Why does the O atom of N-phenylnitrosamine donate its lone pair towards H+ when O is more electronegative than N?

If you look at complete mechanism, this wasn't hard to understand why only oxygen get protonated instead of readily available nitrogen, which is supposedly more nucleophilic than oxygen. OP has given N-phenylnitrosamine as the direct product from reaction between aniline and nitrosonium ion, avoiding the first intermediate I (see following reaction mechanism of formation of diazonium ion). As Waylander correctly pointed out in his comment that all of these steps are reversible until you get diazonium ion (counterion $\ce{X-}$ is chloride ($\ce{Cl-}$) in this case):

As I emphasized in the Blue box of the depicted diagram, nitrosamine can be protonated on either $\ce{N}$ (see in blue box) or $\ce{O}$ (see in the total mechanism). As shown in the total mechanism, protonation is slightly favored towards $\ce{O}$ because charged intermediate is resonance stabilized. Nonetheless, this stabilized intermediate is the one directing to the expected product, diazonium salt (see also Waylander comment), making it more favorable. Also note that protonation on $\ce{N}$ make the intermediate going back to original first intermediate I, which cannot give the any other product.

Note: Also keep in mind that direct conversion from intermediate I to resonance stabilized, $\ce{O}$-protonated intermediacte can be achieved by $\ce{N}$ to $\ce{O}$ proton transfer, which is highly likely.


What is the difference between carbon and charcoal filters?

First of all and I want to clarify something that it often misunderstood, which is that charcoal has carbon but it is NOT carbon.  Charcoal actually have a bit of residual Hydrogen and oxygen in it and is again NOT carbon.

Carbon is a pure element that occurs naturally in a variety of allotropes.  Charcoal is the charred product from pyrolyzing wood.  Carbon is found as graphite and diamond and has no intrinsic filtering properties.  Charcoal is a porous material that can burn or adsorb pollutants. Activated charcoal is charcoal which has been specially processed to have a very high surface area.  This surface is what pollutants adsorb (not a(b)sorb) to to be remove from air and water.  If you said activated charcoal or activated carbon anyone would know what you were talking about, but activated charcoal is more proper as the material is in fact charcoal, NOT carbon.


Why do the reactions of Magnesium in steam and cold water differ?

Magnesium is a very active metal- highest activity of all engineering metals. Your question gives you part of the answer. In cold water it forms a decent protective layer. Given greater activation energy- simple as hot water, that layers is simply damaged the oxidation of the magnesium proceeds unimpeded at a greater rate. Once its in steam the oxidation is extremely rapid.


"Epoxy" here refers to an epoxy resin, which is comprised of oligomers that contain at least two epoxide functional groups. Hardening agents, like polyamines and polysufides, react with the resin via nucleophilic attack on the epoxide (as you desribed). Because the hardening agent has multiple nucleophilic groups, it will react with multiple epoxy oligomers. Likewise, the epoxy oligomer reacts with multiple hardening agents. This allows a complex crosslinked network to form.

Looking at the MSDS for a typical epoxy hardener, "polymercaptan amines" is listed as a "blend." Likely this is just a mix of polysulfides and polyamines that was given an ambiguous name by the manufacturer. These molecules likely contain a varying number of nucleophilic functional groups.

  Stoichiometry at its finest.

The stoichiometry is actually not straightforward here at all. It is possible that the total number of epoxide groups in the resin equals the total number of nucleophilic groups in the hardener, but that isn't necessarily the case. The reason you get to mix the parts 1:1 is because the manufacturer provides an approprite blend of hardeners to provide the desired stoichiometry, whatever it may be.

  But then does the resulting oxide anion attack another epoxide ring?

This reaction certainly competes with attack by the hardening agent. Whether it is competitive enough to be relevant depends on the identity and available amount of the hardening agent. It may even intentionally utilized in the crosslinking process.

As a side note, it is unlikely that an alkoxide anion is involved here. Often times a Lewis acid catalyst is added to facilitate crosslinking, which would provide an alkoxide-Lewis acid adduct. Alternatively, there are plenty of hydrogen bond donors available in these mixtures, so a proton would likely be associated with the epoxide during the ring opening, providing a neutral alcohol upon completion of the reaction.


Why does the Oxymercuration Demercuration mechanism lead to stereorandomization?

The stereorandomization occurs in the formation of the cyclic mercurinium ion intermediate.

When the mercury acetate approaches the alkene, it can do so from either the top face:

or the bottom face:

Since the initial alkene is achiral and the steric factors are the same for both approaches, we get both ions, which are enantiomers.

Each enantiomer reacts stereospecifically with water to produce a chiral product. Note that the position where the mercury sits is still a chirality center in this example, but in many simpler cases it is not.

Finally, $\ce{NaBH4}$ is added to this compound as a second step to reduce the $\ce{C-Hg}$ bond to a $\ce{C-H}$ bond. In most cases, if the mercury was at a chirality center, that position no longer is one because we have two $\ce{C-H}$ bonds present.

  Why do we use mercury anyway?
  It works. It works to put the OH on the more substituted position without rearrangements typical of carbocations.

However, there is ongoing research to replace toxic mercury.


d- l- Nomenclature as well as (+), (-) refer to a physical property of the molecule, its optical rotation. Optical rotation is a result of chirality, a topological property of the molecule. So d-, l- is something that you can generally measure with a polarimeter. If you have a 1:1 mixture of the 2 enantiomers then you will see zero rotation and this mixture is called racemic and named with the dl- prefix. A meso compound also has zero optical rotation since it is achiral but as it is not a (racemic) mixture the dl- notation doesn't make sense but you might see it used sometimes.

The problem now is if you are given a pair of enantiomeric molecules that are mirror images how do you name them so you can tell them apart every time? This is why the D-, L- and R-, S- notations were introduced. D- L- is the older and more naive approach that was later improved by the R-, S- (CIP) notation. Unfortunately as chemistry is a science with long history that is also used by other non chemistry-trained professionals, several times older naming conventions have survived. So you have to know all of them!

So very simply, d-, l- reflect a real physical property of the molecule and D-, L- (and R-, S-) is a way to distinguish between molecules that contain stereogenic centres (and might be or not be chiral). Actually even the R-, S- system has been criticized for not being rigorous enough. I saw a paper where a new naming system was suggested but it is not possible to read this without knowledge of advanced maths (group theory, algebraic topology)


How do you figure out the dissociation energy for ions?

If you are trying to find the heterolytic dissociation energy for a gaseous $\ce{KF}$ molecule then as a rough estimate you can model the ions as point charges and calculate the energy required to separate them from their equilibrium separation (the bond length) to infinity, using Coulomb's Law: $$E = -\int_{2.17\times10^{-10}}^\infty\frac{Qq}{4\pi\epsilon_0r^2}dr$$

Note that modelling the ions as point charges is a major simplification and so this is probably not a very accurate estimate. Unfortunately I can't find any data for the actual energy so I can't say how accurate this is.


What are the intermolecular forces between polychloroethylene?

No, but for polymers of any reasonable size, the dispersion (and excluded volume) forces completely overwhelm any dipole-dipole forces, since they grow with the size of the molecule and dipole-dipole forces do not.  Also, bear in mind that the chlorines aren't all on one side of the molecule first because the way it's manufactured tends to put Cls randomly on each side of the chain, and second because the molecule has considerable conformational flexibility.  That's true even in the solid state, which for polymers is usually substantially glassy with interspersed domains of crystal.


Why is butter harder if it once got warm?

Butter is a water in oil emulsion that contains butterfat that crystallizes at room temperature. The continuous phase of butter is a network of fat crystals. This network surrounds water and air droplets. One of the unique properties of fats are the interactions of fat crystals, and these interactions are directly related to product hardness. More fat crystal interactions lead to a harder fat. During storage, fluctuations in temperature from refrigerated to room temperature back to refrigerated temperatures result in postcrystallization. This postcrystallization strengthens the fat crystal network since more crystals are formed, leading to a stronger fat crystal network.


Preparation: Take the parent compound per se, biphenyl.  The bond keeping the two phenyl rings together leads to one carbon atom we call C1, the other C1'.  Now if you label the other carbon atoms in each  phenyl ring seperately in clockwise direction, you'll get C2, C3, C4, C5, C6 in one cycle; and C2', C3', C4', C5', C6' in the second.  (I haven't a credit to upload a picture yet.)

Now: The larger, the more sterically demanding substitutents on C2, C6, C2' and C6' are (condition), the higher the energetical barrier is to interconvert the conformers into each other.  Eventually, this may lead to atropisomerism, and the isomers are atropisomers.

Note, please, sometimes there may be already a preference for one or the other conformation.  If, for example, you substitute C2 and C2' of biphenyl by nitrogen, free molecules of this compound 2,2'-bipyridine energetically favour a conformer with the nitrogens as far as possible separated (similar to a s-trans conformation).  Offering a cation to bind, to complex to, however may lead to a preference towards the s-cis-like conformation.


Ratio of energy required for electrolysis vs creating water

Theoretically, you will get the same amount of energy out of putting hydrogen and oxygen through a fuel cell as it takes to produce them by electrolysis, and you can reach this result by adding up all the enthalpies of formation involved using Hess' Law and realizing that the fuel cell is just the reverse process of electrolysis.
It is therefore impossible to have a net energy gain by this process as it would violate the conservation of energy.
It is, however possible to lose energy during this process and in the real world, energy is lost during both steps.

Electrolysis:

Takes a net input of energy to produce hydrogen and oxygen
Much of the electricity added is wasted to produce heat due to electrical resistance and overpotential losses.
Current large scale electrolysis systems have efficiencies of at best ~70% [1]
Storage of the gases is also energetically wasteful as they need to be compressed. (or perhaps the hydrogen stored as a metal hydride)

Fuel cell:

Produces a net output of energy by consuming hydrogen and oxygen.
Suffers from the same types of electrical losses as the electrolysis process, resulting in sub-100% conversion efficiency

Fuel cells are still useful devices as they can produce electricity without combustion, without the limitation of sunlight, etc., but they require a source of hydrogen.
Most hydrogen is produced from natural gas because producing it by electrolysis is so inefficient and requires energy from somewhere else.
In the future electrolysis combined with fuel cells might become useful for temporarily storing electricity for load balancing solar panels or other green power sources, and there are some projects using hydrogen generation for grid storage purposes where excess power is stored as hydrogen in underground reservoirs, but they use hydrogen combustion generators to get electricity back as fuel cell technology has not advanced sufficiently. Example in Newfoundland


This is part of the reason I really don't like the way we teach significant figures. What's really important isn't the number of digits you have, but rather the uncertainty in your measurements. I think significant figures are taught as an approximation to the proper uncertainty analysis, but it really seems to be not worth the trouble.

