Why do steel and iron not passivate the way aluminum does?

The main reason is nature of the oxide layer formed on the surface. Iron oxides are not remotely like aluminium oxide. They tend to be flaky and porous meaning that they offer no protection to the surface when formed. And iron oxides tend to be lower density and higher volume than the iron they are made from. Aluminium oxide is not at all porous and is very strong.

Bulk aluminium oxide is corundum, which is one of the strongest minerals (it is just about the hardest other than diamond). Corundum occurs as the gemstones sapphire and ruby when it has the right impurities and they are known for their hardness as well as their beauty. It is also unusually dense for a mineral made up of relatively light elements. So even a thin layer provides strong protection.

It is also worth noting that the aluminium oxide layer can be disrupted (e.g. by gallium or mercury), in which case aluminium does the equivalent of "rust", in some cases quite spectacularly.

It is also worth noting that some steels can be made to have a passivation layer which is why the are "stainless". This is achieved by alloying with metals such as chromium that facilitate the creation of a stronger oxide layer on the surface.


Why does phosphine have a dipole moment and a higher boiling point than carbon tetrafluoride?

The dipole moment of $PH_3$ can be attributed to the lone pair on P, which is directed away from all three P-H bonds. You can imagine that there is a higher electron density near the lone pair, while the electron density is more or less uniform near the nonpolar P-H bonds. This does not happen for $CF_4$, so it has no dipole moment.


I believe the first part of your question is asking why the oxygen with the unpaired electron doesn't find a second one to form a peroxide.

The answer is actually quite simple. It is by design. Next to the nitrogen, you see the 4 methyl groups that give TEMPO the "TEM" part of it's name. These methyl groups give TEMPO a lot of steric bulk around the oxygen, and this bulk makes it very difficult for two TEMPO radicals to form an oxygen-oxygen bond with each other. There is just not enough space. You can build a model to convince yourself of this.


How are octet-rule models supported by thermodynamics and quantum theory?

You can't really compare the energy of $\ce{Cl-}$ to $\ce{Cl.}$, since the two states have different composition. Instead, you're interested in the difference in energy between $\ce{Cl-}$ and $\ce{Cl. + e-}$, where the electron is completely free and resting in a vacuum with no interactions with anything else. We give that electron a reference potential energy of 0. Then the question becomes, how much potential energy is lost by interacting with various atoms, and the thermodynamically favorable state is the one with the most negative potential energy.

If we consider nuclei to be point positive charges and electrons to be negative charge distributed in a much wider space around the nucleus, we find that as that free electron gets closer to the nucleus, it experiences an electrostatic interaction with the positively charged nucleus even if the atom overall is neutral. We can simplistically estimate the magnitude of the potential energy loss based on basic principles of electrostatics, where potential energy is a function of $\frac{q^+q^-}{r}$.

Where molecular orbital theory comes in is when we calculate the charge and the distance r. Simplistically, orbitals that are higher in energy on the same atom are higher in energy because they have a longer average distance from the nucleus (denominator in PE term is bigger) and because there is more electron density from other electrons between them and the nucleus, so the "effective" positive charge of the nucleus is reduced (numerator in PE is smaller). This is called "shielding".

Slater's rules are simple heuristics that provide a fairly good approximation of the effective nuclear charge experienced by an electron based on the atomic number and orbital.

Conceptually, then, the answer to your question is that adding an electron to a neutral atom like $\ce{Cl.}$ is most favorable when the electron experiences the best combination of effective nuclear charge and short average distance to the nucleus. This happens when adding the last electron to an octet because that is the element of that period with the greatest nuclear charge. Adding an electron to the next neutral element Argon is much less favorable because it has to go in the next shell (n=4) meaning that it has a larger r (larger denominator) and has much more electron density between it and the nucleus, so the effective nuclear charge is less (smaller numerator) even though the atomic number of Ar is one more than Cl.

For $\ce{Na+}$, it isn't correct to say that the system has lower energy after an electron is removed, but rather that it takes a lot less energy to remove that electron than an electron from neutral neon, which hopefully makes sense based on the discussion above.


Conversion of an epoxide into an alcohol

Superhydride (lithium triethylborohydride) does this. JACS 95 (1973)  8486 reports complete selectivity for unsubstituted over disubstituted carbon with very high yield. HC Brown JOC 45 (1980) (1) reports complete selectivity for 2-butanol from 1,2 epoxybutane and 97:3 for the reduction of styrene oxide.


Why is pickle salty? How does salt get into pickle?

Your concept of a cell is a little vague. Although the cell wall is made from a diglyceride bilayer, it is not impregnable. In fact, there are many proteins that pass through the cell wall where dissolved material can pass. One of the proteins is called the sodium potassium pump. In a living cell, the protein uses ATP to move sodium out of the cell and drag potassium into the cell. Both of these ions are against the gradient. In a non-living cell, such as the pickle, sodium ions can move across the cell membrane and make the cells salty.


Why tertiary amines can't show chirality?

The short answer to your question

Under conditions that inhibit inversion, an amine that has three different groups attached is chiral.

The above applies not just to tertiary but also to secondary amines.

Now, why inhibit inversion? Why does it make a difference?
As mentioned by Zhe (and mentioned in Newton's answer) in the comments:

Tertiary amines can definitely be chiral. It's just that they epimerize quickly, but they are definitely chiral when there are 3 different substituents. In addition, you can easily create a tertiary amine that does not invert by tying back the substituents.

If you notice, they are chiral for very short periods of time due to the inversion that takes place in amines. In regards to the time scale – Ammonia ($\ce{NH3}$) flips $4 \times 10^{10}$ times every second. Converting that into the time for one inversion that translates into an inversion every $\pu{2.5 \times10^{-11} s}$ which is $\pu{25 picoseconds}$. Such a scale of time cannot be discerned by any of our current methods to identify stereochemistry.
From http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch07/ch7-3.html

In this case, even though a chirality center is present in each molecule, the sample is optically inactive since the optical activity of the two extremes of inversion
averages out because they are enantiomeric.

For why umbrella inversion occurs, see this answer by SendersReagent

Now, what if we restrict this inversion by attaching free-rotation restricting groups? What happens then?
This restriction brings (or rather removes) a fact into the mix. There is no more inversion possible.  This is because rearrangement into $\mathrm {sp}^2$ cannot take place because of the groups added since it becomes sterically inhibited in becoming planar. An example of this would be quinuclidine which has no way of forming a planar structure.


Does the boiling point of an aqueous solution depend on what the solute is?

No, in water the two solutions will not have the same boiling point.

Ethanol is a volatile liquid with boiling point lower than water, and lowers the boiling point.

Water forms an azeotrope with ethanol.  There are positive and negative azeotropes, so when two or more volatile liquids are dissolved, you cannot easily predict whether boiling point will decrease or increase.

There can also be cases, like adding HCl to water, where as solute is added, first the boiling point increases to a maximum (108 °C around 20% HCl) above the boiling point of water, then decrease to below the boiling point of water (48 °C for 38% HCl).

Glucose is non-volatile and increases the boiling point of water.  The boiling-point elevation formula can be applied to non-volatile solutes in volatile solvent.


How to prevent petrol from being flammable?

Nearly every solvent is flammable, some exceptions are halogenated hydrocarbons. But plain hydrocarbons are pretty much all flammable, though some are more dangerous than others like e.g. diethyl ether.

Chloroform is non-flammable, but due to the toxicity I would not recommend to use it. There are some other non-flammable solvents for cleaning mentioned in this presentation from NASA on the search for non-flammable solvents. But it seems that many candidates are being phased out because they are a hazard to the ozone layer.


There are 2 things to consider: What would be the energy difference between before and after rearrangement from initial $2^\circ$-carbocation (I) and each of rearranged carbocations II and III:

It is obvious that the rearrangement of carbocation I to give carbocation II by ring expansion have minimal energy change since both initial and final carbocations are $2^\circ$ and ring expansion does not relieve too much energy because it is cyclopentyl to cyclohexyl. However, the rearrangement of carbocation I to give carbocation III by 1,2-hydride transfer relieves good amount of energy since initial carbocation is $2^\circ$ and final rearranged carbocation is $3^\circ$. Although the carbocation II would undergo further stabilization by 1,2-hydride transfer to relieve extra energy, I to III happens much faster than I to II transfer to give 3 as the major product.


Derivation of the form of the time correlation function in NMR spectroscopy

Your function $G(\tau)= \langle B_x(t)B_x(t+\tau) \rangle$ is an autocorrelation function. If it is only a function of $\tau$, the time delay, and not t, this is called the stationary assumption.

The fluctuating field has  zero average $\langle B_x(t) \rangle =0$. The bracket implies averaging over a long time for a single spin or average over many spins at a particular time, which is the ergodic hypothesis.

The magnitude of the fluctuating fields is the mean square fluctuation $\langle B_x^2(t)\rangle \ne 0$ and is positive thus the mean square is not zero.

It is also necessary to know how rapidly the fields fluctuate and this is where the autocorrelation $G(\tau)= \langle B_x(t)B_x(t+\tau) \rangle \ne 0$ comes in. We compare the field at one point t with another point $t+\tau$ some time later. If $\tau$ is small then the value of $B_x(t)$ and $B_x (t+ \tau )$ are similar and $B_x(t)B_x(t+\tau)$ is large and positive, by comparison if $\tau$ is large the product $B_x(t)B_x(t+\tau)$ is small approaching zero as $\tau$ increases.

Thus the general shape is that $G(\tau)$ is large at small $\tau$ and decays away to zero with increase in $\tau$. Often we assume that the decay is exponential thus $$G(\tau)= \langle B_x^2 \rangle \mathrm{e}^{-|\tau|/\tau_\mathrm{c}}$$ where $\tau_\mathrm{c}$ is a correlation time. This has the correct qualitative form but is difficult to justify on the basis of proper theory.

You state that $\tau_\mathrm{c}$ is caused by rotational diffusion thus it will, in general, be small in fluid solution and large in viscous ones. Of course $\tau_\mathrm{c}$ also depends on the temperature (smaller at larger $T$) and size of the whole molecule's rotation (as a prolate or oblate ellipsoid) or of a  mobile group therein, depending upon which property you are observing. Groups in a protein can undergo 'wobbling in a cone' type motion in addition to rotation of the whole protein. Note also that if the molecule is not spherical then it will have more than one rotational relaxation time so $G(\tau)$ may decay with more than one $\tau_\mathrm{c}$. Hope that this helps!


Volatile and non-volatile solutes in solution

A substance with higher vapour pressure vaporizes more readily than a substance with a lower vapour pressure. In case of a volatile solute this has a high vapour pressure and hence produces vapour. While in the case of a non-volatile solute due to its lower vapour pressure it does not produce vapour. The vapor pressure of a substance is the pressure at which its gas phase is in equilibrium with its condensed phases (liquid or solid).


Nomenclature of organic molecules with more than one functional group

There are some priority rules about naming compounds with more than 1 functional group.

