As is currently being discussed in this question, I'm not convinced that the methyl radical is planar, but even if not planar, it is certainly very close to planarity.  It is also certainly less pyramidal than the $\ce{.SiH3}$ radical as you state.

The $\ce{.SiH3}$ radical is more pyramidal than the $\ce{.CH3}$ radical due to differences in electronegativity between $\ce{Si}$ (1.9) and $\ce{C}$ (2.5) and hydrogen (2.1) and how these differences play into Bent's rule.

In a nutshell, Bent's rule states that, "atomic s character concentrates in orbitals directed toward electropositive substituents" (ref).

Since hydrogen is more electronegative than silicon it will draw more of the electron denisty in a $\ce{Si-H}$ bond towards its end of the bond (the reverse of what happens in the case of a $\ce{C-H}$ bond where the carbon is more electronegative than hydrogen) decreasing the electron density in the silicon atomic orbitals involved in the bond.

If there is less electron density in the silicon atomic orbital used in the $\ce{Si-H}$ bond, then according to Bent's rule, the silicon atom will hybridize so that less s-character is used in these atomic orbitals that now have a lower electron density, and will instead use that s-character in the orbital containing the single electron and stabilize it.  Increasing the p-character in the $\ce{Si-H}$ bonds and increasing the s-character in the orbital holding the single electron will tend to push the molecule towards an $\ce{sp^3}$-like, pyramidal geometry.

The reverse is true in the case of carbon.  Due to electronegativities, there will be more electron density towards the carbon side of the $\ce{C-H}$ bond (compared to the $\ce{Si-H}$ bond).  Due to this increased electron density around carbon, it is energetically less favorable to remove s-character from the carbon atomic orbitals involved in the $\ce{C-H}$ bonds (and destabilize this increased electron density) and transfer it to the orbital containing the single electron.  As a result the carbon atomic orbitals involved in the $\ce{C-H}$ bonds will have higher s-character compared to the analogous silicon case and $\ce{.CH3}$ will be less pyramidal.


I heard aluminium doesn't produce fumes when burnt but the opposite happened, how?

It doesn't. Consumer aluminium foil often is a thin film on thicker polymer film, it is much more durable this way. The most common polymer for films is likely polyethylene, and when heated it produces nasty mix of pyrolysis and oxidation products. While not extremely toxic, they are definitely irritating and unhealthy.


A good way to think of this is by looking at forming butadiene from the MOs of ethene, which are an in-phase and an out of phase combination of p orbitals. If we looked at all the combinations of these two MOs of ethene, we obtain the MOs for butadiene


What would cause a sudden drop in humidity in my home?

the humidity is caused mainly from sources of water in your home such as plants, pets and of course humans. As you breath out water is coming out of you. And the condensation of water on windows is caused by different temperature outside/inside especially at the interface glass/air.
There is no way that heater can reduce humidity in sealed room (in means of decomposition of water as you suggested). Imho if you have CO alarm you have nothing to be afraid of.


How to decide at which end to start the numbering of the parent chain when the options for locant sets are the same?

If numbering from each end of the parent chain gives the same set of locants, the next criterion stipulates that lowest locants are assigned to the substituent that is cited first as a prefix in the name.

On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule f takes precedence over Rule g. If no decision can be attained by application of P-14.4(f), the lowest set of locants is selected according to P-14.4(g) in their order of their citation in the name.

This means in effect that alphanumerical order is used to establish the numbering when a choice is possible since alphanumerical order is used to establish the order of citation of substituent prefixes in the name.

  P-14.5 ALPHANUMERICAL ORDER

  Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved

  Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.

  (…)

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

For example ‘dibromo’ comes before ‘chloro’ since ‘bromo’ begins with ‘b’, ‘chloro’ begins with ‘c’, and the multiplicative prefix ‘di’ does not change the alphabetical order.

However, the example given in the question does not include simple prefixes but the compound substituents ‘1,1-dichloroethyl’ and ‘2-bromo-1-chloroethyl’. Such compound names are considered to begin with the first letter of their complete name.

  P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Therefore, the correct name for the compound given in the question is 4-(2-bromo-1-chloroethyl)-5-(1,1-dichloroethyl)octane since ‘2-bromo-1-chloroethyl’ starts with ‘b’ and ‘1,1-dichloroethyl’ starts with ‘d’.


Why does soup turn purple when lemon juice and salt are added?

A possible answer:
(1) Common table salt is usually "iodized" - i.e. it contains potassium iodide.
(2) Something in the soup could oxidize the iodide to elemental iodine.
(3) Iodine plus starch (from the potato in the soup) gives a deep blue color.


What does the equilibrium equation look like when calculating the solubility from Ksp?

The reason why $K_{sp}$ does not include $[\ce{NaCl}]$ is because it is solid:

$$\ce{NaCl(s) -> Na+ (aq) + Cl- (aq)}$$

The equilibrium constant $K$ would include $[\ce{NaCl}]$:

$$K=\dfrac{[\ce{Na+}][\ce{Cl-}]}{[\ce{NaCl}]}$$

However, the concentration of sodium chloride in solid sodium chloride is constant, thus we can write an equation for the observed rate constant that is based only on the variable concentrations. $K_{sp}$ is this observed rated constant:

$$K_{sp}=K_{obs}=K[\ce{NaCl}]=[\ce{Na+}][\ce{Cl-}]$$


Anion formation in an alpha particle-neutral atom interaction?

If all you have is an atom and an alpha particle whizzing by to kick an electron out of the atom then yes. But how often do you see an atom sitting out on it's own?

If everything happens as before except that now there are other atoms around, when that electron is knocked loose of the first atom, it will likely come to rest in some other atom thereby creating an anion.


Water passing RO does not have the same chloride concentration.
The raw RO water contains almost no ions, aside of those that passed by leakages of the osmotic membrane.
This disbalances the water natural $\ce{CO2/HCO3-}$ pH buffer, leaving the ionic alkaline buffer part before membrane, so water becomes more acidic with corrosion aggressive $\ce{CO2(aq)}$.


Why aren't there any triangular molecules?

There exists triangular molecule, eg. cyclopropane, but they are highly reactive due to ring strain.

  Ring strain results from a combination of angle strain, conformational strain or Pitzer strain, and transannular strain or van der Waals strain.

C-C-C bond angle in rings should be approximately 109.5° degrees which in case of small rings like cyclopropane, bond angles are 60° whereas tetrahedral 109.5° bond angles are expected. The intense angle strain leads to nonlinear orbital overlap of its sp3 orbitals. Because of the bond's instability, cyclopropane is more reactive than other alkanes. Since any three points make a plane and cyclopropane has only three carbons, cyclopropane is planar.

Due to this strain they exists in higher energy state and therefore Because of their high strain, the heat of combustion for these small rings is elevated.

Kindly see Ring Strain, it will help you to understand about these kinds of molecules.


What is the hybridization of diatomic nitrogen, diatomic anything like oyxgen, bromine; hydrogen and chlorine in HCl?

Hybridisation is something you typically deduce from the geometry of the compound you are observing. It is primarily a mathematical concept.

That said, to form the diatomic molecules you asked about, no hybridisation is needed. In fact, you get much better results from assuming unhybridised atoms and then combining their atomic orbitals to molecular ones. Except for hydrogen (which only has 1σ and 2σ), the molecular orbitals of all these diatomic compounds ($\ce{N2, O2, F2, Cl2, Br2, I2}$) are basically identical save the energy levels of the MOs and their populations. Thus, these would all count as unhybridised compounds.

Also, there is no nead to invoke hybridisation to explain the bonding situation in $\ce{HCl}$ either. Instead agan consider a σ bond formed between hydrogen’s 1s orbital and chlorine’s 3p. All of chlorine’s other orbitals will sit there happily filled and unhybridised.


Please note that the following is not meant to be used as an instruction!

According to an article on Detection of Denatonium Benzoate (Bitrex) Remnants in Noncommercial Alcoholic Beverages by Raman Spectroscopy, published in J. Forensic Sci. 2014, 59, 1358-1363
(DOI), the manufacturers of illegal alcoholic beverages often use sodium hypochlorite ($\ce{NaOCl}$, household bleach) to remove the bitter taste of tax-free technical ethanol.

In the study above, the authors used legally available legal alcoholic beverages (vodka, cognac, etc.), and denatured/de-denatured them under controlled conditions:

  The separate batch of purchased alcoholic beverages was
  denatured with 20 mg/L of denatonium benzoate, after which 1.33 mL/L sodium hypochlorite was added.

Spectroscopical measurement of the samples thus obtained clearly showed degradation products of the bitterant.

While hypochlorite treatment might be suitable to remove the bitter taste, this doesn't imply that the result is healthy or recommended to consume.


Do orbitals exist even when they are occupied? Orbitals are mathematical constructs and as such they exist in a mathematical sense regardless of whether they are occupied or not. Whether you can actually say that orbitals exist in reality is a bit dubious. They certainly provide a good description of the observed electron density and behaviour of atoms but they are still only a model. Like everything in science, we develop a model to describe what we observe but that model is not perfect and does not necessarily represent what actually exists. Particularly on a quantum scale, the notion of well defined objects that exist in space starts to become a bit fuzzy.


Is there a relationship between bond strength and polarity?

In terms of acidity, ionic(polar) bonds dissociate more easily than covalent(polar) because the electrons are shared less equally in the ionic bond leading to easier breaking apart of the molecule. An acid for with an ionic bond will dissociate more easily leading to a stronger pH than an acid with a covalent bond.


Is my electrolysis emitting compounds from additional reactions?

TL;DR: Your suggestion that the problem is "simply bits of $\ce{NaOH}$ being carried into the air by the splashing bubbles" is very likely correct.

You have a very simple system, meaning a minimal number of variables to consider in order to eliminate other possibilities. For example, if you were using $\ce{NaCl}$ as the electrolyte, then chlorine gas could be the irritant. But with $\ce{NaOH}$ as the electrolyte there really is no side reaction that could produce such an irritant.

On the other hand, $\ce{NaOH}$ itself can irritate sensitive tissues like mucous membranes, and even skin for that matter. According to one of your comments, you are using a $\ce{NaOH}$ strength of a bit over 1 molar. Droplets of this solution will eventually be neutralized by $\ce{CO2}$ in the air, but at that strength it should stay around long enough to elicit the symptoms you described.


Exactly how the subscripted numbers are given to the different types of boron sheets (borophenes)?

