this (if you had access to a NMR spectrometer) by running a 1H{31P} experiment; that is, a proton with 31P decoupling. This doublet would collapse back to a singlet. You would also notice some changes to the phenyl protons, as they all have some coupling to the 31P also. A 2J coupling of 13.5Hz is very typical for this type of 1H-31P coupling.

The magnitude of 1H-31P couplings varies enormously, depending on bond order, oxidation state of 31P and bonding arrangement.

1JPH of PH3 = ~175Hz.
1JPH of HP(O)(OMe)2 = ~1200Hz
2JPH P(CHx)y are typically 1-20Hz

It is sometimes observed in 31P-1H (and 31P-13C) that high bond order couplings are larger than lower bond order couplings (ie 3JPH > 2JPH)


How to compare the Heat of Combustion qualitatively?

For isomers differences in the heat of combustion are due to differences in the internal energy of the educts. If intermolecular interaction (lattice energy etc.) can be neglected it reflects differences in the internal energy of the molecules.

In your first example C has the smallest heat of combustion since it has the lowest internal Van der Waals strain.

In the second example the order is given by the ring strain, and the Van der Waals strain between the two methyl groups at the cyclobutane ring.


Is E1cB possible in the reaction of cyclopentadiene with acetone in presence of ethanolate?

I believe you were right to conclude E1cB would occur. Both of your considerations are exactly on point. The amount of ring strain is minimal -- if we think that it's OK to deprotonate to form a sp2-like carbon in the first place, then we've already decided that the added ring-strain is not a huge issue. Of course, the added conjugation and the entropic component of eliminating a hydroxide ion also help to drive the reaction forward.

A literature source that confirms our thinking is shown below, for your specific example no less, and it looks like there are many derivatives for which the elimination is observed. [1]

I'll go further to note that Scifinder didn't turn up any reaction results for  for your alcohol product, or any similar derivatives, under equilibrium basic conditions. This suggests that elimination is actually quite likely, and that your answer key is wrong. It is however, possible to isolate the alcohol if acid-base equilibrium between the alkoxide intermediate and solvent/conjugate-acid is minimized.[2]

References

Ragauskas, A. J.; Stothers, J. B. 13C magnetic resonance studies. 119. Tricyclo[3.3.0.0] and [3.3.1.0]octanones from substituted norbornenones via cyclopropanation and homoketonization. Can. J. Chem. 1985, 63 (11), 2961–2968. DOI: 10.1139/v85-491.
Smits, G.; Audic, B.; Wodrich, M. D.; Corminboeuf, C.; Cramer, N. A β-Carbon elimination strategy for convenient in situ access to cyclopentadienyl metal complexes. Chem. Sci. 2017, 8 (10), 7174–7179. DOI: 10.1039/C7SC02986A.


Creating small amounts of cellulosic ethanol from bamboo

A process for creating fuel ethanol from bamboo is described in this original paper and this follow-up. The basic steps these researchers took are:

Bamboo saccharification with sulfuric acid
Removal of color compounds with activated carbon
Separation of acid and sugar (using an ISMB system)
Hydrolysis of oligosaccharides
Continuous ethanol fermentation of saccharified liquid (with yeast strain)
Distillation

There are also other, slightly different processes to do this (e.g. enzymatic hydrolysis, but this is more expensive).

I'm no expert on this, but it's my understanding that creating ethanol from lignocellulosic biomass is still in the research phase. AFAIK there isn't a good and economically viable process yet to create ethanol on a large scale. Several steps still need to be optimized and costs need to be reduced, so I don't think it's very interesting for you to do small scale production.


I think elements can be broken down, but their definition says otherwise, why?

What definition, where? The short answer is given in the IUPAC Gold Book:

  chemical element

  A species of atoms; all atoms with the same number of protons in the atomic nucleus.
  A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under 1, but mostly the term chemical element is used for both concepts.

A longer answer is: Yes, physicists have managed to split atoms further down than to nuclei and electrons. It is also possible to "transmute" an element into another, such as creating Plutonium from Uranium. This involves a lot of machinery and energy, whereas splitting a substance into its constituent elements can be as simple as heating or dissolving in an acid and could be done in an afternoon with a few glass beakers.

Looking at it from a more theoretical perspective, one can (snarkily) define Chemistry as "the Physics concerned with the outer electron shell". Once you work on other areas of the atom (that is, the nucleus) or an electron shell of an atom can no longer be reasonably defined (plasma as in the sun), you tend to leave the area of chemistry (and the applicability of the term "element").


How can 32 g of Gatorade powder contain 33 g of carbohydrates?

Is there a chemical process as some people have suggested that is changing the 32g of powder into 33g of carbs once water is added?

No. Even if there was, it's not relevant in this case: the numbers on the label are an assay of the pure contents of the product before it's mixed with anything, or prepared or altered in any way. Nutrition labels refer to the product as-is.

In the United States, the Food and Drug Administration is in charge of labeling the quantities you're interested in. The relevant document is the Food Labeling Guide, one of many documents which are linked from a more comprehensive list called Labeling and Nutrition Guidance Documents and Regulatory Information. As per the comments (below), I'm noting that it appears as if the product pictured in your post is from the Canadian market and thus subject to nutritional labeling dictated by agencies in Canada. That said, the methodologies used by both the US and Canada are similar.

There are at least 2 regulations detailed in the labeling guidelines that we need to examine here to explain what's going on:

(1) The numbers reported on the container have simply been rounded up or down as per the Food and Drug Administration's Appendix H in the Food Labeling Guide. For instance, they show that for Total carbohydrate and Sugars, anything less than 0.5 g is expressed as 0 g, and quantities greater than or equal to 1 g are to be expressed in the nearest 1-gram increment. This is why of the 33 g of total carbohydrate, only 31 g appears to be accounted for as sugars;

and

(2) Total carbohydrate is calculated in a way that propagates rounding errors introduced as detailed in (1):

  Total carbohydrate is calculated by subtracting the weight of crude protein, total fat, moisture, and ash from the total weight of the sample of food.

In summary: The numbers don't add up because quantifying the components of a food to conform to the rules and regulations for reporting them on a Nutrition Facts label allows for the accumulation of rounding and measurement error.


Decrease in Surface area leads to Increase in load capacity for mesoporous material?

Although I only have access to the abstract of the link you provided (unless you want to spot me $40 ;) they have characterized their glutaraldehyde-functionalized SBA-15 surface using at least 7 different analytical techniques, so I'm pretty convinced that the surface area, pore diameter and hydrophobicity are as they say.

