Oxidizing agent, reducing agent, acid or base

The use of the terms "oxidizing agent" and "reducing agent" in this example apparently refer to the overall reaction. Ammonia certainly can act as an acid in the process of reacting with sodium metal, and as a (Lewis) base during the reaction with chlorine. However, the net effect of the reaction is a reduction or oxidation, rather than a simple proton transfer.


I think you're looking down the wrong path to explain this. Boiling point can often be explained by looking at the interactions between molecules such as hydrogen bonding, dipole-dipole interactions, and london forces, as well as overall molecular mass. I would explain this phenomenon by looking at two data points for each molecule. The dipole of $\ce{SCl2}$ is 0.4 Debye and the dipole of $\ce{PCl3}$ is approximately 0.780 Debye.$\ce{^{1,2}}$ Based on  this alone, I would expect $\ce{PCl3}$ to have a higher boiling point. If you also look at the mass of each molecule, 102.97 g/mol for $\ce{SCl2}$ and 137.33 g/mol $\ce{PCl3}$, it becomes pretty clear which substance one should predict to have the higher boiling point. It's $\ce{PCl3}$.

Ref 1. http://openmopac.net/PM7_accuracy/data_molecules/sulfur%20dichloride_jmol.html

Ref 2. https://cccbdb.nist.gov/diplistx.asp


Electrophilic substitution on disubstituted benzene rings

Electrophilic Substitution of Disubstituted Benzene Rings

When a benzene ring has two substituents, each substituent exerts an effect on subsequent substitution reactions.

  Electrophilic substitution of a disubstituted benzene ring is governed
  by the same resonance and inductive effects that affect
  monosubstituted rings. The only difference is that it’s necessary to
  consider the additive effects of two different groups.

If the directing effects of the two groups reinforce each other, the situation is straightforward. In p-nitrotoluene, for example, both the methyl and the nitro group direct further substitution to the same position (ortho to the methyl and meta to the nitro). A single product is thus formed on electrophilic substitution.

From above, the methyl and the nitro group direct further substitution to the same position (ortho to the methyl and meta to the nitro). This is similar case with chlorine.

If the directing effects of the two groups oppose each other, the more powerful activating group has the dominant influence, but mixtures of products are often formed.
Further substitution rarely occurs between the two groups in a meta-disubstituted compound because this site is too hindered. Aromatic rings with three adjacent substituents must therefore be prepared by some other route, such as by substitution of an ortho-disubstituted compound.

You may also wish to review this table showing positions where common substituents direct:

Basing on the above explanations,  it is easy to predict the products of  electrophilic substitution of disubstituted benzene ring, the only thing you have to know is the effect of each substituent whether it is activating or deactivating. In some cases, observe the positions of the substituents as point 3 mentions steric effects may come into play and substitution may not occur.

On the second note, concerning diazonium ion, I think you have to look at measured dipole moments, experimental data etc I doubt if there is a way of telling whether a substituent is strongly deactivating by merely looking at it

After explaining concepts now lets see what the answer for the question in comments will look like:

Reasons

the directing effects are opposing (nitro is meta while methyl is both ortho and para directing) but the methyl is more activating so we have a mixture:

meta to nitro and methyl
ortho to the methyl but para to the nitro

In this scenario since position 3 is not deactivated by nitro group hence there is also a possibility of a mixture here, also take note of how the nitro group is strongly deactivating at position 4 (more additive effect)

Hope this helps

References

McMurry Organic chemistry
Electrophilic Substitution of Disubstituted Benzene Rings


If ions are floating around in a solution, how do we know they combine to form a compound in a solution?

User EdV comments above get to the heart of the matter. To summarize:

Your first equation is typically how a chemist would write the reaction.

$$\ce{3Pb(NO3)2 (aq) + 2(NH4)3PO4(aq) -> Pb3(PO4)2(s) + 6NH4NO3(aq)}$$

The equation is a "shorthand" explanation. What a chemist reads is:

On the left hand side is that that two solutions are mixed together.

Since $\ce{Pb(NO3)2}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{Pb^{2+}}$ and $\ce{NO3-}$ ions.
Since $\ce{(NH4)3PO4}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{NH4+}$ and $\ce{PO4^{3-}}$ ions.
Since the RHS shows a reaction the two salts can't be in one solution simultaneously. Therefore the salts must have been dissolved in separate solutions and the solutions were mixed.

On the right hand side $\ce{Pb3(PO4)2}$ precipitates and $\ce{6NH4NO3}$ remains in solution.

Also since $\ce{Pb3(PO4)2}$ has a tiny but finite solubility, a very small amount of $\ce{Pb3(PO4)2}$ will remain in solution.
Since $\ce{NH4NO3}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{NH4+}$ and $\ce{NO3-}$ ions.

The "ionic equation" is harder to understand because of its verbosity. Did the $\ce{NO3-}$ ions come into the solution with the $\ce{Pb^{2+}}$ ions or the $\ce{NH4+}$ ions? Being chemists we can puzzle it out, but the equation is less succinct.
$$\ce{3Pb2+(aq) + 6NO3-(aq) +6NH4+(aq) + 2PO4^3-(aq) -> Pb3(PO3)2(s) + 6NH4+(aq) + 6NO3-(aq)}$$


Based on the color alone, the corrosion could be mercury (II) oxide, which is nasty. As with other mercury salts, it can end up in your body in a variety of new and exciting ways, such as by inhalation (that is not to say that elemental mercury cannot do so, it is simply more difficult), which make it more dangerous than elemental mercury. For this reason, do not touch it, even if it ends up being another salt.

What I (and I am no hazmat professional) would do is search around on the Internet for my local hazardous materials disposal facility (it may go under a different name) and would call them. I would ask them about the battery whilst giving as much information as possible about it, even if it seems unrelated. If (and only if) they ask you to bring it in, do the following:

Wear at least two pairs of gloves while handling it, and only do so briefly.
Put it in several layers of baggage, something that you will be okay with losing; and make sure it will not break.
Bring it in carefully, being sure not to breach the membrane. Also, be sure not to touch anything with the gloves used.

In addition, be careful with any of the tools that it touched, they could also be highly contaminated. If you want to learn more about mercury (II) oxide, the Wikipedia article seems relatively accurate (the same could be said about mercury batteries): https://en.wikipedia.org/wiki/Mercury(II)_oxide

Disclaimer: As stated, I am no hazmat professional, and my advice does not take authority over the advice of a professional.


Perplexing claim in Chang and Overby's "General Chemistry"

The  other answer is, I think excellent.

Your question is basically "but what if there isn't a direct proportion between number of gas molecules and volume?" Instead, that was Avogadro's hypothesis in a nutshell: the volume of the gas is determined exactly proportionate to the number of molecules in it.

You are right in principal that you could have such a "false law" where the number of molecules is inversely proportionate to the volume and still have equal number of molecules for $\ce{H2}$ and $\ce{N2}$.

Let's follow your course of logic. That would mean if I get 10 L of $\ce{H2}$, I would then have $k/10$ molecules. Let's imagine I add more and more volume, to get 100 L then 1000 L of gas. At some point, I must necessarily divide the molecules in fractions in your false law. Certainly a molecule must occupy some indivisible volume, since Avogadro believed in the atomistic theory.

Instead, the obvious solution is that the number of molecules must be a measure of the amount of "stuff." So if we have more "stuff," we need more volume for it to occupy. The expression must be directly proportional between the volume and the number of molecules.


A simple validation
The result you quoted is the average translational kinetic energy for an ideal gas.
First, let's sketch out a rough derivation for the average kinetic energy of a particles of an ideal gas using nothing more than high school physics.
The gas molecules just undergo translations, and don't rotate/vibrate. Also, they don't interact with each other.
Imagine a rectangular box which contains an ideal gas; the gas molecules have a speed $v_x$ in the $x$-direction. Since the box has a uniform cross-section of area $A$, a volume element spanned by a gas molecule in the box is $V'= |v_x\Delta t|A$. On average half the molecules move to the right, and half the molecules move to the left, since there is no preferred direction of motion. Thus, the number of molecules colliding with the container walls is
no. of molecules $= \frac{1}{2}.\frac{nN_a}{V}.|v_x\Delta t|A$
The momentum transferred per molecule is $2mv_x$, so the total momentum transferred is $ \Delta P = \frac{1}{2}.\frac{nN_a}{V}.|v_x\Delta t| 2mv_x$
Since force is rate of change of momentum, $F = \lim_{\Delta t \to 0} \frac{\Delta P}{\Delta t} = \frac{nMAv_x^2}{V}$
Pressure is force per unit area, so the average pressure is $P = \frac{nM \langle v_x^2\rangle}{V}$
Now, note that the total velocity is $v^2 = v_x^2+v_y^2+v_z^2$, and since there is no preferred direction of motion, one can assume all three components of velocity to be equal. $v_{rms}^2 = 3 \langle v_x^2\rangle$
Thus, $P = \frac{nMv_{rms}^2}{3V}$ also from the ideal gas equation $P = \frac{NkT}{V}$ we get $\frac{mv_{rms}^2}{3} = kT$
$E_{\text{kinetic}} = \frac{mv_{rms}^2}{2} = \frac{3kT}{2}$ Which is the result you wanted.
Using the Maxwell-Boltzmann Distribution
The same result for an translational kinetic energy can be arrived at using the Maxwell-Boltzmann distribution.
$$f(v) = 4 \pi \left( \frac{m}{2 \pi kT}\right)^{3/2} v^2 e^{\frac{-mv^2}{2kT}}$$
The average kinetic energy is $$\langle E_k \rangle= \langle \frac{mv^2}{2} \rangle = \int_{0}^{\infty}\frac{mv^2}{2} f(v)\mathrm{d}v = \frac{3kT}{2} $$
You can evaluate the integral as an exercise. I have omitted the details for the sake of brevity. It is relatively simple, you can either do it by hand, or look it up in a table or enter it into Mathematica
Now, the general statement of the theorem (using words and not math) us

For a system at equilibrium the average value of each quadratic cobtribution to energy is $\frac{kT}{2}$

An example of a quadratic contribution would be kinetic energy ($\frac{p^2}{2m}$) or the harmonic potential ($\frac{kx^2}{2}$)
A simple statistical mechanics approach
This theorem is strictly valid at "high" temperatures; we require that seperation between energy levels is small, and a lot of states are occupied so that they maybe treated as a continuum. This works well for translational and rotational modes of motion, but is not as reliable  for vibrational/electronic modes. Anyway, note that the analysis given below is under the framework of these assumptions.
note: $\beta = \frac{1}{kT}$
For a one dimensional container the translational partition function derived using particle in the box energy levels is given below
$$q^T = \frac{X}{\Lambda} \qquad \text{where} \ \Lambda = \frac{h \beta^{1/2}}{(2 \pi m)^{1/2}}$$
$$\langle E^T \rangle = -\frac{1}{q^T} \left( \frac{\partial q^T}{\partial \beta}\right)_V$$
again, you can evaluate this derivative as an exercise. (Hint: $\Lambda$ is constant multiplied by $\beta^{1/2}$). The result is given below
$$\langle E^T \rangle = \frac{kT}{2}$$
This is for translation in one direction (say X). The same process can be repeated for Y and Z, and thus for three dimensions
$$\langle E^T \rangle = \frac{3kT}{2}$$
Now let us examine rotations. Once again, I point out that this analysis is valid in the range $T > \theta^R = \frac{hcB}{k} $
Here, B is the rotational constant.
In this range, an approximate expression for the partition function of a heteronuclear diatmic moeclue is $$q^R = \frac{1}{\beta hcB}$$
$$\langle E^R \rangle = -\frac{1}{q^R} \left( \frac{\partial q^R}{\partial \beta}\right)_V = \frac{1}{\beta} = kT$$
Note: I already pointed out that this result is valid at "high" temperatures, however, I did not give you a sense of scale. For hydrogen molecule $\theta^R = 87.6  \ K$ for chlorine molecule it is $ 0.351 K $. Similarly, the approximations made while deriving the translational functional are valid at room temperatures and for everyday macroscopic containers whose dimensions are much greater than the thermal wavelength ($\Lambda$) of the molecules.
A similar analysis can be performed for vibrational modes, however the temperature range over which it is valid is much greater ($> 805 K$ for chlorine and $> 6332 K $ for hydrogen). These conditions aren't within the realm of everyday experience.


