Can our chemical elements be differents on other universes?

I admit this question is far out of my depth, but it's interesting enough that I feel I should give it a shot.

TL;DR - The question is so hard as to be almost unanswerable in detail. Things can be very different, though not arbitrarily different, and the exact line between the two is very difficult to define. There have been some mild investigations on the chemistry of toy universes, however, one of which I describe below.

The problem with thinking about other universes is that there are potentially a lot of different universes which can arise. Famously, string theory currently posits that there are something like $10^{500}$ "valid" universes, each with different details regarding physical laws and physical constants. Furthermore, universes are very complex objects in the sense that there is a ton of interrelation between everything, so even minor changes to details can create wildly different results. Not just that, but there apparently are many parameters to fiddle with, which would create wildly different periodic tables, often apparently breaking any possibility of life. This all leads to fine-tuning problems.

For example, if the attraction between two protons were about 2% larger, then every proton in the Universe would be bound in the form of a diproton nucleus, i.e. the nucleus of helium-2 ($\ce{^2He}$, which is just barely unstable in our Universe), meaning the entire periodic table would collapse to a single element. Apparently there's no very interesting chemistry in that universe! Now, what exactly would have to be tweaked in order to get a 2% stronger attraction between protons while not breaking everything else, I don't know.

So it seems thinking about chemistry in different universes is far too complicated to approach in almost any situation, but still there are some investigations. One particular case I have come across is what would happen in a universe without weak interactions.

Most of Chemistry ultimately has to do with studying different aspects of electromagnetic interactions. However, the stability of atomic nuclei depends immensely on strong interactions (aka strong nuclear force), and to a lesser extent on weak interactions (aka weak nuclear force). The authors in the article linked above analyzed the effects of taking a universe similar to ours, but removing the weak interaction. There are several differences with respect to the evolution of the universe (big bang nucleosynthesis, stellar nucleosynthesis, etc) and the abundance of the elements, but thinking just about their stability, it seems that in such a universe, a large portion of the periodic table would still exist.

Additionally, given that the weak interaction mediates the decay of neutrons in our Universe, its absence means neutrons are completely stable subatomic particles (in our Universe free neutrons decay with a half-life of around 8 minutes). This also means that neutron-rich nuclei are more stable to decay in the hypothetical weakless universe. In our Universe, deuterium ($\ce{^2_1H}$)is the most neutron-rich stable hydrogen isotope, while tritium ($\ce{^3_1H}$) is stable enough, with a half-life of about 11 years, to be chemically important (heavier hydrogen isotopes can essentially be neglected). However, in the weakless universe, it seems that even hydrogen nuclei as rich as $\ce{^6_1H}$ are indefinitely stable, and there are possibly even more neutron-rich isotopes. See the following figure:

(Source)

For comparison purposes, the most neutron-rich stable isotopes for each element in our Universe are $\ce{^2_1H}$, $\ce{^4_2He}$, $\ce{^7_3Li}$, $\ce{^9_4Be}$, $\ce{^11_5B}$, $\ce{^13_6C}$, $\ce{^15_7N}$ and $\ce{^18_8O}$.

Due to this added nuclear stability, the periodic table in the hypothetical weakless universe could actually have more stable elements in it. In our Universe, elements beyond about the 110th are quite unstable, being produced naturally in small amounts, if any, and they will not likely have great significance in Chemistry. However, highly neutron-rich nuclei could make ultraheavy elements more accessible, even in natural conditions such as supernovae.


What is the (conceptual) difference between binding energy and adsorption energy?

For solids, the binding energy is what makes the crystal (if you will) hold together. It is what makes the individual constituents of the compound bind together.

Adsorption however is an entirely different process: Here, some molecules interact with the surface of another compound. Since (strictly speaking) this is no chemistry but mere physics (there is no formation or breaking of bonds) there are other interactions at play, mainly van der Waals-type interactions.

As these are weaker (it's easier to wipe off the water from your windshield than rip out a silicon atom) and aren't based on the same principles as the binding energy, they get a different name.

I hope that cleared things up for you.


If you look at the list of ingredients, you will find that Coca-Cola and many other brands contain phosphoric acid (food grade - don't worry). This is probably the dominant cause of acidity. Citric acid is another common component. Of course, the carbon dioxide is also acidic, but weakly, so there should be very little difference between carbonated and flat.


What is the difference between these two formulas for pH of weak acids?

The purpose of this $c^\circ$ is to ensure dimensional consistency. Keep in mind that Wikipedia can be edited by anyone, although the content is often of very good quality, sometimes the volunteer writer misses some points and assumes that the reader might be aware of his/her symbols. In your link, the writer does not explicitly define $c^\circ$.
The German version is ensuring that the equilibrium constant is dimensionless and all the quantities which are mathematically operated upon are dimensionless numbers. Therefore $c^\circ$ must be 1 mol/L, if $c_0$ has molar units.
$$c(\ce{H3O+}) = c^\circ\cdot\sqrt{K_a\cdot c_0/c^\circ} \tag{1}$$
Note the quantity under the square has been made dimensionless. In order to attach a unit to the square root term, $c^\circ$ has been multiplied.
If you want to dig deeper, there is something called quantity calculus, which I quote again from my previous answer

Are the units of mole of oxygen molecules the same with the units of mole of nitrogen molecules?
Calculus here is not the integration / differentiation, but rather the
Latin calculus implying a method of calculation.
There is a very nice article "Quantity Calculus: Unambiguous
Designation of Units in Graphs and Tables" by Mary Anne White in the
Journal of Chemical Education. Please read this if you are seriously
interested. Search on Google Scholar and it is free to download from
there.
In quantity calculus Each physical quantity as the product of a
numerical value and a unit:
physical quantity = numerical value × unit
This approach was introduced by British scientists and many leading
physicists used it. Now there is there is nothing less or nothing
more. Therefore your ambiguity arises from introducing another factor
such as "oxygen" or "nitrogen". The unit mol does not know whether it
belongs to oxygen or nitrogen.
As explained in the comments, suppose we write L symbolizing the
height of a tree, then I can only write, L = 10 m. For mathematical
purposes, I will not introduce "tree" anywhere in this equation. The
tree is already incorporated in L (in your mind) but not in the
mathematical equation. One can also write L/m =10. Now you have a pure
number on both sides.


Sorry @ JavaScriptCoder, but I've to agree with aventurin. I'd say idoform test should have been done in a certified laboratory because of safety concern. However, I think fractional distillation might be helpful for this identification (or simply verification). OP can do this at home since he/she has done simple distillation at home. Ethanol/water binary system give an azeotrope, which boils at $\pu{78.1 ^\circ C}$ and gives $95\%$ ethanol. Methanol, on the other hand, make no azeotrope with water, and boils at $\pu{64.7 ^\circ C}$. The temperature range of $\pu{13.4 ^\circ C}$ need only about ≤20 theoretical plates to separate. If you get first distillate below, say, $\pu{67 ^\circ C}$, then you got methanol or something other than ethanol. So, don't drink it.

For more information, read the Wikipedia page having Azeotrope tables.


why does silver chloride darken when exposed to electromagnetic radiation

The energy in the electromagnetic radiation decomposes AgCl into its components, silver and chlorine. This produces finely divided silver particles, which look dark because yet solid silver (in the form of an ingot, for example) has a typical metallic 'colour', silver powder is dark.


Can someone please explain to me where the flaw is in my proposed synthesis mechanism?

The steps for conversion of Benzene to 1,3-dichloro-5-(propan-2-yl)benzene are shown below (Scheme).


Adiabatic compressibility factor of ethyl alcohol

A calculation using the NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, and assuming a temperature of $T=15.5\ \mathrm{^\circ C}$ and a pressure of $p=1\ \mathrm{bar}$, yields an isothermal compressibility of ethanol of
$$\varkappa_T=-\frac1V\left(\frac{\partial V}{\partial p}\right)_T=1.0795\times10^{-9}\ \mathrm{Pa^{-1}}$$
and an adiabatic (or isentropic) compressibility of
$$\varkappa_S=-\frac1V\left(\frac{\partial V}{\partial p}\right)_S=9.1749\times10^{-10}\ \mathrm{Pa^{-1}}$$
The number of digits is the default setting of the software and does not necessarily correspond to the number of significant digits.

By way of comparison, the results for a temperature of $T=20\ \mathrm{^\circ C}$ and a pressure of $p=1\ \mathrm{bar}$ are
$$\varkappa_T=1.1130\times10^{-9}\ \mathrm{Pa^{-1}}$$
and
$$\varkappa_S=9.4729\times10^{-10}\ \mathrm{Pa^{-1}}$$
The data in “Pressure and Temperature Dependence of Liquid Density”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL include a value of $\varkappa_T=1.119\times10^{-9}\ \mathrm{Pa^{-1}}$ for $T=20\ \mathrm{^\circ C}$. Therefore, it seems a likely supposition that the corresponding calculation result of $\varkappa_T=1.1130\times10^{-9}\ \mathrm{Pa^{-1}}$ is reasonable but the number of digits is excessive.


Spontaneity of an exothermic reaction when temperature is decreased

The key is $\Delta G = -RT \ln K$.

The $T$ specifically is causing your reasoning to be flawed, as it complicates the link between temperature dependence of the equilibrium constant and $\Delta G$.

$-RT \ln K = \Delta H-T\Delta S$

$-R \ln K = \frac{\Delta H}{T}-\Delta S$

In this form you can see that (at least to the extent that $\Delta H$ and $\Delta S$ are temperature independent), the sign of $\Delta H$ (meaning whether the reaction is endothermic or exothermic) determines whether or not temperature change favors the forward or the reverse reaction. You can also see that the sign of $\Delta S$ is irrelavent.

For an exothermic reaction, $\Delta H$ is negative.  If $T$ decreases, $\frac{\Delta H}{T}$ and becomes a larger magnitude negative number, $-R\ln K$ decreases, $R\ln K$ increases therefore $K$ increases and the reaction shifts forward.


Function of 'organic' vs regular pipe cleaner

Congratulations on being environmentally sensitive!

The problem with sink drains is the device that lifts and lowers the stopper. It is a rod that protrudes into the drainpipe above the U-trap; it catches hair and provides a bacterial nest and eventually slows water drainage.

Older drain cleaners were based on alkali to dissolve hair; some used aluminum to foam and break up the hairy mass. Bleach plus alkali works well too, and kills bacteria. The newer, less aggressive formulas can use thioglycolate to dissolve hair (also used on human legs!).

In the overall scheme of life, with all the other bodily wastes going down the drain, a little drain cleaner may not be a big problem, and where it is important, the government usually has a statement or a regulation.