All the subscripted numbers are to denote different phases.
In the physical sciences, a phase is a region of space (a thermodynamic system), throughout which all physical properties of a material are essentially uniform. Distinct phases may be described as different states of matter such as gas, liquid, solid, plasma or Bose-Einstein condensate.
Distinct phases may also exist within a given state of matter. As shown in the following diagram for iron alloys, several phases exist for both the solid and liquid states.
For Borophene, α, β, χ, δ and ψ are different types of structures based on their coordination number(CN) like,

α (CN = 5,6)
β (CN = 4,5,6)
χ (CN = 4,5)
δ (CN = 3,4,5)
ψ (Single valud CN)

Now each different structures can have more than one phases. When any particular type of structure has more than one phases, those are denoted by numbers.
For example, α has 7 different structures. Now the first one is named α. After that total of 6 phases are found. One of which is a very similar structure with α. Hence it was named α'. The difference between α and α' is that there is no corrugation in α whereas α' has a corrugation resulting in better stability than α. However, the other 5 phases are named accordingly from α₁ to α₅ based on their earlier prediction. You can find the different phase structures of α Borophene from the following image -
Ref:

https://en.wikipedia.org/wiki/Phase_(matter)
Xiaojun Wu, Jun Dai, Yu Zhao, Zhiwen Zhuo, Jinlong Yang, and Xiao Cheng Zeng. Two-Dimensional Boron Monolayer Sheets. ACS Nano 2012, 6, 8, 7443–7453. https://doi.org/10.1021/nn302696v


Why does ligand substitution occur when OH- is a better ligand than NH3?

This is an equilibrium reaction. Using LeChatelier's principle, you can figure you that if you increase the concentration of NH3 sufficiently, you shift the equilibrium so far to the right that you get the hexamine as the main product. At neutral pH in water, the concentration of hydroxide is quite low (100 nanomolar). You would expect this reaction to be pH-dependent, and of course depend on the concentration of ammonia in solution.


Understanding sodium hydroxide drain cleaner's exothermic dissolution of aluminum; does the aluminum need to be oxidized before it dissolves?

If you want to understand the reaction of $\ce{NaOH}$ on Aluminium, you may describe it this way. First, the aluminium reacts spontaneously with water, like sodium or calcium, producing gaseous Hydrogen $\ce{H2}$ and an metallic oxide or hydroxide. With sodium or calcium, the hydroxide produced by this reaction quits the metal surface, and get dissolved, leaving a fresh surface of metal ready to react with more water. The reaction with sodium is:

$$\ce{2 Na + 2 H2O ->   2 NaOH  +  H2}$$

With aluminium, it is not the case. The reaction is:

$$\ce{4Al + 6 H2O -> 2 Al2O3 + 6 H2}$$

But the aluminium oxide is not soluble into water and it makes a waterproof layer on the surface of the metal, which prevents further contact between the metal and water. So the reaction stops, due to this protecting effect of $\ce{Al2O3}$.

There are two ways of allowing the reaction to continue:

Dissolve $\ce{Al2O3}$ as soon as it is produced by adding $\ce{NaOH}$, as @Uhoh tells it in its first equation.
Prevent the $\ce{Al2O3}$ oxide layer from adhering on the metal surface by adding some mercury salt in the water. In this case the mercury ions $\ce{Hg^{2+}}$ may go through the oxide layer, reacts with the metal and produces a mercury deposit which forms an alloy with Aluminium. The reaction is:

$$\ce{2Al + 3 Hg^{2+} -> 2 Al^{3+} + 3Hg}$$

The $\ce{Al2O3}$ layer cannot adhere on the surface of this alloy Al-Hg. So the
 reaction with water continues as long as there is some mercury atoms left on the
 metallic surface.

So, to go back to @Uhoh's question, it is not $\ce{NaOH}$ that acts on Aluminium. It is water. The action of $\ce{NaOH}$ is to remove the oxide layer.


HDAC inhibitor and how it works on a cellular level

Do you have a more specific question? The best I can give you is the mode of action that is generally seen across most biological mechanisms:

An HDAC inhibitor would cause inhibition by binding to HDAC, thus causing a conformational change that disallows the protein from being able to transfer  acetyl-groups to a histone-residue (cysteine I believe?).

Usually, if a protein is something that is abundant in cells, an inhibitors/activators work directly through bonding to the target to cause conformational changes. There is probably also another inhibitor that can bind directly to the HDAC-gene to cease its expression (but I don't see this being beneficial in a growing cell - but maybe in mature cells).


How do the fields of medicinal chemistry, pharmaceutical chemistry, pharmacy, and pharmacology differ?

I'm in a medicinal chemistry department, the floor below is the pharmaceutics department. We are both located in our university's college of pharmacy. The pharmacology department is under the college of medicine in same building as biochemistry, microbiology, etc. There is little interaction between departments here and that's not good.

But we can break the differences in each discipline down into some order based on drug development. Medicinal chemistry and pharmacology come first, with the discovery of molecules that bind to receptors and other proteins and have biological effect. This used to be based almost entirely on natural products, and those still play a large role, but high throughput screening and computer modeling are becoming more important for discovery of drug candidates. While pharmacology focuses on the effects of drugs and other molecules on cells and the body, medicinal chemistry has traditionally focused on the chemistry behind them. A medicinal chemist might look at a natural product with desirable activity, and ask "how can I make this better?" and apply organic synthesis to produce new versions of that drug, with better activity, reduced toxicity, and longer half-life. Of course the biology cannot be ignored, and has become more important in recent decades. Our department used to be pure synthesis, but now only 2 of our 8 labs are pure synthesis, and 2 almost entirely biological.

Once the medicinal chemists have developed a drug candidate to the point they think it can be used in the real world, it moves on to pharmaceutical chemists. They not only have to refine the synthesis of the drug to produce it efficiently and reliably on large scale, they have to develop the formulation the drug is delivered as. If it's a tablet, they have to determine what to make the bulk of the tablet from, how to press the tablets, how to store them, etc. A large part of pharmaceutics is stability, how long can the product sit on a shelf and not degrade? How does temperature and humidity affect degradation? What does it degrade into and are those products dangerous? Pharmaceutics is also concerned with drug absorption and excretion. They often try to develop "extended release" formulations so that a tablet takes a longer time to break down in your gut, releasing the drug over a longer time and keeping an effective concentration of drug in your blood longer, thus requiring fewer pills and less variation in concentration. Pharmaceutics is a lot of organic chemistry, physical chemistry, analytical chemistry, and quality control.

Pharmacy is the end product of all of these. After a drug is developed and approved the pharmacist is responsible for distribution. While doctors prescribe drugs, they usually have very little training in how the drugs work and interact. The pharmacist needs to know more than just how to sell pills, they must have at least a working knowledge of the pharmacology behind the drug, be aware of potential interactions and side effects, and advise the patient on the best practices for using the drug. They must also be aware of the rules and regulations, and make sure the controlled substances under their care are not stolen or lost or abused. Substance abuse is actually the biggest workplace hazard for pharmacists.

I have some knowledge on these things because I'm in a med chem department, my father is a chemist for a pharmaceutical company, and my brother just got his pharmacy license, so we see all parts of the process.


NaOH being added to Formic Acid (Please catch my error!)

It looks like you misused the acid dissociation constant, which is apparently the $K_a=1.80×10^{-4}$ figure you have.

Properly, you should do the following:

1)  Let $x$ be the hydrogen ion concentration $[\ce{H^+}]$ from the formic acid that actually disdociated.

2)  Since you ended up with $0.469 \text{M}$ total acid the un-dissociated acid $[\ce{HCOOH}]=0.469-x$.

3)  The formate ion comes from both the salt ($0.00521 \text{M}$) and the dissociated acid ($x$), so $[\ce{HCOO^-}]=0.00521+x$.

4)  Plug these into the equilibrium relation

$\frac{[\ce{H^+}][\ce{HCOO^-}]}{[\ce{HCOOH}]}=K_a$

and solve for $x$.  The equation reduces down to a quadratic equation so you should have an easy formula for $x=[\ce{H^+}]$.

5)  Get the pH from the result of Step 4.

Good luck!


Terminology can be confusing especially when some terms are only vaguely defined.

"Sugar" is one of those vague terms. There is a whole family of sugars recognised by chemists. In casual use what people usually mean is sucrose, a specific sugar molecule made from two other sugars, fructose and glucose, joined together. this is produced from sugarcane or sugar beet and is the dominant form of sugar seen by non-chemists.

But is sugar the molecule or the crystal? Both. Many pure chemicals form crystals, indeed crystallisation is a very common way of producing purer chemicals. Those crystals consist of a regular 3D array of a single molecule (for a pure chemical). So "sugar" describes both the molecule and the crystals of that molecule which consist of a large number of them arranged in a regular structure.


What is the favoured geometry of chlorine trifluoride?

Short Answer

The structure on the left is "preferred" because the structure on the right cannot exist.  You cannot put 4 electrons in a p-orbital.

Detailed Explanation

In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized.  This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons  or bond to a ligand.  Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.

The drawing on the right has two lone pairs placed in one p-orbital.  A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.

The question really becomes, does $\ce{ClF3}$ prefer to exist with

2 axial fluorines and 1 equatorial fluorine

or

1 axial fluorine and 2 equatorial fluorines?

We can use Bent's rule to answer this question.  Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital).  An electronegative substituent will draw electron density away from the central atom.  Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.

Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals.  Therefore, your structure on the left with 2 axial fluorines and 1 equatorial fluorine correctly depicts the preferred arrangement of fluorines and lone pairs in $\ce{ClF3}$.

A final note on the bonding in $\ce{ClF3}$, the two axial fluorines bond with the chlorine p-orbital to form what is referred to as a hypercoordinated or hypervalent or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons.  See these earlier answers for more detail on hypercoordinate bonding.

Why does F replace an axial bond in $\ce{PCl5}$?
How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?
Hypervalency and the octet rule


How does a spectrophotometer measure absorbance based on the intensity of detected light?

There are two possibilities:

If there is only one optical pathway, you record i) the blank sample for all wavelengths of interest and store the detector's intensity information in function of $\lambda$ in a working memory.  Light passing the blank is then assumed to equate $I_0$. Then you record ii) the wavelength dependent intensities $I_\mathrm{abs}$ with your sample of interest, and report the subsequently corrected data as spectrum $\mathrm{Abs} = f(\lambda{})$.

.or.

Your instrument permits the simultaneous recording of sample and blank to record both intensity data and reporting the already corrected absorption in one run.  Be sure, though, to assert which of the two paths (= two sample holders) is about your sample, and the blank (which may depend on spectrometer and software).