What's really going on is this: when you say $A = 34$ oz, you're actually saying that $A$ is closer to 34 oz than it is to either 35 oz or to 33 oz. In other words, you mean $A = (34 \pm 1)$ oz. Likewise, when you say $A = 34.0$ oz, you're really saying $A = (34.0 \pm 0.1)$ oz. And in the final example you give, when you say $A = 0.000023$ oz $= 2.3 \times 10^{-5}$ oz, you're actually saying $A = (2.3 \pm 0.1)\times 10^{-5}$ oz.

The more precise way to do things is to use the uncertainty you have for each measurement, and use the rules for propagation of uncertainties to get the uncertainty in your final answer. One advantage of using the uncertainties is that your uncertainty won't always be 1 unit of your least significant digit, but the rules for propagating significant digits essentially imply that. Using sig figs usually gets close to the answer you would get from properly using uncertainties, but there are some cases where the result is really weird.

To answer your main question directly, you ignore those placeholder zeros because they don't affect the fractional uncertainty of the quantity you're looking at. Keeping them would give you an answer very different from the proper uncertainty analysis.

  but isn’t that value something only a very fine tuned scale could measure?

Not necessarily. The cost of an instrument is much more closely related to the relative precision of its output, rather than the absolute precision. If all you're trying to do is weigh things with masses between $1 \times 10^{-5}$ and $1 \times 10^{-4}$ oz, then you don't care at all if your scale can measure 1 oz. But if you have objects that are 1 oz and $1 \times 10^{-5}$ oz, and you care about the difference, then you'll need a really precise scale.


Let's ignore the safety aspect here, and focus on the 'if i bathe the jug in water so that all the antimony can leach into the water and wash the jug one more time, can all the antimony be removed that way'. In short, no.

There have been a number of studies on antimony in PET bottles. They look at various factors that influence the rate of leaching of Sb from PET bottles. These factors include temperature, UV, colour of bottle, pH, bottle volume and time. There is evidence that leaching is still occurring well beyond 6 months. One study suggests that leaching levels out after about 800 days. Another study calculates that the total levels of Sb in the plastic are very unlikely to be completely leached out.

Factors that have significant influence of amount of Sb leached include:

time stored (major influence)
temperature (higher temps have major influence)
volume of bottle (smaller volume bottles have higher SA/vol ratio and increase leaching)
type/origin of PET (significant differences between the colour of bottles used in one study)

I'm not aware of a study that attempts to find the conditions whereby Sb no longer leaches into the water, but effectively you would need to bathe the the bottle at high temperature for several days at a time, repeatedly, over months. That's just not going to happen.

Some useful references for you include:

doi.org/10.1016/j.watres.2007.07.048
doi.org/10.1016/j.scitotenv.2009.04.025
doi.org/10.1021/es061511+


Is it correct to say that SOME iodine undergoes sublimation at 1 ATM

To answer part 1 of your question:

It is correct to call the change form solid to gas below the equilibrium vapor pressure sublimation. And there is no separate term for vaporization of a solid above the equilibrium vapor pressure; it is still sublimation.  Similarly, a boiling liquid is still perfectly correctly said to be evaporating.

To answer part 2 of your question:

At temperatures below the melting point of iodine, it will sublimate until the equilibrium vapor pressure is reached without going through a liquid phase in that process.

Once the iodine is heated to it's melting point, it will continue the same process of vaporizing to it's equilibrium vapor pressure. Because it is now a liquid, we just call that evaporation rather than sublimation.

If the iodine is further heated to it's boiling point at 1 atmosphere, then it will continue to rapidly evaporate as it boils away.

Regarding the thermodynamic relationships between these processes as mentioned in one of your comments (which I think you should just edit into the question), what you have said is correct. This is a path-independent thermodynamic process, so with respect to the ΔH of the phase-change processes, it doesn't matter how you get from the solid phase to the vapor phase. ΔH for the direct solid-vapor transition (sublimation) is equal to the ΔH of melting plus the ΔH of evaporation.

I hope I have addressed all your points, please ask for clarification in the comments if I've left anything unclear.


Order of solubility of sodium halides and potassium iodide

The order b-a-d-c is correct, if the solubility is in gramm per liter. If it is in mole per liter, the order is a>b>c>d, as the saturated solutions of NaBr is the highest ($8.78~ mol L^{-1}$), then KI ($8.48~ mol L^{-1}$), then NaCl ($6.19~ mol L^{-1}$), and NaF ($0.95~ mol L^{-1}$)


By using the magnetic susceptibility, calculate the number of unpaired elections in the complex

Let's start with "what is the species?"

While you write it $\ce{HgCo(NCS)4}$ it's probably better written as $\ce{Hg^{+2} [Co(NCS)4]^{2-}}$.

You can take the two metal ions separately as far as electronic structure.

$\ce{Hg^{+2}: d^{10}}$ So we don't expect any unpaired electrons.

$\ce{[Co(NCS)4]^{2-}}$ Well, $\ce{Co}$ must be $\ce{Co^{+2}}$, so it's $\ce{d^7}$ on paper. Next you'd need to consider the shape of the complex and possibly where the ligand is in the spectrochemical series and the likely orbital occupations.

   Here, $\ce{Co^{+2}}$ is tetrahedral, so it doesn't matter where NCS falls on the series. You expect a high spin complex with 3 unpaired spins.

Note: (NCS) can be a tricky ligand, since it can coordinate with either N or S atoms, but the crystal structure is clearly N-coordination.

I mention this, because while the Hg ion is, on paper, separated from the Co complex, in the crystal structure, there's clearly some Hg-S interactions.

So let's get to your question about the calculation. Yes, if you need an integer, you'd round up. But since you have many, many more than one molecule, your answer implies there are multiple magnetic species.

   This is clear from the electronic structure. Since the calculation above gives three unpaired spins in an isolated $\ce{Co^{+2}}$ complex, there must be some magnetic interactions between multiple species in the solid state. Indeed, this is observed at different temperatures. I'm hardly an expert in magnetic coupling here, but it certainly implies Hg must be involved as well.


If acyl chlorides are reacted with aqueous ammonia, will carboxylic acids also be formed?

Acyl halides react with water but usually not violently. The acyl halides are normally not soluble in water so the reaction is slow. Then, when the reaction proceeds, acid is formed and the pH decreases, reducing the reactivity of water.

Ammonia and amines are much more nucleophilic and react much faster with the acyl halide. This reaction is so much faster that it can be carried out in water. In this context it is called the Schotten–Baumann reaction, which normally proceeds with high yields.

An excess of ammonia is required at least to absorb the HCl formed, but normally a huge excess is used. The acyl halide is more expensive than ammonia so we want to have it completely transformed to the amide. The excess of ammonia will also help to dissolve the acid, if it is formed in the reaction.

The other points:

To separate acids and amides, extract with alkali. The acid will go to the aqueous phase and the amide to the organic phase.  Or, if they are small, recrystallize it. The amides are less soluble than acids.
Huge excess of ammonia or simple amines is normally used.
The reaction of water is negligible under these conditions.


Doubts on Diffusion/ Effusion of gases at different pressures

"But we also know that, volumes of a gas diffused per unit time gives us rate of diffusion too. And in the question above, the volumes (in case they were given instead of moles) would have been different because of different pressures."

If you wanted to measure the effused gas in terms of volume, you had to somehow calculate this volume from the number of moles effused. This volume has nothing to do with the volume or the pressure inside the vessel from which the gas effused.

[edit]

From the number of particles that escaped to the right, you could calculate a volume, given an arbitrary pressure (e.g. standard pressure). With this you could express the rate of diffusion in units of volume per time unit.


Finding the equilibrium constant of your overall reaction is straightforward, as you've done in the comments, and we have $$K_\text{tot} = K_1K_2K_3 = \left(\frac{3k_+}{k_-}\right)\left(\frac{2k_+}{2k_-}\right)\left(\frac{k_+}{3k_-}\right) = \frac{k_+^3}{k_-^3}.$$ But this is insufficient to conclude that $k_1 = k_+^3$ or that $k_2 = k_-^3$. Indeed, let us consider what is stated when we have, as in your last reaction, rate  constants for the overall reaction. Such a scenario presumes an overall rate law of the form $$\frac{\text{d}\ce{[A_3]}}{\text{d}t} = k_1\ce{[A_0][B]^3} - k_2\ce{[A_3]},$$ and there is no a priori reason to suggest that this is likely. (Often this is possible in certain limiting regimes, but not in general.) To investigate $k_1$ and $k_2$, we will have to solve our system fully. We can write out the rate equations for our system as
\begin{align*}
\frac{\text{d}\ce{[A_1]}}{\text{d}t} &= 3k_+\ce{[A_0][B]}-k_-\ce{[A_1]}-2k_+\ce{[A_1][B]}+2k_-\ce{[A_2]}\\
\frac{\text{d}\ce{[A_2]}}{\text{d}t} &= 2k_+\ce{[A_1][B]}-2k_-\ce{[A_2]}-k_+\ce{[A_2][B]}+3k_-\ce{[A_3]}\\
\frac{\text{d}\ce{[A_3]}}{\text{d}t} &= k_+\ce{[A_2][B]}-3k_-\ce{[A_3]},
\end{align*}
a coupled system of differential equations, which can be solved using matrix methods---diagonalize the coupling matrix, find the eigenvectors and eigenvalues, and solve---but these calculations are often quite messy, so I will leave them to you if you're interested.

Once $\ce{[A_2]}$ and $\ce{[A_3]}$ are known, we can plug them into the last differential equation and compare it with the presumed form of the overall rate law to determine $k_1$ and $k_2$.


As Zhe explains in the comments, the answer is D. Even though small changes in pressure are not expected to significantly affect the chemical potential of a solid, it is a function of pressure:

$$\mu(\mathrm{s}) = \mu^\circ(\mathrm{s}) + \int_{p^\circ}^{p}V_\mathrm{m}\,\mathrm dp \qquad T = \mathrm{const}$$
This means that a small change in the chemical potential of the solid also contributes to the total free energy change of the system when the pressure is altered. Usually this response is ignored as $V_\mathrm{m}$ is orders of magnitude greater for gases than for condensed phases.


How to easily distinguish salts physically? Edit

Draw a matrix of cations vs anions, mark the possible combinations, and calculate the probabilities.

The chances are 4/12 for a sulphate, 3/12 for a carbonate, etc.

I wouldn't start testing for sulphides ;-)


The strengths of oxidizing and reducing agents are indicated by their standard electrode potentials. Let's consider  standard potentials of some redox couples.

$ E^0(\ce{Li^+/Li})= -3.04V$,

$ E^0(\ce{K^+/K})= -2.92V$,

$ E^0(\ce{Cu^2+/Cu})= +0.34V$,

$ E^0(\ce{F_2/F^-})= +2.87V$

The values above are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential.


