Can endergonic reactions be driven by heat?

I think you can look at this entirely from the perspective of entropy (forget about the Gibbs free energy for a moment). You can divide the universe into two parts, the system and its surroundings.
When a change happens in your system, and some heat is exchanged with the surroundings then the change in entropy of the surroundings is: $$\Delta S_\mathrm{surr}=-\frac{\Delta H}{T}\tag{1}$$
(minus sign because if $\Delta H$ is positive, that means heat is taken in from the surroundings so the entropy of the surroundings must go down).
The change in entropy of the system itself is: ${\Delta S_\mathrm{sys}}$
(which you are referring to as $\Delta S$ in your question).
Now the total entropy change of the universe is:
$$\Delta S_\mathrm{univ}=\Delta S_\mathrm{surr}+ \Delta S_\mathrm{sys}=-\frac{\Delta H}{T}+\Delta S.\tag{2}$$
If you have a reaction where $\Delta H$ is positive, then the first term (i.e. $\Delta S_\mathrm{surr}$) is negative. If the entropy change in the system is also negative (i.e. $\Delta S_\mathrm{sys} < 0)$ then the total entropy change of the universe becomes negative. Of course, the universe is an isolated system, so the second law forbids any change that reduces the entropy of the universe.
Now what happens when you increase $T$? Well… nothing. Assuming that the signs of enthalpy and entropy change of the system stay negative regardless of rising temperature, the total entropy change is going to stay negative. All you do by raising $T$ is decrease the magnitude of the first term.
If you look at the third equation, and multiply both sides by $T$ and change signs, you will basically get the expression for Gibbs free energy. In the end, the sign of Gibbs free energy basically tells you if the total entropy of the universe is going down or not when you do that change in the system.
So even if you raise the temperature, an endergonic reaction will not go, and heat won't flow into the system because that would mean reducing the total entropy of universe.

As an aside, I should mention here that all of this applies to one system. In real world, there are many things you can do to drive an endergonic reaction. For example you can couple it with another reaction/change that makes more entropy than it consumes, and that can drive the endergonic reaction. You can supply electricity (think of it as the entropy that was produced during the generation of electricity feeding the endergonic reaction).
Lastly, even if a reaction has $\Delta_\mathrm{r} G < 0$ that does not mean that you won't get any product. It just means that the equilibrium lies to the side of the reactants. The values of $\Delta_\mathrm{r} G,$ $\Delta_\mathrm{r} H,$ $\Delta_\mathrm{r} S$ that we write consider only the reaction happening, it does not include the entropy of mixing that comes from mixing of the products and reactants. So, even though the reaction is disfavoured, you might get a tiny amount of the product.


why there is no simple rule to predict the possible ionic charges of element in middle of periodic table

TL;DR - The upshot is that there just happens to be a simple approximation governing small atoms but as the atoms get bigger and more complex, those approximations become less good and there is not another simple rule that works.

In more depth, the issue is a combination of two factors: distance from the nucleus and full/half-full valence orbitals.

In the first place, valence (outermost) electrons that are further away from the nucleus are more weakly bound than valence electrons that are closer. This means that the energy required to add or remove an electron is less, and so the charge given by a given electron matters less to the stability of the atom. Because of this, the further down the periodic table you go, the less the "normal" charges matter. As an example, Germanium is two rows below Carbon and is naturally found to have charges of +4 and +2 (as well as sometimes having +1, +3 and -4!).

The second factor is that elements find a special stability when their valence orbitals are either full (preferable) or half-full (less preferable, but still better than neither). So an element like Cr, which has almost no hope of gaining or losing enough electrons to look like a noble gas, "wants" to gain lose the 4 3d electrons, getting it down to a full 4s orbital. The tricky bit is that unlike the p orbitals, not all of the d orbitals have the same energy, so there are some that are more likely to be removed than others. There just isn't a convenient rule of thumb about which is which (nature is governed by physics, not approximations).


How to reverse damage done by nail polish remover without acetone to a laptop keyboard?

The solvent appears not only to have removed paint on the keys but also to have caused solvent crazing in the plastic itself. Regrettably, there is probably not much that can be done to actually remove the crazing, which may have penetrated too deeply to remove by buffing, but you might be able to hide it with a substance such as an oil that has the same index of refraction as the plastic.

If the appearance is unacceptable, you might replace the keyboard, use a plug-in keyboard, or get a keyboard cover, such as the Kuzy Russian Keyboard Cover, specific to the Mac Pro. Individual key stickers would probably not adhere well.


Yes you are correct, at least two products are formed (ref. 1, ref. 2).  Unfortunately, neither of these references provide the actual product distribution between the isomers.

I say that "at least two products are formed" because depending on your reaction conditions 3 products might be expected.  NBS brominations can be carried out over a range of temperatures, from cold to somewhat above room temperature.  The activation energy required to rotate about an allylic bond is only ~15-16 kcal/m (see here).  Therefore, your 1-bromo isomer could exist as a cis-trans mixture, unless the reaction is performed at very cold conditions.

In the analogous case of NBS bromination of 1-octene, a mixture of 3-bromo-1-octene (~18%) and 1-bromo-2-octene (~82%) - both cis and trans are produced.

$\ce{RCH2CH=CH2  +  NBS \rightarrow RCHBrCH=CH2  +  RCH=CHCH2Br}$ (cis & trans)


According to the Merck Index${^{[1]}}$

  isobutanol is miscible with ethanol

1) O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 953


Yes they make sense but the same molecular formula can represent different compounds, for the first case $\ce{C3H6O3}$ may rappresent:
Glyceraldehyde:
or lactic acid (and its optical isomers):
$\ce{C4H8O4}$ leads to even more possibilities see the three different tetrose

Generally these compounds may be created when the sum of the formal charges of each atom in the molecular formula is equal to zero. Oxygen, hydrogen normally have formal charge of -2 and -1 carbon +4 so:
$$-2\times n + (-1 \times n \times 2) + (+4 \times n)=-4\times n +4\times n=0$$


What is the difference between pH and pKa?

$\mathrm{p}K_\mathrm{a}$ is the intrinsic property of an acid. It does not depend on how much acid there is in a solution‡. For example, acetic acid has a $\mathrm{p}K_\mathrm{a}$ of $4.76$. This $\mathrm{p}x$ means $-\log x,$ we can calculate $K_\mathrm{a}$ to be $10^{-4.76}$ is $1.7\times 10^{-5}$, which means that at equilibrium, $\frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = 1.7\times 10^{-5}$. On the other hand, $\mathrm{pH}$ is related to the actual amount of $\ce{H+}$ ions, so if you had more acetic acid, the solution would be more acidic ($\mathrm{pH}$ is lower). This is why on Wikipedia, you see $\mathrm{p}K_\mathrm{a}$ given for acids, rather than a $\mathrm{pH}$, because $\mathrm{pH}$ is amount dependent. Similarly, in experiments, we are more interested in $\mathrm{pH}$ because that tells us the actual acidity of a solution.

Also know that though $K_\mathrm{a}$ may always be constant,‡ percent ionization isn't; as there is more weak acid, they dissociate less.

‡ Just like the ideal gas law, aqueous equilibrium constants based on concentrations are for ideal solutions, but they is very accurate for most work.


Does carbonation permanently increase the acidity of soda water?

Both water samples ( assuming pure water ) would theoretically end up slightly acidic with $\mathrm{pH} \approx 5.6$, but the carbonated one after long time. Tap water would have its baseline $\mathrm{pH}$ determined by its $\ce{CO2(aq)/HCO3-(aq)}$ $\mathrm{pH}$ buffer.
This is valid for an ideal case, as side processes - especially for tap water -  will affect the result. As examples come to mind microbial/algae consumption of carbon dioxide, forming compounds affecting $\mathrm{pH}$, or precipitation of carbonates.


Does inversion of configuration necessitate reversed optical rotation?

if optical rotation of optically pure 2-bromobutane is some (+X)
  degrees, will optical rotation of the product be some (-Y) degrees?

No, there is no correlation between the configuration at a chiral carbon and the sign of optical rotation.

For example (from the Aldrich catalogue), the following chiral bromo compound, (R)-(+)-2-Bromo-3-methylbutyric acid,

has an optical rotation of [α]22/D +21°, c = 37 (c is the sample concentration in grams/100 ml solution) in benzene.  The inverted hydroxyl analogue, (S)-(+)-2-Hydroxy-3-methylbutyric acid,

also rotates light in the (+) direction.  Its optical rotation is [α]20/D +19°, c = 1 in chloroform.


The major ingredient in Diet Coke is water (H$_2$O). The second ingredient is carbon dioxide (CO$_2$), about 30 psig, or three times atmospheric pressure. (I measured it with an automobile tire gauge.) Since CO$_2$ is soluble about 1.7 g/Kg at 20$^o$C at atmospheric pressure, Ref 1, this is about 5 g/L, or ~0.5%, using a density for Diet Coke of 1 g/L Ref 2.

So, in the list of ingredients for Diet Coke, Ref 3, everything will be between a maximum of 0.5%, down to about 0.01295% ("46 mg caffeine/12 fl oz" on the label of my 2L bottle).
And, as Matt Black noted, the recipe varies from country to country, and from time to time (and is noticeably different in cans vs fountain delivery, according to those with highly developed preferences), but the big impacts are provided by
#1. water
#2. carbon dioxide
#3. artificial sweetener
#4. caffeine and
#5. flavoring, which might include a little more acid, like citric or phosphoric.
Seltzer is just #1 plus #2; many caffeine-free beverages just leave out #4.
Ref 1. https://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html
Ref 2. http://web.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/36.34.html
Ref 3. https://www.coca-cola.ca/brands/diet-coke; see also https://www.huffpost.com/entry/coke-zero-sugar-taste-difference_n_597f71d7e4b00bb8ff3875d9 for varieties of Diet Coke


I think the last step  is for the Hydrolysis.  After reaction with $\ce{LiAlH_4}$, the amide is reduced to amine and thus when it is reacted with a ketone it forms an imine($1G$) like this below,

which quickly tautomerises to form the most stable Enamine as this

After  that when it is reacted with $\ce{CH_3Br}$, the nitrogen's lone pair conjugates and the nucleophilic attack occurs from the adjacent carbon(like Michael addition) with attached methyl group. Thus you have the intermidiate substituted proiduct as,

But this imine  is hydrolysed in the presence of $\ce{H^+/H_2O}$ to give you back the original ketone but with an extra methyl group attached to it. So I think your final product $1H$ should be this,

and also you get the amine after hydrolysis which is actually the  same as $1F$.


Reaction stoichiometry for calculation of Kp

Each time one molecule of salt is decomposed, it produced $\ce{2 NH3}$ for $\ce{1 CO2}$. The number of molecules $\ce{NH3}$ is always double from that of $\ce{CO2}$ in the gas phase. Now the partial pressure of a given compound is proportional to the number of molecules of this compound in the gas phase. As a consequence, the partial pressure of $\ce{NH3}$ is always twice as high as the partial pressure of $\ce{CO2}$


Why does covalent bonding satisfy atoms?

You seem to forget that electrons have spins. And that there are two possibilities for the spin : up or down. No other possibilities exist. Now electrons have the tendency to make pairs with one spin up and on down. Sorry for the image, but it looks as if spins up and down represent two different sexes. And that the electron have the tendency to make "couples" of "up" and "down" together, as in the traditional way of being together. On the average, there exist the same amount of electrons "up" and of electrons "down".

