How do significant figures work on this computation?

Assuming the precision of the given volume, $V_1 = \pu{400 mL}$, is only to 1 significant figure (i.e. only the hundreds place digit of "4" is precise), then your final answer would be rounded to $V_2 = \pu{300 mL}$.*

If we were to say that $V_1$ had 3 significant figures, then the final answer would round to the ones place, resulting in $V_2 = \pu{342 mL}$. However, the value would technically need to be written as "$\pu{400\!. mL}$," including the decimal place to denote the precision going out to the ones place, in order to imply 3 significant figures instead of 1.

In general, zeros (to the left of the decimal place) not surrounded by non-zero values are not denoted as "significant." For further rules and guidelines for identifying what numbers are significant, how many significant figures there are, and how addition/subtraction or multiplication/division impact the final answer's significant figures, check out this web page from Northern Kentucky University (recommended search terms for future information on this subject would be "understanding significant figures" or "significant figures rules").

*If difficulty was also found in the process of where to begin rounding the answer from, the key is finding the "rounding digit," which, if you have only 1 significant figure to work with, will be the furthest numerical value to the left. See this page for further tips.


This was a tricky synthesis. There were several issues. One is that the aryl glycine is possibly easy to epimerize, though the ortho substitution may help prevent that. One other major issue is that the thiophene is reducing and not compatible with any oxidative conditions and probably not compatible with electrophiles.

The amino acid moiety can be synthesized in several ways to set the stereogenic center. One is epoxide ring opening, but here I opted for an asymmetric Strecker reaction.

The six member ring was created via a [4+2] cycloaddition. This is probably the sketchiest portion of the synthesis.

The thiophene moiety was created using Paal-Knorr reaction with Lawesson reagent (B).

Detailed route:

imine formation
asymmetric Strecker with 5% PHTA (partially hydrolyzed titanium alkoxide) and 5% ligand (L). 1 eq tBuOH, tol. (Org. Lett., 2010, 12, 264.) Also, possible with Hoveyda type ligand: (JACS, 2000, 122, 2657.) This exact substrate is reported with good ee.
Hydrolysis with HCl. Similar route reported in (JACS, 1999, 121, 4284.) Though there, the authors Boc-protect at the end. We'll attempt to trap a different way...
formalin, toluene, pTsOH. reflux. Product imine is A.
LDA; TMSCl on methyl vinyl ketone to make electron rich diene.
Hopefully, a little heat will coax this [4+2] cycloaddition with A to take place.
Tsuji-Trost allylation is established for these types of TMS-protected enol ethers. Conditions are $\ce{Pd_{3}(dba)_{2}}$, DPPE, allyl methyl carbonate, THF. (Chem Lett., 1983, 1325.) We'll carry both diastereomers forward.
$\ce{O3}$; DMS.
Lawesson reagent (B).


A Tool to atomize liquid in the milliliter range

You might want to look into using an inkjet (salvaged from an old printer). They can produce a regular, directed stream of small droplets and have a capacity of at least several mL. You can power them and control their behavior electronically. There are probably inkjet printers available that print onto a fixed platen, so that would take care of the production aspects.


What is the difference between an ion-induced dipole attraction and just an ion-dipole attraction?

It is useful to note the definition of a van der Waal force: weak electric forces attracting neutral molecules to each other in gas thus causing a gas to deviate from an ideal gas.

So, there are three types of these forces

Dipole-dipole force
Dipole-induced dipole force
London dispersion force

It becomes clear that forces involving ions do not fit van de Waals forces as the molecules involved must be neutral.

An ion-induced dipole force is not a dispersion force, either (as dispersion forces are a subset of van der Waals), and with ion-dipole forces, ion-induced dipole forces have a category of it's own

A dipole-induced dipole force is a van der Waal force but not a dispersion force


$\ce{LiCl}$ and $\ce{LiBr}$ are ionic compounds with a significant covalent character. This is due to the strong polarising power of $\ce{Li^{+}}$ cation. This polarising power is due to its large charge to mass ratio.

$\ce{Li^{+}}$ pulls the electron cloud of the halogen, hence resulting in a covalent character. The covalent character is more in $\ce{LiBr}$ than $\ce{LiCl}$ as $\ce{Br^-}$ is larger in size, hence valence shell is held comparatively loosely.

This is what essentially happens.


Predicting melting points of substances based on their molecular/atomic structure

With regard to organic compounds, you might be interested in group contribution methods, e.g. the Joback method.

These methods divide a molecule into functional groups or similar small building blocks and estimate the melting point (and other thermodynamic properties) based on the contributions of the individual groups.


The picture below from the UC Davis ChemWiki summarizes why that lone pair is not available. It also explains how pyridine can function as a base.

The lone pair in the $sp^2$ orbital is perpendicular to the $p$ orbitals and $\pi$ bonds in the molecule. the electron pair from the nitrogen that participates in aromaticity is the one in the $\ce{C=N}$ bond.
To answer your first question about hybridization, we have a conservation law that guarantees that when orbitals mix (for any reason), we need to get the same number of orbitals out as we put in. The law is the Conservation of Angular Momentum, which does apply in quantum systems.
Nitrogen, in its valence shell, has four orbitals: $2s,2p_x,2p_y,2p_z$. The $sp^2$ hybridization is achieved by mixing two $p$ orbitals and one $s$ orbital. Because of the conversation of angular momentum, four orbitals in means four orbitals out. Thus, the $p$ orbital that was not mixed is still there (and is the only one still there).
$$[s,p,p],p\implies[sp^2,sp^,sp^2],p$$
We cannot have two $p$ orbitals at an $sp^2$ hybridized atom because then we would have 5 orbitals (or we would not have $sp^2$).

Now, let's create a general rule for aromatic heterocycles.

Look at pyrrole (also from ChemWiki);

In the structure of pyrrole, the $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair is in the $p$ orbital. In general, if there is a double bond drawn to the nitrogen atom (like in pyridine), then the $\pi$ bond uses the $p$ orbital and the lone pair is in the $sp^2$ orbital and does not count toward aromaticity. If there is no double bond drawn to the nitrogen atom, usually there is a $\ce{N-H}$ bond (or some other bond to the nitrogen). This $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair uses the $p$ orbital and counts toward aromaticity.


Alkane nomenclature - choosing the parent chain - maximize or minimize substituents?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44 SENIORITY ORDER FOR PARENT STRUCTURES

  A parent structure is defined (P-15.1) as a parent hydride, for example, methane, a functionalized parent hydride, for example, cyclohexanol, or a functional parent compound, for example, acetic acid. For choice of the preferred IUPAC name based on a senior parent structure, see P-45.

  (…)

  P-44.3 SENIORITY OF ACYCLIC CHAINS (THE PRINCIPAL CHAIN)

  In an acyclic compound consisting of individual atoms, alike or different (an acyclic chain), the chain on which the nomenclature and numbering is based is called the 'principal chain'. When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached.

  The principal chain:

  (…)

  (b) has the greater number of skeletal atoms;

  (…)

  P-45 SELECTION OF PREFERRED IUPAC NAMES

  P-45.0 INTRODUCTION

  Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. (…) Application of
  the criteria in this section will generate the preferred IUPAC name.

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has
  the maximum number of substituents cited as prefixes (other than 'hydro/dehydro') to
  the parent structure.

In accordance with P-44.3 (b), the principal chain is a decane.
It could be either “5-(2-ethylbutyl)-3,3-dimethyldecane” or “5-(2,2-dimethylbutyl)-3-ethyldecane”.

Because of P-45.2.1, the preferred principal chain corresponds to “5-(2-ethylbutyl)-3,3-dimethyldecane”, which has a principal chain with three substituents (5-, 3-, and 3-), whereas the principal chain of “5-(2,2-dimethylbutyl)-3-ethyl” has only two substituents (5- and 3-).


Current and reversible heat in battery reaction

How is the current related to nF? current and nF have different dimensions?

$nF$ is the charge $C$ transferred during the reaction, while current $i$ is the rate of charge transfer ($dC/dt$).

The following is a derivation, now edited to be more rigorous. The equation you provide is an expression for electrical Joule heating, which follows in the steady-state from the expression for the power (rate of work) generated by the battery. From Ohm's law

$P=\left(\frac{dw_{rev}}{dt}\right)=-i^2R=-i E_0 \tag{1}$

where $E_0$ is the (open-circuit) electric potential, $i$ the current, $w_{rev}$ is the reversible electrical work. Note that $E_0$ is the potential when reversible work is done.

Now assume a steady state, with constant P and T constraints, with work done and heat generated cancelling (so that the internal energy $U = constant$), that is

$Q_{rev} = -w_{rev} \tag{2}$

and therefore also power and rate of heat dissipation equal:

$\left(\frac{dQ_{rev}}{dt}\right)=-\left(\frac{dw_{rev}}{dt}\right)=i E_0 \tag{3}$

Now according to the Nernst equation,

$\Delta G=w_{rev} =-nFE_0 \tag{4}$

In the steady-state the dissipated heat is equal to the electrical work, so that, combining (2) and (4), we have that

$\frac{Q_{rev}}{nF}=-\frac{w_{rev}}{nF}=E_0 \tag{5}$

which leads, combined with (3), to the equation in the book:

$\left(\frac{dQ_{rev}}{dt}\right)=i \frac{Q_{rev}}{nF} \tag{6}$

Aside

Note I apply the opposite sign convention for work from that in the OP (apologies) - in the convention I use work is positive when performed on the system (charging a battery is positive work, discharging negative work). This does not affect the result of the derivation, since I apply the same sign convention for heat (for an exothermal process heat is negative).

On $\Delta H$ and an alternative (longer) derivation

For a process at constant p, it is common to encounter the expression

$$\Delta H = Q_p$$

However if there is non-PV work, the more general form of this equation is

$$\Delta H = Q_p + w_{non-pV}$$

This expression is general and applies when the process is carried out either reversibly or irreversibly. In differential form,

$$d\Delta H = dQ_p + dw_{non-pV} \tag{a1}$$

Note also that in the preceding derivation non-pV work is electrical.

When the process is carried out reversibly a maximum amount of work is done and

$$w_{non-pV,rev} =\Delta G = \Delta H - T \Delta S = \Delta H - Q_{rev}$$

This leads to the following expression at constant T, equal to that in the linked problem (but note the different work sign convention here):

$$ d\Delta H = dQ_{rev} + dw_{rev} = d\Delta G + Td\Delta S \tag{a2}$$

where I dropped the "non-pV" subscript since the work is assumed to be electrical.

