At first you need to identify, if there are any well know functional groups.

https://en.wikipedia.org/wiki/Functional_group

If there are any, then they will be used to create the final name, and this is done according with the rules of IUPAC nomenclature.

https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry

In your first example.
There is a benzene ring, which is a very common structure in organic chemistry. There are other more complex derivatives of it too, for this example or similar looking molecules the name would refer to either benzene or phenol functional group, this is a little nuanced.

So, in this case we know it is a benzene we also know, there are 3 methanes attached to it or more specifically methane derived groups called methyl groups.(Methane is alkane and a functional group of alkanes is called an alk(-yl) group).

3 methane groups = trimethyl

The last thing is to say where they are attached. This can be found by counting around C atoms on the ring 1, 2 and 3. Ring has no beginning and no end so the starting point is arbitrary, by convention the first group has number 1 and the rest follows.

1,2,3-Trimethylbenzenes

The second molecule.
Here, the idea is to pick the back bone of the molecule and then attach to it smaller and smaller parts like a tree with its branches.
So the trunk of the tree would be 5 C atoms so pentane.
Here the functional groups attached is also methyl group on 3th atom of the backbone pentane, so that would give 3-methylpentane
but..
there is still a double bond that has to be included in the name which  changes pentane to pentene, and lastly its nice to say where is it exactly therefore pen-2-ene

Putting it all together gives us,
3-Methylpent-2-ene


Is there a relationship between density and viscosity of a fluid?

They are not directly related as such. Density is defined as mass/volume units $\ce{kg/m^3}$. Water has a value 1000, (liquid) iron 7000, (liquid) lithium 516 , ethanol 789 and carbon tetrachloride 1632 all in units of $\ce{kg/m^3}$.

Viscosity is the rate of transfer of momentum exerted on a plane of fluid when a second plane of fluid a parallel distance $x$ away is moved with viscosity $v$.  The force exerted is $F=\eta \frac{v}{x}$, where $\eta$ is the viscosity.

The fundamental process that takes place when a liquid changes shape is that one layer of it has to slide past another. If we suppose that the liquid could be divided up into cells, then one cell has to 'hop' past another so that the liquid can move. The more difficult that this is the greater the viscosity. Thus to 'hop' a cell has to have an activation energy, that is it has to accumulate in itself enough energy to overcome resistance from other cells that prevent it from moving. Now defining just what a cell is is somewhat vague, but it is larger than a single molecule. But leaving that aside, the activation energy does reflect the inter-molecular interaction and shapes of individual molecules. Spheres with little inter-molecular interaction, e.g. small van-der-Waals forces, no hydrogen bonds, will have a lower viscosity than an awkward shaped molecule such as glycerol with the ability to form lots of hydrogen bonds.

Viscosity has units $\ce{Pa~s = N~m^{-2}S^{-1} = kg~m^{-1}~s^{-1}}$ but more commonly centipoise where $\ce{1~cP = 10^{-3} Pa~S}$. Some values  (in CP at 25 $\ce{^oC}$) are water 1, $\ce{CCl4}$ 0.97, iodoethane 0.56,  ethanol 1.07, cyclohexane 0.89, cyclohexanol 57.5, 1-propanol 1.945, ethylene glycol 16, glycerol 934.


Flammability (NFPA) - how is it defined?

You have found one of biggest flaws in the  NFPA "Fire Diamond". The NFPA rates the flammability of the compound as a fuel. ClF3 is not a fuel. It is an oxidant. ClF3 is playing the role that oxygen normally does. It causes fuels (reductants) to burn by oxidizing them rapidly and exothermically. Oxidants are incapable of creating fire in absence of fuel regardless of temperature. Fires started by ClF3 continue to burn after being smothered because oxygen isn't necessary to continue the oxidation process. And since ClF3 is denser than air, it gets trapped under the sand. Many other powerful oxidants, like fluorine and perchloric acid, have low flammability ratings. Oxygen also has a flammability rating of zero. They do not burn; they cause other compounds to burn.

The scale for the flammability rating has to do with flash point, which is the temperature at which a fuel forms an ignitable mixture with air. Since ClF3 doesn't burn (isn't a fuel), and doesn't react with air (N2, O2, CO2, Ar, etc, which are all inert towards oxidation), it does not have a measurable flash point.

The danger posed by ClF3 is summed in the yellow (reactivity or stability) diamond and the white (physical and chemical hazards) diamond. Your msds rates ClF3 as a 3 in reactivity/stability, which is appropriate. Wikipedia's NPFA rating article, even mentions chlorine trifluoride:

  3 -  Capable of detonation or explosive decomposition but requires a strong initiating source, must be heated under confinement before initiation, reacts explosively with water, or will detonate if severely shocked (e.g. ammonium nitrate, chlorine trifluoride).

The white triangle will have appropriate hazard information. For ClF3, it has the symbols OXY, which means it is a strong oxidizer, and W, which means it reacts with water.

I have not found the specific reaction that ClF3 undergoes with sand (mostly SiO2). I expect that it involves ${\ce{SiO2 -> SiF}}_6^{2-}$, except that this half reaction is not an oxidation.


Nomenclature of alkenoic acids - which group has priority: the alkene or the acid?

The carboxylic acid moiety takes precedence over the alkene moiety, so the molecule has to be numbered starting from the carboxylic acid carbon. Hence, methacrylic acid, for example, has the systematic IUPAC name of 2-methylpropenoic acid.


Reacting calcium with warm water and magnesium oxide with carbon dioxide, what are the products?

Purpose of ionic equation is to describe ions in (aqueous) solution, without the need to bother about details.

The first equation looks reasonable, you could possibly end up with $\ce{Ca(OH)_2}$, which is not very water soluble, but it's IMHO the type of detail you should not bother with.

The second one is more difficult, though. At first, there is no solution to speak about, therefore the request for ionic equation itself is invalid. The reaction written is correct, as long as it describes possibility. If it would really happen is question of reaction conditions, therefore totally outside the scope of homework. As such, I'd say, correct.


What are D_b and D_a in the Generalized Fugacity Coefficients table?

$D_b$ and $D_a$ are deviation terms which must be used in the place of $D$ to obtain the fugacity coefficients for fluids whose critical compressibility factor is not 0.27. Replace $D$ with $D_a$ if $z_c > 0.27$ and with $D_b$ if $z_c < 0.27$.


In general no.

What you're describing is known as inverse design, that is, designing something with particular properties.

I agree completely with the comment above, that you can find many things from databases. But when you mix together multiple components, into new inter-metallic compounds or alloys, the results do not necessarily match the individual components.

Consider, for example, "Galinstan", the various alloys of gallium, indium, and tin. While each of these have low melting points, the eutectic point for the ternary alloy is $\:\mathrm{-19 ^{\circ} C}$.

There are many other examples, but the point remains that without careful experiments or theory for materials design, it's tricky to do inverse design.


"Useful work" does not include expansion work. For a proof of why $w_{\text{by,add}} \leq \Delta G$, you may want to first refer to my answer to this question: Why does the Gibbs free energy only correspond to non-expansion work? then come back here.

In that proof, I used equalities by assuming that the process is reversible. If we drop this assumption, then by the Second Law,

$$\begin{align}
\mathrm{d}S &\geq \frac{\mathrm{d}q}{T} \\
\mathrm{d}q &\leq T\mathrm{d}S \\
dU &\leq T\mathrm{d}S - p\mathrm{d}V + \mathrm{d}w_{\text{add}} \\
dG &\leq dw_{\text{add}} \\
\Delta G &\leq w_{\text{add}}
\end{align}
$$

Equality holds if the process is reversible, or more generally, if both the initial and final states are equilibrium states. Now this seems like an obvious contradiction of the equation in the very first paragraph. The key lies in the definition of the work - whether it is done on the system, or by the system. In my proof, I have been referring to the work done on the system; however, here we are interested in the work done by the system, since that is the work that can be "extracted". The work done by the system must be equal to the negative of the work done on the system: $w_{\text{by,add}} = -w_{\text{add}}$. This leads to the final result $w_{\text{by,add}} \leq \Delta G$.

Levine's Physical Chemistry 6th ed. writes:

  In many cases (for example, a battery, a living organism), the P-V expansion work is not useful work, but $w_{\text{by,add}}$ is the useful work output. The quantity $-\Delta G$ equals the maximum possible nonexpansion work output $w_{\text{by,add}}$ done by a system in a constant-$T$-and-$p$ process. Hence the term "free energy". (Of course, for a system with P-V work only, $\mathrm{d}w_{\text{by,add}} = 0$ and $\mathrm{d}G = 0$ for a reversible, isothermal, isobaric process.) Examples of nonexpansion work in biological systems are the work of contracting muscles and of transmitting nerve impulses.

If you want a direct answer as to why your explanation isn't right, it's because you can't equate $\Delta H$ with the work done. Work done is given by $w = -\int\! p\,\mathrm{d}V$. You can only equate $\Delta H$ with the heat transferred at constant $p$. Apart from the First Law ($\Delta U = q + w$), heat and work are generally unrelated.


Is ΔS of a system related to temperature and change in enthalpy?

If the system experiences a change at constant pressure, the change in enthalpy of the system $\Delta H_{system}$ is equal to the heat Q transferred from the surroundings to the system.  So the heat received by the surroundings is $-\Delta H_{system}$.  Typically, the surroundings is regarded as an ideal constant temperature reservoir, and the reservoir is considered as being "internally reversible."  So, if T is the temperature of the reservoir, the change in entropy of the surroundings is $\Delta S = -\frac{\Delta H_{system}}{T}$.  The system is typically not regarded as experiencing an internally reversible change, so the same kind of approach can not be applied to the system.


Can you in reality give neon a charge of +8?

Of course you can take all the electrons off an atom - it is then called "fully stripped" in atomic physics. You don't need to do it to an entire mole, mind you. In accelerators one would send energetic neon ions through a background gas or a thin foil, and the interactions will result in various charge states coming out, up to and including fully stripped.


Can freezing temperatures cause chalk to fragment?

Chalk cannot at given conditions explode chemically, spontaneously turning into other compounds.
It could locally "explode" mechanically as consequence of mechanical stress caused by thermal dilation due changes to temperature. But calling it explosion would be very much exaggerated. It would be rather just cracking.
There is a way such cracking would be more probable, if the chalk is porous, soaked by water. The internally forming expanding ice could crack the chalk easily  But it would not happen at the first freezing event. It would need probably repeated freezing and melting, similarly like long term freeze weathering of rocks. The speed of such process depends on porosity..
Note that such cracking would be rather progressive than sudden, with cracks or crevices developing progressively, with occasional small pieces or blocks  falling away. Sudden break of the whole block into small pieces is highly non-probable


Rate of a reaction depends on the temperature conditions. So to make a reaction proceed significantly, we have to take the reactants up to the level of activation energy. One more thing to note here is that low temperature is not a fixed temperature, for example - consider 2 reactions A and B, now if low temperature for reaction A is 400 Kelvin, then it is certainly possible that for reaction B low temperature is say 800 Kelvin.