Let's take H atoms. Half to the H atoms have spins up, and half down. If you produce somewhere a lot of free H atoms, they will collide. When they have the same spin, nothing happen after the collision. When they have different spins, they will join and stay together, make a covalent bond, which makes a H2 molecule,  after the collision. After a while all H atoms are groups in H2 molecules

In the next atom, Helium, there are two electrons on the first layer. They are of opposite spins and Helium will not join with any other atom, and not produce any molecule.

For heavier atoms, form Li to Ne, there is enough room for 8 electrons. The first four have the same spin. The next four have the opposite spin. Hopefully you may understand your self that Li makes LiH, Be makes BeH2, etc. But Oxygen has 6 electrons, 4 of a given spin and 2 of the opposite spin. In Oxygen, there are two doublets which cannot form covalent bonds. Only 2 can do it.

I will not develop this theory too long. You may do it your self. It is not too difficult to do it.


Ignoring initial H+ Concentration For Addition of Strong Base to a Buffer Solution

Suppose you have a weak acid and its corresponding base in solution. Suppose the solution has the pH = 5.

HA + H2O = A- + H3O+

Now we add a strong base to the solution. What will happen?

The pH will increase, suppose it will increase to pH = 6.

The weak acid-base system will immediately respond to the pH change of the solution and a new equlibrium will be established defined by the new pH and the equilibrium constant of the weak acid-base system. The equilibrium will be forced to the right, i.e. more A- will be fomed at pH = 6 compared to what we had at pH = 5.

Ok, what has happened to the hydrogen ions present at pH = 5 or those that were formed up to

pH = 6?

They have reacted with the strong base forming water.

Do we think the concentration of the hydroxide ions from the strong base has decreased at pH = 6?

No.

Why?

At pH = 6 the hydrogen ion concentration is lower than at pH = 5. This is equal to say that the hydroxide ion concentration is higher at pH = 6 than at pH = 5. The difference in the amount of hydroxide ions present at pH 6  and 5 will be equal to the amount of hydroxide ions consumed in the transfer from pH 5 to 6. What you put in, you get back in this case. The net effect is that the hydroxide ion concentration of the strong base has not changed at the equilibrium of the process.

I suggest you also look at my answer for another question "Buffer and strong acid".


Does fumaric acid form maleic anhydride upon heating?

Fumaric acid does form maleic anhydride due to partial combustion. From wikipedia article of fumaric acid:

  Fumaric acid converts to the irritant maleic anhydride, upon partial
  combustion.

Also, dehydrating fumaric acid with a  dehydrating agent also forms maleic anhydride. This paper describe how to convert $\ce{C^{14}}$ labelled fumaric acid to maleic anhydride.

  Maleic anhydride-2-$\ce{C^{14}}$ is readily obtained from fumaric
  acid-2-$\ce{C^{14}}$ through dehydration using phosphorus pentoxide
  getting 96% yield under controlled condition.


Which thermodynamic ensemble does Gaussian's BOMD functionality sample?

@pentavalentcarbon: As of now (Gaussian16), I believe that BOMD calculations can only be run in the NVE ensemble.

However, you can do ADMP (also available with Gaussian and uses an extended Lagrangian scheme), and offers equivalent functionality to BOMD. Thermostatting is available here, but the caveat is it implements only the simple velocity rescaling algorithm. On performing simple tests this seems to be "nearly" canonical on some systems I have worked on. i.e small ion-water clusters.


A way to calculate orbital energy of an electron

You are correct that the ionization energy of an electron depends on the orbital. There is no easy formula, because the electrons in atoms are governed by quantum mechanics and for multi-electron atoms, the equations are un-solvable algebraically.

That said, there are some reasonable tabulations, which have been verified by calculations and by spectroscopy, e.g., http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf (from a quick Google search).

Both ultraviolet photoelectron spectroscopy and x-ray photoelectron spectroscopy can be used to eject electrons from various materials, and you can thus identify elemental composition (and other properties) from the known orbital energy levels


Cause of m/z = 56 peak in 4-methyl-1-pentanol

Unfortunately I did not find a high resolution spectrum of this compound, which would have quickly answered your question.

Nevertheless, ionized alcohols and even more primary alcohols have a main fragmentation pathway which is the loss of water to yield an ionized alkene. In this case, the corresponding alkene would be 4-methyl-1-pentene, at m/z 84. The peak is present although rather small in the EI spectrum. From there, it is quite obvious, when one looks at the fragmentation spectrum of 4-methyl-1-pentene (NIST Webbook spectrum), that it shares a lot of similarities with the 4-methylpentan-1-ol EI spectrum. (See for instance peaks at m/z 69, 56, 43, 42, 41.) From there, it is quite likely that the m/z 56 peak is ionized isobutene ($\ce{C_4H_8}^{\bullet +}$) arising from a retro-ene type rearrangement.


How do I model non-covalent interactions in Gaussian?

As hBy2Py, Greg, and others have pointed out in their comments, the cation is just a counter ion for your benzothiazole-based cyanine dyes.

I suggest to be reasonably lazy and perform the calculations on the cyanine dye only.

Even for calculations in solvents of sufficient polarity (from an experimental photochemical perspective, acetonitrile seems to be a good choice), I would just ignore the cations.


What is the scientific term to describe the operation of a bong?

I think the term describing the initial process of separation of the solid phase (dust particles) from the gas phase via bubbling is a wet scrubbing process.

On an industrial scale, it's more efficient to spray liquid phase, whereas on a laboratory scale bubbling in a compact glass gas scrubber or a gas washing bottle is more convenient.
In order to increase efficiency of the lab apparatus, it's preferred to

keep the surface between gas bubbles and liquid phase as large as possible;
increase time of contact.

This is usually achieved by producing many smaller bubbles by using a spiral tubing (larger particles), frit of long columns filled with Raschig rings.

Figure 1. Left: Büchi miniScrubber, a glass gas scrubber (Source: Büchiglas — Gas scrubber). Right: Corning PYREX® 125mL Gas Washing Bottle with Extra Coarse Fritted Disc (Source: Thomas Scientific)

Eventually collected insoluble dust particles tend to settle out from the solution; this a basis for sedimentation, another separation technique.


Why does Oxygen act as Nucleophile over here?

You may be seeing the impact of the aromatic ring.  Putting a negative charge on carbon in your structure disrupts the aromatic coupling.  So carbanion contributions are reduced and the oxyanion contribution, where the oxygen is nucleophilic, is favored.


In any element after Argon [1s22s22p63s23p6],
the repulsion force between the elctrons increase at a great extent and also the atom becomes very stable. So the elctrons tend to move in far 4s orbital first and after filling it completely , fill the 3d orbital. About chromium and copper, they are exceptions. Any half filled or fully-filled orbitals are more stable then rest of the configurations, so in Chromium one electron of 4s comes to 3d orbital and a stable 3d5 is obtained, similarly in Copper one electron from 4s comes to 3d orbital and forms stable 3d10.

I hope this answers your question,
If you have any further doubts you can ask me :)


When SiO2 changes state, what forces/bonds are being overcome/broken?

$\ce{SiO_2}$ is a network covalent compound that has an extremely high melting and boiling point, because many silicon-oxygen bonds have to be broken in order for it to achieve the necessary freedom. To clarify, SiO2, which has a tetrahedra network lattice formation, shows that each silicon is actually bonded to 4 oxygens; each oxygen is bonded to 2 silicon. These excess bonds aside from the ones of SiO2 are broken.


Rate-determining step of Claisen condensation

Let me get one thing out of the way: the hydride ion is not a strong base. However, once a hydride is protonated to form hydrogen the acid-base equilibrium breaks down which makes it seem like a much stronger base than it is.

Your actual question — whether the rate determining step shifts from enolisation to nucleophilic attack — is harder to answer as it depends primarily on the reaction procedure. It depends on how you perform the Claisen condensation and what you want to condense.

If, for example, you want to perform a hetero Claisen condensation (i.e. couple two different esters), then you would typically make sure to deprotonate quickly and kinetically first before adding the second ester as Zhe implied in a comment. For example, you might use a very strong base such as $\ce{BuLi}$ or $\ce{LDA}$ for complete deprotonation followed by a waiting period before adding another ester. In this case, obviously you are already dealing with a fully deprotonated product and the question becomes moot.

However, you might be interested in a homocoupling product. For that, you might decide to add $\pu{0.5eq}$ of a strong base to a solution of the ester. Here, it depends on whether you add the strong base rapidly, (meaning the acid-base equilibrium is practically immediately established, in turn meaning that now the attack becomes rate determining) or whether you add it slowly and dropwise, in which case you, the experimentor, are rate-limiting deprotonation and causing it to remain the rate-determining step.

Finally, in response to a comment you gave under your question: You cannot use a protic solvent if you intend to use a strong base. Protic solvents are characterised by a relatively high acidity, meaning they will be deprotonated long before the ester ($\mathrm{p}K_\mathrm{a} \approx 25$). This will completely inhibit the reaction.


Can a solution of FeCl2 go bad over time?

Water evaporates. Water dissolves the fixed gases ($\ce{N2}$, $\ce{O2}$, $\ce{CO2}$). A well-sealed bottle of water will last for years - that's assuming that we understand that 'bottles' are made out of heavy glass*, and seals are permanent (in the sense that opening the bottle is irreversible). That's not a particularly useful way to store stuff you need access to. (In other words, most bottles allow only limited storage life.) $\ce{O2}$ can oxidize $\ce{Fe(II)}$. Unless working under (and stored under) inert atmospheres, the titer will inevitably change over time. Even when working in inert atmosphere, there will be evaporative losses. All that said and assuming that there wasn't any accident involving the reagents, nor changes to their storage conditions (temperature, location, etc.) then the decline in $\ce{Fe(II)}$ should be roughly proportional to the number of times the bottles are accessed. Assuming that that is regular, with no recent upsurge, then the change in titer should be gradual over time; no sudden change should have occurred.(*please don't tell me that they're storing reagents in plastic! yet still any changes should be gradual, not sudden.)


Making elemental sulfur from pyrite FeS2

You may just need to heat it.

$$\ce{FeS2 ->[\Delta] FeS + S}$$

In the wikipedia article of pyrite, it is written that:-

  Thermal decomposition of pyrite into $\ce{FeS}$ (iron(II) sulfide) and elemental sulfur starts at 540 °C.

A good experiment for recovering sulfur from iron pyrite is called ' steam-pyrite interactions'.  The following is an excerpt from the original journal:-

  The reactions of natural iron pyrite with steam, hydrogen, and carbon monoxide have been studied in a fixed
  bed tubular reactor at 900-1 100°C. A new process has been developed for the recovery of elemental sulfur
  by the reaction of pyrite with steam. In this process, the yields of elemental sulfur and sulfur evolved as hydrogen
  sulfide and sulfur dioxide were, respectively, 84.5 and 13.7% of total sulfur under the following
  conditions: particle size, below 100 ASTM mesh; temperature, 1000°C; reaction period, 1 5 min; space velocity,
  33.2 X lo4 cc/cc/hr; and partial pressure of steam, 1 atm. Under these conditions, the effluent gas also
  contained an appreciable amount of hydrogen. The calcine contained ferrous oxide, ferrosoferric oxide,
  silica, and 1.3% sulfur. The residual sulfur could be further brought down to a minimum of 0.2% in the calcine
  at 1 100°C using a 30-min reaction period. The incorporation of hydrogen in the steam-iron sulfides reactions
  at 1 100°C increased considerably the yield of hydrogen sulfide (59.5%), but only a small amount of metallic
  iron (1.6%) could be obtained.


