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An introduction to "oxidation state"/"oxidation number"
I just can't get my head around oxidation states. They're really confusing! Why do I need to assign a number to some element?

What are oxidation states?
How and why should we assign oxidation states?
Are oxidation states real?

TL;DR;

Oxidation States are useful tools used in nomenclature, redox chemistry etc.
There are a few simplistic rules to determine the oxidation state of elements in an atom; but they're not applicable to all the cases.
There are instances where the assignment of oxidation state isn't useful, or where the oxidation state is ambiguous.
We're easier off assigning, rounding off or estimating oxidation states ourselves in some cases, such as intermetallic compounds.

What is an oxidation state?
Oxidation state (OS), or oxidation number, is an integer or decimal number assigned to an element in a chemical species. This assignment follows certain rules and guidelines, to make it a useful tool whenever we need it. Oxidation states can be negative, positive, or zero.
According to Comprehensive definition of oxidation state
(IUPAC Recommendations 2016), p 833:

The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear
bonds.

All those fancy words up there mean that, when you're assigning OS, you take all of the bonds in the species to be completely ionic. No purely ionic compound exists, and even, say, the $\ce{Cs-F}$ bond in $\ce{CsF}$ has some covalent character. So you have to understand that OS does not happen. It doesn't refer to any real-life phenomenon.
Why do we assign OS's, then?
Oxidation states are a powerful tool for chemists, and save us a lot of headache. You might question "why do we need to give an element a number anyway?" which arises because you usually only learn how to give elements those numbers, but no real-life application becomes necessary or evident.
OS is a tool, and is not the only tool, wherever it's being used. It's like the shovel that digs the ground. You can also do the digging with a spoon. Why do you choose a shovel?
Assigning OS
The Basis
The basis of assigning OS, in simpler cases, is just good old electronegativity. There are exceptions where this would not work, but you don't need to worry about them at this stage.
You're usually given the values of electronegativities, and those aren't what you should worry yourself about. There are different scales for measuring electronegativity, and IUPAC recommends the Allen scale because it's the only scale independent of OS.
OS Assignment Using the (Molecular) Formula
You're given a formula of a compound you've never met, and you're asked to determine the OS of some, or all of the elements present. This is a classic scare if you don't understand what you're doing.

Determine the OS of all the elements in $\ce{H2C2O4}$.

Like most of the problems you face, there are some "unknown" and some "known" values. You need to identify a starting point, go on from there until everything sorts itself out.
Here's what we know: We need to use the electronegativities to find out what the OS's are. O is the most electronegative, C is next, and H is the least electronegative.
Here's what we want to know: We have three OS's here that we need to determine. C, H and O.
As it happens, the elements in the rightmost and leftmost parts of the periodic table are the most and least electronegative elements of all, respectively. So we expect their stance to be more clear: The ones on the left, the alkaline metals, tend to lose electrons and the ones on the right, tend to gain. As a general rule of thumb, the closer an element is to the middle of periodic table, the fuzzier and harder determining its OS is.

Always try to remember the location of the element in the periodic table, and start from one that's the rightmost and/or the leftmost. Don't try to quantify this rule, since it's merely a guideline. Sometimes, deciding the OS of oxygen becomes difficult!
Thus, we would expect oxygen or hydrogen to be easier to determine here. That is the case: the electronegativities tell us that oxygen must have a negative OS, and hydrogen a positive one.
Time to quantify this. You remember from the intro to electronic configurations that O's electronic configuration ends in $\mathrm{2p^4}$, and that if it ended in $\mathrm{2p^6}$, it would've given oxygen a relative amount of stability. It would've done the same here if it could grab the two electrons from the carbon, but in reality, that hasn't happened. But anyway, when assigning OS, you assume that the bonds are ionic, and the oxygen has already stolen carbon's two electrons.
If a neutral oxygen with an electronic configuration ending in $\mathrm{2p^4}$ gains electrons to become one ending $\mathrm{2p^6}$, how many electrons has it gained? Two. Thus, the OS of oxygen in $\ce{H2C2O4}$ is −2.
You can apply the same process for hydrogen. It has only one electron in its neutral form, and it's about to lose it. How many electrons can it lose?
Determining carbon's OS is more tricky. Carbon is in group 14, and it's not obvious what its OS should be. It's the closest element to the middle. Here's a workaround: Instead of trying to find out which OS's carbon can have, we calculate how much is left as carbon's share.
$$\color{green}{\text{The sum of the OS's in the species is equal to the charge} \\ \text{of the species, be it a molecule, an ion etc.}}$$
This is a rule you have probably also met in your textbook. The charge of the species is an indication of excess or lack of electrons, and should definitely affect what OS's we assign to the elements. Why? Because OS is itself some form of counting electrons.
($\Rightarrow$ if they're just counting electrons, why do we see fractional oxidation states? See OS Assignment Using the Lewis Structure below)
$\ce{H2C2O4}$, oxalic acid, is a neutral molecule. So the sum of OS's must equal zero. Note that we count hydrogen twice, because two hydrogens have two electrons in neutral form, and they will lose two electrons. Likewise, four oxygens will, in a hypothetical purely ionic oxalic acid, gain eight electrons overall.
$$\begin{equation}
2(+1) + 2(\mathrm{OS}_\ce{C}) + 4(-2) = 0 \\
\mathrm{OS}_\ce{C} = +3
\end{equation}$$
Most Common Patterns in the OS of elements
If you consider all the possible species that can be synthesized, there are great many bizarre trends and anomalies in OS assignment. Refer to the Further Reading section to learn more about them.
However, for most educational purposes, OS's for a number of elements follow simple guidelines worth mentioning:

Fluorine's OS is always −1 (except in $\ce{F2}$ itself).
Halogens below fluorine typically adapt a −1 OS as well, but always check when there are more electronegative elements such as fluorine and oxygen present.
Alkaline metals adopt a +1 OS.
Alkaline earth metals adopt a +2 OS.
You're almost always better off determining the OS of transition metals using the sum rule.
If oxygen is not the leftmost element present, or there are alkaline and alkaline Earth metals present, don't start by assigning an OS to oxygen.
If there are any of the elements in guidelines 1–4; don't start by assigning an OS to hydrogen.

Determine the OS of all the elements in $\ce{NaO3}$.

There are only two possibilities: Start from Na, or start from O. Both are, as said, in left and right parts of the periodic table. So which?
If we take oxygen, and assign the "normal" OS, −2, we see that Na has to be +6 because the sum should equal zero. Something doesn't sound right. As guideline 6 says, we shouldn't start from O when there's Na present in the compound. So we take the other route.
$$\begin{equation}
+1 + 3(\mathrm{OS}_\ce{O}) = 0 \\
(\mathrm{OS}_\ce{O})= \frac{-1}3
\end{equation}$$
Check your understanding
Find the OS of all the elements in these compounds:

1.$\ce{NaAlF6}$  2. $\ce{C2O2F2}$
3. $\ce{C8O8}$ 4. $\ce{Au2O3}$

OS's That Don't Exist
The upper bound for OS's that you will meet is +7 for some manganese $(\ce{MnO4-})$, or chlorine $(\ce{ClO4-})$ compounds. OS's more than that are only found in exotic compounds. Thus, if you get a, say, +9 OS, you should start doubting the route you took.

Determine the OS of W in $\ce{H4W2O13}$.

Three possible starting points. We rule out tungsten because it's the nearest to the middle. Of the other two, hydrogen would lose 1 electron, and oxygen would tend to gain two. Right?
Well, if you solve it for tungsten, considering the neutral overall nature of the compound, you'll get an OS of +11! That's definitely not right, since the highest observed OS is +10 (See Further Reading). This could only mean one thing, since hydrogen can't adapt another OS or tungsten would have an even higher OS. This means that some of the oxygens are in the form of peroxides, superoxides etc.
Had we known that the structure of the compound, more elaborately written, is $\ce{(H2O)(O)(O2)2WOW(O2)2(O)(H2O)}$, we wouldn't have made the mistake of thinking all the oxygens have the normal OS of −2.
Hence, it becomes necessary to be able to deduce OS from structures.
OS Assignment Using the Lewis Structure
At times you need to determine OS in an electron dot diagram, or Lewis structure. This is sometimes useful/needed over determining from the molecular formula because

either the molecular formula doesn't give us the whole picture (see below), or
it gives us impossible answers or no way to go further, like the tungsten compound's case, or
it's already what the examiners gave us!

Determining OS from Lewis structure is even easier than deducing it from the molecular formula, because you already see the bonds. It's the same process in the IUPAC definition above: Ionic approximation

The electrons in a bond go count as the more electronegative atom's. (There are exceptions to this, see Further Reading)
The electrons in a bond between the same atoms are shared equally. So in a $\ce{C-C}$ bond, each C atom gets one electron from the bond.
The OS of each element is the number of valence electrons minus the sum of lone pairs and electrons it gains from bonds.

A good example used to illustrate this is acetone.

Determine the OS of all the elements in acetone, $\ce{C3H6O}$.

The answer is in the spoiler below (no peeking ಠ_ಠ):

 $\mathrm{OS}_\ce{O} = -2, \mathrm{OS}_\ce{C} = \frac {-4}3, \mathrm{OS}_\ce{H} = +1$

There's one oxygen in acetone, and it's bonded to the carbon:

Assigning the rules, we get

$\ce{C1}$ has four valence electrons, but there are "seven electrons around it". Both electrons from the $\ce{C-H}$ bond belong to carbon. Thus, $4 - 7 = -3$.
$\ce{C2}$ has four valence electrons, but only two electrons are around it. One from of each of the two $\ce{C-C}$ bonds. That makes $4 - 2 = +2$.
$\ce{C3}$ is just like $\ce{C1}$, $\mathrm{OS} = -3$.