Even without regulations tho, it would be kind to the environment to limit our use of toxic chemicals. Using a lot of ineffective product may be worse than using a little of a more aggressive product. I would suggest two approaches:

1) Pour just a little of the effective product down the drain, but only when it drains slowly (you can measure the drain speed by filling the bowl to a certain line, then timing how long it takes to empty). This little bit (10-20 mL?) will saturate the clog because it is higher than the bottom of the U-trap. Give it some time (many minutes, even hours); see if it has opened up. If not, repeat. This may go against product recommendations to use a lot of product, but I suspect that the bottom of the U-trap is not blocked (any water flow usually cleans that out). This will reduce your environmental impact by using your brain.

2) Use a mechanical method. a) A stiff copper wire with a tiny hook at the end is useful for digging out hair, and often opens up the drain with just a lot of hot water. b) Cut a hollow rubber ball in half to make a tiny plunger. Place it over the drain (with the stopper lifted a little) and press it up and down to push the clog all the way down the drain.


Will dioxygen difluoride set fire to sand?

It is hard to be sure as it is too reactive to test
FOOF has been made, though no sane person would want to repeat the most recent synthesis (see the paywalled article reporting it) as it involves passing oxygen a fluorine through a red hot metal tube. An earlier paper that tested its reactivity stored it as a solid at 90K to stop it from, well, exploding. Derek Lowe, in his hilarious blog "Things I Won't Work With" describes some of the (very low temperature) reactions:

And he’s just getting warmed up, if that’s the right phrase to use for something that detonates things at -180C (that’s -300 Fahrenheit, if you only have a kitchen thermometer). The great majority of Streng’s reactions have surely never been run again. The paper goes on to react FOOF with everything else you wouldn’t react it with: ammonia (“vigorous”, this at 100K), water ice (explosion, natch), chlorine (“violent explosion”, so he added it more slowly the second time), red phosphorus (not good), bromine fluoride, chlorine trifluoride (say what?), perchloryl fluoride (!), tetrafluorohydrazine (how on Earth. . .), and on, and on. If the paper weren’t laid out in complete grammatical sentences and published in JACS, you’d swear it was the work of a violent lunatic. I ran out of vulgar expletives after the second page. A. G. Streng, folks, absolutely takes the corrosive exploding cake, and I have to tip my asbestos-lined titanium hat to him.

Many other reactivity experiments were not possible because they are just too impractical with something so dangerously explosive at temperatures any normal person might want to conduct reactions at (anything much above liquid nitrogen is bad for FOOF).
So, I suspect, we don't know whether it would set fire to sand as nobody has ever been able to make enough of it to test that reactivity, or, at least, test it and survive.
FOOF's–far more benign–relative chlorine trifluoride, which is commercially available and has several uses, most certainly does set sand on fire. As rocket scientists know. This account is from John D Clarke's Ignition:

It is hypergolic with every known fuel, and so rapidly hypergolic that no ignition delay has ever been measured. It is also hypergolic with such things as cloth, wood, and test engineers, not to mention asbestos, sand, and water—with which it reacts explosively.

So, given ClF3 is far less reactive than FOOF, it seems likely that FOOF, too, would set fire to sand. But with any luck we will never know for sure as nobody will be so monumentally, suicidally insane to try making enough of it to find out.


Poly(3-hexyltiophene-2,5-diyl) (P3HT) reactions with inorganic salts in aqueous solution

I'll point to a very recent paper on oligothiophenes for solution-gated FET sensing that suggests there are no chemical reactions. (This article popped up in my feed - there are many others on solution-gating organic electronics.)

$\alpha,\omega$-dihexyl-sexithiophene thin films for solution-gated organic field-effect transistors. Appl. Phys. Lett. 108, 073301 (2016)

The authors look at many factors influencing the device performance:

  Finally, excellent transistor stability is confirmed by continuously operating the device over a period of several days, which is a consequence of the low threshold voltage of DH6T-based SGOFETs. Altogether, our results demonstrate the feasibility of high performance and highly stable organic semiconductor devices for chemical or biochemical applications.

The devices in this article use KCl and phosphate buffer. They vary the pH and ionic strength.

I haven't seen anyone suggest chemical reactions, and I'm not sure how it would occur. Certainly the oligo- or polythiophene could degrade over time, but most articles argue these are stable for long-term use.

The authors do note a small drift on the time scale of days:

  both changes may possibly be attributed to the passivation of trap states at
  the semiconductor-electrolyte interface by water molecules penetrating into the thin film

I can't speak to the drift you see, but I'd make sure that:

You aren't seeing evaporation of solvent over time (e.g., from device heating) that's changing your electrolyte concentration.
You use a proper reference electrode
You try removing water via vacuum and repeat the experiment (i.e., you drive off any water penetrating the semiconductor film)

Certainly organic semiconductors are porous, and while mostly hydrophobic, it's not unreasonable to expect that after days of use, some water may migrate into the film.


To which state of matter does the flame belong to?

A candle flame is the product of the oxidation of wax to $\ce{CO2}$ and other small organic compounds. Since these small compounds are energetically lower, energy is released during the reaction, in the form of light and heat. So really, we start out with gas.

However, flames can also conduct electricity, and the conductivity varies with the location in the flame. The linked science fair project summary states that the lowest resistance (highest conductivity) was found at the edge of the bright yellow part of the flame the conductivity was the largest, meaning that the concentration of ions was the highest. The lowest resistance value found was $80~\Omega$, which is still very large in comparison to the resistance of plasma, which is generally considered to be $0$.

So, summarizing my answer: A flame is a gas with lightly ionized portions, but does not qualify as a real plasma.


Why is it wrong to use the concept of hybridization for transition metal complexes?

Tetrahedral complexes
Let's consider, for example, a tetrahedral $\ce{Ni(II)}$ complex ($\mathrm{d^8}$), like $\ce{[NiCl4]^2-}$. According to hybridisation theory, the central nickel ion has $\mathrm{sp^3}$ hybridisation, the four $\mathrm{sp^3}$-type orbitals are filled by electrons from the chloride ligands, and the $\mathrm{3d}$ orbitals are not involved in bonding.

Already there are several problems with this interpretation. The most obvious is that the $\mathrm{3d}$ orbitals are very much involved in (covalent) bonding: a cursory glance at a MO diagram will show that this is the case. If they were not involved in bonding at all, they should remain degenerate, which is obviously untrue; and even if you bring in crystal field theory (CFT) to say that there is an ionic interaction, it is still not sufficient.
If accuracy is desired, the complex can only really be described by a full MO diagram. One might ask why we should believe the MO diagram over the hybridisation picture. The answer is that there is a wealth of experimental evidence, especially electronic spectroscopy ($\mathrm{d-d^*}$ transitions being the most obvious example), and magnetic properties, that is in accordance with the MO picture and not the hybridisation one. It is simply impossible to explain many of these phenomena using this $\mathrm{sp^3}$ model.
Lastly, hybridisation alone cannot explain whether a complex should be tetrahedral ($\ce{[NiCl4]^2-}$) or square planar ($\ce{[Ni(CN)4]^2-}$, or $\ce{[PtCl4]^2-}$). Generally the effect of the ligand, for example, is explained using the spectrochemical series. However, hybridisation cannot account for the position of ligands in the spectrochemical series! To do so you would need to bring in MO theory.

Octahedral complexes
Moving on to $\ce{Ni(II)}$ octahedral complexes, like $\ce{[Ni(H2O)6]^2+}$, the typical explanation is that there is $\mathrm{sp^3d^2}$ hybridisation. But all the $\mathrm{3d}$ orbitals are already populated, so where do the two $\mathrm{d}$ orbitals come from? The $\mathrm{4d}$ set, I suppose.

The points raised above for tetrahedral case above still apply here. However, here we have something even more criminal: the involvement of $\mathrm{4d}$ orbitals in bonding. This is simply not plausible, as these orbitals are energetically inaccessible. On top of that, it is unrealistic to expect that electrons will be donated into the $\mathrm{4d}$ orbitals when there are vacant holes in the $\mathrm{3d}$ orbitals.
For octahedral complexes where there is the possibility for high- and low-spin forms (e.g., $\mathrm{d^5}$ $\ce{Fe^3+}$ complexes), hybridisation theory becomes even more misleading:

Hybridisation theory implies that there is a fundamental difference in the orbitals involved in metal-ligand bonding for the high- and low-spin complexes. However, this is simply not true (again, an MO diagram will illustrate this point). And the notion of $\mathrm{4d}$ orbitals being involved in bonding is no more realistic than it was in the last case, which is to say, utterly unrealistic. In this situation, one also has the added issue that hybridisation theory provides no way of predicting whether a complex is high- or low-spin, as this again depends on the spectrochemical series.

Summary
Hybridisation theory, when applied to transition metals, is both incorrect and inadequate.
It is incorrect in the sense that it uses completely implausible ideas ($\mathrm{3d}$ metals using $\mathrm{4d}$ orbitals in bonding) as a basis for describing the metal complexes. That alone should cast doubt on the entire idea of using hybridisation for the $\mathrm{3d}$ transition metals.
However, it is also inadequate in that it does not explain the rich chemistry of the transition metals and their complexes, be it their geometries, spectra, reactivities, or magnetic properties. This prevents it from being useful even as a predictive model.

What about other chemical species?
You mentioned that hybridisation works well for "other compounds." That is really not always the case, though. For simple compounds like water, etc. there are already issues associated with the standard VSEPR/hybridisation theory. Superficially, the $\mathrm{sp^3}$ hybridisation of oxygen is consistent with the observed bent structure, but that's just about all that can be explained. The photoelectron spectrum of water shows very clearly that the two lone pairs on oxygen are inequivalent, and the MO diagram of water backs this up. Apart from that, hybridisation has absolutely no way of explaining the structures of boranes; Wade's rules do a much better job with the delocalised bonding.
And these are just Period 2 elements - when you go into the chemistry of the heavier elements, hybridisation generally becomes less and less useful a concept. For example, hypervalency is a huge problem: $\ce{SF6}$ is claimed to be $\mathrm{sp^3d^2}$ hybridised, but in fact $\mathrm{d}$-orbital involvement in bonding is negligible. On the other hand, non-hypervalent compounds, such as $\ce{H2S}$, are probably best described as unhybridised - what happened to the theory that worked so well for $\ce{H2O}$? It just isn't applicable here, for reasons beyond the scope of this post.
There is probably one scenario in which it is really useful, and that is when describing organic compounds. The reason for this is because tetravalent carbon tends to conform to the simple categories of $\mathrm{sp}^n$ $(n \in \{1, 2, 3\})$; we don't have the same teething issues with $\mathrm{d}$-orbitals that have been discussed above. But there are caveats. For example, it is important to recognise that it is not atoms that are hybridised, but rather orbitals: for example, each carbon in cyclopropane uses $\mathrm{sp^5}$ orbitals for the $\ce{C-C}$ bonds and $\mathrm{sp^2}$ orbitals for the $\ce{C-H}$ bonds.
The bottom line is that every model that we use in chemistry has a range of validity, and we should be careful not to use a model in a context where it is not valid. Hybridisation theory is not valid in the context of transition metal complexes, and should not be used as a means of explaining their structure, bonding, and properties.


