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How to cancel once-mentioned element in thermochemical reactions
If I have three thermochemical reactions in a series(in order to find the enthalpy change) how can I cancel an element having not included in the reaction that's being asked about?
and only mentioned in one of the reactions ( either products or reactants)
e.g: $$\ce{NO + O3 \rightarrow NO2 + O2}$$
     $$\ce{O3 \rightarrow 3/2 O2}$$
     $$\ce{O2 \rightarrow 2O}$$
The reaction that is being asked: $$\ce{2NO2 \rightarrow 2NO + O2}$$

Woild you be so kind as to format
mathematical expressions and equations using $\LaTeX$? Thank you very much in advance.

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We can make use of this formula: $$\Delta H = \Sigma ~\Delta H_f~(\text{products}) - \Sigma ~\Delta H_f~(\text{reactants}) $$
Take $\ce{O3 \rightarrow \frac{3}{2} O2}$ into consideration. Ozone molecule is the less stable form of O. If you know that how much the $\Delta H$ for that reaction is, you can in fact predict the $\Delta H_f$ for $\ce{O3}$. If you're interested, wikipedia has documented some of the enthalpies of formation. Notice that the most thermodynamically stable form of the element is considered to have $\Delta H_f = 0$.
Now that the $\Delta H_f$ for ozone is determined, you can apply the formula above for the first reaction: ($x$ is $\Delta H$) $$x = (\Delta H_f \ce{(NO2)} + 0) - (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3}))$$
So, from the equation above we can conclude that $$x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $$
Now write the master formula for the last reaction: $$(2 * \Delta H (\ce{NO}) + 0) - 2 * \Delta H_f (\ce{NO2})$$
What did we conclude from the last reaction? $x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $ Assuming that $x$ is at hand, do you really need to know the $\Delta H_f$ for $\ce{NO}$ & $\ce{NO2}$?

Also, you could reverse the asked reaction. The difference in the enthalpy change that it would have with the reaction one will be equal to the formation of $\ce{O3}$. But you should be careful about the negative and positive enthalpy sign.

If you're just interested in the reaction enthalpy of the reaction
$$\ce{2NO2 <=> 2NO + O2}$$
and one set of equations for the calculation of $\Delta H_f(\mathrm{products})$ and/or $\Delta H_f(\mathrm{reactants})$ doesn't work out, just use another one.
It does not matter! The Hess Law is invariant of the path.


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How does "breathable" siloxane water sealant work?
(This is an interdisciplinary question, between molecular chemistry, materials science & construction. I think Chemistry SE is the right fit, but please move it if needed.)
Silane/siloxane solutions are applied to porous building materials (e.g. brick or concrete) to repel water. Unlike a solid coating, these sealants are "breathable", allowing water vapor to exit instead of being trapped, which would cause slow structural damage. How does that work, molecularly?
I presume these compounds affect surface tension in some way, reducing the ability of liquid water to permeate, without completely sealing the pores, but I don't understand it.
Optional: also, how does this sealant affects the permeability of other liquids (polar vs non?) and other gases?

According to How silicone water repellents work,
"Because silicones have a very open molecular structure, water vapor trapped in the substrate can easily slip through. Water molecules, however, are too large to enter."
If I remember correctly, siloxanes' Si-O bonds are not intrinsically hydrophobic, but the long molecular chains with outward-facing hydrophobic methyl groups, and thermal molecular motion, prevent liquid water from seeping through, which presents less of a barrier to water vapor and other low-molecular-weight gases.

The info about methyl groups is helpful, but that quote is nonsensical. Liquid & vapor, water molecules are exactly the same size.

Not true! Water molecules are not just H2O but are strongly held to adjacent molecules by hydrogen bonds; see http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html. There is electron delocalization, implying, in effect, a "supermolecule". The much more diffuse water vapor has "broken" much of that bonding.