Fractional OS's
Didn't we say earlier that OS assignment is some kind of counting electrons? Then how come we encounter fractional OS?
Now you know the answer. Some elements in some compounds show two different oxidation states, and when computing the OS from the molecular formula, you actually compute the average OS. None of the carbons in acetone have an OS of $\frac {-4}3$.
Familiar examples of this 'average' OS are carbon in acetic acid, which is computed to be zero but is in fact −3 and +3, and iron magnetite, which is $\frac 83$.
Further Reading
The school lessons are over, but the fun is actually just about to start. This is the part where, you know, you see what you've been taught isn't accurate. Read on if you're interested. It might get confusing for first-timers.
Be Liberal in Thinking
OS is a useful concept. Unfortunately, it's hypothetical. There's this mindset that there should only be one OS for each element in a compound out there. That's simply not the case.
$\ce{S2O3^2-}$, thiosulfate anion, is one such compound where the two approaches we just introduced produce conflicting results. The average $\mathrm{OS}_\ce{S}$ is +2 for sure, but is it +5 and −1, or +6 and −2?
Turns out, studies are conflicting on this matter. The final say here is provided by IUPAC recommendations linked above:

5.3 Nominal OS

In systematic descriptive chemistry, OS is used to sort out compounds of an element; in electrochemistry, it represents the electrochemically relevant compound or ion in Latimer and Frost diagrams of standard (reduction) potentials. Such purpose-oriented OS may differ from the OS per definition and are termed nominal, or more specifically “systematic” or “electrochemical”.

Consider thiosulfate, the $\ce{S2O3^2−}$ anion: The sulfur–sulfur
bond is practically a single bond, which yields OS = +5 for the central S and −1 for the terminal S atom by both algorithms. The electrochemical OS of sulfur in thiosulfate is their average, +2. On the other hand, a systematic OS of +6 may be chosen for the central S atom, as if the sulfur–sulfur bond were approximated ionic, to emphasize the similarity of the peripheral O and S atoms obtaining OS = −2 as in sulfate.

Exceptions
These are the cases where the methods above don't reliably work in determining OS:

The most glaring example is when the more electronegative atom is bonded reversibly as a Lewis acid.
We originally introduced electronegativity as the basis to decide OS. Truth is, the real basis is the energy level of molecular orbitals, and this, in most of the cases, corresponds to the more electronegative atom getting all the fish. It's difficult and/or impractical to use complicated MO schemes to determine OS, so we stick to electronegativities.
In the case of Lewis-acid ligands, however, this backfires and gives us the opposite of the correct answer.
Non-innocent ligands are ligands whose OS is ambiguous because they're very redox-active. This article does a nice job of explaining $\ce{H2}$ and nitrosyl non-innocent ligand briefly (§ 10 & 11).
Nominal OS's; like thiosulfate above.
Metallic compounds; there are some obvious cases such as TiO, but it gets more complicated – Source
Intermetallic compounds, such as Zintl phases, where it's safe to assume an OS of 0 for redox purposes or compounds with delocalization over different atoms  – Source (§ 13)
Compounds with a close electronegativity such as Se and C in $\ce{Se-C}$ bonds or H and P in $\ce{H3PO3}$. – Source

Exotic OS
These are compounds whose certain elements demonstrate unusual OS's:

Alkalides, which are salts containing alkali metal anions with an OS of −1. Lithides are currently not known, but some kalides and natrides are stable in room temperature.
The highest known OS belongs to Ir in $\ce{[IrO4]+}$ which is +9, and potentially +10 for Pt in $\ce{[PtO4]^2+}$. The lowest known OS is −5 for boron in Zintl phases. – Wikipedia
Platinum hexafluoride is a strong oxidizing agent that can oxidize $\ce{O2}$, giving dioxygenyl, with an unusual OS of $+\frac 12$ for O.
Alkaline earth metals are observed to show unusual +1 OS in some complexes. – This article reports the synthesis of complexes with Mg–Mg bonds.

Useful Links
[1]: List of Oxidation States of Elements
[2]: High-valent Iron
[3]: Using OS to analyze redox reactions; a tutorial
[4]: ChemGuide OS tutorial

That's an excellent start to exotic OS's, but it may be worth adding a disclaimer that it's hardly an exhaustive list. Lastly this is not a fault with your answer at all, but I think it's important to note that OS's are only formalisms and that they do not necessarily represent the true bonding picture. In $\ce{PtO4^2+}$ the positive charge on platinum is not going to be anywhere near +10. (I know you said it at the top, but it might be worth repeating.)

The IUPAC gold book actually defines another procedure for assigning oxidation states, see this question, which leads to the anomalies for fluorine in HOF, see here and here.

What started out as an aid to help balance redox equations seems to have turned into a religious ritual. It is much simpler to use the simple rules and for more complicated molecules take a serious look at the chemical -molecular structure. For organic molecules work with the functional groups that react and ignore the rest.

@orthocresol why is it when I Google PtO42+ the 1st 3 articles all say Pt is +10 oxidation?

@NealConroy, I didn't say that the oxidation state wasn't 10. I said that the oxidation state is a formalism. Sure — we can follow the rules, and at the end we get this number 10 out of it. That's great! But it doesn't really mean much more beyond that.


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Schrödinger's equation and spin quantum number
As spin quantum number cannot be derived from Schrödinger's  equation, it cannot predict opposite electron spin.
I mean to ask that how do we obtain the information conveyed by the spin quantum number when we don't speak of it in this theory.

As far as I know, in nonrelativistic quantum mechanics the existence of spin is a postulate, i.e. we assume it's there (and the reason for assuming it's there is because we saw it).

Regarding your edit, exactly what information are you talking about? The idea is that we postulate that spin is a form of angular momentum with quantum numbers 0, 1/2, 1, ... Angular momentum is a topic that is very well-described by the Schrodinger equation, and spin is nothing but a form of angular momentum, so all the rules that apply to spin quantum numbers are simply derived from the theory of angular momentum. This includes the Pauli exclusion principle which can also be derived in the non-relativistic picture. The only postulate that needs to be made is the existence itself of spin.

and once we assume (for example) that an electron has spin-1/2, it automatically follows from the rules of angular momentum that it possesses "up" and "down" states, because the allowed values of $m_s$ for a particle with $s = 1/2$ are exactly $+1/2$ ("up") and $-1/2$ ("down"). For more information on angular momentum you might want to consult Atkins Molecular Quantum Mechanics chpt 4 although any decent QM textbook will have a section on it.

Good question.
This was one indication that there had to be something more than just the Schrodinger equation. Another problem which was perceived quite quickly is that the Schrodinger equation treated time differently than the spatial dimensions which is not in the spirit of special relativity. Thus, Paul Dirac constructed what we now call the Dirac equation. When you solve the Dirac equation in a Coulomb potential, spin, and hence the fine structure of the hydrogen atom energy levels in an electric field, falls out naturally.
This proves to be fairly technical and I won't try to do any of the mathematics for this (I don't completely understand it all), but you can find solutions to the Dirac equation for some common cases here.
As orthocresol points out, we first observed the intrinsic angular momentum of the electron in the Stern-Gerlach experiment, so people guessed what this would look like at first (this is how it was used in the Schrodinger equation) and Dirac came along and explained that people had guessed correctly. This can also be seen from the fact that the Dirac equation reduces to the Schrodinger equation in the non-relativistic, low energy case.

Why do we keep using the term non relativistic for schrodinger equation?

@lily The Dirac equation can be thought of as a relativistic version of the Schrodinger equation. By relativistic, I mean relating to the theory of special relativity. That is, it is accurate when particles are moving close to the speed of light.

That might seem stupid..but as far as I understand it, electrons can never get their speeds close to that of light..so why do we need to have a relativistic picture of that

It is true that valence shell electrons generally don't travel that close to the speed of light, but core electrons often travel at speeds quite close to the speed of light. Generally, the core electrons of heavier elements travel faster. For instance, the reason that gold looks gold and part of the reason mercury is a liquid (at room temperature) is due to relativistic effects. See here and here.

Dirac didn't explain spin. Spin is in the Dirac equation because it was introduced in the equation. Moreover, the Dirac equation is not a valid relativistic wavefunction equation and early attempts to build a consistent relativistic quantum theory based on it have been abandoned.

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Disappear alpha, concentrations
I have a doubt that torments me constantly ... And I hope your answer so that you can solve it. When they give me $K_{a}$ or $K_{b}$, the concentrations and they ask me to find alpha, when can I discard alpha?
Example:
$K_{b} = 1.79·10^{-5}$, $(NH_{3}) = 0.74M$, $\alpha =$
Okay, with these data I arrive at the following expression:
$K_{b} = (0.74·\alpha ^ 2) / (1-\alpha)$, here, in the "$1-\alpha$", can I discard?
I think the answer is no, but could you tell me why not? And in what cases can I do it?

I assume that by "despise" you actually mean discard the value of alpha. As in making the approximation: $$(1-\alpha ) \approx 1$$
To answer your question we need to consider the meaning of alpha and the value of the equilibrium constant.
$\alpha$ represents the degree of ionization, i.e., the amount that dissociates from 1. If you say that $\alpha$ is 0.1, for example, then what you are saying is that for every 1M of concentration 0.1M dissociates. If you have a different concentration than 1M, then the amount that dissociates is $\alpha M$ where M is the concentration of your starting substance.
The equilibrium constant, K, tells you something about the relation between products and reactants. If K is greater than 1 then products are favored and if K is less than one reactants are favored (or, the reaction does not "like" to go towards the right).
Look at the value of your constant $K_b = 1.79 . 10^{-5}$. It is small. That should tell you that very little amount of your starting reactant will dissociate to yield products. Thus, we would expect the value of $\alpha$ to be small when compared to one. If this is the case then we can say that $$(1-\alpha ) \approx 1$$
If you solve the equation with the approximation you find $\alpha = 0.004906$ . If you solve the equation without approximation you get $\alpha = 0.004918$
Thus, it seems that the approximation is justified particularly if you don't need great precision in your answer.
So, when do you approximate? There is no a set answer to this. But, if your equilibrium constant is small ($ < 10^{-4}$) usually) and the concentration of your initial substance is big enough you can try and see what happens. The rule of thumb that I was told was to accept the approximation if the final result is less than 5% of the value of the approximated value. In this case, we obtained 0.004918 over a value of 1. This represents 0.5% and the approximation is justified.
Make sure you ask your professor what is the criterion he or she uses to justify an approximation.

Re; "So, when do you approximate? There is no a set answer to this. ..." // That really isn't correct. There is the concept of significant figures. The ammonia was noted as 0.74 molar. So the ammonia concentration is only good to +/- 0.005 molar.  That is 0.68%. This is about the same as assuming that $1 - \alpha \approx 1$, so the assumption is ok. This sort of assumption is used all the time in chemistry to simply the mathematical analysis. // This is different than in a math class where 0.74 means 0.740000000...

@MaxW You are technically correct.However, I have yet to find any instructor in General Chemistry that goes through that analysis of significant figures. The usual criterion, at least what I have heard most often, is the 5% criterion, that is, if your approximation is 5% or less of the value on which the approximation ws made it is taken as ok.

I was taught significant figures when I was in high school.