The ~50% increase in surface adsorption in spite of the 13-fold decrease in surface area just illustrates the fact that surface adsorption is a function of two parameters: the affinity of the compound for the surface in question and the area of the surface available for adsorption.  It seems that for the system you've described, these two parameters simply switched from high-surface area / low adsorption affinity to low-surface area / high adsorption affinity, with the adsorption affinity "winning out" over the decreased surface area.


The proper explanation for the general formula of the DU?

My answer is based on V. Pellegrin, J. Chem. Educ. 1983, 60 (8), 626 which might be inaccessible for you by a pay wall.

Starting with Eulers characteristic for planar graphs wich states that adding the faces and vertices of a planar graph and substracting all the edges must be 1 $$F+V-E=1$$ which can be rewritten as $$F=1+E-V$$
Now in more chemical words we can redefine this by saying that edges are bonds and vertices are atoms and the $F$ will be our $DU$ from now on. If we know the structural formula we could stop now, count all atoms and bonds and calculate what we want to know. Those are two examples from the paper:

But what if all we have is a sum formula from let's say mass spectrometry?

Then we have to calculate how many bonds this molecule will have based on the knowledge of how many atoms the molecule contains and how many valence electrons each atom has.

Each molecule consists of $$N=\sum_i n_i$$ atoms of element $i$ whereby $n_i$ denotes the amount of each element. Each atom has to be assumed in it's lowest valency state and has $v_i$ valence electrons which would make in total $$\sum_{i=1}^N n_i~v_i$$ valence electrons. As every single bond contains two electrons, the amount of all possible bonds is half the amount of all valence electrons, namely $$\sum_{i=1}^N \frac{n_i~v_i}{2}$$

Let's try it with the mass formula $\ce{C9H7NO}$:

The molecule consists of $N=18$ atoms.
The atoms are grouped in $4$ groups containing the elements $i=\{C,H,N,O\}$
Each element is contained $n=\{9,7,1,1\}$ times.
The lowest valency of each contained atom is $v=\{4,1,3,2\}$
The sum of all valence electrons is $\sum_{i=1}^N n_i v_i=9\cdot4+7\cdot1+1\cdot3+1\cdot2=48$
The sum of all possible bonds is $\sum_{i=1}^N \frac{n_i~v_i}{2}=48/2=24$

Now let's insert those numbers into the equation $$DU=1+E-V=1+24-18=7$$ Because I knew the moleule that I have chosen we can compare

and see that the calculation was correct ($5$ for the double bonds + $2$ for the ring closures).

As it was correct, we can combine it and derive your equation.
$$\begin{align}
F=&~1+E-V\\
DU=&~1+\text{number of bonds}-\text{number of atoms}\\
DU=&~1+\sum_{i=1}^N \frac{n_i~v_i}{2}-\sum_{i=1}^N n_i\\
DU=&~1+\sum_{i=1}^N \left(\frac{n_i~v_i}{2}-n_i\right)\\
DU=&~1+\frac{1}{2}\sum_{i=1}^N (n_i~v_i-2~n_i)\\
DU=&~1+\frac{1}{2}\sum_{i=1}^N n_i~(v_i-2)\\
\end{align}$$


As defined in an excerpt from this Wikipedia entry:

  "In chemistry and physics, Dalton's law (also called Dalton's law of
  partial pressures) states that in a mixture of non-reacting gases, the
  total pressure exerted is equal to the sum of the partial pressures of
  the individual gases."

Where this assumption falls apart for your question is that $\ce{NH3}$ and $\ce{HCl}$ readily react with each other, both in the gas and aqueous phases.  This alone renders Dalton's Law mute.


Ionizable and non-ionizable in this context refers to the observation that the counterions of a complex ionize in solution and can thus be precipitated and the ligands of the complex cannot and are thus non-ionizable.

There are two things at work here:

All complexes exist in an equilibrium with the free ligands. In the case of diamminesilver(I), $$\ce{Ag+ + 2NH3 <=> [Ag(NH3)2]^+}$$ the equilibrium is overwhelmingly toward the complex and no AgCl will be precipitated if chloride is added. However, if we add iodide instead, we will actually observe an AgI precipitate because AgI is far less soluble than AgCl and there is sufficient $\ce{Ag+}$ that we see some precipitate. Therefore, there is actually always a little bit of $\ce{Ag+}$ free in solution.
The Tollen's test is a redox reaction. Unlike the case above where we're precipitating an ionic compound and thus need to form an ionic bond, electron transfer to the silver is quite easy and once reduced, it is unfavourable for the silver to remain in solution. Even more extensive ligand coverage in something like $\ce{[Fe(CN)6]^{4-}}$ doesn't prevent redox reactions from occurring. Even some metalloporphyrins undergo redox reactions at the metal centre.


Why do cations have higher ionization energies than neutral atoms?

Consider a simple thought experiment. Your ionisation process is essentially removing a (negatively charged) electron from a remainderatom.

$$\ce{Atom^n+ -> Remainderatom^{(n+1)+} + e-}$$

Throughout this process, you are separating charges, thus there is an electrostatic force driving the electron back to its remainderatom.

If the remainderatom has a charge of $2\ (\Rightarrow n=1)$, the attractive force is twice as strong in simplistic theories than if remainderatom’s charge is $1\ (\Rightarrow n=0)$.

Thus, it is twice as hard to remove a second electron.

This thought experiment makes use of a number of simplifications. For example, it does not consider electron energy levels (orbitals), it does not consider orbital contraction upon ionisation and much more. But it does a good job at explaining the general trend.


Suggest possible structures for the compound PF3(Ph)(NMe2) using the NMR data

I agree that the trigonal bipyramid structure is what we are dealing with, so let me focus on that structure.

The nmr data shows that 2 of the fluorines are equivalent and one is different. This results in 3 possible isomers as shown below.

Both $\ce{^31P}$ and $\ce{^19F}$ are spin 1/2 nuclei, so each of these isomers would produce phosphorous and fluorine nmr spectra with splitting patterns and integration ratios consistent with the reported data. That is, the phosphorous will split each different fluorine into a doublet, the two equivalent fluorines will split the singular  fluorine's doublet lines further into triplets.  The one non-equivalent fluorine will split the other two equivalent fluorine doublet lines into doublets.  Similar analysis of the phosphorous signal leads to the expectation of a doublet (from the singular fluorine) of triplets (from the two equivalent fluorines).

Two additional pieces of information:

Fluorine prefers the axial position in pentacoordinate phosphorous
(see here)
Because the orbitals involved in axial and equatorial bonding in pentacoordinate phosphorous compounds are different, the axial and equatorial coupling constants will be different.  Generally, $\ce{J_{PF(eq)}}$ > $\ce{J_{PF(ax)}}$ because there is increased s-orbital density in the equatorial bond (see here, for example)

The larger reported P-F coupling constant (959 Hz) is associated with the intensity = 1 fluorine signal.  This suggests that structure I, with one equatorial fluorine (with the larger $\ce{P-F}$ coupling constant) and 2 axial fluorines (with the smaller coupling constant and as preferred electronically - point 1 above) is the structure being observed.