Aldehyde has higher priority than ketone (I don't know why but I think it is because of aldehyde always take place at the end of carbon chain)
Alkene has higher priority than alkyne, because -ene suffix comes before -yne

So the answer for your question is:

2-oxobutanal
3-oxobutanal
pent-1-en-4-yne


Are Daniell cells rechargeable (reversible)?

The Daniell cell is not rechargeable. According to this Wikipedia article:

  In the Daniell cell, the porous barrier cannot prevent the flow of
  copper ions into the zinc half-cell. Hence, recharging (reversing the
  current flow by an external source of EMF) is impossible because, if
  the zinc electrode is made to become the cathode, copper ions, rather
  than zinc ions, will be discharged on account of their lower
  potential.


When HF loses proton the conjugate base formed is F- Fluoride anion,Since fluorine is small and compact atom having lower affinity to keep electrons because of lack of d orbitals thus fluoride ion is unstable conjugate base compared to iodide ion which is stable.Happy conjugate base(stable), more reactive acid. Hence HI is more acidic than HF.

Even though fluorine is highly electronegative(tendency to drag shared electrons in a covalant bond),it has lower affinity to keep this electrons(lower electron affinity) due to its smaller size .


Number of hydrogen bonds of organic compounds?

Hint you could write down general formula for eg for alcohols it $C_nH_{2n}O$ and also  due to covalent bonds the hydrocarbons arent dissolved in water.


aldehydes are easily oxidised but ketones are not

In the oxidation of aldehyde, the hydrogen attached to carbonyl compound is replaced by an $\ce{-OH}$ group resulting in the formation of carboxylic compound. In case of ketones, bulkier alkyl groups ($\ce{-R}$) should be replaced by an $\ce{-OH}$ group to form carboxylic compound.

We know that hydrogen abstraction is much easier than that of bulkier alkyl groups. Likewise, after the oxidation of the aldehyde, the carbon chain containing the $\ce{C=O}$ group remains same but due to loss of $\ce{-R}$ group during oxidation, the length of the carbon chain decreases in ketones.


According to Palstics International's Chemical Resistance Chart, the A-rated plastics (no solvent attack) towards acetone are:

ECTFE (Halar®): transparent films available
Fluorosint® PTFE: white
HDPE: transparent films available
Nylon®, Type 6/6: white
PP: clear sheets available
PPS: opaque white
PTFE: white

Among these, polypropylene appears to be the only plastic that satisfies all criteria.


How to round significant figures correctly?

In rounding with significant figures, you want to avoid undue precision, yes, but it's also important to minimize error. It's this error minimization which is the primary goal of all "round to nearest" approaches.

So in short:

\begin{align}
0.383 - 0.3825525 &= 0.0004475\\
0.3825525 - 0.382 &= 0.0005525
\end{align}

$0.0004475$ is less than $0.0005525$, so you should round up to $0.383$, as that's the nearest value, and will minimize any error you introduce by rounding.

Even & Odd?

So what's this about the even/odd rule? Is there any validity there?

This technique is called "Banker's rounding" and is also attempting to minimize error.

When a number falls exactly in the middle of the two values to be rounded to (that is, when the digit to round is "5" exactly), each of the two difference is equally far away. So to a first order, both will result in an equal amount of error.

But picking a consistent direction (e.g. "always round up") introduces a subtle bias in the sums over the long run. This is because the errors are biased.

.     |  Always  |  Up   | Banker's | Banker's |
Value | Round Up | Error | Rounding | Error |
------+----------+-------+---------+------+
2.0   |     2    |  0.0  |    2    |  0.0  |
2.1   |     2    | -0.1  |    2    | -0.1  |
2.2   |     2    | -0.2  |    2    | -0.2  |
2.3   |     2    | -0.3  |    2    | -0.3  |
2.4   |     2    | -0.4  |    2    | -0.4  |
2.5   |     3    | +0.5  |    2    | -0.5  |   <<<<<
2.6   |     3    | +0.4  |    3    | +0.4  |
2.7   |     3    | +0.3  |    3    | +0.3  |
2.8   |     3    | +0.2  |    3    | +0.2  |
2.9   |     3    | +0.1  |    3    | +0.1  |
3.0   |     3    |  0.0  |    3    |  0.0  |
3.1   |     3    | -0.1  |    3    | -0.1  |
3.2   |     3    | -0.2  |    3    | -0.2  |
3.3   |     3    | -0.3  |    3    | -0.3  |
3.4   |     3    | -0.4  |    3    | -0.4  |
3.5   |     4    | +0.5  |    4    | +0.5  |   <<<<<
3.6   |     4    | +0.4  |    4    | +0.4  |
3.7   |     4    | +0.3  |    4    | +0.3  |
3.8   |     4    | +0.2  |    4    | +0.2  |
3.9   |     4    | +0.1  |    4    | +0.1  |
4.0   |     4    |  0.0  |    4    |  0.0  |
------+----------+-------+---------+-------+
63.0  |    64    | +1.0  |    63   |  0.0  |   - Totals

With "always round up" you get a consistent $+0.5$ bias per decade, and this is consistent even if you extend the range to more decimal places. You can balance the errors for $0.00...01$ to $0.49...99$ with those for $0.99...99$ to $0.50...01$, but there's that dead center $0.50...00$ (aka $0.5$) which is the odd man out - consistent rounding up (or down) will introduce a bias because of this entry.

So what are you to do? Well, introduce a rule which rounds that entry up half the time and rounds down the other half. Bankers' rounding is the most common such rule. This rule says that if the digits to be rounded off are $5$ exactly, then round such that the last digit of the number after rounding is even. (Note that this is equivalent to a rule to round up if the digit to the left of the $5$ is odd, but round down if the digit to the left is even - two different ways of saying the same thing.)

0.255 -> 0.26 ... because 6 is even (or round up because 5 is odd)
0.265 -> 0.26 ... because 6 is even (or round down because 6 is even)
0.275 -> 0.28 ... because 8 is even (or round up because 7 is odd)
0.285 -> 0.28 ... because 8 is even (or round down because 8 is even)

But again, this is the rule only if the number to be rounded falls exactly between the two possibilities. If the number is closer to one or the other, then round to the nearest.

0.2453   -> 0.25 (Round to nearest)
0.2553   -> 0.26 (Round to nearest)
0.2651   -> 0.27 (Round to nearest)
0.275000 -> 0.28 (Falls exactly between 0.27 and 0.28, apply even/odd rules)
0.275378 -> 0.28 (Round to nearest)
0.285000 -> 0.28 (Falls exactly between 0.28 and 0.29, apply even/odd rules)
0.285023 -> 0.29 (Round to nearest)

Note also that you should only round once, and don't cascade your rounding:

0.14555 -> 0.1    (Round to nearest)
0.14555 -> 0.1456 -> 0.146 -> 0.15 -> 0.2  (WRONG!)


The dissociation of hydrogen on UO2 and PuO2 surfaces: Homolytic or heterolytic?

This reference$\ce{^{[1]}}$ shows what can happen even in a single oxide, in this case magnesia, with different metal environments.

Magnesia is, of course, strongly ionic, so the dissociation of hydrogen to form $\ce{O^{2-}-H^+}$ and $\ce{Mg^{2+}-H^-}$ on the surface might well be expected.  Putting the magnesia as a thin film on silver does not change this.  But on gold the hydrogen dissociates in a redox fashion, forming only $\ce{O^{2-}-H^+}$ as the electrons are injected into the gold.  The lowered Fermi energy of the gold causes this transfer as the electrons "fall" into the lower energy state.  It's an example of gold acting as an electronegative element, as it does in compounds such as $\ce{CsAu}$, ultimately caused by the effects of relativistic mechanics on the electron structure of gold.

In the case of bulk uranium and plutonium dioxides, electron donation or withdrawal must come from the oxide itself.  With uranium (IV), the 5f orbitals are at relatively high energy, and where they have electrons those electrons are readily available for bonding and for donation -- but conversely, empty 5f orbitals are relatively poor electron acceptors.  So we get metal hydride formation like the $\ce{Mg^{2+}-H^-}$ bonds in plain magnesia or magnesia/silver.

With plutonium in the same +4 oxidation state, enough increased effective charge and orbital contraction have occurred to make the plutonium 5f subshell a better electron acceptor, making the $\ce{O^{2-}-H^{+}}$ bonding mode more favorable.

References:

Hsin-Yi Tiffany Chen, Livia Giordano, and Gianfranco Pacchioni, "From Heterolytic to Homolytic H2 Dissociation on Nanostructured MgO(001) Films As a Function of the Metal Support", J. Phys. Chem. C, 2013, 117, 20, 10623–10629.

2.
Chemical Forms of Uranium". Argonne National Laboratory. Archived from the original on 22 September 2006. Retrieved 18 February 2007.  Link


Why is the boiling point of sugarcane juice lower than the boiling point of water?

A more plausible explanation is that it is not water that is boiling.  The  "organic substances" and maybe the "nitrogenous substances" in the juice could include some more volatile components that may pass selectively into the gas phase at a lower temperature than most of the solution.  Such a vapor would not be the pure organic or nitrogenous substance; it would carry off some water vapor as the vapor pressure of the water is a large fraction of an atmosphere.  The temperature at which such selective vapor begins to form is properly called the bubble point.
Of some relevance is the effect of the sugar itself on the boiling point.  In the absence of other solutes this effect is very small, cfailing to reach $5°C$ boiling point elevation until we have nearly 80% sucrose in the solution (by mass).  Sucrose has a large molar mass smd water has a high heat if vaporization, both of which factors attenuate the boiling point rise at concentrations relevant to sugar cane juice.  This lack of an effect from the main nonvolatile solute would enable small amounts of more volatile components to yield a bubble point lower than the boiling point of pure water.

From Ref. [1]
Reference
1.
Willism Heffner and Hinanshi Jin, "Building a Low Cost, Hands-on Learning Curriculum on Glass Science and Engineering using Candy Glass", MRS Online Proceeding Library Archive (January 2009) 1233,
https://www.researchgate.net/deref/http%3A%2F%2Fdx.doi.org%2F10.1557%2FPROC-1233-PP03-06.


How to know the formula of different compounds?

In my experience if someone says something like tin chloride they are usually insinuating you know the most common oxidation state that metal is found in when complexed with the chloride atoms. This is a bad practice of them if they haven't explained further but I digress... In this example with the metal center of the complex being tin, the most common oxidation state is +2. This means the formula in its neutral state would be $\ce{Sn^{+2}}$ with 2 chlorides with a (-1) charge making the molecule neutral at $\ce{Sn(II)Cl2}$.

Different effects cause  common oxidation states for specific elements. These are outlined further with Mendeleev's Periodic Law. It breaks down into how shielded the nucleus is by electrons, and how big the radius of the atom is.

A further explanation of tin's common oxidation state calls upon knowledge of the inert pair effect, where elements to the right of the second and third row transition metals such as: In, Tl, Sn, Pb, Sb, Bi form compounds that are less reactive than one would think where the oxidation state is 2 less than the expected group. This is caused by the $\mathrm{ns^2}$ electrons being relatively inert. The cause of that nonreactive state is because of two reasons specifically:

Larger-than-normal effective
nuclear charges in these elements.
Lower bond energies in their compounds.