Are bisulfates and bicarbonates soluble in water?

According to it's Wikipedia page, calcium bicarbonate is not a known solid compound, but it does exist in solution, presumably prepared by dissolving calcium carbonate and bubbling carbon dioxide through the solution until bicarbonate dominates the following pH-dependent disproportionation:

$$\ce{2HCO3- <=> H2O + CO2 + CO3^2-}$$

It apparently has a solubility of 16.6 g per 100 mL of water at room temperature. which is about a 1 M solution.

Magnesium bicarbonate similarly exists only in aqueous solution, though it is much less soluble at 0.077 g / 100 mL.

As solids, these salts probably fall victim to the following equilibrium( $\ce{M}$ is a generic divalent metal:

$$\ce{M(HCO3)2 <=> H2O + MCO3 + CO2(g)}$$

The carbon dioxide escapes, preventing the system from ever staying at equilibrium.


Why CO2 is not reducing the pH of NaOH as the way it is expected

There are a so-called acid error and an alkaline error with glass electrodes. Which means that the pH electrode shows a systematic bias when your solution is too acidic or too basic and if it contains lot of sodium ions. The value which you quote for +13 are unreliable. So if you are bubbling carbon dioxide into NaOH solution, you are indeed consuming NaOH and converting it into sodium carbonate. The Wikipedia article on glass electrode is poorly written however if you search the term "alkaline error of glass electrodes" you will hundreds of better results.

May be you should follow the pH changes spectrophotometrically using a proper indicator (dyes) in NaOH.

Also keep in mind that you shouldn't store NaOH in glass containers because glass starts to dissolve in NaOH (very slow process though)


Why does water vapour deviate much more from ideal gas behaviour as it is cooled compared to nitrogen gas?

Well, an ideal gas has no attraction force between its molecules and does not have a boiling point, as you cannot make it liquid. Also, the sum of  molecule own volumes is supposed negligible wrt the gas volume.

Water does not form anything close to an ideal gas because of its hydrogen bonds.  You have to heat water to 100 °C to make it boiling to overcome these bonds between molecules of liquid water. Molecules of water vapour interacts with each other as well, especially at  high pressure.

If water had not had these bonds, it would have reportedly boiled at -120 °C. And if water had not been polar, what makes intermolecular attraction as well, its boiling point would  have been comparable with nitrogen.

Nitrogen is a non polar gas, with minimal molecular attraction, compared to water, due weak van Der Waals force, which leads to nitrogen boiling at -196 °C.


The key missing piece is identifying when the system is in equilibrium. When you suddenly increase the volume, the system is temporarily not in equilibrium. In this situation, a few things will happen:

The rate of evaporation won't change for the reason given
The vapor pressure will decrease. As a first approximation, we can use the ideal gas law to see how. $PV=nRT$ can be rearranged as $P=nRT/V$. The number of moles ($n$), gas constant ($R$), and temperature ($T$) will stay the same (this last one in particular isn't really true, but it doesn't change the qualitative result). Then, if the volume increases, the pressure will decrease.
Since the pressure has decreased, the rate at which gaseous molecules bump into the liquid decreases. The ones that do bump into the liquid will condense into the liquid at the same rate. The total condensation rate is the product of those two factors, so the condensation rate will decrease.

Now, since the evaporation rate is the same as it was before, but the condensation rate will go down, we will get a net evaporation. As more liquid becomes gas, the vapor pressure will increase. As the vapor pressure increases, the condensation rate will increase, for the reasons discussed above. Once the vapor pressure is high enough that the condensation rate equals the evaporation rate, then we will have a dynamic equilibrium.


Alcoholic potassium permanganate for oxidation

Practical considerations, mainly that toluene is non-polar, inorganic. Aqueous $\ce{KMnO4}$ wouldn't be miscible so the oxidation can only take place in the interface. By using alcoholic $\ce{KMnO4}$, we can get the $\ce{KMnO4}$ into the organic layer much more easily, thus making the reaction much faster.

To form an ester, you generally need a dehydrating environment. Plus, the oxidation of toluene with $\ce{KMnO4}$ in neutral conditions will produce base which neutralizes the benzoic acid, thus deactivating it making it unable to undergo esterification.


Coordinate covalent bonds do not exist.  They are all plain old covalent bonds.  We use "coordinate covalent bonds" as a means of counting electrons from atoms that go into bonds; here nitrogen contributes five electrons to four bonds so we figure that the nitrogen must have contributed two electrons to one of the bonds.  But the actual bonds formed between the nitrogen and each hydrogen atom are all alike.  Ammonium ion has the same tetrahedral structure as methane, and for methane we do not need coordinate covalent bonds to count the electrons.


How to study Sodium hypochlorite stability with materials

Sodium Hypochlorite  is soluble in water and reacts with all organic solvents. Sodium hypochlorite is extensively used in house-holds as a bleaching agent and it is a chemical compound with formula $\ce{NaClO}$.

Household sodium hypochlorite bleach was introduced to Americans in 1909 and sold in steel containers, then in glass bottles. In the early 1960s, the introduction of the plastic jug brought a cheaper, lighter, and nonbreakable packaging alternative. It reduced transportation costs and protected the safety of workers involved in its shipping and handling. Additionally, the thick plastic did not permit ultraviolet light to reach the bleach, which improved its chemical stability and effectiveness. In recent years, how-ever, plastic containers have become an environmental concern because of the time it takes the material to decompose in a landfill. Many companies that depend on plastic packaging, including bleach manufacturers, have begun to reduce the amount of plastic in their packaging or to use recycled plastics. In the early 1990s, Clorox introduced post-consumer resins (PCR) in its packaging. The newer bottles are a blend of virgin high-density polyethylene (HDPE) and 25% recycled plastic, primarily from clear milk jug-type bottles.

Read more: http://www.madehow.com/Volume-2/Bleach.html


Diastereotopic methylene protons in achiral molecules

Well, I can't speak to your NMR simulator and computational tools but from my perspective, and that of ChemDraw, the hydrogens at C1, and C3, in diol 1 are diastereotopic. While diol 1 is meso and achiral, introduction of deuterium in 2a renders both C1 and C2 stereogenic. Consequently, 2a is a diastereomer of 2b where the deuterium has been moved to the geminal C1 position. You may want to look at  this.<b


No. of photoelectrons vs frequency and intensity

The number of photoelectrons per incident photon doesn't really vary with photon energy (frequency). Particularly a photon won't eject multiple "photoelectrons." By definition a photoelectron is a loosely bound electron (i.e. not in an atomic or molecular orbital) which is ejected in a one-to-one basis by a photon and thus satisfying the equation $$ \rm{KE}_{electron} = \rm{h}\nu - \phi$$ where:

$ \rm{KE}_{electron}$ is the kinetic energy of the electron
$\rm{h}\nu$ is the energy of the incident photon
$\phi$ is the work function for the substrate (how much energy it takes to remove electron from substrate).

In general the substrate would be a metal and the electron would be in the metallic (electrically conducting) layer.

If the electron is bound in an atomic or molecular orbital then the
technique is Electron Spectroscopy for Chemical Analysis (ESCA).
If the photon is too energetic then it penetrates too deeply to eject an photoelectron.

The gist of the above conditions is that photoelectrons are created with relatively low energy photon just above the needed threshold given by the work function, $\phi$.

Another aspect of this is that a photon can only interact with one electron at a time. However if the photon is greater than $2\phi$ in energy it is possible that the electron can receive enough energy to knock yet another electron out of the metal. These two electrons wouldn't be "photoelectrons" since they don't fit the equation. They would be noise in the experiment and the experiment would need to be designed to minimize counting such electrons. So the ejection of multiple electrons by a photon is possible, but not multiple photoelectrons by definition.


The major product of 1,4-pentadiene NBS bromination

The Main reason for looking at the stability of the intermediate is Hammond's Postulate. Now the intermediate has the electron density spread all over it's body. SO it's reasonable to look at the activation energy of the transition state itself.

When the Br radical attacks at the end position, you have a much sterically favored transition state and the other two bonds are in conjugation and hence we would expect the activation energy would be much lower for this and not for the other transition state where the Br attacks at $3^{rd}$ position.

Hence we can conclude that conjugated diene is most probable product !.

Please correct me if I am wrong.


"Since change in enthalpy ∆H = Q + VΔP, it seems that enthalpy and heat flow (Q) are not strictly the same thing."

First, note that, when you write "∆H = Q + VΔP," you are assuming the only type of work is PV-work.  That's a reasonable assumption for these discussions.  It's just important to make that assumption explicit.

Given this, the quoted statement is correct.  Note, however, that many chemical reactions are carried out in an open container, such that the system stays at constant pressure (∆P = $0$) throughout the reaction. Thus, under this condition,  ∆H = Q + VΔP = Q.  I.e., the reason ∆H is a particularly useful thermodynamic function is that, under the common condition of ∆P = $0$, it corresponds directly to an important quantity: the amount of heat that flows into or out of the system.

  "Is it not therefore possible to have a reaction which absorbs heat (+Q), but has a decrease in pressure, such that the product -VΔP > Q, thus both absorbing heat and being exothermic (or if exo-/endothermicity is determined by Q, then an exothermic reaction with positive ∆H)?"

Here it comes down to a definition of terms.  The IUPAC Gold Book defines an "exothermic reaction" as "A reaction for which the overall standard enthalpy change ΔH⚬ is negative" ( https://goldbook.iupac.org/html/E/E02269.html ), where "standard" means P = 1 bar. Thus it doesn't cover how one would define exothermic or endothermic if the process is not at constant pressure.*   Having said that you could, in principle, have a reaction that takes place, say, at constant volume, and that releases thermal energy (Q < $0$) but, because of a sufficiently large and positive VΔP term, has ΔH > $0$. But, in practice, $\mid{Q}\mid$ would have to be very small in order for the VΔP term to dominate.

  "As a sort of corollary, is there an example of a thermodynamic process that changes only the pressure (thus $V∆P)$ with no heat flow or $PV$ work? (Q = $0$ = PΔV)"

Yes.  Consider, for instance, an isolated system, which is a system in which no flow of heat or work is allowed, and that is closed (no flow of matter).  The pressure could increase in an isolated system if, for instance, a chemical reaction is taking place in that system that evolves gas, and/or for which (were we to remove the adiabatic constraint) Q would be negative.