Angular Momentum Of S-subshell of an atom?

Your confusion arises from the wrong choice of equation for the angular momentum. Your equation

\begin{align}
\lVert \vec{l} \rVert = m \lVert \vec{v} \rVert \lVert \vec{r} \rVert \ ,
\end{align}

where $\vec{l}$ denotes the angular momentum vector ($\lVert \vec{l} \rVert$ is its length, i.e. the angular momentum), $\vec{v}$ the velocity vector, $\vec{r}$ the radius vector  and $m$ the mass, does already entail some assumptions that are not necessarily valid. It derives from the more general equation

\begin{align}
\vec{l} = \vec{r} \times  \vec{p} ,
\end{align}

where $\vec{p}$ is the momentum vector which under certain circumstances can indeed be written as $\vec{p} = m \vec{v}$ (there are a few subtleties that I don't want to touch here, keyword: canonical momentum, since they don't relate to the problem at hand) and $\times$ denotes the cross product between two vectors. If you now take the absolute value of the angular momentum vector and use the properties of the cross product you arrive at

\begin{align}
\lVert \vec{l} \rVert &= \lVert \vec{r} \times  \vec{p} \rVert \\
&= \lVert \vec{r} \rVert \lVert \vec{p} \rVert \sin \theta \ ,
\end{align}

where $\theta$ is the angle between the vectors $\vec{r}$ and $\vec{p}$. So, the factor that is missing in your equation is the $\sin \theta$ term, whose absence means that your equation implicitly assumes that $\theta = 90°$ since $\sin 90° = 1$. And indeed if this would be the case, then for $ \lVert \vec{l} \rVert = 0$ to be true either $ \lVert \vec{r} \rVert$ or $ \lVert \vec{p} \rVert$ had to be equal to zero. However, since this assumption isn't made anywhere in the quantum mechanical description of an atom the $\sin \theta$ term has to be kept and there is another way to achieve $\lVert \vec{l} \rVert = 0$, namely $\vec{r}$ and $\vec{p}$ being either parallel ($\theta = 0°$) or antiparallel ($\theta = 180°$) to each other because then the $\sin \theta$ term is equal to zero.

Of course, for the quantum mechanical description you have to replace the vectors by the quantum mechanical operators and the absolute values of the vectors by the expectation values of the operators.


I would expect that the amino phenol is more polar than the phenol. This due to the fact that $\ce{NH2}$ donates electrons to the Benzene ring, making the difference in charge between the Benzene Carbon and the Oxygen stronger.

The Chlorine atom is less electro negative than oxygen (if I'm not mistaken) so that would mean: Naphatlene < Chlorobenzene < Phenol < Amino Phenol.


Why Na/C2H5OH not preferred in the reduction of carboxylic acid, despite being used for the reduction of esters

As the comments point out: the sodium in ethanol generates sodium ethoxide, just as sodium in water generates the hydroxide.  The ethoxide ion, acting as a strong base, deprotonates the acid to a salt and thwarts the reduction.  Get rid of the easily removed proton by using an ester and you unleash the reduction reaction you want.


Initial rate of reaction from the graph?

It seems like you have the right idea to me,

For the "Initial Rate," you're taking a curve and doing a straight approximation of the curve over a small portion, as you appear to be doing.

Your logic also appears correct to me, a decrease in concentration should correspond to a negative rate.

My guess is that the question, (a poorly worded one?) is actually referring to the rate of product creation, and asking you to assume that there is a 1-1 relationship between the decrease in reactant, and the appearance of product. Thus if your reactant is disappearing at $0.02~M/s$ (as indicated by $rate= -0.02 ~M/s$) , then your product should be appearing at $0.02~M/s$, making A the correct answer.


Question:

Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic
solution:
As a general rule:

Strong acid + Strong base $\rightarrow$ Neutral solution

Strong acid + weak base $\rightarrow$ acidic solution

Weak acid + Strong base $\rightarrow$ Basic solution

That's not true. The only time in which these rules are true is when you have an equimolar concentration of the acid and base in question, and the acids and bases must both be monoprotic and monobasic, or diprotic and dibasic, etc. Otherwise there are too many variables to consider for these guidelines to always ring true. Consider mixing 1.0 molar $\ce{H2SO4}$ with 0.10 molar $\ce{NaOH}$ ... I guarantee you that the pH of that solution is not "neutral."

d) $\ce{K(H2O)6+}$ is the conjugae base of a weak acid, therefore the soltuion is basic.

I didn't know that potassium hydrated to a significant extent in water; even so, what the solution manual has written down is not the conjugate base of hydrated potassium cation - the weak acid.

However, all these answers imply that water is acting as a strong
base. How is this possible when water pH is $\sim 7$?

How do these answers imply that water is acting as a strong base?
Also regarding how water can be a strong base in general - it just depends on your solvent system.


Spontaneity of reversible chemical reactions - Both sides?

You may look here for more information.

We know the formula $\Delta G=\Delta H-T\Delta S$ and also know that if for the left reaction $\Delta H_\text{r}=c$, then for the other-way reaction $\Delta H_\text{r}=-c$ (because $\Delta H_\text{r}=\sum\Delta H_\text{f} \text{(product)}-\sum\Delta H_\text{f}\text{(reactant)}$)

Same for the entropy change $\Delta S$: If for the left reaction $\Delta S_\text{r}=c$, then for the other-way reaction $\Delta S_\text{r}=-c$ (because $\Delta S_\text{r}=\sum\Delta S_\text{f}\text{(product)}-\sum\Delta S_\text{f}\text{(reactant)}$).

So, for the reverse of a reaction, since both $\Delta H_\text{r}$ and $\Delta S_\text{r}$ are opposite, the whole $\Delta G_\text{r}$ will be the opposite, hence at equilibrium $\Delta G=0$, so that neither way is thermodynamically favoured.

In fact, how much a reaction (warning: reversible reactions isn’t the same as reversible processes, not to be confused) is reversible and the direction it proceeds is determined not only from thermodynamics (I would say much more from other factors), but from Le Chatelier’s principle (temperature, pressure, concentration), catalysts and of course activation energy.

So, we have seen the relationship of $\Delta G_\text{r}$ of a reaction and its reverse and the fact that at equilibrium $\Delta G=0$.

It must be noted that:

$\Delta G_\text{r}$, $\Delta H_\text{r}$, $\Delta S_\text{r}$ refer to “reaction” (r)

and

$\Delta G_\text{f}$, $\Delta H_\text{f}$, $\Delta S_\text{f}$ refer to “formation” (f) (of chemical compounds and elements)


Why is ice cubical in structure at very low temperature?

First, a little clarification. The kind of ice you see at ambient pressure, ice I, exists as stacks of bilayers of water molecules. If they are stacked in an ABAB sequence, they form hexagonal ice, and cubic ice in an ABCABC sequence. The difference between the two allotropes is just the stacking of the molecules, so in fact they are structurally quite similar. The structure of cubic ice is analogous to the diamond structure of carbon, with hexagonal ice being analogous to lonsdalite. This picture is complicated by the fact that ice is generally proton-disordered, i.e. the molecules can have different orientations that break the crystal symmetry, subject to certain rules (the "ice rules").

There is evidence that ice crystallises as cubic ice in the upper atmosphere, namely the angle subtended by solar haloes and the angles formed in nascent snowflakes, for example. However, more recent evidence suggests that cubic ice doesn't not exist as a pure phase, but only in stacking-disordered hexagonal ice (http://pubs.rsc.org/en/content/articlelanding/2015/cp/c4cp02893g#!divAbstract). Curiously, there is also evidence that the lonsdalite form of carbon doesn't exist in reality, and is just stacking-faulted diamond (http://www.nature.com/articles/ncomms6447).

This is all very well, but does not tell us why hexagonal ice forms in preference under certain conditions, or cubic ice (?) under others. If you were to perform density functional theory calculation (essentially approximately solve the Schrodinger equation for a many atom system), you would find that hexagonal and cubic ices have almost identical energies, and are essentially degenerate at zero temperature, which makes it even more confusing why hexagonal ice is almost exclusively seen on Earth. I don't think there is a convincing answer to your question, since the best calculations do not show why the hexagonal structure of ice is lower in energy than cubic, but it could be due to a number of reasons: orientational disordering, which complicates modelling ices, zero point energy or free energy considerations, or a breakdown in the approximations used to perform such calculations. In any case, it looks like cubic ice doesn't exist under terrestrial conditions.


Why are elements on the right side of the periodic table nonmetallic and gas at room temperature?

The elements on the left-side of the periodic table are relatively electron deficient (i.e., they have few valence electrons), and due to their comparatively low effective nuclear charges (the net positive charge of the protons minus the shielding core electrons below the valence level), their electrostatic hold on these electrons are weak. Consequently, the electrons of metal atoms tend to be highly delocalized, not forming localized covalent bonds or possessing tightly held lone pairs. In metals, the electrons are effectively shared between many atoms, loosely bound in what is often described as an electron "sea," and the atoms of metals arrange to achieve maximally efficient packing in a crystal lattice, through which the electrons (particularly the most extensively delocalized electrons in the s and p orbitals at the highest main energy level) move freely. The tight packing of the crystal lattice accounts for the density and hardness of metals, while the extensive delocalization of electrons partially explains their malleability and conductivity.

Elements further right on the period table, on the other hand, have higher effective nuclear charges and stabilize electrons more effectively, leading to localized covalent bonding and the formation of molecules (vs. the ionic bonding and crystal lattices of metals). Electrostatic intermolecular forces of attraction are generally weaker than ionic bonding or delocalized metallic bonding, so non-metals and molecular compounds are more likely to exist in the liquid or gaseous phase at comparatively low temperatures than metals or ionic compounds.


How to evacuate a glass vessel containing powders?

I often high-vac powders on Schlenk. Issues only arise when going between positive and negative pressures though to minimise any risk I often lightly plug the sidearm of the glass stopper/adapter(?) with cotton wool. Glass wool would also suffice and I am certain there are already stoppers/adapters which have a porous mesh build in to prevent powder from entering the vacuum system.


Which gas among radon and hydrogen chloride has higher ideal behavior?

Some molecules are highly polar and thus are subject to electrostatic (dipole-dipole) interactions, which are typically far stronger than the interactions of induced dipoles (commonly referred to as van der Waals forces). See where this gets you.


How to determine the acidity or basicity of dissolved salts?

The $\ce{KNO2}$ is basic, the HF is acid, the $\ce{(CH3)3NHBr}$ being an ammonium salt is acid too, the last one is an alcohol, I would consider it neutral for your purpose. But you are missing one step to get your answer quite right.
Last part, in the pH world, everything turn around two ions H+ (or H3O+) and OH- and the equilibrium between those two. A substance is considered acid when it either give a H+ as disolving or takes an OH-. On the other hand a substance is considered basic when it give an OH- or takes a H+. Like HF is an acid (H+, F-) and NaOH is basic (Na+, OH-). For KNO2, it can take an H+ to go back to HNO2, so it is basic. For organic compounds, alcool groups are usually considered inerte as they don't ionate in normal conditions.
Now, maybe I'm wrong but your question sound like if you put all those in a bathtub what'S the final result? Is it right or not?