How can I find the eutectic point of gallium and aluminum?

The Al-Ga system is reasonably well studied, and there are a number of articles in the Calphad (Calculation of Phase Diagrams) literature. For example, Al-Ga-Zn System: Reassessments of the Three Binary Systems and Discussion on Possible Estimations and on Optimisation of the Ternary System, M. Mathon et al., CALPHAD 24(3) 253-284 (2000) has the thermodynamic analysis yielding the fairly simple model for the Al-Ga system. These parameters may be used in one of the many phase diagram software packages, such as ThermoCalc. Or, one could write a simple simulator in Python, yielding something like:

The eutectic is way over on the Ga side. The paper listed above also references a number of experimental papers with measured eutectic points.

For other systems of interest, one easy place to start is the ASM Alloy Phase Diagram Database, assuming your institution has access. The phase diagram entries have a set of references, often containing both experiment and calculation papers. If you have access to Calphad, Journal of Alloys and Compounds, and Journal of Phase Equilibria and Diffusion, those are good places to search for a given binary or ternary system.


How does the room gets heated by leaving a fridge open?

Also the radiator emits out the heat in the surrounding. Thus the heat absorbed from the room is released in it and the must neither get cooled nor heated.

You have forgotten a few parts to your system in how a refrigerator pumps heat. You have assumed a spontaneous steady state cycle of:

$$\text{radiator}\underset{\text{conduction}}{\overset{\text{heat pump}}\leftrightharpoons} \text{fridge}$$

But that is not true to move heat from a low-temperature body body to a high-temperature body requires an input of work which is added to the heat dumped into the high-temperature body.

In this case work is done by the compressor which takes in electrical energy and transforms it into work and heat. Since electrical energy is 100% efficient at heating any electrical energy consumed by the compressor is eventually converted into heat either by ohmic heat, efficiency losses or work being converted into heat at the high-temperature body. Since your high-temperature body and low-temperature body are allowed to conduct/convect/radiate/transfer heat between the two sinks, heat overall is added to the room.

  Just before posting this question I have tried to experimentally verify this and observed no change in temperature of the room.

There are a number of problems with this, and why you may not have observed anything.

Rooms are not closed nor isolated systems thus are subject to many thermal variations from light (radiation), drafts (convection), and conduction of heat through walls to/from the outside which can make your results unreliable.
A the heat capacity of a room is very large and it will take a while for your fridge to impart a change that a thermostat/thermometer can detect.
The energy consumed by a refrigerator is relatively small* and it will again take a while for your fridge to impart a change that a thermostat/thermometer can detect.

All and all, such an experiment would require better control than your average room/kitchen. If you did do such a controlled experiment you would observe that the temperature of the room would rise with time due to the heat losses of the compressor.

 *I understand some parents/utility customers (possible yours) may find this statement controversial. Kids do not quote me at home.


What is the difference between fractional freezing and fractional crystallization?

There is no difference. Strictly speaking, "freezing" does not necessarily include crystalline solids, and may include amorphous materials that precipitate out of solution. But, this does not really matter in the context of these eutectic (or similar) systems.

I suspect the latter is referred to as "partial freezing" because the solid is water, whereas it is more general in the former. We rarely speak of water crystallising to ice, but rather water freezing to ice. This is more of a linguistic issue than scientific.


Why does the viscosity of honey decrease when mixed with other viscous fluids?

Honey is a homogeneous substance, a solution of around 98% sugar and water.

Mustard is a heterogeneous mixture, containing about 1:1:1 oils, proteins and water, all in very small particles. The proteins act as a stabiliser to the small oil droplets. They cant slide past each other easily, and act more like a dense network of small solid particles. Like a heap of wet sand.

Now if the aqueous phase gets much larger due to the honey, the oil+protein droplets loose direct contact, and you no longer have the solid-like behaviour of wet sand.

Your aqueous honey-phase however now contains perhaps 25 or 30 percent of water instead of 15 or so in pure honey, and that lowers its viscosity a lot.

You have a major phase with reduced viscosity, and a disperse phase which is no longer continous (not "percolated"), and so the whole thing flows rather easily.

(Btw. the viscosity of mustard is not that high, however it usually shows a yield stress, that is it does not flow at all e.g. under its own weight. You open a mustard glass and the inside looks just the same as it did when you last closed it. Honey however is just a boring Newtonian fluid, no shear thinning, no thixotropy, nothing.)


Percent composition of nitrogen in fertilizer

I got 17.542% by mass nitrogen.

An explanation for beginners, a refresher for the rest of us:

To make sure anyone reading knows what molar mass means: The mass of one mole of something, in our case the mass of a mole of nitrogen, or the mass of a mole of ammonium sulfate, etc. A mole is just a way of saying a really big number which is actually $6.022 * 10^{23}$. It is a word we use instead of saying that big number, like when we say a dozen of eggs, but we know it means $12$ eggs. The same concept applies to a mole of eggs (that's a lot of eggs). To calculate the molar mass you need to add the atomic masses of each atom present in the molecule. If it is water, then we must count hydrogen's atomic mass twice (two hydrogen atoms per molecule), oxygen only once, since the formula is $H_2O$. Atomic mass can be found on any respectable periodic table, be sure not to confuse the atomic mass with the atomic number! The atomic mass always represents the weight in grams of a mole of that element. This convention makes our calculations much easier.

Ultimately we need to know how many $mol$ of nitrogen exist in this bag of fertilizer, so we can find out the mass of nitrogen in said bag (atomic mass can be used both ways because it is a ratio of grams per mole). To find that we need to do this process for both of the compounds separately because they have different molar masses. If there are two $chickens$ per $hen$ and we have $4_{mols}$ of $hens$ we should be able to tell that we have $8_{mols}$ of $chickens$. The same concept is used to determine how many $mols$ of nitrogen we have, based on how many nitrogens we expect per molecule of each substance.

Here is how that calculation might look in dimensional analysis in case it surprises you:

$81_{grams} * \frac{1_{mol}}{12_{grams}}$

Notice units of grams cancel each other out, and the remainder is in moles. This is how you would find the number of moles of carbon there exist in a pure sample of 81 grams.

Notice we used the ratio $12$ $\frac{grams}{mol}$ which is carbon's atomic mass.

I highly recommend writing out the entire problem using dimensional analysis so that you can be absolutely sure you have the answer correct for whatever problem you are solving. You will know this is true when all the units cancel correctly.

To solve the problem it may be good to take a step back from the calculator and consider what we know. What is the molar mass of these two substances? It should be the summation of the atomic masses of all the atoms in the molecule. Just like a water balloon's mass is not only the mass of the water, but also the rubber. Similarly the mass of water is not just the mass of the hydrogen, but the oxygen as well.

There is $500_{grams}$ of ammonium sulfate, but if we are trying to figure out the moles based on this information, we will need to do some dimensional analysis. Use the molar mass of ammonium sulfate which has units $grams/mol$ to multiply this ratio we found by the total mass of ammonium sulfate $(500_{grams})$ in such a way that the $_{grams}$ cancel each other, leaving $mol$ units. This is how many moles of that molecule exist in a 500g sample. We need to know this because the relationship of nitrogen per molecule is not a gram to gram relationship. You could of course, go another route and figure out the percent by mass of nitrogen per molecule and your final calculation should be equivalent. You should get about $21$% nitrogen by mass in a molecule of ammonium sulfate, and therefore 21% of the mass of that substance should equal the mass of nitrogen in that substance. It will be different for the other substance in the bag.

Once you know the total moles of nitrogen, convert back into grams of nitrogen, make sure your units cancel. From there you know the mass of nitrogen in the fertilizer bag, and the total mass of the fertilizer bag.

find the answer: what percent mass is the nitrogen atoms compared to the total mass of the fertilizer bag (all the atoms/molecules)?


Find increase in OH- ion concentration after base is added

The problem is that you are trying to subtract the values, whereas all you need to do is find an increase in ratio $\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1}$ knowing that $\mathrm{pOH} = - \log{[\ce{OH-}]}$ and $\mathrm{pOH} = \mathrm{p}K_\mathrm{w} - \mathrm{pH}$:

$$\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1} = \frac{10^{-\mathrm{pOH_2}}}{10^{-\mathrm{pOH_1}}} = 10^{\mathrm{pOH_1} - \mathrm{pOH_2}} = 10^{\mathrm{p}K_\mathrm{w} - \mathrm{pH_1} - \mathrm{p}K_\mathrm{w} + \mathrm{pH_2}} = 10^{\mathrm{pH_2} - \mathrm{pH_1}} = 10^{12 - 7} = 10^5$$


Hartree product and the Slater determinant

The Hartree product can definitely be used for approximate calculations; as you might know Hartree developed his method for atoms in 1927 (for reference, the Schrödinger equation was discovered in 1926) and he implicitely used a Hartree product for the wave function, although the connection to the exact solution of the many-body Schrödinger equation via the variational principle wasn't know. It was Slater in 1930 (and, independently, Fock) to introduce the determinant form which bears his name. When using the Hartree product one has to account for the Pauli principle "by hand", by allowing only two electrons to occupy the same spatial orbital.
By starting with a Hartree product, enforcing the Pauli principle and applying the variational principle one obtains the Hartree equations, which are similar to the Hartree-Fock ones but are missing the exchange term, usually indicated by $K$; this term is, however, smaller than the  electrostatic term $J$ so Hartree calculations are meaningful and give useful results, although of course very approximate ones.

Kohn-Sham DFT calculations use in principle the Slater product form, but in practice, depending on the exchange-correlation functional used, they lead to equations which may be more similar to the Hartree ones (i.e., without the exchange term) than to the Hartree-Fock ones. In the original LDA (local density approximation) functional the exchange term is dropped and replaced by a functional of the electron density which is much simpler to compute. As a result, DFT-LDA is similar in complexity to the Hartree method, not to the Hartree-Fock (a rather significant simplification). However, most modern functionals for molecules (as opposed to solid state calculations), e.g. B3LYP, re-introduce the exchange term (with a scaling coefficient).