The dropping mercury electrode (DME) is a working electrode made of mercury and used in polarography. Experiments run with mercury electrodes are referred to as forms of polarography even if the experiments are identical or very similar to a corresponding voltammetry experiment which uses solid working electrodes. Like other working electrodes these electrodes are used in electrochemical studies using three electrode systems when investigating reaction mechanisms related to redox chemistry among other chemical phenomena.

Source: dropping mercury electrode
Massive thanks to Ed V for advice in the comments prompting my research


Can you create a buffer with a strong acid?

I think the issue with buffer systems is that they are resistant to changes from strong acids/bases in their normal state.  For example, biological systems often are buffered so that they have some resistance to changes in pH, which can wreck havoc on biochemistry.

Recall the basic buffer equation:

$\ce{HA <-> H+ + A-}$

where $\ce{HA}$ is a weak acid, and there is an excess of $\ce{A-}$, the conjugate base. Because $\ce{HA}$ is weak, it is a relative poor proton donor, but $\ce{A-}$ is a relatively good proton acceptor.  So if you now add a strong acid, the $\ce{H+}$ it donates reacts with the $\ce{A-}$ to produce $\ce{HA}$, shifting the equilibrium to the left, rather than to the right as one might expect.  Or perhaps, preventing as much shifting to the right as might be expected. This gets to your issue of buffering capacity.

This line of reasoning does not work with strong acids/bases because they donate/accept protons with gusto, and thus are not resistant to shifting the equilibrium.  You would need to be way at the extremes of pH for there to be a small change to pH, as your graph shows.  In contrast, buffers are resistant to pH change in their normal state.


Effect of steric effect on acidic and basic strength

Ortho tert butyl benzoic acid will be more acidic than the other one .because former is more bulkier so it will rotate the COOH group out of the benzene ring plane to a more extent .
For NH2 group Ortho to CH3 or C(CH3)3 .Ortho effect never applies NH2 can never go out of the plane .rather this time CH3 and C(CH3)3 goes out of plane .so hyperconjugation effect is lost only inductive effect is present .still tert one will be comparatively less basic because of STERIC INHIBITION OF PROTONATION (SIP) . Due to hindrance nitrogen can't donate it's electron pair easily to a proton


(1) There are several web sites giving details of this reaction and so this need not be copied out here.

(2,3) Rather than describe a complex reaction scheme it is easier to understand a generic autocatalytic reaction. The general autocatalytic reaction of species A with catalyst B has the form $\ce{A + B $\to$ P + 2B}$ and the rate equation is deceptively simple and is $da/dt =- kab$ if $a$ and $b$ are the concentration A and B. The important point here is that species B is both reactant and product. As more B is produced than that there initially, the reaction accelerates as it proceeds. This is the reason for the time lag: the reaction rate is initially small but as reaction proceeds the extra B produced speeds it up. Eventually A runs out and so the reaction stops. (The colour change you mention is just a measure of the concentration of permanganate. Note also that from the Beer-Lambert law describing transmission of light the absorbing species concentration is related exponentially to light transmission (and so absorption) and this probably complicates what you observe by eye).

If $x$ is species A concentration, the rate equation is, $\displaystyle \frac{dx}{dt}=-kxb=-kx(a_0+b_0-x)$ or $\displaystyle \int\frac{dx}{(a_0+b_0-x)x}=-kt+c$ which can be integrated using partial fractions;

$$\frac{1}{(a_0+b_0-x)x}=\frac{1}{(a_0+b_0)}\left[\frac{1}{x}+\frac{1}{a_0+b_0-x} \right]$$

which produces two log functions when integrated. The initial conditions are $x = a_0$ when $t$ = 0 then the integration constant is $\ln(a_0/b_0)$ and the rate equation,

$$ \ln \left(\left| \frac{xb_0}{a_0(a_0+b_0-x)}     \right|  \right) =-(a_0+b_0)kt$$

and the absolute value is added because the log cannot be negative. Rearranging produces $\displaystyle x = a_0\frac{a_0+b_0}{a_0+b_0e^{-(a_0+b_0)kt}} $.   This and also the increase of $B =a_0+b_0-x$ are plotted.

Figure Autocatalysis $\ce{A + B $\to$ P + 2B}$. Initially species A decreases slowly as the concentration of B is small. As the reaction proceeds, more B is produced and although A is reduced overall, the rate increases and A is consumed even more rapidly. At longer times, the concentration of A becomes so small that even though that of B is large the reaction rate is slow.


Acid catalysed synthesis of ethyl ethoxymethylenemalonate

Ok, I will try my hand on the mechanism. In my opinion the reaction should work with a catalytic amount of $\ce{H+}$ since the key step for the reaction is the hydrolysis of the orthoester. This is an entropically very favourable reaction since from one reactant molecule you get two product molecules.

The resulting molecule is extremely reactive. So it can easily react with the enolized malonate. The enolization might be facilitated by the $\ce{ZnCl2}$ and/or the cat. $\ce{H+}$.
(I'm not entirely sure whether the $\ce{ZnCl2}$ really plays the role I've given it below in this reaction. So, if anyone knows good arguments against it, please let me know.)

The protonated product can then take part in an $\mathrm{E}1$ elimination reaction to yield ethyl ethoxymethylenemalonate. The carbocation which results when $\ce{EtOH}$ leaves the molecule is stabilized by the neighboring ethoxy group. The hydrogen on the central carbon atom is very acidic, since it is placed between three carboxylate groups, so it can even be removed by rather non-basic molecules, e.g. by an $\ce{EtOH}$, a malonate or an othoester molecule.

So much for the acid catalysed mechanism. Now for the question, why one doesn't simply use a base to facilitate the addition of the activated methylene compound. Well, this is due to the fact, that orthoesters are quite stable against basic hydrolysis. So, under basic conditions you don't get the carbonyl component of the addition reaction.


Does ICP-OES detect molecules in suspension?

It will depend on the size of the particles which are in suspension. If the particles are very large compared with the capillary feeding the solution into the spray chamber then the capillary will block up and you will detect nothing and end up very unhappy with a broken machine.

If the particles are much smaller than the capillary then they will be able to enter the spray chamber and then be present in the small droplets which enter the plasma torch. As a result it will be possible to observe the particles by means of the ICPOES machine.

With a normal solution a droplet enters the torch.
The droplet dries out to make a solid particle
The solid particle melts
The molten droplet then boils in the heat of the plasma
The molecules formed are broken down to atoms in the heat of the plasma
The atoms are then ionized.

The inclusion of solid particles in the liquid fed into the ICPOES will give as the droplets boil to dryness additional solids particles. These solid particles may modify the behaviour of the other elements in the torch. I suspect that it may be similar in some ways to the processes which occur in the fireball as it cools after a low altitude nuclear detonation where things can adsorb onto existing particles. If you look in the literature at studies of fallout particles, the particles from the nevada tests have radioactivity deposited on minerals which are similar to those found at the test site.

I have found that if you increase the salt content of what you are feeding into either a ICPOES or ICPMS that the intensity recorded for a given concentration of an element (not found in the added salt) will decrease. This is a possible sign that some of the element is being lost onto the particles formed by the dehydration of the salt solution.

It is best if the ICPOES machine is only ever fed with samples which have been filtered through a very fine filter. I tend to use a syringe filter with holes which are about 0.5 micrometers. This helps to keep the fine particles away from the machine.


Product of an intramolecular aldol condensation

The two protons are of similar pKa and you can get deprotonation at both sites with equal ease, however it is important to understand that most intramolecular aldol reactions are reversible and thus give rise to the thermodynamic product i.e. the most stable product. [6]-ring formation is always preferred over [8]-ring formation 1, 2. If you prepare the [8] product by other means, and subject it to the aldol conditions you will get the [6] product. The only exception to this is when you set up the reaction to be under kinetic control e.g. by using LDA, then you can get a mixture.


Why flame color of salts is determined by metal

Not all flame colours are determined by the metal, some anions also colour flames (see the linked question.)

Also, not all metals impart a notable flame colour (platinum, for example, is often used as a substrate material to do flame tests because it doesn't cause interference: small amounts of the substance of interest are deposited on a platinum wire which is then inserted into the flame).

What actually causes colour are the electron transitions where excited electrons fall back to lower energy levels emitting light in the process. Flames cause enough excitation that this process is common. Some metals have strong electronic transitions that match wavelengths of visible light. The bright yellow characteristic of sodium come mostly from electrons falling from the 3p to the 3s levels in sodium atoms and have a wavelength around 589nm. This is the same bright yellow that is visible in low pressure sodium street lighting (and is caused by exactly the same electronic process).

Other metals have different electronic structures and their emission spectra therefore have different colours. But not all metals have bright lines in the visible (the strongest lines in the spectrum of mercury, for example, are in the UV so won't give the same notable visible colour in a flame test). Many metals do, though, which is why flame tests are often useful. And even when you can't see the colour, instruments can, so spectroscopy is a useful analytical tool even when your eye can't see much.


How can an electron being a wave have such property as spin?

An electron is not a wave nor a particle. An electron is usually described as a quantum object with some wave-like properties and some particle-like properties.
Some of them, like particularly a spin, do not have direct counterpart in our familiar macro world. It is a kind of a mysterious, specifically quantum property of all elementary particles and  atomic kernels, that interact with angular momentum and interchange with it its values.
Non-relativistic quantum mechanics has no theoretical idea/model about the observed spin phenomena, that would come e.g. from resolving the Schrödinger wave equation. It was  added just as an ad hoc extension. OTOH, the relativistic Dirac wave equation brings spin as the natural part of the quantum mechanical model.
Note that the classical idea about a spin as a rotation related angular momentum of a fast spinning small ball is wrong. Considering the electron mass, energy and the limit of the maximum electron size, an electron simply cannot rotate fast enough to have such amount of classical angular momentum, related to its eventual spinning.
Both orbital and spin contributions to the total angular momentum have their counterparts as orbital and spin magnetic momentum, if quantum objects have nonzero electric charge or charge nonhomogenity (like neutrons ).


You need to take the longest chain first, i.e. the basis of the hydrocarbon skeleton is hexane. It has a ethyl group at position 3, so the IUPAC name will be 3-ethylhexane.


What is the reasoning behind the assumptions of Transition State Theory?

This is deeper discussed in Rate Constant Units and Eyring Equation, but I am going to post a very short take-home message here.
First, the three assumptions you have cited are not complete. They completely gloss over the most important one, i.e. nuclear and electron motion can be separated, energies (translation, vibration, rotation) can be treated additively.
Having gotten that out of the way, reactants are in constant thermal equilibrium with the transition state structure. This actually means that the reactant structure(s) is in equilibrium with all possible structures that can be reached from it. It is hence more correct to say that the reactants are in thermal equilibrium with all possible transition state structures.
This also basically ensures, that all these structures can be treated within a single, combined Boltzmann statistic, where the energy is distributed according to all the states of the assembly/molecule.
The next assumption is that the transition state will be crossed only once (point of no return), which means that when an assembly/molecule reaches this (high) energy state, it will form the product. This basically sets the limit to the Boltzmann statistic as all states following the transition state are not included anymore.
Putting both together we arrive at
\begin{align}
  \ce{A + B <=> &AB^\ddagger -> P}\\
  \ce{P <=> &AB^\ddagger -> A + B}.
\end{align}
However, you see that you can look at it from both channels, but that'll become to involved here. Please read the other post first.
All these assumption basically lead up to that the transition vector can be treated classical as translation.
In conclusion, the equilibrium part is necessary to write the combined Boltzmann statistic; the transition state collapses into products part is necessary to only include everything up to and including the TS, but nothing more into the statistic.