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Why do we use the external pressure to calculate the work done by gas
I read in a textbook that in the case when we have a gas in a cylinder fitted with a massless frictionless piston being held with an external pressure $p_1$, and when the pressure is reduced to become the value $p_2$, the gas pushes up against the piston and then the work done by the gas for a small change in volume is calculated by:
$$\mathrm dW=p_2\,\mathrm dV$$
Here, is what I don't conceptually understand. If the gas's molecules was under some pressure $p_1$ which is equal to the external pressure in the static state then after the external pressure became lower than the internal pressure, shouldn't the work done by the gas be the difference between the two pressures?

If the piston is frictonless and massless, then, if you do a force balance on the piston, you must have that the force per unit area that the gas exerts on the inside face of the piston will always be equal to the external force per unit area that one imposes on the outside face of the piston.  The sudden drop in pressure on the outside face of the piston causes the gas to undergo an irreversible expansion.  During an irreversible expansion, the local pressure within the cylinder becomes non-uniform, so that the average pressure of the gas differs from the force per unit area at the piston face.  As a result, the ideal gas law (or other equation of state) cannot be applied globally to the gas in the cylinder.  In addition, during an irreversible expansion, there are viscous stresses present in the gas that allow the force per unit area at the piston face to drop to the new lower value while requiring that force to match the external force on the outer face.  So the work done by the gas on the piston is equal to the external force per unit area times the change in volume:  $$W = \int{P_{ext}dV}$$  This equation is always satisfied, irrespective of whether the expansion is reversible or irreversible.

Is perfect gas viscous?

no...............

@Anonymous There is no such thing as a perfect gas.  Viscous stresses are a feature of the behavior of all real gases, which the perfect gas concept is incapable of capturing.  And for irreversible expansions, it is viscous stresses that make the difference between the perfect gas equation applying within the gas, and the actual stresses present that enable the compressive stress of the gas on the massless piston to equal the external pressure.  You are aware that, in an irreversible deformation, an ideal gas does not satisfy the ideal gas law, right?  It applies only to equilibrium states.

@ChetMiller I was answering the learner

Consider a thought experiment; imagine playing your favourite sport competitively against someone. Imagine  losing the game as time runs out (bleak scenario sorry). The work is what you actually put in to catch up. If they are good it becomes even harder to catch up. Winning is just how much better you were with your work ethic over theirs. Work is done against the external force. The external force is the resistance to which effort must be put in to overcome it. The external force is your opponents score that you must achieve. Achieving this is work. The resultant force or resultant pressure (internal minus external) merely causes an acceleration and tells you how much time it will take to achieve a particular distance change. If your internal pressure is much larger than your external pressure before equilibrium the faster it will take to reach your desired volume change due to a bigger acceleration. The resultant is just how much better you are over your opponent and this will show by the massive scoreline difference (acceleration) after the game. If the system is quasi-static the internal pressure is only infinitismally larger than the external pressure which means the resultant is infinitismally small which means the acceleration is really tiny, which means you were only slightly better than your opponent because the scoreline was almost tied. Your work, however, is still how much you had to put in to match your opponent therefore the work is based on the external pressure just as work against friction is based on the frictional force and work against an incline is based on the resisting forces of inclination because this is what you need to overcome. In free expansion a gas expands without an opposing external pressure. Even if the original internal gas pressure was 2 or 10 atm, the work is still zero because no opposing force is present. If your opponent doesn't show you cannot call it work. Work is a measure of how much hell someone gives you before you win. So you use their measure not yours. This is why we use external pressure. Resultant force doesn't indicate work, it indicates how easy or hard you overcame that work (or resistance) as measured by an acceleration. In ideal gas behaviour a piston is considered massless therefore resistance by inertia of the piston is Zero. The only resistance is therefore the external pressure and thus this is what you must work against.

The derivation of pressure-volume work is as follows, in reverse:
$W = P\Delta V$
$P = F/A$
$\Delta V = A\Delta d$
Subbing those last two into the first:
$W = (F/A) (A\Delta d)$
Or:
$W = F \Delta d$ which is the definition of work.
So, the work done by the gas in the cylinder is indeed independent of the outside pressure, it is a function of the force exerted by the gas and the displacement of the piston. But, the work done on the piston is different. It is the sum of the work done by the gas inside and that done by the gas outside. The gas outside is exerting a force inward while the distance moved by the piston is outward so that work is negative compared to the positive work done by the gas inside the cylinder. So the gretaer the difference between inside and outside pressure, the greater the net work done on the piston. But still, the work done by only the inside gas on the piston is a function of its pressure and the $\Delta V$.

This is even my confusion that in physics when we write the work done by gas we use its pressure and we don't calculate the work done in irreversible processes because in such processes pressure of the gas at any moment is not defined but in chemistry when there is irreversible process we actually find the the work done by the surrounding on the gas which we take as "-Pext*dV" (-sign is taken to get the right result, you can derive it taking initial volume as the origin and Final volume as end you will get the same result) and reverse its sign to find the work done by gas, as use it in FLOT, for e.g, if there is irreversible compression lets say work done by surrounding on the gas is "+W" i.e W amount of energy is transferred from the surrounding to the system so , it also means that system must also have done -W amount of work on the surrounding to receive W amount of energy  from the surrounding..as there is no other thing from which energy can be exchanged. ...I don't know why we don't do this in physics like we do to find work done in irreversible processes in chemistry, but I think this is not wrong even in irreversible process you cannot calculate work done by gas directly by you can calculate how much you are supplying and it will do the same amount of work on you but of opposite sign to take it from you.

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Why does iodine get oxidized to iodic acid and not periodic acid by nitric acid?
Nitric acid oxidizes iodine to iodic acid, $\ce{HIO3}$ and not periodic acid, $\ce{HIO4}$, which is of a higher oxidation state. This happens although other nonmetals are oxidized to their highest oxidation state. How can this be explained?

Can u pls give a balanced chemical equation for the reaction of I2 with conc HNO3.

@Yashwini Sorry. This was a long time back. I am majoring in a completely different field for more than three years

Nitric acid is not strongly oxidizing enough to get iodine to +7.  Relativistic effects tend to make the $s$ subshell valence electrons more stable than the others in heavy atoms1. So to some extent, iodine acts as if it has only five valence electrons. To pull out those last two electrons and get periodic acid, electrochemical oxidation is commonly used with a very high oxidation potential.
References

Relativistic Effects in Chemistry: More Common Than You Thought,
Pekka Pyykkö, Annual Review of Physical Chemistry 2012 63:1, 45-64 (link)
https://en.wikipedia.org/wiki/Periodic_acid

The equation is : $\ce{3 I2 + 10 HNO3 -> 6 HIO3 + 10 NO + 2 H2O}$

Nitric acid isn't strong enough, but ozone is, isn't it? Still, it doesn't oxidises it to periodic acid... can you explain why?

It may be easier to get iodine(VII) by using an alkaline medium.

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Determining whether something is an acid, base, or amphoteric
I have to find whether these compounds are acids, bases, or amphoteric: $\ce{CO3^{-2}}$, $\ce{HPO4^{2-}}$, $\ce{H3O^{+}}$ and $\ce{NH4^{+}}$.
I know the first can obviously act as a base, but others can gain or lose $\ce{H^{+}}$, so I think they are amphoteric. However, according to my book, I'm wrong. Can you help me clear my doubt? Thank you very much!

Can $\ce{H3O+}$ and $\ce{NH4+}$ really accept another $\ce{H+}$? What would be the molecular formula for those compounds?

An amphoteric substance can act both as an acid and a base. That is, an amphoteric substance can donate protons and also accept protons.
$\ce{CO3^{-2}}$ does not have any protons so it cannot act as an acid. It is the conjugate base of $\ce{HCO3^{-}}$.
$\ce{HPO4^-{2}}$ is the conjugate base of $\ce{H2PO4^{-}}$. It can accept a proton to go back to $\ce{H2PO4^{-}}$ or it can donate a proton to become $\ce{PO4^{-3}}$. Thus, it is amphoteric.
$\ce{H3O^{+}}$ is the conjugate acid of $\ce{H2O}$. It can donate a proton to go back to $\ce{H2O}$ but it cannot accept any other protons. Thus, it not amphoteric.
$\ce{NH4^{+}}$ is similar to $\ce{H2O}$. In this case it is the conjugate acid of $\ce{NH3}$. As such it can donate a proton to go back to $\ce{NH3}$ but it cannot accept any protons. It is not amphoteric.
So, out of your list only $\ce{HPO4^-{2}}$ is apmhoteric.


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When, how & why do atoms become stable. If they're always bonding, do pure elements even exist in the real world? What happens to unstable atoms?
I really can't get my head around the fact that an element has to always become stable by bonding.
Doesn't that mean that you can't actually find a pure element, and that it doesn't exist in it's true form? Because it will always be bonding? Does it mean that we are always looking at bonded atoms in the real world, and the actual true form doesn't exist?
And what happens if there is, say, one atom left out and no other atoms left to bond with. Will it explode or die because it is unstable?
EXAMPLE: chlorine forms ionic bonds to become Chloride. But if it needs at always do that, doesn't it mean that we are always dealing with chloride? How can we ever have just chlorine when it always turns into chloride? And how long does it take to bond?
If what I've said is true (though I doubt it), then shouldn't we just study it in it's actual/final bonded form and have the bonded forms on the periodic table, rather than the one that hardly exists?
I've started to study for GCSE Chemistry but can't go further than the first chapter of ionic bonds etc. because this question keeps bothering me too much and I've become unable to study without knowing the reason behind this, and the consequence of atoms not bonding.
I'm sorry if it's long but it feels like I'll go crazy and can't thank anyone ENOUGH who is able to answer this question. I can't find it anywhere on the internet :'(

Your statement is not correct. Elements do not always become stable by bonding. All the noble gases (last group in the periodic table) do not usually form any compounds. You can get silver, gold, iron and a lot of other metals in their elemental form. You can find clorine gas as well.

Atoms can form bonds with same type of atoms, therefore biatomic molecules (halogens, oxygen, nitrogen..), molecular solids (sulfur, phosphorus,..), many metals are stable under ambient condition in pure elemental form.