Can a stereospecific / stereoselective reaction produce a racemic mixture of enantiomers?

You haven't mixed up your terms: a stereoselective reaction is defined as one which preferentially forms one stereoisomer over another. Likewise, all stereospecific reactions are also stereoselective, also stereospecificity has stricter criteria then plain old stereoselectivity.

However, you have to remember that the term stereoisomer encapsulates both enantiomers and diastereomers. The hydroboration reaction is stereospecific because reacting one particular diastereomer of the alkene (either cis or trans) leads to one particular diastereomer of the product. It doesn't necessarily lead to a particular enantiomer. Basically, hydroboration leads to the formation of a racemic mixture of a particular diastereomer.

The alkene you have chosen, 2-methylbut-2-ene, is not a good example substrate that can actually show how the reaction is stereospecific. Since one of the alkene carbons is symmetrically substituted, there's no diastereomerism present in the products. If we use (2E)-3-methylpent-2-ene as an example substrate, then it becomes clearer:

Note how neither of the (2R,3R) or the (2S,3S) stereoisomers are formed. However, the two products that are formed are enantiomers of each other.


Does an acetal give a positive Tollens test?

The description of the hydrolysis of an acetal is accurate. The trouble is that the initial protonation event happens to such a small amount that the reaction does not proceed. We can estimate where the equilibrium of that protonation lies by looking at the pKa's of the relevant acids [Evans pKa table]. In basic solution, water (pKa 16) is the strongest acid available to protonate the acetal. The protonated acetal will have a pKa close to protonated methanol (pKa -2). So the equilibrium for that protonation will lie to the water side by approximately $\ce{10^18}$. As Jori is alluding to in the comment, the rate moving forward is going to be exceedingly slow because the concentration of the protonated intermediate will be exceedingly low. This is where the rule of thumb that acetals are stable to base comes from. See also: Why are acetals stable to bases and nucleophiles?

If the Tollen's test was performed under acidic conditions, then you would be correct that the acetal would test positive for an aldehyde. A hemi-acetal would also test positive.


Is there relation between “Plane wave basis” “Correlation-Consistent” basis and “Fourier basis”

This was supposed to be a comment, but there is too much text, so it will be an "answer".

  Is there a relation between "Plane wave basis"
  "Correlation-Consistent" basis and "Fourier basis"?

Well, mathematically, they are entities of the same type that serve for the exact same purpose:

they all are concrete examples of a well-known idea of linear algebra & functional analysis, the idea of a basis;
and they all are used to expand square-integrable functions.

Does this counts as a relation?

  I was looking for 'a kind' of basis (instead of Fourier) for some work
  and I personally gave the name for my basis as "correlation basis".
  When I typed my thoughts in google to my surprise I got the link
  mentioned in my question and I tried to understand at what context it
  is using the basis. From their I tried to relate the three basis to check whether I can use it for my work.

Are you just afraid of a terminological ambiguity out there? If so, don't worry too much. First, only the phrase "correlation-consistent basis", not just "correlation basis" has a particular well established meaning in quantum chemistry community. This is reflected even in the names of correlation-consistent bases that all starts with "cc" prefix. Secondly, "correlation-consistent basis" has a well established meaning only in quantum chemistry community concerning electron correlation and complete basis set extrapolation, so that unless you doing quantum chemistry, you shouldn't be interested in correlation-consistent bases.

Electron correlation is interdependency of the states of individual electrons in a many-electron system. Consider the simplest many-electron system, a two-electron one, and the following two events:

$1$ - the event of finding electron-one at point $\vec{r}_{1}$;
$2$ - the event of finding electron-two at point $\vec{r}_{2}$.

Then mathematically the idea of electron correlation can be simply expressed as the inequality of the joint probability of $1$ and $2$ on the one side and the product of unconditional probabilities of $1$ and $2$ on the other side
$$
    \Pr(1 \cap 2) \neq \Pr(1) \Pr(2) \, .
$$
Similarly, in terms of electron densities, we could indeed write
$$
    \rho_2(\vec{r}_{1}, \vec{r}_{2}) \neq \rho(\vec{r}_{1}) \rho(\vec{r}_{2}) \, ,
$$
where

$\rho(\vec{r}_{1})$ is the probability density of finding electron-one with arbitrary spin at $\vec{r}_{1}$ while having electron-two with arbitrary spin at arbitrary position.
Similarly, $\rho(\vec{r}_{2})$ is the probability density of finding electron-two with arbitrary spin at $\vec{r}_{2}$ while having electron-one with arbitrary spin at arbitrary position.
And $\rho_2$ is the joint probability density of finding a pair of electrons one and two with arbitrary spins simultaneously at $\vec{r}_{1}$ and $\vec{r}_{2}$ respectively.

I think $\vec{r}_{1}$ and $\vec{r}_{2}$ can indeed be thought of as random variables and that essentially electron correlation is about these variables being in a statistical relationship with each other (they are unlikely to be close to each other and likely to be as different as possible). But if you think that a special basis set is needed to account for electron correlation, that is wrong: taking electron correlation in account has very little to do with a basis set choice. There is no need to use a correlation-consistent basis set to treat electron correlation: Pople basis sets can be used to treat it, as well as correlation-consistent one.


Problems with creating sodium hydroxide from sodium (hydrogen) carbonate

You are correct: Taking baking soda and heating it to $850~\mathrm{^\circ C}$ will generate sodium oxide and liberate carbon dioxide and water. To generate sodium hydroxide, all you would need to do is wait for the heated soda to cool down and then add water. Beware! The reaction of sodium oxide and water is strongly exothermic. React the two carefully!

A red-glowing spoon is a certain indicator that the temperature of the flame was larger than $500~\mathrm{^\circ C}$. A blue flame is usually an indicator for complete combustion. In the context of bunsen burners, that is typically larger than $1000~\mathrm{^\circ C}$. I cannot say whether your gas flame reached that temperature or not.

Note that once you have created sodium hydroxide you should handle it carefully. Especially do not attempt drinking it and keep it off your hands. Proper safety equipment (goggles, gloves) would be preferable.

Concerning the industrial process, the question is usually not so much what the problems with a certain method are, but rather what the benefit of a certain method is — especially if that benefit relates to cash. Electrolysis of a sodium chloride solution is a very cheap process with sodium chloride and water both available at almost no cost and electricity also typically cheap. Acquiring sodium carbonate is probably already more expensive than the required sodium chloride and heating to $850~\mathrm{^\circ C}$ is certainly much more expensive than letting a current flow through the solution. Hence why the chloralkali process is preferred in industry.


Which hydrogen does the nucleophile attack in E2 reaction?

Well done to narrow the options down to E2 and SN2 and recognize that E2 is likely favored over SN2.

As the comments suggest, in this case, it doesn't matter which one is deprotonated. The reason is that deprotonation at either beta-position gives the same product. If you are unsure, draw out both options.

In future similar examples, you may come across reactants with multiple possible reactive positions that are not equivalent. For example, consider 3-chlorohexane (instead of 3-chloropentane, as in the original problem). In this case, again there are two possible sites to be deprotonated (Ha and Hb). Looking at the two possible products, both are disubstituted alkenes, so they will be close in stability. There isn't much difference in the groups nearby Ha and Hb, so the kinetics of the deprotonation are probably similar. So which is the "correct" answer? In reality, both are correct. They may not be formed in exactly equal amounts, but they will probably both be formed in substantial amounts.

There are a couple of key lessons here for those studying organic chemistry:

1) Draw out all of the possible products that come from mechanisms that you think make sense. This might be lots of drawing! Nevertheless, it is much easier to realize the differences if everything is drawn out in front of you.

2) Organic chemistry is messy. Often times multiple mechanistic pathways will be viable and multiple products will form. Much of the creativity in organic chemistry is finding ways to ensure that only one mechanism can occur or one (desired) compound is formed as the major product.


Does DMT-Acetate become Polar in H2O, with added NaCl?

Does NaCl cause DMT acetate (or any acetate) to become polar in $\ce{H2O}$?

In aqueous acidic solution, that is in the presence of a sufficient amount of acetic acid, N,N-dimethyltryptamine is protonated. The acetate is the counter ion of the ammonium salt. The compound does not become more polar by adding sodium chloride!

  Will the DMT acetate have migrated into any of these two polar layers?

Are you having two polar layers? I doubt it. You're having a top unpolar layer, the aqueous layer and some gunk.

Take the separated aqueous layer, neutralize carefully, extract with an unpolar solvent of your choice, dry over sodium sulfate, remove the solvent in vacuum. If necessary, recrystallize and determine the yield.

Please keep in mind that DMT is a controlled substance in many countries!


Why is o-nitroaniline not more basic than aniline?

Yes it is true that SIR is observed in o-nitro aniline so it will not show -M effect.

But I think you're confused in the resonance of the lone pair of NH2 group with the benzene ring which in fact occurs in both cases. The thing that makes o-nitro aniline less basic is the -I effect which makes the lone pair on - NH2 group less available for donation .

In fact the complete order of basic strength of nitro substituted anilines is:
[Aniline>m-nitro aniline>p-nitro aniline>o-nitro aniline ]


Why does diazo coupling preferentially occur at the para position?

Both steric and electronic factors play a role in this reaction.  Resonance structures such as those shown below favor electrophilic attack (the diazonium ion is an electrophile) at the ortho and para positions. Since there are twice as many ortho positions as para positions in N,N-dimethylaniline, on a purely statistical basis we might expect a 67:33 ortho/para product distribution if only electronic effects controlled the reaction.  Since diazonium coupling usually produces a majority of the para coupled product, steric factors must be playing a role as well.  The somewhat bulky dimethylamino group does make it more difficult for an electrophile to approach the ortho position and the diazonium ion is relatively large.  These two steric factors combine to reduce the amount of the ortho isomer and the para isomer predominates.


Does the number of H+ ions in solution go up on dilution of a weak acid?

Let's say we have got a solution of an acid with a pK$_a=5$. The resulting pH dependent on the concentration of that acid would be like:

$\hskip2in$

Examples: $0.1~\ce{M}$: pH$=3.0$; $0.001~\ce{M}$: pH$=4.0$; $0.00001~\ce{M}$: pH$=5.2$; $0.0000001~\ce{M}$: pH$=6.8$

Depending on the pH resulting from these example concentrations we can have a look at percentage composition of the whole acid base system. (Degree of association $\alpha'$ in blue, degree of dissociation $\alpha$ in light orange.)