Aha. This comment is exactly what I needed to know. In chemistry classes last millennium, hydrogen bonds were always mentioned in passing, not important enough to merit an entire lesson. But actually they are. http://blogs.scientificamerican.com/lab-rat/2011/08/02/hydrogen-bonds-why-life-needs-water/

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Why does phosphorus tribromide act as a Lewis acid (electron acceptor)?
This question is purely out of curiosity and to help me remember the "why" of a mechanism to help me remember the "what/how".
In organic chemistry we recently studied the  Sn2 Substitution of Primary Alcohol with PBr3:

My question is, why is it favorable for PBr3 to act as a Lewis acid (i.e. accept electrons from the oxygen to form a bond) when it already has a full octet and no formal charge?

As an off-topic side note, I thought of a joke while studying this mechanism: a man wants to buy beers for himself and two of his pals, walks up to a bartender and says, "I'll have PBR—three please." He and his friends die of internal caustic wounds.

Thanks @DelPate, I was aware of that, but I'm still unclear why the phosphorus PBr3 is attacked by the oxygen in the primary alcohol above.  We're taught that "something negative attacks something positive," but while the oxygen can clearly act as a nucleophile (because of its lone pairs), it's unclear to me why the phosphorus would "want" to act as an electrophile. That is, how is it the "something positive" in this mechanism when it's already an overloaded octet?

While I was typing out that clarification question in the comment above, something occurred to me.  Is it because of a dipole moment between the phosphorus and the bromines that makes the phosphorus slightly positive?

Could be a reason . But I feel its more about the molecule than the P atom . Since Br is a good leaving group , P doesnt mind a nucleophilic attack . And the ultimatum is " Nature loves stability " , P-O bond is stronger than P-Br .

I guess I should make this the answer unless you have something to add .

Go for it! I like that explanation.

What do you mean by 'overloaded' octet? P has eight electrons in PBr3.

Oops, totally right @jerepierre!  Corrected the question above.

The why of a mechanism is always the best approach.
The reasons for this attack are:

$\ce{P}$ can expand its octet due to presence of vacant $d$-orbitals.
The $\ce{P-Br}$ bond is polar due to the electronegativity difference between the atoms and hence P develops partial positive charge favouring a nucleophilic attack on it.
Looking at the molecule as a whole, $\ce{Br}$ is a good leaving group so $\ce{P}$ doesn't mind a nucleophilic attack on it by the lone pair of oxygen.
"Nature loves stability" and  $\ce{P-O}$ bond is stronger than $\ce{P-Br}$ so that might be the driving force for this reaction mechanism.

No, phosphorus can not expand its octet and its d orbitals are energetically too far removed to take part in bonding.

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Is the following paragraph about steam distillation accurate?
I've been struggling with the various methods of purification of organic compounds. If a question comes up in my examination, asking me to explain steam distillation in my own words, I'd like to know if this applies and if this is accurate.
i) Steam distillation is utilized to separate volatile constituents which are immiscible with water but steam volatile.
ii) Steam from a steam generator is passed through a heated flask which contains the liquid to be distilled.
iii) The volatile component separates and thusly, one obtains a mixture of steam and the volatile organic compound.
iv) This is then condensed and separated by means of a separatory funnel.
In steam distillation, the liquid boils when the sum of vapour pressured due to organic liquid and that due to water become equal to the atmospheric pressure.
i.e Patm=Pw+Pc
Since $Pc<P$ ; liquid vaporises at temperature lower than its boiling point. That is, if water insoluble substance and water are the components, the mixture boils at a temperature close to, but below 373K.
Example: Aniline-Water
Please be highly specific. There should be no error, however minute.
Many thanks.

You have it right on all counts. But, to clarify...
i) Steam distillation is used to separate steam-volatile organic components of a mixture from non-volatile ones (not from each other, however). It is most useful for higher boiling, relatively non-polar organic compounds that are immiscible with water.