Also related: References for Transition State Theory and Quantum Transtion State Theory


Implication of completely non relativistic Hamiltonian

Your reduced equation is just the normal hydrogen Hamiltonian written in atomic units, that is with $\hbar=\frac{1}{4\pi\epsilon_{0}}=m_{e}=e=1$ (e is the charge of the electron, which they include for some reason despite it being 1 in atomic units). They also make the substitution $p=-i\hbar\nabla$ where, again to be consistent with atomic units, $\hbar=1$.

The form you have normally seen of the Hamiltonian is already spinless and non-relativistic (which are essentially synonymous for quantum mechanical operators). For something as light as Hydrogen, relativity (as in Einstein's theory of Special Relativity) is known to have very limited impact, and so it is neglected in forming the Hamiltonian.

There may be other questions on the Chem or Physics SE related to relativity in QM, but I would suggest unless you are particularly interested in relativity to simply focus on the QM side of things for now until you have that clear and can build on that knowledge.


Effect of temperature on equilibrium constant in terms of entropy change

The key point is that the entropy of the surroundings changes, not that of the system (at least for a small change in $T$), so only [B] and [D] are relevant, as you correctly determined.
You can write
$$\log(K)=-\frac{\Delta H}{RT}+\frac{\Delta S}R$$
This of course is one form of van 't Hoff's equation. It tells you, assuming $\Delta H$ and $\Delta S$ are constant, that the temperature dependence is due to $\Delta H-$ the heat transferred to/from the surroundings (entropy increase/loss of the surroundings):
$$\log(K)\propto-\frac{\Delta H}{RT}=\frac{\Delta S_\text{surr}}R$$
Now increasing $T$ makes the change in the entropy of the surroundings smaller.
When $\Delta H\gt0$ (endothermic), ie $S_\text{surr}\lt0$, this makes the reaction more likely to happen. When $\Delta H\lt0$ (exothermic), ie $S_\text{surr}\gt0$, this makes the reaction less likely to happen.
So both [B] and [D] are correct.


It is unlikely that a potential applied to a sodium electrode in water would prevent the spontaneous oxidation of sodium by water.  Consider the following reduction potentials:

$$\ce{2Na+ + 2e- -> 2Na(s)}\ E^0=-2.71\ \mathrm V\tag{1}$$
$$\ce{2 H2O + 2e- ->H2 + 2 OH-}\ E^0=-0.83\ \mathrm V\tag{2}$$

The reduction of water will always occur before the reduction of sodium ions.  If we reverse equation (1) so we can combine the two and get the full redox equation for your system, we get:

$$\ce{2 Na(s) + 2 H2O <=> 2Na+(aq) + H2 + 2 OH-}\ E^0= 1.88\ \mathrm V\tag{3} $$

A positive potential indicates a spontaneous reaction under the given conditions so thermodynamically, it will be near impossible to prevent the reaction from being spontaneous.

How impossible?  We can do a thought experiment and ask the question at what pH would this reaction be non spontaneous? In other words, when would the Nernst Equation for this system evaluate to zero:

$$E=E^0-\frac{RT}{nF}\ln\frac{\text{Products}}{\text{Reactants}}\tag{4}$$

Entering our values, converting Ln to Log and substituting $\mathrm{pH} = -\log[\ce{H}]$ (and assuming all other concentrations are fixed at 1 M or 1 atm):

$$E=1.88+0.059(14-\mathrm{pH})\tag{5}$$

The reaction would not be spontaneous until the pH reached 45 units.  Something that can't be done in water.


Yes, they do exist and were characterised spectroscopically, see here (and there is a note on similar clusters for sodium):

Blanc, J.; Bonačić‐Koutecký, V.; Broyer, M.; Chevaleyre, J.; Dugourd, P.; Koutecký, J.; Scheuch, C.; Wolf, J. P.; Wöste, L. Evolution of the electronic structure of lithium clusters between four and eight atoms. J. Chem. Phys. 1992, 96 (3), 1793–1809. DOI: 10.1063/1.462846.

They are not, however, "stable" enough to exist in solid state or even to obtain a pure gas, rather complicated arcane trickery is used to work with (very small amount of) them.

Lithium has rather complicated chemistry, easily forming polyhedral structures when bound to proper partner. For example, methyllithium is a tetramer. The simplest rationalization is that its s- and p-orbitals have very close energies and thus low-energy vacant p-orbitals participate in formation of multicentre electron-deficient bonds.


Reaction between a vinegar/salt/hydrogen peroxide mixture as an etching solution and copper

The etching solution itself is fine to dispose of down the drain. However, if you use it to etch copper, the dissolved copper ions are quite toxic, and should not be disposed of through the sewer system.

You can bypass the problem by collecting the etch solution, and submerge aluminum foil. $$\ce{3Cu(CH3COO)2(aq) + 2Al(s)->2Al(CH3COO)3(aq) +3Cu(s)}$$React with the aluminum until the bluish color disappears, then collect the solid copper + remaining aluminum to throw in the trash. You can now flush the solution down the drain.


In the moment from the video that you referred to, Sal was using the equation $\rm pK_w=pH+pOH$ to find the $\rm pOH$ of the solution. $\rm pK_w$ is equal to 14, since $K_w$ is equal to $1\times 10^{-14}$, giving us the new equation $14=\rm pH+pOH$. We know that the $\rm pH$ of a 1M solution of $\ce{HCl}$ is zero from earlier on in the video. This gives us that $14=0+\rm pOH$, or in words, that the $\rm pOH$ must be equal to 14.


In sp² hybrid orbital, why does the p orbitals only have 3 sides instead of 4?

First, an $sp^2$ orbital does have two lobes. One is large, and one is tiny. Secondly, $sp^2$ means "a 'mixture' of one $s$ and two $p$ orbitals". Like the following:

So, you plug in three orbitals, and you get out three more orbitals. When drawn in place, you get this:

There are a total of six lobes here, but they are distributed amongst three $sp^2$ orbitals.

The number of lobes need not stay constant--it's the number of orbitals that does, since each orbital corresponds to two electrons, and there's no way for new electrons to be added/removed (yet). What's happening is that the wavefunctions(orbitals) of the electrons are "mixing up" to distribute the electrons in a way that stimulates bonding (one per orbital). The lobes have a significance, but it's much deeper.


First of all as @chipbuster says $\ce{HF}$ in diluted solutions in water is nearly completely dissociated and therefore shouldn't be called weak. Wikipedia describes this nicely and cites several sources for this claim.
It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to fluorine anions with hydrogen bonds, in what is called tight ion pairs. It prevents from detecting the true strength of $\ce{HF}$ with methods like acid–base titration - they show dissociation of ionic pair. Similar effect is present in ion exchange resins.
Strength of this acid is revealed in more concentrated solutions, where $\ce{HF}$ molecules replace hydronium ions, creating bifluoride anions, freeing them - it's homoassotiation mentioned by Wildcat.
More to the point - does it mean that $\ce{HF}$ isn't weaker acid than other hydrogen halides? The answer isn't simple, and acidity itself depends on many factors.
In terms of $\ce{K_a}$ in diluted aqueous solutions $\ce{HF}$ is probably still weaker than $\ce{HCl}$ and the rest but getting exact value is greater problem than normally (generally values of $\ce{K_a}$ for strong acids aren't precise). Decrease in acidity between hydrogen halides is described in answer to this question
For concentrated solutions one needs to use Hammett $H_0$ function which gets lower than -10 in pure $\ce{HF}$ (pure sulfuric acid has -12) - it means it's very strong, although not superacidic.
See also:
The Hammett Acidity Function $H_0$ for Hydrofluoric Acid Solutions


Why does oxalic acid have exactly two water molecules of crystallization?

As @Ivan Neretin states, the crystal structure determines how much water can be accommodated. For example, $\ce{CuSO4}$, cupric sulfate or copper(II) sulfate, may be found as a mineral with three, five or seven molecules of bound water:

Bonattite (trihydrate mineral)
Bluestone (pentahydrate)
Boothite (heptahydrate mineral)

Of course, it can also be heated to form the anhydrous salt.

One could say different crystal forms have room for different amounts of water, or the converse: by adding or removing water, the crystal structure is changed.


Activated complex theory vs. consecutive reactions

The key difference is that transition states occur at a maximum of the potential energy curve for the reaction whereas intermediates occur at a local minimum. Take this example of the reaction profile for an $\ce{S_{N}1}$ reaction:

You will see that the products and reactants occur at minima on the curve and the intermediate also occurs at a minimum, albeit a higher energy one. In between the minima are located the maxima where you find the transition states.  Unless the activation energy is zero (see this question for rare examples) there will always be a transition state located between any two minima.

Transition states are usually represented using dashed bonds to show bonds in the process of being broken or formed as opposed to being fully formed as in intermediates or products. Additionally the Hammond postulate says that the transition state will most resemble the stable species closest to it in energy, in this case the carbocation, and this can be used to help predict the structures of transition states.

Transition states are very short lived because they immediately 'roll downhill' on the potenetial energy curve to reach an intermediate or product. By contrast some intermediates are actually quite stable and can be isolated.


A "function of proportionality" in a rate law

If by "fancier" you mean "uglier", then absolutely yes. Real chemistry is messy, much more than we can fully handle. The simple rate laws we learn in undergraduate physical chemistry are just approximations.

I've written elsewhere that, in general, a full description of chemical kinetics for any overall reaction requires studying an entire tree of simultaneous chemical reactions. There can be dozens, even hundreds of intermediate species involved, all mutually linked in complex ways, and with more than one final product ("by-products"). The difficulty in performing this full description has lead to the whole field of chemical reaction network theory. I hope you like matrix algebra.

To make chemical kinetics tractable in most cases, two major approximations made are:

Forget about the whole reaction network, and look at only the overall kinetically fastest path from reagents to products
Forget about all the steps and intermediates in the fastest path, and describe only the bottleneck - the slowest step

Even with these two crushing approximations, it's possible to describe many useful reactions with a fair level of accuracy. This is partly luck, and partly unavoidable (due to simplicity naturally arising in randomness). This level is what most undergraduate and even graduate work sticks to. Even when it's a bad fit, we'll sometimes stubbornly fit a system into a particular rate law, do some variational analysis, and hope the error isn't too great.

