text:
How to dissolve a spider with household chemicals?
Somehow a spider got into the little window that shows the water level on my electric kettle. The kettle doesn't come apart very easily, and it looks like I can't get this part off without breaking it.
I'd hate buying a  new kettle just because of a little spider but, I also don't want spider juices with my coffee.
Would bleach/some other household chemical dissolve the spider? Is there some other methodology you can recommend?

I want to unsee this question. Also, chitin is quite durable, so the prospects are bleak.

Boil the water a few times and the spider won't give off any more juices. Just like a bag of tea. And the empty husk will be a nice decoration.

I've heard of a guy who had a very similar problem in a bigger scale: a dead cow in a cesspit. He also looked for a way to dissolve it.

Just wash it out with some cold water, should be easy. Why try to kill it?

I'd wonder about mechanically breaking up the spider. Is there a large opening in the top in which you could insert something. I'm thinking about using a water pick, or a can of dust remover, or a plastic line through the top hole of the indicator.

As entertaining as this is, and even though chemists might be likely to know better than most how to dissolve away the little bugger, this is probably a better fit for http://lifehacks.stackexchange.com/.

Piranha acid will do the trick. However, it is unfortunately not accessable with household chemicals. And the valid question remains what your kettle will look like after a piranha application (although given it’s probably plastic you could be lucky).

You're not the old lady that swallowed a fly and trying to avoid your lyrical fate, are you?
"Biological" washing powder might do the trick - plenty of proteolytic enzymes - probably take several days at least.  Then you'd have to wash out the resultant spider soup.
I hope you like coffee with a lot of body to it.

I would try with warm caustic soda. Possibly with bleach as well.
Alternatively, ants.

Depends on kettle construction. If there are any metal parts, caustic soda (sodium hydroxide) could react in an  undesirable fashion. There is a literature on the use of NaOH (usually boiling) to extract proteins and purify chitin derivatives. Be careful. Could end up with goo or glue.

Crap in, crap out ... Worm castings (dung for all practical purposes) are rich in chitinase, and it is all natural! This link is an example of a liquid product - whole bottle for $13.95 You would need to sanitize as a final step. The worms eat cow dung to produce their primo bug dissolving poop. If you are averse to this, study how you may be able to extract chitinase from plant matter. Wikipedia states "Chitinases occur naturally in many common foods. Bananas, chestnuts, kiwis, avocados, papaya, and tomatoes, for example, all contain significant levels of chitinase, as defense against fungal and invertebrate attack." You may be able to make a do it yourself crude extract considering the reported method by Sanchez-Monge et al (Clinical and Experimental Allergy,1999, Volume 29, pages 673–680) is as follows:
"Canary banana (Musa paradisiaca L., an hybrid of M. acuminata Colla  M. balbisiana Colla) fruits were peeled, cut in short pieces, freeze-dried, ground in a mortar with cold acetone, and then defatted with the same solvent (3x, 1 : 5 (w/v); 1 h; 48 C). The dried residue was extracted with PBS buffer (0.1 M sodium phosphate, pH 7.0, 0.15 M NaCl; 1x; 1 : 5 (w/v); 1 h; 48 C), and after centrifugation, the supernatant was dialysed against H2O and freeze-dried(crude extract). Alternatively, the PBS-extracted supernatant was brought to pH 3.5 by addition of 0.1 M HCl to obtain a protein preparation enriched in chitinases [18]. After centrifugation, the supernatant of the acidic treatment was dialysed against H2O and freeze-dried." Be careful with acidification. Acid may react with metal. A do it yourself pH test can be done with red cabbage juice.

Your can may be able to osmotically chip at him. Ethanol will dehydrate the tissue. Then water will swell the tissue. Repeat the cycle. This will also extract lipids.
Only molecules left would be proteins. Other posts talk about proteases. But at this point detergent should liberate proteins.


text:
Work done by gas while expanding against constant external pressure?
This is the graph for the same between pressure (y) and volume (x):-

Here, why do we assume the gas pressure to be same as the external pressure everytime in during the process of expansion?

Presumably, because the gas is contained in something that cannot maintain a pressure differential, which would mean that some exothermic process is causing the gas to expand at constant pressure. Hang in there, this looks like the middle of an explanation of something.

Because you are talking about reversible process, which is per definition quasi equilibrium.

In a reversible process, the pressure of the gas is uniform throughout, and the force per unit area that the gas is exerting on the piston face is equal to its pressure, as determined from its equation of state.  So the work is equal to the force integrated over the displacement, or, equivalently, the uniform gas pressure integrated over change in volume.
In an irreversible process, the pressure of the gas is not uniform throughout, but, at the piston face, the force per unit area that the gas exerts on the piston face must match the force per unit area that the piston face exerts on the gas, by Newton's third law.  But, in this case, because the gas pressure is not uniform throughout (and also because viscous stresses contribute to the force per unit area of the piston), the force per unit area at the piston face cannot be determined from the equation of state.  However, if the force per unit area at the piston face is imposed by specifying external conditions on the system, the work that the gas does can still be calculated as the integral of the external pressure integrated over the volume change.
So, to summarize, in both reversible and irreversible processes, the force per unit area exerted by the gas on the piston face is equal to the externally applied pressure and the work is equal to the external pressure integrated over the volume change.  However, in the case of a reversible process, the uniform gas pressure can be determined from the equation of state of the gas, but, for an irreversible process, it cannot (and needs to be determined by controlling the external pressure in some way).

I believe pressure is not constant in reversible reaction and is constant in irreversible reaction. Why is that ? I want to know

You don't need to, but it's a lot simpler if you do. Otherwise, you'll need to compute work as $\int PdV$. Note that this is $PV$ when $P$ is constant.

Well? Only for simplicity? Nothing else?

Sure, this is a contrived scenario. You can do anything you want to. But for a simple gedanken, I don't see what you want some pathological mess of a function for $P$. You're probably also looking for a system that is microscopically reversible, so you trying to change it as little as possible. In that case, keeping $P$ constant means you can perform the changes very slowly.

According to me it we do not assume the gas pressure always equal to external pressure in an irreversible process.We select external pressure for calculating the work done because it is the only pressure against which gas had done work by creating a small change in volume until both the internal as well as external pressure becomes equal.

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text:
Change in internal energy of a gas at constant volume
According to the first law of thermodynamics, $u=q+w$, where $u$ is changing in internal energy, $q$ is heat liberated and $w$ is the work done in the process.
Now at constant volume, $w=0$, hence $u=q$. Since $q$ is $n\cdot C_v\cdot T$, where $n$ is the amount of substance in mole, $C_v$ is the molar heat capacity at constant volume and $T$ is the temperature change. $u$ comes out to be $n\cdot C_v\cdot T$. However, this expression is valid for all other cases too whether volume is constant or not.
Same is the case with $H=n\cdot C_p\cdot T$. Why so?

The change in internal energy should be written as $\Delta U=nC_v\Delta T$, not $nC_vT$.  This equation is valid for any temperature change (irrespective of whether the volume or pressure changes) only for an ideal gas.  The equation for the change in enthalpy should be $$\Delta H=\Delta U+\Delta (PV)$$  For an ideal gas, this equation reduces to
$$\Delta H=nC_v\Delta T+nR\Delta T=nC_p\Delta T$$  This equation is valid for any temperature change only for an ideal gas.

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text:
How does a glow stick work
I understand that when the glow sticks snaps, it produces light, and that process is called triboluminescence. But I don't understand how that works

As noted in the answers, triboluminescence is not the correct explanation.

Light from a chemical reaction is chemiluminescence. Triboluminescence is the emission of light caused by rubbing, scratching, or similar frictional contact of a substance. Two quartz pebbles rubbed together will create triboluminescence.

See this and the links therein:http://chemistry.stackexchange.com/questions/35527/how-do-things-glow-in-the-dark/35537#35537

A glow stick is composed of two different compounds (normally diphenyl oxalate and hydrogen peroxide), separated by a thin, breakable, barrier. When you bend the glow stick, the barrier breaks, causing these two compounds to be mixed together. When this happens, hydrogen peroxide causes the diphenyl oxalate to break down into phenol and peroxyacid ester. The peroxyacid ester then breaks down into $\ce{CO2}$, transferring energy to some sort of a dye that then releases it to produce light.

Simple answer: It is a simple chemical reaction.   Any and all chemical reactions produce some form of energy output as a result of changing factors.
In the example of a glow-stick, you have 2 separate chemicals contained in the tube, separated by a thin barrier.   When you snap the glow-stick, it breaks that barrier separating the two and allows them to mix together, creating the chemical reaction.
The energy output of that particular slow reaction is production of visible light.  This is why glow-sticks are never reusable.  Once the reaction is done.  It can't be recreated without fresh chemicals.
Fire spreads because its chemical reaction results in an output of a large amount of heat.  That heat is what is required to begin the reaction in the first place, so the radiant heat from the fuel source currently undergoing that reaction causes nearby fuel sources to heat up enough to also ignore.
Water puts out fire because of the huge evaporative properties it has.  When water hits an object on fire, it immediately evaporates.  Evaporation has the (handy in this case) side effect of removing a large amount of heat energy from whatever is around it...   Removing the heat stops the chemical reaction from continuing, thus putting out the fire.


text:
Confused about the direction in which the bond will break

Why does the bond break in different fashion in the two cases?
For case X, a 3° carbocation is formed, bit why isn't this the scenario for case Y?
A 3° carbocation is anyday more stable than a 1° carbocation, isn't it?

Under basic conditions, epoxide opening follows a $\mathrm{S_N2}$ mechanism (since there is a good nucleophile). Hence attack at less substituted carbon. This answers the question: http://chemistry.stackexchange.com/a/8468/16683 but not duplicate.

The reactions happen on different centres because they follow different mechanisms.
Epoxides are actually surprisingly stable. They don’t undergo spontaneous ring-opening reactions by themselves unless the oxygen atom has been protonated previously. That is what the $\ce{H+}$ in reaction path X is there for. After having protonated the oxygen, the ring-opening reaction is charge-neutral and thus releiving the Baeyer strain to generate a tertiary (more stable) carbocation is favourable. This carbocation can be captured by a nucleophile in an $\mathrm{S_N1}$-type mechanism.
This does not happen under basic conditions as removing a negative charge would not help the epoxide opening. However, under these conditions solvent molecules such as methanol are easily deprotonated creating much better nucleophiles. These nucleophiles are then able to attack the epoxide under $\mathrm{S_N2}$ conditions. Hereby, an anion attacks a neutral molecule to generate an anion, so the ring-opening is again charge-neutral. However, $\mathrm{S_N2}$ attacks are more likely to happen on less hindered, primary carbons.
Taking this into account, path X that follows an $\mathrm{S_N1}$ path, generates the most stable carbocation hence the nucleophilic attack happens on the tertiary carbon. Path Y that follows an $\mathrm{S_N2}$ mechanism includes an attack on the less hindered, i.e. primary carbon.