Part B Experiments:

The trigonal bipyramid structure could be confirmed by heating the sample while in the nmr spectrometer.  As Berry pseudorotation commences the 2 fluorine signals should coalesce and determination of a barrier height around 35-40 kcal/m would be expected.

Also, compounds II and III must have similar energies and both would be expected to be present at equilibrium.  If we heated our sample so that pseudorotation took place, and then upon cooling we saw an additional set of signals, that would be consistent with II or III being our starting compound.  No new signals upon cooling would support structure I.


How is indigo carmine used in Lowenthal-Permanganate titration as a redox indicator?

I did some brief looking but given that the carmine is also pH indicator that remains blue at below 11.4 pH it would appear as though your sulfuric acid is preventing the reduction from occurring in your indicator. Under acidic conditions KMnO4 is a strong oxidizer where carmine is a weak oxidizer under these conditions. However I am not sure how the carmine gets reduced to yellow upon equilibrium. Perhaps the pH changes. It appears as though permanganates acid conjugate is strong so its a weak base. Just throwing out ideas there though.
http://chem.lapeer.org/Chem1Docs/IndigoCarmine.html
https://www.sciencedirect.com/topics/earth-and-planetary-sciences/permanganate


Bandgap of a semiconductor for photocatalytic water splitting

The original purported advantage of semiconductor|liquid junctions proposed by those in the field of photoelectrochemistry arose from the hypothesis that charge carriers would have a shorter distance to travel than in solid-state photovoltaic systems, thereby allowing lower quality materials to be used in an integrated solar energy system and still achieve high conversion efficiencies. In theory, one would also be able to avoid forming metallurgical junctions, say those p-n junctions made by using different dopants and creating a well-defined interface to separate charge carriers. Instead of forming these solid-state structures, you can define the electric field by selecting a suitable redox couple in solution. The specific energetics of the semiconductor and redox couple will then define the maximum power obtainable from the given system, so this is why the valence band must be below the energy of the OER and the conduction band above the energy level corresponding to the HER. By natural induction, this means that the bandgap must meet or exceed 1.23 eV, but this is only a thermodynamic constraint.

In reality your system’s bandgap must actually exceed 2.0 eV, this is due to the kinetic constraints based in part on the OER (and to a lesser extent the HER ) and losses encountered within the light absorber. Of course, for a single light absorber there is a tradeoff between the percentage of utilized solar flux and the bandgap.

This explains why many efforts are focused on using dual light absorbers, with one dedicated to the anode and another for the cathode. In this tandem structure they can efficiently utilize more of the solar spectrum. If you were to stack these light absorbers and use a 1.2 eV and 1.8 eV materials (with each one carrying out one of the two half-reactions required to split water), you could in theory get twice the efficiency compared to a single absorber with a bandgap of 2.2 eV.

Now are these semiconductor|liquid junctions going to beat a dedicated photovoltaic solar panel/electrolyzer any time soon? Maybe, but probably not. Is water splitting really worth doing? Maybe, but probably not. Is it a beautiful idea? Definitely.


Are there any other reagents except HNO3 and H2SO4 for Nitration?

It's getting on for 3 decades since since I've done organic, so a proper organic chemist can probably give a more complete answer, but yes, there are other ways to do this. In the $\ce{HNO3}$/$\ce{H2SO4}$ system the "real" nitrating agent is nitronium, $\ce{NO2+}$, so anything that can form that will nitrate benzene. One example is simply to use a nitronium salt such as nitronium tetrafluoroborate, $\ce{NO2+ BF4-}$. As for others I would be surprised to find that something like $\ce{NO2F}$/$\ce{BF3}$ doesn't work, but I don't know for certain - it's been a looooong time!


Kinetic isotope effect - why transition states taken as isoenergetic

When you deal with energies, absolute energies never matter. Energy only makes sense when you consider it relative to some reference point. For example, for standard values, we consider standard states. You're perfectly fine declaring that both TS's have the same energy because you're not comparing them to each other; you're comparing them to two ground states.


Do we consider the spatial orientation of a molecular orbital while assigning plane of symmetry?

The reason that orbitals are not considered in the symmetry of a molecule is that the orbitals cannot decrease the symmetry of the molecule. If we look at the canonical delocalized molecular orbitals of any molecule, everyone of them is symmetric or antisymmetric with respect to a symmetry operation of the molecule, with those symmetry operations determined only by the positions of the atoms, not the electron density.
Looking specifically at carbon dioxide, the simplistic picture of $\ce{CO2}$ as being like allene with lone pairs instead of the C-H bonds is not an accurate view of the electron density. If one looks at the delocalized pi-type orbitals, both span all three atoms in either the xz or yz planes (using the standard convention of the z axis being aligned with the molecular axis). Thus, both orbitals are symmetric with respect to the xy plane of symmetry of the molecule (which passes through the C).
Similarly, the lone pairs do not occupy separate sp2 orbitals that have specific alignment in a plane perpendicular to the p orbital involved in pi bonding. Instead, the create a combined electron density that can be deconvoluted into an sp-type orbital along the z-axis and partial contributions from the px and py orbitals, such that the total has no bias along the x or y axis.
If we were to look at the total electron density of $\ce{CO2}$, we would therefore see only a pill shape that does not vary as one rotates around the z-axis.
To visualize this, I recommend looking at all of the calculated delocalized molecular orbitals of $\ce{CO2}$. You can do this on the molcalc.org website if you do not have other software for that purpose.


The OH- and the H+ will react, neutralize each other to form H2O, H+ + OH- = H2O. Hence the ions remaining is: 0,0029... - 0,0025... = 0,00048... which will be OH- ions. I believe you can proceed on your own from here. Just calculate the concentration and then the pH.


Best way to clean a gold-plated yoyo bearing?

In most of the case distilled water is enough, but you should be sure that there are no salts inside your object if you can perform a Conductometry test on the washing water follow this procedure: Wash the item inside a buck for 3 minutes with distilled water measure Conductometry, take another bucket with the same amount of water wash the item for 3 minutes, measure Conductometry repeat these steps until conducibliy drop down.

This is a Conductometry test I've made while restoring an ancient metal object. You can see that I needed 5 washing to get good result.

If you have high conductivity you should identify the salts with some spots test to understand where they came from.

If you have grease you can use a surfactant like Plysorbate 20 2% in distilled water not more!
Eventually if there to much salts and grease you can use: Thiourea solution 8-10% Sodium thiosulphate 30% EDTA 10 % (reference here).