*Indeed, the IUPAC definition doesn't even seem to cover how one would define exothermic or endothermic if the reaction took place under constant P $\neq$ 1 bar, which seems quite limiting!  Further, it only defines the term for a "reaction", which IUPAC seems to take to mean a "chemical reaction" (typing "reaction" into the search field redirects you to "chemical reaction"), thus excluding important processes like phase changes.


Transition Metals colours and electron absorbtion

I've heard someone mention that the electrons do indeed fall back to ground state but, you don't see that part of the emission spectra since the energy is dispersed everywhere and not just towards your eyes. This is why the colour comes from the absorbance spectra.

The reason is does not stay in the excited state is simply because it is unstable. It will gain energy and then release it as mentioned above, then repeat.


For molecular orbitals, how is an energy of zero set or defined?

The answer technically depends on the method. In most cases, meaning HF or DFT with standard basis sets, it is infinite separation of all bare nuclei and all electrons. One exception known to me is VASP (a periodic, solid-state code), which uses pseudopotentials to avoid having to model the core singularity and core electrons. There, the zero-point is in reference to some atomic state.

Note that the MO energies are basis-set dependent, which is most pronounced and most easily understood for core electrons.


Which steps are most critical in driving glycolysis foward?

Those steps you mentioned are not merely critical steps, in glycolysis but are however the control points of the process and I would prefer to call them rate controlling steps. There is nevertheless one critical step which is believed to be the first committed step in glycolysis (not really sure if there are more).

There are three irreversible steps in glycolysis, and the differences
between glycolysis and gluconeogenesis are found in these three
reactions. The first of the glycolytic reactions is the production of
pyruvate (and ATP) from phosphoenolpyruvate. The second is the
production of fructose-1,6-bisphosphate from fructose-6-phosphate, and
the third is the production of glucose-6-phosphate from glucose.
Because the first of these reactions is exergonic, the reverse
reaction is endergonic. Reversing the second and third reactions would
require the production of ATP from ADP, which is also an endergonic
reaction.

Having said that, there are factors whih affect the process of glycolysis:
Enzymes
Control of hexokinase
The enzyme that catalyses the first reaction is hexokinase. The substrate of hexokinase is not necessarily glucose; rather, it can be any one of a number of hexoses, such as glucose, fructose, and mannose. Glucose-6-phosphate inhibits the activity of hexokinase; this is a control point in the pathway.  Hexokinase is inhibited by high levels of its product, glucose-6-phosphate. When glycolysis is inhibited through phosphofructokinase, glucose-6-phosphate builds up, shutting down hexokinase. (However, the liver contains a second enzyme that phosphorylates glucose-glucokinase).
Control of phosphofructokinase
The phosphorylation of fructose-6-phosphate is highly exergonic and irreversible, and phosphofructokinase, the enzyme that catalyses it, is the key regulatory enzyme in glycolysis. Phosphofructokinase is a tetramer that is subject to allosteric feedback regulation. fructose-2,6-bisphosphate an important allosteric activator of phosphofructokinase (PFK), the key enzyme of glycolysis; (it is also an inhibitor of fructose bisphosphate phosphatase (FBPase), which plays a role in gluconeogenesis.)
Control of Pyruvate Kinase
The final step of glycolysis is also a major control point in glucose metabolism. Pyruvate kinase (PK) is allosterically affected by several compounds. ATP and alanine both inhibit it. (It would be wasteful to breakdown more glucose is there is already vast amounts of ATP). Alanine is essentially pyruvate with an amino group. (converted to pyruvate via a transaminase enzyme). Therefore, a high level of alanine indicates that a high level of pyruvate is already present, so the enzyme that would make more pyruvate can be shut down. Fructose-1,6-bisphosphate allosterically activates pyruvate kinase so that the incoming products of the first reactions of glycolysis can be processed.
Substrate
The reaction in which fructose-6-phosphate is phosphorylated to give fructose- 1,6-bisphosphate is the one in which the sugar is committed to glycolysis. Glucose-6-phosphate and fructose-6-phosphate can play roles in other pathways, but fructose-1,6-bisphosphate does not. After fructose-1,6-bisphosphate is formed from the original sugar, no other pathways are available, and the molecule must undergo the rest of the reactions of glycolysis.
It is frequently observed that control is exercised near the start and end of a pathway, as well as at points involving key intermediates such as fructose-1,6-bisphosphate. This is the summarised version of control factors of glycolysis:

Remarks
To sum up, the reason I explained your mentioned reactions not to be critical is because (with reference to reaction 1 and 3):

Even if hexokinase is inhibited, glycolysis can still take place via catalysis by glucokinase in liver.

There are other feeder pathways for glycolysis even if the glucose sugar is not present, the saccharides (maltose, lactose, trehalose, and sucrose, mannose and galactose) can still be used to produce fructose 6-phosphate.

e.g $$\ce{Fructose + ATP  (Mg2+) → fructose 6-phosphate + ADP}$$

The last reaction would be bypassed by transaminase (is not important since the reaction would have been completed):

So these are major controlling steps, meaning glycolysis can still take place  (if committed step is reached) though at reduced rates until optimum conditions are restored. The second part of your question can’t not answer it because it is a different concept. Hope this helps
References

Harpers illustrated Biochemistry
Lehninger Principles of Biochemistry
Biochemistry (Campbell and Farrell)


Will it become more clear for a reader if you use an an abbreviated, "coded" form to express what you have actually done?

If not, why waste your time to encode it and waste the time of the reader to decode it (and maybe still leave space for ambiguity)?

I'd be fine with a report that precisely describes the experimental procedure and the observations.


Why is the 2-position of pyridine electron richer than the 4 position

When you deal with an aromatic ring, pairwise covalent bond structures miss a lot of key features.  You need to consider the entire conjugated ring as a unit and take account of symmetry.

Consider almost a benzene ring.  We may describe its $\pi$ orbitals as linear combinations of the out-of-plane atomic $p$ orbital on the carbon atoms.  Let $p_k$ represent the wave function of one such atomic, with the positive lobe on a specified side of the ring plane. The same orbital but with the positive lobe on the"wrong" side of the ring plane is rendered as $-p_k$.  The subscript $k$ refers to atom number $k$ as determined in the standard IUPAC scheme.

We then describe the canonical, occupied $\pi$ orbital as the familiar linear combinations:

$\psi_1=(1/\sqrt{6})\times (p_1+p_2+p_3+p_4+p_5+p_6)$

$\psi_2=(1/\sqrt{12})\times (2p_1-p_2-p_3+2p_4-p_5-p_6)$

$\psi_3=(1/2)\times(p_2+p_3-p_5-p_6)$

Now suppose that atom $1$ is changed from carbon to nitrogen to make pyridine.  Now that atom, being more electronegative, acts an an electron sink on occupied orbitals having an interaction with that atom -- meaning $\psi_1$ and $\psi_2$ are vulberable to thus $\pi$-electron withdrawal but $\psi_3$ which has a node at atom $1$ is "protected" by its symmetry.  Then positions $2$ and $3$, with some of their $\pi$- electron density on the protected $\psi_3$ orbital, are less susceptible than atom 4 whose $\pi$ electrons are all on the vulnerable $\psi_1$ and $\psi_2$ orbitals.

Now, going a little beyond the question, suppose we were to N-oxidize the pyridine, as with a peroxy acid.  In pyridine N-oxide the oxygen acts as a $\pi$ donor so that the nitrogen (or more accurately, the N-oxide functional group) is now effectively an electron source instead of a sink.  The "vulnerable" orbitals are now electron enriched and this characteristic is most strongly imparted, by conjugation and symmetry as argued above, to position 4.  Then electrophilic substitution, enhanced (or enabled?) by the N-oxidation, is most strongly so at position 4.  So, the primary electrophilic substitution product, after subsequently reducing the N-oxide, is the 4-substituted pyridine.  Which is what we see empirically with a wide variety of pyridine N-oxide electrophilic substitutions.


Could compressed refrigerant be stored feasibly in residential setting?

I happen to have a bottle of refrigerant in my house and another in my car (besides the stuff already being stored in my AC units).

While this is definitely more of a physics question, or perhaps a question for the HVAC guys on the Home Improvement SE... to generally answer your question, there are many factors for cooling a home, like size and location, for example. But let's say your average 2-ton window-unit AC pumps 24,000 BTU's per hour... and 4000 BTUs requires 1 gallon of refrigerant per minute.... you need 6 gallons per minute (being released through an evaporator coil) to match a window unit AC.


In principle you could measure it by getting a sample of blood and putting it in contact with gaseous O2 at various pressures.   Whatever external pressure of O2 results in the amount of O2 dissolved in the blood neither increasing nor decreasing would equal p(O2) in the blood.

As for how it's actually measured, I believe the usual answer these days is spectroscopically.  That is, you shine light through a sample of blood and look for the absorption at certain frequencies which are typical of hemoglobin with O2 bound to it.  The amount of absorption is proportional to the concentration of oxygenated hemoglobin, and your software has some internal chart that converts that to partial pressure.

As for why it's measured that way, in part it's probably just tradition, and medicine changes very slowly and conservatively, for obvious reasons (you're experimenting with health and life).  In part, though, it may be that it directly relates to something important: the partial pressure of O2 in the atmosphere, which is about 160 mmHg.  That's the concentration of O2 in the lungs.  What fraction of that reaches the blood at various points in the circulation is a natural indicator of how well oxygen is being taken up, transported, and delivered.  That is, you can directly relate this particular measure of O2 concentration to the form of concentration usually used to describe a mixture of gases, in this case the air you breathe, which is the origin and destination of the blood gases.


Equivalent hydrogens in 2,3-dimethylbutane

The hydrogen atoms in a methyl group are equivalent to each other. There are 4 methyl groups and therefore there are 12 equivalent hydrogens. 2,3-dimethylbutane is symmetric as you stated, and so on one side you have 6 equivalent hydrogens, this implies that there are 6 more equivalent hydrogens on the opposite side.


Is there a type of detergent/surfactant that evaporates at STP?

Volatile surfactants

There are many molecules that are both surfactants (in water) and volatile, but of the ones I can think of, none can be used safely in home laundry applications.

Hydrocarbon derivatives.  The examples you found by googling are in this class.  3,5-dimethyl-1-hexyn-3-ol is the principal component of Surfynol 61 (probably named Surfynol 61 because it is a surface-active (surf) alkyne alcohol (ynol) with six carbons (6) and the alkyne moiety at carbon one (1).  Other examples would be medium-chain fatty alcohols like hexanol, octanol, cyclohexanol, etc.  These molecules reduce surface tension of water and have OK-ish cleaning power (not as good as larger but non-volatile alternatives like SDS), but their volatility is also their downfall: their use in home environments creates a strong fire and explosion hazard when used in automatic heated dryers.
Fluorocarbon derivatives.  I found the same article you probably did about volatile fluorocarbon surfactants.  The most powerful at reducing the surface tension of water seems to be nonafluoro-tert-butyl alcohol, which reduces water surface tension by 78%!  However this experiment required saturating the water with the vapor of this compound.  And 1 gram costs >$80.  And toxicity is a concern, despite this compound apparently being researched as a component of or precursor to artificial blood mixtures in the 1970s.
Other halocarbon derivatives.  These include solvents like perchloroethylene, chlorobutanol, etc.  These are all highly toxic, and with some of them, there is a flammability/explosion risk as with hydrocarbon derivatives.