A good book to use referencing this subject further is Descriptive Inorganic, Coordination, and Solid-State Chemistry by Glen E. Rodgers. This is the book I own and used for when I studied periodic trends, and inorganic chemistry as an undergraduate. It's a great book for this subject, among others.

Also there are commonplace names that are used as somewhat of a slang for certain compounds used in research frequently. These can be learned through experience, or you can just google common names... All formal names are dictated by the rules set forth by the IUPAC. Learning those rules would be good idea. Also keep in mind that in some research projects you will encounter molecules that have informal nicknames too.


What is the reason for the blue color of an ammoniated electron?

Both of your questions could benefit from more background, so please forgive me if I understood something wrong. However, I'll try to answer as best I can:

The excess electron in liquid ammonia is not really independent, otherwise, as you stated correctly, the electron would not absorb a certain wavelength but act as a black radiator/absorber like in a plasma (for example the sun). However, how the electron is bound and how the transitional states are formed is not completely understood yet. In this paper, I found the following sentence: "The excess electron in liquid ammonia ("ammoniated electron") is commonly viewed as a cavity electron in which the s-type wave function fills the interstitial void between 6 and 9 ammonia molecules." And then they go on to present a different hypothesis...
In a flame, the spectrum is recorded in the gas phase and formed not by the salt but by the individual atoms (even if you put the salt into the flame). Thus the spectrum depends only on the elements contained in the salt, not on the type of salt. This is different, if the spectroscopy is performed in the liquid phase of alkali metal halides. There the spectrum depends on the mole ratio of the anions and cations.


How are the vapors transported?

The additional question from the comments:

  Why [is] the amt. of vapours taken up proportional to vapour press. Why directly proportional? Why not nothing else?

First things first, let's get a definition (Wikipedia, italics added):

  Vapor pressure [...] is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

So basically, before you start the experiment (see below) the vapors have already formed, meaning that thermodynamic equilibrium has been achieved. Then you start bubbling through dry gas, which then mixes with the vapor gas already in the bubbles. If you set the gas flow appropriately, you get a near-equilibrium situation (see the graphic below) and the gas that exits the bulb assemply has exactly the same amount of vapor in it as the vapor "layer" (which can not be thought of as a layer anymore because of the bubbling) above the liquid surface.

We do not have an equilibrium situation, which is why so much care has to be taken that we get as close as possible to the equilibrium situation. (Also, we do not have a closed system, so everything starts to get slightly problematic at this point. But to a first degree of approximation, this will hold up fine.)

This is also why it is directly proportional to the vapor pressure: There is no other mechanism that moves solvent molecules from the condensed phase to the vapor phase. Of course you could add a supersonic vibrating membrane and force more solvent molecules into a vapor phase, but that is not what you want to do, because you want to quantify the relative lowering of the vapor pressure.

Is relative vapor pressure lowering a thing?

  Why "The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution"?

I guess the question you really want to ask is: Why is the vapor pressure of a solution lower than the vapor pressure of a pure substance?

The answer is not simple, but the explanation of the vapor pressure lowering effect can be given by assuming a simple model: Imagine that in the solution, there are two types of molecules; the solvent and the solute. In the pure solvent, there is only one kind. Statistically speaking, that means that there is also a smaller amount of solvent molecules at the boundary between the bulk phase and the vapor. Since solute molecules don't go over into the vapor, there is a smaller area of solvent molecules that can go over into the vapor, which means that the vapor pressure of the solution is smaller when compared to the pure solvent.

What about the proportionality constant?

  How can we add $\propto p_s$ and $\propto p_0−p_s$ when we don't know what the proportionality constants are or even if they are equal?

While we don't know the magnitude of the proportionality constants, we know that they are equal. The proportionality constant is most likely dependent on ambient pressure, temperature and flow rate of the gas. These three variables are being controlled during the experiment and kept constant for all glass bulbs. (Well, I don't know about keeping ambient pressure constant. But if you perform the experiment fast enough, that shouldn't pose much of a problem.)

The Ostwald-Walker Process

  It would be better if you could explain in detail the process because I'm not getting it properly from here.

Why do we do this?

First, we have to remember: Why are we doing this process; what do we want to find out? The answer is: We want to quantify the relative lowering of the vapor pressure.

How do we do this?

Here's what happens during the process:

Dry gas arrives at the first set of bulbs, filled with a solution.
The dry gas gets saturated with solvent vapor.
The solvent-vapor-saturated gas arrives at the second set of bulbs which contain the pure solvent.
Because of the vapor pressure lowering effect (here now applied in reverse) the vapor pressure of the pure solvent is higher. As such, the gas that enters is not saturated. When it exits the second set of bulbs, it is saturated again.
The pure-solvent-saturated gas enters the desiccator and loses all solvent vapor to it.
Dry gas exits the apparatus.

This might still be hard to understand, so here a graphic that shows you what happens to the solvent vapor partial pressure in the gas:

At time 1 the gas enters the first bulb. It exits the first cluster of bulbs at about time 4. Then it travels to the next cluster of bulbs (containing the pure solvent) and loads up there, up until time 8. At time 9 the gas enters the desiccator where it loses all the solvent vapor and exits dry at time 12.

Since we weigh all components, we know exactly how much mass we lost from solution and solvent, and how much mass the desiccant gained (both those numbers [the total of the mass loss of course] should be equal):
$$ \Delta m_\text{solution} + \Delta m_\text{solvent} = \Delta m_\text{desicc.} $$

The following proportionalities hold:

$ \Delta m_\text{solution} \propto p_s $
$ \Delta m_\text{solvent} \propto (p - p_s) $
$ \Delta m_\text{desicc.} \propto p $

The relative lowering of the vapor pressure can now be calculated, it is simply
$$ \frac{p-p_s}{p} = 1 - \frac{p_s}{p}$$

If you know the vapor pressure of the pure solvent, you can simply plug it in the equation, get the proportionality constant and from that calculate the vapor pressure of the solution.

Conclusion

In conclusion, I can only say that I hope to have clarified things for you. I myself spent quite some time researching this method, as I have never heard of this question before. Thank you for the opportunity!


Why isn't it commercially suitable to dissolve sulfur trioxide in water to produce sulfuric acid?

Part of the difficultly with this question is that it is asking about what is commercially viable as opposed to what is merely possible. It is, of course, quite possible to dissolve SO$_3$ in water, but doing so creates a runaway chemical reaction. So let's go through each of the points and explain why they don't make commercial sense.

(A) The contact process produces SO$_3$, as well as several other noxious gasses anyway, so they is no reduced risk as regards spillage.

(B) The density of the product is totally irrelevant, as you get the same product in both cases.

(C) Sulphur is a solid at standard temperature and pressure (STP), so one can transport a large amount using a standard container and transport vehicle. As a gas at STP, one can transport far less (many millions of moles less) SO$_3$ at STP for the same size container, so an unfeasibly large container would be needed unless a highly pressurized or very low temperature container is used. Either method is considerably more expensive, so less commercially viable.

(D) The last step of the contact process is to react oleum (H$_2$S$_2$O$_7$) with water. This is in fact relatively safe to do if done slowly and water is extremely cheap, so the process is not expensive

(E) The reaction of oleum and water is exothermic. However, in practice water is added relatively slowly to the oleum (both of which are controllable liquids), which controls the process.

(F) As said earlier, the reaction of water with SO$_3$ gas cannot be easily controlled. It runs away with itself, and the water starts to boil. This throws up lots of small droplets (an aerosol) of conc. sulfuric acid. This is highly dangerous, can increase the pressure in the reaction chamber, and also makes it difficult to cheaply extract the product from the reaction chamber.

Therefore, on economic grounds, the answer is indeed C and F.


According to the Oxford dictionary, a machine is:

  An apparatus using mechanical power and having several parts, each with a definite function and together performing a particular task.

Based on that definition, I think some proteins and enzymes could definitely be considered machines. A good example for that, in my opinion, would be ATP Synthase and all of the proteins that participate in the oxidative phosphorilation process.


You should investigate several pH and study the effect of time on the fluorescence spectra. One cannot quote single values without mentioning the wavelengths of excitation and emission.
pH can alter the state of  ionization and it can also cause hydrolysis of the sugar unit.
You need to collect three different types of spectra at each pH, say from pH 2 to 12.
At each pH, first measure the absorption spectrum at those pH values. This is the easiest experiment. Do you see a difference in absorption spectra at different pH values? Note the trends.  If yes, pH is affecting either the structure (decomposing) or the state of ionization.
Next, collect the excitation spectrum at a fixed emission wavelength (you have to estimate that). The excitation spectra should look similar to the absorption spectrum in shape.
Once you know the correct excitation wavelength at each pH, collect proper emission spectrum by using the lamba max of excitation.
So the answer is that be ready to see the changes in fluorescence spectra since there is no "the" fluorescence spectrum of a given molecule in a solution. You have to specify the solvent, pH, and excitation wavelength for experimental measurements.


What is the temperature needed for a reflux using a mix of solvents as the reaction medium?

Trial and error, but most often the boiling point of the lower-boiling solvent.
The absolutely correct answer would be to look at a phase diagram, determine whether you have a positive or negative azeotope (or non at all), find out where on the phase diagram you are and then derive the temperature. That would tell you approximately at what temperature your mixture will boil – after you factored in boiling point increase due to dissolved compounds.
In practice, you need to set your oil bath to some 5 or 10 degrees higher than the boiling point of the lower-boiling solvent anyway. So you can just ballpark it, selecting a temperature that would keep the lower-boiling solvent refluxing. Come back after 15 minutes, if it is not refluxing, increase by 5 or 10 degrees.


Attack on cyclic intermediate during NGP

Inductive reasoning in organic chemistry can be somewhat hard at times, given all of the exceptions that can pop up, for example the SN2-like mechanism of opening an epoxide in basic conditions versus the acidic conditions.

For the epoxide opening in base; as you know sterics wins, and in acid the stability of the reactive transition state is our reasoning for the product observed. I believe for the NGP reactivity that stabilization of transition states is an acceptable platform to reason the important question as to exactly how this reaction happens in the first place.

lets look at the NGP example you've pulled from Wikipedia
(Awesome btw! reading organic chem on wiki for fun?!)

I have a hard time fully following your question but here's the best answer I could come up with:

I see a symmetrical transition state where the reactive-intermediate charged cyclopropyl-S-heterocycle can probably just be attacked from either side in a SN2 like fashion. If you prefer charge polarization to be our main reasoning in the S-hetero cycle, given that it is a symmetrical substrate I believe the same product will form either way due to symmetry.

Did you read the part of that wiki article below that depicts NGP reactivity by an Alkene? The driving factor there appears to be pi-stabilization of the carbocation in the transition state.

This is a physical organic chemistry question that is not trivial. I applaud your curiosity.


Water always contains a certain amount of $\ce{OH-}$ and $\ce{H3O+}$ ions. The pH is the negative decadic logarithm of the $\ce{H3O+}$ concentration (pH = - log(c($\ce{H3O+}$)). As chemical equilibria might change with temperature so does the concentration of $\ce{H3O+}$ and therefore the pH.