Since H is a state function it must be equal for reversible and irreversible processes, and we can equate (a1) and (a2), leading to the following general expression:

$$ d\Delta H = dQ_p + dw_{ele}=  d\Delta G + Td\Delta S $$

which leads finally to

$$ dQ_p = - dw_{ele} +  d\Delta G + Td\Delta S  \tag{a3}$$

Taking the time derivative of this equation results in the final expression in the linked problem (making sure to account for differences in the work sign convention):

$$\dot{Q} = \dot{Q}_\text{rev} + \dot{Q}_\text{irrev} = IT\,\frac{\mathrm dE_0}{\mathrm dT} + I(E-E_0) \tag{a4}$$

To obtain the equation in this problem it is only necessary to apply the reversibility condition starting from either (a3) or (a4). Starting from (a3),

$$ dQ_{rev} = - dw_{ele,rev} +  d\Delta G + Td\Delta S  $$

But $ dw_{ele,rev} =  d\Delta G  $ which leads to (somewhat trivially)

$$ dQ_{rev} = Td\Delta S  $$

Taking the time derivative and inserting the Nernst expression for $\Delta S$ (twice!)  gives

$$\dot{Q}_\text{rev} = \left(\frac{dQ_{rev}}{dt}\right) = IT\,\frac{\mathrm dE_0}{\mathrm dT} = I\frac{Q_{rev}}{nF}$$

which is the desired expression. The same result can be arrived at by applying the reversibility condition to equation (a4).


If you are certain that there are no other halogenated byproducts or reagents remaining in your solution, and that any levels of bromine and iodine would be negligible compared to chlorine (fluorine would not be detected), then one of the total halogen analysis (TX) methods would be a good choice.

If this is an ongoing analysis that you need to set up, I suggest the Carius halogen method. Wikipedia gives a good overview here:

  The Carius halogen method in analytical chemistry is a method for the
  quantitative determination of halogens in chemical substances. A
  known mass of an organic compound is heated with fuming nitric acid in
  the presence of silver nitrate contained in a hard glass tube known as
  carius tube, in a furnace. Carbon and hydrogen present in the compound
  are oxidised to carbon dioxide and water. The halogen present forms
  the corresponding silver halide (AgX). It is filtered, washed, dried
  and weighed.

Alternatively, depending on the equipment available to you, you could also do an oxygen bomb combustion of the sample with a bicarbonate absorbant, followed by either titration of the resulting chloride or ion chromatographic (IC) analysis. In the case of the IC finish, you can usually just use the IC eluent as the adsorbant.

If you are only going to do this analysis once, or only rarely, then I would strongly suggest a commercial TX anaysis using Standard Methods - For the Examination of Water and Wastewater, Method 9020. Method 9020B is for the analysis of total organic halogens (TOX) in water, which is not appropriate for your analysis as a whole because it includes a step for extracting trace amounts of organics from water onto activated carbon. But, then a TX analysis is performed on the activated carbon, which is what you would want to do for your compound rather than for the activated carbon. This is typically about an $80 analysis.

Regardless of the TX technique used, you would of course then assume that all of the halogen came from the ethyl chloride and calculate the ethyl chloride concentration from there.


How do I start studying chemical nomenclature?

Chemical nomenclature is mixture of historical traits, empirical knowledge and little bit of systematics.
And as you mentioned, it is exhaustive to learn all at once.

To understand the nomenclature, you must understand the systems themselves. Because it's their structure and reactivity which is reflected in the name.

This makes an unfortunate cyclic dependency. To start studying any given chemistry branch, you must know its language. But without knowing the science behind, the language does not make sense. It's like that and you probably cannot do anything about it, just cut this circle and crunch through the nomenclature. It will start to make sense, once you get sufficiently deep. Additionally, the "why" is good indicator. Once you have enough knowledge to understand the answer why it is named so, than you can probably answer yourself.


Second electron affinity of sulfur and oxygen

The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.

It is the tendency of an atom to gain electron. It is numerically equal to the negative of electron gain enthalpy.

Although fluorine has the highest electronegativity, chlorine has the highest electron affinity and this is because the considerable repulsion in the tightly packed $2p$ subshell of fluorine.

While electron gain enthalpy is enthalpy change when an isolated gaseous atom attain an electron to form a monovalent gaseous anion.

Chlorine ($\ce{Cl}$) has more negative electron gain enthalpy than fluorine the reason is same  as above.

Source:

Wikipedia and chemistry reader


Mechanism of ring formation of an epoxide under acidic aqueous conditions

This is a compilation from the comments (some of them deleted by now) showing the thought process in finding the reaction product and eliminating some of the steps that might seem plausible. None of the insight is mine.

Formation of a cation

  I think the H+ will first attack the oxygen of the epoxide

Yes, that is the first step.

No attack by a nucleophilic oxygen at this point

  [...] and another OH− will be added to the most substituted carbocation, which is the secondary carbon of the epoxide. This will form a glycol,but don't know where to go from there.

This does not happen.

  H3O+ means H+ in water so OH- is available.

The solution is acidic, so there is very little hydroxide available.

  H20 will act as a nucleophile and attack the carbocation, this will result in a H+ being released from +(H2O)

This does not happen.

  Why is the H2O not attacking before the formation of 5 member ring?

Compared to hydroxide, water is not a strong nucleophile. ANBENZENE comments: "[...] the molecule will always try to generate a stable carbocation [...]".

Rearrangement to make a more stable carbocation

As Waylander states in a comment: "Have the double bond electrons attack the protonated epoxide to give the most stable cation. Remember, forming [5] rings is favorable".

This is an intramolecular reaction, going from a carbocation on a secondary carbon to one on a tertiary carbon. Also, a five-membered ring is formed and a three-membered ring was opened.

Final step: water reacts with the carbocation

Finally, the oxygen on water does attack a carbocation, resulting in the second hydroxyl group found in the product.


Unless you have a way to separate CO2 from the rest of the gases, you won't be able to measure it properly. Which isn't easy. You could put a substance that absorbs carbon dioxide inside the container for the beans. However, the substance needs to be able to indicate that it has reached capacity. And also, the volume of CO2 that it can absorb must be known.

You could also set up a respirometer. You could make some assumptions about how the oxygen is consumed. Nevertheless, it should give you a good estimate. There are several videos and tutorials explaining them. And if you still have questions, you can ask Bio Stack Exchange.

The comment about the van-der-Waals equations: I think that you were probably thinking on using the ideal gas law to estimate the number of gas molecules in a container.
$$pV=nRT$$

According to the equation above, you can measure the number of particles if you know the pressure and the volume. This is true for ideal gases. For a gas molecule to behave like an ideal gas particle, it would have to have no volume and have no interactions with other gas molecules. Real gases, however, do not behave like an ideal gas.  Therefore, a better model to describe real gases is needed. The van-der-Waals equation is one of these attempts.


Can the "rate coefficient" of a chemical reaction really be called a "rate constant"?

Yes the rate coefficient can be referred to as a "rate constant".
This is in fact a very, very common name for the rate coefficient.  Encyclopedia Brittanica has an entry for rate constant.  Wikipedia has a reaction rate constant page.  Google's N-gram viewer shows that "rate constant" is used more often in the English written corpus by a factor of more than five, relative to "rate coefficient".
The "constant" in rate constant refers to the fact that it does not vary with concentration of the reactants or products.  However, as you note, it could depend on other variables, such as temperature, pressure, catalyst, light, etc.


How come ∆G ≤ 0 is used for spontaneity in electrochemistry, not ∆G ≤ W(other)?

Most likely the source is wrong, or it has purposfully omitted the correct equation to pseudosimplify a problem. It might be that an electrochemical reaction was under discussion, not a cell. A further possibility is pursued in  the appendix. I recommend adding a source where you came upon the quoted exchange to avoid a straw man argument; even though in this case  I have observed (and possessed) the misconception first hand.
One definition of spontaneity
I define the true spontaneity condition as such when there is a tendency toward net 'something different' (chemical reaction, expansion etc) to establish state 2 instead of some original state 1. An example could be a chemical reaction where there is a higher reaction extent than at equilibrium, so a net reverse reaction takes place.
You are correct that the true spontaneity equation (from Clausius's inequality) in the thermodynamic limit is, for net-constant pressure and temperature in the system,
$$\operatorname{d} G-\delta w_\pu{other}\leq 0.\tag1\label1$$
When electrochemical work is the only component besides expansion work, this implies (due to $|\operatorname{d} n_\pu{e}|=|\nu_\pu{e}|\operatorname{d}\xi$)
$$\Delta_\pu{r} G_{T,P} +|\nu_\pu{e}|FE\leq0\tag2\label2$$
where $\Delta_\pu{r} = \partial/\partial \xi$; here $\xi$ is extent of the reaction. The Greek $|\nu_\pu{e}|$ signifies an absolute value of the stoichiometric coefficient of an electron in some half-reaction. Equation $\eqref2$ assumes that only one net reaction occurs. The term $|\nu_\pu{e}|FE$
should be a good indicator of electrical work.
It might also be that they are discussing an electrochemical reaction, not the cell itself. We can have the process
$$\ce{Zn(sln) + Cu^2+(sln) -> Cu(sln) + Zn^2+(sln)}\tag3\label3$$
without harnessing its electrical work via an external circuit. So $\operatorname{d} G\leq 0$ would hold for spontaneity.
Appendix: An alternative
A different definition of spontaneity might be in play. Namely, the term spontaneity is also used to mean a large enough standard equilibrium constant (especially in biochemistry).
$$RT\ln \frac1K = \Delta_\pu{r} G^\circ_{T,P} \tag4\label4$$
Equation $\eqref4$ is technically a definition of the standard equilibrium constant. When non-negative absolute temperatures are quaranteed, the LHS of equation $\eqref4$ will become non-positive for all $K$ big enough, i.e., $K\ge1$ . That implies
$$\Delta_\pu{r} G^\circ_{T,P} \leq 0. \tag5\label5$$
Note, however, that $K\ge1$ doesn't necessarily imply much about the reaction balance itself because the equilibrium constant comprises of activities (not concentrations), and because the stoichiometric coefficients in the denominator may drown out the coefficients in the numerator. (Equation $\eqref6$ assumes that surrounding fugacity is equal to an agreed fugacity in the standard state, denoted here and elsewhere by '$^\circ$'.)
$$K(\pu{in solution, solvent \ce{A}}) = \left[a(\ce{A})_\pu{eq}\right]^{\nu_\ce{A}}\prod_i \left[a(\ce{B_$i$})_\pu{eq}\right]^{\nu_i}. \tag6\label6$$
Reactants–products (other than solvent) are designated by $\ce{B_$i$}$. But still, for $K$ big enough (2nd definition of spontaneity), equation $\eqref5$ will hold by definition. Also keep in mind $\Delta_\pu{r} G^\circ_{T,P} \neq \Delta G$ (even their dimensions are different!). So the source is presumably still at fault for poor notation.


How does intramolecular hydrogen bonding cause the molecules to be separated from each other?

Molecules with inter-molecular hydrogen bonds tend to associate with one another, while molecules with intra-molecular hydrogen bonds tend to associate with themselves.

Ortho- and para-nitrophenol are often used as examples of molecules with intra- and inter-molecular hydrogen bonds respectively (see picture).

Because of their molecular structures the ortho isomer can form an intramolecular hydrogen bond with itself, whereas the para isomer can only hydrogen bond with another molecule of p-nitophenol.  One would say that the para isomer associates with other p-nitrophenol molecules through hydrogen bonding, whereas the ortho isomer associates with itself through hydrogen bonding.  Because the para isomer associates with other molecules of p-nitrophenol through hydrogen bonding, it takes more energy to break up this association and cause the molecule to melt or boil.  On the other hand, the ortho isomer melts and boils at lower temperatures because there are fewer intermolecular forces (such as the hydrogen bonding) holding the ortho molecules together.