No, the temperature gradient is not uniform across all three layers.

To see why, start from the Fourier law for heat flow, shown in my solution to this problem.

Since at steady state the heat flux $q$ must be equal for all layers, we have that

$$q_{1}=q_{2}=q_{3}$$

And since the layers are equal in thickness,

$$ k_1\Delta{T_1}=k_2\Delta{T_2}=k_3\Delta{T_3} $$

Clearly, if the $k$ are not equal, the $\Delta{T}$ cannot be equal either, so the profile will not be that in (1).


How to treat skin corrosion by sodium hydroxide?

We use a lot of chemicals […]

Before you start to use potentially hazardous compounds and products, get their Material Safety Data Sheets (MSDS) and read them.
Learn about storage, handling, disposal and emergency measures in case of

eye contact
skin contact
inhalation

For a $50\%$ sodium hydroxide solution, a typical MSDS looks like this and states:

  Skin Contact:

  In case of contact, immediately flush skin with plenty of water for at least $15$ minutes while removing contaminated clothing and shoes. Cover the irritated skin with an emollient. Cold water may be used. Wash clothing before reuse. Thoroughly clean shoes before reuse. Get medical attention immediately.

Forget about the ammonia! Use lots of water!


It is a Fortran format specifier.

It means

$n$: repeat the input $n$ times, once for each weight
F: input is a floating-point number
10.8: the number is of total width 10 (including the decimal point), and has 8 digits to the right of the decimal point

This means the number cannot have more than one digit to the left of the decimal point, which makes sense since weights should all be $< 1.0$.


How many significant figures should I report calculating volume concentrations?

We cannot say how many significant figures you should be reporting because:

We cannot be sure about the precision of your volume measurements; and
We cannot be sure about the precision of the original solution’s concentration.

Concerning volumes: most of the time when working with such volumes in the lab, I would use measuring cylinders. These come with a rather poor precision but are adequate for most purposes. However, considering we need $\pu{300ml}$ and $\pu{900ml}$, we would probably want to work with very large cylinders so the error would be in the range of $5{-}\pu{10ml}$. It is possible to use much more precise equipment: in analytical lab courses, we would use pipettes that measure exactly $\pu{25.0 \pm 0.1 ml}$ and dilute using measuring flasks at a similar level of precision.

If, as it seems from the values you plugged, the initial solution was $\pu{4M}$ (1 significant figure), I would argue the result should be given to 1 significant figure too – introducing a large error. Arguably, a proper error propagation calculation would probably give you a better precision in the final result.


I think this pretty much comes down to three things:
1) The estimates of the atmospheric half-life of $\ce{SO2F2}$ has changed over time.
2) The current estimate of 30-40 years isn't a particularly long time relative to some other players (i.e. methane's atmospheric lifetime is around 6-8 years and that of carbon dioxide is around 50 years, though there is poor agreement on that one).
3) The emissions relative to the other greenhouse gas players is small enough to offset the fact that it is "potent" in this aspect.

From the sulfuryl fluoride Wikipedia page:

  Based on the first high frequency, high precision, in situ atmospheric
  and archived air measurements, sulfuryl fluoride has an atmospheric
  lifetime of 30–40 years, much longer than the 5 years earlier
  estimated.

That addresses my first item.  From the same page regarding my second and third items:

  Sulfuryl fluoride has been reported to be a greenhouse gas which is
  about 4000-5000 times more efficient in trapping infrared radiation
  (per kg) than carbon dioxide (per kg). It is important to
  note, however, that amounts of sulfuryl fluoride released into the
  atmosphere (about 2000 metric tons per yr) are far lower than the
  amounts of $\ce{CO2}$ released by hydrocarbon-burning vehicles, industry, and
  other processes (about 30 billion metric tons per year).

That means that annual carbon dioxide emissions will trap 3000 times as much heat as will sulfuryl fluoride.  So the pest control people may not have worded it quite right, but the bottom line is that sulfuryl fluoride emissions have a negligible effect on the planet's heat budget.


Why do sodium halides react so differently with sulfuric acid?

A factor you didn't touch upon which may have some weight in determining the reactions is the acidities of the hydrogen halides formed by reactions of the first type (which is fundamentally a coarse qualitative analysis of the free energy of reaction). In aqueous solutions, the order of acidity is $\ce{HF} \ll \ce{HCl} < \ce{HBr} < \ce{HI}$. While concentrated sulfuric acid is obviously not the same as water, perhaps it is fair to say the relative acidities of the hydrogen halides are approximately the same, given they are both polar protic solvents with very high dielectric constants.

As you go down the family, the acid formed in reactions of the first type become stronger, and there is less thermodynamic drive to follow such a reaction pathway. $\ce{HI}$ is somewhat close in acidity to $\ce{H2SO4}$ since neither conjugate base is very coordinating. Meanwhile, as you say, the reducing power of the halides increases, and since sulfur is in a high oxidation number in $\ce{H2SO4}$, redox reaction pathways become more attractive. A detailed analysis of free energies of reactions should agree with the change in the mechanism since it is unlikely these reactions are kinetically controlled.

Edit: Another neat consequence of the acidity strength argument is that it explains why the reactions stop at the bisulfate anion instead of going all the way to the halide sulfates; $\ce{HSO4-}$ is a much, much weaker acid than $\ce{H2SO4}$, so there is no free energy to lose out of consuming it to produce a much stronger acid (in the case of $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ at least). $\ce{HF}$ is similar in acidity to $\ce{HSO4-}$ (in water), so it's not as readily explainable in such a qualitative analysis.


Why do different substances containing saturated hydrocarbons burns with different flame?

LPG is propane, butane or a mixture of both. Paraffin candle wax is $(CH_2)_n$

typically $C_{31}H_{64}$ (other, nonparaffin, waxes are also used in candles, like stearin, beeswax, etc.)
However, LPG is a gas from the start and can mix with air (i.e., oxygen) before it begins to burn, while candle wax has to be heated by the flame before it can begin to burn, so the outer periphery of candle wax vapor (the part of the vaporized candle not yet heated up to ignition) is very small - a dark zone surrounding the wick. The possibility of contact with oxygen is limited to this area.
You will notice a blue zone around the wick also - perhaps obscuring the dark zone closer to the wick. This is a hydrogen-rich flame; under ideal conditions, you can burn off all the hydrogen and leave the carbon (put a glass into the flame: it will get coated with black soot). But if you don't cool the flame with a glass, the carbon which would otherwise deposit will burn yellow.
So, "insufficient supply of oxygen" is an OK answer, but maybe a better one is "scanty mixing of candle wax vapor with oxygen". I suspect that if you boiled candle wax rapidly in a closed container with a vent, and ignited the gases coming out of the vent, they would burn much more like LPG and less yellow.
The colors are dependent not only on line spectra of the atoms, but also on black body radiation from solid carbon. In the candle flame, by the time the carbon get access to oxygen, its temperature is not as high as in a LPG flame, and the possibility of solid carbon particles burning increases the amount of yellowish light emitted. Kind of like charcoal burning, but on a much smaller scale.


Mercury vapor release from amalgam fillings

All solids have a certain vapor pressure, and if this vapor pressure exceeds the atmospheric pressure, then evaporation of the solid occurs. Some of the solid turns into gas. More specifically, it's the uppermost layer of the solid that evaporates.

  What would cause vapour release from elemental mercury? Would it be
  pressure, friction...

Friction could definitely increase the kinetic energy of the mercury particles and increase the solid's vapor pressure. An increase in pressure could also increase the temperature and therefore kinetic energy of the solid's constituent mercury atoms.

  could one stop the vapour release through cooling the mercury?

Yes, that would decrease the kinetic energy of the particles and lower the solid's vapor pressure. Here's a chart relating temperature (in Kelvins) to vapor pressure. At physiological temperatures, we're still talking about an extremely low vapor pressure.

http://www.physics.rutgers.edu/~eandrei/389/NISTIR.6643.pdf


Okay so say you have two carbon atoms and you want to make a sigma bond with their s orbitals. They won't want to since their orbitals are filled but VSEPR theory tells us that carbon can rearrange and hybridize its orbitals to allow such bonds to happen. Normally you'd have your 3 pi orbitals along the 3 spatial axis, but the s orbital, holding 2 electrons, can hybridize with the p orbital that has no electrons to get two sp orbitals that have one electron each. This sp orbital will be along an axis we'll call z. This leaves us with two p orbitals orthogonal to our sp orbitals which may overlap laterally to give two pi bonds.

If we have two p orbitals overlapping to make two pi bonds, the third p orbital leaves us a problem. If we don't hybridize the orbitals, the third p orbital is empty and cannot make a bond. Even if you had a suitable lewis acid to try to give it some electrons, the bond would't be a pi bond since the orbital overlap would be along the inter nuclear axis. If we do hybridized then we end up with the situation described earlier.

Perhaps there could be a strange configuration where the inter nuclear axis is equiangular from the three base axis and the empty p orbital is filled by some electron donor and the bonding p orbitals overlap at right angles. I wouldn't really call it a pi bond at that point and anything in that position would likely collapse into a more stable state.

Basically the geometry of the p orbitals means that we can really only have two of them overlapping at any one time in the basic sense of a pi bond. This is assuming we are not counting the cases where a single p orbital participates in bonding with more than one atom such as in graphite, though I wouldn't really call that three distinct bonds.


How to predict (know) the products of the reaction SnO+NaOH?

I've discovered the answer, at least I guess...

The given reaction is

$$
\ce{SnO + 2NaOH -> Sn^{2+} + O^{2-} + 2Na+ + 2OH-}
$$

The hidroxy breaks in $\ce{H2O + O^{2-}}$. This released oxigen will get together with the oxigen from the $\ce{SnO}$, to form $\ce{O2^{2-}}$.

$$
\ce{SnO + 2NaOH -> Sn^{2+} + \underbrace{O2^{-2}}_{\textrm{nox = -4}} + 2Na+ + H2O}
$$

The oxidation state of this molecule is -4. Performing the double replacement reaction we get

$$
\ce{SnO + 2NaOH -> SnO2^{-2} + 2Na+ + H2O}
$$

And finally,

$$
\ce{SnO + 2NaOH -> Na2SnO2 + H2O}
$$

What is already balanced.


Bridgehead carbocation: Violation of Bredt's Rule

The reaction in question is known and has been studied fairly extensively, both experimentally and computationally, and is accepted to go via the bridged carbocation that you propose.

To give some data, one experiment  measured the relative reaction rate of a series of such structures towards solvolysis in 80% aqueous ethanol (basically the reaction you describe above):

The intermediate carbocation is known as a non-classical carbocation- a carbocation that does not conform to the standard organic chemistry definition of a planar sp2 intermediate (as we often see with SN1 reactions).