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Calculating the Molarity of an Ion?
I'm completely stuck on this one - I was doing an experiment in a lab and now I cannot find the molarity of an ion in a solution. I've looked everywhere on the web but after 2 hours of searching I couldn't find anything. Maybe I am looking in the wrong direction.
I mixed three solutions: 1mL of $\ce{0.1M ~CuSO4}$,  $\ce{0.28 M ~~NH3}$ and  water.
This solution turned into
$$\ce{Cu(H2O)4^{2+} + 4NH3 = Cu(NH3)4^{2+} + 4H2O}$$
This turned the solution from light blue to dark blue-ish purple. The final mL of the mixed solutions was 25 mL.
Now I am supposed to calculate the molarity of the $\ce{Cu(NH3)4^{2+}}$ ion. I am pretty sure that I only have to calculate the molarity of $\ce{Cu^{2+}}$ but I am not entirely sure. If you guys could give me some help on this I would be very grateful.

you need to know more information like the volume of the 0.1M solution and the volume of the final mixed solution to calculate the molarity of Cu(NH3)4^(2+)

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I'm sorry! I did forget to add the volume - I will add it right now.

Is the 1 ml also for the ammonia solution?


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Can we make Gold in a Lab?
Can we make Gold in a Lab? Is it possible to make changes in metals that will create Gold? and How?
I have tried to search something related to this but didn't get any good results.
Thanks in Advance

Depends on the lab actually. http://www.scientificamerican.com/article/fact-or-fiction-lead-can-be-turned-into-gold/ You need to be in the fanciest most advanced labs in the world to create gold from lead and prove you can! That gold won't be worthwhile though.

@MARamezani gold was made from mercury over 90 years ago http://link.springer.com/article/10.1007%2FBF01505658

Thanks @DavePhD! That was good to know, but I wasn't bringing in records, just put there an example. I don't say that article is fully credible, though.

it seem that making gold from lead is more difficult than from mercury

I once read (in a pop sci magazine for kids, so I have no idea how valid this info is) that the easiest metal to turn into gold is...platinum! Unfortunately, you're significantly decreasing its value by doing so...

@chipbuster That's what I heard too!

Yes, gold has been made from mercury.
See "Transmutation of Mercury by Fast Neutrons" Phys. Rev. 60, 473


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How can you manage SCF convergence problems?

"Failure to locate stationary point, SCF has not coverged"

This is the most annoying message after Optimization run ( even with a good primary start geometry like UFF we ay see that) .
The only solution I know is to use the output log file for another run but it is very time consuming.
Is there any way to force GAMESS to automatically do that ?( I mean makes another automatic run on current output results ).

There can be many reasons for the SCF to not converge.

The initial guess wasn't great, and the convergence is slow. Increasing the number of SCF iterations can help here.
The SCF oscillates. This can happen if there are two orbitals close in energy. Using damping, mixing (i.e., the next iteration is 50% the old and 50% the updated iteration), Fermi broadening, etc. can help here.
The SCF strategy (e.g., DIIS) is causing problems. Usually DIIS can greatly accelerate convergence, but in some hard-to-converge systems, you need a lot of iterations with another strategy (e.g., Newton-Raphson)

Without more information from your particular file, it's impossible to know how to help. Many times, you may simply want to increase the maximum number of SCF iterations and it'll work fine.

Thank you Geoff your suggestion for using DAMP=.T. solved my current problem. Sometimes I have seen using Direct SCF +/- DIIS  helps also, but not always.   What I am looking for is a systematic approach for non-convergence problem.  Can we reach a conclusion on it for future reference ?

Also I don't know if there is somewhere in Avogadro for setting Fermi broadening.

I don't think GAMESS has an implementation of Fermi broadening. Different QM packages have different SCF convergence tricks.

There is also no "systematic approach" because the approach will vary depending on the program and depending on the type of SCF convergence issue. You have to take a look at the SCF iterations and diagnose for yourself (or ask others).

Moreover, the problem may depend on the basis set chosen, the geometry, and the molecule properties. Some basis sets may be inappropriate for some molecules (and switching to other basis set may fix the problem), some geometries may be more prone to problems (and changing bond lengths may solve the issue), if you do not consider the correct multiplicity or electron distribution it may also fail, and you should change those, allowing orbital mixing may also help in the same way. Sometimes, computing the structure of the same molecule with less electrons will facilitate convergence and then you may use the result for a final full-electron calculation. The same may happen if you include (or avoid) excited states. In some cases I've seen "naturally unsolvable" problems were the structure was oscillating between two states with very similar energy, maybe tautomers. And it may also happen that there is some chemistry going on with bonds breaking and forming (I saw this on some long-chain radicals which might break down into pieces)... A possible approach may be sometimes to build the molecule incrementally.
The point is that there are too many things that may be going on and that each case is special, so you'll have to look into your problem, see the evolution of energy and error, visualize the evolution of the geometry with energy, look at the values of the basis set for your atoms (some of them may lead to non-orthogonality if a coefficient for one of your atoms is too small) and decide on your specific problem.
Right now, there is no silver bullet that can solve all problems.