Remember to dry very carefully the object with a hair dryer, NEVER leave the object wet for more than a couple of minutes!

Eventually you can use a protective like Soter or Zapon these normally cause in terms of colour change a $\Delta E$ of 2-4 so ask to your self if you really need it for a gold object.

In my opinion avoid ultrasonic cleaning for composite materials! If you really care about the object ask for a metal restorer!


What's the definition for mole before Avogadro's constant?

These constants have been named in honour of these people. They did not even know the existence of mole concept.

In order to search how early precursors to Avogadro's number's were found, read about Loschmidt constant and then move to Avogadro's story. There were many $independent$ methods to determine Avogadro's constant. Loschmidt was trying to estimate size of the "air molecule" in 1865.  Number of particles in air in a given volume.

The determination of Faraday's constant does not require the knowledge of Avogadro' ideas or his laws. This is a modern misconception. However, it does require the concept of gram equivalents (precursors to atomic mass or formula weights). In Faraday's time, measuring current/voltage and masses of substances was not a big deal. By electrolyzing a solution of silver salt, you can deposit a known mass of silver on the electrode. I am showing a page how Faraday's constant was found. As you can see we never needed to invoke Avogadro's law or numbers and constant. Faraday did such neat experiments! Clever and simple.

Electro-chemistry: General theory, Volume 1, pg 4, By Robert Alfred Lehfeldt, T. S. Moore. Longmans, Green and Company, 1904.


Why don't salt crystals attract each other?

Sodium chloride ("salt") is made of a giant lattice of ions: Na+ and Cl-.

The crystals overall do not have any net charge as the the number of chloride ions is equal to the number of sodium ions. So, overall, there will be no electrostatic force between crystals as they are neutral in the bulk compound.

On the surface, crudely and without getting into the detail of the crystal structure, the ions will alternate on the surface at an atomic scale. You might imagine that this creates atom-sized patches where there will be net positive or negative charges. Would these attract another crystal? Would these attract the patches of charge in another crystal if they were adjacent. Maybe, but only if the surfaces were both atomically flat and lined up exactly. In reality the surfaces are rough and there is no net charge so large salt crystals won't stick because of this effect.

So the intuition that some crystal made of ions should have strong electrostatic attraction to another similar crystal is wrong because of electroneutrality and the detail of the physical surface.


Does the tetrahedral structure of the alkanes contribute to their lower reactivity?

Hybridization is at the center of the answer.

Due to poor P-P overlap, it only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about a double bond involving ${sp^2}$ hybridized carbons, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond involving ${sp^3}$ hybridized carbon atoms with highly directional orbitals that result in strong bonds. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane. Alternatively one can view ethene as a two-membered ring (ref. 1), no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring. It is easy to imagine that such a ring system would contain a significant amount of strain (ref. 2).  Either way (and by the way these two views of ethene are equivalent and complementary) - using a poorer overlap argument or a higher strain energy argument, we see why alkenes (and even more so alkynes) would generally be more reactive than alkanes.

From this argument we would expect that if we were to somehow increase the strain in an alkane, then we should increase its reactivity. A good example of such behavior can be found in cyclopropane.  Placing 3 "tetrahedral" carbon atoms into a ring dramatically increases the strain energy in the system (it changes the carbon hybridization too, see ref. 3) and the reactivity increases dramatically.  Indeed, cyclopropane adds bromine much like an alkene (ref. 3)

Since someone above commented that ${sp^3}$ $\ce{C-H}$ bonds are "very stable and typically do not want to react". Let me just point out that $\ce{C-H}$ bond energies vary as follows

\begin{array}{|c|c|c|c|} \hline
\ce{C-H} & \text{Bond Strength}  \\  \text{bond type} & \text{(kcal/mol)} \\ \hline
\ce{sp^3 C-H} & 101 \\ \hline
\ce{sp^2 C-H} & 111 \\ \hline
\ce{sp C-H} & 133 \\ \hline
\end{array}

These bond dissociation energies (ref. 4) don't fit the observed reactivity pattern.  They indicate that an ${sp^3}$ bond is actually easier to break than an ${sp^2}$ or ${sp}$ $\ce{C-H}$ bond


How do I (further) lower the pH of a pyrrolidine solution?

A pH of ~ 4 would correspond to a solution of pyrrolidinium chloride.  Use less pyrrolidine or more hydrochloric acid to arrive at that point.

The commercial concentrated hydrochloric acid is 37 % (w/w), this would be a concentration of ~ 12 M, and 164 ml of that is slightly less than 2 mol HCl.  It's very strange that you are seeing a basic pH.


What does the "total energy" of a molecule mean for the Chem3D MM2 minimized energy calculation?

General definition of energy

Usually, the total energy of a system is subdivided into kinetic energies of all particles and potential energies of the interactions between them, where the later terms are most often limited to the strongest Coulomb interactions only. In atomic units we get then:

$$
    E
    =
    - \sum\limits_{\alpha=1}^{ν} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^{2}
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{\alpha=1}^{ν} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{\alpha=1}^{ν} \sum\limits_{\beta > \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $β$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.

Now, the zero of energy corresponds to the case when all the terms are zero. Kinetic energy is zero when velocity is zero, while the Coulomb interaction energy is zero when particles are infinitely separated from each other. Thus, the zero of energy corresponds to the case when all the particles are infinitely far away from each other and not moving. In accordance with that choice of the zero of energy, the energy of any stable molecular system in the modeled environment (usually, an isolated molecule in the gas phase) has to be negative.

Molecular mechanics definition of energy

Molecular mechanics (MM) operates with a different kind of energy. To start from, usually, it is just the potential energy. Secondly, in MM we think about a molecule in terms of atoms and bonds (rather than nuclei and electrons) and the potential energy is divided into contributions from bond stretching, angle bending, torsional rotations around single bonds, non-bonded interactions, etc.
$$
    E = E_{\mathrm{stretch}} + E_{\mathrm{bend}} + E_{\mathrm{torsion}} + E_{\mathrm{nb}} + \dotsc \, ,
$$
where the first term, for instance, is typically proportional to the sum of the squares of extensions from the equilibrium length for all the bonds:
$$
    E_{\mathrm{stretch}} = \sum\limits_{\mathrm{bonds}} k (r - r_{\mathrm{e}})^2 \, .
$$
Now, we could take the energy corresponding to the equilibrium length $r_{\mathrm{e}}$ as the zero of energy and proceed similarly for the other bonded terms, which would mean that (ignoring usually small non-bonded interactions) the MM potential energy is zero when the molecular system is completely "unstrained" (i.e. when all geometrical parameters have their "natural" equilibrium values). However, different molecules are bonded differently, and thus, the zero of energy would, in general, be defined differently.  Each potential energy will thus be computed relative to a different zero of energy which makes the comparison of energies for any two different molecules meaningless.