A nice example of a reaction whose kinetics is slightly more complex than the standard "nice" rate laws is the radical reaction between hydrogen and bromine.

$$\ce{H2 + Br2 -> 2HBr}$$

According to experimental data, it's possible to obtain the following rate law (which may still be approximate!):

$$\mathrm{r_{HBr}=\frac{k_1[H_2][Br_2]^{3/2}}{[HBr]+k_2[Br_2]}}$$

The deduction is nicely described in this site. As you can see, depending on the reaction parameters, the denominator can potentially be simplified down to a single term, with some degree of approximation, and a simple-ish rate law could be used. But in the general case, it's messier. Because of these reaction networks, even the concept of a "reaction order" does not exist in general. An example of a natural but extremely difficult reaction to describe kinetically in full would be the combustion of a moderately sized alkane, such as octane ($\ce{C8H18}$).

When taking full chemical networks into account, there appear cases with pretty unique kinetics. For example, if you haven't heard of oscillating reactions, they're a topic of considerable experimental and theoretical study. With a particular set of coupled reactions, it's possible for chemical systems to display the hallmarks of mathematical chaos. Even if individual steps can be described by simple kinetics, the global process cannot. There's a nice discussion of oscillating kinetics in this Chem.SE question (and possibly others).

I'm sure there are other ways to get strange rate equations. Some reactions just don't follow kinetic theory in its simplest form. This can lead to weird conclusions, such as negative activation energies. Furthermore, rate laws almost always assume macroscopic amounts of matter. When a tiny amount of reagents are involved, the discrete nature of matter calls for statistical corrections to chemical kinetics, because an integer number of molecules can't follow continuous analytical curves.


How can I find the Van der Waals constants (a and b) without any critical values?

I would approach this a little differently.  I would integrate the pressure between $V_D$ and $V_B$, such that the average pressure is 17.76 bars:  $$\frac{\int_{V_D}^{V_B}{PdV}}{(V_B-V_D)}=17.76$$or,
$$17.76(V_B-V_D)=RT\ln{\frac{(V_B-b)}{(V_D-b)}}-a\left[\frac{(V_B-V_D)}{(V_B-b)(V_D-b)}\right]$$
This would provide an equation for expressing a in terms of b.  So the pressure could then be expressed exclusively in terms of b.  Then, to determine the value of the constant b, I would set $P(V_B)$ equal to $P(V_D)$ and solve the resulting equation for b.


Tap water contains calcium and magnesium ions; more so if the water is hard. These ions can bind to soap over time and make an insoluble soap curd.

Is the sponge clean? If not, another reason is that the surfactants in the soap that make the foam might be forming micelles around whatever grime is in your sponge, so that there aren't many left to make the foam.


Is hydrazine a better nucleophile than ammonia?

Yes, hydrazine is a better nucleophile than ammonia. This is the case in both water (a protic solvent) as well as acetonitrile (an aprotic solvent). A quantitative approach may be found in the Mayr nucleophilicity parameters (Ctrl-F "ammonia" and "hydrazine"):

$$\begin{array}{ccc} \hline
\text{Nucleophile} & N\text{ parameter in }\ce{H2O} & N\text{ parameter in }\ce{MeCN} \\ \hline
\ce{N2H4} & 13.46 & 16.45 \\
\ce{NH3}  & 9.48 & 11.39 \\ \hline
\end{array}$$

The larger, the more nucleophilic. It also makes sense that the protic solvent water reduces the nucleophilicity of both species.

For more details I suggest reading the articles cited in the database entries. The most relevant one is: J. Org. Chem. 2012, 77 (18), 8142–8155. Inside there is a good discussion of the methodology.

As for why hydrazine is a better nucleophile, the jury is still out. In general there is a phenomenon whereby nucleophiles with lone pairs on two adjacent atoms are more nucleophilic than expected, which is called the α-effect. [Note that I deliberately avoid saying that hydrazine is more nucleophilic "because of the α-effect", because that merely begs the question. The α-effect is not (yet) a theoretical explanation, but rather an empirical observation.]

I don't have the interest to summarise reviews on the α-effect (at this point in time, at least) so I will quote Mayr from the article above:

  The α-effect, which has been investigated for various reactions of nucleophiles including acylations, Michael additions, SN2 reactions, and nucleophilic aromatic substitutions has been the topic of several reviews, but its origin and extent are still discussed controversially. [...] Several theories on the origin of the α-effect have emerged, which include the destabilization of the ground state by electron repulsions, the stabilization of the transition state, thermodynamic stabilization of the products, and solvent effects.

See also: Is there a general consensus on the causes of the alpha-effect? Unfortunately, that question does not have a proper answer, and I believe the direct answer to that question is "no". As Jan's comment indicates, too, the explanation must go beyond "hydrazine has two lone pairs but ammonia has one", because that alone doesn't actually explain anything.


Rust is by definition the transformation of iron metal to iron oxide (and iron hydroxide as an intermediate). You cannot "rust" other metals, although you can form oxides and degrade them etc. It's just not called rust when you convert, e.g. $\ce{Cr}$ to $\ce{CrO3}$. So there is only one rusting process.

1) Water is not a catalyst in the reaction that forms rust. A catalyst by definition is not consumed during a reaction (its bonds are not broken). Water can be and is the medium that carries $\ce{O2}$ to the iron (otherwise $\ce{O2}$ from air can react with the metal), and it is consumed in the process by dissociation. The fact that it is reproduced does not make it a catalyst. First iron atoms are oxidized by $\ce{O2}$ as in:

(a) $$\ce{Fe + O2_{(aq)} \rightarrow Fe^{+2}_{(aq)} + 2e-}$$

Then the free electrons are quickly absorbed by free protons from water by:

(b) $$\ce{2e- + 2H+_{(aq)} \rightarrow H2}$$

It is highly unlikely for $\ce{H2}$ to react with $\ce{O2}$ to form water -- that is a combustion reaction, so high heat or pressure would have to be added. Rather, water is reproduced by a different reaction (d), below.

2) "Dry" rusting is not possible because the process requires reactions with water's ionization products $\ce{H+}_{(aq)}$ and $\ce{OH-}_{(aq)}$ (notice the aqueous states of all ions). The process continues by production of iron hydroxides:

(c) $$\ce{Fe^{+2}_{(aq)} + 2OH-_{(aq)} <=> Fe(OH)2_{(s)}} $$
(d) $$\ce{Fe^{+2}_{(aq)} + 4H+_{(aq)} + O_2_{(aq)} <=> 4Fe^{+3}_{(aq)} + 2H2O_{(l)} } $$
$$\ce{Fe^{+3}_{(aq)} + 3OH- <=> Fe(OH)3_{(s)}} $$

This is why more acidic media produces rust faster; it pushes (c) and (d) to the right because (b) consumes free electrons and renders more $\ce{Fe^{+2}}$ for further reaction.

Finally the iron oxides "dry" to form the actual rust compound, which is a hydrated oxide: $\ce{Fe2O3\cdot nH2O }$.

3) You can see by now that water is the required solvent because the reactions of interest all involve $\ce{H+}$ and $\ce{OH-}$.

One could say there exists a "rust triangle" where "time, oxygen, and water" are required as an analog to the "fire triangle": heat, oxygen, and fuel.


the particle in 1-D box momentum and total energy(P.E + K.E (x))

As this is obviously a homework problem, I'll just give you a hint:

Observables can be measured simultaneously with arbitrarily small uncertainties if their operators commute. Compute the commutator of the momentum and particle-in-a-box Hamiltonian operators; if the commutator is zero, the answer to your question is "yes".


Energy cycle centered around methane production / consumption

Vehicles that run on natural gas already exist, and have done for many years - you can find CNG (compressed natural gas) cars and larger vehicles around the world.

But there's very little synthesising of methane because there are very few places with a sensible carbon price, so it's just cheaper to extract natural gas and burn it.

Research is ongoing on methane synthesis - this is usually done under the name "power to gas", which can refer to hydrogen or methane. Either way, it usually starts with the splitting of water to get hydrogen, and then is sometimes combined with methanation of the hydrogen. The latter is an exothermic reaction, so combining the two processes can give pretty good efficiencies (see e.g. this paper).

Hydrogen / methane for cars is looking like a lost battle: electric cars have almost certainly already won; it's just going to take a few years until they completely dominate. However, there is still the possibility that larger vehicles (heavy goods vehicles, boats, aircraft) might settle on hydrogen or renewable methane as their decarbonisation option.

By the way, just because a fuel-cell might use rare-earth metals, doesn't make it unsuitable for mass market. Most rare-earth metals aren't rare, despite the name. Fuel cells have lots of issues, and fuel-cell vehicles really haven't achieved much yet, but that's to do with lack of reliability and longevity, rather than scarcity of components.


Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule.

Sort of. In calculating oxidation numbers we pretend that covalent bonds are ionic. Perhaps it is valid to say we are pretending the entire molecule is an ionic compound.

In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule.

More accurate, we find the that oxidation states of atoms change going from reactants to products. Charge conservation, coupled with the way we assign oxidation states, guarantees that the sum of oxidation states is constant. So if one atom shows an increase in oxidation state, at least one other has to show a decrease in oxidation state.

Does this mean an actual electron was transferred given that oxidation numbers are fictitious?

It is not clear what "an actual electron" means. Electrons are indistinguishable, you can't paint one red and follow its path during a reaction. If the electron transfer (according to oxidation numbers) is intramolecular, it becomes a bit of a philosophical question whether there was an electron transfer. For example, is there an electron transfer during keto-enol tautomerism?

I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

It is not clear what "physical electron jumps" are.
For some reactions, it is clear that you have an electron transfer because the only thing that distinguishes reactants from products (pairwise) is the number of electrons:
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$
If this reaction is run as two half-cells, this is even more clear-cut.
For other reactions, the changes are more subtle. If a iodine radical combines with a methyl radical (as part of a chain termination reaction), it is formally a redox process:
$$\ce{I. + .CH3 -> I-CH3}$$
However, the electronegativities of iodine and carbon are quite similar, so it is not intuitive that an electron transfer happened. As Maurice points out in the comments for a similar case, you would have trouble writing this in terms of half-reactions.
In the reaction of methyl lithium with a carbonyl, the oxidation numbers change. You might not see it as a electron transfer and would rather think of it as methyl transfer. However, because carbon is more electronegative than lithium, the oxidation numbers of the carbonyl carbon and the methyl anion change when a bond is formed between them.

Does a change in oxidation number in a redox reaction mean that an actual electron transfer occured?

If I answer this like a certain politician: it depends upon what the meaning of the phrase "actual electron transfer" is. So it is mostly a question about how we talk about redox reactions rather than a question about the underlying chemistry.


How to stop a lye + oil + water reaction in aluminium container?

Slowly add vinegar. DO NOT TURN THE FLAME BACK ON.