$\hskip2in$

Examples: pH$=3.0$: $\alpha\approx 1.0~\%$; pH$=4.0$: $\alpha\approx 9.5~\%$; pH$=5.2$: $\alpha\approx 61.8~\%$; pH$=6.8$: $\alpha\approx 98.4~\%$

One can see that low acid concentrations lead to higher pH values than higher acid concentrations which automatically result in an increasing degree of dissociation.

So for me this is no big intuition thing but rather a small sequence of actions.

Lets say it in math:

The proton concentration for your system can be described by $$c(\ce{H3O+}) = \sqrt{c_0(\ce{HA})~k_a+k_W}$$ (adding $k_W$ for very low concentrations) and the equation for the degree of dissociation is $$\alpha = \frac{k_a}{k_a+c(\ce{H3O+})}$$ which can be combined to give $$\alpha\left(c_0(\ce{HA})\right)=\frac{k_a}{k_a+\sqrt{c_0(\ce{HA})~k_a+k_W}}$$

One could expand this system a little more (using the exact pH equation including the self-ionization of water) and come up with a thorough scheme like the following.

$\hskip2in$


Why are isomers of parathion less active acetylcholinesterase inhibitors than paraoxon?

Background
Paraoxon is an irreversible inhibitor. It forms an adduct with the catalytic residue Ser203 of AChE, along with release of p-nitrophenol as leaving group. The resulting covalent phosphoester conjugate is quite stable, but also reversible when using an appropriate oxime before the complex ages. See Franklin et al. 2016 for more details.
To facilitate the discussion below, here are the key molecules:

Regarding the OP's questions
Parathion is known to be metabolized to Paraoxon in the liver, likely though the action of cytochrome P450. (Chambers et al., 1991)
A phosphorothioester bond is indeed expected to be more susceptible to hydrolysis than a phosphoroester bond. Such a P-S labile linkage is ~106-fold more labile than P-O at neutral pH, according to Kuimelis-RG and McLaughlin-LW, 1995. However, Xu-Y and Kool-ET, 1998 noted the importance of unfavorable interactions of a studied enzyme with the sulfur atom, “because of the increased bond lengths or relatively poor hydrogen bond accepting ability of the sulfur relative to oxygen.” Indeed the P-S bond length is ~0.4 Å longer than P-O bond. (Knight et al., 1991)
The enzyme’s active site environment, that is beyond how good the leaving group is may help better understand the inhibition results of the molecules listed above.
Structural environment in AChE
The co-crystal structure of AChE with paraoxon (see Franklin reference above) deposited with RCSB as entry 5HF5.pdb provides the context of how the residual adduct binds to the AChE active site. The Franklin paper provides nice descriptive panels of this structure, but I thought I’d prepare the following image to help give more context of this interaction. In the following panel of the co-crystal structure, the residual adduct covalently linked to Ser203 is colored with cyan carbons:

a) The oxygen of P=O accepts a Hydrogen bond from multiple amides. Given that a sulfur atom does not form as good Hydrogen bonds as an oxygen, then this enzymatic environment will not provide as good stabilization to a sulfur as it does to an oxygen. In addition, as a sulfur atom has a larger radius than an oxygen atom, there may not be enough room for a sulfur atom. These two factors may begin to explain why parathione is not as good a substrate as paraoxon.
b) One of the ethoxy oxygen accepts a hydrogen bond from His447. Having a sulfur atom at that position, such as with Compound II, would bring it too close to the imidazole, leading to steric hindrance. In addition, such a sulfur atom would not form as a productive hydrogen bond as the oxygen. Similarly, the other ethoxy forms close van der Waals interactions with Phe297 and Trp236, there may not be enough room to accommodate a larger sulfur atom. This may begin to explain why Compound II is not as good as paraoxon.
c) Similar to the above, the environment may be too restrictive to accommodate Compound III for the productive formation of an adduct.
In summary, the immediate environment of the active site may be more optimized for paraoxon than the thiol analogs; this may help explain the published SAR of the active isomers of parathion.
Edited for clarity.


There are a number of factors that can influence the rate of an $\mathrm{S_{N}2}$ reaction.  Solvent, leaving group stability, attacking group nucleophilicity, steric factors and electronic factors.

In the series of compounds you've presented, all of these parameters are held constant except for the steric and electronic factors.  Considering only steric factors for a moment we would order the compounds as P>S~R>Q.  But now let's consider electronic factors.

Considering the allyl system in compound R, the transition state looks something like this

image source

The p orbitals in the adjacent double bond overlap with the p orbital at the $\mathrm{S_{N}2}$ reaction center (remember that $\mathrm{S_{N}2}$ transition states are roughly $\ce{sp^2}$ hybridized) and stabilizes the transition state through resonance delocalization.  Note that the transition state is electron rich, there is a full unit of negative charge in the transition state.

When a carbonyl is placed next to the $\mathrm{S_{N}2}$ reaction center, we replace the $\ce{C=C}$ double bond in the above drawing with a $\ce{C=O}$ double bond.  Again there is overlap and resonance stabilization of the transition state.  But in the carbonyl case the stabilization is even greater than with the $\ce{C=C}$ double bond because the carbonyl carbon is positively polarized and inductively stabilizes the electron rich (remember that negative charge) transition state even more.

image source

While we would expect the allyl case (R) to be faster than the ethyl case, in your series of compounds apparently it is not faster than the methyl case.

(Note: often the allyl system will react a bit faster than the methyl analogue, but the two rates are usually close)

However, the additional inductive stabilization from the carbonyl allows it to react faster than the methyl compound so our final reactivity order becomes S>P>R>Q.


Metal and reactions with different temperatures of water

You have two reactions, the metal reacting with water and the hydroxide decomposing:

$\ce{Mg + 2 H_2O \rightarrow Mg(OH)_2 + 2 H_2}$

$\ce{Mg(OH)_2 \rightarrow MgO + H_2O}$

The second reaction, for magnesium, sets in around 350°C.  Below that you can get the hydroxide, above it any hydroxide decomposes and you have the oxide.

Heavier alkaline earth metals, and all alkali metals, have hydroxides stable to higher temperatures, making it more difficult to "direct" the reaction to oxide formation.  In contrast, most transition metals lack low-temperature reactivity with water and their hydroxides decompose easily, so they can give only the oxide.


Determining the Average Bond Enthalpy for the C-F bond

Your approach to use Hess's Law is reasonable!

\[\Delta H_r = -680 - (715 + 2\cdot158) = -1711\ \mathrm{kJ\cdot mol^{-1}}\]

That's the enthalpy for $\ce{CF4}$ - a molecule with four $\ce{C-F}$ bonds.

The average $\ce{C-F}$ bond enthalpy is smaller:

\[\frac{1711}{4}\ \mathrm{kJ\cdot mol^{-1}} \approx  427\ \mathrm{kJ\cdot mol^{-1}}\]


Why do these two IUPAC naming examples not begin numbering the chain closest to the alkyne?

The oversimplified statement “The chain is numbered from the end closest to the triple bond.” applies to simple alkynes with one triple bond. It is not necessarily correct for compounds that include several multiple bonds or other characteristic groups.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), low locants are assigned first to the principal characteristic group that is expressed as a suffix and then to multiple bonds.

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (…)

Therefore, the correct name for the second example given in the question is ‘but-3-yn-2-ol’ (not ‘but-1-yn-3-ol’) since a low locant is given first to the suffix ‘-ol’.

If the compound contains several multiple bonds (double or triple bonds), low locants are given to multiple bonds as a set. If there is still a choice, low locants are given to the double bonds.

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

  (…)

Therefore, the correct name for the first example given in the question is ‘2-methylpent-1-en-3-yne’ (not ‘4-methylpent-4-en-2-yne’) since the locant set ‘1,3’ for the multiple bonds is lower than ‘2,4’.


Why does my filtered water turn thick and almost slimy when refrigerated?

Your setup does not effectively filter nutrients. Nutrients can be problematic in climates like Charleston, SC which are humid.  Humid climates plus a nutrient content in the water create the potential for fungal growth.

Part of living in humid climates is dealing with fungal growth. Fungal growth starts from spores (capable of growing in water) which are probably living in the refrigerator.  Disinfecting the refrigerator should be done by the instructions in the manual which came with the refrigerator.

This should go without saying, the water with the starchy taste should probably be thrown out. Any drinking water left to sit has the potential to grow micro-organisms, so storing water in a refrigerator is a bad idea. Try making ice cubes in the freezer and adding it to water if you want cold water.

In humid climates, anything stored in the refrigerator should be sealed so spores cannot grow.  The refrigerator must be cleaned regularly.

Filtering water on an as-need basis may be best with your setup because a carbon filter is effective at removing anything biological.  Personally, I use a pitcher because I can refill it as needed, let the whole setup dry periodically, and easily throw out saturated cartridges.


What are pharmaceutical uses for Vanillin?

To my knowledge, vanillin nowadays does not have any pharmaceutical uses.

But according to Elisabeth Vaupel in Pharmazeutische Zeitung online, 2002, 38 (SRC), Michael Bernhard Valentini wrote 1714 in his book Museum Museorum, oder vollständige Schau-Bühne aller Materialien und Specereyen, nebst deren natürlichen Beschreibung:

  Sehr erwärmende und zerteilende, anbei aber auch stärkende Kraft, womit sie den Magen stärken, die Winde zerteilen und dem Gehirn, der Mutter und anderen nervösen Gliedern sehr gut tun. Sie treiben den Harn, befördern die monatliche Reinigung, natürliche Geburt und Schwierungen: Treiben auch die Nachgeburt und tote Kinder fort und kommen also dem weiblichen Geschlecht in ihren meisten Krankheiten wohl zu pass. Ingleichen werden sie gegen die erstarrend-machenden giftigen Bisse and andere dergleichen giftige Sachen gebraucht [...] Am meisten aber werden die Vanillen zu Verfertigung der Chocolaten gebraucht, welche sie anmutiger und kräftiger machen. Die Tabaksbrüder brauchen sie auch, den Tabak wohlriechend zu machen.

As a starting material in synthesis, vanillin can be converted to MMDA in 5 steps.