Just want to add that it's not exactly SN1 (implies intermediacy of planar carbocation) - e.g. no alkyl shifts are seen, and rxn proceeds with inversion. My tutor calls it a "SN1-like SN2"..

@orthocresol Yeah, all those effects … I chose to ignore them O:-)


text:
Understanding secular equation derivation
In many basic derivations of the secular equations for the energy of two molecular orbitals from a linear combination of atomic orbitals, we see the following reasoning which I find confusing.
We have a trial wavefunction $\phi = c_a\psi_a + c_b\psi_b$
The energy of the trial wavefunction is
$$\begin{align} \langle E\rangle &= \frac{\langle \phi | H | \phi \rangle}{\langle \phi | \phi \rangle}\\
&= \frac{c_a^2H_{aa}+2c_a c_b H_{ab} + c_b^2 H_{bb}}{c_a^2 S_{aa} + 2c_a c_b S_{ab} + c_b^2 S_{bb}} \end{align}$$
Where $S_{aa}$ and $S_{bb}$ equal 1. Now per the variational principle, we take the partial derivatives with respect to the coefficients $c_a$ and $c_b$ and set them equal to zero to find the optimal coefficients to minimize $\langle E \rangle$.
But instead, things follow a different route. We take the equation and move the denominator over:
$$
\langle E \rangle (c_a^2 + 2c_a c_b S_{ab} + c_b^2) = {c_a^2H_{aa}+2c_a c_b H_{ab} + c_b^2 H_{bb}}
$$
And then it's stated that this equation is differentiated with respect to $c_a$ and $c_b$ such that
$$
\langle E \rangle(2c_a + 2c_b S_{ab}) + \frac{\partial\langle E\rangle}{\partial c_a} (c_a^2 + 2c_a c_b S_{ab} + c_b^2) = 2c_a H_{aa} + 2c_b H_{ab}
$$
with a similar expression for differentiation wrt $c_b$.
Now the partial derivatives are set to 0, such that you end up with
$$
c_a(H_{aa}-E) + c_b(H_{ab} - ES_{ab}) = 0 \\
c_a(H_{ab}-ES_{ab}) + c_b(H_{bb} - E) = 0
$$
which then you can solve for E.
Is my understanding correct that we've obtained the minimum energies in a single step by first rearranging the equations and then differentiating? Of course we do not obtain values for the coefficients. How is the approach valid?
THe final equations are called the secular equations, which for linear systems isa way of finding the eigenvalues of the system by $A - \lambda I = 0$. But here we have off-diagonal terms $-ES_{ab}$, so I do not see the connection to linear algebra here.

Your last paragraph doesn't quite make sense. You're obtaining the eigenvalues from $\det (A-\lambda I) = 0$. It doesn't matter if you have off-diagonal terms because there's no reason why $A$ should be diagonalized. Also, for your specific case, you should consider what $A$ is.

Hmm, I was trying to say that if you have $H\psi = E\psi$, you can obtain $E$ by taking the determinant of $H-EI$. But i don't see the analogy for this situation.

That's not quite right... You obtain the possible values of $E$ as the eigenvalues of $H$ by solving the equation $\det (H - EI) = 0$. The polynomial you obtain is the characteristic equation, and I believe that is the secular equation: https://en.wikipedia.org/wiki/Characteristic_polynomial#Secular_equation

yes that is what i meant. to solve for E.

Doesn't that answer your question about the secular equations? They should be the equivalent of the characteristic equations you get from solving for $E$...

Perhaps there's some gap in my knowledge, and I'm not asking the right question in my last paragraph. Nonetheless, I'm more interested in the reason why the main derivation i described works.

It doesn't matter how you obtain an expression for the partial derivative, as long as you obtain one, and set the partial derivative to zero. The method you described above is probably easier than differentiating the fraction. All the steps that you did are mathematically valid, so your final result is valid. (The "differentiate both sides" step is valid because the equality holds for all values of $c_a$ and $c_b$.)
If you want to, you can differentiate the fraction and you will end up with the same result (i.e. the secular equations). I actually typed the maths out before, so here you go: https://chemistry.stackexchange.com/a/42636/16683 The notation is a tiny bit different, but you should be able to understand it.

As for the linear algebra parallel, the secular equations are only analogous to $Ax = \lambda x$ if you make the additional assumption that $S_{ab} = 0$, in which case $H-ES$ is directly analogous to $A - \lambda I$ (since the overlap matrix is now equivalent to the identity matrix). In this case, the secular equations reduce to $Hc = Ec$.
If you do not make that assumption, then the secular equations take the form of $Hc = ESc$ and the parallel you should draw is to the generalised eigenvalue equation $Ax = \lambda Bx$. The eigenvectors $x$ are said to be the "eigenvectors of $A$ with respect to $B$". There is a lot more information online if you are interested - simply google "generalised eigenvalue".


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Steam Pumpkin Launcher
I am doing a pumpkin launcher project and my group is torn between using an air cannon, trebuchet, centrifugal arm, and a steam cannon. I want to take water fill the tank x % with water and heat it to above 100 °C.
When I release the pressure, if all of the water is above 100 °C, shouldn't it all vaporize? I need to know if I am right and what math would back me because I can't for the life of me remember/find the equations to do so.

Related: What happens if I make a hole in a container with superheated water?

This somehow made me recall a certain folk song. "They found him in the wreck, with his hand on the throttle, and scalded to death by the steam." I mean, be careful. No, all of it would not vaporize, but a sizeable portion would.

You're essentially suggesting making an autoclave/pressure cooker and then poking a hole in it. Take a look at OSHA reports to get an idea of what could happen--these failures happened with industrial equipment that costs thousands of dollars that are designed to remain closed during operation. Your idea is certainly possible, but if you're asking internet randoms for help, maybe think about if you have the knowledge and skills necessary to pull this off safely.

If you mean superheat as in make it extremely hot gas (which I don't think you do), there is a practical problem of containing and then releasing all that pressure. If you're superheating the water in terms of it not boiling while it is above boiling point, yet the pressure is approximately atmospheric, there's practical limits to that. The further from boiling the water is, the less stable the phase is. If you want to increase it enough to propel medium sized objects, I'd say you'd probably be safer with home made nitroglyercin...

@chipbuster has the right idea, but I'd be more dogmatic. Your question is so basic that it is obvious that you do not have the necessary knowledge and skills to safely use superheated steam.

I guess that a steam cannon is probably the least effective of the above alternatives. And it is probably the most dangerous since steam can cause severe burns.

If you heat water to a temperature above $T=100\ \mathrm{^\circ C}$, you need an increased pressure to keep it liquid. This is called superheated water.
When you release the pressure, the pressure drops to the atmospheric pressure (typically $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$) and the hot liquid water immediately starts to boil. The temperature drops to the boiling point ($T=99.974\ \mathrm{^\circ C}$) at the new pressure.
At this temperature and pressure, the specific enthalpy of steam is $h_\text{steam}=2675.5\ \mathrm{kJ\ kg^{-1}}$, and the specific enthalpy of liquid water is $h_\text{liquid}=419.06\ \mathrm{kJ\ kg^{-1}}$. Thus, the enthalpy of vaporization of water is very high: $\Delta_\text{vap}h=h_\text{steam}-h_\text{liquid}=2256.5\ \mathrm{kJ\ kg^{-1}}$. Therefore, the water does not completely flash to steam. The vaporized fraction $x$ may be estimated using the following enthalpy balance:
$$\begin{align}
h_0&=x\cdot h_\text{steam}+(1-x) \cdot h_\text{liquid}\\[6pt]
x&= \frac{h_0-h_\text{liquid}}{h_\text{steam}-h_\text{liquid}}
\end{align}$$
where $h_0$ is the initial specific enthalpy of the superheated water. This value depends on the initial temperature and pressure of the superheated water and can be obtained from so-called steam tables. (If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha.)
Example
Initial temperature: $T_0=120\ \mathrm{^\circ C}$
Initial pressure: $p_0=2\ \mathrm{bar}=200\,000\ \mathrm{Pa}$
Initial specific enthalpy: $h_0=503.81\ \mathrm{kJ\ kg^{-1}}$
New pressure: $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$
New temperature: $T=99.974\ \mathrm{^\circ C}$
New specific enthalpy of steam: $h_\text{steam}=2675.5\ \mathrm{kJ\ kg^{-1}}$
New specific enthalpy of liquid water: $h_\text{liquid}=419.06\ \mathrm{kJ\ kg^{-1}}$
Evaporated fraction:
$$\begin{align}
x&= \frac{h_0-h_\text{liquid}}{h_\text{steam}-h_\text{liquid}}\\[6pt]
&= \frac{503.81\ \mathrm{kJ\ kg^{-1}}-419.06\ \mathrm{kJ\ kg^{-1}}}{2675.5\ \mathrm{kJ\ kg^{-1}}-419.06\ \mathrm{kJ\ kg^{-1}}}\\[6pt]
&=0.038\\[6pt]
&=3.8\ \%
\end{align}$$
This means that approximately $3.8\ \%$ of the superheated water flashes to steam when the pressure is released. (This may not sound like much; however, it means that $1\ \mathrm{kg}$ of water would generate about $63\ \mathrm l$ of steam.)


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Can other molecules than H+ cause a kind of acid reaction?
In our body there are plenty of other positive ions like $\ce{Na+, K+}$ and $\ce{Ca^2+}$. Can they have a similar acid reaction as $\ce{H+}$ ions can do? If not why is just $\ce{H+}$ considered as acid?

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

Could you elaborate 'cause it seems it's not clear what answer do you want?

My main question was actionally to find out if other molecules/atoms than H+ could be as harmfull. For instance, when a body has a too high pH (or too acid) this could harm the body. I don't know exactly how the acids influence the body, but perhaps it is degenerating proteins or changing metabolism. And the question is could molecules which are not primairily considerd as acid cause the same harmfull effects as acids. And second can Na+ K+ and Ca+ be considered as a Lewis acid?