Most likely, the alkali concentration is high enough that you've dissolved away the passivating oxide on the aluminum, and the exposed aluminum metal is reacting with water to form $\ce{Al(OH)4-}$ and hydrogen gas.  Adding vinegar should bring the $\mathrm{pH}$ back down into a range where the aluminum can passivate again, stopping the reaction.

The batch of soap is probably already ruined, so I wouldn't worry about that.

In the future, I would recommend using a cast iron or mild steel (not stainless) pan for this procedure.  Plain iron is remarkable for its resistance to alkali, in my experience.


Why do Grignard reagents react with epoxides but not THF?

The difference between THF and epoxides is perhapse bigger than you might think. Due to the three-membered-ring of an epoxide, a lot of energy is 'stored' in ring strain. This destabilizes the stucture and is a driving force for a Grignard reaction (or any substitution on an epoxide).

The THF is a five-membered-ring, which is far more stable. This bond is simply an ether, that happens to be in a cyclic structure. Therefore it also reacts like an ether: not easily attacked by nucleophiles.


What analytical methods are used to quantify C, O, N, and H in a sample?

For N:

Total N = N from NO2 + N from NO3 + Total Kjeldahl N

The methods from here are:

Liquid Chromatography: NO2, NO3

Specific electrodes: NO3, Kjeldahl

Spectrophotometry: NO2

This for water including wastewater. I don't know the aplicability for other samples.

Also ICP-MS is worth researching.


How to measure ratio of Na to K in table salt?

To answer your question: no, you cannot do it at home. Some analytical techniques are mentioned in the comments.

However I decided to post because LIKELY people should not worry about the Potassium content of table salt.

Table salt (normally "just" NaCl) containing as much of 28.5 % of Potassium (might be more perhaps, but this is what I have at hand) can be sold free of medical prescription not only in drugstores/pharmacies but also in supermarkets, where it can be found in the special diets section or alongside standard salt.
It is a dietary salt indicated to limit the Sodium intake (for instance in case of high blood pressure).
Its price is much higher that that of common salt. It can be used to replace salt in all kitchen uses. It is less salty and more bitter than the normal salt, and one has to "get the mouth" to it.

I would therefore infer that is highly unlikely that something sold as table salt NaCl could contain so much potassium to worry about. Especially when some of those worrying of it, might be prone to buy Potassium integrators during summer :)

Disclaimer: it is clear that the potassium content of whatever table salt sold in food stores and supermarkets should not be matter of concerns however, as for any diets and integrators (Fe, K, Mg...) there exists health conditions which are incompatible. Before starting a diet or taking dietetic salts one should seek the advice of his/her physician.


Reaction of one mole of (PO4)3- with one mole of H2CrO4

To answer this question, you should stop thinking about the acids and bases reacting with each other, but rather they react with water exclusively.
Firstly, the phosphoric acid and its conjugate bases will have the following reactions with water

$$\ce{H3PO4 + H2O <=> H2PO4^- + H3O^+}$$

$$\ce{H2PO4^- + H2O <=> HPO4^2- + H3O^+}$$

$$\ce{HPO4^2- + H2O <=> PO4^3- + H3O^+}$$

Then there is the chromic acid and its conjugate bases

$$\ce{H2CrO4 + H2O <=> HCrO4^- + H3O^+}$$

$$\ce{HCrO4^- + H2O <=> CrO4^2- + H3O^+}$$

So now you have five equlibrium reactions.  You have eight unknowns, $[\ce{H3PO4}]$, $[\ce{H2PO4^-}]$, $[\ce{HPO4^2-}]$, $[\ce{PO4^3-}]$, $[\ce{H2CrO4}]$, $[\ce{HCrO4^-}]$, $[\ce{CrO4^2-}]$, and $[\ce{H3O^+}]$.

The Ka expressions from the above reactions give you five equations.  Then you have the initial concentrations of phosphate and chromic acids, which will give you two more, then the fact that the number of protons is conserved will give you the eighth and final equation.


Under normal conditions, atoms will bond to one another and produce a molecule, only if the bond formation process is exothermic (e.g. it releases energy) - this translates into the products (the molecule) being thermodynamically more stable that the starting material (the atoms).  The same is true for the adsorption process.  A gas will adsorb on a surface only if this configuration is more stable than a gas and a free surface.

Typically when a gas is adsorbed on a surface, something akin to bond formation between the surface and the gas takes place and we wind up with a chemisorbed species.  There are many ways to study this process.  One simple way is to measure the amount of gas admitted to the system containing the surface to be studied and measuring the change in pressure over time.  When done in a calorimeter, the heat given of during the process can also be measured.  Combining these two measurements(change in moles of free gas and energy released) allows one to understand the basic energetics of the process. The "isoteric method" is one experimental procedure for accomplishing this in a laboratory.  It relies on measuring moles of gas vs. the pressure in the system (an isotherm) at various temperatures.  From these measurements the enthalpy of adsorption can be determined.

The study of the adsorption of a gas on a surface is complicated.  There is not a surface with a monolayer of gas adsorbed on it with free gas above this.  Rather there is a surface and bulk gas and in between the two is a multilayer structure of adsorbed gas where the gas concentration varies throughout.  Here is a link to a recent Cal Tech thesis on the characterization of hydrogen gas adsorption onto various surfaces; take a look at section 1.5 and the beginning of chapter 3 for a nice overview of the adsorption process and the interfaces involved. The rest of chapter 3 is a discussion of the thermodynamics and various experimental techniques used to study the process.  It sounds like you might also be interested in performing calculations on the adsorption process.  If so, section 5.8 delves into computational methods used on one of the surfaces studied.


What is the IUPAC name of [Co(NH3)5H2O]Cl3?

Multiplicative prefixes are not considered in determining the alphabetical order of ligands.
According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), the systematic names of coordination entities are generated by using the following principles:

IR-9.2.2.1 Sequences of ligands and central atoms within names
The following general rules are used when naming coordination compounds:
(i) ligand names are listed before the name(s) of the central atom(s),
(ii) no spaces are left between parts of the name that refer to the same coordination entity,
(iii) ligand names are listed in alphabetical order (multiplicative prefixes indicating the number of ligands are not considered in determining that order),
(…)

Therefore, the systematic name of the coordination entity in $\ce{[Co(NH3)5H2O]Cl3}$ is “pentaammineaquacobalt” since the ammine ligands are named before the aqua ligand according to alphabetical order. Note that, since 1970, the recommended name for water $(\ce{H2O})$ as neutral ligand in coordination complexes is “aqua” (formerly “aquo”).
The ionic composition of the complex salt $\ce{[Co(NH3)5H2O]Cl3}$ may be indicated in several ways:

IR-9.2.2.4 Charge numbers, oxidation numbers and ionic proportions
The following methods can be used to assist in describing the composition of a compound:
(i) The oxidation number of the central atom in a coordination entity may be indicated by a Roman numeral appended in parentheses to the central atom name (including the ending ‘ate’, if applicable), but only if the oxidation state can be defined without ambiguity. When necessary a negative sign is placed before the number. Arabic zero indicates the oxidation number zero.
(ii) Alternatively, the charge on a coordination entity may be indicated. The net charge is written in arabic numbers, with the number preceding the charge sign, and enclosed in parentheses. It follows the name of the central atom (including the ending ‘ate’, if applicable) without the intervention of a space.
(iii) The proportions of ionic entities in a coordination compound may be given by using
multiplicative prefixes. (See Section IR-5.4.2.1.)

Therefore, correct systematic names for the compound $\ce{[Co(NH3)5H2O]Cl3}$ are:

pentaammineaquacobalt(III) chloride
pentaammineaquacobalt(3+) chloride
pentaammineaquacobalt(III) trichloride
pentaammineaquacobalt(3+) trichloride

Note that multiplicative prefixes are usually not used in binary names if there is no ambiguity about the stoichiometry of the compound.


Given the mass of precipitate, how can I calculate the relative amount of the original dissolved compounds?

x = number of moles of chloride
y = number of moles of bromide

Since you know the mass of the original $\ce{NaCl}$/$\ce{NaBr }$mixture and the molar masses of $\ce{NaCl}$ and $\ce{NaBr}$, you can write an equation of the form:

$58.44 \pu{\frac{g}{mol}}~ x + 102.89 \pu{\frac{g}{mol}}~ y = 0.180~ \pu{g}$

Based on this idea, you can write a similar equation for the final products $\ce{AgCl}$ and $\ce{AgBr}$. With that in hand you will then have a system of equations you can solve (for x and y) that can be converted back into the masses of those salts.


Textbook Claim: "... in all cases it is the electrostatic force acting between charged particles that is responsible for all the forms of bonding."

You ask how bonding between two neutral atoms can be electrostatic. In the case of covalent and van der waals the electrostatic basis is not so obvious as it is for ionic compounds. A covalent bond forms when two neutral atoms establish a shared molecular orbital which is much larger than the separate atomic orbitals that the bonding electrons contributed by each atom formerly occupied. Electrostatic repulsion favors electron pairs being in larger orbitals, because this increases average distance between the electrons. In the case of Van der Waals attraction, the neutral atoms are electrostatically attracted because when their electrons are not symmetrically distributed, there is a partial positive charge at one end of a molecule and a partial negative at the other, even though overall charge is zero. Electrons in adjoining molecules will tend to synchronize their movements so that the positive end of one molecule will be near the negative end of the adjoining molecule; this is called induced charge.


Why is it the middle ring of anthracene which reacts in a Diels–Alder?

To provide a reason for the observed regioselectivity, it is helpful to draw anthracene's aromatic π-electron system in alternance of single and double bonds.  In this instance, it is more beneficial than "the ring" symbolizing the delocalised electron system, as this helps you to account for the precise number of π-electrons before the reaction (starting materials), during the reaction (the mechanism), and after the reaction (the product).

For the Diels–Alder reaction, you may imagine two different pathways.  I invite you to draw the mechanisms by yourself:

A reaction that involves carbon atoms #1 and #4 (or #5 and #8).  Possible, by mechanism.  There are five double bonds remaining in conjugation, and you count one six-membered ring in the state of "a benzene ring" (the very left one).  In the very right six-membered ring, there is only a single double bond, too.
Alternatively, a Diels–Alder reaction with carbon atoms #9 and #10.  This is more favourable then the former example, because both the very left as well as the very right ring of the product are in the state of "a benzene ring", each of these rings contain 6 π-electrons, the prerequisite of a Hückel aromatic compound.  It is so much favourable to the former, that this is the reaction observed.

It may be helpful to add that benzene, naphthalene and anthracene are of course Hückel-aromatic compounds; with 6, 10 or 14 π-electrons they fit into the rule of $(4n + 2)$.  Yet gradually, as experimentally found, in this group of three, benzene is the most, anthracene the least aromatic compound.


How are non-carbon stereogenic centers named (S/R)?

but I can't find any canonical reference as to that.