Non-surfactant use

You might consider just using organic solvents that are not (very) surface active as a way to degrease fabric.  However, these may leach dyes out of the fabric or otherwise damage them.  Isopropyl alcohol (IPA) is accessible to home users, and is relatively safe.  Be sure to do the soaking in a well-ventilated area, and do not use heated dryers to remove the IPA!  Instead, rinse out the IPA thoroughly with water.  Depending on local regulations, you will have to dispose of all of the dirty IPA and IPA-contaminated rinsing water as solvent waste.

The best option

Pay to have your clothes dry cleaned!


@Mart 's comment impelled me to return to this question and correct my answer. I've deleted incorrect material and expanded the discussion to, hopefully, provide correct information. There is a good discussion (better than the reference previously cited) of the issue here.

  Reflux is the process of boiling reactants while continually cooling the vapor returning it back to the flask as a liquid. It is used to heat a mixture for extended periods and at certain temperatures...A condenser is attached to the boiling flask, and cooling water is circulated to condense escaping vapors.

If you are refluxing a mixture, as you might in organic synthesis to increase the speed of the reaction by doing it at a higher temperature (i.e., the boiling point of the solvent), then any of the condensers that worked well enough to avoid the loss of solvent and avoid "flooding" would work equally well. When you're refluxing, you want the "reflux ring", the place where the vapor is visibly condensing into a liquid, to be no more than 1/3 of the way up the reflux column.

You have two different basic types of condensers shown, Graham-type condensers (the first 3) and coil condensers (the last two). In the coil condensers (the left condenser in the picture below), the water flows through the coil and the vapor moves up in the larger, outside  area of the condenser, condenses onto the cooled coils, then drips back into the pot. In a Graham-type condenser (the right condenser in the picture below), the water flows around a tube (whether straight or coiled) that contains the vapor/condensed liquid.(picture source) The Graham-type condensers clog (or flood) more easily since they have a more restricted path for the liquid to return to the pot.

Graham-type condensers: The Liebig condenser is simple, but has low cooling capacity and can be fairly easily clogged as the condensed liquid flows back into the flask and blocks the vapor that is trying to escape. The Allihn improves on this design by having a wider bore at the bottom and condensing the liquid on the "bubbles" where it can run down the sides and avoid blocking the vapor. (I've used this to good effect in refluxing many reactions.) The Graham condenser is the same basic design as the other two, but the condensation tube is coiled which provides more surface area for cooling...but also tends to send the condensed liquid right into the path of the vapor trying to move up. It is particularly prone to flooding.

Coil condensers, such as the Dimroth and Freidrichs, have high capacity for cooling with fewer problems from flooding since the vapor condenses on the coils and drips back from the little prominence at the bottom of the coils into the center of the pot. The vapor has an easy time getting past the drops falling into the pot. If you can afford it, this seems like a good choice for most applications. The Freidrichs condensers, which incorporate a cold-finger with the spiral, are higher capacity, quite bulky and heavy. I have seen them used with rotovaps where you are taking a lot of solvent off quickly, but not with an ordinary reflux apparatus. This would be over-kill for a simple reflux reaction situation.

Sorry for the incorrect information (for those of you who looked at this before) and hope this is helpful.


Effect of doubling kinetic energy on phase transitions

The atoms are not locked in place.   They vibrate back and forth.  They do so with plenty of velocity and thus kinetic energy, even though the vibrations switch back and forth so rapidly that the atoms do not travel a large distance in any one direction.


Why x in equilibrium calculations can be "ignored"

In problems like this, the maths can sometimes be simplified by making assumptions so that you do not need to do difficult algebra. You could always do the harder algebra.

In this case, a valid assumption is that when the equilibrium constant is very small in comparison to the concentrations given, then the change in those concentrations (which is represent by $x$) will also be very small and can be neglected. I'll work through your example the long way to show you why this assumption is valid.

$$K_c= \dfrac{[\ce{ClF}][\ce{F2}]}{[\ce{ClF3}]} = \frac{(x)(1.00+x)}{(2.50-x)} = 8.77\times10^{-14}$$

Working through the algebra exactly:

$$(x)(1.00+x)=(8.77\times10^{-14})(2.50-x)$$
$$1.00x-x^2=2.19_{25}\times 10^{-13}-8.77\times10^{-14}x$$
$$0=x^2+(-8.77\times10^{-14}-1)x+2.19_{25}\times 10^{-13}$$

Solving the quadratic formula gives

$$x=\dfrac{(1+8.77\times10^{-14}\pm \sqrt{(-1-8.77\times10^{-14})^2 +4(2.19_{25}\times 10^{-13})}}{2}=[1,\ 2.19\times10^{-13}]$$

The first root $x = 1$ is nonsense. The second is the root you were given (when rounded to the correct number of sig figs).

The value of $x$ is so small compared to the initial concentrations of $\ce{ClF3}$ and $\ce{F2}$ that the concentrations of these species do not change within the magnitude of correct use of significant figures:

$$[\ce{ClF3}]_{eq}=2.50-2.1925\times10^{-13}=2.50 \ \text{to 2 decimal places}$$
$$[\ce{F2}]_{eq}=1.00+2.1925\times10^{-13}=1.00 \ \text{to 2 decimal places}$$

The maths above are equivalent to the assumption that since $x<<1.00$ and $x<<1.00$, $1.00+x\approx1.00$ and $2.50-x\approx2.50$. Whether you do it at the beginning or at the end, you get the same result: $x=2.19\times10^{-13}$. Making the simplification at the beginning is operationally easier and avoid potential algebra and arithmetic errors.

The assumption is not valid if the initial concentrations are close in value to $x$.


What is the difference between a coefficient and a subscript in a chemical equation?

Why can't both of them be written in the same way? Does it matter? Is there a difference?

Yes, there is a difference, and it does matter! The placement of subscripts and coefficients tells you how the substances are put together on the atomic level.

Let's look at the first equation:

$$
\ce{2Li + S -> Li2S}
$$

This one indicates that two moles of lithium combine with one mole of sulfur to form one mole of lithium sulfide. $\ce{Li2S}$ is called a formula unit or empirical formula, and tells you the smallest whole-number ratio of atoms that make up a "building block" of the compound. In this case, the smallest building block of a lithium sulfide crystal would contain two lithium ions for every one sulfide ion. Here's a picture from wikipedia to demonstrate:

From this, we can learn that:

  Coefficients tell us the number of moles, molecules, or formula units of a species

and

  Subscripts tell us how many atoms or ions there are within a compound

Now if we look at the second equation:

$$
\ce{2Al + 3Br2 -> 2AlBr3}
$$

we can see that in this case, it take 2 moles of aluminum to combine with three moles of bromine to produce two moles of aluminum tribromide. If it were instead $\ce{Al2Br6}$, that would indicate a different formula unit - it would imply a different compound.

So to answer your question:

  Why is Lithium in the first reaction written with a small 2, while in the second one Aluminium is written with a big 2

This is because in the first equation, the product has two lithium ions for every one sulfur in the formula unit, while in the second, two moles of aluminum tribromide are produced for every two moles of aluminum, but the formula unit contains only one aluminum atom and three bromine atoms.

For your final question:

  How do I tell which one to use?

You would have to know what the formula of the product is. In these examples, you can predict it by assuming that they are ionic compounds and balancing charges. You would get the correct answer, even though this assumption is not true for the second equation (it is actually a covalent compound, and you would just get the right answer by luck).


The reactivity patterns of $\ce{SO2Cl2}$ and $\ce{SOCl2}$ are quite different. $\ce{SOCl2}$ is a good electrophile, and can be thought of as a source of $\ce{Cl-}$ ions.  These ions can go on to react in their typical nucleophilic fashion. $\ce{SO2Cl2}$ however is often a $\ce{Cl2}$ source, as it readily decomposes giving off sulfur dioxide. Usually much easier/safer to use this than measuring out (and getting into solution) chlorine gas. The chlorination of simple alkanes by $\ce{Cl2}$ gas (or something that makes it in solution) happens by a radical mechanism i.e. $\ce{Cl.}$ not $\ce{Cl-}$.


Are these alternate names for long chain hydrocarbons with complex substituents accepted?

Okay so here is the situation: You have two subsitutions, methyl and 1-methylpropyl as the two substituents here. You wish to know in which order the substituents must be written in the name. You knew that substituents are arranged alphabetically, and both methyl and 1-methylpropyl start with the same letter m.

Consider a dictionary. It has over thousands of words that start with the letter C. But we can still arrange the words in order by looking at the second, third, fourth, fifth, etc. letters of the names. Similarly, if the first letters are the same, look at the second, third and fourth letters.

Here's where we arrive at the conclusion. methylpropyl will be ranked below methyl as per dictionary order. Thus the name of the compound would, in fact, be 2,2,7-trimethyl-4-(1-methyl-propyl)-nonane and the other name is not accepted.

I believe you can continue from here to name the second compound.

Important notes:

The inclusion of di, tri in the name must only be applied for complex subsituents such as in the case of 5-(dimethylethyl)decane. In cases such as 4-ethyl-5,5-dimethyldecane, do not use d as the starting letter. This is becase d is not the part of the name of the subsitituent (methyl) and only indicates how many times this subsitituent occurs in the compound.
Do not add hyphens between methyl and decane such as "methyl-decane". Use "methyldecane" Hyphens must only be used between numbers and subsituents for which the number represents its position in the name.


Organic Chemistry - Effect of methyl substituent in Aromatic Substitution via Benzyne

The Me group has no directing effect in the deprotonation so there are 2 protons (o & p) that can be removed with equal probability by the NaNH2 to give the benzyne. These give 2 possible benzynes (2 and 3 relative to Me). The nucleophile will add with equal facility to either end of either benzyne giving a mixture of 3 isomers in 1:2:1 ratio. The major isomer is the m- aminotoluene because it can arise from both of the possible benzynes.


Steam Distillation - Condensate not going through condenser

First of all, make absolutely sure that there is nothing blocking the path between the distillation flask and the final vent, and that all the joints are tight.