What's the definition of multi phase materials? Does it mean the different phases represent different materials?

The definition of a substance that is "multi-phase" is that there is more than one distinct compound in it and the compounds form distinct regions in the substance with different properties.

Examples make this clearer. Salt dissolved in water forms a uniform mixture: a salt solution. There is more than one compound there but they are perfectly mixed.

Fatty organic compounds are not soluble in water but can form small dispersed fatty particles. We call the result milk (I'm simplifying a little as there are other things in there as well). The result has more than one phase: a microscope would reveal very small, dispersed particles of fat floating in water. Milk looks like a single uniform substance, but isn't when you look more closely.

Fibreglass is, essentially, a bunch of small fibres made of glass embedded in a matrix of some polymer. There are two distinct phases with different compositions in the substance.

The characteristic of multi-phase compounds is that different regions of the substance contain different compounds (these regions can be very small as in some metal alloys). In your case the reaction seems to produce two distinct compounds which may be finely mixed together. The XRD pattern shows two different crystal structures. This may result because this specific reaction does not create the compound you wanted under these conditions. Which could be because the compound you want is not stable or could be because the reaction conditions allow more than one product to form. Either way the reaction is giving a mixture not a single compound.


How to calculate the formal charge of sulfate ion?

Your formula is correct. If $V$ is the number of valence electrons originally assigned to an atom, $N$ is the number of nonbonding electrons, and $B$ is the number of bonds ($\frac{1}{2}$ of the number of bonding electrons to be more precise), then the formal charge $FC$ is:

$$FC = V - (B+N)$$

Sulfate ion has two valid structures that you could draw, one with sulfur having a formal charge of zero and one with sulfur having a formal charge of +2. The structures below (from the Wikipedia page, released into the public domain) follow the annoying but allowable convention of replacing lone pairs with bars.

If the structure on the left (#1) is correct, then the sulfur atom ($V=6$) has six bonds ($B=6$) and no lone pairs ($N=0$). Two oxygen atoms (V=6) have two bonds ($B=2$) and two lone pairs ($N=4$), while the other two oxygen atoms have one bond ($B=1$) and three lone pairs ($N=6$). The sulfur atom and two oxygen atoms have $FC=0$ and the remaining two oxygen atoms have $FC=-1$ for an overall charge on the ion of $-2$.

$$FC_{\ce{S}}=6-(6+0)=0$$
$$FC_{\ce{O_{1,2}}}=6-(2+4)=0$$
$$FC_{\ce{O_{3,4}}}=6-(1+6)=-1$$

If the structure on the right (#2) is correct, then the sulfur atom ($V=6$) has four bonds ($B=4$) and no lone pairs ($N=0$). All four oxygen atoms (V=6) have one bond ($B=1$) and three lone pairs ($N=6$). The sulfur atom has $FC=+2$ and the oxygen atoms have $FC=-1$ for an overall charge on the ion of $-2$.
$$FC_{\ce{S}}=6-(4+0)=+2$$
$$FC_{\ce{O}}=6-(1+6)=-1$$

But, wait! A more fundamental question might be "How can the sulfate ion have two very different structures that puts different formal charges on the sulfur and oxygen atoms?"

Formal charge is a nice bookkeeping mechanism, but has no experimentally valid relationship to the actual charge on any atom in most molecules or ions. Formal charge only equals actual charge on monoatomic species. Formal charge falls into the category of models that we use in chemistry that are 1) helpful, 2) produce the correct answer when used correctly, and 3) completely bogus.  Other models in this category include oxidation number, VSEPR, resonance, and electronegativity. The experimental evidence suggests the real structure of sulfate combines features of both structure #1 and structure #2, but would be challenging to draw using the formalisms we have adopted:

All four $\ce{S-O}$ bonds are equal in length (#2).
The length of the $\ce{S-O}$ bonds is shorter than a normal $\ce{S-O}$ single bond and longer than a normal $\ce{S-O}$ double (#1).
The sulfur atom has a partial positive charge (#2) (note that partial charges, unlike formal charges, have some experimental basis).
The four oxygen atoms have equivalent partial negative charges (#2).
The partial negative charges on the oxygen atoms add to more than $-2$ but not anywhere near $-4$ (#1).


This confusion pops up often around here.

The problem usually comes down to insufficient consideration of units.

When you say "rate", what do you mean?  It has units of mol per volume per second, e.g. $\mathrm{\frac{mol}{L\cdot s}}$.  But that isn't enough information.  You need to know moles of what.

First example: if the rate is defined as loss of A

For example, if you say that the "rate" means the disappearance of A, then the units of the rate are really $\mathrm{\frac{mol~of~A}{L\cdot s}}$.  If so, then as long as:

The reaction is $\ce{2 A->B}$; and
The rate as you have defined it can be expressed by $r_A = k_1~[\ce{A}]^2$ ...

Then the rate of appearance of B will be $r_B = \frac{1}{2} k_1~[\ce{A}]^2$

Second example: if the "rate" is defined as appearance of B

If conversely you suppose that the "rate" refers to the appearance of B, then the units of rate are really $\mathrm{\frac{mol~of~B}{L\cdot s}}$, and if so, then as long as

The reaction is $\ce{2 A->B}$; and
The rate as you have defined it can be expressed by $r_B = k_2~[\ce{A}]^2$ ...

Then the rate of disappearance of A will be $r_A = 2 k_2~[\ce{A}]^2$

Third example: a totally separate definition of rate

You say its an elementary reaction.  Let's say that we didn't know that and wrote the reaction as $\ce{200 A -> 100 B}$.  The rate, according to this new example definition, is the rate of the reaction.  The units of the rate are $\mathrm{\frac{mol~of~"reaction"}{L\cdot s}}$. If the reaction is still 2nd-order then the rate of the reaction will be $$r_r = k_3~[\ce{A}]^2$$

According to our reaction, 200 moles of A disappear for each mole of reaction, so $$r_A = 200 k_3~[\ce{A}]^2$$  and 100 moles of B appear, so $$r_B = 100 k_3~[\ce{A}]^2$$

Final thoughts

You will notice that I used different subscripts to distinguish between the $k$ values for the different assumptions.  That's because they aren't equal to each other.  The value of the "rate constant" $k$ depends on how you define the rate.

No matter how we define "the rate" and what we define as the stoichiometry (e.g. $\ce{2A -> B}$ or $\ce{A -> 1/2B}$ or $\ce{200 A -> 100 B}$, the rates must be equal.  Therefore:

$$r_A = 200 k_3~[\ce{A}]^2 = 2 k_2~[\ce{A}]^2 = k_1~[\ce{A}]^2$$

That is,

$$ 200 k_3 = 2 k_2 = k_1$$

But no matter which way we decide to write it, it is still a second-order reaction because "the rate", no matter how we define it, is proportional to $[\ce{A}]^2$.


Does Photoelectric effect occur in a micro-oven while heating food?

No, the photoelectric effect does not occur in a microwave oven.

The workfunction for a metal is the minimum photon energy required to eject an electron from the metal via the photoelectric effect.  Workfunctions are typically in the range of several electron volts (eV).  Potassium, for example, has a workfunction around 2 eV, see here for a list..

The energy in a microwave photon can range from 0.00001 to 0.001 eV; much to low to eject an electron from a metal.

For more background on the photoelectric effect see this earlier SE Chem Q&A


Reaction between chlorobenzene and methyl chloride

You will actually get a mixture of different products, with the product with the highest yield being p-chlorotoluene, or 1-chloro-4-methylbenzene. The other products resulting from the reaction include o-chlorotoluene and m-chlorotoluene, with a significantly higher amount of the former being produced*. In fact, there would also be products with more than one methyl substituent being produced in various proportions depending on the precise conditions being used.

Yes, the reaction will proceed by Friedel-Crafts alkylation: Firstly, the Lewis acid catalyst abstracts the chloride ion from methyl chloride, giving a reactive, electrophilic methenium ion. Then, electrons from the $\pi$ system of the benzene ring of chlorobenzene then attack the electrophilic ion. Finally, a proton is lost to restore aromaticity of the ring, giving the product.

*Refer to the following post to understand why the chlorine atom as a substituent would be ortho/para-directing: Why are halogens ortho para directing even though deactivating?


Armarego's Purification of Laboratory Chemicals suggests using dichlormethane–hexane pair instead of ethanol–water; it also lists crystallization from warm ethyl acetate, ethanol or petroleum ether as alternative [1, p. 342].

  Dibenzalacetone [DBA, bda, trans-trans-1,5-diphenyl-1,4-dien-3-one, 1,5-(bisphenyl)-penta-1E,4E-diene-3-one] [538-58-9] $\ce{C17H14O}$, $M~234.3$, $m~107°$, $111°$, $112°$, $113°$, $120-122°$.
  Purify the ketone by flash chromatography (150 mesh $\ce{Al2O3}$ deactivated with $6\%$ v/w $\ce{H2O}$) using  petroleum ether $40-60°/\ce{EtOAc}$ ($4/1$, v/v), and the yellow solid is recrystallised by layering a concentrated $\ce{CH2Cl2}$ solution with hexane (i.e. $\ce{CH2Cl2}$/hexane, $1:4$, v/v).
  Also recrystallise the ketone from hot ethyl acetate $(\pu{2.5 ml/g})$ or $\ce{EtOH}$.
  It has been also crystallised from petroleum ether b $30-60°$.

References

Armarego, W. L. Purification of Laboratory Chemicals; Elsevier: Boston, MA, 2017. ISBN 978-0-12-805457-4.


Can I remove flesh from seeds with hydrochloric acid?

It depends highly on the seed and the fruit flesh, but using hydrochloric acid to clean off the "jelly" from freshly collected tomato seeds is a recognized laboratory protocol. That protocol uses a 1:1 mix of concentrated HCl and water for 20 min., followed by a 20 min wash in 10% w/v trisodium phosphate to neutralize the acid.

Depending on what sort of flesh is covering the seeds, it may take different amounts of time and different strengths of acid to accomplish the job. You will likely have to experiment a bit. Start "gentle" (as in, less concentrated acid) and work your way up. - You probably don't want conditions that results in complete flesh removal, but rather ones where gentle agitation (e.g with an old toothbrush) after neutralization will dislodge the flesh. - Viability of the seeds will also vary depending on the exact protocol and how tough the seeds are, though it doesn't sound like you need the seeds to be able to germinate afterwards.

As always, when working with concentrated acids and bases, use of proper safety equipment is highly recommended - even for acids and bases they sell in the hardware store.

For the chemical perspective, concentrated acid solutions catalyze the hydrolysis of peptide, ester and glycosidic bonds of the polymers that make up the flesh of the fruit surrounding the seeds. With the polymers degraded, they more easily dissolve in solution. (This is also the reason that our stomachs have hydrochloric acid in them.) The cellulose that makes up the seed coat is recalcitrant to degradation, and so remains untouched to the "mild" acidic conditions.