Although I never tried, I suppose this could affect measurements using pH paper. On the other hand the pH changes might very well be to small to be detectable using this method.

How this affects other fluids / equilibria strongly depends on how each equilibrium is affected by temperature. At higher temperatures equilibria are shifted in the endothermic direction and at lower temperatures they are shifted in the exothermic direction. You will need to look up the details for each equilibrium I suppose.


Why is iso-octane different from other iso-s?

The usage of non-standard prefixes such as "iso" has never been defined with hard, unambiguous rules such as those in the IUPAC recommendations. Instead, their usage has often more been "by convention". In Organic Chemistry (1st ed.) by Clayden et al., the authors write (p 315):

  Strictly speaking, if the standard meaning of ‘iso’ is followed, the name isooctane should be reserved for the isomer 2-methylheptane. However, 2,2,4-trimethylpentane is by far the most important isomer of octane and so, historically, it has been assigned this name.

(This phrase is also found at Wikipedia, where it is copied wholesale without any indication that it's a direct quote. Please don't do this in your work. Even if you provide a citation, you should either paraphrase the information, or clearly indicate that it's a quote, e.g. by enclosing it in quotation marks.)


How to calculate mass produced when a given gas is burned?

You have already balanced the equation, so we can get the stoichiometric  coefficients. From the equation we can say that 1 mole $\ce{CH4}$ gives 2 moles of $\ce{H2O}$ because their coefficients are 1 and 2, correspondigly. So 2.5 moles of $\ce{CH4}$ would give 5 moles of $\ce{H2O}$, that is $\pu{90 g}$ of $\ce{H2O}$.


Your data is not enough to calculate molecular weight. You may have missed the mass of the solid acid you have weighed.
Let's assume you have weighed $a$ gram of acid. Then:
$Molecular Weight = \frac{Weight}{Moles_{acid}} = \frac{Weight}{2 \times Moles_{NaOH}} =  \frac{Weight}{2 \times Molarity_{NaOH} \times Volume_{NaOH}} = \frac{a}{2 \times 0.120 \times \frac{22.0}{1000}}$


Nucleophilic attack at a carbonyl - angle of attack

Let's take the following approach to answer your interesting question:

Use the bent bond model to describe the carbonyl
Calculate the bond angles around the carbonyl carbon
Treat the reaction as an $\mathrm{SN2}$ reaction
Use trigonometry to deduce the angle of nucleophilic approach

The bent-bond model is widely used in chemistry to describe highly-strained systems such as cyclopropane and double bonds (ref_1, ref_2).  In the case of a double bond, we wind up with a two-membered ring (which nicely explains the high heat of formation of such compounds). Here we will use it to describe the carbonyl double bond.  Here is a picture of what the bent bond description of a carbonyl looks like (it's the top left diagram that's of interest, note too that it is a picture of an amide carbonyl).

(image source)

In a simple carbonyl compound like formaldehyde the $\ce{H-C-H}$ bond angle is around 118°.  Using Coulson's theorem (ref_2, ref_4) we can use this bond angle to estimate the hybridization of the carbonyl $\ce{C-H}$ bonds.

$$\ce{1+\lambda^2cos(118)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the $\ce{C-H}$ bond. From this equation we find that $\ce{\lambda^2=}$ $\mathrm{2.13}$, or the carbonyl $\ce{C-H}$ bonds are $\ce{sp^{2.13}}$ hybridized.

We can now sum up the s-character in the 4 bonds around the carbonyl carbon (it must add up to 1) and deduce the hybridization of the 2 $\ce{C-O}$ bent bonds.

$$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-O}^2}=1}$$

Solving this equation we find that $\ce{C-O}$ bent bonds are $\ce{sp^{4.54}}$ hybridized.

Substituting this information back into Coulson's theorem allows us to estimate the $\ce{O-C-O}$ bond angle as being equal to 102.7°.

In an $\mathrm{SN2}$ reaction the nucleophile approaches 180° from the leaving group ("backside attack").  In this case, one of the carbonyl $\ce{C-O}$ bonds is the leaving group.

Using simple trigonometry we can estimate what the angle is for 180° attack of the $\ce{C-O}$ bond relative to the plane of the carbon and oxygen nuclei. Half of the $\ce{O-C-O}$ bond angle is 51.4°, leaving 38.6° to reach 90°. So the angle for backside attack relative to the plane containing the $\ce{C}$ and $\ce{O}$ atoms is $\mathrm{90°+38.6°=128.6°}$

The actual angle for attack has been found to be around 107° (see @Philipp 's comment on the Bürgi-Dunitz angle).  This angle was first determined by extensive, detailed crystallographic investigations and later supported by computational studies. The approach used here is no more than a quick, "back of the envelope" method. Yet it still provides an answer within about 20° of Bürgi and Dunitz's work. For such a quick, crude method, I would consider this result to be in qualitative agreement. It supports the back side attack $\mathrm{SN2}$ model, illustrates how a $\ce{C-O}$ bond participates as a leaving group and how we wind up with a tetrahedral carbon.


The lone pair at the nitrogen is very well suited to form hydrogen bonds. This bond is especially strong, since ammonia has a quite high proton affinity, it is a medium strong base.

Now consider the autoprotolysis reactions in pure water and (liquid) ammonia:
\begin{align}
\ce{2H2O &<=> H3O+ + OH-}\tag1\\
\ce{2NH3 &<=> NH4+ + NH2-}\tag2
\end{align}

Mixing these systems, the reaction of ammonia with water will yield
\begin{align}
      && \ce{2H2O + 2NH3 &<=> H3O+ + OH- + NH4+ + NH2-}\tag3 \\
& (~-~)& \ce{H3O+ + NH2- &-> H2O + NH3}\\\hline
      && \ce{H2O + NH3 &<=> NH4+ + OH-}\tag4\label{eq:ammonia+water}
\end{align}
\eqref{eq:ammonia+water} is incorporated into a huge and strong network of hydrogen bonds. In this case the equilibrium will be very much pushed to the right side, meaning, that almost all ammonia molecules will be protonated.

Therefore it is more appropriate to write:
\begin{align}
\ce{NH4OH (aq) &<=> NH4+ (aq) + OH- (aq)}\tag{4a}
\end{align}

However, it is always advisable to look at the conditions of the whole reaction system.

Note that $\ce{NH4OH}$ is a pretty outdated notation, if you would want to consider one water entity explicitly, $\ce{NH3.H2O}$ would be more appropriate. See the answer by Maurice for some more information: Can acid-base reactions occur in a non-aqueous medium?


According to what criteria was the periodic table of elements ordered?

I do not address the question of history, of Mendeleyev's investigations, that is answered in How was Mendeleev able to develop his table?

Modern understanding of basic principles follows. Let us have a piece of periodic table

  ... A0 B0 ...
  ... A1 B1 ...
  .............

Element "B0" must have the next atomic number after "A0", i.e. one more electron. It is different chemically, although may have (albeit not necessarily) similar physical properties.

Element "A1" must be chemically similar to "A0", although physical properties may differ significantly (as a general rule, ".1" elements are denser than ".0"). It happens that for each element there exist its "analog" in the next period, that has similar chemical properties (based on similar outer-shell electron configuration). This pattern, once noticed, permitted for good predictions (search for Mendeleyev's "eka" elements in Internet for more information).

Periods have different lengths. Analog's atomic number may be 8, 18, or 32 ahead off the prototype's atomic number (the number of electrons expended to "raise" all the configuration one shell up). This difference is due to quantummechanical effects.


Disclaimer: hybridization is a mathematical concept and fails at higher levels.  As per the linked answer, it is clear that the participation of $d$-orbitals in phosphate ion is minimum. But this is just a high-school level, layman's explanation for your question.

  So, could anyone please tell me where my understanding is flawed?

Your reasoning is flawed in assuming that carbon atom has four unpaired electrons in its orbitals after hybridisaton, while phosphorus does not. Actually, both of them do.

In fact, carbon atom in its ground state does have a lone pair of electrons in its $\mathrm{2s}$ orbital. It is only when it gets into its excited state, that that lone pair is broken up, yielding four unpaired electrons. The excited carbon atom now undergoes hybridization as depicted in the figure below, yielding four equivalent covalent bonds as in carbon tetrachloride.

(source)

As you noted, phosphorus also has a lone pair of electrons in its $\mathrm{3s}$ orbital. But, that is its ground state electronic configuration. Once it gets excited, it will have five unpaired electrons, the fifth electron going into its $\mathrm{3d}$ orbital. Now, phosphorus will undergo $\mathrm{sp^3}$ hybridization, to form $4$ sigma bonds. The fifth electron helps form the $d\pi-p\pi$ bond with the neutral oxygen atom.

I hope this helps!


How to study (by heart) all the equations in inorganic chemistry?

In organic chemistry, everything is based around C, H, O and N. Therefore to understand reactivity (somehow), there is just limited number of options.

In contrast, inorganic chemistry deals with the rest of periodic table, and each of the elements has its special properties. But there are general trends as well, but are more difficult to decipher and are often expressed by set of typical reactions, you should learn.

Usually they concern the stability of (oxidation) states, how to go from less stable to more stable one to obtain the desired compound and few processes how tho achieve the least stable state from the one found in nature.

It is difficult to state some simple rules, but there are several most typical anions and cations, and the chemistry of given element is just interactions with them. During the learning it helped me to have "all" the reactions (several hundreds) on a paper and for each one try to understand what happens and if there are others in the set, which are similar, and why. This helps you to spot the similarities. And - it is better to have more reactions then less, because then you have higher probability to spot the trends.


Exact meaning of "degree of crosslinking" in polymer chemistry

Normally, in practice the amount fraction of DVB in the feed is used as a nominal number for the degree of crosslinking. IUPAC provides a definition in a somewhat unexpected place [1, p. 2352]:

Crosslinking: Property of a solid support prepared from polymeric materials with interconnected strands. Often results from the inclusion of multifunctional monomers in the polymerization reaction, e.g. divinylbenzene in polystyrene production. In such cases, the degree of crosslinking is often quoted as the proportion of the multifunctional monomer in the reaction mixture. The extent of crosslinking is important for physical properties of the solid support, such as the propensity to swell in different solvents [18].

Following reference [18], one can find further clarification [2, p. 2278]:

Care is therefore required in defining or interpreting the DVB content of crosslinked polymers since this might be quoted as a percentage of technical DVB used to make the
polymer, or the figure can be adjusted to reflect only the content of actual DVB isomers present. Since for convenience the nominal crosslink ratio or degree of crosslinking of a polymer network is often quoted as the mol% of crosslinker used to prepare the network, defining the actual percentage of DVB isomers employed is more informative since this equates with the nominal crosslink ratio (Fig. 4). Bearing in mind that there is no unambiguous method, and certainly no simple and rapid method, for determining the real crosslink ratio in a polymer network, the nominal figure based on the actual DVB feed is a
very useful parameter.

Fig. 4 Polymerisation of styrene and divinylbenzene to form an infinite polymer network.