$\ce{Ca^2+}$ not $\ce{Ca+}$. And look up Wikipedia: hypernatremia, hyperkalemia, hypercalcemia.

Well, if you are asking like that, $\mathrm{D^\oplus}$ certainly behaves very similar to $\mathrm{H^\oplus}$. And there certainly are lewis acids that are very acidic without containing a proton, e.g. $\mathrm{SbF_5},\mathrm{TiCl_4}$ or $\mathrm{BF_3}$ and they would also be very harmful due to their acidity.
I guess that the basic theories about acidity developed in undergraduate courses (e.g. Henderson-Hasselbach equation, titration, ...) just concentrate on water as the solvent and $\mathrm{H^\oplus}$ to keep the discussion simple, for example pH values will differ depending on the solvent.
Also the mathematical treatment (e.g. pH calculation) doesn't make much sense when working with lewis acids in a synthesis lab ; )

Where does D+ stands for?

$\mathrm{D^\oplus} = \mathrm{^2H^\oplus}$ stands for the deuteron.

There are three (important) definitions of acid around: Arrhenius, Brønsted-Lowry and Lewis. This question seems to focus on Arrhenius/Brønsted acids (their definition of an acid is basically identical save a minor difference) which is why I won’t consider Lewis acids here. Note that under Lewis’ definition, all the positive ions you mentioned are considered acids.
The Arrhenius definition of an acid is:

A substance that dissociates in aquaeous solution to generate $\ce{H+}$ ions.

And Brønsted’s is:

A substance capable of releasing $\ce{H+}$ ions.

Both definitions centre on the $\ce{H+}$ ion (although deuterium and tritium ions would equally fulfil their intents). Why is that? Well, every other cation known in chemistry except for $\ce{He^2+}$ still has core electrons. The electronic configuration of $\ce{Na+}$ is, for example $\mathrm{1s^2, 2s^2, 2p^6}$ which equals neon’s electronic configuration. The hydrogen cation’s is $\mathrm{1s^0}$ — there are no electrons. $\ce{H+}$ can basically be called a ‘naked proton’ — and often is just simply called proton.
The fact that we are dealing with what effectively is only a nucleus without any associated electrons changes the cation’s properties significantly. For example, a crystal structure for $\ce{H+ Cl-}$ analogous to $\ce{Li+ Cl-}$ or $\ce{Na+ Cl-}$ is not known because $\ce{H+}$ on its own is hardly stable enough to form such a structure. Instead, it will immediately try to form a bond with any available element that offers an electron pair — a Brønsted base. Naked protons in solution are unknown, they are always attached to solvent molecules in some way. In water, for example, the smallest structure detected in aquaeous solution is a $\ce{H9O4+}$ type complex: $\ce{3 H2O + H3O+}$ with the proton being shuffled between the four water molecules.
It is this ability of hydrogen to immediately jump to an electron pair and attach to it to form a new, positively charged particle that is the basis for acidity. Since only hydrogen is capable of performing that in that way, it is the only truly acidic particle in both Arrhenius’ and Brønsted and Lowry’s theories.

Acidity only applies to H+ ions. The pH is defined as -log[H+] so only applies to H+ ions. The reason that this is so is that water is H20. It self-ionises to form H+ ions and OH- ions. Since water is neutral [H+]=[OH-] so the pH is 7. It turns out that [H+][OH-]=10^-14 always. So when the concentration of H+ ions increase, the concentration of OH- ions decrease.

This is wrong and most of it has no importance for this question.

@Mithoron How parts do you believe are incorrect or irrelevant?

First sentence is wrong - it seems you never heard of Lewis acids. And the rest is irrelevant.

@Mithoron I am familiar with Lewis acids but I feel that the Arrhenius acid is more relevant to the question as it defines the acid as the proton donor (H+). While not 100% chemically correct I chose to omit the Lewis acid explanation as I believe it would be confusing.

OP's asking if Na+ etc. are acids like H+ and they are according to Lewis theory.

@Mithoron Yes, of course because the are an electron pair acceptor. But he asked why H+ is considered as acid. Hence, I refer to the Arrhenius definition as it explicitly addresses this issue by defining it as a proton donor.

@Mithoron To be fair, the question did not specify which theory of acids and bases is to be employed, so this is not Simon's fault.


text:
Thermodynamics: Solving for Entropy, Gibbs and Helmholtz

A reversible melting of $32.0\ \mathrm g$ of ice at $0\ \mathrm{^\circ C}$ and $1\ \mathrm{atm}$. Solve for $\Delta G, \Delta S_\text{universe}$, and $\Delta A$

I'm also given latent heat of fusion/vaporization, heat capacity at constant pressure, ($C_p$) and the densities of water and ice at the appropriate temperature and pressure,
Solving for $\Delta S_\text{universe}$
from equation sheet: $$\Delta S_\text{vap} = \frac {q_{p,\text{vap}}}{T_\mathrm b}$$ Which is for vaporization and I assume I can transform that into $$\Delta S_\text{melt}= \frac {q_{p,\text{fus}}}{T_\mathrm m}$$
So solving for $\Delta S_\text{system}$ I get
$$\Delta S_\text{system}=\Delta S_\text{melt}= \frac {q_{p,\text{fus}}}{T_\mathrm m}$$
But the question asked for $\Delta S_\text{universe}$ not $\Delta S_\text{system}$. I know that $$S_\text{universe}=S_\text{surroundings}+S_\text{system}$$, so $\Delta S_\text{universe}$ must be $$\Delta S_{universe}=\Delta S_\text{surroundings}+\Delta S_\text{system}$$ But I'm given no information on the surroundings. How can I solve this? Is it possible that the change in entropy in the surroundings constant such that $\Delta S_\text{universe}=\Delta S_\text{system}$?
Solving for $\Delta A$ $$A=U-TS$$
 $$\Delta A = \Delta U -T\Delta S_\text{sys}$$
$$\Delta A = (q-P\Delta V)  -T\Delta S_\text{sys}$$
$$\Delta A = (q-P[\rho_\text{water}m-\rho_\text{ice}m])-T\Delta S_\text{sys}$$
But I don't know how to solve $q$ for this problem. Additionally, I'm not sure if the ice melted completely, so perhaps the change in Volume is less that what I wrote in the equation.
Solving for $\Delta G$
$$\Delta G = \Delta H-T\Delta S_\text{sys}$$
$$\Delta G = \Delta U+P\Delta V-TS_\text{sys}$$
$$\Delta G = (q-P\Delta V)+P\Delta V-TS_\text{sys}$$
$$\Delta G = q-TS_\text{sys}$$
But same as before I don't know how to solve for $q$.

What are the initial and final thermodynamic equilibrium states of the system?

Um I know that S(universe) becomes maximum @ equilibrium. I guess gibbs would need to be negative because the melting happened spontaneously. I don't have much intuitive sense with Thermo.

$$q_{fus}=T_m\Delta S_{syst}$$What does that tell you about $\Delta G$?

dG is Zero?! But the melting is spontaneous! how did you get the equation you just wrote?! what kinds of situations can I say dG is zero?

Ya see, this is why  I asked you what the initial and final equilibrium states of the system are.  Let me give you a hint.  The initial state of the system 32 g of ice @ 0 C and 1 atm.  Now, please tell me what you understand the final equilibrium state of the system to be.

The final equilibrium state would be 32 g of liquid water @ 0 C and 1 ATM since the question describes the system as melting.

Excellent.  Now, since these are two thermodynamic equilibrium states, should the changes in G, S, and A depend on whether the path between the two states is reversible or irreversible?  That said, why do you feel that the transition between the two states would have to be spontaneous (i.e., irreversible)?

!!! I didn't know spontaneous implied irreversible. I had the idea that the phase change was spontaneous because it happened on it's own (which I admit, now that I consider it doesn't make any sense at all), given this, can I say that $\Delta G$ of all Reversible reactions is zero?

Let's keep focused on the problem at hand.  For the constant pressure transition being considered, the change in enthalpy is equal to the heat added.  So $\Delta H=T_m\Delta S$, and $\Delta G=\Delta H-T\Delta S = 0$.  The change in G between any two phase equilibrium states is zero.   Does this make sense now?   Algebraically, how is $\Delta A$ related to $\Delta G$?

$\Delta A=\Delta U-T\Delta S$ and $\Delta G = \Delta H-T\Delta S$ so $\Delta A = \Delta U +\Delta G - \Delta H $

From the definition of $\Delta H$, what is $(\Delta H-\Delta U)$?

$\Delta H = \Delta U - PV$ So $\Delta U- PV -\Delta U= PV$

I can't make sense of this, but, anyway, $$\Delta H=\Delta U+\Delta(PV)$$.  So $$\Delta A=\Delta G-\Delta (PV)$$For constant pressure, $$\Delta A=\Delta G-P\Delta V=-P\Delta V$$where $\Delta V$ is the difference in  volume between 32 g of liquid water and 32 g of water ice.

Thanks for your help Mr. Miller! I think i understand more now. In the original post I neglected to realize $T\Delta S = q$ If  had done that substitution then all my q's would have canceled out.

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text:
Solving for Heat flow in Reversible Isobaric Expansion
This question was given on my Practice midterm:
Calculate $q,w,\Delta U$ and $\Delta H$ for a reversile isobaric expansion from (1.00 Bar, 20.0 L) to (1.00 bar, 40.0 L)
Equations used:
$q_p=C_p\Delta T$ $; C_{p Ideal} = \frac {5}{2}R$ where $C_p$ is heat capacity per mol. so $\frac {C_{p Ideal}}{n} = \frac {5}{2}R$
I solved the problem like this:
$w=-P\Delta V = -2000J$ [correct answer]
$$q_p=C_p\Delta T$$
$$\Delta T = \Delta [\frac {PV}{nR}]=\frac {1}{nR}P\Delta V$$
$$q_p=C_p\Delta T= \frac {5Rn}{2}\frac {1}{nR}P\Delta V= 5/2*P\Delta V = 5/2*2000J = 5000J$$ Which is the wrong answer. The teacher's work goes like this. I dont know where she got n=2mol
$$T_1=\frac {P_1V_1}{nR}=\frac{10^5*20*10^{-3}}{2*8.314}=120.3K$$
$$T_2=\frac {P_2V_2}{nR}=\frac{10^5*40*10^{-3}}{2*8.314}=240.6K$$
$$q_p=C_p\Delta T = (2.00 mol)*3.5(8.314)(120.3K)= 7.00kJ$$
I was originally thinking that she might have made a typo and wrote 3.5 for 5/2 instead lf 2.5, but she magically found n = 2 mols.
Please tell me how to get the correct answer.
Thanks in advance.