When beginning the distillation, you should be able to view the condensation line move up within the apparatus after the mixture begins to boil. If it moves up toward the thermometer, but then stops and never makes it up to the condenser, that might indicate that you need to insulate the apparatus above the distillation flask and up to the thermometer. Try wrapping it with glass wool and foil and see if that helps.


Why is dialysis tubing impermeable to sucrose?

Benedict's reagent detects reducing sugars (free aldehyde group) by reducing soluble blue Cu(II) to insoluble red Cu(I) oxide.  Sucrose aquoous solution is inert unless it is pre-hydrolyzed by heating with strong acid catalyst.

That certainly is a poser.  One possibility (SWAG, not fact) is regenerated crosslinked cellulose is overall not permeable to neutral solutes.  Its size selection is active toward dissociated solutes and proteins that are charged.  One can test this with anti-freeze.  The sodium fluorescene color should freely diffuse.  Tiny molecule ethylene glycol (EG) should be retained.

https://www.chemetrics.com/Glycol
  Test kit and how it works:  Iodate cleavage  of 1,2-diol to aldehyde, then aldehyde detection with Purpald.

One could iodate cleave permeated EG (or propylene glycol, glycerin, etc.) then use Benedict's solution.  Serial dilution of standard EG solution for control and calibration.  One expects any dialysis membrane to leak somewhat.  Major passage of uncharged 1,2-dihydroxylic solutes smaller and larger than than sucrose is the SWAG's falsifying observation.

http://chemsee-foods.com/antifreeze-detection
http://www.chemworld.com/Ethylene-Glycol-Test-Kit-p/k-4815.htm
http://www.hach.com/glycol-visual-test-kit-model-eg-1/product?id=7640218461
  EG in food and fluids tests.
http://www.astm.org/Standards/E202.htm
  Testing standards


What are these crystals on my champagne cork?

They are potassium bitartrate crystals (source).

The crystals form because the potassium bitartrate is not very soluble.

Since solubility is a function of temperature, when wine is chilled the solution can become saturated, causing the precipitate to form in the bottle. Stabilizers are often added to prevent this.

In your case, the precipitate is on the cork, and so what I think has happened is that some wine soaked into the cork, perhaps while the bottle was moved. Then later, the bottle was stood upright or on its side where the wine could not reach the cork. As the wine evaporated back into the bottle, the local concentration of potassium bitartrate increased until it reached the saturation point, at which point the crystals grew. Since the crystals take a very long time to dissolve once formed, they will persist even if wine reaches the cork again.

The good news is, they are harmless (especially good news given that you tasted them - please don't make tasting strange crystals a habit!). They are made of the same substance that cream of tartar, which is a common baking ingredient, is made of.


Are there any free electrons in metal vapor?

Interesting question. There are free electrons and ions in a metal's gas phase, but even at rather high temperatures, they are relatively rare. The formation of free electrons is given by the reaction:

$$\ce {M_{(g)} → M^+_{(g)} + e^{-}_{(g)}}$$

This is an endothermic reaction for all atoms and molecules, and the energy involved receives a special name: the (first) ionization energy. Comparing a chart of ionization energies for the elements, one finds that almost all elements require over $ 500\ kJ/mol$ (or about $5\ eV/atom$) to ionize. Even at $5750\ K$, the temperature of the surface of the Sun, the average thermal energy of a particle is about only about $0.5\ eV$.

Also, a large part of the electrons that did manage to free themselves from their parent neutral atoms would actually be consumed by other neutral atoms, in a different reaction:

$$\ce {M_{(g)}  + e^{-}_{(g)} → M^{-}_{(g)}}$$

This reaction is slightly exothermic for most elements and molecules, and the energy associated with it also receives a special name: the (first) electron affinity.

Calculating the equilibrium concentrations of each species at a given temperature is possible but is a bit of work. However, for most metals, you can expect to find significantly less than one free electron per ten thousand neutral atoms at $5750\ K$, and the relative amount of free electrons decreases exponentially with temperature.


Does iron(III) sulfate react with copper?

In $\ce{FeCl3}$ etching of copper, the chloride ion is extremely important.

Chloride ions coordinate to $\ce{Fe^3+}$, $\ce{Cu+}$ and $\ce{Cu^2+}$.

See Copper Etching in Ferric Chloride Ind. Eng. Chem., 1959, 51 (3), pp 288–290 for the relative concentrations of the particular chloride complexes.

In contrast, if $\ce{Fe2(SO4)3}$ were used, there would only be aqueous ions.

At the copper surface, there is only a one-electron oxidation of $\ce{Cu}$ to $\ce{Cu+}$.  As the above article explains, without coordination of chloride, the $\ce{Cu+}$ would be essentially insoluble.

The rate limiting step of the etching reaction is diffusion of ions from the surface of the copper.  Coordination of chloride provides solubility and enables diffusion.  This is the main reason that $\ce{FeCl3}$ is needed.


For an ideal gas, your equation for $\Delta U$ applies not only to constant volume processes, but to all processes.  It's just that, in experiments to measure Cv, it is very convenient to use a constant volume path because, under those circumstances, no work is done, in which case $\Delta U$ is just equal to the amount of heat added.  So this gives you a direct measurement of Cv.

As far as $\Delta H$ is concerned, the correct equation is $\Delta H=\Delta U+\Delta (PV)$.  But, from the ideal gas law, $\Delta (PV)=nR\Delta T$.  So what does that give you for $\Delta H$?


First off, lets understand how densities of poly(ethylene) are differentiated. The prototypical polyethylene is often represented as a simple repeating ethylene monomer units as shown below:

This is what high-density polyethlyene (HDPE) is constructed of; simple linear monomer units that form closely packed crystal structures. This polymer is synthesized by  free-radical polymerization with catalyses that encourages radical at the end of the chain, keeping the polymer linear. low density polyethylene (LDPE) is free-radically polymerized without this control, which causes branching at random points in the polymer chain. This branching prevents the crystallization that made polyethylene more dense, thus it is more amorphous and lower density. To make a linear low density polyethylene (LLDPE), ethylene is co-polymerized with longer olefins such as 1-butene, 1-hexene or other olefins but with catalysts controlling polymerization similar to polymerization of HDPE. A diagram demonstrating molecular structure differences is shown below:

Source: Plastics Technologies

Bringing this back to about polypropylene, the co-polymers are a different monkey that are used for precisely tuning properties and that is not useful for this discussion. But as you mention the tactility is important. There are two reasons for this. 1.) there is no reason to produce low-density polypropylene as LDPE and LLDPE are cheaper materials that are more mature in their development with plenty of material data available and developed recycling streams, thus there is little incentive to investigate LDPP. 2.) polypropylene is already controlled by its tacticity. Isotactic and syndiotactic polymers crystallize fairly well while atactic polymers do not. Since polypropylenes rigidity and density can be controlled by using certain polymerization catalysts there is again little incentive to produce branched polypropylene


Bond length comparison in substituted phosphorus pentahalide

According to (1) and (2), the values in your table would be as follows (all values in Å):
$$\begin{array}{c c c}
& \ce{PF4Cl} & \ce{PF3Cl2} \\
\ce{P-F_{eq}} & 1.541 & 1.546 \\
\ce{P-Cl_{eq}} & 1.999 & 2.004 \\
\end{array}$$
we have $x < y$ and $a < b$, which means that (b) is the correct answer

According to Bent's rule:

atomic s character concentrates in orbitals directed towards electropositive substituents

Bonds with larger s-character are shorter in nature. Keeping these two points in mind and analyzing the structures, in $\ce{PF4Cl}$, almost all the s-character is directed towards $\ce{Cl}$, which means that $\ce{P-Cl}$ bond length in $\ce{PF4Cl}$ is less than that in $\ce{PF3Cl2}$. Similarily, considering that the $\mathrm{sp^3d}$ hybridization can be considered as $\mathrm{sp^2+pd}$, in the equatorial plane, the s-character shared between the two fluorine atoms in $\ce{PF4Cl}$ would be more than the s-character given to the single fluorine atom in $\ce{PF3Cl2}$, hence the shorter bond length.
Another perspective is that the above set of arguments are largely hand-waving, as Bent's rule is at best empirical and the bond length differences are $\pu{\approx 5 pm}$, which is very small. An easier question would be to compare bond lengths between $\ce{PCl5}$, $\ce{PF3Cl2}$ and $\ce{PF5}$, but for this question, I don't have a 'solid' explanation per se, other than the lengths from the references and the elaboration of Bent's rule above.
References:

Leiding, Jeff, et al. “Bonding in PF2Cl, PF3Cl, and PF4Cl: Insight into Isomerism and Apicophilicity from Ab Initio Calculations and the Recoupled Pair Bonding Model.” Theoretical Chemistry Accounts, vol. 133, no. 2, Feb. 2014, p. 1428. doi:10.1007/s00214-013-1428-7.
French, Richard J., et al. “Dichlorotrifluorophosphorane (PCl2F3): Molecular Structure by Gas-Phase Electron Diffraction and Quadratic Force Field.” Inorganic Chemistry, vol. 24, no. 18, Aug. 1985, pp. 2774–77. doi:10.1021/ic00212a014


You should state the charge balance or the proton balance.

[H3O+] = [OH-] + [CH3COO-]

We take the logarithm of this and get:
log [H3O+] = log([OH-] + [CH3COO-]).

Now, you can plot this in your nice diagram.

log [H3O+] will follow the straight line in your diagram, i.e. log [H+].

The line for log([OH-] + [CH3COO-]) will at very high pH follow the line for log([OH-], but quite soon it will bend off and only follow the line for log [CH3COO-]. This is because [OH+] << [CH3COO-] at lower pH.

When [OH+] << [CH3COO-], you get log [H3O+] = log [CH3COO-]). When the line for log [H3O+] crosses the line for log [CH3COO-], you can calculate the pH of the solution from the diagram.
The lines are crossing each other at right angel: pH = (1.00 + 4.65)/2 = 2.82(5).
Very good, I am encouraging you to use logarithmic diagrams for solving acid-base equilibria.


Reaction of H2o2 on Ozone in jar washing Tank in packaged drinking Water Company

Both ozone and H2O2 are reactive oxygen species. They don't necessarely react with each other, but oxidize organics and act as bacteriocides. Majority of bacteria are killed by H2O2 (which is cheaper and safer). In case some spots were not cleaned ozone is added. Producing enough ozone to clean it all would be difficult. Guaranteeing that H2O2 reached every spot is also hard. this is why they use two chemicals.

0.5ppm (1/2 parts per million) is a relatively low dose, but working with ozone at higher concentrations is dangerous. Ozone is more toxic than chlorine.

So, this is much like mixing dishwashing liquid with laundry detergent. There is no reaction between them.

Note that in a lab you can have O3+H2O2 -> H2O3 + H2O, but you have to run it at low temperature, the yield is low and the product decomposes fast.