Further, the paper lists several reasons as to why the real level of crosslinking is always accompanied with a level of uncertainty:

The effective crosslink ratio significantly decreases with greater quantity of DVB being fed due to growing number of unreacted vinyl groups.
Entanglement crosslinking occurs at higher rates of polymerization in slowly-agitated systems and results in additional mobile crosslinking.

Fig. 5 Permanent entanglement crosslink

Additional crosslinks can also be created as the result of intramolecular side-reactions.

These factors are tricky to estimate quantitatively; a method allowing to do so would be a single pulse excitation (SPE) 13C solid state NMR.
References

Maclean, D.; Baldwin, J. J.; Ivanov, V. T.; Kato, Y.; Shaw, A.; Schneider, P.; Gordon, E. M. Glossary of Terms Used in Combinatorial Chemistry. Pure and Applied Chemistry 1999, 71 (12), 2349–2365. DOI: 10.1351/pac199971122349.
Sherrington, D. C. Preparation, Structure and Morphology of Polymer Supports. Chem. Commun. 1998, No. 21, 2275–2286. DOI: 10.1039/a803757d.


Gibbs of vaporization of a substance at a non-standard temperature

Alright, this is kind of fast but after thinking about it I think the issue with the problem is the definition of $\ce{\Delta S}$. My second try to compute this value involves starting with the form:

$\ce{dS = \frac{C_{p,g}-C_{p,l}}{T}*dT}$

Following integration, I arrive at the form:

$\ce{\Delta S = (C_{p,g}-C_{p,l})ln\frac{T}{T_b}+\frac{\Delta H}{T}}$ where $\ce{\Delta H}$ is determined in the way as in the question above.

Placing everything into the form $\ce{\Delta G = \Delta H-T\Delta S}$:

$\ce{\Delta G = \Delta H_{vap} +(T-T_b)(C_{p,g}-C_{p,l}) - T((C_{p,g}-C_{p,l})*ln\frac{T}{T_b}+\frac{\Delta H}{T})}$ where $\ce{\Delta H}$ is the non-standard enthalpy of vaporization.


Improving efficiency of absorption refrigerator (ammonia+hydrogen)

I've come here for basically the same reason, however I have some suggestions that I've done and I believe they have helped.
And my fridge is basically the same as your, just a bigger model, same manufacturer.
Firstly I should suggest messing with the process (heat etc) would not be a good idea as I would assume the manufacturer would have optimized the design years ago, and I have heard over heating the boiler can cause catastrophic and irrepairable damage .
My fridge has a freezer box (shoebox size, quite useless) within the fridge and a heat sink next to it for the fridge part. I took the door off the freezer and cut the base out of the box, that made a huge difference, infact with it set on high on a cool day of 20c or less items in the bottom of the fridge will freeze.
But as the camping car is used mostly in summer and it's hot that doesn't happen then.
If you need a freezer use the engel.
The fridge is mounted permanently into the camping car with a door on the outside to access the back of the fridge, it has vents top and bottom, on the inside of the bottom door vents I've installed 2 mini computer 12v power fans, they are very week but do set up a positive upward circulation of air over the absorber/condenser. As these use so little electricity they can be connected into your auxiliary battery without fear of draining it.
I think the added circulation helps on hot days.
And finally if you can't avoid parking it with the rear of the fridge on the sun exposed wall try a shade cloth over the area.
Hope that helps.


What dictates the lifetime of a solvated electron in a given solvent?

For those who have never had the pleasure of personally doing this, see this video. It has been known since 1807 that dissolving sodium in liquid ammonia results in a beautiful color. It was originally thought the color was due to some familiar complex instead of a solvated electron. A similar phenomenon happens with other alkali metals in ammonia.

Research in the field says that it takes, at least, forty some ammonia molecules to solvate a given electron. Metstable cavities form and their stability could depend highly on electrostatic interactions to solvate our electron. Ammonia will slowly react by evolving hydrogen gas,

$$\ce{2 NH3 + 2 e- -> H2 + 2 NH2-}$$

Differences in solvent can play a huge role in the stability of our solvated electron. An analogous decomposition occurs in water,
$$\ce{2 H2O + 2 e- -> H2 + 2 HO-}$$

Famously the latter occurs faster than the former, see here in comparison to my earlier link. We could say that these differences in rate reflect different stabilities of our solvated electron. Although, the addition of an appropriate catalyst to our ammonia will result in rapid evolution of hydrogen.

A more quantitative description for the difference in energies is obtained by measuring the UV-vis spectrum for a solvated electron in both water and ammonia. One will quickly notice that the band appears at higher energies in water than it does in ammonia and there are considerable differences in the band shapes (the band is wider in ammonia).

So Why Do They Differ

So here I have to make the disclaimer that no current theories quantitatively reproduce the observed phenomena, such as the absorption spectrum for a solvated electron. This question has no established/accepted answer as it is an ongoing area of research.

The self-ionization of water has an equilibrium constant on the order of $K = 10^{-14}$ and that of ammonia is on the order of $10^{-30}$. Hydronium formation occurs in the case of water and ammonium in the case of ammonia. Hydronium has a considerably lower $\mathrm{p}K_\text{a}$ than ammonium and so it’s reasonable to see why a reaction in water would be more likely with an electron. (Ammonia rarely produces a weakly acidic species, but water often produces a strongly acidic complex.)

A slightly deeper reason may be that water forms more ordered local domains/structures in solution than ammonia and this influences the rate by resulting in a larger entropy of activation when the hydrolyzed electron breaks up these larger structures, recall that $k \propto \exp (\Delta S ^\ddagger /R) $.

Just speculation though.


Conversion of Hessian from internal coordinates to cartesian coordinates

BAGEL (http://nubakery.org/ and https://github.com/nubakery/bagel) has CASPT2 analytical gradients and can run frequency computations.

It's currently being prepared for 1.0 release, so it might still have bugs. On the other hand, it is freely available under the GPL.

As for converting the Hessian, it might be useful to look at the INTDER program, which is (among others?) part of the PSI3 suite of programs available here: https://sourceforge.net/projects/psicode/files/psi/3.4.0/psi-3.4.0.tar.gz/download - PSI3 is under the GPL as well.

The manual page for INTDER is here: https://manned.org/intder and it says:

"Force field transformations between Cartesian and general internal
coordinate spaces up to fourth order, including nonstationary
reference structures.  Both forward and reverse transformations
are possible; hence, by using intermediate Cartesian coordinates,
force fields can be transferred among different representations in
the internal space."

Whether or not it will be easy to coerce the MOLCAS Hessian into a format INTDER likes and then transform it as you want I cannot say as I don't have any direct experience with INTDER - it is typically run as a step in a chain of commands automatically generated by the PSI3 driver routines.


The number of moles produced for each product will be the number of moles times of the limiting reagent times the reaction stoichiometry (the coefficients).

So to solve this you determine the number of moles of $\ce{KClO3}$ and use that as the base for determining the amount of $\ce{P4O10}$ and the moles of $\ce{KCl}$ produced.  Then multiple by the molecular weight per usual to get your answer.


If graphite's just a huge polymer of benzene why isn't it an aromatic compound (polymer?)

Graphite/graphene is aromatic. The first image below is a common crystallographic representation of graphite and does not account for covalent bonding.  The second one shows the molecular structure of graphite as a kekule representation. Graphite is made of planar sheets of carbon atoms in 6-atoms rings with conjugated double bonds. This also satisfies the 4n+2 π electrons criteria needed within the aromatic ring for graphite to be aromatic. This aromaticity is why graphite is not only stable to high temperatures but also a more thermodynamically favorable allotrope than diamond under typical conditions.

Crystallogrphic Representation of Graphite.

Kekule Representation of Graphite.


Mathematical form of four hybrid orbitals

If a molecular orbital $\psi$ (in this case, a hybrid orbital) is constructed from an orthonormal basis set of atomic orbitals $\{\phi_n\}$ via a linear combination

$$\psi = \sum_n c_n \phi_n$$

then loosely speaking, the "amount" of $\phi_n$ in $\psi$ is given not by $c_n$, but rather by $\left|c_n\right|^2$. (See, for example, Griffiths, Introduction to Quantum Mechanics, 2nd ed., Section 3.4.)

Therefore, summing the four values of $c_n$ for each AO has no physical meaning.

Instead let's take wavefunction (b) for example. The coefficient in front of the s orbital is $1/2$, so the "amount" of s-orbital character is simply $\left|1/2\right|^2 = 1/4$. The coefficient in front of the $\mathrm{p}_y$ orbital is $-1/2$, so there is $\left|-1/2\right|^2 = 1/4$ $\mathrm{p}_y$-character.

Armed with this knowledge, it's incredibly easy to see that each orbital has exactly $1/4$ s-character and a total of $3/4$ p-character ($1/4$ from each p orbital), because none of the $\pm$ signs matter (the minus signs are removed by the absolute value).


Which has higher lattice energy: MgO or LiF?

$\ce {MgO}$ ($\approx 3800$ $\pu{kJ mol^{-1}}$) has higher lattice energy than $\ce {LiF}$ ($\approx 1045$ $\pu{kJ mol^{-1}}$) mainly because of the greater charge on  $\ce{Mg^2^{+}}$ ion and $\ce {O^{2-}}$ as lattice energy is directly proportional to the charges of the combing atoms.

Your reasoning isn't incorrect but remember that $\ce{Li}$ and $\ce{Mg}$ show diagonal relationship so the size of their ions has only slight difference in magnitude. Similarly, oxygen and fluorine are placed next to each other in the periodic table so their anions don't have a large difference in size.

Thus, the "charge factor" dominates.


Molecule or molecular ion that spontaneously breaks up in two different ways

This happens all of the time in mass spectrometry following electron-impact ionization. Following ionization, each molecular ion decomposes (or fragments, to use mass spectrometry lingo) into other radicals and ions, usually following multiple pathways. As the pressure inside a mass analyzer is usually near vacuum, each fragmentation event is independent. No collisions or reactions between particles occur. The mass analyzer collects all of the fragments and generates signal that computer software converts to a mass spectrum - a frequency bar graph of the masses of species generated. For some very simple molecules, only one fragmentation pathway may predominate, but for most organic compounds, there are multiple fragmentation pathways.

For example, let's consider the fragmentation of but-3-en-2-one radical cation with the formula of $\ce{C4H6O+}$ and the structure shown below. The mass/charge ratio (m/z) is 70 for this compounds

There are at least two different fragmentation pathways.

The first fragmentation pathway is a loss of a vinyl group ($-\ce{C2H3}$, -27 m/z):
$$\ce{C4H6O+ -> C2H3O+ + C2H3 }$$

The second pathway is a loss of a methyl group ($-\ce{CH3}$, -15 m/z):
$$\ce{C4H6O+ -> C3H3O+ + CH3}$$

Each of these fragmentation pathways is independent. Here is a copy of the mass spectrum from the NIST Chemical Webbook, which is a frequency bar graph of all possible fragments. You can see in addition to the two pathways mentioned above, there are many other possible fragments.