In the case of the norbornyl cation, the carbocation can be observed spectroscopically by treating the parent halide with a superacid such as $\ce{SbF5.SO2ClF}$ at very low temperatures:

One issue with non-classical carbocations is their tendency to rearrange, such that you wouldn't get a single product as predicted, but instead end up with a mixture of products resulting from other classical carbocations formed through hydride shifts on the non-classical carbocation (the exact mix of products would depend upon the conditions of the reaction).

References

All figures taken from Carey, F. Advanced Organic Chemistry - Part A


Oxidation of Unsymmetrical Ketones using Selenium Dioxide

The mechanism of Selenium Dioxide (Riley) oxidation proceeds via the enol tautomer of the ketone (see scheme - image from Wikipedia). If the ketone is assymetric and can enolise to either side then the reaction should give the two di-ketones in proportion to the amount of either enol formed though I cannot find any reference where this has been experimentally demonstrated. In the OP's case of 2-pentanone, a mixture of the two possible products will be formed.


Obtaining magnesium oxide from magnesium hydroxide

Magnesium hydroxide decomposes at 332 °C (630 °F) :

$$\ce{Mg(OH)2_{(s)} -> MgO_{(s)} + H2O_{(g) }}$$

So heat the magnesium hydroxide in a covered crucible. The released water will cause the particles of magnesium hydroxide to "explode" so you want to do this covered.


There are a few options for low-melting glasses that will easily melt at the temperatures of an air-gas torch (i.e. a Bunsen burner).  I suspect that either soda-lime or lead glasses would be the soft glasses of choice for your application (making Prince Rupert's drops).

Soda lime glass should be an adequate choice.  For some applications, it can be prone to cracking if not properly annealed, however.

Lead-based glasses are among those with the lowest melting points of any glass, and probably the lowest applicable to your application.  Annealing is simple and can just be done with the torch itself.  Lead glasses also have a large "soft" temperature range (the range of temperatures below the melting point at which they are still somewhat pliable).  This is seen as a disadvantage in many use-cases, where sagging prior to sufficient cooling/hardening is a problem.  For your work, I would think this would either be a non-issue or possibly could even expand the possibilities for making cool looking drops ;)

Given the availability and affordability of these solutions, further treatment to reduce the melting point just seems like a waste of good glass-working time.


The electron configuration for Tungsten is [Xe] 6s2 5d4 4f14. If it were to lose its electrons, and the answer is that it is diamagnetic, then it would lose its 4 d-electrons.

References

https://www.quora.com/What-are-the-differences-between-paramagnetic-and-diamagnetic-elements
https://www.tabulka.cz/english/elements/configuration.asp?id=74
https://socratic.org/questions/what-are-columbic-and-exchange-energy-and-how-it-it-determined


Mechanism of pseudo buffer pH maintenance

I do wonder what your question is, but I have a hunch. Assuming that what you’re asking is essentially:

  In a concentrated $\ce{HCl}$ solution, the $\mathrm{pH}$ will more or less be the same value even if we add external $\ce{H+}$ or $\ce{OH-}$. Why?

Then there are multiple effects at work:

Aquaeous $\ce{HCl}$ contains a buffering component — the $\ce{H3O+}$ (or closest relative) ion. It is the most acidic ion that can be present is solution, so it ‘buffers down’ the acidity of said solution to what the corresponding $\ce{H3O+}$ acidity would be.
$\ce{HCl}$ in itself is an acid, but at the same time $\ce{Cl-}$ is also a (very weak) base. For high total $\ce{HCl}$ concentrations, the ratio of $\ce{HCl}$ to $\ce{Cl-}$ is at equilibrium. If we add additional $\ce{H+}$ to the mixture, we are effectively shifting that equilibrium back to the $\ce{HCl}$ side, reducing the proton count and thus buffering the $\mathrm{pH}$. Therefore also, if we add a proton scavenger (colloquially also known as ‘base’), then that will shift the equilibrium to the $\ce{Cl-}$ side, and as long as there is still enough $\ce{HCl}$ around, the total proton concentration won’t change notably.

In essence, everything that can act as a Brønsted acid or base will buffer at a specific $\mathrm{pH}$ value, even if that $\mathrm{pH}$ value is far from what one would typically call ‘buffered’.


To extend the life of a cartridge razor blade, should the blade be stored in oil or alcohol?

Most blades are dulled by either the edge folding over itself, or small amounts of rust forming. To stop the first problem, you can strop the blade, but that's not really chemistry. For the second one, anything that stops oxygen and water from reaching the blade will stop it from rusting.
Silica gel does absorb moisture,so that would prevent rust from forming, but this would be temporary. You would have to replace it, and even then it might not stop 100% of moisture from reaching the blade.
Oil might prevent water and oxygen from reaching your blade, but it would make it... oily. If you are okay with that then it would probably be a fine solution.
Alcohol would prevent water from reaching the blade pretty well, as well as disinfect it. This might be the best solution to extend your blade's life. Just make sure you use medical grade alcohol.
Finally, blow drying or towel drying your blade after every use will help a lot as well.


Double bond formation in alkenes during cracking

I don't know if I have understood well your question.
During the process of cracking, homolysis of long chains occurs. Radical intermediates result from breaking the bonds. These intermediates are highly unstable and undergo rearrangements. As a consequence of the rearrangements,  smaller alkanes and alkenes are formed. We have to point out that, due the the high reactivity of radicals, the control of radical reactions is very difficult and we can not readily predict what happens inside the chemical reactor.
I hope I've answered your question.


In compact matter, electrons "trickle" from a source to one atom, forcing off electrons which move to the next atom, ionizing each in turn. The same holds true in the opposite direction, where electrons are removed and the corresponding "hole", or positive ion, is propagated towards the negative electrode. Individual electrons and ions move slowly at the drift velocity, but electromotive force propagates much faster. This is similar to the motion of the particles in the toy, "Newton's Cradle".

In an ionized gas, electrons and ions may move much faster, actually crossing the full volume of the plasma (though electrons move faster, due to their lesser mass for the same charge).


Reaction goes along the lines $\ce{3 I2 -> 2 I3- -> 6 I-}$. The beginning of titration is $\ce{I2}$. The end of titration is $\ce{I-}$. $\ce{I3-}$ forms as an intermediate, but it can be ignored in titration.


The aromaticity of [18]-annulene1 was confirmed by room temperature 1H NMR (60 MHz) in 1964.2 [τ values have been converted to δ values; δ = 10 - τ] When the spectrum was recorded at room temperature the 12 deshielded, exterior hydrogens appeared at δ 8.9 while the 6-interior hydrogens were highly shielded (δ -1.8). Evidence for flexiblity in the ring at room temperature was confirmed by observing a further downfield shift of the 12 hydrogens (δ 9.28) and an upfield shift of the 6 hydrogen signal (δ -2.99) when the spectrum was recorded at -60oC. Recording the spectrum at 110oC caused the two signals to coalesce to a singlet peak at δ 5.45. This value is close to the weighted average of the resolved chemical shifts. Ostensibly, the coalesence is due to the interchange of external and internal hydrogens. When the sample is cooled, the resolved spectra are restored.

1) Sondheimer, F., Wolovsky, R., Amiel, Y. J. Am. Chem. Soc., 1962, 84, 274.
2) Gaoni, Y., Melera, A., Sondheimer, F., Wolovsky, R. Proc. Chem. Soc., 1964, 397.


For safety reasons, I would stick with ethanol or isopropanol, as they are less flammable than methanol (your using hot soldering equipment) and much less toxic, even though methanol is a better solvent for $\ce{H3BO3}$. If it weren't for the expense, due to heavy taxation of ethanol, it would be preferable to isopropanol. From the table below, you'll need almost twice as much isopropanol (though 1-propanol is the solvent listed, it is an isomer of the more common 2-propanol, AKA isopropanol).

  Solubility of $\ce{H3BO3}$ in various alcohols at 25 C, percent by weight:

  Methanol:     22.66%

  Ethanol:     11.96%

  1-Propanol:    7.34%


Why does sp2 hybridization imply adjacent sp2 hybridization?

If an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?

Absolutely not, here are a few examples


Calculating Equilibrium Partial Pressures Given Kp and Mass

Say you start with $A$ moles of $\ce{AsCl5}$, and then you reach a temperature $T$, for which the equilibrium constant is $K_p$.

In order to achieve equilibrium, some amount of forward reaction ($\ce{AsCl5 -> AcCl3 + Cl2}$) happens to consume $x$ moles of $\ce{AsCl5}$. Based on the stoichiometry of the equation, you know that exactly $x$ new moles of the products are formed.

$$K_p=\frac{P_{\ce{AsCl3}}P_{\ce{Cl2}}}{P_{\ce{AsCl5}}}$$
Since for $V$, $R$, $T$ are constant, we can replace the above equation by:
$$K_p=\frac{RT}{V}\frac{n_{\ce{AsCl3}}n_{\ce{Cl2}}}{n_{\ce{AsCl5}}}=\frac{RT}{V}\frac{x^2}{A-x}$$
Solve this for $x$, and substitute in the ideal gas equation to get the answer.


In your example we are comparing the electron releasing/withdrawing properties of a hydrogen substituent and a methyl substituent, and the effect this has on proton ($\ce{H_{a}}$) chemical shifts.

Carbon (2.55) is more electronegative than hydrogen (2.2).  Therefore, the methyl substituent will be more electron withdrawing compared to the hydrogen substituent.  This means that the methyl group will remove electron density from the carbon to which it is attached, compared to the hydrogen substituent.  The fact that this carbon has lower electron density in the methyl case, is further transmitted to the protons ($\ce{H_{a}}$) attached to this carbon. Consequently the $\ce{H_{a}}$ protons will also experience lower electron density in the R=methyl case.

Lower electron density at an nmr-active nucleus results in deshielding (downfield shift). Consequently, the $\ce{H_{a}}$ protons are deshielded in the ethyl ester (R=methyl) relative to the methyl ester (R=hydrogen), in agreement with the spectral data you've provided.


Is the rate determining step the step with the largest Ea?

Yes, the rate determining step is the largest energy difference between any starting material or intermediate on a potential energy diagram and any transition state that comes after it. That transition state will then be the rate-determining step of a given reaction.

The transition state with highest absolute energy may not necessarily correspond to the rate determining step, because if it is transforming from another high energy state, then little additional energy is required to reach this (low $E_a$).


What is the concept behind the calculations for the dilution of solutions?