How to find the molar mass of an iron compound based on its mass and mole fraction in the compound?

Let $\ce Z$ denote the unknown element.
The total amount of iron atom in the compound is,

$$n(\ce{Fe})=\frac{\pu{2.36g}}{\pu{55.845 g\cdot mol^{-1}}}=\pu{0.0423 mol},$$

The molecule comprises of 3 iron atoms and 4 other atoms, so the amount of the unknown atom is,

$$n(\ce Z)=\pu{0.0423 mol}\times \frac43=\pu{0.0564 mol},$$

The molar mass of $\ce Z$ is,

$$M(\ce Z)=\frac{\pu{3.26g}-\pu{2.36g}}{\pu{0.0564 mol}}=\pu{15.97 g\cdot mol^{-1}}$$

So $\ce Z$ is $\ce{O}$. The molar mass is
$$M(\ce{Fe3O4})=\pu{55.845 g\cdot mol^{-1}}\times 3 + \pu{15.97 g\cdot mol^{-1}}\times 4=\pu{231.43 g}.$$


Do electrons hybridize only during bonding?

To expand on orthocresol's excellent comment, you confusion likely comes from a common misconception.

  Electrons and orbitals are not the same things. Electrons are fundamental particles with particle-wave duality. The behavior of electrons can be modeled with wavefunctions. Orbitals are mathematical constructs based on the probability density of the wavefunctions.

Technically, orbitals do not hybridize because orbitals do not exist. Since orbitals are mathematical constructs, they do not correspond to any physically observable entity. Electrons are the physically observable entity. Electrons do not hybridize. Electrons interact quantum mechanically to form bonds when their wavefunctions have appropriately matching energy and symmetry. Electrons then adopt new wavefunctions for their new situation. These multielecton wavefunctions can be intractable.

Hybridized orbitals represent a model that simplifies some of the mathematics by replacing the canonical hydrogenic atomic orbitals with hybridized atomic orbitals that are oriented in directions that match common molecular geometries. Orbital hybridization is a model that is thus only useful when describing bonding. If there is no bonding, hybrid orbitals are a nonsensical model.


Hydrogen bonding leads to deshielding in alcohols. Isolated ethanol in the gas phase has a shift of $\pu{0.55 ppm}$, while the shift is $\pu{0.8 ppm}$ in neat ethanol, so there is a shielding effect due to H-bonding. However, the OH shift is extremely sensitive to self association and/or solvent, resulting typically in greater deshielding. See for instance the table of shifts in reference [1].

According to Hore (in his Oxford NMR book), the charge distribution around the proton is polarized (mainly because the proton is attracted by the H-bond acceptor), effectively deshielding it. In any event, the hydrogen bond acceptor $\ce{O}$ transfers charge to the donor $\ce{O}$. The intervening H apparently gets the hard end of the bargain.

[1]: Fulmer et al, Organometallics 2010, 29, 2176–2179


How to determine whether a given compound is paramagnetic or diamagnetic?

Paramagnetism is a consequence of having one or more un-paired electrons in the outer electronic configuration. e.g. O$_2$, and NO. An an odd number of electrons can clearly produce a paramagnetic ion, radical or molecule. However, molecules with an even number of electrons can also be paramagnetic, the most famous being dioxygen (O$_2$). The experimental observation is that oxygen is attracted to a magnet.(This is best observed with liquid oxygen.) Nitrogen (N$_2$) is not paramagnetic but diamagnetic.  Whether a molecule is paramagnetic or not is only explained using molecular orbital theory. In the case of O$_2$ it is found that the lowest energy state is one that has one electron in each of two orbitals. An alternative would be to expend energy to pair up these two electrons into one orbital. As this requires energy it does not occur. High and low spin transition metal complexes show the same effect of pairing vs. placing electrons into different orbitals.


What were the expectations of Rutherford's gold foil experiment?

The nucleus has a radius roughly 10⁴ times smaller than the size of the atom itself (imagine a sports ball in a stadium). That would mean that its volume were 10¹² smaller than the volume of an atom. Sure the gold nucleus is ~30 times as charged and is ~50 times heavier. But dilute that charge and mass by a factor of a trillion, and suddenly those don't seem so significant anymore.
That's the worst thing about those schematic diagrams that you see for the Rutherford experiment. They blow up the size of the nucleus to prove a point, but those diagrams are definitely "Not to Scale™".


The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

(…)
The stoichiometric name of the compound is then formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes (‘mono’, ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., …). The multiplicative prefixes precede the names they multiply, and are joined directly to them without spaces or hyphens. (…)

So far, the systematic name of $\ce{CaCl2}$ should be calcium dichloride.
However,

Multiplicative prefixes need not be used in binary names if there is no ambiguity about the stoichiometry of the compound (…).

Therefore, the customary name calcium chloride is correct.


To preface this, I'm not sure if this is indeed the correct IUPAC name as per the latest 2013 recommendations. It might well be, but I don't know. However, that's not crucial to your question, which is more about understanding the name that is given on Wikipedia. Here's the structure of amoxicillin:

The amide C=O is in "acetyl", which refers to the $\ce{CH3C(=O)}$ group (see e.g. acetyl chloride, $\ce{CH3COCl}$):

The '24' is probably a typo and should instead be '2'.

Following this numbering scheme you can see that the lactam C=O is in '-7-oxo'.


Aluminium is usually considered to be a metal, as described in the Wikipedia article Metalloids: Aluminium:

  Aluminium is ordinarily classified as a metal. It is lustrous, malleable and ductile, and has high electrical and thermal conductivity. Like most metals it has a close-packed crystalline structure, and forms a cation in aqueous solution.

The reason why, in a few cases, aluminium is referred to as a metalloid, from the article:

  It has some properties that are unusual for a metal; taken together, these are sometimes used as a basis to classify aluminium as a metalloid. Its crystalline structure shows some evidence of directional bonding. Aluminium bonds covalently in most compounds. The oxide $\ce{Al2O3}$ is amphoteric, and a conditional glass-former. Aluminium can form anionic aluminates, such behaviour being considered nonmetallic in character

The line between metals and non-metals is somewhat arbitrary with

  Elements to the lower left of the line generally display increasing metallic behaviour; elements to the upper right display increasing nonmetallic behaviour.

Boron, carbon and silicon (all on the other side of the boundary line) are hard and brittle and are semiconductors, which is a major reason why the majority of resources list Aluminium as a metal, largely due to the metallic properties listed in the first quote.