We can estimate the relative rates of these processes from the frequencies of the peaks in the mass spectrum. The first pathway occurs with a relative frequency of 83.19 (based on the height of the m/z = 43 peak) , and the second pathway occurs with a relative frequency of 1.000 (from the height of the m/z= 55 peak). We can use these as proxy for the relative rates of decomposition. We can also see that the parent ion survives this process with a relative frequency of 0.57779 (from the m/z = 70 peak), which indicates that the parent ion has at least some stability.


Determine # of molecules in an unknown sample

There are attempts to redefine the mol through measuring spheres of monocrystalline silicon.  If you have the macroscopic dimension of your sample and the dimensions of your unit cell (from x-ray diffraction) you know how many atoms if silicon are there.


How does electrophoretic separation work in terms of separating soil into minerals?

The search term: "continuous particle electrophoresis" yielded much more results.

electrophoresis - the movement of charged particles in a fluid or gel under the
influence of an electric field.

source: http://www.youtube.com/watch?v=LNUgcq1M4LI

-

source: http://www.nature.com/articles/srep19911/figures/1

Heard some guy invented it while messing with some clay, but couldint find any video examples of that.


What chemicals cause the acrid smell of burning food?

Take a look at this link about Maillard reactions:

http://www.landfood.ubc.ca/courses/fnh/301/brown/brown_prin.htm

And what occurs if continue burning: pyrolisis

http://www.healthy-food-site.com/burnt-food.html


How does molecular orbital overlap affect energy?

Although the nature of orbital and wave functions can be puzzling, by the end of the day chemical bonds are hold together by the familiar electrostatic forces.  Although the electrons don't have a well-defined position, the forces felt by the nuclei is dictated by the spacial probability distribution of the electron.

To form a covalent bond, electron density has to accumulate between the two atoms.   The energy is lowered because the electron is close to both nuclei.   In a chemical bond such electron-nuclear attraction cancels out the repulsion between nuclei and holds the atoms together.

Orbital overlap changes electron distributions.   Orbital with different signs cancel each other in and lower the relative probability electron appear between the bonding atoms.  In the most extreme case of nodal plane, electron density is completely annihilated.  Because the total number of electron is unchanged during interaction, the lowering of electron density between nuclei is accompanied by an increase of density to the far side.

In other words, when you have orbital with canceling phases interacting, electron density will move away from the bonding area, resulting in higher energy.

On the other hand, a strengthening interaction increase the electron density between nuclei, thus lowering the energy and creating a pulling force between nuclei.


Thermodynamic Derivation of Ideal Gas Law

Classical thermodynamics does not admit that molecules exist thus if the gas law is derived it must be without recourse to any molecular ideas regarding the nature of matter.

The chemical potential of a perfect pure gas at temperature T and pressures p and $p^{\mathrm{o}}$ is defined as
$$ \mu = \mu^{\mathrm{o}} + RT\ln(p/p^{\mathrm{o}})$$
We can choose $p^{\mathrm{o}}$ to be unity then
$$ \mu = \mu^{\mathrm{o}} + RT\ln(p)$$
This is still dimensionally correct as $p^{\mathrm{o}} = 1  \pu{~bar}$ and as we have chosen the temperature, $\mu^{\mathrm{o}}$ is the standard chemical potential which is a function of temperature only.

Differentiating the chemical potential at constant T gives
$$ \left(\frac{\partial \mu}{\partial p}  \right)_T = \frac{RT}{p}$$

To find another expression for the derivative we can use
$$dG=-SdT+ Vdp +\Sigma\mu_idn_i$$
 for $dn_i$ moles of species i gives. For a pure species we can drop the subscript i and replace the summation with $\mu dn$ then
$$ \left(\frac{\partial \mu}{\partial p}  \right)_T=\frac{\partial}{\partial n}\left(\frac{\partial G}{\partial p}  \right)_T= V_m$$
where $V_m$ is the volume per mole.

Comparing these two derivatives gives $V_m=RT/p$ or
$$pV=nRT$$

Thus the definition of the chemical potential implicitly contains the perfect gas law.


How does ice formation glue things together?

The crumpets you have frozen together are very porous and convoluted at a microscopic level. One would then imagine that this means that any water vapour condensing on them as they start to freeze will percolate somewhat into the surface of the material and then freeze. The interior will stay warmer longer than the surface so this will help percolation. (Imagine an ink drop spreading into paper and you get the idea of percolation). On freezing water crystals become connected and thus many of these stick the two parts together. Rather like a sort of very fine velcro. However, most all foodstuff already contains a lot of water surrounding lipids, proteins and carbohydrates. Thus hydrogen bonding of condensed water to these intrinsic waters may also be important at the molecular level.


What makes sulfur-containing compounds such good antioxidants?

There are several factors that combine to make sulfur compounds such effective antioxidants.  First, sulfur compounds often contain S-H bonds (mercaptans) and are excellent radical scavengers because sulfur radicals are thermodynamically more stable than the analogous oxygen radical.  That is to say that the following equilibrium lies far to the right
$$\ce{ROS^{.} + RS-H <=> ROS-H + RS^{.}}$$
where ROS represents a "reactive oxygen species" such as hydroxyl radical ($\ce{HO^{.}}$), superoxide radical ($\ce{O_2^{-~.}}$, etc.  Sulfur compounds (sulfides and disulfides, for example) are also very easily oxidized to their corresponding sulfoxides.
$$\ce{R-S-R +ROS -> R-SO-R + deactivated ~ROS}$$
If multiple mercaptan groups exist within a molecule (for example multiple cysteine residues in an amino acid) they can be oxidized to the corresponding disulfide while reducing an oxidizer.
$$\ce{2~ RS-H + [oxidized~molecule] -> RS-SR + [H-reduced~molecule-H]}$$
This oxidative process provides sulfur compounds with a second pathway to trap or transform, and consequently inactivate, ROS.

In addition to being very effective ROS scavengers, sulfur compounds also bind strongly to many metal ions.  These same metal ions often play a role in the generation of ROS.  By binding to them, sulfur compounds effectively remove them from the ROS generating process thereby further lowering the concentration of ROS in the cell.


Hydrolysis is reaction with water, forming usually 2 different reaction products. It can be also named decomposition by water.
Hydration is (usually) reversible chemical bonding of water. It can be also named water addition.
One action of above can preceed or follow the other. If the hydration step preceeds hydrolysis in the reaction kinetics, the distinction may look vaque from some points of view.
Ammonium acetate undergo hydrolysis.

$$
\begin{align}
\ce{CH3COONH4 &-> CH3COO- + NH4+} \\
\ce{CH3COO- + H3O+ &<=> CH3COOH + H2O } \\
\ce{CH3COO- + H2O &<=> CH3COOH + OH- } \\
\ce{NH4+ + H2O &<=> NH3 + H3O+} \\
\ce{NH4+ + OH- &<=> NH3 + H2O} \\
\ce{2 H2O &<=> H3O+ + OH-}
\end{align}
$$


Is it possible to form an alloy of gold and zinc?

Yes. There are alloys and intermetallic compounds up to $\ce{44\%Au:56\%Zn}$.  This phase diagram is from here.

That said, crystal structure changes with composition, so appearance on solidification might be hard to predict.

Are you trying to emulate Hevesy?


Spacing between adjacent planes of a given set in a simple cubic crystal

When Miller indices are different only in a common factor, the planes are in the same orientation. The higher order reflections share some planes with the lower order reflections. In the case mentioned, every second plane of the (2 0 0) planes coincides with the (1 0 0) planes, whose spacing is double that of the former.


What is the optimal voltage for storing a 3.7 V lithium polymer battery?

The $\pu{3.6-3.8 V}$ range is a good general choice, but it may be battery-specific. The particular voltage for 40% charge may differ for different cell technologies, e.g. various deviations of electrode materials and due to cell aging.
The optimal storage conditions, as you mentioned, are more often expressed as charge/capacity % ratio. Usually, the optimal storage conditions are said to have about 40% of the maximal charge. For longer term storage, it is advised to be kept in a fridge, but not to get below
$\pu{0 ^{\circ}C}$. Note that by the semi-empirical van 't Hoff's rule(*) chemical reactions (= battery deterioration) speed up typically 2-4 times per temperature increase by $\pu{10 ^{\circ}C}$.
Lithium-ion/lithium polymer (Li-ion/LiPo) cells reportedly lose typically 20% of capacity per year when fully charged, but just 4% at 40% charged state.
For practical purposes, sometimes a trade off is made between battery lifetime and battery readiness.

Some background about Li-Ion/LiPo cell chemistry:
The negative anode is formed by lithium graphite intercalate $\ce{Li_\mathrm{x}C6}$, x<=1
The positive cathode is formed most commonly by $\ce{Li_xCoO2}$
The electrolyte is most commonly formed by $\ce{LiPF6}$ dissolved in mixture of dialkylcarbonates $\ce{R1-O-CO-O-R2}$ (**)
About electrode reactions from Li-Ion wikipedia article:

The positive electrode (cathode) half-reaction in the lithium-doped cobalt oxide substrate is
$$\ce {CoO2 + Li+ + e- <=> LiCoO2}$$
The negative electrode (anode) half-reaction for the graphite is
$$\ce{LiC6 <=> C6 + Li+ + e^-}$$
The full reaction (left to right: discharging, right to left: charging) being
$$\ce {LiC6 + CoO2 <=> C6 + LiCoO2}$$
The overall reaction has its limits. Overdischarging supersaturates lithium cobalt oxide, leading to the production of lithium oxide, possibly by the following irreversible reaction:
$$\ce {Li+ + e^- + LiCoO2 -> Li2O + CoO}$$
Overcharging up to 5.2 volts leads to the synthesis of cobalt(IV) oxide, as evidenced by x-ray diffraction:
$$\ce {LiCoO2 -> Li+ + CoO2 + e^-}$$
In a lithium-ion battery, the lithium ions are transported to and from the positive or negative electrodes by oxidizing the transition metal, cobalt ($\ce{Co}$), in $\ce{Li_{1-x}CoO2}$ from $\ce{Co^{III}}$ to $\ce{Co^{IV}}$ during charge, and reducing from from $\ce{Co^{IV}}$ to $\ce{Co^{III}}$  during discharge. The cobalt electrode reaction is only reversible for $x < 0.5$ (x in mole units), limiting the depth of discharge allowable. This chemistry was used in the Li-ion cells developed by Sony in 1990.

In addition to the above, there is another deteriorating process, related to keeping Li-Ion/LiPo cell long time at fully charged state ( typical for laptops permanently connected to AC, unless special coutermeasures are involved ). There is a side tendency of lithium ions in the highly charged cell state to form metallic lithium at surface of graphite-lithium intercalate. Normally, this is reversible. But there is ongoing slow side reaction with the dialkyl carbonate solvent:

The Reaction of Lithium with Dimethyl Carbonate and Diethyl Carbonate in Ultrahigh Vacuum Studied by X-ray Photoemission Spectroscopy
Abstract: The reaction of dimethyl carbonate (DMC) and diethyl carbonate (DEC) with clean metallic lithium in ultrahigh vacuum was studied by the use of X-ray photoelectron spectroscopy with the temperature-programmed reaction methodology. Both molecules are of interest as solvents in ambient-temperature lithium batteries. The solvent molecules were condensed onto the surface of an evaporated lithium film at $\pu{120 K}$, and spectra were collected as the sample was warmed in ca. $\pu{25 K}$ to $\pu{30 K}$ increments. The reaction of either DMC or DEC with lithium was initiated at $\pu{180 K}$, a temperature much lower than their bulk melting temperatures, producing lithium methyl carbonate, methyllithium and lithium ethyl carbonate, and ethyllithium, respectively. At temperatures greater than $\ce{270−300 K}$, the lithium alkyl carbonates start to decompose with $\ce{Li2O}$, elemental carbon, and alkyllithium as products on the surface. Both DMC and DEC are more reactive toward metallic Li than another carbonate solvent, propylene carbonate, which we have studied by the same methodology. Because methyl and ethyllithium are highly soluble in the parent solvent, electrodeposited $\ce{Li}$ is predicted to have poor stability in an electrolyte composed of either DMC or DEC.