I'm not sure the exact question but I suppose you want to find the volume of 1st solution needed to come up with the 2nd solution. I always feel more comfortable with moles for this types of problem. Because mole is conserved (in a way).
So, to make the 2nd solution you need

$$50~\mathrm{ml} \times \frac{1~\mathrm{l}}{1000~\mathrm{ml}} \times 0.45~\mathrm{\frac {mol}{l}}=0.0225~\mathrm{mol}$$

That means you need $0.0225~\mathrm{mol}$ of $\ce{Al(NO3)3}$ from the first solution. You know the molarity and mole number. So the volume is
$$\frac {0.0225~\mathrm{mol}}{0.85~\mathrm{\frac{mol}{l}}}=0.0264~\mathrm{l}=26.47~\mathrm{ml}.$$
But you need $50~\mathrm{ml}$ of the 2nd solution, so you need to add solvent (Water or whatever you used). The volume of extra solvent required is $50-26.47=23.53~\mathrm{ml}$. Now you got the 2nd solution and to check that you have got everything right, you can find the molarity of your new solution.


Deuterium depleted water through fractional distilation

Relying on the difference in vapor pressures of $\ce{HDO}$ and $\ce{H2O}$ is equivalent on relying on the equilibration of

$\ce{HDO(g) + H2O(l) <=> H2O(g) + HDO(l)}$

The fractionation factor $\alpha_{\ce{D}(l)/\ce{D}(g)}$ for this reaction is 1.015 at STP.

Deuterium ($\ce{^2H}$ or $\ce{D}$) is separated from protium ($\ce{^1H}$) commercially, usually via the Girdler sulfide process, which instead of relying on the above reaction, relies on

$\ce{ HDS + H2O <=> H2S + HDO}$

The fractionation factor $\alpha$ for this process is about 2.2.

Steam distillation for deuterium enrichment is certainly possible, but is not nearly as favorable as the industrial process. Given the very low concentration of deuterium in natural waters, the Rayleigh fractionation equation would be applicable:

$$R_{\ce{D}/\ce{H}} \approx R_{\ce{D}/\ce{H}_{init}} \times f^{\alpha-1}$$

In this equation:

$R_{\ce{D}/\ce{H}}$ is the ratio of deuterium to hydrogen in the liquid water
$R_{\ce{D}/\ce{H}_{init}}$ is the initial ratio of deuterium to hydrogen, or about 0.000156.
$f$ is the fraction of the liquid water remaining after distillation.
$\alpha$ is the fractionation factor mentioned above.  However the value of 1.0015 is for the condensation of vapor phase water into liquid; performing the reverse reaction we should use the inverse, i.e. for distillation of existing liquid into vapor, $\alpha = 0.985$

With this equation and with mass balances (the total amount of deuterium and of protium doesn't change in the distillation process) you can find the amount of deuterium in the distillate.

If you distill off 100 grams of an initial 1000 grams of water, then $f=0.9$, $f^{\alpha-1} \approx 1.000156$, meaning the remaining 900 grams of liquid water will have a deuterium-to-hydrogen ratio of 0.0001562 instead of 0.000156.

The mass balance for deuterium would be:

$1000 \times 0.000156 = 100 R_{\ce{D}/{\ce{H}_{distillate}}} + 900 \times 0.0001562$

If I did all my math right, that means your distillate would have a deuterium-to-hydrogen ratio of 0.0001542, or about 1.2% less than the original water.

Hopefully this exercise convinces you that you would have to do a lot of distillation and burn through a ton of water to generate reasonable amounts of deuterium depleted water through simple steam distillation.  There's a reason they don't do it that way commercially.


What is the complete mechanism of reaction for sodium benzoate and ascorbic acid?

From Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst:

$\ce {Cu^{2+} + H_2Asc \rightarrow Cu^+ + HAsc^.}$

$\ce {Cu^+ + O_2 \rightarrow Cu^{2+} + O_2^-}$

$\ce {2O_2^- + 2H^+ \rightarrow H_2O_2 + O_2}$

$\ce {Cu^+ + H_2O_2 \rightarrow Cu^{2+} + OH^- + OH^.}$

Then the hydroxyl radical reacts with benzonic acid as opposed to benzoate.


Why is phenolphthalein at pH 10 non-planar?

Ok, let's look at this structure. I've drawn/calculated (B3LYP/6-31G) a 3D model of this structure and optimized the geometry.

So what do we see here? Actually no two rings are in one plane, but two are nearly planar (~40° of twist) while one is basically at 90° twist. Let's first look at the one with the carboxy group which is twisted a lot. The carboxy group is in ortho position. If we twist the ring to be in a plane with one of the other rings the oxygen would collide with a CH group. The carbons and oxygens are quite big, so is the repulsion between them and the angle has to be quite big for it to work out. Here's a picture of those groups colliding (I did put it at some degree angle because it's easier to see), keep in mind that this representation of atoms is not the truth. The electron "cloud" around them is actually much larger.

Now why do the other two rings have an angle of about 40° between them? If they are planar they can have conjugation which is favorable. Now What happens if we put them into one plane?

You can see that the two hydrogens in the top get very close. In fact, too close. So there are two ways to avoid this. Twist the two rings or make the angle around the central carbon atom between the rings bigger. In a sp2 carbon like we got here the optimal value would be 120°, we could make it bigger, let's say 145 degree:

Now the hydrogens would not clash any more, but it turns out this costs much more energy than we would lose if we twist the two rings to avoid the hydrogens clashing together. What happens in the end is that this angle gets a little bit bigger (~124° instead of 120) and we get that twist between the aromatic rings which is still low enough for good conjugation between them (see first structure).


Electrophilic halogenation of 4-halo-3-(tert-butyl)phenols

The options ABC are correct answers
Reason:

A: Steric effect of halogens will definitely be taken into account since all 3 halogens are acting on the same compound hence the primary factor which will make the product formed for each case be different is the steric hindrance due to their respective sizes.
B: For steric hindrance to be taken into account both atoms need to be bulky and have equal role to play in the first option we affirmed that due to different sizes among halogens there'll be steric hindrance causing different types of products to be formed however that is only true when we have a bulky group( here, tert butyl) providing steric to the group trying to attach(the halogens) to the adjacent c atoms. Hence for us to take steric hindrance as a factor for different products the steric hindrance caused due to the tert-butyl group and halogens is considered
C: The substitution of the halogens is taking place due to the +M effect of the -OH group thus option C must be true for us to proceed even a bit and it is the first requirement and only after this would we take into consideration the steric hindrance. In other words the reaction is taking place due to the electron donating effect of the -OH group since it is donating electrons to the ring via mesomeric and not inductive. And the steric hindrance serves so as to differentiate and predict which product will be formed


Maximum number of atoms in the ground state

Of the $100$, initially some are in $n=1$, and some in $n=2$ and then some in $n=5$ after absorbing 155 nm then some in lots of other levels after photons emitted. Thus, eventually, none will be left in $n = 5$ but spread  in all in the other levels  including the ground state. However, all other possible transitions occur ($n=2 \to 1$ for example) so all end up in the ground state in total $(100-b)$. However, you do not know the initial amount $b$ so I'm not sure that this question can be answered except that the ratio is $b/(100-b)$.


Is there any free GAMESS input file maker for Mac OS?

If you can give me the requisite keywords for TD-DFT in GAMESS, it's not hard to add that to Avogadro

I've never done TDDFT in GAMESS because I use other programs, but given a link to the docs, it's easy enough to enable more drop-down menus.

That said, you can interactively edit the text in the Avogadro input for any quantum program (not just GAMESS).

If you don't want to use Avogadro, the "native" program for GAMESS is MacMolPlt which was developed in the Gordon group itself. I haven't checked, but I'd assume it supports TDDFT options, since it enables pretty much everything else in GAMESS.


Why do tungsten and chromium have different electronic configurations?

Firstly note that the energetic differences between $\mathrm d^m\mathrm s^n$ and $\mathrm d^{m-1}\mathrm s^{n+1}$ configurations are generally not large. Sometimes these are not amenable to simple qualitative rationalisation.

However in this case it is possible to argue along two lines.

Relativistic effects lead to the stabilisation of the 6s subshell and destabilisation of the 5d subshell, which directly favours population of 6s over 5d.
Electron-electron repulsions in the 3d/4s orbitals are generally larger than repulsions in the 5d/6s orbitals, because the latter are more diffuse. Hence chromium chooses to adopt a configuration in which exchange energy is maximised (i.e. electron-electron repulsions are minimised).


How can you explain the solubility of BeF2?

It seems that for BeF2 the hydration enthalpy of Be2+ is really high (due to Be2+ being a small cation and having a high charge density) so the energy required to break the lattice is smaller than that gained from hydration of Be2+ while for something like MgF2 the enthalpy of the lattice is greater than the energy you would get from hydration of Mg2+. In general when dealing with compounds of the group II with F- because of the small radius of F- the lattice enthalpy is great and only the hydration energy released by the hydration of Be2+ outweigh it thus making BeF2 soluble while all the other fluorides formed with the group II elements are quite insoluble in water


Why do we titrate from a colourless solution and why should we remove unused sodium hydroxide from the burette as soon as possible?

Why is it preferred to titrate from colourless to a pink solution rather than pink to colourless?

It is much easier to detect the appearance of a color rather than the disappearance of one.

I really liked the large white ceramic ringstands for phenolphthalein titrations. The bright white background really made it easy to monitor the titration. At first you can just let the base (sodium hydroxide) pour into the solution as your are swirling the solution. You just get a pink cylinder where the base is entering the solution. As you get nearer to the end point then the color will start to go out more into the solution. Finally you do drops at a time, and then a partial drop.

  Also, why is it advisable to remove NaOH from the burette as soon as possible after we are done titrating?

Sodium hydroxide is a strong base and it will etch (dissolve) glass given enough time. If you're using a burette with a glass stopcock the sodium hydroxide can also weld the stopcock joint.

I'll add that you never should pour the unused sodium hydroxide back into the stock solution. Any that you don't use is waste. You want to avoid any possibility of contamination.