The Hartree form of the wave function can be also used to compute expectation values (e.g., to compute dipole moments); you can check by looking a the Slater-Condon rules  that the results you obtain with Hartree functions are, for one-body operators, the same you'd get using the proper Slater-determinant form.


Why is this intermediate borane-hydroperoxide adduct unstable?

Basically an intermediate form, that fulfills the octet rule, would be more stable than the trialkylborane, but only if we suppose that there are no other effects! And if there were no other ffects, it wouldn't be an intermediate form but an isolable compound.
In your reaction scheme you see that $\ce{OH-}$ will leave. You maybe know that $\ce{OH-}$ is - under basic conditions - a good leaving group. Moreover a peroxide is mostly pretty unstable, so it's happy to react and let $\ce{OH-}$ leave. The hydroboration–oxidation reaction gives you the tetrahydroxyborane-anion ($\ce{B(OH)4-}$) and three alcohols ($\ce{ROH}$) as final product. So you'll get an anion that fulfills the octet rule and is more stable than your reactant.

Edit: More details on the hydroxide as leaving group
The reaction finds place under strongly basic conditions. Usually you take $\ce{NaOH}$ to deprotonate the $\ce{H2O2}$ and form $\ce{OOH-}$, so in your solution there are plenty of $\ce{OH-}$ ions. The intermediate also splits off an hydroxide ion which is a good leaving group under this conditions. This means that the ion is stabilized in solution and is not less stable as the intermediate. Even if you can't really compare the stability of those two things because they are not the same... So let's only say, that the hydroxide ion is pretty stable in a basic solution.
As I also said in my original answer above, peroxides are pretty unstable. This is because the two oxygens are right next to each other. As you may know, oxygen is one of the most electronegative elements of the periodic table, so it "fights to get electrons". If you connect two oxygens like in a peroxide, they both fight to get the electron of the other one, so it resembles rope pulling. This isn't the best case scenario for a stable bond.
If you look at hydrogen peroxide, especially in concentrated solutions, it tends to decompose without any external influence.
$\ce{2H2O2 -> 2H2O + O2}$
This reaction releases 98,2 kJ/mol [1] of energy. This means that the products (oxygen and water) are way more stable than the hydrogen peroxide itself. This is mainly due to the very unstable rope pulling-bond between the two oxygens in $\ce{H2O2}$: $\ce{RO-OR}$.
This should explain why the intermediate group is so unstable.


These ring-flips are called 1C4-4C1 transitions and these happen to all of the pyranose sugars in solution when the conditions allow. However, the abundance of each forms depends on their energy. There is a so called 1,3-diaxial interaction between the axial substituents of a cyclohexane analogue. Two axial substituents on the same side are much closer than two equatorial substituens. If these substituents are relatively big, it will be energetically unfavorable to bring them close together due to steric effects. Therefore the molecule will "choose" the conformation that has less energy, so less "large" substituents on the same side. In your case the OH and Me groups are considered the "large" substituents, while the H atoms (undrawn) are the "small" ones. Therefore the conformation you drawn has an unfavorable Me-OH interaction, which will drive the conformation change from 4C1 to 1C4.


How to get or understand a sample in chemistry without knowing its chemical name or compound or formula

Learning the unknown sample is called analysis and is a subject of analytical chemistry.
But your question is probably not about this.
You mean "can i guess something without knowing chemistry at all?"
And answer is yes!
Look at the food salt.
Do you know it without any knowledge of chemistry? Yes, you do!
But how much?
You know what it looks like,
you know its taste,
you know that it's soluble in water...
and probably that's all.
Now suppose you have a sample of something not found yet in human history.
You can easily know it similar to knowing food salt. Just check:
How it looks like?
What it's taste?
Drop it in water - is it soluble?
Now you "know" this substance just like you "know" food salt, without any chemistry!
But I guess you already suspect that there is something wrong with this knowledge.
It's very incomplete.
Chemist can learn more about food salt or your magic unknown sample.
Most of that serious knowledge require understanding of sample structure
at microscopic levels, and using theories that connect the structure to properties.
That was all about "knowing the substance".
Another question is knowing what a substance is, which is rather philosophy than chemistry.
Does a substance exist? What is existance? Does a substance have hidden fundamental purpose? Was it "created" or "appeared by itself"?
These questions are probably more important than chemistry, and you may be the one who one day will guess some answers here.


Alternative to Hanging Drop Crystallization

The process you are describing is essentially the sitting drop crystallisation.

Yes, it is used. Frequently, sitting drop is used in robotic setups, because robots have a much harder time creating hanging drops. With a sitting drop, it becomes simple; just pipet a drop into the sitting drop reservoir.

Gerhard mentioned that the hanging drop method is the preferred one for manual handling since it becomes easier to handle the crystals. Possibly because they sit at the bottom of the hanging drop rather than stick to the ground of a sitting drop’s reservoir.


calculate the number of shots of Captain Morgan’s rum

As Zhe mentioned, you have inconsistent units in your calculation, so your numerical results are not physically meaningful.

Take for instance this expression:

162 = A/(0.55 x 50) – (0.15 x 1)

Here, you have different units for each term; from the values you supplied, the units are

162 mg/dL = A/(0.55 L/kg x 50 kg) - 0.15 g/(L•h) x 1 h

[Note that the units for the elimination rate weren't stated in the question, but a quick search of the literature gave studies measuring it at about 0.01-0.02 g/(dL•h), so I infer the 0.15 value refers to g/L]
If we use uniform units throughout, we get

1.62 g/L = A/27.5 L - 0.15 g/L

from where you get a more reasonable value for A:

A = (1.62 + 0.15) * 27.5 = 48.675 g

Each shot contains

V = 30 mL * 0.40 / 0.789 g/mL = 15.2 g

of ethanol, so you need to drink 48.675 / 15.2 = 3.2 shots of rhum over the course of an hour.

I'd suggest, however, that drinking shots in "rapid succession" implies a negligible drinking time (∆t ~ 0 h). In that case, A = 44.5 g and you need about 3 shots (2.93).


Is delta S the same in reversible and irreversible processes?

For an irreverislbe process with specified initial and final states of both the system and of the surroundings, calculating the change in entropy  between the same two end states requires you to separate the system from the surroundings, and to devise a separate alternative reversible path for each to determine the integral of dq/T for each.  This will give you the separate changes in entropy of the system, the surroundings, and the sum (i.e., the universe).  So the reversible paths between the initial and final states will not typically bear much resemblance to the actual irreversible path that the combination of system and surroundings had suffered in the irreversible process.


I suspended iodine over steel nails. What are these droplets that formed?

Iodine vapors combine with iron to form iron(II) iodide:$$\ce{I2 + Fe->FeI2}$$ In an anhydrous form the compound is dark red-purple. But iron(II) iodide is hygroscopic and deliquescent which means that it dissolves in moisture it absorbs from the air, and forms those droplets. This complicates a bit the chemistry of the droplets - the $\ce{Fe^2+}$ ion exists most likely in a form of various aquo-, hydroxo- and iodo- complexes. Also, when an iodide anion and elementary iodine are present together they form triiodide anions:$$\ce{I- + I2->I3-}$$ Therefore, the droplets are a solution of a rather complicated mixture of iron and iodine compounds.


Can I eat the lemons after I have used them as a battery?

In a lemon battery, the zinc from the galvanized nail is giving up electrons and transitioning into an aqueous state:

$$\ce{Zn → Zn^2+ + 2e−}$$

While zinc is entering the electrolyte, two positively charged hydrogen ions ($\ce{H+}$) from the citric acid with two electrons at the copper electrode's surface and form an uncharged hydrogen molecule $\ce{H2}$:

$$\ce{2H+ + 2e− → H2}$$

While zinc is a mineral our bodies need, I wouldn't recommend using this as a source of zinc. Also, zinc toxicity is a thing, so without knowing the dose you're taking, you risk poisoning yourself.

Lemon Battery Wikipedia


It is a phase change - gold exists as a solid crystal (fcc I believe off the top of my head), as a liquid, and as a vapor (gas).  First order phase transitions exhibit a volume change, and an enthalpy of the phase change.  In the case of gold, both the crystal and the liquid are metallic, with similar numbers of nearest neighbors.

All elements exhibit various phases.  Since pure gold is, well, pure gold, I'm not sure how a phase change in gold could be considered a 'chemical change', particularly in the solid-to-liquid.  The type of bonding does not change, and there are no 'chemical reactions' going on. This is all fairly basic thermodynamics.


Are any bonds broken when NO2 becomes N2O4?

No bonds are broken, but there are several isomers of N2O4.  The unpaired electron can be on either the N or the O, for each NO2 molecule. The long bond between the monomers involves these two unpaired electrons.

See sections 7-1 to 7-3 of Bonding in Electron-Rich Molecules: Qualitative Valence-Bond Approach via Increased-valence Structures for more information.

According to First-Principles Study of the Role of Interconversion Between NO2, N2O4, cis-ONO-NO2, and trans-ONO-NO2 in Chemical Processes J. Am. Chem. Soc., 2012, 134 (31), pp 12970–12978:

there is no barrier to the bonding of two monomers to form the planar isomer with the N-N bond.


Why is Aluminum's 2nd ionization energy higher than Silicon's?

Let us look at the electronic configurations of aluminum and silicon and their respective cations.

$\ce{Al}: [\ce{Ne}]\mathrm{3s^2\, 3p^1}$

$\ce{Si}: [\ce{Ne}]\mathrm{3s^2\, 3p^2}$

$\ce{Al^2+}: [\ce{Ne}]\mathrm{3s^1}$ It involves removing from $\mathrm{3s}$ sub shell

$\ce{Si^2+}: [\ce{Ne}]\mathrm{3s^2}$ It involves removing from $\mathrm{3p}$ sub shell

A $\mathrm{3p^1}$ electron has lower binding energy than that of $\mathrm{3s^2}$ electron.

It is due to more screening effect (as the predecessor sub-shells are higher in $\ce{Si}$) and higher stability attained by having a completely filled $\mathrm{3s}$ sub-shell


Reactivity of group 13 elements with nitrogen

In the case of boron, a reaction with nitrogen does occur under some eye-popping conditions:

Combustion of boron powder in nitrogen plasma at 5500 °C yields ultrafine boron nitride used for lubricants and toners.[1]

The characteristic molecular thermal energy $kT$ at this temperature is roughly half an electron volt, which is enough to at least partially break a significant fraction of the nitrogen-nitrogen bonds.  The boron is also vaporized at this temperature, and both of these factors facilitate the reaction.  Vaporizing the boron also favors forming the fine particles which are desired for the lubricant and toner applications noted above.
Most boron nitride, not requiring the ultrafine particles, is synthesized by metathesis reactions under more typical industrial conditions, which are also given in the WP article noted above.  These commonly involve boron oxide, which both becomes fluid at much lower temperatures than the element and can act as an acid towards basic nitrogen sources such as ammonia.
Cited Reference
1.
Paine, Robert T.; Narula, Chaitanya K. (1990). "Synthetic Routes to Boron Nitride". Chemical Reviews. 90: 73–91.


Can dissolution reactions be endothermic?