I've never used it for that particular experiment, but bromocresol purple has a similar transition range and a very obvious colour change. The trouble with using indicators here is that many are organic compounds and are subject to redox reactions near the potentials at which electrolysis of water is occurring, so they break down or polymerize if they reach the electrodes. Bromothymol blue is known to oxidize at below 1 V and deposit on the electrode. Bromocresol purple tends to polymerize, but it seems to be at a relatively high potential (>1.5 V), so it may fare better.

It may also be possible to limit the diffusion of the indicator to the electrode surfaces. Perhaps using an indicator trapped in polymer beads or a block of agarose gel.


You are confused about several overlapping but mostly unrelated issues.

First, you are right that "benzyl" is the $\ce{C7H8\bond{~}}$ radical and that "phenyl" is the $\ce{C6H5\bond{~}}$ radical.

Second, although "phenol" and "phenyl" sound almost the same, you would probably do well to think of them as entirely separate naming roots.  "Phenol" is the simplest aromatic alcohol.  Simple aromatic alcohols with other substitutions (e.g. 4-chlorophenol) can thus be named as substituted phenols.

Third, "benzyl" and "benzene" similarly sound the same but should be treated as distinct concepts.  Substituted aromatic rings can be named as benzene derivatives, such as chlorobenzene.

Fourth, for shorthand and convenience, "phenyl" and "benzyl" are sometimes used to name substituents.  Note that IUPAC has detailed priority rules for what is a substituent and what is the base molecule, but chemists don't always follow these.  "chlorobenzene" is a correct IUPAC name, and it views the molecule as a substituted benzene.  That is, benzene is the base molecule and "chloro" is a substitution.  Chemists sometimes find it convenient to reverse their perspective, even though the resulting names would not be IUPAC-approved.  From a reverse perspective, the chlorine atom is the "base molecule", and it is substituted by a phenyl group.  Thus you could call the chlorobenzene "phenyl chloride" if you wanted to take this reverse perspective.

Lastly, "bromophenyl" ends in "-yl" and so is the name of a "radical", i.e. a substituent or fragment, not of a complete molecule.  Without any other context, I would think that "bromophenyl" was a $\ce{C6H4Br\bond{~}}$ substituent.  In contrast, "bromobenzene" is a complete molecule, derived from benzene by substitution.


How do CO2 Fire extinguishers procure their CO2

They just buy it from companies producing CO2.
Actual sources may be sequestration from combustion (e.g. power plants) or fermentation (e.g. ethanol production) processes.


What happens if I draw too much current from a home made battery?

Yes, you're right. There will a maximum speed that is determined by the kinetics of several parts of the system.

Electronic Resistance: Pretty straight-forward. All the parts have some inherent resistance. Though this tends to be minial compared to some of the other causes.

Migration: At high currents, the ions become depleted (or in excess) at the electrode's surface. This means that the reaction will not run at the same potential until new ions diffuse from the bulk.

Adsorption: In order for the redox reaction to occur, there is often a step needed where the ions adsorbs to the metallic surface. This is a sort of chemical reaction, so has a rate constant associated with it.

Chemical Reaction: Some batteries can also involve a chemical reaction to form an activated reactant.

Electron Transfer: This is the actual redox part itself. Transferring an electron from the metal to the reductant, or vice-versa.

All of this will play a role, but I mostly think of migration effects when I think of kinetic limitations. Once that surface layer is depleted, all of the other processes stop too.

This is most likely to be worse for a home-made battery. The electrical resistance will probably be higher because there are longer wires, less reliable construction, etc. Distances between the electrodes is longer, so in the long run there's more distance for ions to travel to equalize a charge imbalance. Your electrode surfaces are likely to be dirtier, causing slower absorption.

The book Electrochemical Methods by Alan Bard and Larry Faulkner is a good resource if you want to do a deep dive.


First, the methoxy- group can be demethylated by $\ce{AlCl3}$, beginning with $\ce{AlCl3}$ coordination to the ether oxygen, followed by an $\mathrm{S_N2}$ reaction:

$\hspace{3.6cm}$

$\hspace{0cm}$

The oxygen activates naphthalene's position 1, and the acid chloride will undergo a Friedel-Crafts acylation (as you noted):

$\hspace{.5cm}$

Being a strong Lewis acid, $\ce{AlCl3}$ can coordinate with the carbonyl oxygen. This activates position 4 of the $\alpha$-$\beta$ unsaturated carbonyl towards electrophilic addition:

$\hspace{.75cm}$

Electrophilic addition occurs at naphthalene's spatially adjacent position 8, the carbocation stabilized through the extensive conjugation of the molecule (as well as by the oxygen, as you noted):

$\hspace{1.75cm}$

The spatially adjacent phenyl group accepts the proton and is ejected as benzene:

$\hspace{1cm}$

Addition of water evolves $\ce{HCl}$ and precipitates $\ce{Al(OH)3}$, yielding the final product:

$\hspace{1.8cm}$


Am I using the Nernst equation correctly?

I will show you why $\ce{0.295}$ is correct. :-)

Nernst equation can be written in different ways, I let you choose which you prefer. So for any reaction such as $$\ce{\nu_{Ox}Ox + n e^- \rightleftharpoons  \nu_{Red}Red}$$

One can show that at $\ce{25°C}$,

$$E_{Ox/Red} = E_{Ox/Red}^{°} + \frac{RT}{nF}\ln\left(\frac{a_{Ox}^{\nu_{Ox}}}{a_{Red}^{\nu_{Red}}}\right) \tag 1$$

Which can also be written as, see after how,

$$E_{Ox/Red} = E_{Ox/Red}^{°} + \frac{0.06}{n}\log\left(\frac{a_{Ox}^{\nu_{Ox}}}{a_{Red}^{\nu_{Red}}}\right) \tag 2$$

Where $\log$ is the logarithm in base $\ce{10}$.

So we have $$\frac{RT}{F}\ce{=\frac{8.314 \times 298}{96500}}$$
And we know that $\log(x)=\ln(x)/\ln(10)$ then the coefficient next to $\log$ is $$\frac{RT}{F}\times \ln(10)=\frac{8.314 \times 298}{96500} \times \ln(10) = 0.0591 \approx 0.06$$

To be a little more perfect I sometimes use $\ce{0.059}$ instead of $\ce{0.06}$

So for two electrons exchange the factor coefficient is $\ce{0.059/2}=0.295$


Finding out the composition of seawater/brine?

Your budget should allow you to measure  $\ce{Mg^{2+}, Ca^{2+}, Li+ and Cl-}$ in-house using ion specific electrode analysis. There will certainly be some training and maintenance involved, but anyone with a bit of a chemistry background will be able to learn this technique. Adding to the simplicity, and thus the ease of training, is the fact that your samples are all of a fairly similar composition.

Ion chromatography, which would basically require one set up for anions and one for cations, would break your bank in a hurry unless you have an experienced IC analyst with some time to spare and access to good used equipment.  Frankly, this just doesn't seem like a realistic approach.

One thing to keep in mind is that your budget will buy a lot of analysis by people with expertise in whatever analysis you choose to perform. They will even help you choose the best analytical technique for your needs.

My advice is to look at outsourcing first, particularly if you are going to be analyzing fewer than 10-20 samples per day.  If you will be analyzing a large number of samples, or if you need very quick turnaround times, then I suggest looking into ion specific electrode analysis.


Reference Books on Electrolysis and Chemical Titration process

Any good general chemistry textbook has a chapter on electrolysis. If you wish to add more scholarly work check the Journal of Chemical Education. https://pubs.acs.org/action/doSearch?text1=Electrolysis&quickLinkYear=&quickLinkVolume=&field1=Title&type=within&publication=346464552

In the first part you electrolyze a solution of salt. Check the half cell reactions from the text: What happens at the cathode and what happens at the anode?

Water is reduced at the cathode and -OH ions are produced. This makes your salt solution basic. In the second part, your lab requires you to estimate how much base was produced.*

Essentially you do a titration of NaOH with standardized HCl. The end point is estimated by a pH meter. Look up any general chemistry textbook again for titration of a base with HCl using a pH meter. The technique is called pH titration. You did a very rough version of a pH titration.

Depending on the conc. of NaCl, either chlorine or oxygen is produced at the anode.


Why is magnesium sulfate acidic? Why are my calculations showing that it is basic?

Your answer lies not in the solubility product but in the acidity constants of sulfuric acid and the basicity constants of $\ce{Mg(OH)2}$, though the solubility product may well come into play. You would need to solve a system of 6 equations, one for each of the:

2 $\mathrm{p}K_\mathrm{a}$ values for the acid
Two $\mathrm{p}K_\mathrm{b}$ values for the base
Proton condition or charge neutrality
And as you will be groping in $\mathrm{pH}$, one to be sure you don't exceed the solubility product.

As sulfuric acid is 'stronger' as an acid than $\ce{Mg(OH)2}$ is as a base, you would expect the answer to be slightly on the "acid side" of neutral $\mathrm{pH}$.


Volatile (aroma/scent-defining) compounds from Himalayan Balsam (Impatiens glandulifera)?

There's seemingly a lot published on the invasion dynamics of Impatiens glandulifera, but little on the scent-giving volatile compounds.

All I could find is a poster from a 2010 symposium on Terpenes - Application, Activity and Analysis, organized by the Phytochemical Society of Europe (PSE).

The authors presented a method to sample volatile compounds from the flowers of the neophytic Impatiens glandulifera and the native Impatiens noli-tangere without cutting them off the plant and analyzed the mixture of compounds by GC-MS.

The scents seem to stem from complex mixtures of monoterpenes.

In the flowers of both plants the following terpenes could be detected:

$\alpha$-pinene
$\beta$-pinene
camphene
1,4-cineole
d-limonene
1,8-cineole
$\alpha$-terpinolene
1-terpineol
$\alpha$-terpineol
$\beta$-terpineol

Only in Impatiens noli-tangere were found

(Z)-ocimene
fenchol
geraniol

Only in Impatiens glandulifera was found

$\beta$-phellandrene


What is the mechanism of this pyrylium synthesis?

I tried to work out a mechanism from the information I could gather about the reaction. If someone notices something odd/wrong please comment and I'll fix it.

The steps are either showing the protonation or skipping directly to the protonated forms of the reactants, them being in acidic medium.

Formation of isobutylene

Not very complicated, in acidic conditions we remove the hydroxyl group from the t-butanol to form isobutylene. The latter behaves as a nucleophile in the next steps of the reaction. This step is acid catalysed, so the acid is not consumed.

Formation of Mesityl Oxide

Also relatively simple step. Nucleophile addition to the protonated C=O bond [Adn] followed by elimination of Acetic Acid.

Tautomerism

Shows the Mesityl Oxide tautomerism, important because the tautomer that will react to form the final product is the one in the right (with the double bond in the terminal carbon).

Enol Formation

Here a Enol is formed by a nucleophilic attack of the Mesityl Oxide over another Acetic Anhydride molecule (protonated by the acidic medium). In the last step I added the deprotonation of the more acidic oxygen to respect Pauling's rule of charges by keeping an overall +1 charge in the molecule. The left oxygen is more acidic since the other oxygen can better stabilize the charge with delocalization. I wrote the molecule with wavy bonds because I'm unsure about the stereospecificity of the step, but maybe the Z conformation is favored due to the charges on the oxygens repelling themselves?