How do non-ionic compounds dissolve into water? Into molecules or atoms?

The molecules remain as they are, as discrete particles, they do not break up into single atoms. What can happen to certain compounds when they are dissolved in water is a hydrolysis reaction. But even then, only certain bonds of the molecule are cleaved in a reaction with water, and the molecule does not spontaneously fall apart into the elements it consisted of, because there is no thermodynamic reason to do so.

The same is also true for ionic compounds. Ammonium chloride ($\ce{NH4Cl}$), for example, consists of discrete $\ce{NH4+}$ and $\ce{Cl-}$ ions. $\ce{NH4+}$ is a molecular ion, and it remains as such in aqueous solution and does not break up into single nitrogen and hydrogen atoms/ions. However, the salt can hydrolyze in aqueous solution according to the following equilibrium:

$$\ce{NH4Cl + H2O \rightleftharpoons NH3 + H3O^+ + Cl^-}$$

This is the reason why aqueous solutions of ammonium chloride are weakly acidic.


Particle size of oxides of nitrogen and sulfur pollutants and HEPA filters

If the high efficiency particle air (HEPA) filter says it blocks PM 2.5 that's good as pm 2.5 particles are the ones your lungs cannot easily dispel and cause haze.  Unfortunately sulfur oxides and nitrogen oxides are gasses, so no particle filter will remove them.


How to calculate the boiling point of a solid substance?

you can't...it must be found experimentally (it has) P.S Even if you know the BP at 1 atm, you can never calculate the real boiling point at any given BP, you may come close


Confusion between various formulas of Gibbs free energy

The Gibbs free energy of a multicomponent mixture is a function not only of temperature and pressure, but also of the number of moles of the various species in the mixture:  $$G=G(T, P, n_1, n_2. m_3. ...)$$So, $$dG=\frac{\partial G}{\partial T}dT+\frac{\partial G}{\partial P}dP+\frac{\partial G}{\partial n_1}dn_1+\frac{\partial G}{\partial n_2}dn_2+\frac{\partial G}{\partial n_3}dn_3+...$$or
$$dG=-SdT+VdP+\mu_1dn_1+\mu_2dn_2+\mu_3dn_3+...$$ where the $\mu's$ are the partial derivatives of the Gibbs free energy with respect to the number of moles of each of the species, and are referred to as the so-called Chemical Potentials of the various species in the solution.  The numbers of moles of the species in the mixture change as the reaction proceeds, and that causes the Gibbs free energy to change.


Can you literally separate air from soil?

Your teacher's method of pouring water over soil is an excellent way to demonstrate that there is air in the soil. If you're still unconvinced, it's possible to directly separate the air out. Here's how:

You fill a vacuum chamber with your soil. All of it. So it's all soil and no free space is left. Now you just pump the air to a different chamber. There are several ways to do it so I will not elaborate on the mechanics. This method extracts the air from the intergranular porosity (i.e. the space between the soil grains) to your other container.

  Also, how would the air look like?

Well, it doesn't look like anything, because it is after all air. It is possible to liquefy it by cooling or compressing. It's also possible to measure the composition of the gas you extract out of it and see that it matches the composition of air. It will not be exact, depending on the soil. Some soils have decomposing organic matter (dead bacteria, dead plants, dead other things) so they might have slightly less oxygen and slightly more carbon dioxide, methane, sulfur dioxide, hydrogen sulfide, and a bunch of other things that usually also smell.


Are there molecules that take the shape of every platonic solid?

Yes, for each of the 5 platonic solids, there is at least one molecule that takes its shape.

There exist tetrahedron-shaped molecules. Some instances are $\text{CH}_4$ or $\text{CCl}_4$.

There exist octahedron-shaped molecules. Some instances are $\text{SF}_6$ or $\text{Mo(CO)}_6$.

There exist hexahedron-shaped molecules. An instance would be $\text{C}_8\text{H}_8$ AKA "cubane". Perhaps simple cubic lattices (like NaCl) could be considered "hexahedral molecules" as well?

There exist icosahedron-shaped molecules. An instance is elemental boron, $\text{B}_{12}$.

There exist dodecahedron-shaped molecules. An instance is dodecahedrane ($\text{C}_{20}\text{H}_{20}$).


Is it possible create crystalline solvate of electrons?

At first I would like to advise you reading about electrides, which were mentioned in the comments. Secondly I won't write about electrides because the links given in the comments are much better at explaining them. So I am going to be a little bit abstract with this question. The first problem which you will come through is Coulomb's law:
$$\mathbf F=\frac{1}{4\pi \epsilon_{0}}\frac{Q_{1}Q_{2}}{|\mathbf r|^{2}}\frac{\mathbf r}{|\mathbf r|}$$
If you could somehow prepare even $0.01$ moles of electrons, they would be attracted to the ions which they were separated from very strongly even from kilometers away. The second problem is that electrons can't be simply trapped into any substance because of their wavelike nature and they would tend to "stay" in a localised or in a delocalised state in the molecoules. And here comes the two main things which are preventing such things to exist:
1. Charge distribution
If a conducting body is charged, the charge tends to build up on the edges, and in your situation the repulsive force between the electrons would be so huge, they would "kick" each other off on the edges of the pieces of the substance so it would loose it's electrons.An interesting phenomena such substance would have: Let's suppose the substance containing the electrons is a liquid. All the electrons would be located on the surface of the liquid because of the faraday effect.When the surface of the liquid is disturbed waves appear on it's surface. Let's suppose the electrons somehow can't be kicked off from the surface.
Then there would always be electric current in the liquid because the elecrons would always try to get to the most sharp edge on the surface and the liquid would generate extremely complex magnetic fields because of the currents flowing on the surface of the liquid (I am so excited about the idea of such a liquid, I am so sad I have no idea how it would be possible to make it)
2. Quantum mechanics
The band gap determines if a substance is an insulator or a semiconductor or a conductor. And adding more electrons to something doesn't modify the band gaps. Like if I charge a plastic rod with electrons it won't become a conductor. But there is an alternative solution. If you dope the substance it can modify the band gap energy and it can be more conductive but you can't simply trap electrons in something (F-centers are a completely different story). Imagine irradiating carbon nanotubes with electrons. I wouldn't suspect it becoming more conducting .
Anyway the best source for pure electrons is an electron gun
If you want to deepen your knowledge about the nature of electrons and conductivity I advise you learning some solid-state physics, quantum mechanics and electromagnetism.(None of them is a light weekend program)


What's the difference between reactant, excess reactant and limited reactant in a balanced equation?

Suppose that you want to make a cake. You would first need to go your fridge to check whether you have the necessary ingredients to make one. So you go to your fridge. Luckily, you have all the ingredients: milk, eggs, flour, and frosting. So you decide to make one. You also decide to make the cake as big as possible to avoid the hassle of cleaning/preparing again. Now, what determines how big you can make the cake?

It is the quantity of ingredients of course. But what happens if you  have enough milk to make a 10 cm cake, enough eggs to make a 1000 cm cake, enough flour to make a 1000 cm cake, and enough frosting to make a 10000 cm cake? How big can you make the cake?

You would only be able to make a 10 cm cake. Because you would run out of milk. And since you no longer have milk, you cannot make anymore cake.

Lets go back to your question of excess reagents and limiting reagents. The limiting reagent is the milk in the example above. It is what determines how much of a product you will obtain. It is what limits the amount of product you get (hence the apt name limiting reagent). Everything else are excess reagents: the eggs, the flour, the frosting. They are leftovers; they are in excess.

Now question 2. You cannot tell which reactant will be in excess by just looking at the balanced  equation. Going back to the cake example, the balanced equation is sort of like a recipe for the cake. You cannot tell whether you will have eggs, flour, and frosting left over by just looking at the recipe. You also have to check your fridge. In other words, you have to know what you are starting with.

Now question 3. I think question 3 will be answered in more detail as you study more chemistry. A complete answer involves thermodynamics and statistical mechanics. Nevertheless, going back to the example of the cake: Whether or not you can make cake depends on whether you have all four ingredients. It also depends on how much of each ingredient you have. If you have no milk, you won't be able to make any cake.

But say that you want to make boiled eggs as well. You only need eggs to make boiled eggs so you would could turn all eggs leftover eggs from the cake into boiled eggs. The fact that you can make boiled eggs with the leftover eggs does not change the fact that you can make a 10 cm cake.

This is what the two equations are trying to get at. The major product, will be determined by how much of a certain reactant you have. It doesn't mean that the reaction is solely determined by what it is in excess or not.


If copper has started to corrode, can it stop?

The previous answers doesn't take into account the environment:

  They were operated in conditions of intense humidity/heat and were placed near the sea.

Near the sea means that there are certainly a lot of $\ce{Cl-}$ ions in your metal. Chloride ions are a very dangerous species for metals conservation they tends to depassivate the metal forming complex ions. In the case of copper occurs this series of reactions:
$$\ce{Cu^{+} + Cl^{-} <=> CuCl}$$

$$\ce{Cu^{+} + 2Cl^{-} <=> CuCl2- + Cl- <=> CuCl3^{2-} + Cl- <=>CuCl4^{3-} }$$

That lead to different copper corrosion product. (e.g. atacamite $\ce{Cu2(OH)3Cl}$, malachite $\ce{CuCO3.Cu(OH)2}$ (Did you see some blue-green pigment?), Cupric hydroxide $\ce{Cu(OH)2}$.

These are all dynamic equilibria so your copper with dry-wet, hot-cold cycle will continuously object of degradation until chloride ions are present.

Use distilled water with cotton to clean the salts deposit, dry quickly where you clean.

Reference:

The Open Corrosion Journal, 2009, 2, 71-76
Corrosion Science 46 (2004) 109–135


Is the triplet state of the cyclopentadienyl cation really aromatic?

TL;DR Yes. For cyclic conjugated systems in the $T_1$ state (i.e. lowest energy triplet excited state), $4n$ electrons means aromatic and $4n+2$ means anti-aromatic. The $T_1$ state of the cyclopentadienyl cation, which has 4 pi-electrons, is therefore aromatic.

Aromaticity

The (quite understandable) confusion here arises over the interpretation of the word "aromatic" or "anti-aromatic". These terms can be used differently depending on context, and there is no unified physical description of (anti)aromaticity (although progress has been made in recent years, e.g. nucleus-independent chemical shifts, etc.)

When anybody claims that something is aromatic, it is therefore imperative to check the context, and to figure out what their basis for assigning aromaticity is. In this case, the assignment of the triplet excited state of the cyclopentadienyl cation as aromatic by Costa et al.1 is based on Baird's rules. In turn, looking up Baird's article2 reveals that his assignment of (anti)aromaticity is based on the concept of Dewar resonance energies (DREs).

The concept of DREs has been explained by Baird elsewhere.3 Essentially, it is based on the comparison of idealised cyclic structures with a reference acyclic structure; so, benzene is compared with hexatriene, and cyclobutadiene ($D_\mathrm{4h}$) is compared with butadiene. For the cyclopentadienyl cation, the relevant comparison is with a penta-2,4-dien-1-yl cation.