Look harder ;-)

It seems to be that the lone pair has lowest priority.

OK, I can't find any authoritative reference to that (I don't have access to the original paper), but I've got plenty of others:

Alright, IUPAC says:

  Except for hydrogen, ligancy, if not already four, is made up to
  four by adding "phantom atoms" which have atomic number zero
  (0) and are thus always last in order of preference.

  This has
  various uses but perhaps the most interesting is where nitrogen
  occurs in a rigid skeleton, as for example in a-isosparteine (14);
  here the phantom atom can be placed where the nitrogen lone pair
  of electrons is; then N-i appears as shown alongside the formula,
  and the chiraiity (R) is the consequence; the same applies to N-i6.
  Phantom atoms are similarly used when assigning chirality symbols to chiral sulfoxides (see example to Rule E—4.9).

When ligancy is less than four, we obviously have some lone pairs hanging around (except in Hydrogen--which has been excluded, and some other cases like carbocations--which are planar anyway)

So the lone pair becomes a "phantom atom" and has 0 priority.

Now I wonder what happens to these rules if you have two lone pairs, one of them muonic ;-)

Check the first revision of this post for my original, non-authoritative references


How would I neutralize a PEP (Phosphoenolpyruvic) acid?

For our work, we also need to make 100 mM PEP and we neutralize it using KOH. Traditionally, using Na+ based salts isn't ideal for biological assays so K+ salts are better for our applications.

Alternatively, if you're adding the PEP to a different solution, you can buffer the reaction solution. While the Phosphate does have a good amount of buffering capacity, effective concentrations of the molecule will be low so you ultimately won't be adding much.


Why do glow-in-the-dark substances dim gradually?

There are two potential questions here: one is why is dimming a slow drawn-out process (incoherent, unlike a "switch"). The second question is why different stars may appear to lose their intensity at a different rate.

The light-emission process behind your stars involves phosphorescence. The process involves relaxation of an electronically excited system during which it undergoes a "forbidden" triplet$\rightarrow$ singlet transition. For an individual system, light emission is controlled by the probability of the quantum-mechanical interconversion (or equivalently by the lifetime of the triplet state). It is a stochastic event described by Poisson statistics or "shot" noise.

However, the luminosity of one star represents the collective behavior of a large number of emitters (systems). One star, therefore, appears uniform in luminosity and its time-dependent luminosity decays exponentially.

Variability might be expected in the luminosity of different stars. This is not likely to be QM statistics since, as already noted, you are observing a very large ensemble of relaxing molecules, and one fairly uniform decay constant is expected for all stars, assuming the phosphorescent material is uniform. It seems more likely that luminosity differences among or within stars may occur due to size effects or differences in the concentration of phosphorescent material, or both$^\daleth$. Stickers that are thicker or containing a greater concentration of excited phosphorescent material, due for instance to variability in the manufacturing process, should appear to glow longer simply because they contain more emitters$^\dagger$.

$\dagger$ As helpfully pointed out in the commentaries, more emitters lead simply to the detection limit being reached later.

$\daleth$ As helpfully emphasized in yet another comment, and alluded to in the OP, it is the concentration of excited phosphorescent material that determines the luminosity, and excitation may also depend on the dimensions of the star.


Compiling my comments(now deleted) into an answer.

TL;DR: the products are different due to different stoichiometric ratios of the reactants and is not dependent on the solvent. You can use different solvents so that you can increase the yields.
Hunsdiecker reaction
The Hunsdiecker reaction is a reaction whereby silver salts of carboxylic acids react with a halogen to produce an organic halide.

The reaction was first demonstrated by Alexander Borodin in 1861 by preparing methyl bromide ($\ce{CH3Br}$) from silver acetate ($\ce{CH3CO2Ag}$). Shortly after, the approach was applied to the degradation of fatty acids in the laboratory of Adolf Lieben. However, it is named for Cläre Hunsdiecker and her husband Heinz Hunsdiecker, whose work in the 1930s developed it into a general method and thus the reaction was renamed to Hunsdiecker reaction. The mechanism is given below:

Birnbaum-Simonini reaction
During the discovery of this reaction by Borodin, Angelo Simonini who was working as a student of Adolf Lieben at the University of Vienna, investigating the reactions of silver carboxylates with iodine. They found that the products formed are determined by the stoichiometry within the reaction mixture. Using a carboxylate-to-iodine ratio of 1:1 leads to an alkyl iodide product, in line with Borodin's findings and the modern understanding of the Hunsdiecker reaction. However, a 2:1 ratio favours the formation of an ester product that arises from decarboxylation of one carboxylate and coupling the resulting alkyl chain with the other. It was named Birnbaum-Simonini reaction.

The mechanism is given below:

Simonini reaction
Using a 3:2 ratio of reactants leads to the formation of a 1:1 mixture of both products. This reaction is called Simonini reaction.
$$\ce{3RCO2Ag  +  2I2 -> RI + RCO2R  + 2 CO2  + 3 AgI}$$

What's the matter with different solvents?

The reaction was initially done using carbon tetrachloride. But it was seen that when a different solvent was used, the yield varied slightly. It was seen that the yield of the latter reaction improved when lead tetraacetate was used. So, this solvent was used for this reaction. (See here)


Determine reaction rate from reaction time?

The reactant orders in iodine clock reactions generally turn out to be first order.  Your calculated orders are suggestive of first order dependences, but there is a systematic error evident in the data that possibly stems from the experimental design.  In addition, I don't believe that the manner in which you calculated the orders is correct.  The approach in this experiment is to use the method of initial rates.  Briefly, a differential rate law (which gives instantaneous rate as a function of concentrations) can be written as
rate $=k[\ce{A}]^n [\ce{B}]^m$ for a reaction $\ce{A + B -> products}$.  To find the order of reactant A, for example, one varies the initial concentration of A and keeps the concentration of B constant. This generates a pseudo-rate law in which the concentration for B is treated as a constant(k'):  rate = $kk'[\ce{A}]^n$.  This relation can be linearized as follows:

$$
\begin{align}
\log \text{rate} &= \log kk'[\ce{A}]^n\\
\log \text{rate} &= \log [\ce{A}]^n + \log kk'\\
\log \text{rate} &= n \log[\ce{A}] + \log kk' \tag{I}
\end{align}
$$

Trials are conducted in which [A] is varied and the initial rate is determined.  A plot of log rate vs. log[A] should be linear and its slope should equal n, the order of reactant A. If, on the other hand, we solve EQ I for n we get
$$
\begin{align}
n &= \frac{\log \text{rate} - \log kk'}{\log[\ce{A}]} \\
n &= \frac{\log \frac{\text{rate}}{kk'}}{\log[\ce{A}]}
\end{align}
$$

Thus calculating n as $\log \text{rate}/\log[\ce{A}]$ (your approach) is not fully correct because the pseudo-rate constant $kk'$ is not accounted for.  I plotted your $[\ce{HSO3-}]$ data in the form of $\log \text{rate}$ vs. $\log [\ce{HSO3-}]$ and obtained a value for n (the slope) of 0.84 with $r = 0.997$ (note: the rates must be treated as negative values because reactant is being consumed).  I advise you to attempt such a plot for the data where $\ce{[IO3-]}$ was varied.  Regarding the systematic error, you should think about the fact that you determined average rates experimentally, meaning reactant concentration was actually decreasing during the timed intervals and, therefore, the rates you calculated reflected an average reactant concentration, not an initial concentration.


Conversion of acetaldehyde to butane-1,4-diol

Cross Aldol and add H+

Oxidise Aldehyde to Carboxylic Acid

use any of the reagents

3.Perform Arndt–Eistert reaction

Use LiAlH4 for reduction of acid to alcohol

And finally you get 1,4-dihydroxybutane.

Image courtesy Wkipedia, SigmaAldrich


Equilibrium composition of methane–water system under given conditions

Let
$$moles\ CH_4=1-e_1$$
$$moles\ H_2O=1-(e_1+e_2)$$
$$moles\ CO=e_1-e_2$$
$$moles\ CO_2=e_2$$
$$moles\ H_2=3e_1+e_2$$
where the $e_1$ is the number of moles of $CH_4$ destroyed and $e_2$ the number of moles of $CO_2$ that are produced.
So the equilibrium equations are going to read:$$\frac{(e_1-e_2)(3e_1+e_2)^3}{(1-e_1-e_2)(1-e_1)(2+2e_1)^2}=K_1$$and $$\frac{e_2(3e_1+e_2)}{(e_1-e_2)(1-e_1-e_2)}=K_2$$


how to derive equilibrium concentrations from initial conditions for mass action reactions?

Here is the expression from the question:
$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} = K_{eq}$
Get rid of the fraction:
$(A_0 - [C]_{eq})(B_0 - [C]_{eq}) = K_{eq} {[C]_{eq}}$
Distribute the sums in the product:
$A_0 B_0 - [C]_{eq}(A_0 +  B_0)  + [C]_{eq}^2 = K_{eq} {[C]_{eq}}$
Sort the terms and you have your quadratic equation:
$A_0 B_0 - [C]_{eq}(A_0 +  B_0 - K_{eq})  + [C]_{eq}^2 =  0$


(Copied and edited from comments)

The central atom here refers to the center of attention (i.e. where the given peak assigned), the center of scattering which is often a center of an active site/coordination complex. I.e anything can be a central atom, if you tune your X-ray frequency to their peaks - you choose the central atom. In the given example theoretically you can choose to study C or F, and their closes neighbors. Theoretically you can see their neighbors; C-F distance is much shorter than F..F distance, though.

You should understand the use/limitations of the method: if you have chemically nonequivalent atoms of the same type in your sample, you will have their EXAFS peaks overlap. Both will give you a signal at the same place, which are hard to separate. In practice it means that you can use this analysis for e.g. central metal atom of complexes, but not for carbon atoms of a ligand as you generally have a lot of different carbon atoms in the sample.


Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?

If trans double bonds cause cycloalkenes to be chiral

Keep in mind that a trans double bond in a cycloalkene does not necessarily produce a chiral cycloalkene.  Only when the chain connecting the ends of the double bond is short enough so that the trans double bond is twisted will a chiral conformation exist.

So in large trans-cycloalkenes the double bond is not twisted and cannot produce chirality.

Remember too, that even if the trans double bond is twisted and a chiral conformation results, the molecule may still be optically inactive if the chiral conformation rapidly interconverts with its enantiomeric conformation. This latter point was discussed in one of your earlier questions where it was pointed out that

  At room temperature, trans-cyclooctene is the only trans-cycloalkene
  where the two enantiomeric conformations are stable enough (e.g. the
  barrier to reach the planar conformation is high enough) to be
  isolated.  If we lower the temperature to 0°C, then we can isolate the
  enatiomers of trans-cyclononene.