As far as the 3.5 is concerned, $C_v=\frac{5}{2}R$, $C_p-C_v=R$, so $C_p=\frac{7}{2}R$.
Regarding the 2 moles, were you told either the number of moles or the initial temperature?

You're completely correct. I did not have the entire question. The other information was written on a different PDF. I'm sorry for wasting time !!


text:
Is pressure equalization "diffusion"?
There is another question on the site which is:

Two containers, A and B, connected by a closed valve were evacuated and the Chlorine gas (Cl2) with a pressure of 200kPa was put into A and Oxygen (O2) with a pressure of 150kPa into B. a)If Container A has a volume of 10L and container B has a volume of 5L, find the total pressure within the system if the valve were opened and the gases allowed to mix b)Which of the gases would diffuse through the valve at a faster rate?

It has been a long time since I took chemistry and blundering around with Google didn't really seem to answer my question about a technical point about question (b) for this problem.
Obviously when you first open the valve between containers A and B the pressure is going to equalize. Since A has a higher pressure, gas will flow from A into B. Is this diffusion?
In a weird way the pressure equalization does fit the basic notion of diffusion - molecules mover from an area of higher concentration to an area of lower concentration. Thus if both containers had oxygen, then container A would be more concentrated (e.g mol $\rm L^{-1}$) since it has a higher pressure. Thus the molecules will move from A to B because there is a lower "concentration." However this doesn't really seem to fit my notion of either  diffusion or effusion. So is there a better proper term for this or is the phrase "pressure equalization" about as good as can be done? Or am I just wrong?
Obviously after the pressures equalize, then the oxygen from the oxygen/chlorine mixture will diffuse into the chlorine faster since oxygen molecules have a lower mass.
The gist is that the question would be simple to answer if containers A and B had the same pressure. But the difference in pressure seems to make the question turn on a very exacting definition of diffusion.

Its not diffusion, it's effusion. The difference is that with diffusion the particles/molecules are migrating through a permeable medium where as for effusion the particles/molecules are migrating through a pore/hole.
For example, a scent travelling across a room is diffuision, a scent from the room traveling outside is effusion. Also fun fact: the helium escaping through a balloon skin is actually effusion not diffusion.
helpful source
Side note: I am starting to flashback to all the times my thermo professors said diffusion when they meant effusion.

Perhaps one way to think about this is that free space diffusion can occur without pressure differences and even when the fluid, gas is homogeneous. Albeit you may not be able to differentiate one particle from another. Diffusion in free space is a passive process of mixing that does not require a pressure gradient and just relies on the movement of molecules due to Gibbs Free Energy.
Pressure differences caused by concentration gradients, compartments separated by membrane at different pressures, etc. just add to the potential for mixing.

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text:
Is it safe to store muriatic acid, hydrogen peroxide and acetone?
I am recently getting into PCB etching. So I use 10 M HCL solution, 3% hydrogen peroxide mixed in 1:1 ratio for my etchent. Currently, I am storing my HCL and hydrogen peroxide in my shed which can vary in temperature from -10 degrees Celsius to 25 degrees Celsius. The shed is outside my house. I also store my etchent next to HCL and hydrogen peroxide in the shed as well. Will it be safe to store all of these three chemicals inside a shed?
I do have metal tools in there but I don't care if it rusts. I am just afraid of hydrogen gas that might form because of HCL fumes. Also, I store my acetone indoor inside a plastic container because I heard that its flash point is -20 degrees Celsius. I don't want it to spontaneously catch on fire if I did store it in the shed. So are these storing method adequate or do I need more precautions.

You probably have gasoline in the shed which is about as dangerous as acetone. 3% hydrogen peroxide can be bought in a drug store, so no biggie. The 10M acid is a problem to me. It probably is in a glass bottle. What if you knock something over in the shed which breaks the bottle? Since you're asking the question I seriously doubt you're taking the proper safety precautions. Also kids then to play around in sheds.  So the shed needs to be locked at least.

PPE and ventilation are a must. Your info on acetone is way off - mine regularly withstands 40 Celsius (or higher). You'll get more expert advice that what I can offer here, but do be careful. Also: my glassware and all chemicals (including things like bug spray and the like) are in double-locked, explosion-proof cabinets.

I dont have gasoline in my shed and no one even comes in here anyway. The 10M acid is in a polyethylene jug not a glass bottle. Also the info about acetone's flash point is from MSDS.

Not a danger, but household 3% hydrogen peroxide slowly decomposes, and decomposes more rapidly when mixed with almost anything (phosphates being one exception). The peroxide's shelf life is limited, particularly once opened, or mixed with HCl, or with large temperature swings. Unless it takes up too much room, I'd store it in my house.

You don't need to take any special precautions when storing these chemicals together compared to how you would store them individually. They should all preferably be stored in their original, properly labeled containers with properly fitting lids.
Regarding your concern for hydrogen gas forming from the HCl fumes, that is simply not a reaction that will take place from only the compounds you've listed, or in combination with anything else you would typically find in a utility or storage shed. You will want to avoid breathing HCl fumes while working with it of course, but as far as a storage concern even this will not be an issue for you.
I commend your extreme caution here in general, although your fears of the acetone igniting from sitting in a hot shed are unwarranted. The flash point of a compound is the lowest temperature at which it can be ignited with an ignition source, like a spark. This is commonly confused with the autoignition temperature, which is the lowest temperature at which a compound can spontaneously combust without an ignition source. So, while you really don't want your acetone near a flame or spark source at any temperature, it's autoignition temperature is $\pu{465^oC}$ and so you don't have to worry about it igniting just from sitting in a hot shed.

I do recall researching on what can cause storage bottles to fail with age, and my recollection included the obvious (like the chemical nature of what is being stored, heating as your shed is likely not air-conditioned and the action of direct or diffused sunlight) together with other things I would not have expected.
For example, a polymer may be effected from the build-up of electrostatic discharge on parts of the surface area of the container (the top, for example) from dust (dirt) particles driven by wind drafts. A source notes:

Most plastic materials are electrical insulators and have the ability to support high static build up. Static is produced by charge separation caused by the movement of one material over another. Static build up can result in issues including:
•Increased handling and contamination issues during transport, storage and packing

With respect to contamination, see 'Effect of Electrostatic Charge on the Contamination of Plastic Food Containers by Airborne Bacterial Spores'. So, upon opening,  charged dust particle contaminants could enter a vessel, leading to a possible decomposition reaction.
A possible chemically more direct effect of electrostatic forces per this source, to quote:

formation of polyelectrolyte complexes through electrostatic interactions between the protonated amino groups of chitosan and the negatively charged side-chain groups in the other biopolymer at the operating pH [26,29].

This source  cites some new unexpected causes of failure of polymers with aging, to quote:

However, non-biodegradable polymers can be degraded/fragmented by various mechanisms: physical, such as heat and light, and chemical, such as oxidation, ionic radiation, and hydrolysis. Certain air pollutants such as CO, sulphur dioxide (SO2), nitrous oxide (NO) and ozone (O3) can also play a major role in the degradation of polymers [71]. The effect of the aforementioned degradation mechanisms to the polymer is to embrittle and fragment it into smaller pieces.

So ostensibly, storing in an outside shed may also, unexpectedly, introduce a role for air pollutants, as well, attacking the outside of the container. This may result in embrittlement, which together with physical moving of the vessel, could eventually cause leakage (even if you were storing something inert!).
Now, from higher temperature exposure along with a possible contamination of metal oxide riched charged dust particles entering the H2O2, I would expect a reduced life span with some oxygen evolution.
More interesting is there any expected impact on the HCl? If the hydrochloric acid is muriatic, already an Fe presence. Adding more metal ions from dust, I would expect only a minor loss in the acid strength, with limited to no gas creation. However, a possible increased attack of the container from within, proceeding via a REDOX reaction forming radicals, but only if there is a good amount of air available in the container along with a transition metal(s) ions and especially light. Here is a source on the alluded to  transition metal/oxygen based REDOX:

Transition metals are efficient catalysts of redox reactions and their reactions with dioxygen are not spin restricted. Therefore it is likely that the "autoxidation" observed for many biomolecules is, in fact, metal catalyzed. In this paper we discuss: 1) the quantum mechanic, thermodynamic, and kinetic aspects of the reactions of dioxygen with biomolecules; 2) the involvement of transition metals in biomolecule oxidation; and 3) the biological implications of metal catalyzed oxidations. We hypothesize that true autoxidation of biomolecules does not occur in biological systems, instead the "autoxidation" of biomolecules is the result of transition metals bound by the biomolecules.

Related comment on acetone with no presence of dioxygen, likely no chemistry-based issues. Otherwise, still not too likely that one will see breakdown products (including gases, like CxHy). However on opening, one has a mixture of acetone and air, together with any possible ESD, is not a completely safe scenario.


text:
Does a double bond or a triple bond have a higher priority in nomenclature?
I have this structural formula:

Should it be named pent-1-en-4-yne or pent-1-yn-4-ene?

The relevant section in the 2013 IUPAC Recommendations (Blue Book) reads:

P-31.1.1 [...] Locants as low as possible are given to multiple bonds as a set, even though this may at times give 'yne' endings lower locants than 'ene' endings. If a choice remains, preference for low locants is given to the double bonds. [...]

In this case, the two possible numbering schemes lead to the same locant set of (1,4).

However, according to the italicised portion of the quoted text, the double bond is considered to have the higher priority, and the name would correctly be pent-1-en-4-yne.
Note that the final "-e" from "-ene" is dropped if the following suffix begins with a vowel or "y," and is retained if the suffix begins with a
consonant. Additionally, spaces are only used in very specific cases, like in the carboxylic acid ending "-oic acid." Otherwise, dashes are used to separate different functional groups and numbers in the name.

According to R-$3.1.1$ of IUPAC, Commission on Nomenclature of Organic Chemistry. A Guide to IUPAC Nomenclature of Organic Compounds (Recommendations 1993), 1993, Blackwell Scientific publications, Copyright 1993 IUPAC:

The presence of both double and triple bonds is similarly denoted by endings such as "-enyne", "-adienyne", "-enediyne", etc.
Numbers as low as possible are given to double and triple bonds as a set, even though this may at times give "-yne" a lower number than "-ene".
If a choice remains, preference for low locants is given to the double bonds.
Only the lower locant for a multiple bond is cited except when the numerical difference between the two locants is greater than one, in which case the higher-numbered locant is cited in parentheses (see R-$0.1.4$).