The question thus arises as to how this comparison can be made. The way that is commonly taught, at introductory level, is to use Hückel molecular orbital (HMO) theory. So, let us use cyclobutadiene as an illustration. Note that within HMO theory the singlet and triplet configurations are of the same energy, as electron-electron repulsions are not taken into account.

The resonance integral $\beta$ is negative. So, singlet cyclobutadiene is anti-aromatic because it has less π-bonding energy than the singlet state of its acyclic counterpart butadiene. However, triplet cyclobutadiene is aromatic because it has more π-bonding energy than triplet butadiene does.

There are two things of note here: firstly, the Jahn–Teller distortion in cyclobutadiene isn't relevant for the purposes of finding whether it is aromatic or not; the aromaticity is assigned based on a hypothetical planar structure with no bond length alternation. Secondly, the aromaticity is determined with respect to the corresponding electronic state of the acyclic reference. That is to say, the triplet state of cyclobutadiene is compared with the triplet state of butadiene and not the singlet state of butadiene.

Perturbational MO theory

Qualitatively, we have reached the correct conclusion. However, this was not the approach that Baird used. We will restrict our discussion now to that of triplet states, since it is relevant to the question. Baird used a more sophisticated method involving perturbation theory. That is, Baird used MO theory to calculate the stabilisation/destabilisation that would occur upon bringing together two radicals to form either the cyclic compound or the acyclic compound.

This is confusing, so a bit more explanation is in order. The triplet state of butadiene can be described by the promotion of a π electron to a π* electron. Upon this promotion, the C=C bond twists until it has a dihedral angle of 90°, which corresponds to the breakage of the π bond.

In the triplet state of butadiene (1 ($T_1$)), the π systems of the allyl and methyl radical fragments are orthogonal to each other; they are simply joined by a C–C σ bond. Hence the total π-bonding energy of 1 ($T_1$) is simply equal to the sum of the individual π-bonding energies of 2 and 3.

On the other hand, the triplet state of cyclobutadiene (4 ($T_1$)) can still be considered to be an allyl fragment plus a methyl fragment.4 The difference is that, due to the planarity of the ring, the two radical systems are forced to overlap in a π manner (depicted by the red lines).

Now, if this overlap is stabilising in nature, then the π-system of 4 ($T_1$) is more stable than the π-system of 1 ($T_1$), and therefore we can say that 4 ($T_1$) is aromatic. Conversely, if the overlap is destabilising, then it is anti-aromatic.

Finally, we get to the crux of the answer. (I make no apology for taking so long to get to it - I thought the buildup was pertinent.) Baird first constructed the MOs of the radical fragments 2 and 3, and labelled each MO with their symmetry under a reflection. The mirror plane is drawn in light blue; antisymmetric MOs are in red and symmetric MOs in green.

Then, Baird identified two types of orbital interactions in the cyclic structures that could lead to either stabilisation or destabilisation. The first type is called a Type I interaction, and it involves the overlap of the two SOMOs on the two fragments. In this case, since the two SOMOs are of different symmetries, they cannot overlap and therefore the Type I interaction leads to neither stabilisation nor destabilisation.

The second type of interaction (I leave the reader to guess what it was called) involves the interaction of the SOMO on one fragment with doubly occupied or empty orbitals on the other fragment. In this case, the SOMO of the allyl radical does not find any symmetry match on the methyl radical, and so there is again neither stabilisation nor destabilisation. However, the SOMO of the methyl radical can overlap with both symmetric MOs of the allyl radical; this is a stabilising interaction.

Overall, therefore, the π-type overlap between the two radical fragments leads to a favourable interaction. As such, triplet cyclobutadiene is considered to be aromatic.

For the cyclopentadienyl cation, one can construct an analogous diagram, starting from an ethene radical cation and an allyl radical.

Again the SOMOs do not match in symmetry, so the Type I interaction is necessarily zero. And again, the Type II interaction is stabilising; this time, both SOMOs find symmetry matches on the partner fragment. Based on this analysis, the triplet state of the cyclopentadienyl cation is indeed aromatic.

Triplet benzene

At this stage one may wonder: when two orbitals interact, the only possibility of destabilisation is if both orbitals are fully filled. Since both interactions involve orbital overlap between not-fully-filled orbitals, how can any of them be destabilising, and how can anything be anti-aromatic? To address this issue, and for the sake of completeness, let us analyse the case of benzene. We expect the triplet state of this 6π system to be anti-aromatic. The interaction diagram may be constructed from two allyl radicals:

First, we see that by symmetry, the Type II interaction is identically zero. Now, we turn to the Type I interaction. Surely the overlap of two SOMOs must be a bonding interaction? The astute reader will notice, though, that we are talking about a triplet state of benzene. The overlap of the two SOMOs still forms a bonding and an antibonding orbital, but if we were to put both electrons into the bonding orbital, we would obtain a singlet state.

As such, the Type I interaction leads to both the bonding and antibonding MOs having one electron each. Since the antibonding MO is more antibonding than the bonding MO is bonding, this is a net destabilisation. Hence, triplet benzene may be assigned anti-aromaticity.

Further reading

The concept of excited state aromaticity has been reviewed by Rosenberg et al.5, which is a good read (I found the explanation more accessible than that in Baird's original article2).

Notes and references

Costa, P.; Trosien, I.; Mieres-Perez, J.; Sander, W. Isolation of an Antiaromatic Singlet Cyclopentadienyl Zwitterion. J. Am. Chem. Soc. [Online early access]. DOI: 10.1021/jacs.7b05807.
Baird, N. C. Quantum organic photochemistry. II. Resonance and aromaticity in the lowest 3ππ* state of cyclic hydrocarbons. J. Am. Chem. Soc. 1972, 94 (14), 4941–4948. DOI: 10.1021/ja00769a025.
Baird, N. C. Dewar resonance energy. J. Chem. Educ. 1971, 48 (8), 509–513. DOI: 10.1021/ed048p509.
Technically, there is one hydrogen less on both fragments, but the C–H bonds do not contribute to the π framework so they can be ignored - after all, we never bothered taking the C–H bonds into account, not even in the Hückel approach. The point is that whether it is a methyl fragment or a methylene fragment, the π system remains the same (likewise for the allyl fragment).
Rosenberg, M.; Dahlstrand, C.; Kilså, K.; Ottosson, H. Excited State Aromaticity and Antiaromaticity: Opportunities for Photophysical and Photochemical Rationalizations. Chem. Rev. 2014, 114 (10), 5379–5425. DOI: 10.1021/cr300471v.


It's not exactly true that there are 5 "tastes". There are 5 generally accepted classes of gustatory transducers, each comprising many different types. The perception of taste is a product of not only the output of these transducers, but also information from the olfactory system, visual cues, textures, sound, etc.

The gustatory system itself is based (like the olfactory system) on overlapping sensors with partial selectivity. So a given compound, sucrose for instance, may stimulate several sweetness sensors (and probably even some from other classes) to varying degrees, and the particular combination of the responses is what gives sucrose its unique taste.

The types of molecules that stimulate the different sensors is not known clearly, but we can make some broad generalizations:

Sour - These sensors are thought to be primarily ion channels sensitive to pH—acidic solutions produce a sour taste.

Salty - Also thought to be mostly ion channels sensitive to small metal cations. Different ions have somewhat difference response. ($\ce{Na+}$ doesn't have the same response as $\ce{K+}$) There is some debate about the actual mechanism though.

The rest are g-protein coupled receptors. They change shape when they bind a molecule which causes a chemical signalling cascade.

Sweet - Not known precisely, but many compounds containing hydroxyl or carbonyl groups trigger these receptors.

Umami - Mostly triggered by amino acids, particularly glutamate. Inosinate and guanylate are known to enhance the effect of glutamate.

Bitter - This one is rather strange. A very wide variety of compounds stimulate these receptors, even with very different structural features. This taste is thought to have evolved to protect us against toxic compounds as many poisonous alkaloids and such taste bitter. Denatonium is an example of a bitter compound (one of the most bitter known).

As for whether any other classes of transducers exist, it's entirely possible. Some researchers have identified candidates for receptors for things like fats. Also, there some receptors like TRPV1 (which responds to capsaicin) are known that respond to compounds found in food, but aren't usually considered taste receptors.


Confusion in solution for this Question regarding mole concept

16 g of dioxygen contains 16 g of oxygen atoms. To find out how much carbon dioxide is produced, you can either use a realistic description of the reaction, or you can go atom by atom. Either path gives the same solution, but the one chosen by the textbook is not modern chemistry, and makes some of us angry or sad.

My question is why not to use the reaction $\ce{C + O2 -> CO2}$?

You should because that chemical equation describes the reactants and products much better than what the textbook has. It does not describe the mechanism (i.e. the steps to arrive at that net reaction), though.

What would the difference be in the answer.

Same numerical answer (unless you get confused because the reaction does not match the actual reactants).


Uncertainty in measuring the mass of the palm oil and the base in saponification

It would effect primarily the mass obtained (yield) and probably the quality, too, but that depends on how you measure this. Given that you use some procedure to purify your crude product that is equivalent in all cases independent on the mass of sodium hydroxide used, you simply will get less purified product.

Palm oil is a natural product and composed of a mixture of compounds. These compounds are know as fats, which are esters of propane-1,2,3-triol (glycerine) and fatty acids. Saponification is the hydrolysis of these esters. In that case sodium hydroxide is the strong base used, and the obtained soap is a mixture of glycerine, and the sodium salts of the fatty acids.

In principle for complete saponification you need one equivalent of base for each ester group.
$$\ce{RCOOR' + NaOH -> RCOONa + ROH}$$

Extending this scheme to palm oil, for the fat or triglyceride you need three equivalents of base:
$$
\ce{\underset{triglyceride}{H5C3-(OOCR^{1/2/3})3}
   + 3 \underset{base}{NaOH}
   -> \underset{glycerine}{H5C3(OH)3}
   + 3 \underset{salts of fatty acids}{R^{1/2/3}COONa}}
\tag{1}\label{eq:fat}$$

Since it is a natural product you probably have a number of environmental factors influencing the composition of your specific sample of palm oil. Wikipedia cites the following approximate composition:
\begin{array}{lr}
\text{Type of fatty acid}    & \text{Abundance}/\%\\\hline
\text{Myristic saturated C14}       &  1.0  \\
\text{Palmitic saturated C16}       & 43.5  \\
\text{Stearic saturated C18}        &  4.3  \\
\text{Oleic monounsaturated C18}    & 36.6  \\
\text{Linoleic polyunsaturated C18} &  9.1  \\\hline
\text{Other/Unknown}                &  5.5  \\\hline
\end{array}

As you can imagine this composition of the palm oil has an uncertainty. You probably want to look up more accurate values and the errors of the employed techniques. However, the mass of base needed can therefore be only a crude recommendation. In one case be too much, resulting in leftover (unreacted) base, while in another case it might be too little, resulting in only partially or not at all hydrolysed fat.