How can tautomers occur during a SCF calculation? The coordinates of the nuclei do not change so how can there be different tautomers? How can bonds break or be formed, long chains breaking down? That should be impossible during a single point calculation. Do you maybe refer to a geometry optimization here? Because that is something completely different!


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What is sprayed on chromatography paper when developing?
When developing the chromatography paper, what chemical will you use to spray on the chromatography paper?

Are you trying to stain particular classes of compounds? Some reagents are quite specific. Most of the reagents I used in the lab were tailored to be used in tlc with silica (or aluminium oxide) on aluminium (or glass) plates. These reagents aren't necessarily compatible with paper.

It depends on what you will use the chromatography paper for. Different testable substances will require different mediums. However, for a plain-simple chromatography without staining/substance identification, aluminium oxide (or silica) coated on a plate is used, however paper can also be used.


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Am I right in giving the following reason for some questions relating to the p-block?

I'm seeking clarification as I believe this to be the answer to why boron compounds are electron deficient.
While  bonding, the $2s^2$ electrons of Boron are excited to the $2p_y$ orbital, they engage in bonding. However, even after bonding, there exists an empty $2p_z$ which leaves an incomplete shell, thus these compounds require two more electrons and so, they are electron deficient.
(As in the case of $\ce{BCl3}$)
$\ce{CO2}$ cannot put out fire caused by the reaction of saline hydrides and water as it will be converted into formate ions during the process. (although, I'd appreciate some help on chemical equations if possible)

Moreover, if this is accurate, does the same logic apply to $\ce{NH3}$?
Grade 11.


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Fentons Reagent tips: H₂O₂ + FeSO₄
I've been having trouble with reducing COD (Chemical Oxygen Demand) below approx. 500 ppm by using Fentons Reagent.
For 100 gm of the sample (initial COD 3000 ppm) I use 5 gm of 30% $\ce{H2O2}$ & 0.1 gm of $\ce{FeSO4}$. pH is adjusted to between 3 to 5. Temp. is approx. Room Temp. & I've tried reaction times up to 6 hrs.
Do these sound reasonable choices?
Any advice?

I'm not an expert in Fenton chemistry but it appears that difficulty in getting the Fenton reagent to reduce COD to below 500 ppm is not unusual.

Does your particular wastewater have  chelants such as phosphates, EDTA, formaldehyde, and/or citric/oxalic acids?
Perhaps at long reaction times all the iron you add is being precipitated as iron (hydr)oxides.  Could you lower the pH further?

Reaction rates with Fenton’s Reagent are generally limited by the rate
  of $\ce{.OH}$ generation (i.e., concentration of iron catalyst) and less so
  by the specific wastewater being treated. Typical $\ce{Fe:H2O2}$ ratios are
  1:5-10 wt/wt, though iron levels < 25-50 mg/L can require excessive
  reaction times (10-24 hours). This is particularly true where the
  oxidation products (organic acids) sequester the iron and remove it
  from the catalytic cycle. Fenton’s Reagent is most effective as a
  pretreatment tool, where CODs are >500 mg/L. This is due to the loss
  in selectivity as pollutant levels decrease...In addition to free radical scavengers,
  the process is inhibited by (iron) chelants such as phosphates, EDTA,
  formaldehyde, and citric/oxalic acids. Because of the sensitivity of
  Fenton’s Reagent to different wastewaters, it is recommended that the
  reaction always be characterized through laboratory treatability tests
  before proceeding to plant scale.

My tip would be to increase the concentration of both $\ce{H2O2}$ and $\ce{FeSO4}$, and, if possible, to make sure the pH is 3 (or maybe even lower).  If that doesn't work, do you think iron chelators could be present in your particular wastewater sample(s)?