You will find the answer in probably any textbook on chemical thermodynamics, or in a thermochemistry chapter in most physical chemistry textbooks, or in Wikipedia's enthalpy of dissolution page:

  Dissolution can be viewed as occurring in three steps:

  Breaking solute-solute attractions (endothermic), see for instance lattice energy in salts.
  Breaking solvent-solvent attractions (endothermic), for instance that of hydrogen bonding
  Forming solvent-solute attractions (exothermic), in solvation.

  The value of the enthalpy of solution is the sum of these individual steps. Dissolving ammonium nitrate in water is endothermic. The energy released by solvation of the ammonium ions and nitrate ions is less that the energy absorbed in breaking up the ammonium nitrate ionic lattice and the attractions between water molecules. Dissolving potassium hydroxide is exothermic, as more energy is released during solvation than is used in breaking up the solute and solvent.

So, it can have any sign. The linked page also gives data for common compounds in water, some of them endothermic, some of them exothermic. Ammonium nitrate, $\ce{NH4NO3}$, is a good example of a salt which dissolves in water in an endothermic reaction: if you dissolve some $\ce{NH4NO3}$ in water, the drop in the temperature will be noticeable.

Regarding why the substance is soluble why its dissolution is endothermic, you have to remember that the reaction takes place if $\Delta_\mathrm r G$ is favourable (i.e. negative), and $\Delta_\mathrm r G = \Delta_\mathrm r H - T \Delta_\mathrm r S$. Overall the free energy must be negative for dissolution to occur (on a thermodynamic basis; kinetics are another issue), not the enthalpy.


2017 Nobel Prize; What does Cryo-electron microscopy data actually look like?

This is a summary of the account by Peter Brzezinski, Professor of Biochemistry, Stockholm University, Member of the Nobel Committee for Chemistry from the Scientific Background document.

  [Brzezinski:] Jacques Dubochet developed methods for preparation of samples for cryo-EM studies of biomolecules in water. Joachim Frank developed methods for structural determination of biomolecules from analyses of ensembles of particles in solution. Richard Henderson demonstrated that it is possible to obtain atomic resolution structures of biomolecules using cryo-EM.

To answer the questions of the OP, I will use some of the excellent figures in the document.

  [OP:] What does the data that Cryo-electron microscopy produces when examining the structure of "biomolecules in solution" actually look like? Is it electron transmission images of the molecules, or a diffraction pattern produced by single molecules, or something different?

For larger particles such as viruses, it looks like the actual objects. Here is an example:

As you can see, the virus particles are not oriented (not a crystal, different molecules rotated differently). Other studies used two-dimensional or three-dimensional crystals. No diffraction images are used (in contrast to single molecule X-ray diffraction using free electron lasers). When the objects are large and have high symmetry, orientation of the molecules is easier than for small objects lacking symmetry. Thus, success in structure determination started with highly symmetric virus particles (pioneered by Henderson) and progressed to smaller objects requiring more powerful algorithms for orienting images (developed by Frank).

  [OP:] Are the molecules still in liquid solution when they are measured, or are they actually frozen by the time the data is collected?

Samples are in vitrious ice. For the technique developed by Dubochet, no cryo-protectant is necessary (different from cryo-cooling crystals for X-ray diffraction):

Advances in detectors

The image posted by the OP mentions advances in detector technology. Traditionally, electron microscopy uses film or CCD detectors. In both cases, signal from electrons are "converted" into visible light, introducing noise. The new type of detectors are "constructed from Monolithic Active Pixel Sensors, based on Complementary Metal Oxide Semiconductors, known as CMOS technology". They directly detect the electrons, giving images with less noise.

Disclaimer

For brevity, I skimmed over the contributions of the laureates and did not mention any of the many other contributors. The article I reference above is excellent; I recommend it as a starting point to a more in-depth study of the topic.


Can we prepare alkyl Fluoride by Finkelstein reaction?

In the future I recommend consulting March's Advanced Organic Chemistry before asking such questions. It is a priceless reference work for organic synthesis methods. The following text is result of brief consultation with the book.

The equilibrium shift in Finkelstein reaction is governed by relative nucleofility of halogenide-ions. In general, in polar solvents iodine is the strongest nucleophile, because it is poorly solvated, while in nonpolar solvents chlorine is the strongest solvent as it is smaller.

Fluoride in non-polar solvents should be the strongest nucleophile of all halogenide-ions, but it so strongly solvates in protonic solvents that it almost do not migrate in organic phase even if phase-transfer catalists are present. Strong basicity of fluoride does not help either as beta-eliminations becomes significant side-reaction.

This article claims to develop practically meaningful procedure of clean fluoro-de-halogenations.


First, there is very very little chance it will remain same. Density will surely change with temperature.

Secondly, even if there is a small chance that it will remain same, then also there is an even greater chance that it will change.

So we will have to look at the general case and conclude that molarity depends on temperature.


'In MO theory, electrons are treated as spreading throughout the entire molecule'- isn't it exhibited in VB theory also?

In MO theory, you construct a molecular orbital by a linear combination of atomic orbitals. Because those atomic orbitals are part of different atoms, in this way, the valence electrons are associated with all the atoms in the molecule. So even if you consider a molecular orbital that is bonding between, let's say, atom 1 and atom 2 in the molecule, that molecular orbital has also contributions from all the other atomic orbitals (besides those that arise from atom 1 or atom 2).

In contrast, in VB theory, a bond between atom 1 and atom 2 can only have atomic orbitals of atom 1 and atom 2. The bond between two atoms is then considered to arise from two weakly coupled atomic orbitals. I have to point out though that modern valence bond theory now complements molecular orbital theory and no longer adheres to the valence bond idea that electron pairs are localized between two specific atoms in a molecule but that they are distributed in sets of molecular orbitals which can extend over the entire molecule.

Given the discussion below, I would like to add to my answer a simple example. Let us consider the $\ce{H2O}$ molecule. Within the framework of Hartree-Fock and using a STO-3G basis set, the $2\sigma$ orbital of water can be expanded in terms of the atomic orbitals as follows:
$$
\psi_{2 \sigma} = (-0.4330)\; \phi_{\textrm{O},1s} + (-0.4686)\; \phi_{\textrm{O},2s} + (0.2369)\; \phi_{\textrm{O},2p_y} + (0.2685)\; \phi_{\textrm{H1},1s} + (0.2685)\; \phi_{\textrm{H2},1s}.
$$

From this expansion you can see that multiple atomic orbitals contribute to the same molecular orbital. This is a molecular orbital that forms, in a way, a multi-atom bond between O and the two H atoms. You could also say that this molecular orbital contributes to both of the O-H bonds (in fact, more molecular orbitals than just the $2\sigma$ do).


Most activated position for Electrophilic addition in substituted biphenyls

Dipole moment is what matters in substituted biphenyls.

The dipole moment in the first case is from left ring to right ring due to -OH groups predominant +M effect to the ring.(the alkene being an EWG(electron withdrawing group)), further enhances this effect. So effectively, electron density is more in the right hand side (I know it is counter-intuitive but it is how it is).
So EAS takes place preferably/majorly in the right side ring.

In the second case, NO2 shows -M effect and this dominates that of the alkenyl group. So dipole moment is from right to left ring and EAS takes place preferably/majorly in the left side ring.

Note: If there is any group(which is symmetric, or else we need to consider electronic effect due to assymetric grp also) between the C-C bond linking the two rings together, then the answer is as it "intuitively seems/as you expect". Try to rationalize this yourself.


Chemicals that form coloured complexes with Pb2+

In the old book from Hopkin and Williams, Organic Reagents for Mineral Analysis. London (1933), the following information can be found in the Chapter "Diphenylthiocarbazone, or Dithizone".

Diphenylthiocarbazone is a black powder, insoluble in water, but soluble in organic solvants, giving green solutions. Its main application is its reaction with lead. Lead ions produce a red color to its CCl4 solution. To 10 mL Lead aqueous solution, 1 mL KCN 10%, and two drops of ammonia are added. Add 2 mL of a dilute solution of diphenylthiocarbazone in CCl4 (50 mg/L). Stir. A reddish color must appear if the sample contains 0.001 mg Pb. A similar color appears without lead, due to the solubility of the reagent in CCl4. Addition of KCN removes totally any excess of reagent in the aqueous phase.

The only drawback about this method is the necessity to use KCN, which is forbidden today. I have found in the same book a small article about benzidine, where they say that lead solutions give a blue color when mixed with benzidine, if a solution 0,05% benzidine in acetic acid 10% is used.

Lead can also be detected and quantitatively determined by 8-hydroxyquinolein. But the book mentions only the following bibliographic  references :
V. Marsson, L. Haase, Ueber die Bestimmung des Bleis mit Hilfe von o-Oxychinolein, Chem. Ztg, 52, 993, (1928)
V. Hovorka, Sur le dosage du plomb au moyen d'ortho-oxyquinoléine, Coll. Czech. Chem. Comm, 9, 191 (1937)


What is the difference between a bond, force, and interaction?

Here is how I think of the situation.

The term "bond" is generally reserved to describe an attractive, stabilizing interaction between atoms or groups in the same molecule or in different molecules, that has a magnitude of at least several kcal/mol.  Everything from hydrogen bonds at the low end up to full chemical bonds at the high magnitude end are included.
The terms "force" and "interaction" can be used to describe any attractive or repulsive situation, of any magnitude, between atoms or molecules. So these terms would be used to describe the van der Waals "interaction", where the "forces" involved are much less than those involved in bonding.  However, the terms "force" and "bonding" can also be used to describe situations involving larger magnitude interactions such as, "bringing the atoms together led to a bonding interaction."

"Force" and "interaction" can be used to describe any situation - attractive, repulsive, high energy, low energy; whereas "bonding" is typically reserved to describe stabilizing, higher energy interactions.


Why green leaf turns brown in Brine solution?

It is most likely the $\ce{Na+}$ is formed a complex salt with chlorophyll (bulky group) subsequently displacing $\ce{Mg+}$ cation, even though $\ce{Na+}$ doesn't have extra electronic shells for complex ion stabilisation, instead the chlorophyll (through the $\ce{N}$ atoms with lone pairs) provided extra stabilisation of the positively charged $\ce{Na+}$ cation.

Chlorine dissociated from $\ce{NaCl}$ to form aqueous chloride ions $\ce{Cl-(aq)}$ forming a typically greenish-brown solution you noticed during your experiment.


Citric acid is relatively soluble in ethanol (62g/100g) whereas sucrose is relatively insoluble in ethanol (1g/170ml). There are many different kinds of sugars and sucrose is probably not the primary one in this extract; branched sugars... oligosaccharides (which have generally higher solubilities than sucrose) are likely, and I'm sure there are a few other organic acids involved with a crude, fruit extract, but the different solubility characteristics should be relatively helpful for extracting citric acid from sugars and some other organic acids.

So, I would try extracting the citric acid from the dried fruit or dried water extraction with 99%+ ethanol. I would then allow the alcohol to evaporate to help concentrate the acid.

Dilute citric acid (from a concentrated extract) can be precipitated with calcium hydroxide as practically insoluble calcium citrate, which is then reacted with sulfuric acid to form (food grade) citric acid and calcium sulfate (gypsum) as a recoverable by-product.