I would express equivalent weight $M_\mathrm{eq}(\ce{E})$ of unknown element $\ce{E}$ using the number of equivalents $x$ and corresponding molecular weights $M$:

\begin{align}
M(\ce{ECl_x}) &= x \cdot M_\mathrm{eq}(\ce{E}) + x \cdot M(\ce{Cl}) \tag{1} \\ \to \quad M_\mathrm{eq}(\ce{E}) &= \frac{M(\ce{ECl_x}) - x \cdot M(\ce{Cl})}{x} \tag{1a}
\end{align}

To find $x$, let's apply to the balanced reaction, assuming it is complete and there is no chlorine left:

$$\ce{E (s) + x/2 Cl2 (g) -> ECl_x (g)}$$

The amounts $n$ of gaseous products are bound by stoichiometry:

$$n(\ce{Cl2}) = \frac{x}{2}n(\ce{ECl_x}) \tag{2}$$

At constant temperature and pressure $n_i V_i = \mathrm{const}$. The problem says the volume of the system reduces by $1/3$, which means:

\begin{align}
V(\ce{Cl2}) &= \frac{3}{2}V(\ce{ECl_x}) \tag{3} \\
\to \quad n(\ce{Cl2}) &= \frac{3}{2}n(\ce{ECl_x}) \tag{3a}
\end{align}

Now, comparing/equating (2) and (3a) it's easy to see that $x = 3$:

$$\frac{x}{2} = \frac{3}{2} \quad \to \quad x = 3$$

You already correctly determined molecular weight from vapor density $\rho_\mathrm{v}$:

$$M(\ce{ECl_x}) = \rho_\mathrm{v} \cdot M(\ce{H2}) = 68.75 \cdot \pu{2.016 g mol-1} = \pu{138.60 g mol-1} \tag{4}$$

All that's left now is to do the math with (1a):

$$M_\mathrm{eq}(\ce{E}) = \frac{\pu{138.60 g mol-1} - 3 \cdot \pu{35.45 g mol-1}}{\pu{3 equiv}} = \pu{10.75 g mol-1 equiv-1}$$

I'm not sure this makes any sense from the chemistry prospective as this would suggest that element $\ce{E}$ is sulfur and so the gaseous product is $\ce{SCl3}$. I would've expected the compound to be something like $\ce{BCl3}$, but in this case the vapor density should've been $58.12$.


Non-wetting fluids for precise volume measurements

For a very precise volume measurement, you might want to consider using mercury because of its high density. (You would actually measure the weight of the fluid.)


Most stable conformational isomer of 3-methoxycyclohexan-1-ol

Sometimes, the intuitive rule of thumb that favours conformations with $\ce{OH}$ substituents in equatorial positions miserably fails.

For 1, a cyclization product of geranyl acetate, one might assume that the favoured conformation in solution and in the crystal is 1eq. However, this is not the case! Both NMR spectroscopy and X-ray crystallography indicate that 1ax is the actual conformation.

In the case of 1, the reason most likely isn't a (stabilizing) intramolecular hydrogen bond that would form a seven-membered ring, but the sterical interaction of the methyl groups. The unfavourable interaction of the 1,3-diaxial methyl groups in 1eq is avoided in the conformer with the axial $\ce{OH}$ group.


Forming hydrogen sulfide from electrolysis of sodium thiosulfate

Yup, you're forming sulfide.

Render the thiosulfate ion as $(\ce {S-}-\ce {SO3-})$.  Then at the cathode:

$(\ce {S-}-\ce {SO3-})+2e^- \rightarrow \ce {S^{2-}}+\ce {SO3^{2-}}$

Both products are basic, with sulfide ion being strongly so,  and they will react further with protons; so the sulfide ion gives some hydrogen sulfide.  Sulfide ions can also combine with silver ions, from the anode, giving a black precipitate.  The other product, sulfite ion, can be reduced further, possibly forming still more sulfide ion/hydrogen sulfide.

So the reaction is not safe.  But kudos for using a fume hood.


Electrolytic conductivity vs metallic conductivity in a galvanic cell

The resistance you'll find in your system basically corresponds to that of the solution (plus some charge-transfer resistance if your kinetics are somewhat sluggish).

Remember that for current to circulate, you need a closed loop; so the movement of electrons from anode to cathode is only part of the big picture. An equal ammount of current will be transported in the solution by ions approaching/leaving the electrode's surface. So, if you want to minimize your resistance effects, two intuitive approaches come to mind: i) make your ions move faster and/or ii) have a lot of ions moving at the same time. That means that the diffusion coefficient and the concentration of your ions (hence, the molar conductivity) will have direct effect on the solution resistance.

Since metals/alloys are, normally, excellent electronic conductors when compared to electrolytes (which, in turn, have ions as charge carriers), the overall cell resistance is pretty much due to the latter.


F-center defects in crystals are due to unpaired electrons and can be caused by ionizing radiation; these defects can accumulate over thousands of years. This can be used to find the age of minerals with retroactive inclusions, or to intentionally color gems.

An electron gun, as in a cathode ray tube (CRT), will also do the job. Older televisions had CRT's, and to prevent electrons building up on the front of the tube, they were siphoned off to ground. "Static" electricity on the faceplate of a CRT can be felt; it is due to the build-up of electrons.

Of course, one could question if electrons in the above examples are bound to a particular atom, are interstitial, or are free to move, but as mentioned above by @MaxW, an ionized gas (plasma) does have particles with excess electrons, as well as those with a deficiency. In interstellar space, an ion could last for years.

Also, radical anions are easily produced, for example the reduction of naphthalene by an alkali metal.


Reaction mechanisms for organohalides from alcohols

Nucleophilic substitutions, regardless whether proceeding via $S_N1$ or $S_N2$ require a decent leaving group. $\ce{HO-}$, however, is a lousy leaving group!

To overcome this, several methods exist:

Perform the reaction under acidic conditions, where the alkanol is protonated. Your leaving group now is $\ce{H2O}$. In addition, the corresponding anion of sulfuric acid as a bad nucleophile.
React the alkanol with an acid chloride to form an ester and replace the ester group with a halide. Consider $\ce{PCl5}$ to be an acid chloride of phosphoric acid.
Form the acid halide in situ by reaction of phosphorous and iodine.


Comparing acidic strengths between benzylammonium ion and phenol

The acid strength of each compound can be explained, but the acidity order is much more difficult to compare, because the two compounds are only remotely connected. It is misleading to conclude that the mere presence of a phenyl group somehow connects these molecules. The similarity of the pK$_a$s is likely a coincidence.
The question needs a clear definition of pK$_a$. We can ask what the pK$_a$ of phenol is, or the pK$_b$ of phenoxide. When we ask for the pK$_a$ of benzyl amine, we want the pK$_a$ of benzylammonium ion, not the pK$_a$ of benzyl amine going to C$_6$H$_5$CH$_2$NH$^-$. There is a bit of flexibility (or sloppiness) in the terminology here, and also a conflict in the values presented: the question quotes 9.33 for benzylammonium ion (I found 9.34 - no big deal), but the comment by the OP gives 8.82 for benzyl amine. The values should be identical because we know what the OP means. It appears that 8.82 is incorrect.
The difference between these molecules is far more important than any similarities. For example, one is a neutral molecule that you can put into a bottle with 6 x 10$^{23}$ more; the other is a charged piece of a molecule that you can hardly bring next to another one.
The phenyl ring has an enormous effect on the acidity of phenol, due to the resonance effect. It stabilizes the oxygen anion by more than 5 orders of magnitude, compared to, e.g., methanol (pK$_a$ = 15.5) or benzyl alcohol (15.4) or ethanol (15.9).
(Interesting that phenylethanol (pK$_a$ = 14.81) is more acidic than benzyl alcohol. This suggests that the methyl group in ethanol is more electron-donating than the benzyl group in benzyl alcohol; this agrees with the discussion in #3 below, for phenethylammonium ion.)
The phenyl ring has a much smaller effect on the amine molecule-ion; no resonance, just inductive. The pK$_a$ of benzylammonium ion (9.34) can be compared with, e.g.:

methylammonium ion (pK$_a$ = 10.66) Replacing the phenyl of benzylammonium ion with H reduces the K$_a$ by a factor of 21.

This is equivalent to saying that benzyl amine is less basic (pK$_b$ = 4.66) than methylamine (pK$_b$ = 3.34); therefore the protonated benzyl amine will give up an added proton more readily than methylamine. We can conclude that methyl is more electron-donating than benzyl.

ethylammonium ion (pK$_a$ = 10.71) Replacing the phenyl of benzylammonium ion with a methyl group reduces the K$_a$ by a factor of 23. Thus, ethylamine, like methylamine, is more basic than benzyl amine because methyl is more electron-donating than phenyl. (But we knew that.)

Phenethylammonium ion has a pK$_a$ = 9.73. Putting another CH$_2$ group between phenyl and nitrogen reduces K$_a$ by a factor of 2.5. This indicates that a benzyl group is less electron-donating than a methyl or ethyl group, in agreement with the comparison of the inductive effect between benzylammonium vs methyl- or ethylammonium ions and benzyl alcohol vs phenylethanol.

Benzyl amine and phenol are the neutral actors in this drama, and might react nicely in a one-to-one ratio. What would be the pH in solution? The graph shows the ratio of ion concentration to neutral molecule concentration for phenol and benzyl amine, separately, over the pH range of 5 to 9. They show contrasting slopes which cross at pH 7.33. At this pH, the ionic fraction of each compound is 0.21% of the whole - it doesn’t matter what concentration you started with if everything is soluble (benzyl amine is very soluble, but phenol, mp 43 C, is soluble in water slightly less than ~0.1 M in water at room temperature). Just add the same number of moles of each neutral compound to get 0.21% reaction and pH 7.33. What a coincidence! Practically the same pH as pure water.


Why Does an Area in Phase Diagram Considered as Having Degree Of Freedom=1?

The Gibbs phase rule is $f = c - p + 2$, where $f$ is the degrees of freedom, $c$ is the number of components, and $p$ is the number of phases.  Your diagram says $P = \mathrm{cst}$, which I interpret to mean that pressure is constant.  That specification lowers the number of degrees of freedom by one, so $f = c - p + 1$.

Inside the "center" region, liquid and vapor phases coexist.  Thus there are two phases, and (as with every point on this particular diagram) two components, thus $f = 2 - 2 + 1 = 1$.

Your question is why.

  Aren't we able to change the concentration and the temperature at the same[ ]time without changing the [p]hase?

Essentially, no, you can't do that.  Inside the center area, liquid and vapor coexist at equilibrium.  And -- this is the key point -- that equilibrium is another constraint that lowers the degree of freedom.  The compositions of the both the liquid and the phase vapor are not independent of temperature, given the constraint that the liquid is in equilibrium with the vapor.

The composition of the liquid phase will be given by the lower curve.  And the composition of the vapor phase is given by the upper curve.

Temperature as the single degree of freedom

So if you decide to use temperature as your (first and only) degree of freedom, then the composition of both the liquid and the vapor is fixed.  Say you pick an arbitrary temperature $T_{arb}$, which must obviously be between $T_{bB}$ and $T_{bA}$.  This temperature could be represented by a horizontal line on your chart.