What is this "actual true form" of an element that you are speaking of? If you're hoping to go around finding carbon atoms, oxygen atoms, ... you are out of luck, apart from the noble gases.
However, carbon atoms can bond to themselves, and the result is usually called graphite or diamond, which is pure carbon. Granted they aren't individual carbon atoms, but they're still pure carbon because all the atoms inside are carbon. $\ce{O2}$ is pure oxygen and is 20% of the air you are breathing right now. It isn't individual oxygen atoms, but both atoms are still oxygen, which makes it pure oxygen. Is that a sufficiently "true form" for you? If not, too bad, because to be honest: the oxygen atoms out there don't really care what you think. They're going to adopt their most stable form, and that involves bonding with other atoms. (Again noble gases being an exception.)
If there's one lone atom in the middle of nowhere, then it just sits there by itself (unless it's radioactive, which is another matter entirely). So, a better way of looking at it would be that atoms themselves are already stable, but they could be even more stable if they form bonds with each other. Atoms aren't alive and so they can't die, and it's not going to get depressed because it's lonely. Don't anthropomorphise atoms!
Anyway, it's called the "Periodic Table of the Elements" and the element is chlorine, not chloride. Why? Well, chemistry isn't restricted to how you find an element in real life. Just because most of the chlorine on Earth exists as chloride doesn't mean that that's all you can do with it. You can jolly well turn it into chlorine gas if you like (and tons of other things). Chemistry is really all about figuring out what you can turn it into. And what you can turn it into is decided by what the element is, not the form in which you find it naturally.
You're asking, shouldn't all chlorine just turn into chloride the moment you make it? Well, in whatever book you have been studying from, you should have noticed that chloride has a partner, for example a sodium cation $\ce{Na+}$. That's because chlorine took an electron from sodium to produce $\ce{Cl-}$ and $\ce{Na+}$.
Now if you make your chlorine gas, keep it in a glass container, and not let it go anywhere near sodium, where is it going to get that electron from?

You will not see elements in their pure form- most of the time. Let's take the case of sodium, naturally it will always be found bonded to something else, oxygen, chlorine, fluorine, etc. Even if you buy sodium metal it usually comes in a bottle of oil because exposure to air will rapidly oxidize it in just a few seconds. It will also readily bond to water in the air which is very energetic.
If you have a compound of sodium chloride, you could separate the two by heating them to extreme temperatures, enough to break the bonds between the two. You would be left with ionized chlorine and sodium gas in this case. One electron from the sodium would be left with the chlorine.
An exception is the noble gasses, they are an exception. They don't tend to bond to anything so they do exist in elemental form, but for that same
reason we never studied much about them in chemistry. Also metals can be found in elemental form, but they form metallic bonds so I'm not sure if that would count as it's "true form" to you.
Now for the "if one atom is left out" part. Let's consider a box full of hydrogen and oxygen in almost perfect proportion to form water with no limiting reagent, except there's a few atoms short of hydrogen, what will happen to the leftover oxygen that has nothing to bond with? It would simply bond with the other oxygen atoms to make O$_2$. Okay so what if there's just a single oxygen atom left with nothing to bond to? I don't know exactly what a single oxygen atom would do in this case since this isn't a realistic situation nor is it relevant, but I can tell you it won't explode. In the real world bonds would be breaking and forming all the time in this system, I could go a lot more into detail about this but to keep it simple and brief, it would probably end up like this: $H_2O + O^{2-} \rightarrow 2OH^-$
So yes, you are right, most of the time you won't come across elements in their elemental form. Atoms like to bond to become more stable. In my experience, early on elements are studied in elemental form to learn the mechanisms of how atoms bond. But as I got into more advanced topics of chemistry, we would move away from that and study compounds the same way they would naturally exist in the real world. Hope this helped, let me know if I left anything out.

I see elemental oxygen, nitrogen.. every day.


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Quantitative information from silica gel TLC plate?
I have a general question about thin-layer chromatography using silica gel.
Can we get quantitative information from a TLC analysis? Is this mainly just the Rf or can we tell more? I'm guessing purities, the number of compounds, and identity are qualitative aspects.

read this and see if it helps with your question. http://www.chemguide.co.uk/analysis/chromatography/thinlayer.html


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When 3-chloro-2,2-dimethylbutane reacts with water what are all of the products and followup questions
I would like to know what all the products are. I know that in the classical Sn1 reaction, we have 3,3-dimethyl-2-butanol. For the alkyl shift in a Sn1 reaction, we also have 2,3-dimethyl-1-butene and 2,3-dimethyl-2-butene.
I have other follow up questions: This is my first time learning about organic chemistry as a layperson. I have no access to university resources like equipment. How do chemists separate all of this in a lab setting?
Also, I'm curious about these alkyl shifts. How do you explain them? From what I understand when the carbon is attached to four other carbons and it is adjacent to a carbocation, we can have the methyl group move over to the adjacent carbocation. This is used when a hydride shift is not favorable due to higher order carbocations being more favorable.


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Random order of melting points of group 2 elements

Compare the melting points of $\ce{BeF2, MgF2, CaF2, SrF2, BaF2}$.

Actual order is  $\ce{ BeF2 > CaF2 > SrF2 > BaF2 > MgF2}$.
I have worked out the following things while comparing their melting points.
a) Melting points of ionic compound increases as the ionic interactions between cation and anion in the metal lattice increase. In this case, smaller cations will result in stronger interactions. That gives this order :  $\ce{ BeF2 > MgF2 > CaF2 > SrF2 > BaF2}$.
b) None of these compounds is a covalent solid. If it were, its melting point would have been higher.
c)The order of melting point also depends upon the crystal lattice structure.
(i) Be and Mg crystallise in hcp, Ca and Sr in fcc and Ba in bcc.
(ii) The cations will occupy the lattice points and since fluoride ions are smaller than cations, they will occupy the voids.
(iii) The more is the coordination number, the more will be repulsions between similar ions. In the lattice, when two cations are in contact, there will be repulsions. But, the repulsions will be less when the size of carion will be small since in that case, the electron cloud of one cation will experience nuclear attraction by the other cation.
(iv) Hcp and fcc have the same coordination number.
(v) Based on this logic, the order should be:
         $\ce{ SrF2 > CaF2 > MgF2 > BeF2 }$.
(vi) Although, the coordination number of Ba is very less, it size is
          very big. So, we cannot say with certainty if it would have the
          lowest melting point.
Now, I don't know how to reconcile these two points, the factors are totally opposing. Where to put Barium? This made me think that the order cannot be predicted with certainty just based upon theoretical analysis of their lattice structures.
It is important for me to know whether the points that I have mentioned correct or not. Are there more factors that could influence their melting points? And finally, if the order can be predicted with certainty purely on theoretical basis, how would that be done?

So are you asking about the elements or are you asking about the fluorides?

The fluorides...I understand that the fluorine ions will just occupy the voids made by the metal lattice,..so how does it matter if we analyze the crystal structure of the metal lattice or the metal fluorides. Correct me if I am wrong

related http://chemistry.stackexchange.com/questions/53447/melting-and-boiling-point-trend-in-group-ii

Your assumption that "that the fluorine ions will just occupy the voids made by the metal lattice" seems not to be correct to me.

@aventurin Correct me.


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Why does oxygen atom allow positive charge on its surface when its electron pair is in resonance with benzene ring?
Oxygen atom is a highly electronegative element.
Why does oxygen atom allow positive charge on its surface when its electron pair is in resonance with benzene ring?
The reason can be that its octet is complete so oxygen is not disturbed by its positive charge.

Oxygen having a positive formal charge is not all that uncommon: see $\ce{H3O+}$.

@orthocresol  but also it is not that stable as it dissociates and gives H+ ion , chlorine atom also bears positive charge in chloro benzene resonance (+M) and anisole also shows +M effect.


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Variation of compressiblity factor with temperature

While I certainly understand the order of temperatures, I can't find a reason for the curves to intersect at one common point.
Why do the curves intersect at one point? or do they really intersect at a common point or is it a coincidence that they are intersecting in this diagram?

It is plausible that the isotherms intersect, it just means that the ratio of the real volume to the ideal volume are the same at those conditions; the ratios can be the same while the absolute values are different.

two lines intersecting is understandable but why do all the lines intersect at one point? I am trying to reason out why all the lines intersect at one common point.

Why all the isotherms intersect would require a more detailed analysis. Maybe for that system, that particular pressure could be the critical pressure so some fixed behavior of the system is expected. I happen think the intersections are coincidental given the information presented.

Remember the mathematical definition of $Z$:
$$ Z = \frac{pV}{nRT}$$
Given all the $Z$ graphs I've ever seen -- the single-point interaction of all these isotherms is just coincidence (and kind of misleading). There is nothing in thermodynamics that suggests for a given gas, there exists a $p$ for which $Z$ is constant, regardless of temperature. Even for constant $V$ and $n$, you have that $Z \propto \frac{p}{T}$, so changing $T$ can in no way leave $Z$ constant.
You can find graphs that look similar even for different gases at the same $T$:

Nonetheless, there is no reason for a given material to have the same $Z$ at a given $p$ regardless of temperature:

Generally speaking, if you zoom-out far enough, and have significantly different temperatures (like 100s of Kelvin apart), some $Z$ graphs may appear to have a uniform $(p,Z)$ point, but in reality the intersections are not uniform for any physical/mathematical reason. For this reason, graphs like that can be a little misleading.

There is another graph in my textbook titled "Variation of Z with different pressure at different temperature" where at $P=0$, $Z$ is not equal to 1 for all the temperatures. Is that possible?

What's the gas?

The textbook does not mention any specific gas.

$Z$=0 at $p$=0, what you see near $p=0$ is really just very low pressures. I suppose for some gases at certain temperatures this is possible but not common, because generally, low pressure lowers potential energy to ~0, which is what an I.G. can be described as.


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How can I find the eutectic point of gallium and aluminum?
I've been interested in liquid metals for a while, but I've stayed away from mercury for obvious health reasons. I have a sample of gallium and I've seen a liquid metal mixture of gallium and aluminum. I'd like to make this myself, and I want to achieve the lowest melting point I can, but I can't find much information on the eutectic point of these metals. Is there a way I can calculate it?

You'd need some nice software to find  theoretic values for eutectic point

This article (http://iopscience.iop.org/article/10.1088/0026-1394/45/1/011/meta;jsessionid=515C0CB11B5E1C9C968249B25D19677F.c4.iopscience.cld.iop.org) measures 26°C and cites 1.67% (from Lyakishev N 1999 Phase Diagrams of Bi-Metal Systems
(Moscow: Russian Academy of Science) (in Russian)) as the eutectic for Ga-Al system.