To really purify the stuff, an ion exchange resin or some other chromatographic method would be the last step.


Is energy required to form bonds [phase change]

Bond formation always lowers the energy of the system (or bond formation is a consequence of lower energy, take your pick.)

Indeed, you may have to add energy, because presumably certain bonds must break in order to rearrange the atoms.

Adding energy won't guarantee formation of much higher energy isomers, because the atoms have a large amount of kinetic energy, and so they can just as easily turn around and go back whence they came.

Sometimes the kinetic energy can be dissipated by solvent, collisions with inert gasses, etc. and then you may end up with a measurable quantity of the higher energy species.


Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution.
It's unit in S.I. is: $$\ce{S.m^2.mol^{-1}}$$


Understanding and usage of Arrhenius equation

What activation energy would lead to a doubling in the reaction rate
  for a 10 degree increase in temperature, using the Arrhenius equation?

You’re right it is not generally true. Let’s set up an equation to solve the problem.

The Arrhenius equation is
$$k=A\operatorname{e}^{\frac{-E_\text{a}}{RT}}$$

In this case

$$2=\frac{A\operatorname{e}^{\frac{-E_\text{a}}{R\cdot(T + 10\ \mathrm{K})}}}{A\operatorname{e}^{\frac{-E_\text{a}}{RT}}}$$

or

$$\ln 2 = \frac{ - E_\text{a}}{R\left( T + 10\ \text{K} \right)} - \frac{ - E_\text{a}}{RT}$$

let’s pick a temperature around room temperature, 300 K is a nice even number, but you can re-solve the problem with any temperature you like.
$$0.69 = \frac{ - E_\text{a}}{R \cdot 310\ \text{K} } - \frac{ - E_\text{a}}{R \cdot 300\ \text{K} }$$
Using $R=8.31\ \mathrm{J/(K\ mol)}$ will produce an answer in J/mol.  Solving we find
$$E_\text{a} = 53\,568.49\ \mathrm{J/mol}$$
or,
$$53.6\ \mathrm{kJ/mol}~~(12.8\ \mathrm{kcal/mol})$$

So around room temperature a reaction that has an activation energy around 54 kJ/mol (or 13 kcal/mol) would show a doubling in rate for a 10 kelvin increase in temperature.


Differences between KOH and phosphorus pentoxide as drying agents?

I found this page, which recreates information from Purification of Laboratory Chemicals, by Perrin and Armarego. It provides a qualitative ranking of drying agents, listing phosphorous pentoxide higher than KOH. The two drying agents work by different means:

KOH

Anhydrous KOH is hygroscopic. It absorbs water from the atmosphere. It much better at doing so than more common hygroscopic drying agents like calcium chloride or calcium sulfate (drierite). A pellet of KOH left out in a beaker or dish open to the air will have absorbed so much water vapor within an hour that it will now be a solution of KOH (that will continue to be hygroscopic, at least until the solution is 50% water). Calcium chloride and calcium sulfate only continue to absorb water until they reach their maximum hydration.

P4O10

Phosphorous pentoxide (really P4O10) is an anhydride of phosphoric acid. Phosphorous pentoxide reacts with water to form phosphoric acid:
$$\ce{P4O10 + 6H2O -> 4H3PO4}$$
Phosphoric acid is also hygroscopic and absorbs water to form a solution.

Since P4O10 is water reactive it is more difficult and dangerous to handle, especially for large groups of students. I might not use P4O10 if the substance being dried is very acid sensitive. Eventually, the vacuum will start volatilizing the acid. I might not use KOH for compounds that are incredibly base sensitive, although it is unlikely that KOH will sublime. In an undergraduate lab with multiple students, KOH should be fine if you want to save money and hassle.


Reaction of barium peroxide forming barium carbonate

Before the precipitation of $\ce{BaCO3}$ starts, the following equilibrium is arranged:

$$\ce{BaO2 + 2H-O-H <=> H2O2 + Ba^{2+} + 2OH-}$$

$\ce{BaO2}$ is but very slightly soluble in $\ce{H2O}$ because the then occurring hydrolysis produces $\ce{OH^-}$, which raises the pH, and hence stops hydrolysis (and dissolving) of further $\ce{BaO2}$. Simultaneously, by the formed $\ce{OH^-}$, the back reaction is fostered. However, these two blockades may be overcome by adding dilute acid, since this latter neutralizes at once the alkaline product $\ce{Ba(OH)2}(aq)$, and so the equilibrium may be continually shifted to the right side.

Hence, it doesn't matter whether the acid is more or less diluted (nor whether it is strong or weak), provided that acid is continually supplied to the extent that it is consumed during the dissolution/hydrolysis process of $\ce{BaO2}$. An aqueous solution of $\ce{Ba(OH)2}$ is an efficient gas detection or gas scrubbing method for $\ce{CO2}$; hence, by bubbling  $\ce{CO2}$-gas through it for a long time, the $\ce{Ba(OH)2}(aq)$ will easily be transformed into $\ce{BaCO3}(s)$. Of course, a gas frit will be used to spread the gas, and the reaction mixture should be constantly stirred.

In the textbook of Sarda/Handa/Arora$^1$ this process with $\ce{CO2}$ is shortly mentioned (without any explanations except that ice-cold water should be used, and that $\ce{BaCO3}$ may be removed by filtration, since the desired product is, of course, $\ce{H2O2}$).

$^1$ Sanda/Harda/Arora, Chemistry, part-2, New Dehli, 2016, p. 885.


Chemistry behind Gale's coffee maker in Breaking Bad

Yes, there is a scientific basis, but I think you can't do anything with the apparatus shown in the figure. Here you can find much information about it. I will try to summarize.

It seems that extracting coffee at a lower temperature makes it tastes better. Gale thinks that the amount of quinic acid is the key variable that makes good coffee. In fact, some studies (McCamey, D. A.; Thorpe, T. M.; and McCarthy, J. P. Coffee Bitterness. In "Developments in Food Science." Vol 25. 169-182. 1990.) report that bitterness of coffee is due to quinic acid but is not the only substance involved (see this link). I don't have any reference about the dependence between temperature and method of extraction and quantitative of quinic acid extracted.

By the way, the apparatus should extract the coffee in a more effective way. The vacuum pump seems to suggest the use of the vacuum to make the water boil at a lower temperature. But the connections in my opinion make no sense, so I will try to describe it:

All start with an autoclave that is connected with a Gast vacuum pump. This is connected with an Allihn condenser (not the right thing to do because in there you should connect the water used to refrigerate), and then to a heated Erlenmeyer flask that is connected with a strange steel cylinder (I don't know what it is, maybe a filter), then to a Florence flask and then to another steel cylinder (...). The inspiration comes from a Florence Siphon but this is much simpler and is generally something like this:

Here you can read and here you can watch how it works.

Of course this is not so nice and intriguing as the previous apparatus and is not a geeky thing that makes us think we are dealing with a great chemist, so the authors thought to make the apparatus more appealing but completely useless!


How to find the possible valence states of elements in the lower end of the d block and the f block?

[For silver], why don't the electrons in the 4d orbital move to the 5p orbital?

In the case of silver, this is likely because the energy separation between the 4d and 5s orbitals are quite large (absorbing in ultraviolet region, which is why silver appears white/silver), so the energy separation between 4d and 5p would be even larger.

In general, when you write down the electron configuration you are writing down the ground state, i.e. the most stable configuration. While it is possible in principle to move electrons from 4d to 5p, it will not give you the most stable configuration as moving that electron requires a large amount of energy.

For inferring the likely valence states, you can only use orbitals with unpaired electrons, and/or move the electrons that are not separated by large energy differences between different orbitals to make hybrid orbitals with unpaired electrons. In this case, as you cannot move 4d electrons to 5p (which would make some kind of spd hybrid orbital), you can only have a valence state of 1 from the unpaired electron in 5s.

  Is it possible for any elements [to move their electrons between $n$d and $(n+1)$p orbitals]?

Yes, but only for elements nearer the top of the periodic table (and not at the very top, as you will then not have any accessible d orbitals). Period 3 and period 4 (I think) would be most likely to do this. For example, phosphorus can have a valence state of 5, and sulfur can have a valence state of 6, both by involving their accessible 3d orbital, to make hybrid $\ce{sp^3d}$ and $\ce{sp^3 d^2}$ orbitals respectively.

  When the f orbital is [completely] filled with 14 electrons, should we take it in consideration to find the valence states?

In general, no. When the f block is completely filled, the electrons in the f orbital are generally assumed to be core-like, i.e. stable, or low in energy, therefore will have a large energy gap to the other electrons in 6s, 5d and 6p (or 7s, 6d and 7p), and do not tend to participate in bonding/valency. There are possibly one or two exceptions, but I cannot recall them right now.

  When Z=58, can you show me the electronic [configuration] and how do we form the valence state(s)?

Z = 58 is Cerium, which has the following electronic configuration: $\ce{[Xe] 4f^1 5d^1 6s^2}$.

On paper, it will appear that due to the diffuseness of the orbitals 4f, 5d and 6s, it is probably unusual to form covalent compounds to cerium, and you would expect a valence state of 3 (if you only use the 5d and 6s electrons), or a valence state of 4 (if you also use in addition the 4f electron).

It was found the 4f is actually near the energy of 5d and 6s, giving rise to a variable electron structure and a valence state change of 3 to 4 when cerium is cooled or compressed. Moreover, it is possible to form organocerium compounds with a valency of 3. By the way, this is not something that you can deduce simply by looking at the electronic configuration. You have to find this out empirically, using experiments!


Does mass equal to moles times molar mass take into account coefficients?

What do you mean by $\pu{3 moles}$ of $\ce{ 2NaN3}$? The coefficient (the 2) is the number of moles of that molecule that you have.

If you're looking at an equation with $\ce{ 2NaN3}$ written in it, and you want that 3 times, then you're going to have $\pu{6 moles}$ of $\ce{NaN3}$, so $\text{mass}~= ~6 \cdot 65$ .


Until 2005 or 2006 the best source was Chemical Market Reporter.
It had the form of a newspaper and was edited once a week.
The last pages listed market prices for commodities, some speciality chemicals and also agricultural products.

Hundreds of products were listed and you had some idea of what were the options you had for designing an industrial process.
You could easily decide which solvent to use for an extraction, which metal for a reduction or which raw material for a synthesis.

The publication was on subscription base but it was available in some University libraries or you could get an older copy from a contact in the big chemical industry.

Now this has been converted on a pay per product information scheme.
You can enter in www.icis.com and ask a report for one product.
If you want to compare prices you have to buy the information for every product.
It is almost impossible to have a complete picture of the economic trend in the Chemical Industry.


Isn't water a Lewis base in the ammonia–water reaction?