A crudely (over)nsimplified calculation:
Assuming palm oil is made out of 100% triglyceride of palmitic acid, we know the molecular mass is $M(\text{palm oil})\approx\pu{807 g/mol}$, we also know $m(\text{palm oil})=\pu{1 kg}$, $m(\ce{NaOH})=\pu{142 g}$.
\begin{align}
  M &= \frac{m}{n} &
  \Leftrightarrow&&
  n &= \frac{m}{M} &
  \implies&&
  n(\text{palm oil}) &= \frac{\pu{1kg}}{\pu{807 g/mol}} \approx \pu{1.24 mol}
\end{align}

From $\eqref{eq:fat}$ we know we need three equivalents of the base to produce three equivalents of soap. Ergo $n(\text{soap}) = n(\ce{NaOH}\text{ needed}) = \pu{3.72 mol}$. With $M(\ce{NaOH}) = \pu{40 g/mol}$ we know we need $m(\ce{NaOH},\text{ needed}) = \pu{149 g}$. We have $m(\ce{NaOH})=\pu{142 g}$, or $n(\ce{NaOH})=\pu{3.55 mol}$, ergo we only produce that much of soap. The molecular mass of our soap is $M(\text{soap})=\pu{278 g/mol}$. Therefore we will obtain $m(\text{soap}) = \pu{987 g}$.
I'll leave it to you to do the more precise calculations, since you are the mathematician of the both of us.


You miss the fact that the $p$ orbitals lying immediately above the occupied orbitals are also valence orbitals.  These unoccupied orbitals then allow the formation of unoccupied antibonding molecular orbitals whereas the corresponding bonding MOs are occupied.  Such a $p$ orbital contribution is then responsible for both the partially covalent bonding of these elements in compounds and the metallic bonding of the pure elements in their condensed phases.
In the case of metallic bonding in the elements, we see this effect in the alkaline earth metals as well as the Group 12 metals.  In the case of alkaline earth metals, if these were bonded only through the $s$ orbitals they would not be bonded at all; but the real metallic bond structure mixes in the otherwise unoccupied $p$ orbitals and this gives more net bonding (as measured by hardness and by melting and boiling points) than that in the alkali metals that precede them.


How pure NaCl is typical table salt? How is it purified?

Table salt is often iodized (iodine is an essential micronutrient, and this is one way to introduce it into the diet in the right amount). According to Wikipedia,

  Edible salt can be iodised by spraying it with a potassium iodate or potassium iodide solution. 60 ml of potassium iodate, costing about US $1.15 (in 2006), is required to iodise a ton of salt. Dextrose is added as a stabilizer to prevent potassium iodide from oxidizing and evaporating. Anti-caking agents such as calcium silicate are commonly added to table salt to prevent clumping.

Elsewhere, Wikipedia mentions other anti-caking agents, like sodium ferrocyanide, sodium aluminosilicate, and magnesium carbonate. And they point out that salt may also be "fortified" with iron or other micronutrients.

The certificate of analysis for "ultra pure NaCl" from American Bio shows TRACE amounts of aluminum, arsenic, bromides, heavy metals, iron, magnesium, phosphates, and sulfates. Those impurities may well be expected in table salt, too.

From all that, you'd think that table salt is fairly impure. Let's look at the numbers to put the amounts of these impurities in perspective. The box of Morton iodized salt in my kitchen gives me the amount of sodium per serving. If all of the sodium is in the sodium chloride, I estimate the purity as

$$\frac{0.590~\mathrm{g}~\ce{Na+}}{1.5~\mathrm{g}~\text{table salt}}\frac{1~\mathrm{mol}~  \ce{Na+}}{22.989769280~\mathrm{g}~\ce{Na+}}
\frac{1~\mathrm{mol}~\ce{NaCl}}{1~\mathrm{mol}~\ce{Na+}}\frac{58.44277~\mathrm{g}~\ce{NaCl}}{1~\mathrm{mol}~\ce{NaCl}}\cdot 100\% = \fbox{99.99% NaCl}$$

where I'm assuming the label information is actually a bit more precise than reported (exactly 1.5 g of table salt contains approximately 590 mg Na). This is just a BOE (Back of the Envelope) calculation; I should round my answer to three or possibly two significant figures.

In fact there are sometimes other sodium salts present (anti-caking estimates such as Tivolex contain sodium) so this purity is a slight overestimate. One salt company (WA Salt supply) lists "typical" purities for the table salt it provides at 99.72%; this salt contains uses Tivolex. Morton Salt claims that their iodized table salt contains 0.04% dextrose and "less than 0.5% calcium silicate", so the mg Na per serving on the label is a little bit higher than it actually is; I should get something above 99.46%, considering that additional information.

Still, that's pretty pure. The assay for "ultra pure" NaCl linked above gives 99% to 100.5% NaCl. Sigma-Aldrich has reagent-grade NaCl (>98%), ACS grade NaCl (>99.5%) and TraceSelect NaCl for doing heavy metal analysis that is more than 99.999% pure.

Is 99% purity "pure enough for high level experiments in chemistry?" There is no one answer. It depends on what those experiments are. How many significant digits do you need in your results? Are the results going to be affected by the presence of tiny amounts of iodide, iodate, glucose, calcium, or silicate, or the other impurities listed above? Maybe, maybe not. If a specific impurity interferes, find a way to lock up, remove or otherwise neutralize its effects.


What are some general tests that would determine an unknown compound?

As usual, it depends on what is available.

Mass spectrometry, $^{1}\ce{H}$ and $^{13}\ce{C}$ NMR, IR, UV/VIS spectroscopy, would be obvious choices in organic chemistry nowadays. XRD of a suitable crystal would be nice to have!

Back in the days, (undergrad) laboratories didn't have sophisticated analytical instruments and you were happy to find a melting point apparatus, a refractometer and an IR spectrometer.

You would have used IR spectroscopy to identify funcional groups, possibly supported your hypothesis by using staining reagents for TLC, like Dragendorff's reagent and then started to synthesize derivatives from which you knew that they show sharp, well-defined melting points.

You would have converted

amines to sulfonamides
carbohydrates to osazones
ketones and aldehydes to 2,4-dinitrophenylhydrazones or semicarbazones

and compared the melting points of the derivatives with tabulated values. Those were the days…


Confusion in unit cells of crystal system

First, a note: the classification of crystal systems, or the reduction of possible lattice types into primitive Bravais lattices, has nothing to do with “space available” or “hindrance”. It is purely a mathematical property having to do with the symmetry of the lattices.

To answer your question, the base-centered cubic lattice is not a Bravais lattice, because it is equivalent to a simple tetragonal lattice:


In general, we can describe a quasi aromatic compound as a compound, which is ionic in nature with a counter ion, and the $\pi$ electrons in such compounds follow Huckel's rule ($4n+2$).

In other words, quasi aromatic compounds are those in which the charges present on the molecule are a part of aromaticity of the compound. A few examples of such compounds are depicted in the diagram:

However, it has a deeper and broader meaning. For a more in-depth explanation, please read the given references.

References:

T. M. Krygowski, B. Bankiewicz, Z. Czarnocki, M. Palusiak, “Quasi-aromaticity—what does it mean?” Tetrahedron 2015, 71(30), 4895–4908 (https://doi.org/10.1016/j.tet.2015.05.074).
E. Kleinpeter, A. Koch, “Characterization and quantification of quasi-aromaticity by spatial magnetic properties (TSNMRS),” Tetrahedron 2015, 71(33), 5275–5284 (https://doi.org/10.1016/j.tet.2015.06.019).


In electrolysis, why does each atom wait to turn into gas until they reach a particular electrode?

$\ce{H2O}$ is not simply split apart by electricity, as you say. No ! What happens on one electrode is not related to what happens to the second electrode. Let's start by discussing what is happening on the negative electrode, the cathode.
The negative electrode behaves as if it contains plenty of electrons ready to react with anything able to do it. It could be positive ions. But in neutral solutions it may be water that does react with these electrons, according to :
$$\ce{2 H2O  +  2 e^- -> H2 + 2 OH-}$$ So, some bubbles of gaseous Hydrogen $\ce{H2}$ are produced on the cathode and the solution become basic because of the appearance of $\ce{OH-}$ ions in solution.
On the other electrode, the anode, which is the positive electrode, something must occur to produce electrons. Once again, it is water that manages to be decomposed in order to produce electrons. It is done this way : $$\ce{ 2 H2O -> 4 H+ + O2 + 4 e-}$$ Here some oxygen bubbles are produced on the anode. And the solution becomes acidic, if it was neutral in the beginning.
A last consequence of these chemical transformations is the fact that the solution becomes acidic at the anode and basic at the cathode. Of course, by simple mixing the $\ce{H+}$ and the $\ce{OH-}$ ions neutralize one another, producing water. So the only visible effect of the electrolysis is the production of $\ce{H2}$ at the cathode and $\ce{O2}$ at the anode, as if $\ce{H2O}$ has been split. Only "as if".


Flame test: Is the metal atom or the metal ion responsible for the flame colour?

The phenomenon that is witnessed during a flame test is an "atomic emission".
This statement may seem inappropriate, since it is a solution of metal ions (and not atoms) that is tested. The reason for calling it atomic emission lies in the process occurring in the flame. One of the steps of the process is an atomization step. That is, the flame converts the metal ions into atoms. When a solution of sodium chloride is placed in a flame, for example, the solvent evaporates, leaving behind solid crystalline sodium chloride. This evaporation is then followed by the dissociation of the sodium chloride crystals into individual ground state atoms - a process that is termed atomization. Thus sodium atoms are actually present in the flame at this point rather than sodium ions, and the process of light emission actually involves these atoms rather than the ions.


Why is OH group activating towards electrophilic aromatic substitution?

The oxygen in the $\ce{-OH}$ group in phenol has unpaired electrons that are capable of being donated to the ring. Although oxygen is electronegative and does exhibit some I- effects, the electron sharing that oxygen is capable of (shown by when drawing the resonance structures) has a stronger effect and results in $\ce{-OH}$ being a good ring activator. I've included the resonance structures below.


Nifedipine is a weak acid.

$pK_b = 10.07$
$pK_a = 3.93$
$pK_a = 4.3$


Understanding the bond dissociation energy of poly-atomic molecules

Whoa!!!

The C-H bonds in $\ce{CH4(g)}$ are not the same as the C-H bonds in $\ce{CH3(g)}$. The $\ce{CH3(g)}$ "molecule" is electrically neutral but it is a radical with an unpaired electron in one orbit. In fact the physical orientation of the molecule changes to a planar configuration.


What is the difference between Dielectric and Capacitive Polarizable Continuum Model?

Short answer: It matters most that you are using a modern solvation model than which one

Slightly longer answer: If you have the time, try a few solvation models, compare with experiment and see which gives you the best results. In my experience, modern models do fairly well for TDDFT calculations. Be careful about the atomic radii used and whether the computational program generates a reasonable solvation surface.

Also keep in mind that not every program supports both D-PCM and C-PCM. Many packages just implement one solvation model.