The Al-Ga system is reasonably well studied, and there are a number of articles in the Calphad (Calculation of Phase Diagrams) literature. For example, Al-Ga-Zn System: Reassessments of the Three Binary Systems and Discussion on Possible Estimations and on Optimisation of the Ternary System, M. Mathon et al., CALPHAD 24(3) 253-284 (2000) has the thermodynamic analysis yielding the fairly simple model for the Al-Ga system. These parameters may be used in one of the many phase diagram software packages, such as ThermoCalc. Or, one could write a simple simulator in Python, yielding something like:

The eutectic is way over on the Ga side. The paper listed above also references a number of experimental papers with measured eutectic points.
For other systems of interest, one easy place to start is the ASM Alloy Phase Diagram Database, assuming your institution has access. The phase diagram entries have a set of references, often containing both experiment and calculation papers. If you have access to Calphad, Journal of Alloys and Compounds, and Journal of Phase Equilibria and Diffusion, those are good places to search for a given binary or ternary system.


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What is the specific functional group in MDMA that caused addictive effect on user?
First of all, the reason I'm asking this is because I'd just done an experiment detecting MDMA in a urine sample with GC-MS for my internship program. I'm having difficulties figuring the functional group which cause the said effect as in my question. I've read two journal Shulgin and Nichols and I thought they explored more of the hallucinogen effect from N-methylation.

I'm voting to close this question as off-topic because it would be much better suited for Biology.SE, and is a very poor fit for Chemistry.SE.

Firstly, MDMA is widely regarded not to be physically addictive, which simply means that the compound (actually the absence thereof) will not cause potentially life-threatening symptoms on discontinuation, as is the case with alcohol or morphine. Note however that this does not make any statement on the addictive potential of the substance, as e.g. cocaine also falls into this category of substances, which are called psychologically addictive.
In general psychological addiction (which is not necissarily associated to substances, but can also involve habits such as gambling or eating) is widely believed to be strongly associated with the dopamine system and the overstimulation that dopamine facilitates during the addictive actions.
Which brings me to the second point:
The active site of the molecule. In a way, looking at both structures you will notice a set of similarity between dopamine and MDMA. It is not that simple however. The effects of MDMA are quite complicated and due to its status as illegal substance research is kind of limited.
Known effects are:

Known and experimentally confirmed increase in dopamine levels, although the mechanism is not known.
Serotonine release from vesicles to synaptic cleft (has effects on dopamine levels through mediation/coupling)
MAO inhibition (decreases dopamine and serotonine degradiation - among others)

Other effects like oxytocin response are also proposed and may also have effect on dopamine levels. But in this case the effect could as well be a down-regulation.
Which structural component is responsible for all of this? All of them, as you can tell from the complexity of the physiological reaction on the substance, especially if compared to similar compounds with in some regards drastically different physiological effects. Think of (meth-)amphetamine(very high dopamine response, next to no serotonin effects) or mescaline(direct agonist on specific serotonin and dopamine receptors, as well as dopaminergic effects). (It would really blow the reach of this question, but taking a look at other compounds, such as 2-CB or 3,4-CTMP will make this picture even more complicated, but also more complete...)
These substances can give you at least some hints at two honorable mentions in terms of 'addiction-causing functional group':

(N-Me-)Phenethylamine
3,4-(+M)-substitution pattern on the aromatic ring

A small sidenote on the difference between physical and psychological addiction: I am not 100% sure about the current research in that topic, but afaik in both cases there is in overexpression of receptors that causes the addiction, only that in the case of a psychological addiction the effects of the receptor-agonist is not live-threatening. In that sense the terminology which kind of suggests that psychological addiction is not associated with any changes in brain structure is somewhat misleading.

It's not really any particular functional group that causes addiction. For example, the only functional group in ethanol is the alcohol. While mdma has a methylene dioxy group and an alkyl amine.
Both of the above mentioned drugs are addictive. And their addictiveness can be associated to biochemical changes in the body. Such as the over/under expression of certain genes.


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Inability of alkyl fluorides to be reduced to an alkane in the following reaction
Alkyl halides except fluorides on reduction with $\ce{Zn}$ and dil. $\ce{HCl}$ gives alkane. Example:
$$\ce{CH3Cl + H2 -> CH4 + HCl}$$
Why cannot alkyl fluorides be reduced?

Other answers may be more elegant, but it boils down to the carbon-fluorine bond being too strong, in fact stronger than carbon-carbon or carbon-hydrogen bonds.  The required bond breaking simply cannot be done under normal conditions.  Similarly Grignard reagents (formed with a much more powerful reducing agent) are much less common with fluorine than with heavier halogens, and the bad things chlorofluorocarbons do to our ozone layer come from breaking $\ce{C-Cl}$ rather rhan $\ce{C-F}$ bonds.


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Why does the crystal sublime more quickly at the corners where there is a higher vapour gradient?

The crystal responds to the larger vapor gradients near the edges by
  sublimating faster there and thus rounding the crystal.
J. Nelson, J. Atmos. Sci., 1998, 55, 910-119.

The change in shape according to the gradient makes intuitive sense to me but I can't for the life of me figure out why it does this. Maybe I'm being really foolish here, but surely a higher concentration of vapour at the corners would decrease the rate of sublimation there.

The diagrams here are actually super helpful in answering your question. Thank you for including those figures.

The vapor gradient reflects the fact there is much more space at the corner than in the middle of a face. The vapor builds up near the surface but because there is much more space near a corner, the vapor pressure drops off, where as the vapor pressure near a point on the middle of a face is reinforced by sublimation at points near it.


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Different types of carbides
I am confused between different types of carbides such as
$\ce{CaC2}$, $\ce{Mg2C3}$, $\ce{Al4C3}$, $\ce{BaC2}$ or $\ce{BeC2}$.
And if they are different, why they are called carbides?

Could you expand a little? Is there anything in particular associated with this set of carbides that is causing your confusion? Otherwise I have to admit I can't really see what you are asking - They are called carbides because they are a binary compound of carbon and a more electropositive element, but I feel that this is not really what you want to know

@IanBush actually I want to know whic carbide on hydrolysis produces which product

Strongly related: http://chemistry.stackexchange.com/questions/34099/why-do-different-elements-form-different-types-of-carbides

BaC2 and CaC2  on hydrolysis gives C2H2    Mg2C3 givesC3H4   Al4C3 and Be2C gives CH4

Carbides are a specific class of carbon-containing compound; there is not a single compound called "carbide".  According to this Wikipedia article:

In chemistry, a carbide is a compound composed of carbon and a less
  electronegative element. Carbides can be generally classified by
  chemical bonding type as follows: (i) salt-like, (ii) covalent
  compounds, (iii) interstitial compounds, and (iv) "intermediate"
  transition metal carbides. Examples include calcium carbide
  ($\ce{CaC2}$), silicon carbide ($\ce{SiC}$), tungsten carbide
  ($\ce{WC}$) (often called simply carbide when referring to machine
  tooling), and cementite ($\ce{Fe3C}$), each used in key industrial
  applications. The naming of ionic carbides is not systematic.

How to know which carbide on hydrolysis produces which product

@hey You should edit your question to actually ask that.

@pentavalentcarbon so what should I do now

@hey , I agree with pentavalentcarbon that your comment is a separate question. I think a cleaner solution than editing the question at this point is to follow the guidelines for an answered question (it's your question, you get to choose what to do with an answer), then just ask a separate separate question. Be sure that you include all other details you would like to know regarding the topic so that you don't have to keep asking related questions. Also, if you edit your question, my answer becomes deficient unless I delete/modify it.


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Niobium and (no) aqueous chemistry?
When I was doing my Bachelor thesis I worked a lot in Niobium ampoules as we were doing solid state experiments at high temperatures with a lot of fluoride. As this attacks glass we chose Niobium instead and it seems to be quite unreactive even at those high temperatures (working in a helium atmosphere to prevent oxidation).
So when I was finished with my work I descided to try some Niobium salts only to discover that there is no real aqueous chemistry of Niobium, Tantalum and a few other elements in the PSE. They don't form any Nb(III)-ions in solution besides complex-anions for example and certainly no real hexaqua-complex. Vandium above however and also Dubnium below (at lower pH values) show a lot of aqueous chemistry and also quite often pure aqua-complexes.
This surprised me a bit first as nearly all reactions involving Nb, Ta and a few others (I think W, too) usually end with the Oxide (pentoxide) which is super stable. So I though a bit about that and came up with another example. Not only does this resemble the chemistry of Polyvanadates like the reddish decavanadates and all the stuff that forms if you acidify ammonium metavanadate but also the reactions of silicic acid which, as you try to protonate silicates often form polysilicates and SiO2.
Then I remembered the deep red colores of polychromates. If you acidify potassium chromate with HCl which is 70% water you will get the dichromate solution which is orange if you use nitric acid which has 30% of water you will get the trichromates, tetrachromates and higher, red polychromates. And if you use conc. sulfuric acid you will recieve the trioxide. I assumed for a long time that it was a simple dehydration where a H2O will be taken from a H2CrO4 to form CrO3 but then I looked at the structure of CrO3 only to discover that it is a long chain of joined tetrahedra so a high polychromate.
If you look at the structure of Niobium(V)oxide it's pretty much the same only with octahedra. So I imagine a "niobate" as it can be dissolved in basic environments will do the same elimination of water to form long chains or networks which are then called oxides. And in this case those niobates and tantalates are so basic that anything pH 8 (or whatever that point is) is considered a strong acid for it (much into comparison of chromate for example) and this is why a pH7 aqueous chemistry of Niobium or Tantalum can't work.
If this was the case I am not sure why it's exactly in the middle with those few elements. I imagine if a metal is quite electronegative and has a short bond towards the oxygen it will pull the oxygen so close that the positive charge and the hydrogen influence each other (much like with the chemistry of Beryllium) and it will be easer to deprotonate it. So in my case the lower electronegativity of the Nb and Ta in comparison to the V(III) should increase the electron density on the oxygen more making it more basic.
And if I compare the electronegativities those elements are some of the least electronegative ones (there isn't much Sc chemistry known and I think Ti, Zr and Hf don't have that much aqueous chemistry either).
But is there another maybe better explanation to why only those few elements show pretty much now real aqueous chemistry at all ?


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Why is a silent electric discharge used for the preparation of ozone from oxygen?
I learnt that ozone is prepared by passing oxygen through a 'silent' electric discharge. Can someone please explain what is meant by a 'silent' discharge here, and why that helps? I couldn't get a satisfactory answer while Googling.

Dielectric barrier discharge did not help?

Ok, It says what the discharge is, but why is it used for ozone prep?