(Original text does not contain numbering.)

According to Rule 1, it would not be of the format pent-1-ene 4-yne or pent 1-yne 4-ene, but rather pent-1-en-4-yne.
Therefore, $\ce{CH#CCH=CHCH3}$ is named (3E)-pent-3-en-1-yne according to Rule 2, but the molecule in question ($\ce{CH2=CHCH2C#CH}$) is named pent-1-en-4-yne according to Rule 3.
Rule 4 is irrelvant here.

In addition to the correct answers please also note that the unsaturations in the carbon chain have no preference over the chain length, when looking for the principal chain, according to the current rules. So, the preferred name of the substituted structure

is, interestingly, not 3,3-dipropylpent-1-en-4-yne, but 4-ethenyl-4-ethynylheptane (see Longest chain IUPAC rule).

Are you sure about this? Because in school I have been taught the opposite, i.e. to choose the double and triple bonds in the chain regardless of length. Please confirm it and provide some references.

@Kartik This is how it used to be, but the IUPAC naming rules were changed in the last few years. See, e.g., this answer by Loong.

Ok .... so I guess that many Indian Books will need to be updated ... I will try to contact their authors and notify them


text:
If sigma bonds are stronger than pi bonds, why are the BDE's lower and the bonds longer?
If sigma bonds are stronger than pi bonds, why then do we say that the triple bonds which have 2 pi bonds are stronger than single bonds?

Some points to clarify:

Sigma bonds is indeed stronger than pi bonds.
Therefore, sigma bonds have a lower energy than pi bonds (although sigma bonds do indeed have a higher dissociation energy than pi bonds).
Double bonds (usually) contain $1$ sigma bond and $1$ pi bond.
Triple bonds (usually) contain $1$ sigma bond and $2$ pi bond.

From this, you can see why the triple bond has a higher (dissociation) energy than a single bond: because a triple bond contains three bonds, one of which is the sigma bond also present in the single bond.


text:
Are groups other than carbonyl reduced by Clemmensen reduction?
According to my textbook, Clemmensen reduction reduces only aldehydes and ketones to alkanes. But in this question:

$\ce{Zn-Hg/HCl}$ has reduced both the ketonic and nitro group. Correct answer given is D.
So, what exactly is this reduction's mechanism? Is it an exception? If not, why isn't it stated in the books alongside the carbonyl groups?

The paper - Diez-Cecilia, E., Kelly, B., Rozas, I., One-step double reduction of aryl nitro and carbonyl
groups using hydrazine, Tetrahedron Letters (2011), doi: 10.1016/j.tetlet.2011.09.112 - discusses Wolff Kishner reduction of nitro compounds. I still find it surprising, though, that Wikipedia doesn't mention even once at least the possibility of reduction of nitro compounds by Clemenssen reduction, let alone their industrial use or whatever.

@AshishAhuja This may be of use https://www.sciencedirect.com/topics/chemistry/clemmensen-reduction and this as well https://www.organic-chemistry.org/namedreactions/clemmensen-reduction.shtm

@AshishAhuja Seems like the reduction here is specific to aryl nitro groups, see here https://www.masterorganicchemistry.com/2018/10/08/nitration-baeyer-villiger/ and you might also want to take a look here https://www.researchgate.net/publication/257908880_Reduction_of_aromatic_nitro_compounds_to_amines_using_zinc_and_aqueous_chelating_ethers_Mild_and_efficient_method_for_zinc_activation maybe one of the references of this paper might have some related information.

@AshishAhuja  Found it! https://onlinelibrary.wiley.com/doi/10.1002/9780470638859.conrr063 the abstract is enough and here is Bechamp reduction https://en.wikipedia.org/wiki/Bechamp_reduction.

@AshishAhuja This as well confirms the doubt. https://pubs.acs.org/doi/pdf/10.1021/ie50513a019?src=recsys

@Rishi thanks, I had a look at the papers you linked. I also found this and this. I don't think it is specifically called Bechamp reduction, since that seems to be using iron. It looks like such reductions only apply to nitroarenes, albeit none of these seem to be using exactly the same Zn-Hg/HCl mixture. You can post an answer quoting the papers if you'd like (you can find them on SciHub).

@AshishAhuja Yes, I think I made a mistake, it may not necessarily follow a mechanism similar to Béchamp. But as mentioned in this Zinc and acid has been mentioned and even in this the same has been mentioned. I also forgot the fact that Clemmensen uses the amalgam :/ . Also, I don’t think I will be able to write an answer. Sorry! :(

@AshishAhuja You want source for what? That Clemmensen isn't magical reduction that has perfect selectivity? On contrary, it's an old method using aggressive reagent and reactions like Wolff–Kishner and Mozingo reduction tend to be used instead.

@Mithoron I was looking for a source for the reduction of a nitroarene to aniline using Zn-Hg/HCl, and maybe it's mechanism. I've found papers (after the bounty) which reference the same reaction with zinc + mild acid, so I'm satisfied with that. You say that Clemmensen is an aggressive reagent, but I am yet to find a source which uses the classic clemmensen reduction Zn-Hg (or zinc dust) with a conc. acid to reduce nitroarenes.

https://chemistry.stackexchange.com/questions/29192/how-can-an-aromatic-ketone-be-reduced-in-the-presence-of-a-nitro-group https://chemistry.stackexchange.com/questions/140426/how-does-al-hg-react-with-carbonyl-in-presence-of-concentrated-hcl\

@Mithoron the first question talks about clemmensen reducing the nitro group but without a reference, I don't see how the second question is relevant. Also (second paragraph): "Moreover,  when  attempted  in  the  present  work, the concurrent reduction of carbonyl and nitro groups with  Zn,  using  Clemmensen  conditions  (as  described  in  reference 6), was not achieved", which is exactly what this question is asking about.

https://chemistry.stackexchange.com/questions/60959/are-groups-other-than-carbonyl-reduced-by-clemmensen-reduction

I would look at it in the following way: The oxidation state of the carbonyl carbon is +II, while the oxidation state of the nitrogen in the nitro group is +III. So nitrogen is in a higher oxidation state and it is also more electronegative. Thus, isn't it sensible that it gets reduced first before the carbonyl group? In fact, reduction of a nitro group is also possible using much milder reducing agents (e.g. only zinc dust).
In my experience, most questions of chemoselectivity can be answered by "thinking about it". However, if you would still like to have a good book for that, I would recommend Warren "Organic Synthesis: Strategy and Control" which contains a very nice treatment of chemoselectivity problems.

Just a correction: " The oxidation state of the carbonyl carbon is +II" oxidation state of carbon in aldehyde is +I instead; +II occurs for ketones only. Doesn't change the fundamental reason of your answer though.

@GaurangTandon I don't get that. The oxidation state of the carbonyl carbon in the above reaction is +II. This wasn't meant as any general statement.


text:
What is the functionality(reactive sites) of phenol molecule?
My class 12 chemistry textbook says that phenol molecule has three reactive sites or it's functionality is three. But I am confused whether it is so or not, because, what I know, phenol has three double bonds, it must have a functionality of six. Besides, it also has an -OH which is yet another reactive site, so in all making it seven. Please help so that I may have a clear understanding of how to judge functionality of a monomer molecule.

Could you quote the original text?

"The functionality of ethene, vinyl chloride and styrene is 2 because these molecules have 2 reactive sites whereas the functionality of glycerol and phenol is three because they have three reactive sites."

Could you quote the whole section/paragraph?

"Functionality of a monomer is the number of bonding sites in it. For example, the functionality of ethene, vinyl chloride and styrene is 2 because these molecules have 2 reactive sites whereas the functionality of glycerol and phenol is three because they have three reactive sites."

Maybe it's talking about this polymer of phenol, where indeed phenol only opens the ortho- and the para- sites.

Phenol has $3$ double-bond-equivalent, but not $3$ localized double bonds:
Look at its resonance structure:

Courtesy Wikipedia.
Therefore, to a electrophile, there are $3$ reaction sites, which are the ortho- and the para- positions.
A nuecleophile does not attack phenol.

Why oxygen atom is not a reaction site?


text:
Oxidation states of oxygen
I'm given a question:

Oxidation number of $\ce{O}$ in $\ce{BaO2}$ is $x$ and in $\ce{OF2}$ is $y$; then the value of $x+y$
  is what?

Now my main question is that if $\ce{F}$ has $-1$ valency in $\ce{OF2}$ then $\ce{O}$ must have valency of $+2$. But is that possible since $\ce{O}$ mainly has oxidation states of $-1$ & $-2$ or am I making any mistake?

That is possible, and O can take any value from -2 to +2.

one more thing - oxidation number includes the +- signs with it?

Yes. ${}{}{}{}$

Does valency include the signs too?

No.${}{}{}{}{}{}$

it is one of the exceptional oxidation state of Oxygen.another exception is O2F2 where oxidation state of o is +1

The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegative atom acquire a positive charge.
Depending on this hypothesis oxygen have 5 oxidation states.

In all the oxides,oxygen has an oxidation state of $-2$. Eg. $\ce{CO2,CO}$
In all peroxides (oxygen-oxygen linkage), oxygen has an oxidation state of $-1$. For example, consider $\ce{H2O2}$, here $\ce{H}$ is less electronegative so it will acquire a charge of $+1$ and to balance the $2$ positive charge of 2 H-atoms,each oxygen atom will acquire a charge of $-1$.
In all superoxides ($\ce{KO2,CsO2,RbO2}$), oxygen has an oxidation state of $-\frac{1}{2}$,this is because $\ce{K,Cs,Rb}$, being elements of the first group and less electronegative than oxygen acquire a charge of $+1$, to balance it, each oxygen atom acquires a charge of  $-\frac{1}{2}$.
In one of the exceptions $\ce{OF2}$, the fluorine being more electronegative acquires a charge of $-1$ and to balance the $-2$ charge of 2 fluorine atoms oxygen acquires a charge of $+2$.
As last, there is $\ce{O2F2}$, similarly here to balance the $-2$ charge on 2 $\ce{F}$-atoms each oxygen atom acquire a charge of $+1$.