Back to your question.

  does a second trans or maybe cis double bond induce another chiral
  plane

The key here is that, as mentioned above, if the second double bond is also twisted, then there will be the potential for an increase in the number of chiral stereoisomers (of course inserting a double bond will always increase the number of stereoisomers because we have additional cis and trans double bond possibilities, but I think your question is focused on chiral stereoisomers, not geometric stereoisomers).

Look at the above figure on the left, of a trans double bond in a small ring.  We are sighting down a trans double bond. To alleviate ring strain, the double bond if forced to twist.  It is this twisting out of plane that infers chirality upon the double bond.  Its insertion into the ring will produce an additional chiral plane and the R or S chirality of the new double bond will  need to be specified in the name of the compound.

Note the figure on the right of a cis double bond in a small ring. To alleviate strain here, the $\ce{C_{ring}-C=C}$ angle is pinched down; there is no out of plane twisting motion, the plane of the double bond still exists.  Therefore, in the cis double bond case, there is no chirality associated with the double bond; no additional chiral isomers will be produced.

  and hence four stereoisomer are possible for the given cyclodiene?

Not always.  Just as in more common compounds with 2 chiral centers, meso stereoisomers are possible.  As an example, if the known E,E-cycloocta-1,5-diene
adopted a conformation as pictured below, then it has a plane of symmetry and would be an achiral meso compound where one trans double bond has the R configuration and the other double bond the S configuration.

(image source)


Calculating the number of ligands of complex ions in d- block

1st-row $d$-element + 2nd-row $p$-element - usually octahedron except for $\ce{Ni(II),Cu(II),Co(I)}$ (often a square or a distorted octahedron with two weakly bound ligands) and $\ce{Zn}$ (usually a tetrahedron). 1st-row $d$-element + 3+-row $p$-element - usually a tetrahedron.

2nd/3rd row $d$-element are somewhat more complicated as they allow higher coordination number up to 9 ($\ce{ReH9^{2-}}$), but the situation is so much influenced by electron configuration, ligands and metal-metal interactions here that simple rules loose a lot of meaning.

In short, the coordination number is usually a compromise between 18-electron rule and size of ligands. For example, $\ce{V(CO)6}$ should dimerize, giving c.n. of 7 to vanadium, but can't do so because of size constrains. Somewhat similar happens with bulky alcoholats of titanium, that tend to include $\ce{Ti(IV)}$ in octahedral coordination, but lower c.n. is observable in case of tritox (tris-tert-butyl) alcoholat.

Though it is generally not recommended to guess the c.n., it is always better to google for some reference compounds.


lowest energy conformation of butan-2,4-diol

I presume it is in the gauche form that its energy is minimun because of possible hydrogen bonding between the hydroxy groups


Please remember that nucleophilic substitution at a saturated carbon atom does not directly correlate with pKa.

The pKa of a substance (in the Bronsted-Lowry definition) relates to the substance's ability to have strong interactions with $\ce{H^+}$. Thus, this interaction is mainly electrostatic in nature; since $\ce{H^+}$ is a very small, weakly polarizable ion, according to hard-soft acid base theory, strong interactions with this ion are largely electrostatic in nature, i.e. proportional to charge density.

However, nucleophilic substitution at a saturated carbon atom occurs mainly through orbital interactions, not through electrostatic interactions. Namely, the highest occupied molecular orbital (HOMO) of the nucleophile interacts with the LUMO (sigma antibonding orbital) of the substrate (usually alkyl halide).

Since higher energy HOMOs interact better with LUMOs, molecules or ions with higher energy HOMOs are more nucleophilic.

With the specific case that you gave ($\ce{EtO-}$ and $\ce{EtS-}$), consider that sulfur is a larger atom, and the electrons in HOMO in $\ce{EtS-}$ are going to be less tighly held by nuclear charge than electrons in the HOMO in $\ce{EtO-}$. Alternatively, you can consider that the highest energy level orbital in S is 3p while that of oxygen is 2p --> electrons are more tighly held and lower in energy in $\ce{EtO-}$ --> $\ce{EtO-}$ is less nucleophilic.

Thus, if you consider the relative energy levels HOMOs in related molecules and ions, you get the general trend below:

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.


It describes the entire region to the right of the vertical dashed line through the critical point.


Does the mixture of a ligand exchange reaction contain all the complexes obtained in the part reactions?

Yes, all those complexes from all those reactions will be present, and some more. Essentially Loong did all the mathematics in his answer to: Why is ligand substitution only partial with copper(II) ions and ammonia?

Let's shorten this a little bit and write the general reaction equation:
\begin{align}
  \ce{[Cu(H2O)_{(7-n)}]^2+ + n NH3 &<=> [Cu(NH3)_n(H2O)_{(6-n)}]^2+ + n H2O};&
  n&\in\{\mathbb{N}; 1,\dots,6\}
\end{align}
Therefore we can write the general equilibrium constant:
\begin{align}
  K_n &=
    \frac{{\left[ \ce{[Cu(NH3)_n(H2O)_{(6-n)}]^2+} \right]}}
   {{\left[ \ce{[Cu(H2O)_{(7-n)}]^2+} \right]\left[ \ce{NH3} \right]^n}}\\
%  K_\mathrm{B} &=
%    \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}
%    {{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}}
%  &&= \prod_{n=1}^{6} K_n
\end{align}

Since all reactions are in equilibrium, all possible species will exist, although they might not be quantifiable. The overall equilibrium depends on the concentration of the involved species, especially ammonia.

If $[\ce{NH3}]$ is low, then a complex with smaller $n$ will be more stable, if you go to higher concentrations, larger $n$ will be preferred.

In all of this it should be mentioned that there are also complexes with vacant coordination sites $\square$, like anything in between
$$\ce{[Cu(H2O)5\square_1]^2+},\dots,
  \ce{[Cu(H2O)3(NH3)2\square_1]^2+},\dots,
  \ce{[Cu(NH3)5\square_1]^2+},\\
  \left(\ce{[Cu(H2O)4\square_2]^2+},\dots,\ce{[Cu(NH3)4\square_2]^2+}\right),$$
and possibly others.


Does molarity change if you decrease volume?

No, solute will not decrease if volume will decrease.But molarity will increase for sure, cause molarity is the amount of solute in litre of solution.

Consider container A in which you took 1 litre of water and added two spatulas of NaCl and in another container B you took 0.5 litre of water and added two spatulas of NaCl. Now which container has greater amount of solute , the second one that is container B ,hence molarity of container B solution will be greater than container A.

Hope it will work to clear concept.


Equivalent concept is an archaic unit of measurement that was used in chemistry and the biological sciences in the era before researchers knew how to determine the chemical formula for a compound.
An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of another substance in a given chemical reaction. The mass of an equivalent is called its equivalent weight.
In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

react with or supply one mole of hydrogen ions ($\ce{H+}$) in an acid-base reaction.
react with or supply one mole of electrons in a redox reaction.

I know that it is a bit difficult to understand by the means of only definitions, so let me give you some intuition about it.
The basic idea of equivalent concept is that,

$1$ eqiv of a substance reacts with exactly $1$ equiv of another substances to produce $1$ equiv of each product.

Let us start with the example of Acid-Base reaction, so for acids we define one equivalent as the amount of the acid (in moles) that supplies one mole of $\ce{H+}$ ions and for acid we define it as the amount of base (in moles) that supplies 1 mole of $\ce{OH-}$.
For Example:-

$1$ mole $\ce{HCl}$ supplies $1$ mole $\ce{H+}$ ions, so $1$ equiv of $\ce{HCl}$ is equal to $1$ mole $\ce{HCl}$.
Similarly, $1$ equiv of $\ce{H2SO4}$ = $1/2$ mole $\ce{H2SO4}$, $1$ Eq $\ce{NaOH= 1}$ mole $\ce{NaOH}$,
$1$ equiv $\ce{Al(OH)3}$ = $1/3$ mole $\ce{Al(OH)3}$.

Now if you look at the above closely, you will notice that since 1 equiv of acid supplies 1 mole $\ce{H+}$ ions and 1 equiv of base supplies 1 mole $\ce{OH-}$ ions so 1 equiv of an acid should completely react with 1 equiv of base to produce 1 equiv of each product ( eg. 1 mole $\ce{H2O}$, since its a neutralization reaction).
In a similar manner we can define equivalent for all other types of reactions, such that the main idea i.e. "1 eqiv of a substance reacts with exactly 1 equiv of another substances to produce 1 equiv of each product", remains unaltered.
Now lets define equivalent for other things,

Element
The equivalent of an element can be defined as the amount of the element that reacts with or supplies 1 mole electron in a reaction, so if you try a bit you can conclude that 1 equiv of an element is equal to 1/n mole of element where n is the valency of the element. (NOTE:- the equivalent of a substance can be different for different reactions, eg. 1 equiv of $\ce{Fe=}$ 1/2 mole $\ce{Fe}$ if it under goes to an oxidation state $\ce{Fe^2+}$ and it will be 1/3 mole if it undergoes to an oxidation state $\ce{Fe^3+}$).

Ions
Similar to element equivalent of ions can be defined as 1/n mole of ion where n is the charge on ion (try to understand it yourself using concept of electrons used earlier).

Salt
A salt have an equal number of cations and anions, so we need to define an equivalent in such a way that, 1 equiv of salt supplies 1 mole of cations (with +1 charge or 1/n moles of cation with +n charge) and anions (with -1 charge or 1/n moles of anion with -n charge) each so that these cations and anions react completely with another one mole of anions and cations (having -1 and +1 charge) that may have been supplied 1 equiv of another salt/ substance.
Now if you think a bit harder you can get to the point that one equiv of a salt is equal to 1/n mole of salt where n is the total number of charge either on cationic part or on anionic part (eg. in NaCl, total cationic/anionic charge = +1 so, 1 eq of NaCl= 1 mole NaCl, now if you pay your attention, you can notice that 1 mole NaCl will give 1 mole Na+ and 1 mole Cl- which is in accordance to what we said earlier.)
Some more examples for your help: 1 equiv of $\ce{AgCl2= 1/2 AgCl2}$ as it will supply 1/2 mole of $\ce{Ag^2+}$A and 1/2 mole of (Cl-)2, in simpler terms it will give one mole of positive charge and one mole of negative charge.
1 equiv of AlCl3= 1/3 moles of AlCl3.

Oxidizing agent/ Reducing agent
Now if you have come this far without dozing off, so you should be able to define that 1 equiv of an Oxidizing/ Reducing agent is equal to 1/n mole of that agent where n is the number of electrons gained or lost.

Now lets get to the question " Why is equivalent concept is used in chemistry for knowing concentration? ", well you might yourself have got the answer that it is because it is much simpler to apply as you do not need the Stoichiometry of the reaction going on, equivalent concept works fine without it.
And also to know the equivalent mass you simply need to find the mass of one equivalent of a substance using mole concept.
So that was all, I knew. Lets end this with a quote " Science is nothing but simple logic which is often not so simple."