There are a number of excellent reviews on polarizable continuum solvation models. As you're probably aware, they are a really efficient way to handle molecular solvation in calculations, since the details of the solvent are largely ignored. Instead, the solvent is treated as a dielectric environment, and you handle polarization between the molecule and the solvent self-consistently. That is, the solute molecule is treated at normal atomistic detail, creates a solvation shell (usually treated by a set of overlapping spheres) and the electrostatics of the molecule influences the solvent, and vice-versa.

Now, I've never seen "C-PCM" as a "capacitive" PCM. Usually people refer to the "C" as a conductor-style model. That is, in a C-PCM model, the dielectric constant is set to $\infty$ but then you re-scale the charges on the molecule later.

There are more than just these techniques. The main problem is that it's very hard to bench-mark solvation energies.

Some great recent reviews:

"Thirty years of continuum solvation chemistry: a review, and prospects for the near future." Theoretical Chemistry Accounts (2004) v. 112(4) pp. 184-203
"Quantum Mechanical Continuum Solvation Models." Chem. Rev., 2005, 105(8), pp 2999–3094
"Perspective on Foundations of Solvation Modeling: The Electrostatic Contribution to the Free Energy of Solvation" J. Chem. Theory Comput., 2008, 4 (6), pp 877–887
"A Universal Approach to Solvation Modeling" Acc. Chem. Res., 2008, 41 (6), pp 760–768
"The COSMO and COSMO-RS solvation models" WIREs Comput Mol Sci, (2011) 1: 699–709
"Polarizable continuum models." WIREs Comput Mol Sci (2012) 2: 386–404.


Does Dean Kamen's Slingshot really filter virtually everything?

All it does is distill water while keeping heat within the system.  There isn't much detail provided about it but heres what I can say.

  I have a hard time to believe this as: -chemicals could evaporate before water and so get into the distilled water -I have no idea how he keeps the distillation chamber clean (prefiltering?) so it does not clog.

While you are right that chemicals can evaporate with the water this is not usually of concern if the water source is not severly contaminated with volatile chemicals (as opposed to biological and soluble toxins).  If you were to have chemical contamination you could use reverse osmosis or activated charcoal to remove residual contamination.  typically this is not an issue though.

  I know that distilled water cant be drunk as it is without adding salts. I like the project of Mr. Kamen very much, I just want to know the boundaries, as the promises ("Fill in anything that is wet") seem way to optimistic.

Distilled water can be drank.  It may not be the most tasteful water but it does taste better than cholera.  If you want more information his TEDMED Talk is very informative.


Correlation between functional group and viscosity of an organic compound?

There are empirical methods that can give you the viscosity from the structure. For lower temperatures, Orrick-Erbar equation, and for higher temperatures, Letsou-Stiel equation is what you are looking for. Orrick-Erbar maybe more suitable for you, because it is a group contribution method, and by looking at the values assigned to functional groups, you can develop an intuition about how each group affects viscosity. Letsou-Stiel uses acentricity factor, because at high temperatures, secondary bonds become less important.

As for database, NIST may be useful. Or if not, look at some supplier webpage (I use SigmaAldrich for this).


Why wouldn't you see blue or purple when you heat metal and only either red or yellow?

Because hot bodies emit a continuous range of light not a single wavelength

All warm objects emit a range of wavelengths that depend on how warm they are. The source of the light is a variety of mechanisms not all of which involve electronic transitions (some will involve transitions between molecular or solid state vibrations or rotations.) The important point for the perceived colour of the light they emit, though, is that the emitted radiation is continuous.

The chart below show the mix of light wavelengths from bodies at different temperatures:

The peak of the emission curve changes with the temperature. So the mix of light you see does change colour somewhat with the heat of the object but will never be a pure colour like blue or purple. Tungsten light bulbs, for example, reach temperatures of 2.5-3K °C and have peak emissions in the infra-red; their light has more red than blue in the visible so they are perceived as somewhat orange when compared to sunlight (which follows the curve for a body of 5-6k °C, the sun and has peak emissions in the middle of the visible spectrum).

The differences between different sources of black body radiation are harder to perceive than you might think because the eye and brain are good at adjusting what you see to a common perceived colour mix and it is only when comparing different light sources that the differences in the mix of light become easy to see (something also obvious when comparing other types of light bulb to tungsten bulbs).

So a hot object at ~7k °C will emit more blue light than red light but it will still be emitting a lot of red light so you will tend to see blue-tinged white rather than pure blue. Astronomers can, when their eyes are well adjusted to darkness, tell the subtle colour differences of different stars (which range in temperature from maybe 2k °C to maybe 20k °C). But the effect is small.

Of course, if you heat something enough you will turn it into a vapour or plasma and there will be enough energy to excite the outer electrons. The light resulting from transitions of outer electrons is often very narrow in wavelength and can be used to identify the atoms in the vapour. Low pressure Sodium street lights (the bright orange ones) are a result of this effect (the excited outer electrons sodium atoms emit yellow-orange light very characteristic of sodium). Many other elements have characteristic lines that can be just about any specific colour under similar conditions).

But normally when you heat things up you just see the continuous black-body emissions and that can't be a single colour.


Complex Numbers

As you point out, general a wavefunction if complex-valued.  A complex number and be described by two real numbers and hence are of depicted on the 2D complex plane (sometimes called the argand plane).

Numbers in the complex plane can be expressed in multiple forms.  The one usually taught first is as the sum of a real part and imaginary part:

$$\begin{align} z = x + iy\;{} & ; x,y \in \mathbb{R}\\
\Re(z) = x\; {} & ; \Im(z) = y \\
|z| = {} & \sqrt{x^2+y^2}
\end{align}$$

Alternatively it can be described by a magnitude ($r$) and a phase ($\phi$):

$$\begin{align}
z = r e^{i\phi}\; {} & ;r\in\mathbb{R}^{+0}, \phi\in[0:2\pi)\\
\Re(z) = r \cos(\phi) \; {} &; \Im(z) = r \sin(\phi) \\
|z|{} & = r
\end{align}$$

Two signals or functions with the same magnitude are said to be perfectly/exactly 'out of phase' if there phases differ by $\pi$ as $e^{i \pi}=-1, and perfectly 'in phase' if they have the same phase.

Orbitals

Now we can apply this to some orbitals.  Consider a cross-section of the 2p$_{\text{x}}$ orbital in the $x$-$y$ plane through $z=0$:

$$\psi_{2px}(x,y,z) = r(x,y,z) e^{i \phi(x,y,z)} \propto x \exp\left(-\sqrt{x^2+y^2+z^2}\right)$$

The intensity of the colour represents magnitude, $r(x,y,z)$, of the wavefunction and the colour itself represents the phase - red for $\phi=0$, blue for $\phi=\pi$, with a colourless node separating the two lobes.

As the probability density is equal to the square modulus of the wavefunction: $$\rho(x,y,z) = |\psi(x,y,z)|^2$$
we can multiply the wavefunction by a constant phase factor $e^{i\theta}$ and still represent the same particle:

Consider:
$$\begin{align}
\tilde{\psi} {} & = \psi e^{i\theta} \\
\to \tilde{\rho} {} & = |\tilde{\psi}|^2 \\
& = |\psi|^2 e^{i\theta} e^{-i\theta} \\
& = |\psi|^2 \\
& = \rho
\end{align}$$
So both $\psi$ and $\tilde{\psi}$ are valid wavefunctions with either one is, and the wavefunction can only be defined up to multiplicative phase factor, so we can talk about absolute phase, but we can talk about relative phase and phase differences.

Consider the out of phase 2p$_{\text{x}}$ orbital:
$$\tilde{\psi}_{2px}(x,y,z) = \psi_{2px}(x,y,z)e^{i\pi} = - \psi_{2px}(x,y,z) $$

Molecular Orbitals

As we just showed they both have the same probability density, but when we begin to combine them, the phases dictate how much the wavefunctions constructively/destructively interfere.  We can combine linear combinations of atomic orbitals to make molecular orbitals.

Consider adding together two adjacent wavefunctions with the same phase on different atoms:

The adjacent regions of different phase cancel one another out.  When the probability density is examined:

we can see than the chance of the particle being between the two atoms is very low and it is localised around each nuclei.  This would be an anti-bonding molecular orbital.

On the other hand, adding together two adjacent wavefunctions with the opposite phase on different atoms:

The adjacent regions of similar phase add together.  When the probability density is examined:

we can see than the chance of the particle being between the two atoms is very much higher and it is localised in a bond.  This would be a bonding molecular orbital.

This is why we talk about in phase and out of phase orbital when talking about bonding - the regions of in phase overlap of wavefunctions increases the probability that a particle will be found in there, and out of phase overlap decreases the probability.

The physical result is the wide variety of bonding phenomena we see in nature

covalent bonds: local in phase overlap
anti-bonding orbitals: local out of phase overlap
conjugation: extended in phase overlap etc...


Where to find data for Gibbs energy, enthalpy, and entropy?

NIST webbook does have a lot of data, though they are not in any kind of an API form as far as I know.

http://webbook.nist.gov/chemistry/


Enthalpy of formation of ammonium bicarbonate

−2580 cal/g.  Here is a reference that lists heats of formation:

https://engineering.purdue.edu/~propulsi/propulsion/comb/propellants.html


How is the acidic dissociation of acetic acid thermodynamically feasible?

To expand on my comment above, I will reiterate that all chemical reactions are equilibria and this is due to the interplay of enthalpy and entropy. More specifically, it is due to enthalpy and entropy of the individual compounds of the reaction as well as an additional entropy component due to mixing.

Consider the general reaction $$\ce{A<=>B}$$ and let $\chi_\ce{i}$ denote the mole fraction of compound $\ce{i}$ in the solution or gas mixture. The enthalpy and the entropy of the compounds, combined together to form $\Delta G_{compounds}=\Delta H_{compounds}-T\Delta S_{compounds}$ or alternatively $$G=(\chi_\ce{B})\cdot H_{\ce{B}}+(1-\chi_\ce{B})\cdot H_{\ce{A}}-(\chi_\ce{B})\cdot T\Delta S_{\ce{B}}-(1-\chi_\ce{B})\cdot T\Delta S_{\ce{A}}$$ has a particular side of the reaction it favors. This component of the free energy leads to a complete preference; if this were the only factor, the solution would be all $\ce{A}$$(\chi_\ce{A}=1)$ or all $\ce{B}(\chi_\ce{B}=1)$. If we were to plot $G$ as a function of $\chi_\ce{B}$, it would either be linear increasing (reactants favored), linear decreasing (products favored), or flat (reactants and products equally favored).