You’re forgetting $\pm 0$ which is oxygen’s oxidation state in $\ce{O2}$.

Thanks @Jan to mention it but i thought it is too simple that everyone who have studied redox must be aware of it.

If you’re going to list all, then list all imho ;)

i knew about 0 but can you please tell me about the sign + and -

$\pm$ gives $\pm$.

Add $+(1/2)$ for the compound often (a little oversimplified) rendered as $\ce{O_2^+PtF_6^-}$, formed by combining oxygen with platinum hexafluoride.  This reaction heralded the similar oxidation of xenon which busted the notion of "inert gases".

There can also be a $0$ oxidation state in compounds, not just elemental oxygen.  Hypofluorous acid is known.  This molecule with the structural formula $\ce{H - O - F}$ has (in the oxidation state formalism) oxygen gaining an electron from hydrogen but losing one to fluorine.

Oxygen can have 6 different oxidation states. Which oxidation state oxygen is in depends on which element oxygen is bonded to and what ratio the two elements are at:

-2: This occurs in oxides e.g. $\ce{OsO4}$ and $\ce{RuO4}$
-1: This occurs in peroxides  e.g. $\ce{H2O2}$
-0.5: This occurs in superoxides e.g. compounds that contain the $\ce{O^-2}$ ion such as $\ce{KO2}$
0: This occurs in $\ce{O2}$
+1: This occurs in $\ce{O2F2}$ as $2 \cdot -1+2 \cdot 1=0$ and fluorine has a greater electronegativity than oxygen.
+2: This occurs in $\ce{OF2}$ as $+2+2 \cdot -1=0$

Oxygen can have a range of oxidation states including -2,-1,0,+1,+2
So, yes oxygen can have positive oxidation states when present with more electronegative elements than it in a compound.


text:
Balancing the equation
I've got a question:
$$\ce{Cr2O7^2- ->   Cr^3+}$$
This reaction takes place in a acidic medium - how many $\ce{H+}$ are required to balance the equation?
How do I balance this reaction.
Cr had a oxidation state of $+6$ at first then it gained three electrons and now has an oxidation state of $+3$. What does that have to do with $\ce{H+}$?
Additionally, how would I balance this reaction if it were to take place in a basic medium.

$\ce{14H+ + Cr2O7^2- + 6e- <=> 7H2O + 2Cr^3+}\\ce{7H2O + Cr2O7^2- + 6e- <=> 2Cr^3+ + 14OH-}$

This is a basic stoichiometry question. Reaction equations must be balanced due to the law of conservation of mass and charge

First balance the number of atoms both side (except H and O).
$$
\ce{Cr2O7^2- -> 2 Cr^3+}
$$
Since the oxidation number of Cr on the left is $+ 6$ while on the right, it is $+ 3$, since 2 atoms of Cr are involved so reduction has occur. Add 6 electrons  on the left hand side thus,
$$
\ce{Cr2O7^2- + 6e- -> 2 Cr^3+}
$$
To balance the charge we have to add $\ce{H+}$ on either side, since charge on the left is $-8=-2+ -6$ and on the right it is $+ 6$, so add $\ce{14H+}$ to the left hand side. Thus,
$$
\ce{Cr2O7^2- + 6e- + 14 H^+ -> 2 Cr^3+}
$$
To balance the O atoms add $\ce{H2O}$ on the other side,
$$
\ce{Cr2O7^2- + 6e- + 14 H^+ -> 2 Cr^3+ + 7 H2O}.
$$
The H atoms will autometically get balanced.
The next question is what will happen in basic medium, proceed till $\ce{Cr2O7^2- + 6 e -> 2 Cr^3+}$ as earlier.
In acidic medium we add hydrogen to balance the charge, do it here too but also add same number of $\ce{OH-}$ to both sides. As in the example,
$$
\ce{Cr2O7^2- + 6e- -> 2 Cr^3+},
$$
it will become,
$$
\ce{Cr2O7^2- + 6e- + 14 H^+ + 14 OH- -> 2Cr^3+  + 14 OH-}
$$
$$
\ce{Cr2O7^2- + 6e- + 14 H2O -> 2Cr^3+  + 14 OH-}
$$
Add $\ce{H2O}$ on either side to balance the O atoms,
$$
\ce{Cr2O7^2- + 6e- + 14 H2O -> 2Cr^3+  + 14 OH- + 7H2O}
$$
$$\ce{Cr2O7^2- + 6e- + 7 H2O -> 2Cr^3+  + 14 OH-}$$
And that is the answer.

See here about how to write chemical reaction in mhchem notation.

You don’t have to first add $\ce{H+}$ and then additionally add $\ce{OH-}$ — you can just add $\ce{OH-}$ to balance the charge, just remember to do it on the side with excessive positive charge. Balancing with water should then happen on the other side.

i agree,i just added  H+ and then OH- to tell the OP that Why water is formed.

Nice answer. $+1$


text:
Why isn't the orbital angular momentum also considered while calculating the magnetic moments 3d transition elements?
My textbook (Chapter: The d- and f- Block Elements) makes an interesting assertion, however, without any reason to back it up.

Paramagnetism arises from the presence of unpaired electons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e.,
$$\mu = \sqrt{n (n + 2)}$$
where $n$ is the number of unpaired electrons and $\mu$ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of $1.73$ Bohr magnetons (BM).

Now from what I've learnt from my physics classes last week, the magnetic moment of an electron is calculated using the formula:

$$\frac{m}{L} = \frac{e}{2 m_e}$$
Magnetic moment of revolving electron,
$$m = \frac{e}{2 m_e}L$$
In vector form,
$$\vec{m} = - \frac{e}{2m_e}\vec{L}$$

Where $m$ is the mass of the electron, $e$ is the magnitude of charge associated with the electron and $L$ is the angular momentum of the electron.
Now as I understand it, the angular momentum of an electron in an atom is the resultant of the electron's spin angular momentum and its orbital angular momentum.
But (as you can clearly see) my textbook exhibits an unexplained enthusiasm for only the spin angular momentum. It dismisses the contribution of the orbital angular momentum, as inconsequential, when determining the magnetic moments of electrons in the 3d elements.
As for the 'reason' provided by my textbook:

For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and is hence of no significance.

Well, I fail to see how that even counts as a 'reason'. I did ask my teacher about this, but she expressed her reluctance to go into the details; she's of the opinion that my particular query is “not worth clarifying” since it has no significance from the examination point of view (“Just give them what's in the textbook Aaron, nothing more and certainly not anything less”).
Subsequent internet searches have not yield any satisfying explanation (Perhaps I didn't use the right keywords while searching?).
So I'll just break up my question, point-wise, and list them below:

Is the orbital angular really of no significance when calculating the magnetic dipole moment for 3d elements? If it is significant, is there any 'educational' purpose behind that omission in my textbook, or is it just a mistake?

(As stated in the book) What causes the 'quenching'? How is this 'quenching' effect caused?

Do the transition elements of the other periods also have "insignificant" orbital angular momentum contributions? Is it for the same reason as the 3d transition elements.

It is generally quenched for symmetry reasons ( lower symmetry, less orbital contribution)

Simplistically speaking orbital angular momentum is present when some conditions are satisfied:

A set of orbitals are degenerate;
These orbitals can be "interconverted" by rotation about a certain axis;
The set of orbitals is not empty, half-filled, or fully filled.

Physically speaking, the "rotation" of an electron from one orbital to the next generates orbital angular momentum. The rotation must not have an energy cost, which necessitates degeneracy. It also means that the set of orbitals must have at least one electron, as well as one empty space (a so-called hole) for the electron to "rotate" into. The hole must have the same spin as the electron, which means that a half-filled subshell doesn't allow for orbital angular momentum (since all the electrons have parallel spins).
This is most easily illustrated with the three 2p orbitals in an atom. Rotation of a $\mathrm{p}_x$ orbital by 90 degrees around the $y$-axis, for example, turns it into a $\mathrm{p}_z$ orbital.
Any atom with a $\mathrm{p}^1$, $\mathrm{p}^2$, $\mathrm{p}^4$, or $\mathrm{p}^5$ configuration therefore possesses nonzero angular momentum. This is also reflected in their term symbols where $L = 1$. This is said to be "one unit of angular momentum". An atom with a half-filled shell (for example, $\ce{N}$)  or a closed-shell $\mathrm{p}^6$ configuration (for example, $\ce{Ne}$) does not have any orbital angular momentum.
The five 3d orbitals in an atom also satisfy these requirements. It is less intuitive to see this as compared to the 2p orbitals, because we have taken specific linear combinations of the complex 3d orbitals and converted them into real orbitals. Nevertheless one can use quantum mechanics to prove that the 3d orbitals possess angular momentum. The maths can be found in Figgis and Hitchman's Ligand Field Theory and Its Applications (it is very early in the book, but I do not have it with me now and cannot remember offhand where it is). As such, for free 3d metal atoms or ions, the orbital angular momentum has to be incorporated into the calculation of the magnetic moment. Exceptions are, of course, $\mathrm{d^0}$, $\mathrm{d^5}$, and $\mathrm{d^{10}}$ ions (e.g. $\ce{Sc^3+}$, $\ce{Mn^2+}$, and $\ce{Zn^2+}$).