Why are amino acids least soluble at isoelectric point?

In order for molecules to precipitate out of solution, they need to aggregate together. Amino acids that have zero net charge can aggregate together much more easily than those that are charged.
Molecules that have net charge need counterions to aggregate with them to offset the charges or the electrostatic repulsion will be too great.
The concept of isoelectric point is more often applied to polypeptides and proteins than to individual amino acids. In those much larger molecules, the tendency to aggregate when there is no net charge can be quite high. This is especially true if there are any hydrophobic regions that can favorably associate with each other and reduce the interaction with water.
It is important to note, as mentioned in the comments, that this does not mean that all amino acids and proteins are insoluble at their isoelectric point, simply that they tend to be less soluble at that pH than at pH values where they carry net charge.


Is the electric conductivity of chlorine ions influenced by its oxidation state?

It is the similar effect as for molar conductivity of cations of alkali metals, which grows from $\ce{Li+}$ toward $\ce{Cs+}$, in spite of increasing ionic radius.
The point is, the effective radius of the hydrated ions has the opposite trend as the "naked" ion radius trend. $\ce{Li+}$ ion with the smallest radius has the strongest electrostatic field and the largest hydration sphere and has therefore the lowest molar conductivity.
The similar effect I expect for chlorine oxyacid anions. In this case, there is also decreasing degree of delocalization of the negative charge in trend $\ce{ClO4-} \gt \ce{ClO3-} \gt \ce{ClO2-}$.


Can alloys or ceramics have higher density than elemental solids?

Yes, for both alloys and ceramics it is possible have a density that is higher than their densest elemental constituent. This is because every lattice structure of the denser elemental constituent contains interstitial void spaces, or holes, typically tetrahedral holes (surrounded by 4 atoms) or octahedral holes (surrounded by 6 atoms). These holes exist despite the crystal having an optimal packing structure, thus a maximum density for the pure solid.

The densest elemental solids are typically made up of larger atoms toward the bottom of the periodic table, that have larger atomic radii due to electron-electron repulsion, but which have have much higher atomic mass due to their nuclei. This produces larger interstitial voids that can fit smaller atoms without significantly disturbing the lattice constants of the larger, denser component. An increase in mass without concomitant increase in volume will produce a denser composite.

Specific examples are difficult to find, as densities are not always reported. One common material which is denser than the corresponding "densest elemental constituent" is steel. Steel (maximum density: 8.05 g·cm⁻³) is an alloy of iron (density 7.874 g·cm⁻³) and carbon (diamond density: 3.5 g·cm⁻³), although occasionally some chromium (density: 7.19 g·cm⁻³) and nickel (density: 7.81 g·cm⁻³) may also be included.

Another possible, although less clear-cut example is the alloy heavymet, which is used in ATLAS. It is composed of nickel (9.9 atom %), copper (4.0 atom %), and tungsten (86.1 atom %). Heavymet has a density of 19.3 g·cm⁻³, which despite the less dense diluents, may be slightly larger than tungsten's density which is 19.25 g·cm⁻³.


What would happen to a Water/Lard Mix under pressure and high heat?

I'm not sure what aspect you are asking about exactly… but all supercritical liquids are completely miscible with each other, so you would get one supercritical phase with the mixture of the two… supposing the oil doesn't decompose.

However, the smoke point of lard is 182°C while the critical temperature of water is 374°C, so what most probably happens is that your lard will burn, breaking down to glycerol, acrolein and free fatty acids.


Can soap make an liposome around virsuses?

Whether a mixture of phospholipids/lipids/detergents form micelles or liposome is not dependent on the hydrophobic "payload" but on the nature of the amphiphilic molecules.
Detergents form micelles (and bubbles and antibubbles, when there is air around). The micelles are fairly small (radius similar to the length of the detergent), and they don't have space for an entire virus (diameter on the order of 100 nm) or bacteria (diameter on the order of 1000 nm). Detergents that form micelles are able, however, to solubilize membrane proteins (diameter on the order of 10 nm) by forming a band of detergents around the hydrophobic parts of membrane proteins.
Phospholipids mixed with cholesterol form lipid bilayers and liposomes (the latter by extrusion or sonification). The distinct behavior has to do with differently charged head groups and the number of hydrophobic tails.
Some bacteria are killed by detergents because they break down the cell membrane, integrating the phospholipids and cholesteral into the micelle. Viruses can't be killed because they are not alive. If they are surrounded by a lipid bilayer, the detergent can also break this down, rendering the genetic material inside harmless.
No matter whether bacteria and viruses are destroyed by the soap or not, the soap allows the water to wash the germs away, and they end up in the canalization (e.g. coronavirus genetic material in Boston) instead of in people's bodies.


The fluorescence excitation spectra show the change in fluorescence intensity as a function of the wavelength of the excitation light. Does that mean, that at the wavelength of excitation maximum the most of the molecules get excited? Or do they just emit more photons per molecule per unit time?

You have to understand how the excitation spectrum is collected. You have to know (or at least have an estimate) of the emission maximum of a given molecule. For example, if we have quinine, I should know that it emits blue light and maximum wavelength of emission is ~ 450 nm.
Now we will fix the emission wavelength at 450 nm and scan the entire UV-VIS range for excitation. Say, we would excite quinine with 200 to 700 nm. One wavelength at a time, and count how many photons corresponding to 450 nm were emitted. This plot is called the excitation spectrum.
An excitation spectrum basically tells you which wavelength (from a given instrument) is going to show the highest intensity. I emphasized "a given instrument" because an excitation spectrum has to be corrected for the light source intensity. It will vary from instrument to instrument. Nobody corrects it because it is a long procedure as unfortunately, normalization cannot correct it.
Anyway, a corrected excitation spectrum should match in shape with the absorption spectrum provided that there is only one fluorophore in the molecule.
I wanted to connect the excitation spectrum with an absorption spectrum because we can now connect the molar absorptivity with the excitation. Consider molar absorptivity as a measure of probability of excitation. Higher the value of molar absorptivity, more molecules will absorb the light and hence more will emit. So your first interpretation is correct. It is a collective behavior of the all the molecules which are being illuminated in the cuvet.
Also remember that excitation emission is normally a one photon excitation and one photon emission in majority of cases until and unless you are using fancy lasers. So you cannot have one photon in, and multiple photons out in routine cases. In rarer cases, you can indeed have two photon in for excitation and one photon out. You need lasers again.


How to figure out number of atoms in a molecule? e.g. Phosphorus exists as P4

You are asking how to identify the standard state chemical structure of an element; this is obtained from empirical data.

The standard state is defined as the most stable allotrope of the element under standard conditions. For example, graphite is the standard state for carbon. Phosphorus, however, is an exception: while white phosphorus is not the most stable allotrope, it is the easiest to characterize, and is defined as the standard state for phosphorus.

The chemical formula for each element at standard conditions is that of the standard state. Except for the gases $\ce{H2, N2, O2, F2, Cl2, Br2, I2}$, and the solids $\ce{P4, S8}$, all other elements exist in monatomic form in the standard state.


How to manage complex equilibria when studying a mecanism?

If you know what the final equilibrium products are or may be, and you know how much mass you start with, either as elements or molecules, you can calculate the Gibbs energy of the products as a function of (unknown) concentrations (assuming constant temperature and pressure) and then minimize the Gibbs energy with respect to those concentrations. You have to constrain the solution to conserve all of the elements that you began with, and constrain it to have all concentrations be non-negative.


What is the Series Limit in a Spectral line

Every series of wavelengths produced from the hydrogen atom excitations have a certain minimum limit. These minimum wavelengths are called the series-limit for that particular series. Take a look at this picture for example:

The minimum series of the Lyman series would be for the largest transition, which is from ∞$\ce{ -> }$1. Plugging the values into the Rydberg's equation:

$$\frac{1}{\lambda}=R_{∞}\left(\frac{1}{1}-\frac{1}{∞}\right)$$

That should give us a value of 91nm. Which is the series-limit for the Lyman series.

Now you can find series-limits for the other series as well.


How to explain molecular geometry without the help of VSEPR, valence bond, or hybridization theories?

Short Answer: No, you can't. Reason being that so far this is the only way devised, which works most of the time.

Long Answer:
The Molecular Orbital Theory is, in practice, a complement to the Valence Bond Theory, Hybridization and VSEPR.

There are successes as well as shortcomings of both the Molecular Orbital Theory, and the other three.

In general, the VSEPR theory is used to predict shapes of molecules and ions, not the Molecular Orbital Theory. It can be extended to a large number of Covalent compounds, but it has a few limitations as far as ionic compounds and multi-centered bonds are concerned. The MOT tells us that when atoms combine, they form molecules orbitals, which are lower in energy than the atomic orbitals. It is generalizable, and can be extended to even semiconductors and superconductors. It is very useful when it comes to finding the actual energies associated with every single orbital.

The Valence Bond Theory, along with Hybridization and VSEPR, is the one that is generally used to find the shapes of molecules. The reason is simple: The Molecular Orbital Theory tells us the arrangements of delocalized electrons in Molecular Orbitals, whereas, the other theories tell us the possible positions of the combined atoms after considering that the electrons are localized and repel each other.

However, there is the case of multi-centered $\pi$-bonds, also known as banana bonds, and the 4c-3e bonds, which can be explained only by the Molecular Orbital Theory. A famous example would be Di-Borane, or $\ce{B2H6}$.

Thus, both the Valence Bond approach as well as the Molecular Orbital approach are often required for an accurate prediction of the shape of a molecules.

I quote an article from Chemistry LibreTexts:

  Both the MO and VB theories are used to help determine the structure of a molecule. Unlike the VB theory, which is largely based off of valence electrons, the MO theory describes structure more in depth by taking into consideration, for example, the overlap and energies of the bonding and anti-bonding electrons residing in a particular molecular orbital. While MO theory is more involved and difficult, it results in a more complete picture of the structure of a chosen molecule. Despite various shortcomings, complete disregard of one theory and not the other would hinder our ability to describe the bonding in molecules.

Thus, although the VB, VSEPR, and Hybridization theories are outdated and have many limitations, they, along with the Molecular Orbital Theory provide us with a toolbox for almost any molecule possible.

(If there is any place where I have erred, please correct me)


How to draw glucose using a Fischer projection?

Here is a web page that shows how to interpret the position from the structure.
The UP groups go to the left side of the backbone and the DOWN groups go to the right side.


Regioselectivity in iodolactonisation of γ,δ-unsaturated carboxylic acid

There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a thermodynamic consideration, and from a kinetic point of view the entropy factor happens to outweigh the enthalpy factor.