(*) Not to be confused with the van't Hoff's equation - the van't Hoff's rule is about temperature dependency of kinetic rate, the van't Hoff's equation is about temperature dependency of equilibrium constant. But both are related.
(**)( dimethyl carbonate, diethyl carbonate, methylethyl carbonate, ethylen carbonate, propylene carbonate )
---- Feedback to comments
Being fully charged helps to prevent deep discharges, what is not the same as saying it is optimal storage condition. Note that storing cells fully discharge is worse than fully charged.
About half charge is optimal for lifetime and full charge is optimal for the time before needed recharging. But self discharging is not dramatic. LiPo have reportedly self-discharge rate approximately 5% per month.
This discharging is probably not linear, and its rate would be decreasing. Also, consider it as illustrative number only. It may differ for different cell technologies and age. Discharge cutoff cuts the connection before a real 0% charge. So the effective 0% charge > real 0% charge. Generally self discharging rate may increase with battery age, but it can be an individual number.
The usual mode of Li-ion/LiPo charging is by a constant current, until the threshold voltage is reached. Since then, charging switches to the constant voltage charging, until the charging current drops below a threshold value. The point of mode switching is typically between 70 (old cell)-80(new cell)% of the full actual capacity.
All that means the charging current is constant until the charge mode switching, and then it is asymptotically decreasing toward zero, until the current threshold cutoff stops the charging.
For lifetime, more shallow charging cycles is better than fewer deep charging cycles. The former has bigger total provided energy value for the cell lifetime. E.g., just illustratively, if a cell lifetime is 300 0-100% cycles, it has lifetime e.g. 800 30-80% cycles, giving 4/3 times bigger total charge. And it goes further for even more shallow cycles. But, unfortunately, shallow charging strategy is seldom practical.
So, if you know you are going to use the battery shortly after charging, it is the best to fully charge it, or at least to 90%. If you do know it will not be used for some time, it is better to charge the about discharged battery just partially, e.g. just 30-40% of the typical full charging time, depending on the residual charge.
The charging time depends on the battery state. A new one may have bigger capacity and generally different remaining charge. Note that new batteries are shipped partially charged, never empty. So bigger free capacity = longer charging time. The same free capacity, but older battery = longer charging time, because the second stage kicks in sooner due higher internal battery resistance.
The discharging cut-off voltage is vendor and technology dependent. It is usually somewhere between 3.0-3.5 V. Note that voltage as a function of charge is strongly nonlinear. More discharged = steeper voltage fall. The same vendor and technology dependency is for the cutoff charging voltage. The percentage itself is hard to tell, but estimated from the voltage curve just few percents of the capacity, perhaps 2-3%.


Why does ethane release more energy than ethyne when burned?

@a-cyclohexane-molecule has it right. Ethane generates more water. Here is a diagram that helps explain the issue in more detail. The black numbers and arrows are the heats of formation of the three hydrocarbons. Ethane is the only one more stable than the elements from which it is formed. Blue numbers and arrows are the heats of formation of CO2 and H2O from graphite and hydrogen. The red arrows and numbers are the heats of combustion of the three hydrocarbons. Although acetylene and ethylene have a headstart in combustion because the are less stable than ethane, all three compounds produce two CO2's but ethane generates more water than acetylene and ethylene. Ergo, ethane generates the greatest amount of heat. All values are in kcal/mol.


Determine formula of hydrocarbon from combustion data

In order to "solve" the problem the following assumptions are necessary:

The initial 20 $\mathrm{cm}^3$ volume of the gaseous hydrocarbon and 200   $\mathrm{cm}^3$ volume of oxygen are at rtp also.
The burning of the alkane (from part iii) also creates water, but at rtp the volume of water is negligible since it condenses into the liquid phase.
All the gases follow the ideal gas law so that volumes are directly     proportional to moles.

(i) Determine the volume of oxygen used.
The starting volume of oxygen was 200 $\mathrm{cm}^3$, but 50 $\mathrm{cm}^3$ of the final 150 $\mathrm{cm}^3$ was oxygen, so $200 - 50 = 150$ $\mathrm{cm}^3$ of oxygen was used.
(ii) Determine the volume of the carbon dioxide formed.
With assumption (2), the final volume was 150 $\mathrm{cm}^3$, but 50 $\mathrm{cm}^3$ was oxygen
so $150 - 50 = 100$ $\mathrm{cm}^3$ of carbon dioxide are formed.
(iii) The hydrocarbon was an alkane. Determine the formula of the hydrocarbon.
Since each carbon atom in an alkane will burn to produce one carbon dioxide molecule, then 100/20 = 5 carbon atoms must be in the alkane. So the alkane might be $\ce{C5H12}$. (Thus the hydrocarbon could be isopentane, neopentane, or n-pentane.)
$$\ce{C5H12 + 8O2 -> 5CO2 + 6H2O}$$

Problems with $\ce{C5H12}$
(A) Given that the volumes were supposed to follow the ideal gas law, 20 $\mathrm{cm}^3$ of hydrocarbon reacts with 150 $\mathrm{cm}^3$ of oxygen. That gives a molar ration of 2 to 15. But the molar ratio of $\ce{C5H12}$ burning is 1 molecule of hydrocarbon to 8 molecules of oxygen, or 2 to 16.
(B) The initial total volume of reactants was 150+20 = 170 $\mathrm{cm}^3$. The final volume of product was 100 $\mathrm{cm}^3$. So the ration of reactants to products, neglecting water, is 17 to 10. From the chemical equation the ration is 9 to 5, or 18 to 10.

So the answer must be a cycloalkane $\ce{C5H10}$ which would react with right ratio (2 to 15) and have the right ratio of products to reactants (17 to 10).
$$\ce{2C5H10 + 15O2 -> 10CO2 + 10H2O}$$
CHECK:
Assuming 22.4 liters for a mole of the hydrocarbon, then each mole of hydrocarbon would produce 10 moles of water.
$$10 \times \dfrac{0.020}{22.4} = 0.0089 \text{ moles of water}$$
Since each mole has a mass of 18 grams and water has a density of $1 ~\mathrm{g}/\mathrm{cm}^3$, then there is:
$$ 0.0089\times 18 = 0.16 \mathrm{cm}^3$$ of water which is indeed negligible.
Also at rtp water will have a vapor pressure of about 18 torr out of 760 torr so the partial pressure of water is pretty small also in the gaseous phase.
PS: Thanks to DGS for pointing out that the alkane must be a cycloalkane.


Does the mass of sulfur really decrease when dissolved in water and increase when burnt?

Upon reading the answers on Quora (thanks S007 for pointing that out) I realized this trick question is a lousy play upon two somewhat peculiar features of sulfur:

When submerged in water (not "dissolved", mind you, for sulfur does not dissolve), its apparent weight becomes less, thanks to Archimedes and the buoyancy force. This is indeed true of any element and compound, but sulfur has density of about 2, so its apparent weight decreases about twofold.
When burnt, sulfur forms a compound with oxygen which (containing all that sulfur plus oxygen) naturally weighs more than sulfur alone. Again, this is true of any element or compound that can be burnt, which are numerous. If anything, sulfur is "special" in that its atomic weight is 32, and the molecular weight of $\ce{SO2}$ is 64, so the increase is exactly twofold.

The way it stands now, though, the question is quite meaningless. Mass of any element never changes (barring nuclear reactions), that's the law of conservation. If you count compounds, then the mass would increase in any reaction (presuming we started from the pure element) and stay unchanged otherwise. If you count all compounds except gases, then the mass would sometimes increase and sometimes decrease, so there might be an answer, but then sulfur fails both conditions: it does not dissolve in water at all, and it all transforms to gases (that is, "vanishes") when burnt.


How can you predict relative Rf values of cis-trans isomers in thin layer chromatography?

Well, apparently I was wrong.

In fact, the trans,trans isomer is adsorbed more strongly by the stationary phase because it is entirely planar. The cis,trans-isomer cannot adopt a planar conformation because of steric hindrance between the two phenyl rings. The non-coplanar geometry lowers the efficiency of interactions with the stationary phase, making it more mobile.

So the cis,trans isomer has a higher Rf, and the trans,trans isomer has a lower Rf.


What is an example of an isopropyl alcohol resistant caulk?

Not an answer, but a strategy to get one:  I would start by googling 'alcohol-resistant caulk' or 'alcohol-resistant sealant'.  This will give you hits for chemical companies that produce such products.  For instance,

https://www.azom.com/article.aspx?ArticleID=5854
claims to have an epoxy ("UV18MED") that is alcohol-resistant and can be used as a sealant.

https://bowersindustrial.com/product-category/concrete-sealers-caulking/chemical-resistant-caulk/
claims to have caulks that are resistant to alcohol.

Since each of these companies are experts on the properties and performance of their respective products, you should call them, describe your specific application, and ask if they have anything that would work reliably.  That's the best way to be assured the product will do what you want.

One problem you might run into is that these may be B2B companies, i.e., ones that sell in quantities designed for businesses rather than consumers.  If that's a problem, try calling some manufacturers of consumer adhesives and asking if they have anything.

I know this is more work than simply posting a question on CSE (and there is of course nothing wrong with posting such a question), but it's only by speaking directly to the manufacturer that you can be sure the product is designed to work in your application (unless you have a product label that indicates this).

While you're speaking to them, you might also want to ask about leaching -- whether components of their product might dissolve into the alcohol; you would need to judge for yourself how much leaching your system could tolerate.  You'll also need to let them know what kind of mechanical stresses and temperature changes the product must tolerate (e.g., an adhesive that is sealing an aquarium and also holding it together needs to be a lot stronger than a caulk that is merely sealing the top of a sink). The more specific details you can give them about your system, the better.


For an equation
$\ce{Reactant <=> Product}$, at equilibrium, the rate of reactants forming products is the same as the rate of products converting back to reactants. Therefore, the concentration of products and reactants remain constant.


Galvanizing isn't just a protective coating in the sense that it avoids contact between the iron and the oxidant. Zinc oxidizes more easily than iron, and since both conduct electricity, you can have zinc in contact with the iron anywhere, and it will oxidize first. The term for zinc in this case is "sacrificial anode."