Cyclization

Final step is cyclization of the intermediate, by nucleophilic attack of the carbon by the deprotonated oxygen. The intermediate is dehydrated and rearranged to form the final product (releasing a proton to reach aromatic status). This step had an intermediate with +2 charge, which I'm a little skeptical about due to the Pauling's rule of charges, so I also draw the mechanism as a concerted step:

If the rule of charges isn't that strict, maybe the non-concerted path is a closer description of what's happening, but I'm keeping both here for now. If anyone thinks one of the two is the right one, comment and I'll update the answer.


How many products does this reaction yield?

The expected products of mono-chlorination of 2,4-dimethylpentane are depicted in following scheme:

The substrate has 12 primary hydrogens (on 1-,1'-, 5-, and 5'-methyl groups),  2 secondary hydrogens (on 3-methylene group),  and 2 tertiary hydrogens (on 2-, and 4-methine groups), as numbered in the scheme. Altogether, total of 3 different hydrogens are present, therefore, 3 different mono-chlorinated products are possible as sketched in the bottom boxes of the scheme. Although 1- and 1'-methyl groups (and 5- and 5'-methyl groups as well) are equivalent in the structure, mono-chlorination of each would give different stereoisomer (strictly speaking, enentiomers). Therefore, counting stereoisomers, you should get total of 4 different products.


To get directional emission from a molecular substance, just make it into a LASER. Actually, the predecessor to the LASER, the first MASER, was based on microwave emission from ammonia. A $\ce{CO2}$ LASER is quite efficient and fairly easy to make, and the atmosphere of Mars is a natural, solar-pumped laser. So if you want directional emission from molecules, it's not even necessary to align them -- just provide suitable end-reflectors, or even a linear medium with enough gain.

That said, your question is about aligned materials... and there is a strong correlation between alignment and emission. An interesting example is found in topological insulators, which conduct only in certain directions and locations. It is possible to make a LASER from topological insulators. Some of these are as simple as $\ce{Bi2Se3}$, and others are more complex such as bis(methylene) adamantyl carbocation.


We have PV=nRT
(where symbols have there usual meaning.)

You see V is constant as the container doesn't change or deform ( as you mentioned)

so thinking that way PV/nRT=constant
as moles are constant and not leaking
R is universal gas constant .
so ;

P/T is constant
then $P_1/ P_2 = T_1/T_2$


Chemical equation to represent the hydrogen peroxide scavenging experiment, used to determine antioxidant ability of tea

I start by quoting a source (which also goes into great detail on explaining the underlying mechanism of chromophores excitation):

  Photosynthesis is initiated by the absorption of light by the chromophores, which excites the molecules from the ground state to an electronic excited state.

So, simply:

$\ce{Chromophores + hv -> *[Chromophores] -> [Chromophores](e-,h+)}$

with the creation of an electron and an electron hole. Now, absence the H2O2, the action of the electron-hole on OH- on water creates an hydroxyl radical:

$\ce{OH- + h+ -> .OH }$

Also, in the further presence of H2O2, a path to the so-called one-electron reduction of hydrogen peroxide from the action of a formed electron (e-) on H2O2, resulting in an additional hydroxyl radical:

$\ce{H2O2 + e- (aq) -> .OH (aq) + OH- (aq) }$

Now, as the tea acts acts as a radical scavenger, it reacts with the hydroxyl radical usually via an H atom extraction reaction forming water. For example:

$\ce{RH + .OH -> .R + H2O}$

The hydroxyl radical continues to attack the chromophores until there is just simple colorless products like CO2 and H2O. In time, the tea solution becomes colorless.

Now, in the combined chromophores and H2O2 experiment, the UV absorbance by tea solution should measurably display more decline with time due to the increased hydroxyl radical concentration.


Mechanism for hydrolysis of enol ether to aldehyde

Your issue is hardly with the Wittig reaction, although one could very well nitpick. The addition of phosphonium ylid to carbonyl is generally thought to be concerted: see Which is the currently accepted mechanism of a Wittig reaction? for more details.

From what I can tell, you are having more trouble with the acidic hydrolysis of an enol ether to an aldehyde. As has been pointed out by some comments, your proposed mechanism is not really possible. Vinyl $\mathrm{sp^2}$ cations are very unstable and an $\mathrm{S_N1}$-type dissociation of MeOH is very unlikely.1

Instead, you need to use the fact that an enol ether is very much like an enol. It is nucleophilic on the α-carbon, and you can protonate it on that carbon, almost like you are tautomerising an enol back to a ketone. You could also protonate it on oxygen, but that is an unproductive route: nothing really happens apart from the proton just falling off again.

The oxonium ion thus formed is electrophilic, and water can attack it leading to a hemiacetal.

From here on it is standard carbonyl chemistry.

I have deliberately left some stuff quite vague here.

Although it's acceptable shorthand to write $\ce{H+}$ in acidic media, you should know that naked $\ce{H+}$ doesn't exist in water. It's usually bonded to something else. Is it $\ce{HCl}$, or $\ce{H3O+}$?
The identity of the base is left unclear. Is it $\ce{Cl-}$, or $\ce{H2O}$? Hints: Which one of them is a stronger base? Which one is available in greater quantities?

Notes

(1) Just for interest, $\mathrm{S_N1}$-type chemistry leading to vinyl cations has been done before, but with much better leaving groups than MeOH. See: Okuyama, T.; Takino, T.; Sueda, T.; Ochiai, M. J. Am. Chem. Soc. 1995, 117 (12), 3360–3367. DOI: 10.1021/ja00117a006.


Recovery and Disposal of Unknown Copper Compound

Unlikely. Copper hydroxide adsorbs carbon dioxide from air rather
fast, forming basic carbonates. if you want to test, try to boil it.
Copper hydroxyde looses water under mild heating, forming black
copper oxide.

Most likely it is either basic copper carbonate or one of basic
copper sulfates or a mix of both. I highly doubt that it is a pure
copper sulfate, as Mg had to go somewhere.
If you want to be a responsible citizen (which is not actually much
a point with relatively safe copper salts), add some soda, leave
overnight and then filter off the precipitate with filter paper.
If no filter paper is present, cotton white, toilet paper, and
similar should work, just be careful to not tear them. The
precipitate then may be heated until black and disposed any way
suitable.


Competitive protonation between alkene and carbonyl

In the scheme below I have shown the three most probable protonations of the molecule in question. I have considered that a tertiary carbocation is typically more stable than a secondary one and that a resonance-stabilised carbocation is typically more stable than a non-stabilised one.

From left to right, let’s discuss the structures.

The one on the left is not good. The carbocation is only stabilised by being tertiary. It would take a very strong acid to sufficiently protonate in this way, or a good nucleophile that can capture the positive charge on the tertiary carbocation.
The one in the middle doesn’t look as bad initially, because it is resonance-stabilised. However, if you draw the other principal mesomeric structure, you arrive at a positively charged oxygen with a sextet — a very unstable thing to have. It is very improbable that protonation would occur like this.
There is another (undrawn) structure that would protonate the same double bond but on the other side. However, that carbocation would destroy the unsaturated ketone’s resonance and is therefore unfavored.
Finally, the protonated carbonyl. This is clearly the best cation, since it is an all-octet structure and the positive charge can be stabilised by resonance. Futhermore, if you draw the resonance structure, you realise that it localises the cation on a tertiary carbon — the best possible case.

Thus there should be no discussion; protonation of the carbonyl oxygen leads to the most stable cation and is therefore expected to prevail.

(Your lower $\mathrm{p}K_\mathrm{a}$ of carbocations is likely due to the fact that they can often deprotonate to form a double bond giving additional stabilisation.)


There is no simple way to predict beforehand as far as I know. This is a general knowledge type of chemistry question, you will have to know or memorize the colors of common anions and cations. It will be much easier, if you do so group wise and know the colors of the first row of transition element ions.

Any object which absorbs a certain portion of visible light's spectrum, will appear colored to us. For colored solutions, we will always consider absorption of light in the visible region. So if you see a green solution, it is not reflecting green light, it is absorbing a certain portion of the visible spectrum and the transmitted light [from the solution] appears green to us. Keep this concept in mind because a lot of students have this misconception.

You should know that sulfate ion is colorless but copper (II) ion is blue so most likely copper sulfate solution is blue. In part c, you should know the color of potassium ion and dichromate ion and then guess the color. Do the remaining exercise for others.

It will be very difficult to predict if both ions are colored because it is not like that we are mixing two paints, a yellow paint with blue paint will show a green color. You will have to study the absorption spectrum of that salt. The theory of color of ions and that of solids is quite complex. It was developed by physicists. You can start from crystal field theory if you wish to see how the color of transition metal ions is explained but I will not recommend that you explore it if you are in school.


Why is monohydrogen phosphate an incorrect name?

The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

(…)
(…) The prefix ‘mono’ is, strictly speaking, superfluous and is only needed for emphasizing stoichiometry when discussing compositionally related substances (…).
(…)

Therefore, hydrogenphosphate ($\ce{HPO4^2-}$) may be called monohydrogenphosphate in order to emphasize the difference to dihydrogenphosphate $\ce{H2PO4-}$.
Note that, according to IUPAC Recommendations, the systematic hydrogen name of $\ce{HPO4^2-}$ actually is ‘hydrogen(tetraoxidophosphate)(2−)’; however, ‘hydrogenphosphate’ is an accepted simplified hydrogen name.


Reaction between Copper nitrate and ascorbic acid

It is always good to cite the link when you mention that you saw a redox reaction online. It is wrong anyway.

Aqueous Cu(II) is not an strong enough oxidizing agent that it will convert organic matter into carbon dioxide. In other words, it is a weak oxidizing agent. Ascorbic acid is oxidized to dehydroascorbic acid and Cu(II) should be reduced to Cu(I). If you heat the solution strongly, Cu(I) may convert into Cu$_2$O.

Ascorbic acid is indeed an acid, but there is nothing which prevents its conjugate base to behave like a reducing agent. Many organic acids can behave like that.

  I am really confused as to how they determined that to be the product.

By conducting the experiment. There is no other simple way to predict what will happen in a complex redox reaction.


Why do certain compounds dissolve in oil? Why do some types of compounds dissolve in oil, yet not water?

I propose a logical approach, to explain the rule "the like dissolve like". Why don't we admit that all liquids should be able to mix, exactly like gases may be mixed in all proportion ? This is the case for non polar molecules, like oils. And if two liquids don't mix, this is due to the existence of attracting forces between molecules A (hydrogen bonds for exemple), and that mixing them with another stuff B may destroy too many of these bonds. If the mixing operation A + B creates about as many new bonds as those broken during the mixing, the liquids A and B may be mixed. This is the case in the mixture water - ethanol. If the mixing operation A+B does not create enough new bonds between molecules A and B, like water and oil, Nature refuses to break the bonds between A molecules without any counterpart, and prevents the A molecules (water) to be separated from one another in order to accept B molecules, and A and B are not miscible.


Methods: Does failing at intermolecular properties mean the same for intramolecular properties?

I think the distinction that you are trying to draw is (as you have hinted) better expressed in terms of covalent and noncovalent interactions, rather than inter- versus intramolecular interactions. A technique that fails to correctly describe noncovalent interactions will fail whether or not these interactions are between two moieties on the same molecule, or on different molecules. The usefulness of a technique that fails noncovalent interactions is contingent on whether or not those interactions are significant to the particular properties of a molecule that you are investigating.