According to this Wikipedia article:

Dielectric-barrier discharge (DBD) is the electrical discharge between two electrodes separated by an insulating dielectric barrier. Originally called silent (inaudible) discharge and also known as ozone production discharge or partial discharge, it was first reported by Ernst Werner von Siemens in 1857. On right, the schematic diagram shows a typical construction of a DBD wherein one of the two electrodes is covered with a dielectric barrier material. The lines between the dielectric and the electrode are representative of the discharge filaments, which are normally visible to the naked eye.

DBDs can be used to generate optical radiation by the relaxation of excited species in the plasma. The main application here is the generation of UV-radiation. Those excimer ultraviolet lamps can produce light with short wavelengths which can be used to produce ozone in industrial scales. Ozone is still used extensively in industrial air and water treatment. Early 19th-century attempts at commercial nitric acid and ammonia production used DBDs as several nitrogen-oxygen compounds are generated as discharge products.

As the Gibbs energy (G) for the reaction is highly positive , it is made sure that ozone does not decompose back to give oxygen and nascent oxygen and that's why silent electric discharge is used.


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Diatomic and Monoatomic
Why do we represent the molecule of solid metals using the symbol of the element itself(for example ,molecule of iron=$\ce{Fe}$,etc.) whereas we represent the molecule of the elements like oxygen in a different way(we represent molecule of oxygen as $\ce{O2}$,not $\ce{O}$)?
I think that we do so because the metals can exist in atmosphere as single atoms whereas nonmetals do not.
Am I correct?
If I am correct ,then why is it such that metals can exist as single atoms whereas nonmetals need to form bonds in order to exist in the atmosphere?

When we refer to a metal, we refer to the bulk solid, not to a molecule. How many iron atoms are there in a block of iron? If metals do form molecules, we will represent them as such, for example with dilithium, $\ce{Li2}$ (no, not the Star Trek variety).


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Different types of oxide

Among the following oxides in order amphoteric, Basic, Acidic, Neutral
  and mixed oxides respectively:
$\ce{B2O3, SnO2, CO, Pb3O4, CaO}$

I know metals are basic and non metals are acidic .
But  how can we compare??

Compare to what?

@NilayGhosh comparing among themselves

Well starting off with the easiest:

$\ce{CaO}$ is basic, because simple alkali and alkali earth oxides form Hydroxides in water, so they deprotonate the $\ce{H2O}$ to form $\ce{OH-}$ in solution which is a base then. If you follow the rows in the PSE the oxides will change from basic to acidic like most none-metal oxides will be acidic or better they will form acids like $\ce{SO3}$ would form $\ce{H2SO4}$ with water which then behaves like an acid. For the earlier elements in the rows so alkali and alkali earth due to the big difference in electronegativity it's an ionic compound so we are talking about Calcium-cations and Oxide-anions here. But the strongest base in water is $\ce{OH-}$ also called a solvobase (which is the strongest base there is in a solvent). Oxide is a deprotonated $\ce{OH-}$ so it would be stronger than $\ce{OH-}$ thus it will deprotonate the water every time. So all of these ionic oxides tend to be basic.
$\ce{CO}$ is a neutral one. $\ce{CO}$ could be seen as the anhydride of formic acid, $\ce{HCOOH}$ , take one $\ce{H2O}$ away and you will end with $\ce{CO}$ still $\ce{CO}$ and $\ce{H2O}$ will not form formic acid without a catalysts and it has no reaction at all with water thus being neutral.
$\ce{Pb3O4}$ is a mixed oxide you can easily see this if you compare the oxidation states. If this was a regular oxide $\ce{Pb}$ would be +IV and Oxide would be -III but Oxide only exists as -II so this has to be a different case. We could double it to yield $\ce{Pb6O8}$ then we could have 2 $\ce{Pb(+IV)}$. This isn't the case however and its a mixture of a plumbate which exists as one unit $\ce{[PbO4]{4-}}$ with a $\ce{Pb(+IV)}$ in the center and the counter ions are 2 $\ce{Pb(+II)}$, so a $\ce{Pb2[PbO4]}$ which is $\ce{Pb3O4}$.
$\ce{B2O3}$: Boron oxide is pretty much the same as with boric acid, although we can write it as a $\ce{B(OH)3}$ it is an acid as we are approaching the non metals now.
$\ce{SnO2}$: had to look that one up but it is amphoteric forming either stannic acids or stannates.

I hope I explained most of that stuff correctly though.


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The sensitivity of potassium iodide starch paper
I have some questions about the sensitivity of potassium iodide starch paper.
Patients with asthma caused by allergic airway inflammation have high levels of nitric oxide ($\ce{NO}$) in exhaled breath. This is usually measured with chemiluminescence analyzers by measuring the light generated by a chemical reaction of $\ce{NO}$ with ozone. This technique has high sensitivity, and NO can be accurately measured in the parts per billion (ppb) range.
This above method requires a complicated apparatus.
I am now considering an alternative method.
When NO is exposed to oxygen, it is converted into nitrogen dioxide:
$$
\ce{NO + O2 \rightarrow 2 NO2}
$$
When wet potassium iodide starch paper is exposed to an oxidizing gas, such as $\ce{NO2}$, its color changes to purple or blue. The intensity of the color depends on the concentration of the oxidizer.
My questions are:

If, at room temperature, 1 litre of exhaled air is enclosed in a container, inside of which a wet potassium iodide starch paper is placed, would that paper ever get a blue tint if the NO concentration in the exhaled air is only 25 ppb (parts per billion)?
How long would it take?
Would it help if the volume was bigger, for example 10 litres?
Could one improve the sensitivity in some simple way?

If nitric oxide and oxygen reacted quickly there wouldn't be any nitric oxide in an exhaled breath. The body doesn't get all the oxygen. // Is just a qualitative test sufficient? I'd guess quantitative data is wanted.

MaxW: If I understand correctly, the chemiluminescence analyzers depend on a quick flow of exhaled air so that the NO2 is freshly "produced" in the apparatus (by contact between NO and O3, ozone). Some of the freshly produced NO2 then, by changing exitation state, emitts light, which is measured. In my method though, it does not matter if the NO2 is old or not. Regarding quantitativ/qualitative: I would like it to be somewhat quantitative, but high resolution is not required.

MaxW: I should perhaps add that my idea with the container method I describe, is that NO will eventually be converted to NO2 (by contact with the oxygen in the air), and that this NO2 then reacts with the wet paper.

At link it says: At room temperature, NO combines with molecular oxygen to form NO2 at a rate that is dependent on the oxygen concentration and the square of the NO concentration. Therefore, oxidation is extremely slow at very low concentrations. This finding has important consequences for the pathophysiological and pharmacological relevance of NO in the respiratory tract. So my method will not work well becuase of the extremely low concentrations. That could perhaps be fixed by oxidizing with potassium permanganate.

Correct. The point being that ozone is an extremely powerful oxidizer and will react very quickly with NO.


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What is the purpose of the Carothers equation?
I don't really understand the equation. I feel like it is simply demonstrating the kind of exponentiality characteristic for the step polymerization reactions, but it cannot be useful for actual calculations. I mean - according to the equation a system which has executed, let's say, a polymerization reaction with 99% conversion will have an average degree of polymerization of 100 even if the system has fewer than 100 monomers. Am I missing something?

It is an equation useful for describing the behaviour of a large system, not predicting the fate of a few monomers.
Polymerization is a stochastic process—if we only have 80 monomers, the resulting polymer molecules will vary wildly in length if we perform the experiment several times (and logically, could not ever have an average degree of polymerization of 100), and the equation could not provide much insight into the results.
However, 80 monomers is a very tiny quantity.
If we take one of the monomers used to make PET, terephthalic acid, at 166.13 g/mol works out to around $2\times 10^{-23}\ \textrm{kg}$ for 80 molecules, a quantity that is almost impossible to actually measure out so it doesn't really matter that the equation doesn't work.
At a more practical scale, with even 1 g of the monomer, we have ~$10^{21}$ monomers and we will see the average behaviour of the system that the equation describes.


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How do I evaluate the PV term when calculating a change in enthalpy?
As we know enthalpy is defined by the equation $H = E+PV$.
Now to calculate this $PV$ factor for $\Delta H$, Sometime we take $P \Delta V$ and sometime $P_2V_2-P_1V_1$. Why are there so many different approaches?
I am also confused whether $P$ is external pressure or internal pressure.

Hello and welcome to Chemistry.SE! I suggest taking the short [tour] to better familiarize yourself with the site. Regarding you question, it seems a bit unclear what you are asking. I think that giving some examples regarding the different approaches you pointed out would clarify things a bit, and might even lead you to figuring out your question. Best of luck!

The derivation for $\Delta H$ is
$$H = U + PV$$
$$\Delta H = \Delta U + \Delta (PV)$$
The $\Delta (PV)$ term expands to your $P_2 V_2 - P_1 V_1$ term. However, when the pressure is constant $P_2 = P_1 = P$, and $\Delta (PV)$ term becomes $P\Delta V$. The most common situation when the latter is true is when the system is open to the atmosphere and $P$ is the external pressure. In the former case, $P$ is more likely the internal pressure, which changes during the reaction.

If a gas is compressed from state variable p1,v1,t1 to p2,v2,t2 under a constant pressure p3, how will you find change in enthalpy?


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What happens if I increase the amount of water on water based epoxies?
I am using a water based epoxy coating that is composed of a resin and a hardener. There is a specific amount of proportion to use to obtain the optimum cured coating. But I am trying to introduce some dry pigments into the resin then I add the hardener. However adding the pigment increases the viscosity of the mixture and the obtained coating is thicker.
Since I cannot increase the amount of resin or hardener without influencing the optimum condition, I was thinking since it is a water based coating, I could still have the same amount of functional reacting groups and just increase the amount of water in the resin to make the mixing of pigment with the resin easier ?!
But I am afraid during the drying process, the evaporation of the extra added water might influence my coating !
Do you know what could happen in this case ?

This article seems to have some potentially useful information about varying levels of water in a water-based epoxy.


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Molecular orbitals of allylic cations, anions and radicals
I was reading the explanations of electronic configurations of the Allylic Radicals, Cations and Anions in my textbook. I completely understand everything, but there is one thing that confuses me. Namely, it is said that the  antibonding pi3* orbitals of allylic cations, radicals and anions have no electrons. Now my questions is, where are they? Regardless of the type of bonding, there should be 2 pi electrons in an allylic cation, 3 pi electrons in allylic radical, and 4 pi electrons in allylic anion. So the electrons should be somewhere. For example I would think that the allylic radical of pi3* is a molecule with 1 pi electron in each carbon, but because of the absence of orbital overlap they are not shared between the carbons, whereas in the case of pi2 nonbonding they are partially shared and fully shared in pi1 bonding orbitals.