Thanks for the tips. My water sample is expected to have NaCl as an inorganic and only C H O Cl (i.e. no P N S or other elements). In particular it very likely has aromatics, aromatics with sidechains (e.g. Styrene) & -OH groups / Chlorohydrins / Epoxides. No citric nor oxalic acid nor EDTA nor formaldehyde is expected.

I will try increasing both concentrations but I think I'm already considerably above the conc. given at the link you mentioned. About pH I thought 3 to 5 was the optimum so I was keeping it there. Should I keep it lower? e.g. I read on your link that " it may be necessary to perform the oxidation in steps, readjusting the pH upwards to pH 4-5 after each step so as to prevent low pH from inhibiting the reaction." Apparantly reaction rate is too slow at low pH?


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How to predict the color of a pH indicator using Le Chatelier's principle

The question I have is above. Could someone explain why the answer is (A)?
What I think so far:

Does the sodium hydroxide turn the indicator blue because it is a strong base and hence it will react completely to form water,
shifting equilibrium really far to the right?
Does the HCl turn the indicator red for a similar reason?
I am unsure of the sodium acetate effect

Please note that the homework policy on this site requires you
to show some personal effort.

I've added some of my attempts on the question? Could you please help?

IMHO, the equations in the question have been incorrectly written. This is not chemistry, it's rather confusing and playing mean tricks on students. In the alkaline, H3O+ does not exist. It should read H2Z + OH- <--> HZ- + H2O etc.

You're absolutely on the right track!
The keys to the question are

to  order $\ce{HCl}$, $\ce{NH4Cl}$, $\ce{NaOAc}$ and $\ce{NaOH}$ in terms of strong acid, weak acid, weak base and strong base
to realize that in the presence of $\ce{HCl}$, the indicator exists as $\ce{H2Z}$, while it is fully deprotonated to $\ce{Z^{2-}}$ in the presence of $\ce{NaOH}$
in the presence of weak acids or weak bases, the indicator exists in form of each two species. Consequently, a mixed colour is observed.

In the presence of $\ce{NaOAc}$, a weak base, $\ce{H2Y}$ is completely deprotonated, thus no red colour is observed. In addition, a part of $\ce{HY-}$ is deprotonated too: the second equilibrium isn't completely on the left side. Some amount of $\ce{Y^{2-}}$ is present too. As a result, the green colour (yellow + blue) is observed.
It is all about the position of the equilibria.

Alright! I understand points 1 and 2, but why does the indicator exist as two species in the presence of weak base? I think I may have some sort of idea, but I would prefer it if you gave a definite explanation. Thankyou!

@chemistryyo I've tried to address the last point too.

Sorry, I just have another question:
How do we know that H2Y is completely deprotonated. If this was a single equilibrium reaction, there should be some H2Y present

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Relationship between thermodynamic reversibility and reactions
Often we're concerned with changes in thermodynamic properties during reactions, like $\Delta H_p = Q_{reaction}$ (change in enthalpy for an isobaric process). What seems odd to me is that in the derivation of this relationship,
$\Delta H_p = \Delta U + P\Delta V=Q-W+P\Delta V$
we assume the work is given by $W=P\Delta V$, which requires that the process is reversible in the thermodynamic sense. Then the change in enthalpy simplifies to
$\Delta H_p=Q - P\Delta V + P\Delta V=Q$.
Supposedly, measuring the heat transfer during a reaction at constant pressure will give the enthalpy. Even though enthalpy is a state function in general, our case requires that the process is reversible.
To what extent does the restriction of thermodynamic reversibility apply to the reaction? Do we have to assume the reaction happens slowly and can be undone, leaving only an infinitesimal change in the universe? Or does it not matter?
Please correct me if any of my points are wrong. Classical thermodynamics usually doesn't mention kinds of matter, so I'm a bit confused when chemistry comes into it.

Here one assumes that the changes are quasistatic, so slow that there is at each instant negligible departure from thermodynamic equilibrium within the system. In other words, δW = -PdV where P is pressure and V is volume. - Wikipedia . Read this up for a better understanding of your own question .