Secondly, for cationic ring-opening reactions (e.g. reactions of halonium ions, or epoxide opening under acidic conditions), it is usually the more substituted end that preferentially reacts. See e.g. Regioselectivity of bromination of alkenes. This is also explained in Clayden 2ed, pp 436–438.

However, probably the most relevant factor here is stereoelectronic in nature. Stereoelectronic effects refer to effects which depend on the spatial orientation ("stereo") of orbitals ("electronics"). In the case of an SN2 reaction, the key requirement is that the lone pair of the nucleophile (Nu) must be able to reach the σ* orbital of the C–X bond being broken. Since the σ* orbital points out of carbon diametrically opposite the leaving group X, this is equivalent to saying that we need a collinear Nu···C···X geometry in the transition state.

For ring-forming reactions, there are a set of guidelines – Baldwin's rules – which empirically describe whether these reactions are stereoelectronically favoured (i.e. whether the nucleophile and electrophile are capable of orienting themselves in a geometry that allows for reaction). In the case of six-membered ring formation via substitution at a tetrahedral carbon, the reactions are labelled 6-exo-tet or 6-endo-tet depending on whether the leaving group X is outside the 6-membered ring (exo), or inside the ring (endo). [Technically, 6-endo-tet reactions are not ring forming.]

It turns out that although 6-exo-tet reactions are perfectly OK, 6-endo-tet reactions simply do not happen, as the nucleophile cannot stretch itself to reach the C–X σ*.

The opening of a three-membered ring is somewhat intermediate between these two cases. Because the C–X bond is constrained in a small ring, the reaction leans towards being a disfavoured 6-endo-tet reaction, even though formally it is 6-exo (as the leaving group is outside the ring being formed):

The lack of overlap isn't obvious from the 2D diagrams here, but bear in mind that these compounds are not flat – especially in the case of the epoxide, where if the epoxide is in the plane of the paper, the alkyl substituent must be pointing up towards us (or down away from us).

As a result of this, ring opening of the iodonium ion to give a five-membered ring is much more favourable.

A similar example is seen in acid-catalysed epoxide opening.[1,2] The formation of the 6-membered ring (the tetrahydropyran) is not observed at all:

References

Janda, K.; Shevlin, C.; Lerner, R. Antibody catalysis of a disfavored chemical transformation. Science 1993, 259 (5094), 490–493 DOI: 10.1126/science.8424171.
Na, J.; Houk, K. N.; Shevlin, C. G.; Janda, K. D.; Lerner, R. A. The energetic advantage of 5-exo versus 6-endo epoxide openings: a preference overwhelmed by antibody catalysis. J. Am. Chem. Soc. 1993, 115 (18), 8453–8454. DOI: 10.1021/ja00071a067.


Is temperature of crystallisation same as melting point temperature for polymers?

Is temperature of crystallization same as melting point temperature for polymers?

In a word: NO.

If you undrestand what the glass transition, crystallization and melting temperatures are it will make sense why they are not the same. This paper does an excellent job at explaining the three concepts but I will summarize.

The glass transition temperature is the temperature where polymer chains become sufficiently mobile that the material transforms from being brittle to being soft and plastic (neither exo- nor endothermic, but does the heat capacity does increase). The crystallization temperature is the temperature where the molecules are mobile enough to rearrange into ordered arrangements (crystallize not decrystallize; this is an exothermic process; see plot below). The melting temperature  is where the polymer chains can freely move and ordered arrangements are disrupted (this annihilates any prior crystallization; an endothermic process).

Source: Plastic Technology Laboratories


Reaction of tertiary alcohol with NaH

As all types of alcohol can easily reacts with NaH to produce hydrogen gas and sodium salt of alkoxide ion or alcohol. This shows alcohols have acidic hydrogen. It is type substitution reaction. The general reaction is as follows :

When tertiary alcohol say $\ce{(CH3)3COH}$ reacts with $\ce{NaH}$, hydrogen gas is released as follows :
$$ \ce{(CH3)3COH + NaH  ->  (CH3)3CONa + H2}$$

No acetylation of tertiary alcohol occurs under normal conditions

The acetylation means addition of acetyl group to a substrate. In the case of primary alcohol, acetylation is done under normal condition. Reactivity of secondary alcohol is somehow decrease for acetylation. In the case of tertiary alcohol, it is possible in extreme conditions. There is main reason behind.

As we know acetylation is nucleophilic substitution reaction. Here, the attacking nucleophile for acetylation of alcohol is alkoxide ion. More is nucleophilicity, more be reactivity of nucleophile to attack at carbon of carbonyl group of acetyl chloride.

Nucleophilicity order : $\ce{CH3O- > CH3CH2O- > (CH3)2CHO- > (CH3)3CO-}$

This is reason that tertiary alcohols are least reactive for acetylation. You can more understand by the following mechanism :

Oxidation of tertiary alcohol

Tertiary alcohols Undergo oxidation with oxidizing agents such as chromium tri oxide, potassium permanganate, potassium di chromate or nitric acid in neutral, acidic or alkaline medium to give the products. The oxidation may be partial or complete. Tertiary alcohols does not get oxidized by mild oxidizing agents either in aqueous, alkaline or under neutral conditions. They undergo oxidation under acidic condition.In this process of oxidation tertiary alcohols first undergo dehydration to form alkenes. The alkene thus formed then undergoes further oxidation and forms a ketone with one carbon atom less than the alkene. On further oxidation the ketone forms carboxylic acid with one more carbon atom less than the ketone. Thus the oxidation of tertiary alcohol gives a product with less number of carbon atoms than that of the original alcohol taken.


I actually take issue with the question (which was asked of you not by you), as I think this oversimplifies a molecule, that still isn't well understood in the first place.
That is why I find this question incredibly difficult to answer and why I find that Nuclear Chemist's answer dangerous, because it propagates myths about bonding that have been disproved already.
But enough of the commentary, let's clear up some issues.

Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$.

$\ce{ClO2}$ has two $\sigma$ bonds, one lone pair, two π bonds and one odd electron.

There comes a time, where you should realise that hybridisation is not always a useful concept to understand bonding. This time is now.
Your analysis of the bonding is not wrong, but maybe a bit too simple.
The reason for this is that the molecule has a very high symmetry, but it also has an unpaired electron. The problem here is that orbital symmetry cannot match with spacial symmetry, which means that hybridisation doesn't actually work here.
A bit more precise: The electron density of α-spin electrons is different from that of β-spin electrons. The hybridisation of the orbitals to produce either electron density need to also be different. In the very least you would have to consider two different hybridisation schemes at once.
All possible states (that is independent of any hybridisation model) then will be in superposition to form the actual ground state. This ground state will have a complicated bonding situation.
Of course you can use VSEPR theory to predict the shape of the molecule, that is a very good starting point. You will find out that the molecule is bent, and that matches the crystal structure. A bent molecule of the form $\ce{B-A-B}$ always has $C_\mathrm{2v}$ symmetry, it therefore always has π-orbitals. Whether those orbitals are essentially bonding or not, is an entirely different question.
It also does not tell you which atomic orbitals will be involved.
One obvious fault in the question is that there is no involvement of d-orbitals whatsoever in the molecules. The d-orbitals of chlorine are simply not accessible energetically; at least not in the way that it would form stable bonds. Anyone who is jibber-jabbering about 'octet-expansion' has stopped learning about chemistry about 20 years ago. (There are plentiful discussions about this on chemistry.se alone.)
As long as that last part of the question is not meant as a trap (which would be even more awful), it alone shows that the exercise is not well-thought about.
The π-orbitals will therefore be entirely composed out of p-orbitals of the involved elements.
If you look close enough you will realise, that writing a valid Lewis structure of the molecule is not that hard, and you don't even need π-bonds in the first place.
These are solely dictated by symmetry.
In principle that molecule is of a similar system like allyl-systems (with two extra electrons), or ozone (with one more electron). You will need a better than rudimentary grasp at molecular orbital theory to understand it.
Let's assume that things would be easier, and try tor roll with that for as long as possible. I have written a couple of times that terminal atoms can be (as a rule of thumb) regarded as having sp-orbitals.
Therefore the two oxygen will use one sp-orbital to form a σ-bond towards chlorine, the other one is an in-plane-σ-lone-pair. That leaves for every oxygen one in-plane-p-lone-pair, which we will (in the spirit of the approximation) disregard as having any effect on bonding. There is one remaining out-of-plane-p-orbital per oxygen. These will, due to symmetry, participate in π-bonding.
From VSEPR-theory we know the molecule is bent. The approximate hybridisation of the orbitals of chlorine that match that would be sp². Let's just assume that this is the case. Two of these will be the counterpart for the σ-bonds. The remaining will be either a lone-pair, or singly occupied. Already in this very simplified view, the problems start here. Let's further assume, that because it has higher s-character, the orbital will be lower in energy and therefore be doubly occupied. That leaves the odd electron in an out-of-plane-p-orbital, which will participate in π-bonding.
In this simplified and assumption riddled thought-experiment, the π-bonding will form three molecular orbitals pictured below.

Essentially the stabilisation from the π-bonds itself should be small, and could be described as π-backbonding, or is similar to negative hyperconjugation; maybe negative π-backbonding (!?).
This all was deduced from a very reduced and simplified set of theory; the reality is probably by far more complicated. Some of the greatest minds of (theoretical) chemistry have been puzzled about $\bf\ce{ClO2}$, and we have not even started to discuss the three-electron bond they assumed in between. Let's just say that this is not an undergrad study problem.
I know this doesn't really answer your question, and this is not very satisfactory either, but this is science: It often gives you more questions than answers.


Which branches of chemistry and in what order to learn (self study)?

I agree with LDC3 that inorganic chemistry should definitely be your starting point; it builds foundations that apply to all of chemistry, which is why they teach it first; organic chemistry is just a branch, albeit a very large one, of chemistry that can also be very interesting once you know the fundamentals. Unfortunately using conventional textbooks etc, you will most likely be constricted to this order of learning, because of the level of these courses in the education system, where inorganic chemistry is very traditionally taught first, with organic chemistry only afterwards.

  Any short-cuts would be greatly appreciated - I basically just want to get to a point where I can read a wikipedia page on neurotransmitter X, psychoactive Y, or food ingredient Z and be able to comprehend fully what is going on.

I don't know if there are any shortcuts that you are seeking, Wikipedia can be pretty advanced because it is (usually) written by experts who know exactly what they're talking about; and especially on those high-level concepts, there are a lot of "things" (for lack of a better word) that you have to already know to understand fully what the big picture is. Trying to pick and choose would be maybe even more time-consuming than learning things in a progressive order.

Even though this sounds pessimistic, learning chemistry can be rewarding if you work at it, not to mention you have this site and tons of other online resources to help you out along the way.

-Best of luck should you choose to continue, and I hope you do!