Iron(III) acetate will be stable under these conditions. According to this paper, thermal decomposition to iron oxide only starts at 423 K (150 °C). Heating iron and acetic acid together with an oxidation agent (oxygen from air or, like in your case, $\ce{H2O2}$) is actually a suitable way to prepare iron acetate dyes (source).


Cleaning with vinegar and sodium bicarbonate

Your intuition is on point. Pretty much all the "lifehacks" praising cleaning abilities of vinegar and sodium bicarbonate are ineffective. The reaction appears vigorous due to the formation of carbon dioxide gas, but what is left is a diluted solution of sodium acetate:
$$\ce{CH3COOH(aq) + NaHCO3(aq) -> NaCH3COO(aq) + H2O(l) + CO2(g)}$$
Sodium acetate doesn't have any pronounced cleaning abilities. It doesn't act like a surfactant, doesn't drastically increase solubility of the impurities, or participate in complexation helping to dissolve metal-containing compounds.
As for unclogging abilities—yes, this is something that my grandmother used to use, but these days there are much better alternatives.
The idea was to pour some hot water into the system and then put a couple of spoons of baking soda into the drain creating warm concentrated solution in the siphon.
This solution would also be mildly basic due to hydrolysis and help to dissolve some fat and thin hair:
$$\ce{HCO3^-(aq) + H2O(l) <=> H2CO3(aq) + OH-(aq)}$$
After some time (approx. 30 min) it would make sense to add some vinegar hoping the reaction would cause mixing up the constituents of the clog and excessive pressure would help to push the waste further down the drain.


Yes they do. Such compounds are even found in nature, e.g. vernonallenolides.


Why anions of weak acids are weak bases?

Your understanding is correct - on a relative scale, the conjugate base of a weak acid is a strong base, and the conjugate base of a strong acid is a weak base. For example, hydroxide ion (conjugate base of water) is a stronger base than acetate ion (conjugate base of acetic acid) which is a stronger base than chloride ion (conjugate base of hydrochloric acid).


In which country is it allowed to practice non-clandestine amateur chemistry?

Your question is basically too broad to be answered in full detail. I suggest that if you want a more detailed answer about a certain part, you should open a question for that matter separately. That being said, I'll try to summarize my own experience as a hobby chemist.

First of all, are you actually studying chemistry? IMHO, if this is within your possibilities, it is the best way to gain experience and also do your your own research. You will have supervision and the best equipment available. You will get the space to do research in your own areas mostly during your graduate studies, but here (The Netherlands) you will also get some space as an undergraduate.

Consider the following points (in no particular order):

You say that finance is not a problem, but do not underestimate the price of all the chemicals and equipment you will need to do any bit of serious chemistry. Scales, vacuum pumps, glassware, solvents, a dedicated fridge, a specially constructed cabinet to store flammables, a fume hood with explosion proof suction, etc (and I'm not even mentioning NMR, mass spectrometers, IR, all of which are available to you as student). To run any decent lab from your backyard be prepared to invest several thousands of dollars (or euros). This can be done of course a lot cheaper, but then you will be limited in your possibilities or you will have to exchange safety for money.
Almost every chemical you will ever need when doing a normal synthesis (i.e. no explosive making or pharmaceuticals) is legally available in most countries (I heard however that some states in the US are quite notorious w.r.t. hobby chemistry, like Texas). You are even able to obtain chemicals that are clearly dangerous and should not be sold to amateurs, like osmium tetroxide, sodium cyanide (not everywhere), gaseous hydrogen chloride in large 110L pressurized tanks. There are a few exceptions:

Common precursors and preprecursors to illicit drugs in large quantities. Most notably phenylacetone and derivatives, piperonal, piperidine, methylamine, benzaldehyde, nitroethane, etc.
Chemicals that can be used to manufacture explosives in large quantities. Examples include: nitromethane and ammonium nitrate.
Chemicals that are severely dangerous to the environment or can be easily abused to poison people, although oddly enough some of the most hazardous are freely available. Examples are: elemental mercury, a lot of (organo) arsenic and other heavy metal compounds, a lot of radioactive compounds, some cyanides (may be available though from mining suppliers), etc.

Note: you are able to obtain most of the above mentioned chemicals (also in larger quantities) if you start a business and have it legitimately registered to the Chamber of Commerce. This business does not actually have to do anything, it can function as a facade. It still won't work for pharmaceutical precursors or very large quantities of chemicals however, but for a hobby chemist that is mostly no problem.
Note 2: At least here in The Netherlands some pharmaceutical precursors and preprecursors can be legally owned and synthesized, they are just not available commercially because of the high change for abuse. These chemicals however (e.g. phenylacetone) can be used in many other legal synthesis and hence it might be useful to synthesize them yourself.
Be prepared to spend a lot of time on this hobby, both studying theory and doing practical work. Independently doing (serious) chemistry is not easy and you will have to rethink and adjust your setup a lot of times. Also the possibility of failure or accidents is higher without supervision. It is not entirely without risk.


Steric Inhibition of Resonance in ortho-substituted Benzoic Acids

Important information before understanding the reason for steric inhibition of resonance and the ortho effect

In general, resonance stabilizes a molecule. But in the case of carboxylic acids, the conjugate base (anion after losing $\ce{H+}$) there are two equivalent resonance structures.
Take the example of acetic acid, $\ce{CH3COOH}$, and thus acetate ion, $\ce{CH3COO-}$:

The hybrid would look something like this:

It could be easily seen that equivalent resonance structures mostly stabilise the hybrid greatly.
Therefore, equivalent resonance tends to stabilize a molecule/compound more than normal resonance. (in most cases)

Now let's take the case of o-toluic acid.

There are two ways by which resonance can take place:

In (I), it takes place through $\ce{-COOH}$. Its ion is expected to be more stable due to the above result. Also, here $\ce{OH}$ is highly polar and it will tend to lose $\ce{H+}$ to become uncharged and thus be more stable (as compared to $\ce{O-}$).
In (II), it takes place through the ring. Its ion is less stable due to more formal charge and less aromatic character (due to + on the ring). Also, $\ce{OH}$ is less polar than $\ce{H+}$ due to the above point.

Therefore, it will be better to prevent the taking place of resonance through ring.

That is done by $\ce{-CH3}$ group which provides bulky group repulsion and takes the ring out of the plane of $\ce{-COOH}$ and thus providing condition for equivalent resonance in $\ce{-COOH}$.


Applied Potential and Electrochemical Cell Potential

You are mixing two areas, electrode kinetics and electrode thermodynamics. If we have 3 apples and 2 oranges, this is not equal to 5 oranges or 5 apples but there are 5 fruits in total. Electrode thermodynamics tells you a "yes" or "no" answer in the sense that it is saying the reaction is spontaneous or not. Electrode kinetics tells you how fast the reaction will be.

When you wish to electrolyse something, it means that you want do a non-spontaneous reaction i.e., if you leave lithium in the presence of air and moisture, it will spontaneously form LiOH

$$\ce{4 LiOH <- 4 Li + 2 H2O + O2}.--(1)$$

Now, by forcing electric current through molten LiOH, you want to reverse this reaction and want the following to happen

$$\ce{4 LiOH -> 4 Li + 2 H2O + O2}.--(2)$$

The negative Ecell of -2.60 V only tells you that reaction (2) is not spontaneous at all. It has no other meaning whatsoever.  That is, if you wait forever, you will not see any Li spontaneously forming out of LiOH.

Now as you read in the question, they are saying that if you apply exactly +2.60 V potential difference between two electrodes the rate of reaction is very slow. Remember negative sign was just for saying that the reaction is non-spontaneous. However if you apply +2.60+0.60 = 3.2 V of potential difference between the electrodes, the rate of formation of Li from molten LiOH will be quite fast. They don't say how fast but you will not have to wait for a long time.

So +2.60 V is the apple = electrode potential
0.6 V is the orange, it is called the overpotential in formal language.

You can only add them (not subtract) and get the total voltage required to start the electrolysis at an appreciable rate.


You are right: the outer boundary is quite arbitrary. There is no any intrinsic threshold; the probability just gradually decreases lower and lower, but never reaches 0. You may draw a sphere so as to have the electron inside with 90% probability, or 95%, or 99%, or any other value as you see fit.

This, in particular, is the reason why atomic radii are somewhat arbitrary too.


Compound class naming convention: heteropoly-acid vs isopoly-acid

Wouldn't heteropoly-acid and homopoly-acids pair be morphologically more accurate (than heteropoly-acid and isopoly-acids)?

I completely agree with andselisk of this suggestion because the former makes more sense than latter in this regards. For instance, as correctly pointed out in the question, the IUPAC Red Book has utilized the terms "homopolyatomic" vs "heteropolyatomic" in the chapter, IR-5: Compositional Nomenclature, and Overview of Names of Ions and Radicals (p. 68-82, Ref.1). A few examples given for the homopolyatomic cations in the section IR-5.3.2.3 are $\ce{O2^+}$, $\ce{S4^2+}$, $\ce{Hg2^2+}$, and $\ce{Bi5^4+}$ (p.71). Likewise, few examples of heteropolyatomic cations are also given in the following section IR-5.3.2.4, which are $\ce{NH4^+}$, $\ce{H3O^+}$, $\ce{PH4^+}$, and $\ce{SbF4^+}$ (p.71). Thus, it make perfect sense of using the terms "homopolyatomic" and "heteropolyatomic" in the context, since it's the presence of heteroatom in the latter case ($\ce{N, O, P,}$ and $\ce{Sb}$, respectively) that allows the distinction.
Nonetheless, in another chapter in Ref.1 (IR-8: Inorganic Acids and Derivatives, p. 124-141) has utilized the terms "homopolyoxy acids" and "heteropolyoxy acids" to described polyoxy acids:

Examples 1, 3–6 and 14 above demonstrate that homo- and heteropolyoxoacids and their partially dehydronated forms […] (p.136, ref.1)

For instance, the example 1 and 4 mentioned in the excerpt are $\ce{H2PO7^2-}$ and $\ce{H4[SiW_{12}O_{40}]}$, respectively (Section: IR-8.4, p. 135, Ref.1). As we all know that $\ce{H4PO7}$ is an example of isopoly-acids and $\ce{H4[SiW_{12}O_{40}]}$ (Ref.2) is an example of heteropoly-acids. Thus, even if these unconventional names are frequently use in inorganic chemistry literature to date, I think IUPAC is already recognized the necessity to keep consistency in the naming.

References:

Nomenclature of Inorganic Chemistry: IUPAC recommendations 2005 (Issued by the Division of Chemical Nomenclature and Structure Representation in collaboration with the Division of Inorganic Chemistry); Prepared for publication by N. G. Connelly, et al.; RSC Publishing (for the International Union of Pure and Applied Chemistry): Cambridge, United Kingdom, 2005 (ISBN: 0-85404-438-8).
Li Min Yang, Zhi Kui Yin, and Li Qiang Wu, "$\ce{H4[SiW_{12}O_{40}]}$: An efficient catalyst for the synthesis of new spiro-[dibenzo[a,i]xanthene-14,3′-indoline]-2′,8,13-triones," Chinese Chemical Letters 2012, 23(3), 265-268 (DOI: https://doi.org/10.1016/j.cclet.2012.01.002).


How to calculate the standard Gibbs free energy of formation of bromine vapor at room temperature?