In allylic species 3 orbital wave functions are interfering to form 3 molecular orbitals among which the pi antibonding molecular orbital will have the max energy. The first two mo can accomodate 4 electrons. Since pi antibonding are at a higher energy level the first 4 electrons will remain in first two mo only. Similarly in cation and radical electrons will remain in these orbitals hence the pi antibonding will always remain empty.


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How do I estimate ΔG⁰f for Fe2+?
How would one use the equation ΔG° = −nFE° to estimate the ΔG⁰f of Fe2+(aq)? Since Fe2+ appears in multiple entries in the following table, I'm not sure which value to use.

The Gibbs free energy of formation of $\ce{Fe^2+ (aq)}$ would be given by the equation
$$\ce{Fe (s) -> Fe^2+ (aq) + 2e-}$$
since iron in its reference state is a solid. Therefore, this should already point you towards the correct half-equation to use.
It is true that the electrode potentials given are relative, i.e. it is not possible to measure an absolute Gibbs energy of formation, since it is not possible to prepare a solution of $\ce{Fe^2+}$ without any corresponding counterions. Therefore, for the half-reaction
$$\ce{Fe^2+ (aq) + 2e- -> Fe (s)} \qquad E^\circ = -0.44~\mathrm{V}$$
the value of $\Delta G^\circ = -nFE^\circ$ ($n = 2$) actually refers to the value of $\Delta G^\circ$ for the reaction
$$\ce{Fe^2+ (aq) + H2 (g) -> Fe (s) + 2H+ (aq)} \qquad \Delta_\mathrm{r} G^\circ = +84.9~\mathrm{kJ~mol^{-1}}$$
One way of proceeding is to note that the Gibbs energy of reaction can be expressed in terms of the Gibbs energies of formation of all the constituent species, in an exactly analogous manner to enthalpy.
$$\Delta_\mathrm{r}G^\circ = 2G_\mathrm{f}^\circ(\ce{H+(aq)}) +  G_\mathrm{f}^\circ(\ce{Fe (s)}) - G_\mathrm{f}^\circ(\ce{H2(g)}) - G_\mathrm{f}^\circ(\ce{Fe^2+(aq)})$$
Since the Gibbs energies of formation of $\ce{Fe(s)}$ and $\ce{H2(g)}$ are zero this leaves us with
$$84.9~\mathrm{kJ~mol^{-1}} = 2G_\mathrm{f}^\circ(\ce{H+(aq)}) - G_\mathrm{f}^\circ(\ce{Fe^2+(aq)})$$
In order to assign a Gibbs energy of formation to ions, it is conventional to set
$$G_\mathrm{f}^\circ(\ce{H+(aq)}) = 0$$
Atkins' Physical Chemistry, 10th ed. (p 136) writes

In essence, this definition adjusts the actual values of the Gibbs energies of formation of ions by a fixed amount, which is chosen so that the standard value for one of them, $\ce{H+(aq)}$, has the value zero.

Under this convention, $G_\mathrm{f}^\circ(\ce{Fe^2+(aq)})$ would be calculated to be $-84.9~\mathrm{kJ~mol^{-1}}$. Note that by virtue of setting $G_\mathrm{f}^\circ(\ce{H+(aq)}) = 0$, this is still only a relative value.
(Disclaimer: In Table 2C.5 of Atkins (p 976), $G_\mathrm{f}^\circ(\ce{Fe^2+(aq)})$ is given as $-78.9~\mathrm{kJ~mol^{-1}}$. I am still not entirely sure where this discrepancy arises from.)

This is not possible. The potentials are given only as references to the standard reduction potential for hydrogen. As such, they are not absolute.
You can convert between potential and free energy for a full redox reaction.
Actually, your statement has a deeper conceptual flaw. The standard formation free energy of the Fe(II) ion does not make any sense. The standard free energy of formation requires making something from its constituent elements in their standard states, but you cannot make an ion from its constituent neutral atoms. You'll need to account for the energy of the electrons and their binding energy to the ion. Therefore, the cation alone cannot have a standard free energy of formation.

Wow just when I thought this chem homework couldn't make me want to die any more, thanks I guess.

Honestly, the electrochem stuff is pretty straightforward, once you accept that you always want to look at the full redox reaction.


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Terminology: what's the difference between "monomer" and "protomer"?
Using definitions from Wikipedia, a protomer is defined as:

the structural unit of an oligomeric protein. It is the smallest unit composed of at least two different protein chains that form a larger heterooligomer by association of two or more copies of this unit.

In terms of proteins, the term monomer may refer to the amino acid components. However, an oligomeric complex with one subunit can be called a monomer (according to the page on protein quarternary structure).
Are these two terms interchangeable?

Is the prot- in protomer ~ prot- in protein?  If so then the difference between the two -mers is broken up by the word parts "protein piece" and "one piece"

The difference between a monomer and a protomer depends on the associations of the molecule and its 3D structure. A monomer is the isolated unit molecular component (i.e. free or soluble form) of a potential oligomer or polymer. A protomer is the molecule after a conformational change into the shape it takes when incorporated as part of that oligomer or polymer.
Example: α-haemolysin and γ-haemolysin are proteins in staphylococcal pore-forming toxins. These units are components of a haemolytic pore, which is a tubular oligomer (a β-barrel transmembrane channel) constructed from a small number of these molecules arranged together to span a membrane (typically an erythrocyte - or "red blood cell" - membrane). In soluble form, the unit protein is termed a monomer. When assembled within a membrane to form the haemolytic pore oligomer, the protein undergoes a large conformational change when its hydrophobic region associates itself with the membrane, and it is then known as a protomer.
https://en.wikipedia.org/wiki/Pore-forming_toxin#Assembly

Lacking an answer for 3 years, I'm going to attempt to answer this based on internet definitions and logic. Please, I welcome any insight from chemistry experts to improve this answer.
First of all, the Wikipedia link quoted actually gives two definitions for protomer. The one quoted is the "structural biology" definition rather than the "chemistry" definition. On the other hand, it seems as though monomer has one definition according to the Monomer Wikipedia article. However, as pointed out by the OP, another point of confusion arises in the definition of a monomer as an oligomer with just one subunit in protein quarternary structure. So we actually have two definitions of each term out there. Let's recap them and then compare.

Protomer (structural biology): In this case, the term protomer is reserved for the specific case of being part of a hetero-oligomer. The "hetero" part is important. The point is that the oligomer is made of a number of different macromolecules- glued together non-covalently. In this case, a protomer is defined as the smallest repeatable unit within this hetero oligomer, which may be the entire oligomer itself (in which case, confusingly, the oligomer is called a monomer). The term protomer in this case, as pointed out by Wikipedia, serves the sole purpose of disambiguation. When we say "blah-mer", as pointed out by Hanry, we are saying "this thing has blah parts". But what are "parts"? Are they quarks? Electrons? Protons? Atoms? Macro molecules? Or... repeatable sub-units. It seems as though the confusion was between the last two. You can break a hetero-oligomer into macromolecules and you'll get a bunch of different macromolecules. You might get 12 As, 12 Bs and 12Cs. So is this a 36-mer or is it a 12-mer? It depends on what you mean by "mer". If each "mer" is an ABC subunit, then it's a 12-mer. If each macromolecule is a subunit, then it's a 36-mer. The term protomer helps us here. The molecule can be viewed as a 12-protomer, meaning that it's 12 copies of the smallest repeatable subunit: ABC.
Protomer (chemistry): In this case, it looks like the term means something quite different. It looks like a protomer is defined as a specific type of tautomer, one in which the tautomerism arises as a result of variable proton position.
Monomer (standard): A monomer is a single molecule that forms part of a polymer. The polymer can either be a homo-polymer, consisting of repeated copies of that monomer, or it can be a copolymer (analogous to a hetero-oligomer), consisting of different monomers, according to Wikipedia's Polymerization article. The key here is that the monomer belongs to a polymer- meaning "many parts". Typically polymers are distinguished by the fact that they can theoretically continue forever in big long chains.
Monomer (oligomer special case): This is just an unfortunate use of terminology, but it kind of makes sense when you think about it. "Oligomer" means "Few parts". Typically the distinction between oligomers and polymers is that polymers can go on forever- they could keep growing through repeated polymerisation, but an oligomer will have a set number of parts, and then it's fully grown. Because oligomers can have a set number of parts, people decided to further sub-classify them by the exact number of parts. So a hexamer is an oligomer with six parts, for example. Following this naming convention, we have pentamer for 5 parts, tetramer for 4 parts... and all the way down to monomer for one part. Which just unfortunately intersects with the other definition of monomer, being a single part of a polymer chain.

So, now that it's been established that there are different definitions even of each term itself, let's try to compare them:

Protomer (structural biology) vs. monomer (standard): a protomer here is defined as a sub-unit of an oligomer, whereas a monomer is a building block of a polymer.
Protomer (structural biology) vs. monomer (oligomer special case): in this case, we need to think. A monomeric oligomer is one which has one "sub-unit". If by "sub-unit" what is means is protomer, then in this case the monomer, protomer and oligomer all coincide and are describing the same thing. If, though, "sub-unit" means the smallest sub-molecules that are non-covalently bonded together, then the oligomer is, trivially, a homogeneous oligomer and not a heterogeneous one: it consists of just one molecule so all the molecules are the same. And so, strictly speaking according to the above definition, the term protomer doesn't apply here.
Protomer (chemistry) vs. Monomer (standard): these two definitions are not mutually exclusive. By describing a compound as a protomer, in this way, we are just saying that there are different tautomeric configurations of this compound out there. On the other hand, a monomer is just something that can form a polymer chain. It's perfectly reasonable to assume that some tautomer could also be capable of undergoing polymerization, and therefore, would be a monomer. I don't know any examples but maybe someone reading can offer them.
Protomer (chemistry) vs. monomer (special oligomer): The monomer definition here seems to be a bit vague, but in the loosest sense, every single molecule is a monomer. But it seems as though what's meant is that a monomer, in the oligomeric sense, typically doesn't undergo polymerisation with other monomers to form a polymer - it just likes being alone. Still, even with that definition, it's likely that there are examples of things that are protomers, in the chemistry sense i.e. being tautomeric in terms of proton position, and that are also monomers i.e. oligomers that won't bond with each other any further.