That is applicable for free ions. However, 3d metals do not commonly exist as free ions, but exist in coordination complexes with a ligand field. For simplicity we'll consider a perfect octahedral ligand field, which causes d orbitals to split into a $\mathrm{t_{2g}}$ ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$) and $\mathrm{e_{g}}$ ($\mathrm{d}_{z^2}$, $\mathrm{d}_{x^2-y^2}$) set. This means that the five d orbitals are no longer all degenerate, breaking the first condition stipulated above.
At a simplistic level, many textbooks will claim that this leads to a "quenching" of orbital angular momentum, which just means that there is no longer any orbital angular momentum, and as such, the spin-only formula applies. [Note that there are some second-order effects (quantum mechanical mixing of excited states into the ground state) that can lead to deviations from the spin-only formula.]
This is true to a certain extent. However, the five 3d orbitals are not split into individually nondegenerate orbitals; the three $\mathrm{t_{2g}}$ orbitals, for example, are still degenerate. On top of that, one can verify pictorially that it is possible to rotate the $\mathrm{t_{2g}}$ orbitals into one another. In this aspect the $\mathrm{t_{2g}}$ orbitals behave quite similarly to the three $\mathrm{p}$ orbitals in a free atom; this is called the "$\mathrm{t_2}$-$\mathrm{p}$ isomorphism". As such, the $\mathrm{t_{2g}}$ orbitals still have orbital angular momentum associated with them.
Any complex with unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals (i.e. $\mathrm{t_{2g}}^1$, $\mathrm{t_{2g}}^2$, $\mathrm{t_{2g}}^4$, $\mathrm{t_{2g}}^5$) therefore possesses orbital angular momentum and exhibit great deviations from the spin-only formula. These complexes all have a $\mathrm{T_{1g}}$ or $\mathrm{T_{2g}}$ ground state term symbol, so it is commonly said that "$\mathrm{T}$ terms possess one unit of orbital angular momentum". Spin-orbit coupling has to be invoked to obtain a better agreement between the calculated and measured magnetic moments.
As an example of how large the deviations can be, consider the compound $\ce{(NH4)2Co(SO4)2.2H2O}$. This features octahedral Co(II) which adopts a high-spin configuration $(\mathrm{t_{2g}})^5(\mathrm{e_g})^2$. From the spin-only formula, we would expect $\mu_\mathrm{so} = \sqrt{3(3+2} = 3.87$. However, the unsymmetrical population of the $\mathrm{t_{2g}}$ orbitals leads to the generation of orbital angular momentum, and at $300~\mathrm{K}$ the measured effective magnetic moment is actually $5.1$. (Data from here.)

A more detailed discussion can be found in any of the standard inorganic chemistry textbooks. Figgis and Hitchman's book (mentioned earlier) is more advanced and has a very comprehensive treatment of the subject.

Thank you for this amazing answer! I just wonder why doesn't JT distortion break the symmetry of e.g. $t_{2g}^1$ set, whenever it occurs?

It does occur, and $\mathrm{d}^1$ complexes are susceptible to JT distortions. However, usually, it is the unsymmetrical occupancy of $\mathrm{e_g}$ (for example, high-spin d4 Cr(II) and d9 Cu(II)) that lead to large JT distortions. The $\mathrm{e_g}$ orbitals point directly towards the ligands, such that there is a greater change in the energy (i.e. greater stabilisation) upon geometrical distortion

whereas for the $\mathrm{t_{2g}}$ set which point in between the ligands there is not much difference in the energies upon geometrical distortion, so the corresponding distortion is smaller (think of it as not having any incentive to distort). So, it's a decent approximation to treat the complex as having an undistorted geometry. However I am fairly sure that JT distortions will come into play at some point in time when calculating magnetic moments, and I think it is also in Figgis/Hitchman, but I do not have the book with me right now (and I did not read it - it is a complicated matter!).

I try to understand that from a more physical perspective: If the electronic Hamiltonian does not commute with the angular momentum operator $\hat{L}_z$ then the eigenstates are in general no eigenfunctions of $\hat{L}_z$. If I now take some linear triatomic molecule with non-zero $m_z$ (like $\hat{L}_z|0> = m_z \hbar |0>$) and bend it slightly, by an angle $\epsilon$ what will happen? I suppose I still can calculate the expectation value $<0(\epsilon)|\hat{L}_z|0(\epsilon)>$. But what exactly will get? I suppose some real value ($\hat{L}_z$ is hermititan after all) close to $m\hbar$.Any idea?

I suppose in most cases of non-symmetric molecules this expectation value is close to (or =) zero. Since in the symmetrized case $m$ should be usually $=0$ in the first instance (for ground states), right?


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Dissolving Metal Reductions of Organic Compounds
I'm having trouble understanding why dissolving metal reductions, like (for example) sodium with alkenes in liquid ammonia, sodium with aldehydes/ketones in ethanol, or the Bouveault–Blanc reduction, reduce the substrate instead of the solvent. If you put sodium in ethanol (or any protic solvent), the ethanol is reduced and hydrogen gas is released. I can't see how the presence of a ketone (for example) will create a more favorable reaction.
For once, it doesn't seem thermodynamically favorable - reduction of the ethanol will release hydrogen gas, largely increasing the entropy, while reduction of the ketone produces no gas. Even if it is thermodynamically favorable, however, reduction of the ethanol seems much more favorable kinetically, since there are much more solvent molecules than substrate molecules. What is going on here?

Looking at some typical reaction conditions (e.g. http://www.organic-chemistry.org/namedreactions/bouveault-blanc-reduction.shtm) I see that the proton source (e.g. ethanol, isopropanol) is added in stoichiometric amounts and is not employed as the solvent. It would strike me as very odd if the sodium would not just be quenched if it is dissolved in ethanol. In liquid ammonia, dissolved electrons are formed, just have a look at the wikipedia article of the Birch reduction.

There are plenty of examples where ethanol is present at over 10ml/g for this reaction. The key thing is that this is transfer of a single electron from Na initially. Where would this go onto ethanol? Clearly, it is faster to transfer an electron to an unsaturated compound i. e. The carbonyl compound. Once this occurs, protons are transferred. The movement of electrons and protons is fast compared to getting the two requisite H's together to form hydrogen and drive the alternative reaction.


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Why do pregnant women turn toilet seats blue?
My partner is pregnant, and thought perhaps she was going mad when one day she found our toilet seat had turned blue. After some investigation, it seems this is a real phenomena, see here for a nice example, with the image from this link shown below.

One explanation suggested is dye transfer from new maternity jeans, however we find this unlikely for a variety of reasons:

This is late pregnancy and the jeans have been worn for some time without this effect.
My partner assures me she takes her trousers off to go to the toilet, and even her underwear, and there is not a white patch where the underwear would have been (and therefore no dye transfer)

The phenomenon is also described as possibly relating to taking certain pharmaceuticals for Crohn's disease. My partner happens to take (Sulfasalazine) which is used for this (although it's not what she's taking it for) and she's never encountered this before, despite taking them for some time. Another suggestion we've seen is that it is from the dye in a batch of pregnancy vitamin tablets she's started taking, which are indeed blue, although, she's also been taking these daily for a month or so.
However, we don't know the answer, and these are just ideas, so why has our toilet seat turned blue?!

Whoever voted to close this as a medicine question, please provide your reasoning. As far as I know, toilet seats won’t visit doctors (and aren’t concerned about their health).

@bon, It's definitely about chemistry, I want to know what chemical reaction is causing this. If you suggest there's gamma rays shooting from her posterior and ionising the seat to cause this effect, or something like this I'll happily move it to Physics, but its hardly medicine. I originally gave it the tag 'biochemistry'.

Would like to know any reasons for downvoting this question so I can improve it thanks! I don't see any immediate problems with how it is presented, and it is a genuine question.

Unanswerable via internet most probably.

@Mithoron, I'm not sure what you mean?

This is... weird. Excuse my question, but if you are talking about 'dye transfer' then what is the color of your partner's posterior?

@vapid, lol, I meant from jeans=>posterior=>toilet! but we don't think this is what is happening for the reasons given.

OK, I gave it a little though and I think that the most probable answer is that some drug or dietary product is excreted with the sweat, which causes the coloration, as you suggested. Sulfasalazine is known to excrete with bodily fluids, but it is orange. It may be some metabolite of it. The other culprit is that vitamin supplement. I suggest you to take one tablet, dissolve it in water and apply a few drops on a white part of the toilet seat. If the polymer will absorb the dye and turn blue, then that's your answer.

@vapid I'll see if I can give this experiment a go. By the way, on the link I gave in the question, there is a suggestion of a reaction with anti-microbial coating, but no suggestion for what excreted substance might react with it. I'm a little skeptical the tablets are the cuprit however, as she has been taking both of these for some time before the incident without effect.

It may be possible to explain why it didn't happen earlier despite taking those pills for a prolonged time. First we need to find the source of coloration.

The question cannot be answered without a chemical analysis.
However, given that there exist toilet seats with silver based anti-microbial coatings, the bluish grey color may come from metallic silver particles produced by a reduction process similar to the photographic development process.
You could proof the presence of silver by applying Caffenol, an alkaline aqueous solution of instant coffee, washing soda, and vitamin C (ascorbic acid) to the toilet seat. I suggest to use Caffenol-C-M whose preparation is described here. Let the Caffenol react on the toilet seat as long as possible. If grey or black stains are formed that persist after cleaning, your toilet seat contains silver with high probability. Note that the stains cannot be removed anymore.
If the test is positive, the next step would be to investigate why your pregnant partner has reducing powers.

Now that we actually have the baby, finding time to investigate these things is a bit more difficult,  but I'll look at the link you gave.

I do not know the chemical process. I do know it is triggered by a change in hormone levels and appears to be more extreme when vitamin b levels change as well.
Chromhidrosis is a condition that causes visibly blue sweat. However, you would see blue staining on her clothing and bed sheets as well.
Hormone changes can cause temporary pseudochromhidrosis. This condition produces clear sweat which reacts with certain surface bacteria and the result on some surfaces is blue/purple staining.
Only an acid based cleaner will remove it.

The sheets and so on definitely were not affected, but you second explanation is worth investigating.


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Deprotection of carboxybenzyl (Cbz) in the presence of Methyl ester?
Deprotection of carboxybenzyl and methyl ester can easily be achieved by using ammonium formate followed by addition of 10% Pd/C and LiOH/MeOH respectively.
But how we can achieve deprotection of carboxybenzyl group in the presence of methyl ester?

Just leave out the last step, i.e., LiOH/MeOH and you should be just fine, the Cbz group gets removed under reductive conditions which leave the methyl ester unaffected. If you lose the methyl ester, you could just add it again afterwards.

Are you sure that reduction using Pd/C will not cleave the methyl group. As, I know t-Butyl ester is stable against reducing agent like Pd/C but I am not sure about Methyl group. any insight or reference ?

I would at least not suspect esters to be very labile in Pd/C hydrogenation. While this is in principle possible, the carbamate of the Cbz will be cleaved quicker. So, I don't really see that to be a significant problem. Here are two references that look pretty similar to your problem:
T. Maki, T. Tsuritani, T. Yasukata, Org. Lett., 2014, 16, 1868-1871.
G. S. Vanier, Synlett, 2007, 131-135. (see http://www.organic-chemistry.org/protectivegroups/amino/cbz-amino.htm)
One other approach would be to use another reducing reagent for the deprotection that leaves the ester unaffected (NaBH3CN?)