Your reasoning is correct. Perhaps the answer key is mistaken.
For part (a), in the reaction
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
water is donating a proton $(\ce{H+}$ ion) and hence is behaving as a Brønsted acid. Since all Brønsted  acids are Lewis acids, water is behaving as a Lewis acid.
For part (d) as well, the half reactions are:
$$
\begin{align}
\ce{2 NH3 &-> 2 NH2- + 2 H+} \\
\ce{2 Na &-> 2 Na+ + 2 e-}
\end{align}
$$
It's clear that $\ce{NH3}$ is acting as an acid and $\ce{Na}$ is acting as a Lewis base by donating electrons. The $\ce{NH2-}$ ion pairs up with the $\ce{Na+}$ ion to form $\ce{NaNH2}$ and hydrogen accepts the electrons and is liberated as $\ce{H2}.$
EDIT: As pointed out by Mithoron in the comments, the reduction of ammonia is not so simple. Sodium dissolves in ammonia to give a deep-blue solution containing ammoniated electrons (reference here):
$$\ce{Na + \text{(x+y)}NH3 <=> Na(NH3)_x^+ + e(NH3)_y^-}$$
The $\ce{e(NH3)_y^-}$ eventually loses a hydrogen atom to give $\ce{NH2-}$, which then combines with the $\ce{Na+}$ giving the product.


What are some examples of pure substances with no liquid state?

Interestingly enough, when considering not just the set of "typical" or "known" substances, but the vastly larger amount of all possible compounds, almost all pure substances do not have a thermodynamically stable liquid state or gaseous state. To become a liquid, a molecule must gain enough energy to set itself loose from the tug of its neighbours in the solid state. Typically intermolecular forces are fairly weak, and so for simple molecules this energy can be provided at relatively low temperatures.

However, there is no upper bound on how strong the intermolecular forces between two molecules can become; just take an enormous molecule. Even with relatively weak intermolecular interactions between each atom, when summed up across a huge molecule, the required energy becomes vast.

Combine this with the fact that, thanks to combinatorics, almost all possible substances are composed of massive molecules (in a chemical analogue of the frivolous theorem of arithmetic). For example, there are only five possible distinct substances with the formula $\ce{C6H14}$, but over a trillion trillion trillion trillion ($10^{48}$) distinct molecules with formula $\ce{C120H242}$.

The conclusion is that almost certainly the molecules would decompose (breakage of covalent bonds) well before getting close to the extremely high temperatures needed to produce a true liquid state.

Therefore, there should be many, many examples to choose from. A heavy-handed example would be maitotoxin (or many other natural products). Though the theoretical melting point would be very high, the presence of all those hydroxyl groups suggests it would decompose by dehydration around ~200 °C, as is typical of polyols (I'm sure it would actually decompose much sooner).

In the same polyol vein, a much more relatable example would be sucrose (sugar). In the strict thermodynamic sense, sugar doesn't actually melt when heated on a stove. The clear colourless syrup that's formed is actually the first step in caramelization, the complicated thermal decomposition of sugar.

The above cases represent your first suggestion of possibilities. Polymers also fit, with the complication that artificial polymers are almost never actually pure substances but mixtures of very similar molecular chains, varying in size. Something like  your last suggestion is interesting, but examples seem hard to provide. Perhaps dinitrogen pentoxide is close, due to its unusual transition between ionic and molecular forms.


Determining covalent bond strengths without hybridisation

As I already stated, asking this question is wrong on so many levels, but let's just have a look anyway.

We are assuming a set of $3$ rigid $\ce{2p}$ and $1$ rigid $\ce{2s}$ orbitals at carbon. With rigid I mean, hybridisationnote is forbidden - which is wrong, $\text{[very]}_n$ wrong.

As a consequence, a head on overlap between the hydrogen $\ce{s}$ and the carbon $\ce{p}$ results in three bonds, each perpendicular , i.e. $90^\circ$, against each other. This is the feature of the degenerate $\ce{p}$ orbitals. (Answer 1 would be correct.)

As a conclusion from that, the molecule cannot adapt a tetrahedral geometry, because with this geometric arrangement, it is impossible to form a $S_4$ symmetry axis, which is the main feature of the $T_\ce{d}$ or $T$ point groups. Answer 3 would be wrong.

The bond between the carbon $\ce{s}$ and hydrogen $\ce{s}$ also needs to be weaker than the others, since these orbitals are totally symmetric. That means there is a good portion of the carbon $\ce{s}$ orbital already blocked by the other hydrogens. On the other hand, the orbital overlap of the carbon $\ce{p}$ and the hydrogen $\ce{s}$ is quite optimal, since the main contribution of $\ce{p}$ orbitals lie outside of the carbon $\ce{s}$ orbital. (Answer 2 would be correct.)

This $\ce{s-s}$ bond would also be quite wobbly, because neither of the orbitals is directed. Therefore the position of the last hydrogen can easily change without effecting the orbital overlap, and in first approximation the bond strength. (Answer 4 would be correct.)

As a last comment, the molecule in this conformation is not stable, probably not even as an highly excited state. The kinds of thought games actually ruin the more important thing of understanding how orbitals behave. Orbitals are not rigid. They are quite flexible in the end.

Note:
Hybridisation in the generic sense means the linear combination of atomic orbitals of one atom, that contribute to a molecular orbital.


Molecule where an sp3 chalcogen is connected to an sp2 atom

Out of the blue, consider triphenyloxonium cation. It is non-planar thanks to bulkiness of the phenyl groups.

Another possible example is mono-epoxide of allene. Considering oxygen atom as $sp^2$ here is unconvincing since it would involve huge angular strain, much higher than for a structure with $sp^3$ oxygen.


How to determine toxicity of a substance?

No, there is no way to determine toxicity to humans with a "magical method".

Most of our knowledge about toxicity is extrapolated from animal testings and from reported accidental exposures in human.

Also, Nazis carried out a large number of human testing on Jews, and this is not limited to toxicity.


How does a chemical "know" which other species are present and where equilibrium is?

What we got here is a dynamic equilibrium, not a static one. For a static one the molecules would need to know when to stop reacting or when to react in a certain way, in a dynamic equilibrium you don't need to know.

Here's a ELI5 explanation of what's happening:

  Let's say you've got a garden with an appletree and hundred of apples
  are on the ground. You don't want them there so you throw them to your
  neighbor. he's an old man and doesn't want them either, so he throws
  them back. You are much faster than him and there are hundreds of
  apples around you, so you don't even need to move to throw them and
  you can throw them very fast. He on the other hand has only a few
  apples laying around and has to move quite far, so he's throwing them
  back much slower. But what happens after time?

  Well, there will be a point where there are more apples on the old
  mans side than on yours, so you have to run a lot to get to those
  apples while the old man has to move only some steps to reach a new
  apple. Eventually it will level out and both of you will throw apples
  at the same rate. That's a dynamic equilibrium. Because there are more
  apples on the old mans side he can throw them back as fast as you can
  throw them to him, even if you are faster.

Now here the chemical explanation:

Reaction rates are dependent on the concentration. High concentration equals fast reaction, low concentration equals slow reaction. At the beginning you got only starting material, it will react fast to the product. The product will react back but there is extremly low concentration so it's slow. But after some time you will reach a point where the concentration of the starting material is so low that the reaction is so slow, that it matches the speed of the back reaction which got faster because you got more and more product. At that time there will equal amounts react in both directions, making it look like the reaction is standing still.

For example: If we got the reaction $\ce{A + B \rightleftharpoons C + D}$ and the reaction rate of the forward reaction is $r_1=k_1[A][B]$ and the backwards rate is $r_2=k_2[C][D]$ then we will reach an equilibrium $r_1=r_2$ and that's the case if $k_1[A][B]=k_2[C][D]$.


Why is M1*V1 = M2*V2 only used for diluting solutions?

As Satwik has stated in the comments, the equation relies on the conservation of moles principle.

  $$\dfrac{mol_1}{V_1} \times {V_1} = \dfrac{mol_2}{V_2} \times {V_2}$$

Based on the above, given a constant volume of a solution, adding more moles of solute will only make the concentration increase, not decrease.  Adding in more solvent is the only answer in that case.


Why is the pH level of carbonation loss decreasing then increases afterwards?

At the beginning, there was titrated mostly just the citric acid and the equilibrium amount of carbonic acid.

After that, carbon dioxide slowly converted to carbonic acid and $\mathrm{pH}$ has been decreasing.

Finally, the long term process took dominance as dissolved carbon dioxide together with carbonic acid were  leaving solution as gaseous carbon dioxide and $\mathrm{pH}$ have been raising.


What word is to chalcogen and pnictogen what halide is to halogen?

From what I could find in a brief search, "pnictide"(this link includes some of the origin for the name) is used, but the corresponding term used for group 16 seems to be "chalcogenide", which, while it can be used to describe all group 16 elements, typically is used in a way that excludes oxides.


What he said
He did not say that the octet rule is never violated.
On the contrary, he suggested that "the maximum number of electrons in the valency orbit can not [sic] exceed eight". (source)
This means, while he denies that the octet rule is violated by molecules like chlorine pentafluoride, he did not challenge whether molecules traditionally labeled as octet-deficient such as beryllium difluoride indeed violates the octet rule.
He argued for the existence of a two-center one-electron bond.
Why his theory was poorly accepted
Preconceptions?
The success of his theory
To quote Wikipedia:

In the 1940s and 1950s, Rundle and Pimentel popularized the idea of the three-center four-electron bond, which is essentially the same concept which Sugden attempted to advance decades earlier; the three-center four-electron bond can be alternatively viewed as consisting of two collinear two-center one-electron bonds, with the remaining two nonbonding electrons localized to the ligands.

Indeed, the three-center-four-electron explains the bonding in the triiodide anion well.
Also, another success is that this theory, unlike the expanded-octet theory, does not rule out the existence of hypervalent second-period element, such as carbon ("Frozen" Transition States: Pentavalent Carbon et al., J. C. Martin, Science, 05 Aug 1983, Vol. 221, Issue 4610, pp. 509-514, DOI: 10.1126/science.221.4610.509).
The problems with this theory
He was reluctant to accept that ionic characters do indeed play a larger in some molecules traditionally labelled as hypervalent, such as the sulfate anion.
The failures of this theory
In the sulfate anion, there is no two-center one-electron bond (Wikipedia):

For sulfuric acid, computational analysis (with natural bond orbitals) confirms a clear positive charge on sulfur (theoretically +2.45) and a low 3d occupancy.

Also, some of the traditionally-labeled hypervalent molecules involve very electronegative elements such as fluorine for sulfur hexafluoride, oxygen for sulfate, and chlorine for phosphorus pentachloride. A better explanation for these molecules would need to invoke ionic characters instead of the covalent singlet linkage.
For sulfur hexafluoride (source):

The sulfur d orbitals are found to contribute very strongly to the binding energy ($250$ kcal/mol) and to have a  total occupancy of around $0.25\mathrm{e}$, the valence electron distribution on sulfur being about $32\%$ in $3\mathrm{s}$, $59\%$ in $3\mathrm{p}$, $8\%$ in $3\mathrm{d}$, and $1\%$ in $4\mathrm{p}$. The occupancy of the two sulfur $3\mathrm{d}_{\sigma}$ orbitals ($0.16\mathrm{e}$), however, is only one-sixth of what would be required for $\mathrm{s}\mathrm{p}^3\mathrm{d}^2$ hybridization, and the energetic contribution of these orbitals is only two to three times larger than that of the sulfur $3\mathrm{d}_{\pi}$ orbitals. The sulfur $\mathrm{d}$ orbitals are important because they allow strong back transfer from the negatively charged fluorine ligands to the strongly positively charged ($+2.9\mathrm{e}$) sulfur, in turn allowing significant contraction of the $\ce{S-F}$ bonds and greatly increased molecular stability.