A related question is Metal Compounds that bond covalently


i] There is chlorophyll-b (yellow-green) and carotenoids (yellowish).

ii] The green chlorophyll isn't an evolutionary mistake (unless, of course, you take the viewpoint that evolution IS just mistakes). Plant's get plenty of sunlight. The sun is way stronger than indoor lights used to grow plants. They don't really need more energy from the sun. Plants actually have some pigments to dissipate light because they would otherwise absorb too much.

iii] Plants might absorb too much energy. Think wearing all black on a hot summer day, and standing out in the sun ALL day.

iv] Plants don't really use ultraviolet light as far as I know. With the ozone layer blocking most of the ultraviolet rays coming at us, it would be very strange if they did evolve to absorb ultraviolet light. This might also harm the plant. Infrared rays do reach plants, but they don't really do much to help in photosynthesis. Anyways, light has to travel through the cytoplasm of plants which is full of water, and visible light is best at travelling through water.


What is the reason that ice melts with increasing pressure at constant temperature (revelation)?

The potential energy between the O atoms in the H bond has the form of an anharmonic potential an example of the shape is the Morse potential or a Lennard-Jones 6-12. The energy is zero at large O-O separation and moves to a minimum as this distance is reduced (the point at which a the H bond exists with the H atom between the two O atoms, typically 0.15-0.25 nm). As the O-O distance is further reduced the potential rises again to zero then positive potential energy. The H bond thus forms in a lop-sided potential well, with the H atoms between the two O atoms. (I imagine the effect on the covalent bonds would be less as they are stronger and shorter than the H bond.)
Suppose that the distance between the O-O atoms is reduced then the potential rises and becomes closer to zero or repulsive (positive) potential. Suppose that the pressure causes this to happen just a little and is such that the H bond energy is now only slightly below zero. At this point the thermal energy (as molecular vibrations) could in places be enough to increase the energy above zero and so break the H bond, at which point the solid would start to melt. The structure beginning to break down could further collapse rather like a building collapses by bending and twisting if it structural girders are removed.


We define anode to be the electrode where oxidation takes place.

Oxidation signifies loss of electrons and Reduction signifies gain of electrons.

For example, in this electrochemical cell,

Copper has more reduction potential than Zinc. So it will be easy for copper to reduce (or gain electrons) and if copper is reducing (or gaining electrons) then there must a source of these electrons. These electrons are supplied by Zinc (which gets oxidized).

At anode, $Zinc(s)$ in losing electrons and converting into $Zn^{+2}(aq)$ ions and at cathode, $Cu^{+2}(aq)$ ions are gaining these electrons lost by $Zinc(s)$ and converting into $Cu(s)$.

These reactions occur because they are spontaneous $(\Delta G =-ve)$. Spontaeous reactions occur in nature by themselves without any external driving force. A system aims for least energy $(\Delta H = -ve)$ and maximum entropy $(\Delta S = +ve)$ and its the combination of both these factors that result in spontaneous processes. $(\Delta G = \Delta H - T\Delta S)$.


Do tetrachloroaluminate salts hydrolyse?

NaAlCl4 will hydrolyze according to the following reaction $$\ce{AlCl4^- + \ 2H2O ->  Al(OH)2^+  +  2 H+ +4 Cl-}$$and if there is enough water it could goo further away like $$\ce{AlCl4^- + \ 3H2O ->  Al(OH)3  +  3 H+ +4 Cl-}$$ But the reaction will never produce $\ce{NaAl(OH)4}$ as you think. The substance $\ce{NaAl(OH)4}$ can only be produced in very basic solution, and here the solution is acidic.


The parameter a in front of √C is empirical. It varies from solute to solute.
And yes, the dissociation is complete at infinite dilution.
Dissociation is the separation of ions when dissolving a salt in water. If NaCl is dissociated, it yields two independent ions, one Na+ and one Cl−
Ionization is what happens when an atom or a molecule looses one electron, without being modified or broken. If $\ce{NaCl}$ is dissociated, it gives two independent ions, one $\ce{Na+}$ and one $\ce{Cl−}$
If $\ce{H2O}$ is ionized, it gives an ion $\ce{H2O}$+ and one electron, but the atoms of the molecule are not separated from one another.


Several things that may help your situation, listed roughly in the order in which I'd try them:

1. Application of the sample to the TLC plate

TLC is most usually ran by applying a small spot of the sample to the plate, and allowing it to run up. Analytically, this gives the greatest resolution, and as such for things such as monitoring reactions or identifying components in  column chromatography fractions, spots are the most widely used.

In some cases, the TLC plate is used to purify components of the sample, this is called preparative layer chromatography, or PLC for short. In PLC, bands are applied to the plate, to allow greater quantities of the sample to be applied and hence separated.

When applying bands to the PLC plate, it's crucial to ensure that the band doesn't get too close to the edges (as it has in the images you've uploaded), as a rule of thumb, at least 1/2 cm from each side should be left clean.

The reason for this is that the solvent doesn't run perfectly level up the plate, this is most evident at the edges of the plate. If you re-run your TLC plates with a shorter band, you should see a decrease in the curving effect that you're observing.

2. Quantity of sample loaded onto the TLC plate

Looking at the images you've provided, you may also have loaded too much sample onto the TLC plate, saturating the silica and reducing the separation.

You could try either increasing the size of the plate, increasing the thickness of the silica layer (special PLC plates are available with thicker layers of silica, allowing more material to be loaded), or simply just diluting your sample and loading less on.

3. Solvent system

Although you say that "this solvent system has been used to run similar wax extracts", its always worth trying different eluents with your chromatography, unless your sample is identical to the prep you're following.  A different modifier (formic acid or trifluoroacetic acid often work well), or different solvent mixtures (methanol in dichloromethane is often a lifesaver) might help tighten up the bands.

Finally, and possibly most of all, make sure you completely dry the plate out before visualising it (leave it sat on the side for a few minutes or use a hot air gun).


There are two isomers of pyran, 2H-pyran and 4H-pyran.

In principle, it is possible to describe the corresponding saturated compounds using the prefix ‘hydro’, which yields the systematic names ‘3,4,5,6-tetrahydro-2H-pyran’ and ‘2,3,5,6-tetrahydro-4H-pyran’. However, these two names actually describe the same compound. In this case, the correct choice is ‘3,4,5,6-tetrahydro-2H-pyran’ because ‘2H’ is lower than ‘4H’ (note that indicated hydrogen atoms have priority over ‘hydro’ prefixes for low locants).