Schrodinger's Equation and Wave Function

I suspect the problem is not mathematical. It is in the meaning of the wave function. Well. I will now show you how I explain it qualitatively in my high school classes. I state that the electron is like a violin string, but a string with three dimensions that is vibrating in the fourth dimension. As I know this last words are not understandable, I start with one dimension and pass slowly to two and three dimensions.
Let's start from a horizontal string whose end is fixed on a wall, and the other end is in your hand. If you put some energy to the string, it will vibrate in a stationary vibration. The string looks like half a sine curve, with the maximum in the middle of the curve. If you want to describe the curve with words and without drawings, you can say that the curve is like a half-sinusoid whose amplitude has the same sign all over its length, part time positive, part time negative.
Now if you put more energy in the string, it will vibrate in another mode, with a node in the middle. When one half of the string is vibrating upwards, the other half is vibrating downwards. With still more energy you can have it vibrate with two nodes. To describe the shape of the string as before without drawings, you can say that the string is a sinusoid with one or several nodes. So that you need one number $n$, the number of nodes, to describe its movement : $n$ is the first quantum number. $n = 1$ for the first vibration, because the only point not moving is the end of the string in the wall. $n = 2$ for the case with one node in the middle of the string.
Now let's consider the string in two dimensions. This is not a string any more, but a two-dimensional plate like a drum. If you hit it softly in the middle, it will vibrate with all its points moving simultaneously upwards then downwards. The only points of no vibrations are the periphery of the drum. In two dimensions, points of no vibration are replaced by lines of no vibration. Here there is only one such line : the number of nodal lines is $n = 1$.
If you put more energy by hitting the center of the plate, you may produce a new vibration mode if the center of the plate is moving upwards (or downwards) when the outer part is moving in the opposite direction. In this mode no vibration occurs on the small circle between these. This circle is a nodal line. There are two concentric circles on this mode. So  $n = 2$. We may describe its amplitude by drawing curves of equal amplitude on the plate, like level curves on a map.
But it is going to be more difficult, because there is another type of vibration on the plate. If you hit the plate not in the center, but elsewhere, you may produce a plate vibration where one half of the plate moves upwards and one half downwards. Here the nodal line is a diameter. And there are two nodal lines : the diameter and the periphery. Here $n = 2$. As a consequence, two vibration modes exist with the same quantum number. We must introduce a second quantum number, which is called $l$, and which is the number of straight nodal lines. So this type of vibration is given by $n = 2$ and $l = 1$. Hopefully you understand that other vibration modes do exist with the drum, with arbitrary values of $n$ and $l$ < $n$
Let's summarize. An object with one dimension vibrates in another dimension, and one quantum number is enough to describe its vibration. An object with two dimensions is vibrating in the third dimension, and two quantum numbers are required to describe its vibration. The number of dimension is equal to the number of quantum numbers.
Now let's go to the electron, which has $3$ dimensions, and which may be considered as if vibrating in a fourth dimension, called $\Psi$. As this dimension cannot be represented in our world, all we can do is to give different numerical values to three quantum numbers for all vibrating modes, and represent surfaces which correspond to equal numerical values of the local "vibration" of the electron. These surfaces of equal value of $\Psi$ are spheres for $l = 0$, the famous dumb-bells with two lobes for $l = 1$, because of $1$ nodal plane. Here is how it goes.
Here is how I explain the wave function to my high school students. They usually have some difficulty understanding. But all have told me that this explanation had helped them sudying quantum mechanics later on in college.


How to determine Langmuir adsorption isotherm constants and use the models

For langmuir isotherm:

Let say we are using linear langmuir isotherm.

The formula is : $\frac{1}{q_e} = \frac{1}{q_m} + \frac{1}{q_mKL}$

find the value of $q_e$. Using the formula $q_e = \frac{(C_i-C_f)V}{\text{mass}}$ of adsorbent, and then plot the graph $\frac{1}{q_e}$ versus $\frac{1}{C_f}$. It would be easier by using excel. From the graph, we can calculate the slope. The linear graph will produce $y=mx+c$ equation. m=intercept=1/qe, hence qe=1/m. C=1/qm.KL. From the C value, we can actually find KL.

From the graph generated using excel, it will shown the R2 value. This value is the important indicator to determine which isotherm model fit the system. The highest R2 value will fit the system. Let say R2 value langmuir is 0.8 and R2 for freundlich is 0.9. So, we can conclude that, Langmuir model fit your system.

For Freundlich isotherm

Plot the graph log qe versus log cf. From the graph you will get linear equation, y=mx + c and the R2 value. Compare the R2 value with langmuir's R2. Highest R2 value indicate that those model fit with the system.

Cf=final concentration
Ci= initial concentration


Hydrolysis is a broad chemical term that means reacting with the solvent water in some way. So for examples esters can be hydrolysed according to the following reaction:

$$\ce{H3CCOO-C2H5 + H2O -> H3CCOOH + HO-C2H5}$$
(Ethyl acetate and water react to form acetic acid and ethanol.)

Dissolution is the process of something being dissolved in water, intermixing with the water molecules so that each atom/ion/molecule of the substance is more or less completely surrounded by water molecules. This process is usually understood to happen without any reaction, thus the term hydrolysis is not used.

However, the above is often a simplification. For example, some water-free salts such as $\ce{AlCl3}$ or $\ce{CuSO4}$ will react with the water to form the corresponding hexaaquacomplexes $\ce{[Al(H2O)6]^3+ + 3 Cl-}$ or $\ce{[Cu(H2O)6]^2+ + SO4^2-}$. These are technically reactions, best exemplified in the aluminium example where the metal was originally coordinated by chlorido ligands and after dissolution is coordinated by aqua ligands.

The only thing that changes is the immediate vicinity of the metal ion in these cases, so speaking of dissolution rather than hydrolysis (essentially the metal cation remains the same as do the corresponding anions) is often tolerated.


What does "the natural bond angle" mean in the context of Universal Force Field (UFF) potential energy calculation and how are bond angles calculated?

The original article on UFF is from 1992, so it is somewhat dated. The selection of "natural bond" values is explained on page 10028:

a. Standard Bond Angles. The natural angles for the group 15, 17, and 18 main group elements are obtained from standard reference structures of the parent hydrides. Thus 0-3 has Bo = 104.5' from HzO, while S3 has Bo = 92.2' from HzS. Exceptions include 0-32, OR, and N-2. The bond angles for 0-32,O-R, and N-2 are fit to (C13Si)z0, methyl vinyl ether, and dimethyl- diazene angles of 146O, 118.3', and 112.3', respectively. Where structural data are unavailable, the natural angles are extrapolated from the element above it in the periodic table. The remaining elements are all assumed to have regular octahedral, tetrahedral, trigonal, or linear structures. The natural angles are collected in Table I.

So yes, the angles are selected from experimental geometries where available, or where not by extrapolation from geometries for compounds in the same group.
References

A. K. Rappe et al.. UFF, a Full Periodic Table Force Field  for  Molecular Mechanics and  Molecular Dynamics Simulations. J. Am. Chem. Soc. 1992, 114, 10024-10039.


After the first step of the reäction (i.e., loss of the halogen), you have the carbocation.  You could draw a resonance structure of this: push the electrons from the C–H bond into making a new C–C bond, leaving you with an alkene and a nearby proton (note the conservation of charge, moving the + from a carbon to the hydrogen).  This is like the reverse of an addition of a hydrohalic acid to an alkene, whereïn you'd have the double bond attack a proton, then have the halogen attack the carbocation.  Either way, you've drawn a very reasonable* resonance structure showing an alkene and H+.  That H+ is pretty much the definition of a Brønsted acid, so that would be a hand-waving explanation of how to see this hydrogen as being particularly acidic.

For the E2 example, we can't use the same method because we can't show this resonance structure (well, we kind of can if you want to get into a really tedious discussion that would be highly debated).  You'd probably want to use the argument about halogens being inductively electron withdrawing, which pulls electron density away from nearby bonds.  This results in less electron density existing in the C–H bond, making the bond longer, and slightly increasing the positive charge on the hydrogen.  Since Brønsted acid theory defines the acid as being H+, the more we can show a partial positive charge on the hydrogen, the more Brønsted-acidic we can argue it to be.

* Note that I mentioned drawing a "reasonable" resonance structure.  That's important, because we could technically show a resonance structure of benzene as having a carbon with a negative charge and a hydrogen with a positive charge, but that wouldn't be an argument for that hydrogen being particularly acidic, as it's a horrible resonance structure to draw.

It certainly is more complicated than this, but I feel like this highlights some of the most important effects without oversimplifying to the point of making anything explicitly wrong or creating a bunch of exceptions.


Strip away a small amount of aluminium with household chemicals

Citric acid will not etch pure aluminum on any reasonable timescale, at room temperature, though commonly-used aluminum alloys, high temperatures, or the presence of salt can change that. However, it may be better to try some other chemicals: here is a decent resource for finding metal etchant solutions. The two that are probably accessible with household chemicals are hydrochloric acid (can be found at hardware stores with pool supplies) and sodium hydroxide (often found at grocery stores as lye). Neither is super dangerous, but you'll want to take precautions to avoid getting them in your eyes, on your skin, or on your clothes and HCl produces hydrogen gas as it dissolves aluminum, so don't put big pieces of aluminum in concentrated HCl unless you have adequate ventilation.

It will probably take some experimentation to get the concentration right: you want an etch rate fast enough that you're not there for hours etching, but slow enough that you can control it. The big problem with wet etching aluminum is that most etchants take a long time to get through the surface layer of aluminum oxide (and the time it takes is not really reproduceable due to differences in thickness and such) so when you put a piece in the etchant, nothing may happen for several minutes, then you look away for a few seconds and turn back to find all the aluminum gone. This can be avoided by adding HF, which happily removes the oxide, but is very nasty stuff that is rather dangerous to work with.

It might be possible to do this by picking a low concentration etchant and checking the piece frequently, but you may run into issues if the surface is dirty (I recommend washing the surface with acetone, isopropanol, and deionized water in sequence) or depending on how the aluminum was deposited. Some deposition techniques produce large variations in film thickness across the surface and you might run into the situation where the aluminum will be completely etched away in one place but is still totally reflective in another.


Why is the binding angle of HOF lesser than that of H2O, despite the lone pairs and size of the flourine?

In $X_2O$ molecules, such as $H_2O$, $F_2O$, $Cl_2O$, the oxygen is at the center of a tetrahedron and therefore theoretically the X-O-X bond angle should be 109.5°. However, this is never the case and it's always a balance between the repulsion of the two lone pairs of the oxygen(they repels more strongly than the bond pairs) and the steric repulsion of the X atoms.

Among the series H, F, Cl, hydrogen is the least electronegative and fluorine the most.
F is very electronegative and thus the electron density of the O-F bond near O is less than that in O-H.
If the electron density is farther away from the oxygen, then the 2 lone pairs of O squeeze the two bonding pairs closer together and the F-O-F bond angle will be less than the H-O-H angle.
Based on the electronegativity, you should expect that the Cl-O-Cl bond angle is less than H-O-H but more than F-O-F. This is not the case and the reason is that chlorine is big enough so that the steric repulsion prevails.

In your case you are considering the molecule HOF:
According to the logic above, you may expect that its bond angle is less than $H_2O$ but more than $F_2O$. Instead, H-O-F bond angle is 97° so even less than $F_2O$!

The reason for this unexpected small angle is that the bond length O-F and O-H are very different and their electron density(bond pairs) don't interact much and therefore they repel even less than in the F-O-F case(see picture above, where electron density is represented in blue).


The effects of H2o2 (hydrogen peroxide) with oils

Oils will oxidize (and burn, too). With dilute H2O2, it's not a rapid reaction (in laymans terms)... some oils (like linseed) will oxidize more rapidly than others (like mineral oil).

Essential oils are not really oils... they are more like phenolic and terpenic extracts.


Is it possible for the iodine atom in periodate ester to be octavalent?

I would suggest that a reactive intermediate/transition state was formed, and not an actual compound. I'd say both $\ce{C-O}$ bonds are partially formed in the intermediate

The diagram in the book does not show where is the $\ce{Na+}$ during the reaction (see above picture), probably for the convenience (but therefore, somewhat misleading). Remember, a periodate ester is an ester of periodic acid.  The most common periodate esters are cyclic periodate esters, form as the intermediate in oxidation of vicinal diols such as sugars and glycols. For example: The oxidative cleavage with periodic acid has been successfully applied in the constitutional analysis of sugars (see: Chemgapedia article).