There is need to distinguish polymerization ( like ethylene -> polyethylene, acrylonitril + butadien + styren -> ABS ) and polycondensation ( like dian + phosgen -> polycarbonate + HCl ). Proteins are not exactly copolymers of aminoacids, but their polycondensates.

@Poutnik I welcome you to edit my answer at will with this knowledge. I am eager to know how it fits in as I am myself trying to wrap my head around this complex subject


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Can we chemically characterize a flame?
Can we chemically characterize a flame ?
Recently, while I was lighting a candle, I came across this question; can we find out the chemical composition of a particular flame? If yes then how? In other words, can we know which elements are present in flame, and not just the salts we test using the flame tests?

Well, I guess you'd like description of combustion reactions. It can be done but mechanisms are very complex and of course dependant on specific case.

I'm voting to re-open this question since later questions have been closed as duplicates with this question listed as "the original".  https://chemistry.stackexchange.com/questions/169691/what-is-the-chemical-composition-of-a-flame

If you are using a gas stove with a methane ($\ce{CH4}$) flame, an interesting things occurs if you heat $\ce{CuO}$ in the gas part of the flame. I once  created cupric oxide from heating a penny to glowing red and then shaking in air. Then, I inserted the dark colored penny into the flame zone, and it appeared to be cleaned!
Apparently, the hot cupric oxide is reduced by $\ce{CH4}$ to $\ce{Cu2O + Cu}$. As such, my take on a possible reaction liberating elemental copper, $\ce{Cu2O}$ and carbon:
$$\ce{CH4 + 3 CuO -> 2 H2O + Cu2O + C + Cu}$$
Now, the flame test for Carbon is red-orange and for Copper green/blue, so a mixture of colors could theoretically be observed in this experiment as the methane flame itself participates in element production.
Normally, however, this does not occur.

You can find certain elements in the flames if you are lucky enough and they do not overlap too much. Like in classic flame tests by having the light emitted from the flame pass through a prism or optical grading you can compare the emission spectrum to all known elements. But this won't be a full characterisation probably. Sodium can be seen quite easily with its yellow flame. When I had to distinguish between Strontium, Calcium and Lithium I looked other emission lines like the green ones. So with some tricks you might be able to determine two or three elements at once but I doubt you'll be able to tell whether it's sodium bromide, chloride or sulfate in the end.

but that is for the salts not what are components present in the flame

Then you would be close to something like mass spectrometry where you ionize your compounds and analyze the charged fragments that break off your molecules.


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How to determine the equivalence point of a titration of a mixture.
Today, I'm graphing the titration curve for a mixture, to be precise, this particular case:

In order to get this done, i've been following wisely the mantra of "The strongest species (oxidizing and reducing agent determine the E(V)" and from there, I construct appropriately the Nerst Equation to suit the situation. This has given me positive results until the end point of Fe, where this doesn't seem to work for me.
At this point, my logic is the following; we have Fe3+, Ce3+, and Sn4+, and we don't have any complete redox pair. Given this, the stronger oxidizer is Fe3+, and the strongest (and only) reducing agent is Ce3+; However, when I try to calculate the weighted average for the E(V) associated to these species, the value obtained (1.2) isn't anywhere near 1.05, my theoretical value, but closer to the upper limit of E(V).
For me, it makes sense that it isn't 1.2V as there is another ion that lowers the potential, but I'm not really landing these ideas to determine the second end point nor the upper limit.
I'd appreciate any thoughts on this, thank you.


text:
Mixing two ideal gases
I had total blackout with these ideal gas equations as I haven't needed these in years. If I have two gasses, A and B which both have same volume V and quantity N. We also have temperatures T1 and T2.
Now, I let energy change between these two ideal gases. How do I get the final temperature out using these informations?
I realise we take the ideal gas equation $p = \frac{NRT_1}{V}$ and get the answer from there, right?

I think there should be a heat capacity of gasses somewhere, or else we could not know the final temperature.

@HuyNgo You are right. I suspect that for this question, the heat capacities have to be assumed to be equal, which would be true if both are (for example) monoatomic ideal gases (Cv,m = 3R/2).

Actually, the heat capacities don't have to be equal.

what is the pressure of two gases ? are they the same?

@Physicsapproval  Are you aware that if you know N, R, T, and V, then, from the ideal gas law, you know P?  And why should the initial pressures of the two gases matter in determining the final temperature?

Yes, the heat capacities don't have to be equal, but if they're not equal then their values have to be stated, no?

Because you aren't told, there are three possible cases to consider (well there's more, but these are the natural cases):

Two monoatomic Gases
One monoatomic and one diatomic gas
Two diatomic gases

We should just note that the equilibrium for this system will be the state at which the temperature of each system is equal. We want to find where that temperature ends up between $T_1$ and $T_2$. Let's call this $T_f$. So we have,$$T_1\le T_f\le T_2.$$
Also note that this a constant volume process not a constant pressure one.
$\textbf{Two Monoatomic Gases}:$
The heat capacity of a monoatomic ideal gas is simply $C_v=\frac{3}{2}nR$.
Then, the internal energy of the system at a given temperature is,$$U=\frac{3}{2}nRT.$$
Then, the change in internal energy of one system is the negative of the change in internal energy of the other,$$\Delta U_1=-\Delta U_2$$so that,$$\frac{3}{2}nR(T_f-T_1)=\frac{3}{2}nR(T_2-T_f)$$
Upon rearranging this yields,
$$T_f=\frac{T_2+T_1}{2}$$which gives the (perhaps) intuitive result that the final temperature the gases is simply the average of the two initial temperatures for the fairly restrictive conditions we have here.
$\textbf{One Monoatomic and One Diatomic:}$
The only thing different here is that the heat capacity of a diatomic gas is $C_v=\frac{5}{2}nR$. That is, we pick up an extra $\frac{1}{2}k_bT$ for each rotational degree of freedom per molecule.
It is still gonna be true that changes in energy are equal and opposite so we can skip straight to that step and write,
$$\frac{5}{2}nR(T_f-T_1)=\frac{3}{2}nR(T_2-T_f)$$but we see now that it actually matters now whether the hotter or colder bath had the diatomic molecules. For the expression above, let's say $T_1$ was colder than $T_2$. This gives,
$$T_f=\frac{3}{8}T_2+\frac{5}{8}T_1$$
And by the symmetry of the situation, if the diatomic gas started at the hotter temperature $T_2$, we just switch the coefficients and get,
$$T_f=\frac{5}{8}T_2+\frac{3}{8}T_1$$
This basically tells us that the system with the larger heat capacity needs more energy input than the other.
$\textbf{Two Diatomic Gases}$
Well, this actually ends up being exactly the same as the monoatomic gas case as I'm sure you can convince yourself. So, here the final temperature is simply the average of the initial temperatures as well.
$\textbf{Further Cases}$
Note that I used a linear diatomic which only has two rotational degrees of freedom, but one could easily imagine using water vapor which has three rotational degrees of freedom and combining this with a diatomic or a monoatomic. The form of the answer is always gonna be the same as above though. Just two coefficients in front of the initial temperatures which sum to one.
I'm pretty sure I haven't gone astray, but if I have I'm sure someone will promptly inform me :)

Edit:
It just dawned on me that the question title and the body of the question don't say the same thing. That is, letting two systems of gases equilibrate by heat exchange is different than having them separated and allowing them to mix. I haven't really thought about it, but because the initial volumes of the containers are the same, the answer might be the same as what I write above, but the answers probably aren't the same in general.
Maybe someone else can answer the gas-mixing case.


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Does saltwater damage electronics?
newbie here.
I'm asking a question that is related to hygrometer calibration.
I read in sensitive digital hygrometers, it is stated that saltwater test should not be used to calibrate as it will damage the electronics (resistor?) that is used for measurement.
But how does this ring true? Saltwater does not evaporate, or do they?

Salt water can corrode metals. If your instrument says not to use with saltwater, assume the manufacturer knows the limitations of the device.

@MaxW I think the question here is why saltwater would damage the hygrometer if it only measures humidity in the air with the presumption that salt doesnt evaporate with the water which evaporates to build up the humidity?

I was assuming that "saltwater test to calibrate" somehow wet the device with saltwater to calibrate it. // It could mean measurements over salt water like on a boat. The sea spray would dry tiny droplets of salt water, so the air would have some salt particles in it. In any case the the manufacturer should have stipulated the limitations of the device.

Salt water is an excellent electrolyte, and will accelerate galvanic corrosion reaction between two different metals especially in oxygen rich environments (including dissolved oxygen in water). This may create metal oxide deposits on circuits. The longer the exposure time the more corrosion can occur.
If your device is on, that is, there are circuits with electricity flowing, clearly adding a salt water could also short circuit your device as well.
As to whether any of the above results in irreversible damage is hard to say with certainty as some devices have been dried out and appear to function at least following exposure to common tap water.


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What happens to the solute when a solution freezes?
I came across a situation in which my chemistry textbook gives phase diagram of a pure solvent and a solution with the same solvent.I am considering 2 different , say beakers, one containing some amount of pure solvent , and the other contains a solution formed by a non volatile solute and the same amount of solvent. In this, the author has shown the part of the graph corresponding to solid solvent , and solid form of solution to be identical. It's natural for me to think that a solid obtained by freezing a solution would be different from that obtained by freezing a pure solvent. So they must not have identical graphs. Is my argument correct ? Or is it like this- when the solution is frozen , since the solvent molecules are brought together for making a solid, we simply don't have solvent molecules to solvate  our solute, and the solid is more or like a frozen pure* solvent with solute particles trapped in it?

In you first sentence, do you mean a pure solvent in contact  with a solution containing a solute in the same solvent but which do not mix? Then  solid solvent in contact with the solid solution and/ or liquid solution ?

I am considering 2 different , say beakers, one containing some amount of pure solvent , and the other contains a solution formed by a non volatile solute and the same amount of solvent.

The phase diagram for say pure water and a NaCl/water mixture would be different. The NaCl would lower the freezing point of water. Also the salt would tend to concentrate in the liquid until some lower freezing point was achieved.

If you were to freeze the solution in a normal manner, using something similar to a home freezer, a more and more concentrated solution will form as the majority of the solvent freezes. The solute will be found in a small amount of solution that did not freeze, since the solute allows for freezing point depression. So, if you put both of your beakers into the freezer, one beaker will have a completely frozen block of solvent and the other will have a frozen block of solvent with some volume of highly concentrated solution.