The composition of the liquid phase is given by where that horizontal line for $T_{arb}$ intersects the lower curve on your chart.  And the composition of the vapor phase is given by where that same line intersects the upper curve.

Thus now that we've specified temperature, the compositions of both phases is also specified, and (given that pressure is constant) thus the system is fully specified.  No more degrees of freedom.

(Liquid) composition as the single degree of freedom

Instead of picking a temperature then let's say you pick an arbitrary liquid composition $x_{Aarb}$.  Then, given that vapor and liquid are in equilibrium, then the temperature is specified.  Your specified $x_{Aarb}$ is a vertical line on your chart.  The temperature of the system is given by the value of the ordinate where your $x_{Aarb}$ line intersects the lower curve on the graph.

Now that you know temperature, you can find the corresponding $y_{Aarb}$ by drawing a horizontal line at the newly identified temperature and finding where this horizontal line intersects the upper curve.

Extensive vs. intensive properties

The goal of the Gibbs phase rule is to describe all the intensive degrees of freedom of a system.  It doesn't tell you about extensive degrees of freedom.  The extensive degrees of freedom tell you "how much" of the system you have.  In this system, the extensive degrees of freedom are the absolute amounts of each phase.  (The compositions of each phase are intensive properties.)  In general for $p$ phases there are $p$ extensive degrees of freedom.

As a simple example, consider a single component system (e.g. pure water) at a constant pressure and in equilibrium with its vapor.  The Gibbs phase rule says $f = c - p + 1 = 1 - 2 + 1 = 0$, i.e. that there are no degrees of freedom.  But there can be any amount of water, from one picomole to one examole, and the equation is still true.  Since there are two phases, there can be one picomole of steam and one examole of liquid water, or vice versa.  The Gibbs phase rule isn't going to tell us how much of each phase we have.

The same idea holds in the two-component case that you are asking about.  The total amounts of liquid and vapor can vary in the center region.  The relative abundance of each phase would be given by the "lever rule" as described in this nice electronic lecture note from the University of Omaha.


How to pronounce "potassium manganate(VII)"?

Simple: Just read the roman numerals as a number.

Copper-two sulphate
Potassium manganate-seven
Hexaaquacopper-two.

For negative oxidation numbers use a minus:

Tetracarbonylferrate-minus-two ($\ce{[Fe(CO)4]^2-}$)


How to predict amine donor in biochemical reactions (whether glutamine or aspartate)?

There are several very general concepts that you have to know about in order to see a pattern in biochemical reactions driven by enzymes. A quick answer is given at the very end. Amongst many concepts to understand, two main concepts are as following:
Enzyme are classified with a specific enzyme commission (EC) number, which is a type of ID number based on catalytic function. The first number groups the enzyme into an overall group. In example, enzymes that can transfer functional groups are placed in EC.2. Within EC.2 you can then subdivide into more specific functions, e.g. transaminases that transfer nitrogen containing groups have the the EC number of 2.6, and within this group again you can find enzyme that specifically transfer amine groups. Those enzymes will then have the EC number of 2.6.1 (aminotransferases), and you can then further specify into families, e.g. aspartate aminotransferases EC.2.6.1.1.
Enzymes folds within specific functional families is evolutionary conserved and highly specific. The three-dimensional shape of enzymes within a specific family, e.g. aspartate transaminase; EC 2.6.1.1, is very conserved (the amino acids sequence identity will also be very high)
and enzymes that are grouped into the same family will have highly similar overall structure and thus very similar mechanisms of catalysing reactions. In contrast, enzymes in completely different families, with different foldes and different amino-acid sequence will have completely different catalytic mechanisms.
To predict how some enzyme will catalyse a reaction, you need to find what enzyme it is related to. Here it will also be important to remember that structure is more conserved than sequence. A good starting point is to do a BLAST search with the protein sequence that you are investigating, note the you could also do a blast search against the PDB database to find related published protein structures.
Continuing with the exsample of amino-transferases, you can even subdivide the enzymes into where the catalytic reaction will take place. In example, you have alpha-aminotransferases or omega-aminotransferases, depending on the location on the substrate that the amino will we translocated to and from.
To answer your question:
Now that we covered some very general concepts about enzymes, it is important to know that their structure and sequence are specifically designed to accommodate a particular substrate. In other words, some enzymes will take aspartate as substrate and donate its amine group - while other type of enzymes will take glutamine as substate - and donate its amine group.

As a side-note. In proteins, the amine-group in aspartate would be part of the protein backbone (amide bond / peptide bond), and the side-chain (no amine group) would therefore be the only accessible part of this amino acid. It is therefore a good idea to specify if amino acids are substrates or catalytic residues (i.e. part of the enzyme) when asking about biochemical reactions.


SMILES doesn't define a particular order in which the atoms show up in the string. Hence the position in the molecule where software starts printing will determine the string. E.g. for CO2:

Ketcher prints C(=O)=O (starts with C and considers one of the O's a branch)
MarvinJS prints O=C=O (starts iteration with O, hence no branching)

Both of these representations are valid, they simply start with different elements. Each tool would recognize SMILES string generated by another tool, so there's no need to convert between strings created by different software.
In order to make different tools generate the same string a so called Canonical SMILES was developed. Not every tool can generate it, but if they do - they'll spit out the same string (in theory).


What is the oxidation state of bromine in BrO3?

Bretherick’s Handbook of Reactive Chemical Hazards lists bromine trioxide $\ce{BrO3}$ structure as the one composed of cationic and anionic parts [1, p. 116]:

The solid produced at $\pu{—5 °C}$ by interaction of bromine and ozone is only stable at $\pu{—80 °C}$ or in presence of ozone, and decomposition may be violently explosive in presence of trace impurities [...]. The structure may be the dimeric bromyl perbromate, analogous to $\ce{Cl2O6}$ [2].

so that formula written as $\ce{BrO3}$ is fine being a formula unit, but more correctly bromine trioxide should be written as $\ce{[BrO2]+[BrO4]-}$, implying that bromine is in two oxidation states: $+5$ and $+7$.
References

Bretherick, L.; Urben, P. G.; Pitt, M. J. Bretherick’s Handbook of Reactive Chemical Hazards, 7th ed.; Elsevier: Amsterdam, The Netherlands ; Boston, MA, 2007. ISBN 978-0-12-373945-2.
Pflugmacher, A.; Rabben, H.-J.; Dahmen, H. Zur Kenntnis der Bromoxyde. Über das Bromtrioxyd. Zeitschrift für anorganische und allgemeine Chemie 1955, 279 (5–6), 313–320. https://doi.org/10.1002/zaac.19552790507.


Calculating molar solubility of calcium hydroxide from solubility product

A cursory search suggests the solubilities in g/100 mL on the wikipedia are closer to a consensus, for what it's worth:

solubility 0.2% ($0^oC$):
https://www.cdc.gov/niosh/npg/npgd0092.html
solubility 0.159 g/100 mL
(0.16% by weight) ($25^oC$):
https://www.mdpi.com/1996-1944/10/10/1219/pdf
the value you reported from the National Lime association.

These are reasonably consistent with the values in

https://en.wikipedia.org/wiki/Solubility_table#C

but not with the $K_{sp}=5.5 \times 10^{-6} M^3$ on the wiki page.

Skoog, West and Holler's Analytical Chemistry textbook (7th Edition) reports $K_{sp} = 6.5 \times 10^{-6} M^3$ at 25$^oC$. They provide a ref:

  most of the data were taken from A.E. Martell and R. M. Smith, Critical Stability Constants, Vol. 3-6.

Note the apparent inconsistency recurs (I called this discrepancy an "error" before but am no longer so sure):

http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
http://www5.csudh.edu/oliver/chemdata/data-ksp.htm

It reminds you that you should generally reference your data sources if you want to be helpful to fellow researchers.

The value for $K_{sp}$ in the wikipedia (which is not referenced as far as I could tell) and in a highly respected textbook which references its source is smaller by roughly an order of magnitude compared to values derived using standard equations from reported solubilities in a number of other sources noted above. This inconsistency may be because some of the reported data is for hydrates with higher Mw. Not sure how you could explain such a discrepancy otherwise. If the $K_{sp}$ were much smaller you could argue that the dissociation constant of water affected the solubility.

My current guess is that there is a reasonable explanation for the discrepancy (ie it has a chemical or physical explanation and is not due to human error). It is unfortunate that the solution is not more obvious given that $\ce{Ca(OH)2}$ is a recurrent example in problems (textbook calculations or experiments) teaching solubility to undergraduates.

See also:

Solubility of calcium hydroxide determined by titration


Having problems running DFTB using dftb.org parameter files on Gauss 09 (G09)?

First, at least for C.01, dftb=readparameters, in the route section, does not work dftb=read does but is not necessary.

Skip a line after molecular descritption and include your parameter files there. Make sure you include all pair interaction files, not just bonded atoms.  It should look something like this:

@C:/G09W/C-C_mio.skf
@C:/G09W/C-H_mio.skf
@C:/G09W/H-C_mio.skf
@C:/G09W/H-H_mio.skf

(I included the mio because we are running more than one parameter set, not a requirement just a name)

The real glitch comes from my final hint.  The DFTB parameters generally have more than 500 points, but the first line in the code is 500 (not 519, or 520, etc).  G09 Rev. D.01 doesn't care about this, B.01 and C.01 do.  You simply need to change 500 ro reflect the correct amount of points and add in the correct order for your atomic numbers (this part is explained on Gauss' website) and you are good to go.

Hopefully this helps anyone trying to run DFTB on G09 for the first time, and helps save some time.


You will be better off if you join a real lab. Researchers can help you get started in a safe manner. Chemistry experiments killed many professionals. Here is a list of things you would need.

A chemistry lab needs at least this:
Hood with 650 cubic feet per minute air flow. You CANNOT make it with a small pump.
Water.
Chemistry waste disposal. Flushing it down the drain is not a good approach.
Freezer and fridge for chemicals only.
Fire extinguisher.
Rotovap. Basic one costs 5000 dollars.
Source of vacuum.
Glassware.
Emergency shower.
Analytical instruments. NMR costs 100 000 dollars to buy, plus you will need to refill liquid helium once in a while. You can also ask chemistry department to run you samples for 10-20 dollars per piece.

Lab rules strictly prohibit working alone. In case of emergency someone should be around to save you.

Bottom line: you will poison yourself without a hood and set garage on fire. Find a university professor and ask to join their lab. If you show decent knowledge of chemistry they might agree to help you.