Experimental evidence for presence of cyclic periodate esters (rightfully with hepta-valence iodine) can be found in: Buist, G. J.; Bunton, C. A.; Hipperson, W. C. P. The mechanism of oxidation of α-glycols by periodic acid. Part X. The oxidation of pinacol, and a general discussion of the stability of periodate esters and their role in the mechanism of oxidation. J. Chem. Soc. B 1971, 2128–2142. DOI: 10.1039/J29710002128.

Also, two forms of periodic acids can be found in the corresponding Wikipedia article.


Low or high enthalpy of combustion in relation to stability

I have redrawn your two structures A and B in their most stable conformations. Note that A has all equatorial methyl groups while B has two equatorial groups and one that is axial. The estimated difference in the heats of formation and combustion is 1.8 kcal/mol, the same as the difference in energy between the equatorial and axial conformations of methyl cyclohexane. Because A and B produce the same amount of CO2 and H2O upon combustion, the more stable A's heat of combustion is 1.8 kcal/mol less than that for B. Compare your A and B with their enantiomers S and T at this ChemSE site at which time I addressed a similar question.


An R group is used in structural formulae as a placeholder for a range of possible subtituents, e.g. hydrogen, alkyl chains or aryl groups etc. It can also be used as an abbreviation when R is a predefined rest.

An amide is a derivative of an organic acid where an $\ce{OH}$ group is substituted by $\ce{NR2}$ (with R = H or any organic rest, see here, for example). An amine is a base, and not formally derived from an acid, but from ammonia. It has the general formula $\ce{NR3}$, with R being H or any organic rest except X=O (X = C, S, etc.) or any other rest derived from an acid. So amides and amines are different classes of compounds. Here are examples for amides of a carboxylic acid, a sulfonic acid (sulfonamide), and an amine (from left to right):

The compound from the video that you have described is diethyl amine. You can formally derive it from ammonia by substituting two hydrogens with ethyl groups:


Your experimental procedures seem to be perfect. There is just one minor error in your conclusion: you said that no current flowed. I disagree - current flowed, just not as much as you hoped. Harnessing one of these reactions to light up even the teeniest lightbulb would be similar to harnessing a bumblebee to a model airplane.

Getting a visible change upon connecting the battery is a reasonable goal. In one sense, you have already done it, by observing the needle move in the voltmeter. I know, you would like something more. A mechanical device might work well: say, a compass needle next to a coil of wire. When you connect the battery, some current will flow, some magnetic field will be generated, and the needle will move.

Another possibility: if you take a small battery operated clock (either mechanical/quartz or digital) and remove the battery, you should be able to operate it with one of your batteries (you may have to boost the voltage). These devices use very small amounts of current, in the range that your batteries can provide.


What would happen if I mixed a single water molecule in a beaker of hexane?

This is a clever question because the molecular weight of water is only 18.01528 g/mol while the molecular weight of hexane is 86.17536 g/mol. Naturally, you would expect the lower mass to rest on top, but water has a higher density... and we all know that water will sink! But there are other forces at work.

So, supposing that you neglect the fact that one molecule of water would be in a vapor-state, Google says the surface tension of Hexane @ 20 °C is 18.43 mN/m... so placing the water molecule ontop of the hexane would probably cause the molecule to float (like a feather on water).

However, if you suberged the molecule it should dissolve (Google says that the solubility of water in hexane is 0.01% at 20°C). Now, once the saturation point of hexane has been met, such that one more single molecule of water would cause a change to occur, that change would probably result in a condensation and precipitation of the dissolved water, because the reality is, that the hexane will become super-saturated, such that the extra water molecule will cause many water molecules to fall out of solution.

Understanding that you are transferring energy is key to understanding this. Consider pushing a ball over a hill, once the ball reaches the top and begins to fall, it falls into the valley. Once you push a boat to the tipping point, it falls to the bottom.

So the number of molecules necessary for hydrogen bonding together to become more dense than hexane would be the number of water molecules dissolved at the maximum supersaturation point (maybe 0.0105%, +1 molecule) minus the number of molecules at the saturation point (0.01%)

Now having a saturated solution, with water on the bottom of the container, adding another molecule of water would generally cause one of the other dissolved molecules to fall out solution (because there is an equilibrium... some of the water molecules at the bottom are dissolving, while others are falling out of solution).


Find ratio of quantities of water and of argon in airtight container

You should use the ideal gas law to convert between $n_i$ (amount of gas) and $p/T$. For each temperature and pressure pair you can write
$$\begin{align}n_{g,1}=\frac{p_1V}{RT_1}\\n_{g,2}=\frac{p_2V}{RT_2}\end{align}$$
where
$$\begin{align}n_{g,1}&=n_{g,H_2O}+n_{Ar}\\n_{H_2O}&=n_{g,H_2O}+n_{l,H_2O}\\n_{g,2}&=n_{H_2O}+n_{Ar}\end{align}$$
You also know that for each individual gas the partial pressure is proportional to the number of moles (Dalton's law):
$$n_i=\frac{p_iV}{RT}$$
 The "trick" is that at $T_1=\pu{373K}$
$$p_{Ar} = p_1 - p_V$$
where $p_V$ is given as the partial pressure of water at that temperature.
You should be able to show that
$$\frac{n_{H_2O}}{n_{Ar}}=\frac{n_{g,2}}{n_{Ar}}-1$$
and making appropriate substitutions it should not be difficult to arrive at the answer.


Why are halogens base sensitive groups (Wolff-Kishner Reduction)?

You've probably realised already that a possible side reaction is the SN2 nucleophilic substitution of Cl by HO-, which would look like this:

Evidently, this is not the product we actually get. Hydroxide ions are considered 'hard' nucleophiles, as they have a high negative charge density, and they tend to react best with 'hard' electrophiles with high positive charge density. The electrophilic carbon here is reasonably 'soft' whilst the proton adjacent to it is considered 'hard' due to its small size (you can learn more about this here). As a result, hydroxide ions like to act as bases instead of nucleophiles, and the mechanism favoured is not the SN2 shown above, but in fact the competing E2 elimination shown below.

This proton may not look particularly acidic, but formation of a new alkene alongside loss of a reasonably stable chloride leaving group drives the deprotonation.
Another contributing factor is that whilst the substitution has a reasonably small change in entropy, the elimination results in a greater number of species than in the reactants, so has a large positive entropy. This is relevant because Wolff-Kishner reduction is usually done at a reasonably high temperature, and entropy becomes a greater influence on the spontaneity of a mechanism as the temperature increases. This further favours E2 over SN2.
The chloride is considered a base-sensitive group for the above reasons, and this second pathway is inevitable alongside a Wolff-Kishner reduction in a highly basic solution.


Whats the driving force behind simple diffusion? Is it just the count of molecules?

All particles are constantly moving randomly due to their kinetic energy, and are thus constantly colliding into each other and the walls of their container (if any).
Particles in gases move rapidly and randomly and are spaced far apart, particles in liquids slide slowly over each other, and particles in solid vibrate around a fixed position.
Hence, diffusion can be defined as the overall movement of particles, from an area of higher concentration to an area of lower concentration, down a concentration gradient, due to their random motion.
Illustration of particles in a gas:

Source: wikipedia.org/wiki/Kinetic_theory_of_gases#/media/File:Translational_motion.gif
This explanation relies on Kinetic theory and Brownian motion.


When reacting a strong acid with a weak base, would all of the strong acid react with the weak base

This would be a simpler question to answer for a salt of a monobasic acid, like sodium acetate. But, for the specific reaction you ask about:

First, we must consider the conjugate acid of trisodium phosphate: phosphoric acid, which will be formed when nitric acid is added. It has three $\mathrm{p}K_\mathrm{a}$ values for the three decreasingly acidic hydrogens: $2.15,$ $7.20,$ and $12.32.$ These values are measured in dilute water solutions, so they do not apply exactly to a reaction between pure nitric acid and pure trisodium phosphate. But these values do indicate that, by combining pure trisodium phosphate with three equivalents of a strong acid like nitric acid $(\mathrm{p}K_\mathrm{a} < 0)$ you would effectively have a complete conversion to phosphoric acid and sodium nitrate. And, yes, the reaction will be exothermic.

That was the simple answer. But if the reaction is carried out in water solution, which would help to keep the reaction from overheating and allow control of the rate of combining the reactants, we note that the first $\mathrm{p}K_\mathrm{a}$ of phosphoric acid is only $2.15$ units higher than that of nitric acid in water (effectively, hydronium ion with a $\mathrm{p}K_\mathrm{a}$ of $0)$ — in other words, phosphoric acid is $10^{-2.15}$ or about $1\%$ as strong an acid as hydronium ion.

So, after the reaction is complete, we would have about $99\%$ of the phosphate in the form of phosphoric acid and about $1\%$ in the form of the dihydrogen phosphate anion. There would also be small amounts present of monohydrogen phosphate dianion and phosphate trianion, whose  concentrations can calculated as $10^{-\mathrm{p}K_\mathrm{a}}$ for the other two $\mathrm{p}K_\mathrm{a}$ values. So, technically the reaction will be about $99\%$ "complete".

Meanwhile, keep in mind that the sodium ions are "observer" species that do not participate in the acid-base reaction.


Producing ketones from secondary alcohols using hydrogen peroxide

I think it is possible to use hydrogen peroxide. This paper discusses it: http://greenchem.uoregon.edu/PDFs/ResourceID88.pdf

Other alternatives:

Pyridinium Chlorochromate (or PCC for short)
Dess-Martin Reagent
Swern Oxidation: oxalyl chloride in DMSO with triethyl amine

Fétizon Oxidation: silver(I) carbonate absorbed onto the surface of celite (Fétizon's reagent). This can be prepared by adding silver nitrate to aqueous sodium carbonate, in the presence of purified celite.

Oppeneaur Oxidation:  excess ketone reagent (such as acetone) and an aluminium triisopropoxide catalyst

Pfitzner–Moffatt oxidation: dimethyl sulfoxide (DMSO) activated with a carbodiimide, such as dicyclohexylcarbodiimide (DCC)

Corey-Kim Oxidation: Dimethyl sulfide is treated with N-chlorosuccinimide (NCS), and then addition of triethylamine


Why do people commonly refer to Na2HPO4 as Sodium Hydrogen Phosphate, rather than Disodium Hydrogen Phosphate?

Well, turns out that I needed to look at this a little differently.  I looked at calcium hydrogen carbonate and then realized that hydrogen is part of the polyatomic ion.  It's a bit harder to recognize when there are no parenthesis, like in the case of sodium hydrogen phosphate, but the same thing applies there.  Therefore, the name, sodium hydrogen phosphate is correct because there are still only two distinct ions in the compound.