The process does not have to be considered reversible, and, if a reaction is occurring in the system, it is, of course, not reversible.  We are trying to find the change in enthalpy between the following two states:
State 1:  Reactants at temperature $T_0$ and pressure $P_0$
State 2:  Products at temperature $T_0$ and pressure $P_0$
We would like to measure the $\Delta H$ in such a way that, for the process path we select, all we need to do is measure the amount of heat transferred to the system Q.  The process path we select involves keeping the system in contact with a constant temperature bath at $T_0$ and controlling the motion of the portion of the system boundary at which the volume change occurs in such a way that the force per unit area is always constant and equal to $P_0$.  Since, in general, the work done by the system on its surroundings is determined by the integral of $dW=P_{ext}dV$ (where $P_{ext}$ is the force per unit area acting on the portion of the boundary at which the volume change is occurring), we have selected $P_{ext}$ such that $P_{ext}=P_0$; and the total cumulative work done by the system when it reaches its final equilibrium state will be just $W=P_0\Delta V$.
From the first law, it then follows that, for this irreversible path:
$\Delta U=Q-P_0\Delta V$.  So $\Delta U+P_0\Delta V=\Delta H =Q$

Reversible reactions are considered to be ideal, since in our everyday life we don't get to see anything broken come back together by its own. So you have to remember one thing, which I also think it's the answer youre looking for : every reversible reaction we talk about in thermodynamics, happens infinitezimal slowly. For example :
Reversible gas compression in constant pressure, happens by removing the heat, lowing the temperature of the ambient in infinitezimal amounts, so that the temperature of the system changes from T1 to T2. So your assumtion was right, infinitezimal is the key word in reversible reactions.

This confuses reversible reaction (system initially not a equilibrium reaches equilibrium) with reversible process (infinitesimal changes close to equilibrium), see https://chemistry.stackexchange.com/questions/16917/relation-between-thermodynamic-reversible-process-and-reversible-reaction/124602#124602

In chemistry dU = dQ + dW ( not dQ = dU + dW) and dW = -Pdv.The general expression for enthalpy is H = U + PV . Enthalpy is the amount of heat content used or released in a system at constant pressure.
If there is no non-expansion work on the system and the pressure is still constant, then the change in enthalpy will equal the heat consumed or released by the system (q) i.e. (dH = Q).
But,you can calculate change in enthalpy every case throuugh equation (H = U + PV) i.e. dH = dU + d(PV).


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1H NMR Broad peaks
Is there any particular reason that peaks in proton NMR can appear with a larger integral ratio than expected, that is to say one specific proton absorption stretches over a wide ppm range?
I have a very broad singlet that is integrating to nearly 5x expected. The rest of the spectrum is fine in terms of integrating and the position of the peaks. This particular proton is a singlet from 9.95 to 10.25 ppm and is the alpha proton of a carbene hydrogen chloride salt. Either that or it is impure, but so far my spectrum is very clean whenever I run it.
Many thanks! :)

Can you add the structure of your molecule?

One reason for broadening is chemical exchange.
For example, if the proton you are referring to is in exchange with another population of protons, for example from a solvent, the integrated area could represent the total population that is exchanging.
Coordination of a paramagnetic metal ion is another source of broadening.  Integrated area should be unaffected if there is sufficient time between each scan.  However, since the paramagnetic metal ion can cause faster relaxation of nearby protons, it is possible that the integration is distorted by protons near the paramagnetic ion fully (or closer to fully) relaxing between scans, while the other protons do not.  This effect is often used intentionally to enhance detection of protons near a paramagnetic metal center.

Thanks for your answer! I have a palladium centre nearby so I think that might be the culprit

@AngusTheMan Palladium shouldn’t actually be the culprit; both palladium(0) and palladium(II) are diamagnetic …

@Jan usually, but http://pubs.acs.org/doi/abs/10.1021/ja01051a061

@DavePhD Good thing I only said ‘shouldn’t’; bad thing I can’t ninja-edit a usually in there before diamagnetic … ^^'

The most common process that can cause both peak broadening and an increase in the value of a peak integral is exchange broadening between two different protons.  The following diagram represents a typical case where the protons in water are exchanging with the hydroxyl proton in an alcohol.