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Will supercritical nitrogen stay supercritical if we raise temperature and pressure high above critical values?
Nitrogen's critical temperature is -146.96 °C and critical pressure 33.96 bar. If we maintain nitrogen at 220 bar and 400 °C, what will be the state of $\ce{N2}$? Will it be still supercritical? I know that phase diagram says after critical temperature and pressure we have a supercritical phase, but with enough p and T, more exotic states of matter occur.

http://www.wolframalpha.com/input/?i=nitrogen+400+%C2%B0C+220+bar

You'd need more zeros to get something more interesting happen.

@Loong Wolfram is dumb - even for incredibly high values it still says supercrit.

If there is no experimental data then we can only rely on extrapolating from the last  known values, which may or may not be accurate. A good guess, however, would be that it is a supercritical fluid until proven otherwise.

related http://chemistry.stackexchange.com/questions/20009/hypothetical-what-happens-to-water-as-pressure-increases-to-infinity

If we maintain nitrogen at 220 bar and 400 °C, what will be the state of $\ce{N2}$? Will it be still supercritical?

Yes, it will still be supercritical.  The only reasonable definition of "supercritical" is something along the lines of "a fluid phase without surface tension wherein the fluid pressure and temperature exceed the critical point".
Thus, compressed air is a supercritical fluid.  The same is true of compressed nitrogen.
You may be wondering how "supercritical" air is different from regular, non-supercritical, gaseous air.  The answer is, it isn't meaningfully different.
If you research "supercritical fluids", it will be very hard to discover this fact.  That is because the "exotic" properties of "supercritical fluids", such as having densities "in between that of a gas and a liquid", are really only properties of supercritical fluids that are near (i.e. slightly above) the critical point.
It isn't really much of an exaggeration to say that supercritical fluids are really gases.  They have no surface tension.  They expand to fill their container.  Just like gases.  Near the critical point, supercritical gases can have densities that are very unusual for a gas. At these high densities, the gas is very non-ideal, and intermolecular interactions in the gas lead to all sorts of interesting behavior.  Those high-density intermolecular interactions give rise to the interesting properties that people refer to when they say "supercritical fluid".  The strength and nature of those interesting intermolecular interactions depends strongly on density, which is why people refer to "supercritical fluids" having "tunable" properties.  Small changes in density (by changing temp. or pressure) can give rise to large changes in those interactions.
At conditions way past the critical point, those intermolecular interactions don't matter very much any more, and the "supercritical fluid" again starts to behave more and more like an ideal gas.

I know that phase diagram says after critical temperature and pressure we have a supercritical phase, but with enough p and T, more exotic states of matter occur.

Yes, other answers and comments have done a good job of discussing this point.  If you increase the pressure enough, eventually you'll have a black hole!  Of course, long before that happens, you can get other less exotic phases of matter such as various solid phases.  In general, predicting the high-pressure phase behavior of even small molecules like nitrogen is very difficult, and as such, experimental data is required.  Unfortunately, measuring the high-pressure phase behavior of materials is also very difficult, at least past a few hundred gigapascals.  So I'm not sure where the high-pressure "exotic" phases of e.g. solid nitrogen start to form.  But it's not at 220 bar and 400 ° C, under those conditions nitrogen will just be a gas, albeit a "supercritical" gas.

220 bar and 400 °C isn't high for supercritical fluid; in fact it's about critical point of water. You could turn such fluid into something else, if you applied high enough pressure and/or temperature (much higher then these).
At this temperature you'd need more than 10 GPa to solidify it into cubic δ nitrogen (IV) (source). At room temperature you can get "ordinary" nitrogen I ice even below 3 GPa, or the cubic δ phase close to 5 GPa.
On the other hand, to get thermal plasma you'd need much higher temperature - three orders of magnitude and $\ce{N+}$ ions and electrons would be flying around (check this for details).
If you apply both high pressure and temperature you can get, for example, interesting cubic gauche phase of nitrogen, but it's over 110 GPa and 2000 K needed there.

+1 mostly for the nitrogen(IV), learned something new.

I think I'm gonna edit it some more, it's quite interesting topic.

Surpassing the supercritical point -- and the parameters provided by you exceed both the critical pressure and the critical temperature you provided, too -- will still yield the supercritical state; i.e. the absence of a gaseous phase in presence of a liquid phase.  It wont be a plasma, but still a supercritical fluid, as $\ce{scCO2}$ already used for decaffeination and chromatography, for example.


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Debye-Huckel limiting law
I encountered this equation when studying the Electrolyte solutions chapter in my analytical Chemistry textbook, Actually the terms in summation are giving a physical clue of what $I$ is, But what's the point of that $\frac{1}{2}$ ?
I've searched the Internet and a molecular Thermodynamics book that I have but I found nothing but the equation itself.
$I=\dfrac{1}{2}\sum_{i}m_{i}z_{i}^{2}$

The ionic strength I has been defined in this way since 1921. It is just the molality of a 1:1 electrolyte.

$I$ is the ionic strength of a solution, and is a measurement of the concentration of ions in a solution. $m_i$ is the concentration of the ion i, and $z_i$ is the charge of that ion. The $\frac{1}{2}$ in front of the summation is a result of the fact that both positive and negative ions are considered when calculating the ionic strength. $m_i$ could either be a measurement of Molarity or molality, and your book should state which definition you are using (I assume molality due to the lowercase m).

$m$ is mass, $b$ is molal concentration (molality) and $c$ is molar concentration (molarity).

Can you please elaborate? We actually are considering both positive and negative Ions in $z^2$ term so the more of both or one of them results in a bigger ionic strength, Does this formula has a proof or It's just a model of prediction?

If I am correct, the ionic strength is an empirical result. The $\frac{1}{2}$ term is nice because it means that the ionic strength of a 1:1 electrolyte is equal to its concentration. Also, according to this paper, m refers to the molality of the solution.

I prefer to use b for molality, but indeed m is also used. Also, I would add that it's better to formally divide each molality by standard molality of 1 mol/kg so that you end up with a dimensionless quantity.


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Is hydrogen (on its own) considered an acid?
An acid is:

A molecule or other species which can donate a proton

So when hydrogen appears by itself on one side of a dissociation reaction e.g. $\ce{HOCl  ->   H+ + OCl-}$ is the hydrogen an acid? As the hydrogen can donate itself?

Yes in such a dissociation the $\ce{H^+}$ ion is an acid. // A naked $\ce{H^+}$ ion doesn't really float around in aqueous solution but it is rather  the $\ce{H3O^+}$ ion. The $\ce{H3O^+}$ ion is used so much that is has a special name hydronium.

Just as an afterthought to previous answers: while the H+ ion, which you use in your example, is an acid, hydrogen gas (H2) is not.

H2 has (very weak) acidic properties.

Well, you can never have H+ by itself. In aqueous media it's always associated with the lone pairs of water molecules and exists as H3O+. But yes, theoretically speaking H+ is a Brønsted–Lowry acid as the definition of such is to donate a proton and it donates itself to water in the case. It's also a Lewis acid as by donating itself it accepts electrons in its 1s emptry orbital.
But be aware that in water, you never have H+, it's always H3O+ which is the strongest acid that can exist in water solutions, all acids are limited by the strength of H3O+ when dissolved in water. Therefore, a species is an acid in water if it generates H3O+. In other solvents, for example liquid NH3, the corresponding species is NH4+ and an acid in ammonia is defined as species that generates NH4+. This whole paragraph relates to the solvent definition of acids and bases.
I hope all that answers your question, please comment if you want anything clarified.

Nope that explains it perfectly :)  It was written in my book two ways 1) $\ce{HOCl (aq)⟶H+ +OCl-}$ and 2) $\ce{HOCl (aq) +H2O (l)⟶H3O+ (aq) + OCl- (aq)}$ but I didn't connect the two before reading your answer.


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Understanding the difference between physical and chemical changes

Which of the following is a chemical change?
A. Hydrolysis of a dipeptide.
B. Stretching of a balloon.
C. Freezing of water.
D. Breaking a stick.

I'm pretty sure that the answer is (A), but I'm not 100% positive.
I know that, for (C), you can freeze water, and then you can melt it back to liquid water.
For (D), you can break a stick, and the stick is still now two sticks.
For (B), however, I think that it is a physical change, but I'm not entirely certain.
Can someone please explain this for me?

It seems you're getting it more or less right.

Yes, (B) is a physical change. Stretching the balloon simply rearranges the polymer strands in the balloon material, making them straighten out instead of being kinked up. When you let the balloon go, they just pull back to their former kinked shape.
Your reasoning on (C) and (D) is solid.
The answer is indeed (A). See Wikipedia's page on hydrolysis for more information.

The question was upvoted because it has some interesting features. On the one hand, it suggests that the four answers have significant differences (which they do). But it also suggests that the world consists of two kinds of people: chemists and physicists.
Actually, there is a third kind: non-scientists.
But each of the changes has a component that could be of interest to a chemist, and also one that could be of interest to a physicist, hence the usefulness as a question. Here, one must put on the mathematical hat. Answer A changes the chemical composition of the starting material, so is certainly a chemical change, but the physical properties of the final product will be different (mass, density, melting point), so could possibly be of interest to a physicist, or, perhaps to a chemical engineer, who sort of resembles a physicist in some ways. The mathematician would rate the degree of chemicalness as about 99%.
Answer B would seem to be a physical change: we assume the balloon returns to its original shape when the stretching force is removed. But if that force is repeated, and/or increased, the balloon does not return to its original shape, and the process may be better explained by a chemist than a physicist. Hence the uncertainty of the OP. A serious mathematician could rate one small stretch to be about 99.99% physical while multiple large stretches could be as much as 2% chemical.
Answer C, the freezing of water, depends on the purity of the water. If it is distilled, air-free, then my mathematician would say 100% physical change, but also of considerable interest to chemists. But since the chemical composition does not change, we'll give it to the physicists.
Answer D, breaking a stick, must have some molecules sliding over some others, but surely some must break - perhaps a few thousand in $10e^{23}$. The mathematician says that's insignificant, and he's right: it's a physical change, at least to non-scientists, who make up the vast majority of people in the world.
And that's my point, after all the verbiage: the question requires a level of analysis which is not too deep, because nitpickers of the chemical or physical persuasion could probably find something of interest in all four answers and claim a different answer. But we hope to acquire a depth of reasoning and explanation that enables us to slice thru the question at a depth of reasoning that enables the quickest separation and answer. Otherwise our answer will be marked incorrect and require extensive verbal defense, which will still be regarded as inadequate.
Perhaps a dash of philosophy would help.