All locants are omitted in compounds or substituent groups in which all substitutable positions are completely substituted or modified, for example, by hydro, in the same way. Thus, in this case, the systematic name is simplified to ‘tetrahydro-2H-pyran’. Furthermore, omission of indicated hydrogen is permitted in general nomenclature if no ambiguity would result. Thus, in this case, a correct name is just ‘tetrahydropyran’.

Anyway, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), preferred IUPAC names of saturated heteromonocyclic compounds are formed by the extended Hantzsch–Widman system. Therefore, the preferred IUPAC name of the given compound is ‘oxane’.


You would get silver at the cathode, and oxygen at the anode.

Since silver is below hydrogen in the spectrochemical series,it tends to get reduced over hydrogen, similarly, since $\ce{NO3-}$ is above $\ce{OH-}$ , it tends not to get oxidised, therefore giving oxygen.

$\ce{Ag+ + e- -> Ag}$ (at cathode)

$\ce{ 4OH- -> O2 + 2H2O + 4e-}$ (at anode)


The tertiary alcohol has more R groups compared to primary & secondary alcohols, there by removing the C-O Bond from a tertiary alcohol is more easier. (the tertiary carbocation is more stable)


How to figure out if the reaction proceeds via SN1 or SN2 mechanism?

As you have the information that the yield depends on solvent polarity. It is the key point to be considered in analysis of type of reaction. In case of a $\ce{S_{N}1}$ reaction, the polar solvent favors the reaction. $\ce{S_{N}1}$ is a two stepped reaction, in which the 1st step is carbocation formation, such that there appears the dipole dipole interaction between the carbocation and polar solvent which favors the reaction.

Whereas in the case of $\ce{S_{N}2}$ reaction, the dipole dipole interaction develops between polar solvent and the nucleophile rather than with intermediate product. It is because the charge is dispersed in two centres in intermediate while charge is centered in single center in nucleophile. This deactivates the $\ce{S_{N}2}$ reaction and the reaction cannot proceed forward smoothly.

Since you have another specific information that the higher the solvent polarity, the higher the yield. Since higher solvent polarity favors the $\ce{S_{N}1}$ reaction, the higher will be the yield through a $\ce{S_{N}1}$ mechanism. So the reaction with tendency with higher solvent polarity and higher yield is a $\ce{S_{N}1}$ reaction.


Heat of combustion of stereoisomers of 1-tert-butyl-2,4-dimethylcyclohexane

Which of the compounds below have the smallest heat of combustion?

The question is asking, which of these 4 compounds is the most stable.  The most stable compound will have the lowest heat of combustion.

In the cyclohexane ring system, bulky substituents prefer the equatorial position rather than the axial position.  This is due to steric (this can also be described in electronic terms since it is due to the repulsion between interpenetrating electron clouds) destabilization. For example, in methylcyclohexane, the methyl group can be in an axial or equatorial position.  In the axial position, it is destabilized by the steric (electronic) interactions it has with the axial hydrogens on carbons 3 and 5.

These interactions are nicely described by cyclohexane A-Values.  The A-Value measures the energy difference (in kcal/mol) between having the substituent in the axial and equatorial positions. The larger the A-Value, the larger the preference the substituent has for the equatorial position. The A-Value for a methyl group is 1.7 kcal/mol. The A-Value for a t-butyl group is so large (>4 kcal/mol) that the t-butyl group is effectively locked into the equatorial position. t-Butyl substituted cyclohexanes do not undergo chair-chair interconversion - this would place the t-butyl group in an axial position, hence they are conformationally locked into the chair with the t-butyl in an equatorial position.

It is this last piece of information that helps simplify our problem.  Here are the structures of the 4 cyclohexanes redrawn in the chair conformation.

Each structure has the t-butyl group in the equatorial position.  Compound 1 has one methyl group in an axial position; this adds 1.7 kcal/mol to the energy of the molecule. Molecule 2 has two axial methyl groups, therefore it is 3.4 kcal/mol higher in energy.

I'll leave molecules 3 and 4 to you to figure out.  Once you determine their relative energies, which one of the 4 has the lowest energy relative energy?

Note: There are exceptions to the additivity of A-Values. See near the bottom of this page for a few examples.


Deciding higher priority order between a triple bond and tertiary carbon

From the Wikipedia article on CIP rules.

  If an atom A is double-bonded to an atom B, A is treated as being singly bonded to two atoms: B and a "ghost atom" that is a duplicate of B (has the same atomic number) but is not attached to anything except A.

Ghost atoms are attached to resolve the triple/double bond.

So according to CIP rules the two compounds can be resolved as

 The pink atoms are ghost carbons

From here we can see that ethyne has a greater preference over isobutyl as after the first point of difference we move over to the next highest order carbon.In ethyne the second carbon is connected to $\ce{C2H}$ whereas the next highest order carbon in isobutyl is connected to $\ce{H3}$.


Mechanism for chloromethylation of benzene with formaldehyde and HCl

This reaction is chloromethylation, similar to Blanc chloromethylation, but using $\ce{AlCl3}$ as co-catalyst instead of $\ce{ZnCl2}$. These are reactions belonging to a group related to Friedel-Crafts reactions, but characterized by usage of protonation instead of coordination with molecular Lewis acids

Mechanism of your reaction (not accounting for influence of co-catalyst) can be written as:

(source)

$\ce{AlCl3}$ coordinates to $\ce{HCl}$ creating adduct ($\ce{HAlCl4}$) which is much stronger acid.


Fundamentals of mathematics and physics for physical chemistry

There are three levels of scholarship. An analogy is that someone has looked at the ocean from far away and he feels that he has seen it all (standard undergrad mindset). Next stage, is when you walk on the beach and your feet become wet and dirty with sand. You have a feeling how ocean water and the beach feels like (fresh MS/PhD). The higher level of scholarship is when you are deep diving into the ocean and realize there is too much to learn and the knowledge you have is quite limited (a mature scientist who has spent his life in research).

Since you are saying that you have not taken mathematics and physics courses, it may not be the right time to begin a proper physical chemistry course.

Couple of nice books:

Mathematics in Chemistry An Introduction to Modern Methods by Hecht

Mathematics for Physical Chemistry by Robert G. Mortimer

Get hold of them from a library and get a feel for it (the first level of scholarship). You don't need to master all to open an undergraduate physical chemistry book. You rather learn it along the way. See if you like the subject in general.


Are there other carbon free gasses that can be used for energy transport, apart from hydrogen?

Yes, ammonia has been cited.

Per a source, 'Ammonia as a Hydrogen Source for Fuel Cells: A Review', to quote:

  Like hydrogen, ammonia is carbon free and can be produced from any energy resource. However there are also some significant advantages in terms of storage and transport.