If the rate of exchange is fast on the nmr timescale, then a single, sharp peak of increased amplitude and a weighted average of the individual chemical shifts is observed.  If the rate of exchange is slow compared to the nmr timescale, then two distinct peaks, both with accurate integrals, are seen.  In the case where the rate of exchange is comparable to the timescale of the nmr experiment, then a broadened signal with integral intensity equal to the two combined signals and located at a weighted average of the two separate signals is observed.
One way you can test to see if a small amount of water in your sample is the culprit is as follows.  Add a small amount of $\ce{H2O}$ to your sample and see if the peak for your sample shifts to a new location and the integral value further increases.

Thank you for your answer! I don't think exchange is at play here, I do have a paramagnetic palladium centre nearby so maybe something is going on with that?


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Concentration of expanding gases
At $t = 0$ we have 10 moles of $\ce{NO2}$ and 10 moles of $\ce{N2O4}$ satisfying this reaction in a 0.75 liter tank (the barrier is closed at this time):
$$\ce{N2O4 <=> 2NO2}$$
The system is in thermal contact with the environment.
At time t=t0 the barrier is released and the gases are free to expend to the other tank ($V_2=1$ liter)

The question is what will be the concentration of $\ce{NO2}$ when they reach equilibrium.
What I tried is calculating the chemical equilibrium before based on the concentration quotient $\text{K}_c$ (13.33) and then to use it as reference after the system reaches equilibrium. I got a weird result (negative number) Where have I gone wrong?

You Kc value is correct , the answer comes positive ( on solving the quadratic which I hope you got ). So I guess there is some calculation mistake on your part . Can you include your steps ?

This is a pretty straight forward equilibrium problem in disguise. If we assume the initial state represents equilibrium, we can determine the equilibrium constant by plugging in the concentrations to the law of mass action. We can get the concentrations because we have the number of moles and the volume of the container.
$$\ce{N2O4 <=> 2NO2}$$
$$K_c=Q_{eq}=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$
Okay, now we let the gas expand. Being in thermal contact with the surroundings is important, since it means the temperature will not change (which means the equilibrium constants does not change). Now we will find that we are not at equilibrium any more. We will probably find that $Q_c < K_c$, so equilibrium needs to shift toward $\ce{NO2}$.
Now, we can set up the ICE (initial, change, equilibrium) table:
          [N2O4]           [NO2]
I         5.71 M           5.71 M
C         -x               +2x
E         5.71 -x          5.71 + 2x

Substitute into the law of mass action at equilibrium:
$$K_c = \dfrac{(5.71+2x)^2}{5.71-x}=13.33$$
Rearrange and solve for $X$. Being a quadratic equation, you should get two roots (two solutions for $x$), and as is always the case for an equilibrium problem, one of them will be nonsense. For example, you could get one value of $x$ to be larger than the initial concentrations!

This being a homework question , shouldn't you have waited for OP to add the working/steps/calculation ? I even commented this above .

I think it's fine. The asker showed effort, Ben focused on the problem description and the buildup to reach the answer, and the actual final computations were not performed, so the asker still has work to do on their part.

I guess the build-up was not required just the last paragraph would have been great . Anyway nice answer .

Even though the OP may only need the last paragraph, someone else looking at this question may need a little more.


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Lowest freezing point room temperature liquid?
What substance that is liquid at STP has the lowest freezing/melting point? I think it's either 2-methylpentane or 3-methylpentane, but I get different numbers from different sources.

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After a bit of random searching through small molecules, I found this very large dataset with 28 645 experimentally measured melting points. Sorting them by increasing melting point and checking the first few entries individually, it seems that N,N-diethylmethylamine holds the record, with a melting point of −196 °C (!) and a boiling point of 63 to 65 °C.
I also mentioned another possibility for very low melting liquid substances at room temperature, namely ionic liquids. Even though the species present are much larger and have ionic interactions, presumably the ions pack so extremely poorly in the solid that freezing is very unfavourable.  Aldrich already offers multigram quantities of ionic liquids melting at −88 °C, and ongoing research offers even lower melting alternatives, such as the ones in Reiter et al. [1] which purportedly do not crystallize until −150 °C. This limit may well be improved upon in the next few years.
If mixtures are allowed as well as pure substances, then a combination of small organic molecules or ionic liquids will very likely reach even lower temperatures than the examples above without freezing.
Reference

Reiter, J.; Jeremias, S.; Paillard, E.; Winter, M.; Passerini, S. Fluorosulfonyl-(Trifluoromethanesulfonyl)Imide Ionic Liquids with Enhanced Asymmetry. Phys. Chem. Chem. Phys. 2013, 15 (7), 2565. DOI: 10.1039/c2cp43066e.