What is the reason for anomalous expansion of water?

So, the first question to ask would be: how "anomalous" is change in the thermal expansion coefficient?  It turns out that for most real materials the coefficient of thermal expansion is not a constant, and takes on both positive (expansion) and negative (contraction with increasing temperature) values.  For silicon as an example, see ioffe,  Above room temperature Si has a positive (but not constant) CTE, below about 100K silicon's CTE is negative.

The second question is why would this occur?  Taking a liquid or a solid (ignore gases for the moment), an individual atom (or molecule) sits in some kind of potential well which is what is causing the material to be bound together. At absolute zero, these atoms sit at the very bottom of the potential well.  As the temperature increases, heat is added and the atom starts rattling around in the potential well, and is able to explore more of the well.  There is no particularly good reason to believe that the shape of the well, across the entire temperature range of existence of the phase, will be such that a single linear coefficient will describe where the atom will most likely sit in the well at a given temperature. On top of that in a solid is the equilibrium concentration of point defects that is dependent on temperature.  As an example, Google for Simmons and Balluffi and their experiments combining x-ray measurements of the lattice parameter (where the atoms sit in the potential well) and thermal expansion (change in length of a heated bar) to determine the point defect concentrations vs temperature for various solids.


Springmühl indeed reported Anthracene Blue (orig. "Anthracenblau" (German)) from an intermediate with brutto-formula $\ce{C14H7O3}$. It seems like this method only worked well at a lab scale.
According to Springmühl's Lexicon der Farbwaaren und Chemikalienkunde [1, p. 111-112], Anthracene Blue is obtained as a byproduct of oxidation of anthracene to anthraquinone by fuming nitric acid. When gas (hydrogen in original text) evolution stops, resulting solution is neutralized with potash and heated under pressure in a sealed glass tube for 8 hours. Dark-green precipitation is then filtered off, and deep-blue solution containing Anthracene Blue is then treated with sodium chloride to precipitate the product in a form of blue water-soluble powder. But, treated with tannic acid, Anthracene Blue forms dark-blue precipitate and cannot be dissolved neither in water, nor in alcohol; but slightly soluble in acidified solutions.
Original German text:

Springmühl, der Entdecker des Anthracenblau, stellte dasselbe zuerst dar, indem er eine Verbindung von der Zusammensetzung $\ce{C14H7O3}$, welche er bei der Oxydation des Anthracens neben Anthrachinon erhielt, mit rauchender Salpetersäure behandelte und auf das erhaltene Product Wasserstoff in statu nascendi einwirken liess, die resultirende Lösung mit Kali versetzte und unter Druck erhitzte. Es bildete sich eine tiefblaue Färbung in der zugeschmolzenen Glasröhre und nach 8stündiger Erhitzung fand sich ein dunkelgrüner fast schwarzer Niederschlag in derselben, während die Lösung tiefblau gefärbt war und das Anthrancenblau enthielt, es liess der Farbstoff aus filtrirter Lösung sich mit Chlornatrium niederschlagen und mit Gerbsäure konnte der in Wasser wieder gelöste Farbstoff völlig als dunkelblaues Pulver gefällt werden, welches weder in Alkohol noch Wasser, hingegen leicht in Säuren sich löste.

Later synthetic process was changed to adapt it for industrial production. Beilstein catalogue from 1925 lists Anthracenblau-WR as 1,2,4,5,6,8-hexahydroxy-9,10-anthraquinone $\ce{C14H8O8}$ [2, p. 569], which can be obtained from 3-hydroxybenzoic acid with concentrated $\ce{H2SO4}$ in the presence of $\ce{H3BO3}$ at $\pu{260 .. 280^\circ C}$. The product is then poured into water and boiled. Alternatively, 3,5-dihydroxybenzoic acid can be used. There are also plenty other synthetic techniques slightly deviated from the mentioned above.
These days Anthracenblau is also known as commercially available dye Anthracene Blue (S)WR.

Reference

Springmühl, F. Lexicon der Farbwaaren und Chemikalienkunde: practishes Handbuch für Kenntiniss und Prüfung aller in der Industrie und im Gewerbebetrieb verwendeten Farbstoffe und Chemikalien; Leipzig: Gustav Weigel, 1877. (in German, Archive.org)
Beilsteins Handbuch der Organischen Chemie, 4th Ed.; Prager, B., Jacobson, P., Schmidt, P., Stern, D., Eds.; Springer: Berlin, 1925; Vol. 8. (in German, Archive.org)


Why can't Carbon Dioxide be liquid at 1 atm?

This is the phase diagram from a better source:

The pressure axis is the vertical one at the left. One bar (which equals 0.986923 atmospheres) is near the bottom, as shown on the diagram. So carbon dioxide is a gas at all temperatures above −78.5 °C when the pressure is one bar. Solid carbon dioxide (dry ice) sublimes at one bar (about one atmosphere), i.e., goes directly from solid to gas.

If you want liquid carbon dioxide, you need a temperature, pressure pair in the liquid region, e.g., −40.0 °C and 20 bar.


How to work out equations with period 3 oxides?

The first thing to say is that you final equation doesn't balance. In fact it is impossible to balance your equation using hydroxide because you can never conserve charge with only neutral products.

You can write a reaction that forms $\ce{Al(OH)3}$: $$\ce{Al2O3 + 3H2O -> 2Al(OH)3}$$

However, beyond about pH 6 the $\ce{Al(OH)3}$ will start to dissolve to form $\ce{[Al(OH)4]-}$. Once you get to around pH 7.5 all of the solid has essentially dissolved and you are left with solution of $\ce{Na+}$ and $\ce{[Al(OH)4]-}$ (see footnote). So unless you are very sparing with your addition on $\ce{NaOH}$ you will get the tetrahydroxo product in solution.

$$\ce{Al(OH)3 + OH- -> [Al(OH)4]-}$$

Footnote:
The aluminium species present in aqueous solution are strongly dependent on the pH, particularly around the neutral point. There is a nice graph here showing the different species. The notation is $\ce{Al}$ corresponds to $\ce{[Al(H2O)6]^{3+}}$ and then $\ce{AlH_{-n}}$ denotes the loss of n protons from the hexaaqua complex. This was originally found by @Jan for this question.


Can pthalic acid or 1,2-Benzenedimethanol be possible products of Cannizzaro reaction of phthalaldehyde?

The Cannizzaro reaction is the disproportionation of an aldehyde, which do not have an $\alpha$-hydrogen, to an equimolar mixture of corresponding primary alcohol and carboxylic acid salt. For example, original Cannizzaro's experiment:
$$\ce{2C6H5-CHO ->[aq NaOH] C6H5-CH2OH + C6H5-COO^-Na+}$$
Since there is no $\alpha$-hydrogen, the mixture does not undergo an aldol condensation, which is the characteristic of this reaction. The Cannizzaro reaction is an example of a typical redox reaction, which is considered to be one of the most important synthetic reactions of organic chemistry prior to the discovery of $\ce{LiAlH4}$ in 1946 (Ref.1).
The proposed reaction mechanism is as follows (Ref.1):

According to Ref.1, the hydride transfer must be from one molecule of benzaldehyde to the other molecule (when the reaction is performed in $\ce{D2O}$, no deuterated $\ce{CD2-O}$ is found in benzyl alcohol). The formation of benzyl benzoate was suggested and supported (e.g., Ref.2), evidence in Ref.1 doubts this proposal.
The Cannizzaro reaction of phthalaldehyde is studied in aqueous base ($\ce{NaOH}$) to obtain sodium o-hydroxymethylbenzoate as the product. The studies include the kinetics of the reaction, which have been measured over a range of $\ce{NaOH}$ concentration at $\pu{40 ^\circ C}$. This kinetic study proposed the hydration of phthalaldehyde to hemi-ketal monooxoanion is one of the first steps in the mechanism (hemi-ketal formation is depicted in the insert box of above image). The dioxoanion formation is depend on the concentration of $\ce{NaOH}$ solution (Ref.3). The hydride transfer from mono- or di-oxoanion follows, which is the rate determining step (determined by the isotope effect). The observed rate is based on following equation:
$$k_\mathrm{obs} = \frac{a[\ce{OH-}]}{1 + b[\ce{OH-}]}$$
where $a= \pu{(1.38 \pm 0.02)\times 10^{-3} M-1s-1}$ and $b= \pu{0.99 \pm 0.02 M-1}$ (the reaction follows this equation for $\pu{0.05-2.0 M}$ $\ce{NaOH}$). There are no other product has been observed.
I understand your concern of intra- versus inter-molecular hydride transfer. If intermolecular transfer occurs, at least 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid should form as minor components. Since hydride transfer is the rate determining, intramolecular hydride transfer is much faster than that of intermolecular version. Therefore no chance of minor product formation is possible since all other steps are in equilibrium.
Note: I assume your suggestion of forming phthalic acid and catechol as byproduct is mistaken for 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid (phthalic acid). Keep in mind that there is noway of breaking two aromatic-CHO bonds and making two aromatic-O bonds to form catechol molecule.
References:

C. Gardner Swain, Arnet L. Powell, William A. Sheppard, Charles R. Morgan, "Mechanism of the Cannizzaro reaction," J. Am. Chem. Soc. 1979, 101(13), 3576–3583 (https://doi.org/10.1021/ja00507a023).
Arthur Lachman, "The Benzyl Rearrangement v. Cannizzaro's reaction," J. Am. Chem. Soc. 1923, 45(10), 2356–2363 (https://doi.org/10.1021/ja01663a017).
Robert S. McDonald, Christine E. Sibley, "The intramolecular Cannizzaro reaction of phthalaldehyde," Canadian Journal of Chemistry 1981, 59(7), 1061-1067 (https://doi.org/10.1139/v81-157).


How many protons are required to satisfy hydroxyamine's (NH₂OH) lust for protons?

Well, hydroxylamine is technically amphoteric, meaning it can act as either a Bronsted acid or base. The acidic proton is presumably the one on the oxygen (although the resulting salts are apparently explosively unstable), while the nitrogen is the proton acceptor. I wouldn't expect the oxygen to be basic (which, unless I've misunderstood, seems to be what your question implies), at least not in any measurable way or for any appreciable length of time, given hydroxylamine's instability and tendency to spontaneously decompose in myriad ways. When treated with acid, it forms hydroxylammonium salts, which disproportionate upon heating.

Note: this book was my primary reference; it contains a concise but informative summary of the reactions hydroxylamine undergoes.


Why aren't neutralisation reactions reversible?

Neutralization reactions are reversible. In theory, at least, even if not so much in practice, all reactions are reversible.

If you look at things on a molecular scale, there's a principle called microscopic reversibility. In effect, there's no physical reaction mechanism you can create that's a one-way door. Maxwell's demon doesn't exist, and you can't get around that no matter how many steps you chain in whatever order, or which particular reagents you use.