Bond Angles of Hypofluorous Acid - Gas and Solid Phases

It is due to hydrogen bonding in the solid but not gas phase, in analogy to what occurs for the water bond angle.

From a VSPER point of view, the oxygen atom of one molecule would form a hydrogen bond with the hydrogen of another molecule, decreasing the repulsion of the lone pairs to be closer to that of the covalently bonded pairs, so the geometry would be closer to tetraherdral in the solid phase than in the gas phase.


Do chemists refer to water as "dihydrogen monoxide"?

No, it's not. The "dihydrogen monoxide" name is used as part of a hoax. In the scientific community, there are chemical names for water, and which one is used in the literature generally depends on how it interacts with something else (hydroxic acid and hydrogen hydroxide were two I heard most often in acid-base reactions).
IUPAC, the standards committee that sets standard names for chemical structures, suggests "oxidane" as a starting point for the construction of other names for chemicals that are derived from water. However, they do not suggest "oxidane" itself be used to refer to plain water.
Most chemists would use "water", even when writing scientific papers.


In a typical covalent bond, such as is found between carbon and hydrogen in
methane, each atom is considered to contribute one electron to the two-electron,
two-centre bond which is formed. The metal-ligand bond is often envisaged as a second type of covalent bond in which we still have a two-centre, two-electron bond, but where
both of the electrons come from one of the atoms or from a molecule. This type of
bond is known variously as a coordination, a dative covalent or a donor-acceptor
bond.

It was recognized early on that the formality of dative covalency in coordination
compounds presents some difficulties. Many inorganic compounds are conveniently
thought of as ionic salts in which there is an essentially complete charge separation
between cationic and anionic species. Compounds of the group 1 and group 2 metals
tend to be readily categorized as ionic. Dissolution of a salt of a group 1 or a
group 2 metal results in the formation of solutions containing solvated cations and
anions - only in concentrated solutions are there significant cation-anion
interactions beyond simple ion-pairing. However, the ionic model does not appear
to be suitable for the description of the properties of many transition-metal
compounds. For example, the compound $\ce{K4[Fe(CN)6]}$ dissolves in water to give
solutions containing solvated potassium ions and the $\ce{[Fe(CN)6]^{4-}}$ ion, rather than
solvated potassium, iron(II) and cyanide ions. The interactions between the cyanide
and the iron(II) centre appear to result in longer lived species than result from
simple electrostatic interactions of the type observed in sodium chloride. This is, of
course, the sort of argument which led to the development of the description of
coordination compounds in terms of donor-acceptor interactions between the
ligands and the metal centre.
Let's have a look at the consequences of the formation of a donor-acceptor
bond in a little more detail. If the donor-acceptor bond is completely covalent,
then there is a net transfer of one unit of charge from the donor to the acceptor as
a direct consequence of the equal sharing of the electron pair between the two
centres. This result leaves a positive charge on the donor atom and a negative
charge on the acceptor atom. The limiting ionic and covalent descriptions of a
complex cation such as $\ce{[Fe(H2O)6]^{3+}}$ are shown below.

As commented above 'ionic' structures are not in accord with the
chemical properties of coordination compounds. However, there are also a number
of objections to the covalent description. The charge distribution is such that the
iron(III) centre bears a three minus charge, whereas the oxygen atoms of the water
ligands each bear a single positive charge. This would be unrealistic in view of the
electronegativities of these elements ($\ce{Fe}$, 1.8; $\ce{O}$, 3.5), which predict that the $\ce{Fe-O}$ bond should be polarized in the sense $\ce{Fe^{\delta +}-O^{\delta -}}$. The problem was addressed by Pauling, who recognized that, in reality, it was not appropriate to describe most bonds as being purely 'ionic' or purely 'covalent'.
He developed a description of the bonding between a metal and its ligands which included considerable ionic character in the metal-ligand bonds within a basically covalent regime. In order to determine the amount of ionic character within a given metal-ligand bond, Pauling framed his electroneutrality principle. In this, he opined that the actual distribution of charges within a molecule is such that charges on any single atom are within the range -1 to +1.
It can be seen how this works by reconsidering the ion $\ce{[Fe(H2O)6]^{3+}}$. An ionic
description results in a +3 charge on the metal centre, whilst a covalent description
gives the metal a -3 charge. Now the electroneutrality principle suggests that the
ideal charge on the metal centre is zero. That would be achieved if the iron centre
gains a total of three electrons from the six oxygen donor atoms; in other words, if
each oxygen loses one half of an electron rather than the whole electron which the
fully covalent model demands. Pauling describes this situation as 50%
covalent (or 50% ionic).

Pauling's model gives a good indication that the bonding situation in transition metal complexes is rather complex since the bonds are highly polarized (An 'accurate' view of the bonding in transition metal complexes can only be gained by more sophisticated methods such as the molecular orbital method).
Thus, Lewis structures are not particularly well suited to depict the bonding situation in coordination compounds because Lewis structures are intended for covalent compounds only.


Chelation-controlled aldol: why is the α substituent pseudo-axial?

Having the methyl group psuedo-axial means that when the nucleophile approaches, the transitiation state is chair-like. The energy penalty for having the methyl group initially in the less favourable position is compensated for the low energy TS.

The ring-flipped version, in which the methyl group is psuedo-equatorial, would open via a high energy twisted TS (the opposite of the argument presented above, i.e. the low energy conformation of the starting material doesn't compensate for the high energy of the twisted TS).

Looking at the Evans' paper specifically, the TS he draws has an addition isopropyl group on the beta-carbon, which is equatorial (this would further stabilise the starting material, since the isopropyl would be far higher in energy if axial compared to the methyl).


The major product is the Zaitsev one i.e 1-methylcyclohexene. The OH needs to be protonated first in order for it to leave and moreover it is a very fast leaving group (also we have the case that $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ are very weak bases and considering that only elimination reactions occur tells us that it must follow an E1 elimination). When the OH is protonated and it leaves, we form a secondary carbocation, but that can be further more made stabilized via an hydride shift from the adjacent carbon producing a tertiary carbocation. Now either of the base $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ can capture the hydrogen from the adjacent carbon leading to the desired Zaitsev product. (Also the stereochemistry isn't necessary to worry about here, as the reaction proceeds through E1.)


It does meet Huckel's rule.  When you have $m$ carbon atoms in a conjugated ring with a positive charge of $q$, there are $m-q$ pi electrons.  Here $m=3, q=1$ therefore the number of pi electrons is $m-q=2=4×0+2$.

A Huckel model calculation indicates that cyclopropenyl cation has roughly as much stabilization as benzene, but it's distributed over three instead of six carbon atoms.  Thus aromaticity is especially powerful in a cyclopropenyl cation ring.   This explains how a positive (formal) charge can be stabilized on as few as three carbon atoms and the ring holds together despite steric strain.


After searching on Google about Isomerization of Glucose to Fructose, you can find so many sources that explain the process. I think that this one is easy to understand among them. It explains that

  The isomerization of glucose to fructose is part of the glycolysis cycle that converts glucose to pyruvate. The way this is done is to isomerize the aldehyde (hemiacetal) glucose to the ketone (as a hemiacetal) fructose,and make another phosphate ester. The isomerization takes advantage of the ease of breakage of a C-H bond which involves a carbon next to a carbonyl carbon.

This gives a clear Idea on the mechanism of Glucose Isomerage to fructose. :)

Also see this.


Calculating entropy change of surroundings

When you evaluate the change in entropy of an entity (in this case the surroundings) between an initial and final (equilibrium state), you separate that entity from everything else, and then you only look at how that entity has changed between the two states, disregarding the actual process path.  In the case of the surroundings (treated as an infinite reservoir), it had received a certain amount of heat between the initial and final states, and its internal energy and entropy increased as a result.  In the formalism of thermodynamics, changes in an  ideal infinite reservoir are always regarded as taking place reversibly, irrespective of other irreversibilities in the actual process.  So if the actual process is irreversible, the irreversibility (entropy generation) is all considered to take place in the system, rather than in the surroundings (in the case of an ideal infinite reservoir).

You may also find my response in the following link pertinent to this question:  Thermodynamics entropy and temperature


The principle that "heavier nuclei are more stable" is a very crude approximation. Theoretical explanations of nuclear stability are complex. While the transuranium elements become progressively more unstable with increasing atomic mass, this trend is not predicted to continue indefinitely; there is an island of stability around copernicium and flerovium, and the as-of-yet unsynthesized copernicium-291 has been predicted to have a half-life of 1200 years or so. You may be interested this paper for details; figure one is a nice illustration of how stability is affected by neutron/proton number towards the end of the periodic table.


Where to find a look-up table for mechanical properties of thermoplastics?

I suggest MatWeb:

  MatWeb's searchable database of material properties includes data sheets of thermoplastic and thermoset polymers such as ABS, nylon, polycarbonate, polyester, polyethylene and polypropylene; metals such as aluminum, cobalt, copper, lead, magnesium, nickel, steel, superalloys, titanium and zinc alloys; ceramics; plus semiconductors, fibers, and other engineering materials.

There are 84,173 entries in the Polymers tab in the physical properties search page at the time of my writing this. Note that you can select up to 3 properties at a time to narrow your search (or 10 if you pay for the service: MatWeb is a freemium deal, so be sure to look into whether or not your needs are satisfied by the free/pay versions.)


How to calculate current density in an Electrolysis Process?

Do I use the total surface area of both Cathode and Anode or just Anode?

You use the cross-section, just like it were a wire.

  Cross-section of both Anode and Cathode or just Anode?

Just one. Like in a wire, you would measure the cross-sectional area once, not cut the wire into many piece and add up all the areas.


What are some examples of a clear, liquid, UV photoactive polymer?

Methacrylates make nice starting materials for the lightinduced polymerization.

In methyl methacrylate (MMA), camphorquinone (a typical initiator) has an absorption maximum around $\lambda$ = 473 nm.


At what amount is digesting soap unhealthy?

I'm less familiar with safety regulations elsewhere in the world, but in the US, "chemical manufacturers" (which includes soap, sugar, salt, and just about everything else you can imagine) have to produce Materials Safety Data Sheets (MSDS) and make them publicly available.

Out of curiosity, I did some digging for soaps. Proctor and Gamble is a huge consumer producer, including dish detergent. So there are MSDS sheets for soap.

The MSDS gives oral LD50 values. That is, the lethal dose required to kill half the population.

The LD50 for soaps seem to vary by animal, but judging from a few MSDS I pulled up, they're ~2 g/kg or so. (That value was for rats.)