However, there is an additional entropy component due to mixing, which leads to an additional free energy component. We can write the entropy of mixing as $$\Delta S_{mixing}=-\mathrm{R}[\chi_\ce{A}\ln(\chi_\ce{A})+\chi_\ce{B}\ln(\chi_\ce{B})]=-\mathrm{R}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$
where we have used the fact that $\chi_\ce{B}+\chi_\ce{A}=1$. We can then write the free energy of mixing as $$\Delta G_{mixing}=\mathrm{RT}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ We can see graphically what these two mixing equations would look like

Combining all this together, we can write $$G_{total}=(\chi_\ce{B})\cdot H_{\ce{B}}+(1-\chi_\ce{B})\cdot H_{\ce{A}}-(\chi_\ce{B})\cdot T\Delta S_{\ce{B}}-(1-\chi_\ce{B})\cdot T\Delta S_{\ce{A}}+\mathrm{RT}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ We can take the derivative of this function with respect to $\chi_\ce{B}$ and set it equal to zero to determine where the miniumum free energy is $$\frac{\delta G_{total}}{\delta\chi_\ce{B}}=0= H_{\ce{B}} - H_{\ce{A}}- T\Delta S_{\ce{B}}- T\Delta S_{\ce{A}}+\mathrm{RT}[-\ln(1-\chi_\ce{B})+\ln(\chi_\ce{B})]$$ $$H_{\ce{B}} - H_{\ce{A}}- T\Delta S_{\ce{B}}- T\Delta S_{\ce{A}}=-\mathrm{RT}[-\ln(1-\chi_\ce{B})+\ln(\chi_\ce{B})]$$

This second equation is the crux of all this. While we cannot solve this for an exact value without knowing the enthalpy and entropy of the compounds, we can definitively say that $\chi_\ce{B}\ne1$ and $\chi_\ce{B}\ne0$. Why? Plugging either of the values into the equation will result in the right side being $\infty$, meaning the differences in enthalpy and entropy of the components would also have to be $\infty$ for the equation to be satisfied. Thus, the minimum free energy will never be attained by letting the reaction go to completion; it will always be more favorable to allow at least some mixing even if one side of the reaction has a much lower enthalpy and higher entropy of its compounds.

Thus all reactions are technically equilibria. I say technically because the mixing free energy might be negligible compared to the overall free energy and so the reaction could appear to have gone to completion to the extent that we can observe experimentally.

So for your reaction, the reason it doesn't all remain as $\ce{CH3COOH}$ and water is that the entropy of mixing makes it favorable for some of the $\ce{CH3COOH}$ to dissociate.

(For a brief version of this argument and where I got the graph, see http://rkt.chem.ox.ac.uk/tutorials/equilibrium/entropy_mixGas.html)


How to take water-dispersed PEDOT:PSS and disperse it in an organic solvent?

Even if you switch to an organic solvent, if the polypyrrole is in water, it will perturb the PEDOT:PSS layer.

Your best bet would be to use a poly-alkyl pyrrole or other conjugated polymer that's soluble in something that doesn't dissolve PEDOT:PSS.

Let's say you switch the solvent to DMF or DMSO and it dissolves the PEDOT:PSS. Then you spin-coat that new mixture onto ITO. Once you drop the aqueous solution of polypyrrole, it will dissolve the PEDOT:PSS.

I think you're looking at this the wrong way. Rather than trying to change the PEDOT:PSS (which will be tricky - PSS is anionic) you should consider changing your polymer. Alternatively, get rid of PEDOT:PSS and use another hole injection layer. (PEDOT:PSS is acidic and will etch the indium out of ITO.)


How to determine oxidising and reducing agents in the reaction between ammonia and hypochlorous acid?

You're on the right track with assigning oxidation numbers to each element. And you've assigned them successfully.

To successfully complete the task you need to define a few terms:

What's oxidation?

What's reduction?

What's the oxidizing agent? Try thinking of this in terms of what oxidation is, and then try figuring out what an oxidizing agent would do.

What's the reducing agent?


Is there any possibility of tautomerisation to en-ol form?

If you're asking specifically about the tautomerisation, then I made this comparative computational study using B3LYP theory at the 6-31g* level, and the results are:

So as is evident, the bridge causes the alkene to be dramatically strained. You can compute the Boltzmann ratio and find that the tautomerisation in the first structure is completely negligible at room temperature.

If you're talking about the formation of an enolate ion, the results shouldn't differ much, because the alpha carbon still needs to become sp2 hybridised, in order to delocalise the carbanion into the carbonyl group. And this, as we know, is the reason for the increase in acidity of protons adjacent to carbonyl groups.

All this is of course to the best of my knowledge.


To Debunk the 9/11 conspiracy, what is the science behind melting of the towers architecture?

Jet Fuel type A (Jet A) burns at approximately 2,230 °C. 0.2% carbon steel has a maximum forging temperature (i.e., the point at which it'll melt) at about 2410 °C, while 1.5% carbon steel is at about 1920 °C. This also doesn't take into account effects that happen with how aerosolized jet fuel will burn, which is, much hotter, and strange effects that'll happen as everything else has it's surface area very spontaneously increased, subjected to a large amount of kinetic energy at once, and then lit on fire.

Not to mention everything else burning at the same time. Finely shredded aluminum is very flammable.

Also: As a final note: The steel didn't actually have to melt, persay. It only had to weaken enough that the mechanical forces of the skyscraper that it was supporting became too great for the softened steel, at which point it'll all collapse in a chain reaction.

TL;DR: Conspiracy theories are BS.


How many signals would be seen in total in the 13C NMR spectrum

3 different methyl signals: (1, 16) (40, 41) (27, 28, 29, 33, 34, 35)

4 different methylene signals: (17) (2, 15) (3,14) (37, 39)

1 aliphatic CH signal: (36)

2 different aliphatic quaternary C signals: (26, 32) (38)

4 different triple bond C signals: (24, 30) (25, 31) (4, 13) (5, 12)

8 different aromatic carbons: (7) (10) (6, 8) (9, 11) (18) (21) (19, 23) (20, 22)

22 different signals

Just looked at a model, methyls 40 and 41 are different; one is on the same side of the 6-membered ring as the benzene ring, the other is on the same side of the 6-membered ring as the hydrogen at C-36

23 different signals!


Is an ionic bond more like a covalent bond or an intermolecular force?

An ionic bond could maybe be described as an inter-ionic force. All electron interactions are most accurately described by wavefunctions and quantum mechanics, but in practice we use successively more detailed approximations for convenience, stopping at the lowest level of detail that suits our needs at the time.

At the lowest-detail end of the spectrum, you have ionic bonds - we usually model these as simple Coulombic (a.k.a. charge-charge or electrostatic) interactions. This is good enough to describe bonding in most cases where the bond is between a metal and a non-metal. We model the bond as a complete electron transfer from one atom to another, which results in an electrostatic charge. The Coulomb force between the charges holds the ion together, and so we call it a molecule. What we really mean is "these atoms are now stuck together."

As we move along the periodic table and the metal atoms become less metallic and the non-metals become more metallic, at some rather arbitrary point we say "the ionic approximation is no longer good enough" and start seeing the bond as covalent. What we really mean is that the electrons are not fully transferred between atoms, and so charge-charge interactions don't explain bonding well enough. At that point we can no longer ignore quantum mechanics, and have to acknowledge that the electrons exist in molecular orbitals, not atomic orbitals.

That is the main difference between ionic and covalent bonds. As for dipole-dipole interactions, it again goes back to what the electrons are doing. In a molecule with a dipole moment, what this means is that within the molecule there are polar covalent bonds, and the geometry is such that the electron density on the surface of the molecule is lopsided - one side has more electrons than the other. The dipoles can align and will be attracted to each other via the Coulomb force, but this is not the same as how we model an ionic bond - in that case, one atom gives up one or more electrons completely. In a molecule with a dipole moment, one (or more) atoms just gets "more" of the electron density. Then the two separate molecules, each with their own dipole, are attracted to each other.

We call ionic bonds "intramolecular" forces because they are what hold molecules together - without ionic bonds, you couldn't have a salt crystal, for example. When you break apart sodium chloride, you get sodium ions and chloride ions.

We call dipole interactions "intermolecular" because they are what make separate molecules stick together. When you pull apart two water molecules, you still have two water molecules.

Thinking about it this way, an easy way to distinguish between intermolecular and intramolecular forces is to ask yourself whether you still have the same substance after pulling things apart. If you do, it must have been an intermolecular force. If you don't, it must have been a bond or an intramolecular force.

note - For very large molecules like proteins it gets a little tricky - you can have forces between sites on the same molecule that act like what we normally describe as intermolecular forces.


Cyclisation of acetone on reaction with conc. H2SO4

You may have noticed that your question received a downvote. This was probably because you did not show any effort toward a solution of the question. I posted in a Comment some clues to assist you in a solution. In the interim an Answer was posted that showed the compounds formed in the reaction of acetone with H2SO4 but it did not illustrate a mechanism for the formation of mesitylene, which was your question. Here is my contribution to your understanding.
The reaction between the keto form of acetone 1a and its enol 1b forms aldol 2. Acid-catalyzed dehydration of 2 via the enol 3 leads to mesityl oxide 4. The loss of water from 3 may be stepwise but, to save space, I have presented the loss of water in a single operation. The third unit of acetone is incorporated via the vinylogous enol 4b to form the dienone 5a. Enolization of 5a forms 5b' that allows rotation about the red single bond in 5a and 5b' to form conformation 5b". This species may cyclize by an allowed thermal electrocyclization to form cyclohexadiene 8 which is readily dehydrated by acid to mesitylene 9. Alternatively, enol 5b' may reketonize to 6 and undergo cyclization to cation 7 and thence on to mesitylene 9. continued...

Addendum: An old name for acetone was mesite, the source of mesitylene. Read about it here and here.


The "amount of molecules in moles", such as 7 moles, provides no information concerning the mass of an individual molecule.

This is like asking, "if I have the amount of eggs in dozens, how do I calculate the mass of an egg?"


In a Koppers-Totzek reactor, why do I get CO and H₂ instead of CO₂ and H₂O?

The reaction occurs in two stages: in first $\ce{C}$ oxidized to $\ce{CO2}$ and then $\ce{CO2}$ is reduced to $\ce{CO}$.

The process of having carbon react with oxygen and water vapor is called steam-oxygen conversion. It is used, because in reality, due to the high reaction temperatures, the allothermic principle is hard to provide.

$$\ce{C + O2->CO2}$$
$$\ce{C + 2H2O->CO2 + 2H2}$$
$$\ce{C + H2O->CO2 + H2}$$
$$\ce{C + CO2->2CO}$$

The equilibrium concentrations of $\ce{O2}$ and $\ce{H2O}$ in the gasification products at all temperatures are negligible. Steam-oxygen gasification products are a mixture of $\ce{CO2}$, $\ce{CO}$ and $\ce{H2}$.

Before carrying out the reaction of the shift
$$\ce{CO + H2O->CO2 + H2}$$ from a mixture of $\ce{CO2}$, $\ce{CO}$ and $\ce{H2}$ must be separated $\ce{CO2}$. To this must be expended work division (again burn some $\ce{C}$).


Why does Carnot Engine have the curves as isothermal and adiabatic?

By definition, a Carnot cycle is comprised of 2 reversible isotherms and 2 reversible adiabats.

Regarding your question about isochoric heating or cooling, if you are going to carry this out reversibly, then you can't do it with just one constant temperature reservoir.  For the heating or cooling to be reversible, the temperature difference between the system and the reservoir(s) must be tiny over the entire path.  To do this, you need a sequence of constant temperature reservoirs running from the starting temperature to the ending temperature.  After the system has equilibrated with each reservoir in the sequence, you move on to the next reservoir.  That way, the temperature difference between the system and the reservoir is tiny along each step of the process path.