What you can do, though, is set things up so it's exceedingly unlikely that the reaction proceeds in the other direction. That's what's happening with reactions we normally deem "irreversible". There's something about them which makes it difficult, in practice, for products to go back to reactants.

One common feature is the size of the equilibrium constant. If a particular reaction has an equilibrium constant that's really big, the equilibrium for that reaction is going to be vastly tipped to the product side. So much so, that the reactant side of the reaction will not be present in appreciable amounts. (For example, this is why combustion is viewed as irreversible. The reaction is so favorable toward the products that only parts per million or parts per billion of the reactants will be present at equilibrium.)

Another method for promoting "irreversibility" is for the products to partition to another phase. For example, if the reaction forms a gas or a precipitate, that product is removed from the reaction volume, and the concentration of the products drops, pulling the reaction toward the products. Note that even in this multi-stage setup, each step is reversible. That evolved carbon dioxide molecule could re-dissolve in solution, and then react backwards, but it's terribly unlikely for it to do so. You've basically fixed one of the products at a very low concentration, which draws the equilibrium toward the product side.

Precipitates work much the same way, as the activity of the solid is fixed by surface behavior, and does not increase because you increase the amount of solid you add. (The rate of reaction may increase by adding more solid, but the equilibrium of the reaction does not.)

So in your examples, the reason you get hydrolysis or neutralization is due to how you set up the reaction conditions. Have a large excess of HCl? You drive things toward neutralization. Have a way of getting rid of HCl (for example by driving it off as a gas, or reacting it with a base)? You drive things toward hydrolysis.


What is the difference between ideal stages and transfer units in a packed bed gas absorber?

A transfer unit gives the change of composition of one of the phases equal to the average driving force producing the change
Source: http://www.separationprocesses.com/Absorption/GA_Chp04c.htm

So to phrase it differently, for some ideal zone in the absorber that has a particular value to the average mass transfer driving force, there will be a fixed change in the phases present at that ideal zone.
NTU is somewhat analogous to the total number of theoretical stages if the operating and equilibrium line are straight and parallel.
There is no direct physical interpretation of what a transfer unit. If you design any appreciable size packed bed tower, you will find that the vendors of the packing will use a lot of empirical correlations for their own products to determine separation efficiencies and other related parameters.


Formation Of Gemdiol

Carbonyl compounds react with water to give gem diols. This reaction is catalysed by acid. The reaction is reversible reaction.

Gem diols are highly unstable compounds hence equilibrium favours the backward direction. The extent to which an aldehyde or ketone is hydrated depends on the stability of gem diol.

Stability of gem diols depend on the following factors:

(i) Steric hindrance by +I group around α-carbon decreases the stability of gem diols. +I group decreases stability of gem diol and hence decreases extent of hydration.

(ii) Stability of gem diols mainly depends on the presence of –I group on α-carbon. More is the –I power of the group more will be stability of gem diols.

(iii) Intramolecular hydrogen bonding increases stability of gem diols. –I groups present on carbon having gem diol group increases strength of hydrogen bond.

More is the strength of hydrogen bond more will be the stability of gem diol.

Overall stability : (CH3)2C(OH)2 < CH3CCl(OH)2 < CH3CN


How to dissolve grass/leaves without affecting enclosure?

Plant matter is mostly cellulose which is insoluble in most solvents so it is unlikely you can "dissolve" the material.  You can react the material.  Two possibilities are nitric acid and sulfuric acid but there are a number of safety precautions you need to follow.


Why is my sodium acetate not resulting in any crystals?

The solubility of sodium acetate is interesting (that means confusing!). Sodium acetate exists in two forms: anhydrous and trihydrate. When compounds exist in two forms like this, the solubilities, given in grams per 100 mL H20, are predictable: the hydrated form is much more soluble, because it is only partially the compound, and the other part is extra water, the solvent.

However, sodium acetate displays a different behavior:

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

Sometimes different hydration stages of compounds have different temperature ranges of stability and solubility, but the behavior of sodium acetate is unique. The precipitate that forms so impressively upon cooling is probably the trihydrate, because the amount of sodium acetate in solution according to the recipe does not seem (in my quick estimate) to exceed the solubility of the anhydrous compound, and so the solution produced might be metastable. For some length of time. But, we may be able to initiate crystallization by scratching the glass container, or stirring it, or allowing a drop to dry completely to produce some solid, then using that as a seed for the rest of the solution. Or freezing it.

What I have done is to heat some (pure) sodium acetate to form a concentrated solution (calculated to be supersaturated at room temp), then cooled it: no ppt. Then I touched the surface of the solution with my fingertip and BOOM! Crystallization proceeded like a shot, producing a mass of crystals that filled the container and solidified the former liquid mass. (Did I say it was supersaturated?) I was surprised at the mass of crystals compared to the amount of material I had added to the water at the beginning, but did not think about it till now.

It turns out that the solubility curves of trihydrate and anhydrous sodium acetate are quite separated:

https://www.google.com/search?q=solubility+of+sodium+acetate&client=safari&rls=en&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjt-un3m4XlAhXrY98KHQExDiEQ_AUIESgB&biw=1004&bih=846#imgdii=qbOI-c2mtjJb3M:&imgrc=lRafmM5w0nsXhM:

The formation of trihydrate at low temperatures must be a difficult process, and the brown goo obtained here might signify some crystallization inhibitor is present. Maybe a slight excess of NaHCO3 or HCH3CO2 is holding things up - you could try adding a drop of the vinegar, or a few grains of NaHCO3 to see if that changed anything. But If you have a glass rod, scratch the side of the glass container - that works almost all the time.


What is the difference between SBA-15 and MCM-41?

The difference between your two materials is primarily in pore geometry: for MCM-41, the pores are hexagonal; for SBA-15, they are round. Further, there are apparently differences in wall thicknesses.

  Using high-resolution transmission electronic micrograph (HR-TEM) observation, one can clearly see the pore geometry of the MCM-41 and SBA-15 mesoporous silicas to determine that their pore shapes are hexagonal and round, respectively. With the perpendicular orientations of the nanochannels to the electron beam, parallel line images of the (100) and (110) repeating spacings were observed. In the SBA-15 mesoporous silicas, there are byproducts of the granular silica and disordered mesostructures, attributed to the weak hydrogen interactions between Pluronic 123 blockcopolymer and the silica species. There are also many different and significant +π disclination defects in SBA-15 and MCM-41 surfactant-silica composites. The SBA-15 with a thicker silica wall is more stable under irradiation by high-energy electron beams compared to MCM-41, which has thinner wall thickness. Some carbon nanostructure impurities were found in some carbon films on the metal grids.

Reference: Lin, Tang, and Lin. Detailed Structural Characterizations of SBA-15 and MCM-41 Mesoporous Silicas on a High-Resolution Transmission Electron Microscope. Journal of the Chinese Chemical Society 49(6) 981-988 (2002). Abstract here.


Total number of stereoisomers of 1,2-dimethyl-4-(propan-2-ylidene)cyclopentane

Here is a similar molecule to illustrate:

Two of the ring carbons are asymmetric (chiral). This molecule itself (as shown) is chiral. If you change one of the chiral carbons, you get a molecule that is not chiral (mirror plane along the double bond and perpendicular to the ring).


How does the splitting in this molecule take place?

According to given data, you may consider expected peak patterns are not complicated such as $\mathrm{ABX}$ or $\mathrm{A_2B}$  patterns (more reading look here. This is simply a  $\mathrm{A_2X}$  pattern ($\delta_X = \pu{6 ppm}$ and $\delta_A = \pu{1 ppm}$ without geminal coupling of two $\ce{H_b}$ protons), and hence, peak is a $dt$ for $\ce{H_b}$ as depicted below with relative heights of the peaks:

Similarly, $\ce{H_c}$ is also a $dt$ resonance with $J = 1 \text{ & } \pu{8 Hz}$ and $\ce{H_a}$ is a $tt$ with $J = 1 \text{ & } \pu{5 Hz}$. OP should able to figure out the intensities of relevant peaks.

A valuable reference:

Roy H. Bible, Jr., Ph.D., In Interpretation of NMR Spectra: An Empirical Approach; Springer Science: New York, NY, 1965 (https://doi.org/10.1007/978-1-4684-8288-1) (Note: Most relavent to this question is chapter 4 of the book titled "Higher-Order Spin Patterns and Multiple Resonance").


How to compare electrical conductivity of ionic liquids and aqueous electrolytes

No, it is an apple to orange comparison. Ley et al. are changing the frequencies and in the molar conductivity measurements you do not know what was the current frequency. It is usually measured at 1 kHz if I remember correctly. Similarly, S/m cannot be changed to Scm^2/mol. You can indeed calculate the molarity of pure ionic liquids. The molarity of pure water is ~ 55 M. You should know the moles and the volume of the IL, but there is no point in calculating it.
You need to some heavy literature search. People have actually measured molar conductivity of ILs. Also note that many ILs are not water soluble so you have narrow down your choice and goals. Why are you comparing the conductivities to begin with? See this paper "On the concept of ionicity in ionic liquids" in Phys. Chem. Chem. Phys., 2009, 11, 4962–4967.
Also Walden plot might be useful for you.


Is phenolphthalein used in laminar wood construction panels?

Could it be that he meant phenol-formaldehyde resin?

These resins are often used to bond wood chips. They contain unreacted formaldehyde and the resin decomposes slowly, giving off more formaldehyde, which is volatile and very toxic. This is a known health concern. Many companies offer indoor air testing for formaldehyde, you can even buy dubious tests to do it yourself.

My bet would be on such a resin.

There are indeed some resins that use phenolphtalein:

Phenolphtalein might be cancerogenic, even though it was used as a laxative

  Phenolphthalein was commonly used as a laxative for most of the 20th century. The use of phenolphthalein in laxatives has decreased since 1997 when the United States Food and Drug Administration (FDA) proposed to withdraw its classification as an over-the-counter drug (21 CFR, Part 310). Phenolphthalein has been previously evaluated in 2-year carcinogenicity studies by the National Toxicology Program (1996). The major route of human exposure to phenolphthalein is via ingestion, dermal contact, and inhalation of contaminated air originating from process units manufacturing the compound.

In any case, phenolphtalein is a powder that is not very volatile and would not pose a big risk.
Commonly the formaldehyde resin is used for particleboard, I have never heard about the phenolphthalein resin being used for this purpose (but this is far from my area of work)

Of course, there is a simple test. Phenolphtalein changes color in a basic solution and looks like this

Dissolve the material in ethanol and add NaOH (drain cleaner), phenolphtalein will turn fuchsia.


What [if anything] is special about the boiling process of purifying water (excluding temperature)?

There nothing special about boiling that is doing anything.  For example, you can put a beaker of water under vacuum and get the water to boil at room temperature, but that won't kill any bacteria or render any viruses impotent.

It's purely a heat thing.  You want to destroy any pathogens that may cause illness.  Sufficient heat will denature proteins in a virus and kill bacteria, etc.  Boiling simply gives you a visual cue that you've reached a very high temperature (so you don't need a thermometer) and the ~1 minute is just to ensure that you sustain a high temperature long enough to destroy any bacteria or viruses.