Small but modest displacements at negligible forces from DFT

For clarity I will assume Gaussian performs a generic Newton-Raphson minimization (NR), which should suffice to explain the phenomenon.
In NR, the linear problem
$$
\nabla\nabla^\ast E \Delta + \nabla E = 0
$$
is solved, where $\Delta$ is the displacement. In order to arrive at "large" displacements despite "small" (but non-zero) forces ($-\nabla E$), it suffices for the Hessian ($\nabla\nabla^\ast E$) to have eigenvalues close to 0 itself, because a small number is divided by an even smaller number. This happens when the potential energy surface is very flat.


Carbon dating: What to compare the new value against?

Large elements of my answer is drawn from my own notes that I use for teaching General Chemistry II.

What is −0.693?

Radioactive decay processes follow first-order kinetics.  A first-order reaction is one where the rate depends only on the concentration of one of the reactants raised to the first power.  So consider the following reaction:

$$\mathrm{A} \longrightarrow \mathrm{products}$$

The rate can be expressed as the rate of change in the reactant concentration with a change in time such that

$$\mathrm{rate} = -\dfrac{\Delta[\mathrm{A}]}{\Delta t}$$

This simply means that [A] is being consumed (i.e. converted to products) as the reaction proceeds with time.  The corresponding rate law can be written as

$$\mathrm{rate} = k[\mathrm{A}]$$

where $k$ is a rate constant.  We can set these two expressions equal to each other to give

$$-\dfrac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]$$

At this point we must apply a bit of calculus.  Let us write the previous equation in differential form to give

$$-\dfrac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]~~~\mathrm{or}
~~~\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -kdt$$

Integrate over $t=0$ to $t = t$ to give

$$\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int_0^t dt$$

which results in

$$\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$$

or

$$ \ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt$$

where $[\mathrm{A}]_0$ is an initial concentration and $[\mathrm{A}]_t$ is the concentration at some time, $t$.

Now that we have our first-order integrated rate law, we can now discuss half-life.  Half-life (denoted as $t_{\frac{1}{2}}$) is simply the time required for some reactant concentration to drop to half its original value (where the original value is simply [A]$_0$.  So

$$t = t_{\frac{1}{2}}~~~\mathrm{when}~~~[\mathrm{A}]_t = \dfrac{1}{2}[\mathrm{A}]_0$$

Given this relationship, we can rearrange our first-order integrated rate law to solve for $t$ such that

$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt~~~\longrightarrow~~~
t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t} $$

Therefore, we can make some simple substitutions into our first-order integrated rate law at $t = t_{\frac{1}{2}}$ to give

$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{\frac{1}{2}[\mathrm{A}]_0} $$

We can then simplify this equation such that

$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln2\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0} ~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{1}{k}\ln (2)
~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{0.693}{k}$$

So to answer your question, the 0.693 value comes from the $\ln(2)$ term.  As ringo pointed out below (and I will include here to be thorough), you can now solve your first-order integrated rate law for $t$ and substitute in your half-life equation such that

$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}{-k}=t
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}
{
-\frac{0.693}{t_{\frac{1}{2}}}
}
=t
$$

or more cleanly...

$$t = \dfrac{\left(t_{\frac{1}{2}}\right)
\left(\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}\right)}
{-0.693}
$$

How do you get $N_0$?

You could get $N_0$ by examining the activity of carbon-14 in a relatively recent object or plant.  Carbon-14 in a living plant is not expected to have decayed at all (since the half-life is over 5000 years!).  Therefore you could use its activity for $N_0$.  Of course this procedure relies on a few assumptions but carbon-14 dating is an approximation to begin with.


Comparing basicity of imidazole and 2-imidazoline

I was unable to find experimental $\mathrm{p}K_\mathrm{a}$ values of imidazoline. However, the tools one can use to calculate and thereby predict $\mathrm{p}K_\mathrm{a}$ values are pretty good nowadays as can be seen comparing the predicted and experimental values for imidazole.

$$\textbf{Table 1: }\text{predicted and experimental }\mathrm{p}K_\mathrm{a}\text{ values of imidazole and imidazoline}$$
$$\begin{array}{lcccc}\hline
\text{compound} & & \mathrm{p}K_\mathrm{a,1} & \mathrm{p}K_\mathrm{a,2} & \text{ref}\\ \hline
\text{imidazole} & \text{(e)} & \phantom{0}7.13 & 12.70 & [1]\\
& \text{(p)} & \phantom{0}6.97 & 13.40 & [2]\\
\text{imidazoline} & \text{(p)} & 10.17 & \text{N/A} & [2]\\ \hline\end{array}$$

Note: $\mathrm{p}K_\mathrm{a,1}$ refers to the $\mathrm{p}K_\mathrm{a}$ values of the protonated heterocycle. The higher this value, the higher the basicity of the compound in question.

Obviously, the answer set’s statement that imidazoline is more basic is correct; the question remains why that is the case.

Both compounds obviously have the same protonated form, namely it is the imino nitrogen that bears a proton and thus a formal positive charge. Both of these protonated forms are $C_\mathrm{2v}$ symmetric. At first sight, the electronics of the two are pretty much equal. Yet obviously, imidazoline’s nitrogen must experience a greater electron density for some reason.

The key difference is the size of the π system which spans five atoms in imidazole but only three in imidazoline. At first sight, it seems like imidazole should be more basic since the positive charge can be delocalised onto five atoms compared to only three in imidazoline. However, this explanation fails to acknowledge that all π electrons are delocalised across more atoms in imidazole’s case.

With the π system being larger and spanning more atoms, we expect less electron density to be concentrated on each of the two nitrogens. On the other hand, the much smaller π system of imidazoline means that a greater electron density can be found on both nitrogen atoms at any time.

Further effects, such as the greater electronegativity of $\mathrm{sp^2}$ carbon atoms or the $+I$ effect of $\ce{C-H}$ bonds should be neglegible.

References:

[1]: The Evans $\mathrm{p}K_\mathrm{a}$ table

[2]: Calculated by ChemAxon’s Chemicalize service.


Why is HCl used in hydrolysis of sucrose?

The mechanism really isn't much more complicated than what you describe.

The $\ce{H+}$ binds to the acetal linkage. Water then binds to the fructose ring, breaking its bond with the acetal linking $\ce{O}$. Finally, the water molecule on the fructose loses a proton, restoring the catalyst. As Jan mentioned in the comments, the specific acid used is irrelevant so long as it is strong enough to sufficiently acidify the solution.
(Credit to Khan Academy for the image)


TLC: Why is a more polar eluent more effective at displacing the points?

Most likely you are using "normal phase" TLC plates. The stationary phase is silica gel, which has free hydroxyl groups making it a polar surface. In TLC, there is a competition for the plate between the analytes and the eluent. As you make the eluent more polar, the eluent will interact more with the plate leaving the analytes to be carried with the mobile phase.


Combustion reactions are complicated even when they result in simple products
Part of the problem with answering the question "at what point are water and carbon dioxide formed" is that the reactions occuring in combustion are very complicated and not at all well represented by the simple versions shown in the question.
We can say for sure that the picture where all the components are atomised and then recombine is not the way the reaction works. This simplistic idea is part of the reason for disagreement on the reddit thread.
It is an incorrect view for many reasons but mostly because the reaction doesn't all happen at once. Take petrol (that's gasoline for americans) engines. The initiation of combustion is started by a spark and then (relatively slowly in chemical terms) progresses through the rest of the fuel/air mixture (if it were very fast the engine would explode rather than smoothly driving the cylinder, if it were only a little too fast the engine would "knock", damaging the cylinder). So we need to think–even before considering the detailed chemistry–about the (slow) steps involved in the reaction.
The major steps involved in most combustion reactions involve radical chains. Though the full detail is not understood (though this answer gives some more detail), it is thought that the initiating step involves a hydrocarbon reacting with an oxygen molecule to give a hydroperoxide radical and a hydrocarbon radical. Very simplistically this looks like:
$\ce{O2 + CH3(CH2)nCH2-H -> HO2. +CH3(CH2)nCH2.} $
But neither of those radicals is stable or long-lived. In an environment flooded with more hydrocarbons and oxygen, they can both react further to give more products and more radicals. For example, the hydroperoxide may also fall apart to hydroxide radicals which can react with parts of other hydrocarbon molecules to give water and yet more radicals. In the early stages, most reactions create more radicals; later in the process some radicals can combine into stable products (like water and carbon dioxide). Once the reaction is started it can be self-sustaining as the formation of those stable products release a lot of heat, enough to keep creating new radicals and new chains until the source hydrocarbons or the oxygen is depleted.
So, even thinking about the detailed, step by step, chemical reactions suggests that the process is messy. Even early on in the process some final products are being created, though this continues until the fuel/air is mostly used up. if you plotted the amount of final product over time you would see some very early and a rising amount until the reaction had finished. It is pretty meaningless to ask "at which point" the products are formed.
In short, the initiation does not "atomise" the components: it creates radicals (this doesn't need as much energy input as "atomisation"). They can continue to react creating chains of yet more radicals and some chains can stop when stable products are formed, even early in the reaction. The net reaction releases enough heat to sustain combustion while fuel remains. Final products start to form early in the process and continue to be formed until the air and fuel have gone.


What happens to the amino acid when it is in a solution with the pH of its isoelectric point? (zwitterions)

The accepted definition by IUPAC gold book is: The pH value at which the net electric charge of an elementary entity is zero. pI is a commonly used symbol for this kind-of-quantity. It should be replaced by pH(I) because it is a pH determined under that particular condition.

The solubility of an amino acid is the smallest at its isoelectric point and it easily precipitates.

For question 4 you need to be familiar with dynamic equilibirum. The zwitterion accepts and loses protons all the time, but at the isoelectric point, its concentration is the biggest among all the forms of the amino acid, that is, the species that has a protonated carboxylic group, would easier lose that proton than the zwitterion's carboxylic group would accept a proton.


How to find the mass of dissolved AgCl when placed in a solution of NaCl?

The question is most likely badly worded and should have been

  What is the maximum mass of $\ce{AgCl}$ that can be dissolved?

As we all know, $\ce{AgCl}$ is damn insouble in aquaeous solutions — and due to the equilibrium constant it doesn’t get better at all if you add chloride ions (reducing the silver ion concentration). So without having done the maths, I’m going to assume that a small spec of nothingness is all that will dissolve. That spec is so tiny that you will have a hard time weighing less so you can actually perform the experiment you suggested.

Because it’s just a flick of dust, and because the authors likely

knew that, too
assumend you knew that, too

they may have gone with a less obvious wording. I mean it’s clear: You can’t dissolve what you didn’t put in, and to-be-dissolved silver ions aren’t going to magically appear with Alakazam.