How to measure the amount of calcium in a supplement?

For the determination of calcium (at least) two methods are conceivable:

gravimetric analysis
Sulfate is a rather bad choice here. As compared to the carbonate, the sulfate has a much higher solubility. A better choice would be the oxalate with a solubility around $6\,\mathrm{mg\,l^{-1}}$.
photometry
Calcium forms a purple complex with the Arsenazo III dye that can be detected at wavelengths of 660 and 700 nm.
Alternatively, you can use o-cresolphthalein and detect the complex at 545 nm.
The photometric methods are typically used in automated high-throughput clinical analysis of calcium in serum, but given that a photometer isn't available in the students' lab, Karl's comment on EDTA is exactly what you might be looking for. Complexometric titration is the keyword you might want to look for. Here, an indicator that forms a coloured complex with $\ce{Ca}$ is replaced by EDTA. The liberated indicator then shows a different colour.


So what I said in the comment still holds: it could work (particularly in combination with a Dean-Stark apparatus to remove excess water), but there is no way to know for sure unless someone tries it out.

The more usual (and early) method is described in Organic Syntheses and utilizes the reaction of methanol with iodine in the presence of phosphorous

I don't know the exact reaction mechanism (if anyone has a clue, please feel free to edit!) but what happens is that $\ce{PI_3}$ is generated in situ from the elemental iodine and phosphorous and reacts with methanol to yield the methyl iodide and phosphorous acid

$\ce{3 CH3OH + PI3 -> 3 CH3I + H3PO3}$

Phosphorous acid does not act as nucleophile and so the problem of a competing backward reaction, which you would have in your proposed synthesis, is largely prevented.

I'd also like to add some caution: the reaction described in Organic Syntheses is not suited for beginners or in amateur setting. The chemicals involved are nasty: highly toxic, highly flammable, alkylating and pyrophoric. Don't try this at home :)


Enantiomerically pure ketone reaction under acidic conditions gives 1:1 mixture of diastereoisomers?

Carbonyls are able to undergo keto-enol tautomerisation, this is accelerated in the presence of acid/base.

This phenomenon means that alpha-chiral carbonyls are generally regarded as configurationally unstable, as they are prone to racemisation, although in the absence of acid/base the extent to which the alpha-chiral carbonyl exists in its enol form is often negligible (the C=O, C-H combination is ever so slightly more favoured than C=C, O-H in a lot of cases).

The enol form of the carbonyl (the middle structure in my diagram below) is planar. When the enol tautomerises back to the ketone form, the proton may be transferred to either face of the planar structure, leading to both diastereomers being formed, as you describe.


What does it mean if I am unable to write RS configuration for a compound?

I think I get what your question is and will try explain the things to you.

  I was unable to write RS configurations for 1 and 3, since they have carbon atoms with more than one same groups attached to them (they are achiral). Does that mean they are optically inactive?

The crux of your question lies apparently in two problems:

you assume that molecules 1 and 3 are optically inactive, but are unable to visualize this with a center/plane of symmetry
you seem to assume that molecules without chiral carbons are optically inactive

So, let's tackle this:

  Correct me if I am wrong, these compounds don’t seem to have planes or centres of symmetry.

Well, you're wrong. Note that in simple saturated hydrocarbons (with or without halogen substituents), it is usually safe to assume that if no chiral carbons are present, then the molecule will be optically inactive (without the need to validate this assumption by manually drawing out their planes/centers of symmetry). Here are two examples:

 (source)

notice how the plane of symmetry cuts-through through the blue and green atoms, using the remaining two grey identical atoms to obtain a plane of symmetry, to prove the molecule is optically inactive. This was expected, since the molecule had no chiral carbon.

it is possible to visualize the plane of symmetry in 2-chloro-2-methylbutane from its eclipsed Newmann projection, as seen above. The plane of symmetry cuts-through the chlorine atom and the methyl radical in front of it, rendering the molecule optically inactive, as expected.

  Are all molecules without any chiral carbon optically inactive?

No, this is wrong. Common examples like ortho-disubstituted-biphenyls (tl;dr: they are propeller shaped rings), allenes (tl;dr: their terminal groups are in perpendicular planes), etc. have no chiral carbons, yet are optically active. I won't detail these as you're still a "beginner" in this, but you'll definitely come across these examples in the very end of your course on optical activity.

In regards to your edited question, it is unreasonable to draw Fischer projection for achiral carbon atoms. For example, when drawing the Fischer projection of carbohydates, the terminal achiral carbon is written as a $\ce{-CH2OH}$ substituent to the penultimate chiral carbon, and not expanded out into its own Fisher projection.

I hope it helps!


Why aqueous KI is used when collecting Iodine?

Your guess is right. Pure water does not dissolve iodine very well, but a solution of $\ce{KI}$ does, thanks to the formation of $\ce{I3-}$ and other polyiodide anions. And of course we would prefer to have our iodine in a dissolved form when we are going to titrate it.

When it comes to the starch reaction or redox reaction with thiosulfate, $\ce{I3-}$ is as good as $\ce{I2}$.

So it goes.


What are the criteria for a molecule to be chiral?

Mirror images
The very first definition of a chiral molecule is one where it is not superimposable on its mirror image. Therefore, one of the most straightforward ways to determine chirality is to construct the molecule and its mirror image (perhaps via a model), and then to see whether they are superimposable.

Symmetry elements
A fully equivalent definition, rooted in molecular symmetry, is that a species is chiral if it does not have any improper axes of rotation.
An improper rotation is a rotation of the species (say 90° clockwise in the plane of the page), followed by a reflection perpendicular to the axis of rotation (in this case, the page). The improper rotation is generally denoted by $S_{n}$ where $n$ is the number of times you need to rotate to get back to the start, e.g., $n=4$ for a quarter rotation (90°).
So, molecules where you can rotate and reflect them to make them look like the original are not chiral. In other words, it's easier to say when a molecule is achiral. Let's look at some examples of improper rotation that make molecules achiral.
The simplest case to look at is $S_{1}$, but no one calls it that. This is a trivial rotation (a rotation that does nothing) followed by a reflection, or just a reflection, usually denoted by σ. Molecules with planes of symmetry are achiral. This molecule, (2R,4S)-2,4-dichloropentane, is achiral because there is a plane of symmetry that reflects the left side onto the right side.

Another example is $S_{2}$, rotation by 180° then reflection. Generally, we also don't call it that. This is the same as an inversion through the center, which is frequently denoted $i$. Molecules with centers of symmetry are achiral. The following molecule has a center of symmetry. You can also see that rotating it 180° and then reflecting through the page gives the same thing back. For clarity, the mirror image is shown on the right. You can rotate it to see that it's the same.

The above two are the most common symmetry elements that make molecules achiral. Occasionally, you'll run into a higher-order improper rotation axis, like $S_{4}$ in the following example. Rotation by 90° and then reflection through the page gives back the original molecule. The mirror image is shown on the right, and you can rotate it to see that it's the same, thus making this molecule achiral. There are more examples here.

So, looking for the above features will let you know if something is achiral.

Stereogenic elements
To figure out whether or not you even need to consider chirality, you will to find a stereogenic element. Keep in mind that having a stereogenic element doesn't mean that molecule is chiral, because it could have one of the symmetries discussed above and thus, be achiral.
Molecules with stereogenic elements should be considered for chirality, but are not guaranteed to be chiral. In other words, having a stereogenic element is necessary but not sufficient to make a compound chiral.
The main source of chirality that introductory students should worry about is the stereogenic center, usually a carbon with four different substituents:

More generally speaking, a stereogenic center such as this falls into the category of point chirality. Other examples of point chirality include inorganic complexes such as the octahedral $\ce{[Co(en)3]^3+}$ (link, also illustrated below), as well as substituted adamantanes.

Apart from point chirality, here are also two other ways to generate chirality, which are typically seen in more advanced courses.
The first is axial chirality (Wikipedia, IUPAC). This is a twist that can be described as either left-handed or right-handed. These include unsymmetrically substituted allenes, unsymmetrically substituted biphenyls, rings with exocyclic double bonds, and helicenes (image adapted from Wikimedia):

The second is planar chirality (Wikipedia, IUPAC). This is when a plane, which has high symmetry, has its planes of symmetry successively broken by substitution. An example is this borabenzene metal complex. A borabenzene has two planes of symmetry, but these can broken by putting a methyl group on the ring and then binding a metal to the face of the ring.


Why do the first and second ionizations of a diprotic acid produce the same hydronium ion concentrations in solution?

This is saying both the $\ce{H2CO3}$ and the $\ce{HCO3-}$ are both donating $\ce{H+}$ to the water to form $\ce{H3O+}$. What it means when it says "The $\ce{[H3O+]}$ in the two ionization equilibria are one in the same" is to say that both of these reactions are occurring simultaneously in the same system, and that there is only one measure of the concentration of $\ce{H3O+}$ for the system.

It is not saying that both species are donating $\ce{H+}$ to the same extent. As a matter of fact, the $\ce{H+}$ being donated by $\ce{H2CO3}$ will keep the the $\ce{HCO3- + H2O -> CO3^{2-} +H3O+}$ shifted almost entirely to the left, meaning $\ce{HCO3-}$ will be donating an extremely small fraction of the $\ce{H+}$.


Is there a non-violent, practical way of making hydrochloride with only Hydrogen and Chlorine?

The best and practical way is to sell hydrogen and chlorine, and buy yourself some $\ce{HCl}$ (unless it is prohibited in your jurisdiction). However, I would recommend that only if you have some experience in chemistry and understand quite well the dangers thereof.

The technical answer to your question is: build a galvanic cell with platinum anode and graphite cathode, and feed them with hydrogen and chlorine, respectively. It will perform your reaction in a pretty controllable and non-violent way, and also will give you some electricity for free. Not that I expect you or anyone else to be actually doing this, though.

Why would you need gaseous $\ce{HCl}$, really?


Entropy is a measure of unavailable energy?

When the first law was first formulated it was assumed that the most efficient utilisation of a chemical reaction to produce work was to use all the heat  to produce work. The second law was not fully understood and thus at maximum efficiency $-\Delta H$ was expected to represent the maximum amount of work. Many experiments failed to confirm this. We now know that it is the free energy which measures the maximum capacity to do work, $\Delta G=\Delta H-T\Delta S$. Clearly $\Delta G$ and $\Delta H$ are equal only if there is no entropy change between the beginning and end of an isothermal reaction. If this is not the case the work obtainable in a reversible process can be either greater than or less than the heat of reaction.

From the first law the external work performed must be equal to the loss in energy of the system, unless some heat is taken from or given to the surroundings. This is exactly the point first clearly seen by Gibbs. In a reversible isothermal reaction $T\Delta S$ is the heat absorbed from the surroundings and if this is positive the work done will be greater than the heat of reaction.

So 'available/unavailable energy' seems to depend on the particular case. Hopefully some thermo experts will give you a more detailed argument.

edit: It occurs to me that perhaps the textbook was referring to the fact that one can 'convert all the work into heat but not all the heat into work'. In other words not all the random thermal motions of molecules can be made to do work.