So if I had to guess, given, say a ~70 kg adult, we'd be talking about 140 g of soap, minimum to kill a person.

Now.. that's acute doses. I have no ideal if anyone has established long-term toxicity of soaps, and I can't find anything obvious online.


Why is the expectation value of Hermitian conjugate operators $RR^\dagger$ always real and non-negative?

Let $\hat{R}^-|\psi\rangle = |\psi'\rangle$ (a new ket; we don't care what it is). If you take the adjoint of this equation you get

$$\langle\psi'| = \langle\psi|(\hat{R}^-)^\dagger = \langle\psi|\hat{R}^+$$

and hence

$$\langle\psi|\hat{R}^+\hat{R}^-|\psi\rangle = \langle\psi'|\psi'\rangle$$

which by definition of the inner product must be real and non-negative.


Even water below boiling point has high partial vapour tension.

The gas containing mostly vapour, being hot and having low average molecular mass, has significantly lower density than air. Therefore, it raises up.

At contact with air, it quickly cools down and the vapour condensates.

The droplets are small enough to be taken upwards by the strong thermal vertical convection.


The respective speed of five molecules are 2, 1.5, 1.6, 1.6 and 1.2 km/s. The most probable speed in km/s will be?

Following up on user andselisk's answer
...

The problem is weird for several different reasons.

First the solution depends on assuming that the velocity of the gas molecules follow the Maxwell-Boltzmann distribution which is weird for only 5 molecules. The problem also doesn't state that the 5 molecules are a sample from a much larger number of molecules.
Second the answer depends on how you calculate to get a result.

For the Maxwell-Boltzmann distribution there are relationships between the most probable speed $v_p$, the average speed $\langle v \rangle$, and the root-mean-square speed $v_{rms}$. Following the Wikipedia article:

\begin{align}
\langle v \rangle &= \frac{2}{\sqrt{\pi}}v_p\\
v_\mathrm{rms} &= \sqrt{\frac{3}{2}}v_p
\end{align}

Rearranging these two equations we get two different ways to calculate the most probable speed.

\begin{align}
v_p &= \frac{\sqrt{\pi}}{2}\langle v \rangle\\
v_p &= \sqrt{\frac{2}{3}}v_\mathrm{rms}
\end{align}

Thus using the mean (average) speed we get:

\begin{align}
\langle v \rangle &= \frac{2 + 1.5 + 1.6 + 1.6 + 1.2}{5} = 1.58\\
v_p &= \frac{\sqrt{\pi}}{2}\langle v \rangle = 1.40
\end{align}

However using the rms speed we get:

\begin{align}
v_\mathrm{rms} &= \sqrt{\frac{2^2 + 1.5^2 + 2\cdot 1.6^2 + 1.2^2}{5}} = 1.60\\
v_p &= \sqrt{\frac{2}{3}}v_\mathrm{rms} = 1.31
\end{align}

The difference is due to small sample statistics and not really a statistically significant result. However the difference is absolute proof that the sample doesn't follow the Maxwell-Boltzmann distribution.


Do core electrons in bulk material have discrete energy levels or continuous energy band?

What happens with energy levels of core electrons in the bulk, as the
  lattice constant is changed? Do they form a continuous energy band in
  the first place? Why?

Using an organic example, as you move from ethylene to butadiene to hexatriene, you have 1, 2 and 3 bonding and antibonding pi molecular orbitals respectively. If you continue this chain extension and wind up with a very large polyene, the energy difference between all of the bonding (or antibonding) orbitals becomes smaller and smaller.  In addition to this compression between energy levels, each molecule is in a slightly different environment.  Some are near the surface, some deep in the bulk, some have a structural deviation or packing non-uniformity close by.  This serves to broaden the energy width of each orbital.  Between the energy broadening of an orbital and the compression of the energy separation between orbitals, one can imagine a limiting case where all of the bonding orbitals are so close to one another that one continuous band, the valence band, exists. The same would be true of the antibonding orbitals and they would merge into a continuous conduction band. The only remaining separation is the separation between the valence band and the conduction band.  This separation is referred to as the band gap.

In the real world, at least at normal pressures, this limiting case of a single band comprised of all bonding orbitals seems unlikely. Instead, some orbitals where the energy separation between them is large likely form separate bands.  In this case, the HOMO would still broaden significantly (as would the other molecular orbitals) and transform into the valence band, remaining distinct from the other bands at lower energies.

Still, it is difficult to probe the bulk and to understand exactly just what the situation is there. X-ray spectroscopy probes to a depth of only a micron or so, so it is really examining the surface states.  Certainly at least some of the molecules at the surface retain their individual orbital identity and are not part of the band structure.  Therefore it is not surprising that XPS (since it probes only the surface) can detect individual orbitals and serve to identify materials.


Why is the curve of an isothermal process above the that of a adiabatic process during compression?

The two cases you refer to seem to be  for the reversible adiabatic and reversible isothermal change in an ideal gas.

In an adiabatic expansion as there is no heat loss or gain to or from the surroundings ($q=0$) the energy needed for the work of expansion against the external pressure must come from the internal energy of the gas. This results in a lowering of the temperature.

A curve of reversible adiabatic p vs. T change must therfore be below that for a reversible isothermal change. In addition as $\gamma \gt 1$, for example $5/3$ for an monoatomic gas, the adiabatic curve becomes even further below the isothermal one. The slope of the curves is not relevent in determining which is greater.


Titration of H2SO4 w NaOH: Solving for the molarity of H2SO4?

The molarity would be the same whether you have $5~\mathrm{mL}$ of $\ce{H2SO4}$ or a swimming pool full of it.

However, if you wanted to solve for moles of $\ce{H2SO4}$ in $50~\mathrm{mL}$, you would have to multiply the number of moles in the $10~\mathrm{mL}$ sample by $5$.

Note: Make sure you're working with molarity and not moles. I'm in analytical chem right now and often we're multiplying the number of moles in our sample by the total volume of the volumetric flask from which the sample was drawn, so we're doing calculations similar to this.


This occurs during Oxidative phosphorylation and the "proton leak" process is also called Uncoupling of Oxidative Phosphorylation
The mitochondrial electron-transport chain
Electron transport, causes Complexes I, III, and IV to transport protons across the inner mitochondrial membrane from the matrix, a region of low [$\ce{H+}$] and negative electrical potential, to the intermembrane space (which is in contact with the cytosol), a region of high [$\ce{H+}$] and positive electrical potential.

Three of the four electron-transport complexes, Complexes I, III, and IV, are involved in proton translocation.
Two mechanisms have been put across that would couple the free energy of electron transport with the active transport of protons: the redox loop mechanism and the proton pump mechanism.
Uncoupling of Oxidative Phosphorylation
Electron transport (the oxidation of NADH and FADH2 by $\ce{O2}$) and oxidative phosphorylation (the synthesis of ATP) are normally tightly coupled due to the impermeability of the inner mitochondrial membrane to the passage of protons. Thus the only way for $\ce{H+}$ to re-enter the matrix is through the $\ce{F0}$ portion of the proton-translocating ATP synthase.

In the resting state, when oxidative phosphorylation is minimal, the
proton-motive force across the inner mitochondrial membrane builds up
to the extent that the free energy to pump additional protons is
greater than the electron-transport chain can muster, thereby
inhibiting further electron transport.
However, many compounds,including 2,4-dinitrophenol (DNP) and
carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP), have been
found to “uncouple” these processes. In a pH gradient, they bind
protons on the acidic side of the membrane, diffuse through, and
release them on the alkaline side, thereby dissipating the gradient.

The chemiosmotic hypothesis has provided a rationale for understanding the mechanism by which these uncouplers act.
Uncouplers act by dissipating the proton gradient across the inner mitochondrial membrane created by the electron-transport system.

As uncouplers, they function by carrying protons across the inner
membrane. Their tendency is to acquire protons on the cytosolic
surface of the membrane (where the proton concentration is high) and
carry them to the matrix side, thereby destroying the proton gradient
that couples electron transport and the ATP synthase.
In mitochondria treated with uncouplers, electron transport continues
and protons are driven out through the inner membrane. However, they
leak back in so rapidly via the uncouplers that ATP synthesis does not
occur. Instead, the energy released in electron transport is
dissipated as heat.

The presence in the inner mitochondrial membrane of an agent that renders it permeable to $\ce{H+}$ uncouples oxidative phosphorylation from electron transport by providing a route for the dissipation of the proton-motive force that does not require ATP synthesis. Uncoupling therefore allows electron transport to proceed unchecked even when ATP synthesis is inhibited.

Uncoupling of oxidative phosphorylation:  The proton-transporting ionophores DNP and FCCP uncouple oxidative phosphorylation from electron transport by discharging the electrochemical proton gradient generated by electron transport
References

Voet and Voet 4th ed. Biochemistry: Oxidative Phosphorylation
R.H Garret & C.M Grisham Biochemistry 4th ed : Electron Transport and Oxidative phosphorylation


When you talk about the rate of exchange of oxygen with water it might be instructive to realise that actually any aldehyde or ketone in the presence of water is in equilibrium with the hydrate, as shown below.

Source: UC Davis

There are a few ways we know that this is indeed an equilibrium process, but the main method that was used to fully elucidate the mechanism was by using isotopically labelled water containing 18O (the most abundant isotope of oxygen is 16O).
By doing this the scientists could observe the 16O in the starting aldehyde being slowly replaced with 18O from the enriched water.
Given that an equilibrium is involved, there are actually two pertinent questions...

What is the position of the equilibrium?
How fast do we reach the equilibrium?

The table below might be of some use to you. On the left, you can see a variety of carbonyl compounds, and on the right, the equilibrium constants for the hydration.
Two of the compounds you listed (acetaldehyde and trifluoroacetaldehyde) are listed in the table, and you can see that the acetaldehyde is significantly less hydrated at equilibrium than the trifluoroacetaldehyde.

Source: J. Am. Chem. Soc. 1983, 105, 868 (American Chemical Society)

From the table, some trends can be picked out (excluding formaldehyde which is almost exclusively hydrated). There is here a relationship between kinetics and thermodynamics (the more favourable a compound is to form the hydrate, the faster it will form it)

Aldehydes are far more likely to be hydrated than ketones. This is largely though not exclusively a steric effect.

Electron withdawing/electronegative substituents favour hydration. On a very simple level this is due to the fact that they make the carbonyl more electron poor, thus promoting the addition of water via its lone pair.

The order should therefore be 4>1>>2>3   (fluorine is more electron withdrawing than chlorine)