Full coating does help additionally though, by forming a protective layer of zinc oxide. Iron deteriorates when it rusts, so new iron metal is always being exposed as the process continues. However, some metals form oxides that protect the metal underneath from further oxidation. Zinc and aluminum are both examples of these metals.


Which one is nitric Acid, HNO3 by itself, or when it is included in H2O?

In your first reaction, $\ce{H2O}$ acts as a base to abstract an $\ce{H+}$ from the nitric acid. The resulting $\ce{H3O+}$, called hydronium, is the conjugate acid, while the $\ce{NO3-}$, called nitrate, is the conjugate base (this is the molecule of nitric acid, but stripped of its $\ce{H+}$). You could view this as a dilution, in the sense that the resulting solution is less acidic than highly concentrated nitric acid would be, but the process of dilution of acid by mixing with water inevitably involves an acid-base reaction.

As to your second question, the unionized molecules are the neutral water and nitric acid on the reactant side, while the ions are the products resulting from the acid-base reaction. Nitric acid is relatively strong, meaning it tends to ionize almost completely in aqueous solution. Additionally, the less acidic the solution already is prior to addition of the acid, the more extensively the added acid will dissociate into ions (to understand this, please read up on acid-base equilibria and Le Chatelier's principle, as the topics are too lengthy and complex to cover adequately here). This is why the author emphasizes that the dissociation is more complete in dilute solutions. Statistically, however, some proportion of the nitric acid molecules will inevitably not dissociate into ions. The stronger an acid, the greater the extent of dissociation, but there is always some statistical chance that a proportion of the molecules will remain unreacted. For strong acids, however, the overwhelming preponderance of the molecules will dissociate, while only a tiny fraction remain unreacted. Hence, for many (if not most) purposes, it's reasonable when dealing with strong acids to proceed as if the dissociation is truly complete, since the unreacted portion is typically negligible in quantity.


What substances do humans consume that are caloric but neither protein, carb, nor fat?

If you go to https://eur-lex.europa.eu/legal-content/EN/TXT/PDF/?uri=CELEX:01990L0496-20081211&qid=1580028914722&from=EN you will see a fairly extensive list of caloric compounds (screenshot pasted below).  This list is probably not comprehensive -- likely there are many compounds we (which includes our gut bacteria) can metabolize for calories.  Rather, it's limited to componds that are caloric and approved for use in food.  [It also doesn't include compounds that are naturally found in foods, and may be caloric, but are only present in very small quantities, e.g., nucleic acids.]

You can see that, in addition to fat, proteins, carbohydrates, and ethanol, the table lists polyols, organic acids, salatrims, and fiber.

Some explanation of the table may be helpful:

Polyols are artifical sweeteners. They include lactitol, maltitol, mannitol, sorbitol, xylitol, erythritol, glycerol, hydrogenated starch hydrolysates,  and isomalt. [Though note that while erythritol is a polyol, it is listed separately as non-caloric.  And while glycerol, aka glycerin or glycerine, is a sweetener, it is also used in food for other purposes.]

Organic acids include acetic acid (the main component of vinegar, other than water), as well as citric acid, ascorbic acid, and malic acid (the latter three are found in citrus fruits).

Fiber refers to dietary fiber, which is the type of fiber we can digest completely.

Salatrims are "short and long chain acyl triglyceride molecules"; they are a type of low-calorie fat substitute.

Source:  Consolidated text: Council Directive of 24 September 1990 on nutrition labelling for foodstuffs (90/496/EEC) Select: 6 CELEX number: 01990L0496-20081211 Author: Council of the European Union Date of document: 11/12/2008


If a Na+ ion and Cl- ion exist as NaCl,(ionic) it doesn't mean that, that particular Na donated electron to Cl. Na may lose electron somewhere else and Cl may gain electron somewhwere else.
The main point of existence as Na+Cl-, ie NaCl is the electrostatic attraction between them.


Reaction of 2,4-DNPH with aromatic aldehydes

2,4-DNPH is a derivatization reagent for the identification of ketones and aldehydes.
This was extensively used in organic analysis before instrumental methods were readily available.
You determined the boiling point of the problem, which in the case of benzaldehyde is $179\ ^{\circ}\mathrm{C}$ , then you made a solid derivative and determined it's melting point.
In the case of benzaldehyde and 2,4-DNPH that would be $237\ ^{\circ}\mathrm{C}$ .
Then you go to the tables, cross check both values and had benzaldehyde identified.

For aldehydes and ketones there are different options: oxime, semicarbazone, phenylhydrazone, p-nitrophenylhydrazone and 2,4-dinitrophenylhydrazone.
The latter one was usualy preferred because they have higher melting points and so it was easier to get a solid even if not totally pure.


Show proposed mechanism is consistent with the experimental rate law

The final rate equation is generated from substituting $\ce{[NO3]}$ defined in the third to last equation:

$\ce{[NO3]} = \frac{k_1}{k_{–1}}\ce{[NO][O_2]}$

into the second to last equation:

$2k_2\ce{[NO_3][NO]}$

This will give you the final rate equation defined only in terms of the starting concentrations. The $\ce{[NO_3]}$ term is replaced, $\ce{[NO]}$ squared, and $\ce{[O_2]}$ remains. None of the equilibrium constants cancel out. The intermediate $\ce{[NO_3]}$ is not included in the final equation, as it is consumed in the overall reaction.


You can do a normal hydrogenation reaction(H2,Pd) under low pressures as suggested by @Waylander. Alternatively you can use the Wilkinson's catalyst which also is a way to do catalytic hydrogenation but is capable of only reducing the double bond.


Decomposition of alkene in acidified potassium permanganate

gaseous hydrochloric acid or anhydrous,

we have an alkene where each side is attached to the same number of hydrogens —> both “equally substituted”.

Products in this case, 3-chloropentane and 2-chloropentane.


Well, you are right. Esters decomposes at higher temperatures to form several small carboxylic acids and ketones and at even higher temperature, these carboxylic acids decomposes into carbon monoxide, carbon dioxide, carbon, oxygen and/or water. A paper[1] discusses about the thermal decomposition of ethyl formate, ethyl acetate and ethyl propanate at 1248-1638 K, pressure around 1.5 atm at a chamber filled with argon at a concentration of 2000 ppm.

After the experiment, it was found out that 90% of oxygen has been released as oxygen gas and the pyrolysis mechanism was determined to be a six-centered decomposition producing equimolar ethylene and corresponding acid and at drastic conditions, it was known to have a dehydration reaction releasing water and a decarboxylation reaction to form carbon dioxide.

It was found out that EF gave formic acid, EA gave acetic acid and EP gave methyl ketene as the main products which subsequently decomposed into various smaller products at higher temperatures and the concentration ratios were noted.

Carboxylic acid can form white solid as noted in case of formic acid and acetic acid.

References

https://sutherland.che.utah.edu//USCI2013/PAPERS/1A07-070RK-0061.pdf


Concise description of primary organic reaction types mechanisms

The book you are looking for is The Art of Writing Reasonable Organic Reaction Mechanisms by Robert Grossman.

One of the book's publishers offers this description of the book:

  Intended for students of intermediate organic chemistry, this text shows how to write a reasonable mechanism for an organic chemical transformation. The discussion is organized by types of mechanisms and the conditions under which the reaction is executed, rather than by the overall reaction as is the case in most textbooks. The treatment emphasizes unifying principles, showing how common mechanisms link seemingly disparate reactions.
  Each chapter discusses common mechanistic pathways and suggests practical tips for drawing them. Worked problems are included in the discussion of each mechanism, and “common error alerts” are scattered throughout the text to warn readers about pitfalls and misconceptions that bedevil students. Each chapter is capped by a large problem set.
  The author has drawn on his own research and the current literature to ensure that appropriate attention is given to topics across the range of modern organic chemistry. The text is unique in its inclusion of a chapter on reactions mediated or catalyzed by transition metals, an area in which mechanistic understanding is now essential. Relatively new topics such as olefin metathesis and cycloaromatization are covered without giving short shrift to more traditional areas such as carbonyl chemistry. The text assumes a basic knowledge of organic chemistry. It can be used either in a formal course or by students working on their own, and will be particularly useful for graduate students studying for qualifying examinations. It will also be useful to students and researchers in biochemistry, pharmacology, and inorganic chemistry.

Each of the book's chapters are $50 \pm 30$ pages and are packed full of example problems.  Each chapter is organized according to the class of mechanism, not around particular reactants or products.  The chapters are:

The basics.
Polar reactions under basic conditions
Polar reactions under acidic conditions
Pericyclic reactions
Free-radical reactions
Transition-metal-catalyzed and -mediated reactions
Mixed mechanism problems

The author maintains a good web site with additional problems, additional worked examples, and answers to every problem in the book.  You can get a sense of the content of the book by browsing Amazon or by looking some of the answer .pdf files, available for free from Springer.

I found this book very helpful in improving my mechanistic understanding of various types of reactions.


Is cis dipole moment is always greater than trans dipole moment?

No.

The dipole moment of trans-cyclooctene ($\pu{0.82 D}$) is greater than cis-cyclooctene ($\pu{0.43 D}$).
See On the Molecular Geometry of trans-Cycloöctene, J. Am. Chem. Soc. 1958, 80 (8), 1953–1955.

  Blomquist, Burge and Sucsy2 have called attention to the fact that while with the cyclodecenes, the cis isomer has the higher density and index of
  refraction as is typical in acyclic systems, with the cyclooctenes, as first noted by Ziegler and Wilms,3 the reverse is true. From these refractive indices, by application of van Arkel’s dipole rule,4 the prediction could be made that the trans-cyclooctene would have a higher dipole moment than the cis.

  [...]

  The dipole moments of the cis and trans isomers of cyclodecene were previously determined,4 and were found to be normal. The conclusion, drawn from the dipole rule, that the trans-cyclooctene would have a higher dipole moment than the cis represents a situation without analogy.

  [...]

  The cis isomer was found to have $\mu_{cis} = \pu{0.43 \pm 0.06 D}$ which may be compared with the value found for cis-cyclodecene of $\pu{0.44 D}$.4 The trans isomer was found to have a dipole moment of $\pu{0.82 \pm  0.03 D}$, compared with the value zero found for trans-cyclodecene.4

(Van Arkel's rule is: The isomer with the higher dipole moment possesses the higher boiling point, refractive index, and density.)


They do indeed explode. Oxyliquits are probably the most comprehensive example of liquid-oxygen-based primary explosives (which are considered obsolete today). Theoretically liquid oxygen can also be substituted with ozone-oxygen mixture, which in reality has never been utilized due to technical difficulties -- oxyliquits were popular in the beginning of 20th century.


A. The orbital shapes originate from solutions to the Schrodinger equation for hydrogen like atoms. The derivation of the orbitals come from the section for the solution to Schrodinger equation for Hydrogen-like atoms on Wikipedia: https://en.m.wikipedia.org/wiki/Hydrogen-like_atom

The solutions have the same energy because you solve the equation for specific energies. They really are the same, which can proven by doing an integral across the volume.

The different hybrid orbitals depend on what quantum numbers are possible. This question has been answered here: Why an asymmetric geometry with sp3d and sp3d3 hybridization?