If you calculate the equilibrium vapor pressure at 298K (25 C), you can calculate the change in Gibbs free energy of the vapor in going from this pressure to the hypothetical state of 1 bar and 298K using $dG=VdP=RTd\ln P$.  Neglecting the Poynting correction, the free energy of the liquid at 298 and at the equilibrium vapor pressure is 0.  And this is also the free energy of the vapor at 298 and at the equilibrium vapor pressure.


Can a change in internal energy always be expressed as the product of the constant volume heat capacity and the change in temperature?

The statement in the title question "internal energy [can] always be expressed as the product of the constant volume heat capacity and the change in temperature" would only apply to a calorically perfect gas.

If you instead said "an infinitesimal change in internal energy can be expressed as the product of the constant volume heat capacity and an infinitesimal change in temperature", that is only true for a thermally perfect gas.

For real gas, the heat capacities depend upon both temperature and pressure.


It's essentially just the same reaction once and then over again. The basic medium of the Wolf-Kishner reduction is to remove the hydrazone hydrogen, releasing the $\ce{C=N}$ electrons onto the carbon. In this reaction, the hydrazone instead further reacts with the other carbonyl to form an aromatic 6-membered ring.


Why is the Friedel-Crafts alkylation of nitrobenzene disfavoured?

The nitro group is so deactivating that yields are really poor for Friedel-Crafts reactions with nitrobenzene; not even the meta-substituted product is found in good yield. In fact, nitrobenzene is actually used as a solvent in Friedel-Crafts reactions.


Neutral anions in salts and strong acids are similar - is it a coincidence?

Though couched in different terms, the first part this question is essentially the same as this other one:

Does a strong acid's conjugate base act as Brønsted-Lowry base?

No, this is not a coincidence.  The conjugate base of a strong acid will always be a weak base.  The 'neutral salt anions' of your professor's nomenclature are just weak inorganic Brönsted bases.

The reason why $\ce{SO4^2-}$ doesn't appear in this list of 'neutral salt anions' is because it actually can noticeably affect the $\mathbf{pH}$ of a solution when added as a salt.  $\ce{H2SO4}$ is often treated as though it's a diprotic strong acid, but in reality only the first deprotonation is strongly acidic.  This can be seen by examining its acid dissociation constants; from Wikipedia:

  $\mathrm pK_{\mathrm a1} = -3$
  $\mathrm pK_{\mathrm a2} = 1.99$

This $\mathrm pK_{\mathrm a2}$ is about the same as the first $\mathrm pK_{\mathrm a1}$ of $\ce{H3PO4}$ $($$2.148$$)$, and is actually somewhat less acidic than nitroacetic acid $($$1.68$$)$.

So, in some cases sulfate may act more or less as a 'neutral salt anion', when the $\mathrm{pH}$ is sufficiently high that it will all stay doubly deprotonated, but in others it can definitely influence the solution $\mathrm{pH}$. I've actually used sulfate/bisulfate before as an acidic buffer for various projects at work—works great!


Non-carbon based life forms on hot planets

Given that a $\ce{C-C}$ bond dissociation energy is $83\ \mathrm{kcal\ mol^{-1}}$, and $\ce{Si-Si}$ bond dissociation energy is $52\ \mathrm{kcal\ mol^{-1}}$, a silicon based life form is much worse off on a high temperature planet. It's actually for this very reason that there aren't nearly as many silicon based compounds as there are carbon based compounds. Thats not to say they don't exist, however. There are silicon hydrides that in the form of silicon chains, analagous to alkanes, existing up to about 8 silicon long. They are all readily hydrolyzed and are thermally unstable, especially the longer chains.

Silicon is also not quite as versatile as carbon in terms of the types of bonds it can form. Carbon is capable of forming single, double, and triple bonds due to its ability to hybridize its $\mathrm{s}$ and $\mathrm{p}$ orbitals. Silicon is a larger atom than carbon, and as a result it is much more difficult for its $\mathrm{p}$-orbitals to overlap with the $\mathrm{p}$-oribatls of adjacent atoms. Again, this is of course not to say that this doesn't happen, as it does in the case of diselenes and disilynes, but again, these are (very) unstable.

Most often silicon is found combined with oxygen, as in the very stable $\ce{SiO2}$, or as you might know it: glass, sand, or quartz; not exactly ideal for a moving, living creature. Just an idea, but if you're trying to make scientifically feasible life forms on a hot planet, you might just be better off having carbon based life forms in an inert atmosphere. In other words, an atmosphere that doesn't lend to them bursting into flames.

I'm not sure how set you were on making these different races out of other elements, but no element by itself really matches the chemical variability that carbon offers.

EDIT: After doing a little more research, I thought it might be beneficial to include polysiloxanes, or silicone, which consist of $\ce{C-(Si-O)_{n}-Si-C}$ bonds. These actually are actually very interesting because they are extremely elastic, functional over a wide range of temperatures ($-100^\circ \mathrm{C}-300^\circ \mathrm{C}$), resistant to ozone and UV radiation, do not promote bacterial growth (but who knows about extraterrestrial bacteria), have a low chemical reactivity, high electrical resistivity, and most interestingly, have self-healing properties.

I would now change my tune and say that maybe a race of silicon-carbon based life forms isn't too far fetched at all! At least for sci-fi writing purposes.


Why do bubbles form on the side of a plastic bottle?

Your reasoning seems good. A number of things can affect nucleation of bubbles, such as surface defects, which you state, as well as temperature (was one side of the bottle facing the sun, perhaps?), and even sound.

A fun demonstration is to put a cup of carbonated beverage in an ultrasonic bath (e.g. parts cleaner or humidifier). Here, cavitation serves to nucleate bubbles.


How to explain the dipole moment anomaly of methyl fluoride and methyl chloride with respect to their hydracids?

Summary

A systematic examination of the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes with density functional techniques and high-level wave function theory indicates that the charge-separation within these complexes influences the relative corresponding dipole moments for these compounds.  Though fluorine is indeed more electronegative than chlorine, $\ce{CH3Cl}$ had a larger dipole moment due to the longer C-Cl bond with respect to the C-F separation.

Introduction

Recall that a dipole moment can be explained relatively simply by a separation of two charges.  The dipole moment, $\mu$, is dependent on two things: the magnitude of the charge ($q$) AND the separation between the two charges ($r$).  This can be expressed as:

$\mu = q \cdot r$

We know from general chemistry that fluorine is the most electronegative atom on the periodic table.  Therefore, for simple molecules such as $\ce{HF}$ or $\ce{LiF}$, we would expect a rather large dipole moment than if we considered $\ce{HCl}$ or $\ce{LiCl}$.  However, a dipole moment depends on separation as well.  This can lead to some discrepancies in our general trends that we chemists love to use.

Consider chlorine.  The atomic radius for chlorine is larger than it is for fluorine.  A C-Cl bond is going to be longer than a C-F bond.  In the case of methyl fluoride ($\ce{CH3F}$) and methyl chloride ($\ce{CH3Cl}$), we will have a difference in separation of the methyl group and the halogen (larger for the latter).  Looking back at our electric dipole equation, an increase in $r$ translates into an increase in $\mu$.  Perhaps it is this larger separation between the carbon and chlorine atoms that gives rise to these relatively similar dipole moments between $\ce{CH3F}$ and $\ce{CH3Cl}$.

Super Mini Research Project Time

Let us apply electronic structure theory to the problem.  Dave has already provided a link to NIST which is a superb resource for these types of simple computations. However, to be thorough, we will perform our own computations.  The experimental dipole moments for $\ce{CH3F}$ and $\ce{CH3Cl}$ (provided by DavePhD) are $\pu{-1.858 D} \text{ and } \pu{-1.897 D}$, respectively ([1] and [2]).  Dipole moments from electronic structure theory for these complexes can also be found on the NIST website (source3 and source4)

Methodology

Full optimizations and corresponding harmonic vibrational frequency computations are performed using the B3LYP density functional technique (sorry Martin) using the Gaussian 09 software package.  The 6-31+G(d,p) basis set will be employed along with a series of Dunning's correlation consistent basis sets: heavy-aug-cc-pVXZ (where X = D, T, Q, 5).  This basis set denotes that cc-pVXZ was employed for hydrogen atoms and the aug-cc-pVXZ basis set was employed for all heavy (i.e. non-hydrogen) atoms.  A numerical integration grid consisting of 99 radial shels and 590 angular points per shell were employed for all DFT computations.  In addition, full optimizations have been performed using the CCSD(T) level of theory using the CFOUR software package.  The haDZ, haTZ, and haQZ basis sets have been employed for these computations.  All geometries were constrained to $C_{3v}$ symmetry.  Everything is converged to 'overkill'.  Frozen-core approximation applied... 5D 7F spherical basis functions used, etc. etc.

Results

The dipole moments (in Debye) as well as select geometrical parameters ($R$ in Angstroms) for the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes at various levels of theory are given in the table below.  The deviation of the computed electric dipole moment from experiment is denoted as $\delta_{\mathrm{theory}}^{\mathrm{expt.}}$.  Using this deviation as our main measurement of error, we can see that for the $\ce{CH3F}$ complex, the B3LYP/ha5Z level of theory deviates from experiment by only $\pu{0.01 D}$.  A larger deviation from experiment is observed for the CH$_3$Cl complex with a magnitude of $\pu{0.1 D}$ at the B3LYP/ha5Z level of theory.  These magnitudes are qualitatively similar at the CCSD(T) level.  Therefore, with respect to the dipole moment, the B3LYP method with a triple-$\zeta$ quality basis set is necessary to describe this first-order property.

Clearly $\ce{CH3Cl}$ contains a larger separation between the Cl atom and the methyl group with respect to $\ce{CH3F}$ by about 0.4 Angstroms.  If you wanted to back-calculate the charge for these complexes by assuming $R$(CX) (where X is the halogen) is equivalent to $r$ in the first equation, you will see that CH$_3$F has a larger charge (by about 0.25).  However, we still see that CH$_3$Cl has a larger dipole moment.

The final column in the table above ($\mu_{\mathrm{geom}}^{\ce{CH3Cl}}$) denotes the dipole moment of $\ce{CH3F}$ when using the optimized geometry of $\ce{CH3Cl}$ at the corresponding level of theory.  The dipole moment is significantly larger, by about $\pu{1.5 D}$, which is expected due to the larger C-F separation.

Conclusions

Though we as chemists like to resort to using the general trends provided by the periodic table, we must take into account other factors to explain deviations from these trends.  Here we consider the separation of the halogen from the methyl group as an explanation for $\ce{CH3Cl}$ having a larger dipole moment than $\ce{CH3F}$, even though F is more electronegative.  Our computational results qualitatively agree with experiment.  In addition, B3LYP with a triple-$\zeta$ quality basis set is adequate in describing this first-order property according to CCSD(T) results.  Finally, these results indicate that it is the charge-separation in these small molecules that greatly impacts the qualitative ordering of these dipole moments.

References:

Marshall, M. D.; Muenter, J. The electric dipole moment of methyl fluoride. J. Mol. Spectrosc. 1980, 83 (2), 279–282. DOI: 10.1016/0022-2852(80)90051-X.
Wlodarczak, G.; Segard, B.; Legrand, J.; Demaison, J. The dipole moment of CH335Cl. J. Mol. Spectrosc. 1985, 111 (1), 204–206. DOI: 10.1016/0022-2852(85)90083-9.


'Hard' soap vs. 'Soft' soap: Why do they work this way?