TL;DR: According to current IUPAC recommendations, $m/z$ is an abbreviation for a dimensionless quantity.
Use of thompson unit is indeed currently discouraged.

There is an overview provided in Definitions of terms relating to mass spectrometry (IUPAC Recommendations 2013) [1, p. 1516] (reference numbers were updated):

Labeling of mass spectra
The labeling of the $x$-axis of a mass spectrum engendered the most discussion during the creation of this document; however, in spite of a general desire for a better way to label the $x$-axis of mass spectra, there was no broad consensus for any of the proposed changes.
Therefore, this document continues the use of the definitions of the Gold Book [2] and the similar definitions in the Orange Book [3].
The Gold Book recommendation is for the use of $m/z$ as an abbreviation for mass-to-charge ratio, a dimensionless quantity obtained by dividing the mass number of an ion by its charge number [2].
The thomson unit, defined as the quotient of mass in units of u and the number of charges $(z),$ was proposed nearly two decades ago [4], but has not been widely adopted and is therefore not recommended.
Labeling the $x$-axis of a mass spectrum with any unit of mass such as dalton (Da), atomic mass unit (amu), or unified atomic mass unit (u) is strongly discouraged due to the confusion that would result when reporting spectra of multiply charged ions.
The quantity plotted on the $x$-axis of a mass spectrum is a function of both the mass and charge of the ion.
Furthermore, the use of amu in place of u is strongly discouraged in all cases; it has been used to denote atomic masses measured relative to the mass of a single atom of $\ce{^{16}O},$ or to the isotope-averaged mass of an oxygen atom, or to the mass of a single atom of $\ce{^{12}C}.$

Gold Book entry:

mass-to-charge ratio, $m/z$
The abbreviation $m/z$ is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number. It has long been called the mass-to-charge ratio although $m$ is not the ionic mass nor is $z$ a multiple or the elementary (electronic) charge, $e$. The abbreviation $m/e$ is, therefore, not recommended. Thus, for example, for the ion $\ce{C7H7^{2+}}$, $m/z$ equals $45.5$.

For historical purposes the aforementioned letter to the editor regarding the thompson unit [4]:

The ‘Thompson’. A Suggested Unit for Mass Spectroscopists
Recently, there has been a rapid increase in experiments in which multiply charged ions are generated in mass spectrometers. […]
This welcome development has permitted the measurement of molecular weights of compounds having masses in the tens and even hundreds of kilodalton.
These advances exacerbate the problems caused by a common imprecision,
namely, the terms mass measurement, mass range, etc are often used imprecisely when mass-to-charge ratio measurement, mass-to-charge range, etc are intended.
The implicit assumption that the charge state of ions is unity is no longer valid when very-high-charge states are so easily produced.
With this letter come two suggestions.
First, that the longer but more exact term be used whenever it is intended.
Second, that a unit of mass-to-charge ratio be adopted.
After all, this is the quantity which all mass spectrometers measure; mass is a derived quantity requiring an independent measurement or knowledge of charge.
Such a unit would be defined as the quotient of mass, in units of u* and the number of charges, $z.$
The number of charges could be positive or negative, depending on the sign of the charge.
The name Thomson suggests itself in view of J. J. Thomson’s contributions to measurement of this quantity and his preeminent role in the evolution of mass
spectrometry.
Using standard rules for abbreviation, we have $\pu{1 Th} ≡ \pu{1 u/\text{
atomic charge}}.$
For example, the molecular weight of the peptide myoglobin (isotopic average molecular weight $\pu{16950 u})$ can be deduced from measurement […] of a peak at $\pu{998.0 Th},$ provided it is known that the ion bears $17$ charges.
If this suggestion is accepted, other simplifications ensue.
For example, the benzoate anion, mass $\pu{121 u}$ and charge $-1$ atomic units, is $-121$ Thomson not $m/z~121.$
The latter is actually the mass-to-charge ratio of the corresponding (unstable) cation!

* $\pu{1 u} = m(\ce{^{12}C})/12 = \pu{1.660540E-27 kg}$
The unified atomic mass unit, $\pu{u},$ is also known by the name Dalton and
the symbol $\pu{Da}.$

References

Murray, K. K.; Boyd, R. K.; Eberlin, M. N.; Langley, G. J.; Li, L.; Naito, Y. Definitions of Terms Relating to Mass Spectrometry (IUPAC Recommendations 2013). Pure and Applied Chemistry 2013, 85 (7), 1515–1609. DOI: 10.1351/PAC-REC-06-04-06. (Free Access)
IUPAC. Compendium of Chemical Terminology (the “Gold Book”). Online Version (2019-) Created by S. J. Chalk, 2nd ed.; McNaught, A. D., Wilkinson, A., Eds.; IUPAC Recommendations; Blackwell Scientific Publications: Oxford, 1997.
IUPAC. Compendium of Analytical Nomenclature (the “Orange Book”), 3rd ed.; J. Inczédy, T. Lengyel, A. M. Ure, Eds.; IUPAC Recommendations; Blackwell Science: Oxford, 1998. https://media.iupac.org/publications/analytical_compendium/
Cooks, R. G.; Rockwood, A. L. The Thomson-A Suggested Unit for Mass Spectroscopists. Rapid Communications in Mass Spectrometry 1991, 5 (2), 93–93. DOI: 10.1002/rcm.1290050210.


Comparison of basic strengths of hydroxides of alkali metal group

Yes that is it.

To paraphrase the wikipedia article HSAB theory

"Hard and soft (Lewis) acids and bases" (HSAB), also known as the Pearson acid-base concept, HSAB is widely used in chemistry. It assigns the terms 'hard' or 'soft', and 'acid' or 'base' to chemical species.

'Hard' applies to species which are small, have high charge states (the charge criterion applies mainly to acids, to a lesser extent to bases), and are weakly polarizable.
'Soft' applies to species which are big, have low charge states and are strongly polarizable. The concept is a way of applying the notion of orbital overlap to specific chemical cases.

Pearson evidently had revised values in the paper (which is behind a firewall):

Pearson, Ralph G. "Absolute Electronegativity and Hardness: Application to Inorganic Chemistry." Inorganic Chemistry, volume 27, number 4, 1988, pp 734–740. https://doi.org/10.1021/ic00277a030

The data from that paper is on this webpage: http://www.knowledgedoor.com/2/elements_handbook/chemical_hardness.html

\begin{array}{|c|c|} \hline
Ion & Chemical\ Hardness\ (eV) \\ \hline
\ce{Li+} & 2.39\\ \hline
\ce{Na+} &   2.30   \\ \hline
\ce{K+} &  1.92    \\ \hline
\ce{Rb+} & 1.85     \\ \hline
\ce{Cs+} &   1.71   \\ \hline
\end{array}