Basically, yes, by common terms, soft soap has $\ce{K+}$ and hard soap has $\ce{Na+}$.
They are both alkali metals, but $\ce{K+}$ is below $\ce{Na+}$ on the periodic table so maybe we should consider periodicity trends. As you go down a table the atoms become more reactive, so $\ce{K+}$ is more likely to react with polar parts of water than $\ce{Na+}$. Here are solubilities for both (though there are other values that would tell this story, perhaps better than solubility of the product):
[Solubility for $\ce{KOH}$] in $\ce{H2O}$ at 25C: 121 g / 100 mL
[Solubility for $\ce{NaOH}$] in $\ce{H2O}$ at 25C: 100 g / 100 mL
Regarding your question of going straight into hard soap, based on this link it seems doable; he just used $\ce{NaOH}$ (lye) and oils with no mention of $\ce{KOH}$ (potash?) to speak of.
And yes, the positive ions connect to the $\ce{O-}$ of the fatty acid as seen in this diagram.

(img src: http://jan.ucc.nau.edu/~doetqp-p/courses/env440/env440_2/lectures/lec19/lec19.html)


(At what temperature) can salt evaporate?

All substances may have a melting and boiling point. I say "may" because in some instances there can be decomposition, for example, before reaching either temperature.

Sodium chloride has a melting and boiling point. Check Wikipedia:

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

The sidebar lists the melting point at 801 °C (1,474 °F; 1,074 K) and the boiling point at 1,413 °C (2,575 °F; 1,686 K). The values are relatively high being consistent with the ionic nature of NaCl.


Does possibility of Tunneling influence Ground State Energy of PIB with and without finite energy barriers?

You are asking about a double-well potential where the two wells are separated by a finite square wall. This model is very well studied as it is the simplest model which involves tunneling through a barrier, so it provides a conceptual understanding to tunneling along a reaction coordinate.

The potential is visualized below along with the two lowest energy solutions:

Note, however, that in order to make a good comparison to the particle in a box, one must think about two times the particle in a box energy. This is because the double-well potential is really two infinite square-wells with a finite barrier between them. So, this tells you that in the limit that the barrier in the middle becomes infinitely large, there will be two ground state solutions with equal energy. Namely, the particle could be on the left or on the right.

So, as we lower the barrier and the particle is allowed to tunnel between the two wells, this degeneracy will be broken. These are the two wavefunctions which you see in the diagrams above. Namely, the solutions very closely resemble a symmetric and an antisymmetric linear combination of the ground state of a particle in an infinite box.

So, this has the effect of splitting the degeneracy into two states, one of which is higher in energy and one of which is lower in energy than particle in a box.

Why should the ground state of this double well be lower in energy than the ground state of a particle in a box where the particle stays on only one side of the barrier?

One way to think about this is that tunneling allows the particle to "leak" into the other box, which has the effect of making the effective box size of the particle larger.

This is all very qualitative, but the details of the math aren't really all that much more enlightening than simply looking at the qualitative features of the problem. The most important point I think is to recognize that tunneling through this barrier has the effect of splitting an otherwise degenerate state into a lower and higher energy state.


What will be the possible major product(s) on heating this primary alcohol i.e. 9-fluorenylmethanol with (conc) H2SO4?

What I feel is that forming phenanthrene would be possible because of its high stability due to resonance. Initially, $\ce{-OH2}$ leaves out forming a 1°carbocation. Now the $\ce{\beta C-\gamma C}$ cleaves so as to make way for the better 2°carbocation which is also thermodynamically favourable.


What is the derivation of the formula for finding number of pure and hybrid orbital?

All we are doing is counting orbitals, and these formulas are generalizations of counting the same way that multiplication is generalization of repetitive addition. The derivation is exactly as you present it. For the first formula, we are multiplying the number of hybrid orbitals per carbon atom by the number of carbon atoms. For the second formula, we are multiplying the number of non-hybrid orbitals per carbon atom by the number of carbon atoms and adding the number of hydrogen atoms (since each hydrogen atom only has one orbital that cannot hybridize**).
If you took benzene $\ce{C6H6}$ and counted all of the hybrid orbitals, you would count 18. Each carbon atom is $sp^2$-hybridized, so each has three hybrid orbitals. We could count each individually:
$$1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 18$$
Or you could group by each of the six carbon atoms, which eventually simplifies to $3\times 6$
$$(1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) = 18\\
3 + 3 + 3 + 3 + 3 + 3 = 18\\
3 \times 6 = 18$$
Your formula has some limitations. First, it only works for hydrocarbons (compounds with only carbon and hydrogen atoms), and it assumes all carbon atoms have the same hybridization. Here is a hydrocarbon where the carbon atoms do not all have the same hybridization. This is hex-1-en-5-yne and it has two $sp^3$-hybridized carbon atoms, two $sp^2$-hybridized carbon atoms, and two $sp$-hybridized carbon atoms.

Your formula could be extended to:
$$N_{\mathrm{HO}} = 2n_{sp} + 3n_{sp^2} + 4n_{sp^3}$$
Where $N_\mathrm{HO}$ is the number of hybrid orbitals, $n_{sp}$ is the number of $sp$-hybridized carbon atoms, $n_{sp^2}$ is the number of $sp^2$-hybridized carbon atoms, and $n_{sp^3}$ is the number of $sp^3$-hybridized carbon atoms. For hex-1-en-5-yne, we could count:
$$N_\mathrm{HO} = 2\times 2 + 3\times 2 + 4\times 2 = 18$$
There is a more general formula for hydrocarbons. Let's look at it this way. Each carbon atom has four orbitals (either hybridized or non-hybridized). If there are no $\pi$-bonds, then the number of hybrid orbitals is four times the number of carbon atoms. Each $\pi$-bond requires to non-hybridized $p$-obitals, so each $\pi$-bond reduces the number of hybrid orbitals by 2. Here is a general formula for hydrocarbons:
$$N_\mathrm{HO} = 4n_\mathrm{C} - 2n_\pi$$
Here $n_\mathrm{C}$ is the number of carbon atoms and $n_\pi$ is the number of $\pi-bonds$. For benzene, $n_\ce{C} = 6$ and $n_\pi = 3$, so
$$N_\mathrm{HO} = 4\times 6 - 2 \times 3 = 24 - 6 = 18$$
For hex-1-en-5-yne, we have the same number of carbon atoms and $\pi$-bonds, so we get the same result. For the number of non-hybridized orbitals ($N_\mathrm{NHO}$), we have a general formula as well. Each hydrogen atom contributes one non-hybridized orbital ($n_\ce{H}$), and each $\pi$-bond contributes two.
$$ N_\mathrm{NHO} = n_\ce{H} + 2n_\pi$$
For benzene and hex-1-en-5-yene, $n_\ce{H} = 6$ and $n_\pi = 3$, so
$$ N_\mathrm{NHO} = 6 + 2\times 3 = 12$$
This is the same as your second formula.
** I expect one of my physical/theoretical chemist colleagues to come along and inform us that under certain circumstances, hydrogen atoms may in fact act as if they have hybrid orbitals, but for most organic compounds we can ignore this behavior.


Freezing breaks the cell structure of an apple, not individual chains of short polysaccharides you can feel the taste. What happens is probably enzymes which the apple contains in cell structures go out of these structures and split long polysaccharides (which have no taste) into short sugars which you feel as sweet.

In other words, the apple digest itself by its own enzyme systems after breaking the cell structures.


What is the direction of charge transfer between Aluminum and aliphatic amines physisorbed on top?

Amines could bond to the aluminum surface via a Lewis acid-base reaction between the electron rich amine nitrogen and the electron poor Al atom on the surface (perhaps formed by loss of a surface hydroxyl group as hydroxide ion). The direction of charge transfer would be from the amine nitrogen to an aluminum atom. The resulting "Lewis salt" would have a partial positive charge on nitrogen and a partial negative charge on aluminum (like an aluminate ion or an amine-aluminum chloride adduct).


I am not sure that I understand your question, but in terms of establishing the valency in this case the key point is that in d block elements the energy level of the outer s electrons are very similar to that of the d electrons in the shell below, so the valency in this case will be 3.


Answer

This article provide two reactions for the thermal decomposition of HMTD, depending on reaction conditions.

  (at or below 150°C, air or vacuum)

$$\ce{C_6H_{12}N_2O_6 -> 3CO_2 + N(CH_3)_3 + NH_3}$$

  (160°C to 180°C, vacuum)

$$\ce{C_6H_{12}N_2O_6 -> 3CO + N(CH_3)_3 + 1/2 N_2 + 3/2 H_2O + 3/4 O_2}$$

References

Oxley, J.C., J.L. Smith, H. Chen, and Eugene Cioffi. “Decomposition of Multi-peroxidic Compounds: Part II. Hexamethylene Triperoxide Diamine (HMTD).” Thermochimica Acta 388, no. 1–2 (June 18, 2002): 215–225.


from "Stuff Matters" by Mark Miodownik:

Paper yellows with age for two reasons. If it is made from cheap low-grade mechanical pulp, it will still contain some lignin. Lignin reacts with oxygen in the presence of light to create chromophores ( meaning, literally, 'color carriers'), which turn the paper yellow as they increase in concentration. This type of paper is used for cheap and disposable paper products, and is why newspapers yellow quickly in light.
It used to be fairly common to increase the textural quality of paper by coating it with aluminum sulfate, a chemical compound that is used primarily to purify water, but what wasn't appreciated at the time was that this treatment creates acidic conditions. This causes the cellulose fibers to react with hydrogen ions, which results in another form of yellowing. It also decreases  the strength of the paper. Large number of books from the 19th and 20th centuries were printed on this so-called acid paper and can now be easily identified in book shops and libraries by their bright yellow appearance. Even non-acid paper is susceptible to this aging, just at a slower rate.


When a thin lead sheet is hammered, what happens to internal energy

When you hammer a sheet of lead, you are adding energy in form of work into it, increasing it's internal energy. If this experiment were to be done in complete vacuum, it'd retain most of the energy, however if this experiment were to be on air, it'd lose a lot of energy in the form of sound.


In some cases, such as [F-H-F]- (biflouride), there is a symmetric hydrogen bond, with both H-F bonds equally strong.

However, just because the two atoms bonding to the H are the same doesn't mean it is a symmetric hydrogen bond.  For example water dimer does not have a symmetrical hydrogen bond, the H is closer to one O atom than the other.

There can be cases where the hydrogen being exactly in the middle is a potential energy minimum, and other cases where the middle is not a local minimum but there are two minima offset from the middle.

See Symmetry of hydrogen bonds in solution for more information.


Is aromaticity of the 4 aromatic amino acids affected by pH?

There are two nitrogen atoms in the imidazole ring in histidine, but the $sp^2$ nitrogen is significantly more basic than the $sp^3$ nitrogen, as the electron pair in the latter is delocalized over the ring. Protonating the $sp^3$ nitrogen would destroy the aromaticity, but that would require very strongly acidic conditions, possibly not attainable in aqueous solution. Protonating the $sp^2$ nitrogen does not affect aromaticity because the electron pair is approximately perpendicular to the plane of the ring, meaning it cannot overlap with other orbitals in the ring and therefore does not participate in mesomerism.

Tryptophan also has a $sp^3$ nitrogen atom in an aromatic ring, but like in the imidazole ring it is not very basic.