Great answer.  Quick unrelated side question: why is "N,N-diethylmethylamine" not called "N-ethyl-N-methyl-ethylamine"?  (And you are right that e.g. Sigma-Aldrich uses the same name as you)...is that a colloquialism or the bona fide correct name?

@CurtF. I'm not sure, I think the second is formally more correct (IUPAC actually would recommend N-ethyl-N-methylethan-1-amine) but the first is just more concise and easier to pronounce.

The true melting point for diethylmethylamine is yet unknown; the original reference for this "melting point" value is J. Org. Chem. 22, 840-843 (1957), which states that the material became a glass at liquid nitrogen temperature. Also of interest is that the same exact data points are given for isopropyldimethylamine and diisopropylmethylamine, but have apparently never been misconstrued as melting points for those compounds!

When dealing with such a huge dataset it's hard to avoid the small fraction of incorrect reports. The -196C=77K value is suspicious since it is the boiling point of liquid nitrogen. Wikipedia claims 130K as the melting point (or is this the glass point?) which is a bit more reasonable but still very impressive!

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Zeolite formation
I read in a book that zeolites are formed by Aluminium replacing a silicon atom in Silicon dioxide. It says there is a negative charge formation. How does this happen? Thanks.

There's no magic in this - just the beauty of science.

Imagine a lattice of $\ce{Si^{(+4)}O4}$ tetrahedrons ($\ce{Si}$ in the centres, $\ce{O}$ on the corners) connected through the edges corners (= the $\ce{O}$ atoms)
Replace some of the $\ce{Si^{(+4)}}$ with $\ce{Al^{(+3)}}$

As a result, the whole lattice becomes a huge polyanion.
The charges are balanced via cations, such as $\ce{H+}$, $\ce{Na+}$, etc. embedded in the pores of the lattice.

I'm not sure what the standard nomenclature is, but the first time I read $\ce{SiO4}$ lattice, I was confused because if $\ce{Si}$ is +4, how can there be four oxygens instead of 2?  I guess the answer is that the tetrahedra are linked, so each oxygen atom is "shared" by two Si atoms...my point is, is there a better term than "$\ce{Si^{(+4)}O4}$ tetrahedrons" or not?  Maybe it's just me...

@CurtF. As it is in Silicones ?

http://en.m.wikipedia.org/wiki/Silicon_dioxide#Structure something like this ?


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Mixing Different Solutions with Different Wavelengths to Get Different Colors?
I recently was told that the reason you can mix colors (blue and yellow make green because wavelengths 605nm + 450nm = green) was based on wavelength (which seems very obvious when you think about it but please humor me).
This is kind of cool but I don't get it - is this based on the wavelength range that's transmitted?  If  605nm + 450nm = green then how would you find that exact wavelength of green? Would it be the average of the two, or is there some constant that I am missing?

You have to distinguish two sorts of colour mixing:

Subtractive mixing is observed when you mix two coloured solutions ot put two colored films on top of each other and shine white light through it. Here, $\mathrm{\color{yellow}{yellow}}$ and $\mathrm{\color{blue}{blue}}$ yield $\mathrm{\color{green}{green}}$.
Additive mixing is observed when you overlay two coloured light sources. In the RGB model, $\mathrm{\color{red}{red}}$ and $\mathrm{\color{green}{green}}$ yield $\mathrm{\color{yellow}{yellow}}$, the combination of $\mathrm{\color{red}{red}}$, $\mathrm{\color{green}{green}}$ and $\mathrm{\color{blue}{blue}}$ yields white.

As far as the wavelengths are concerned: Each "colour" is represented by a wavelength range, rather than an exact value. The fact that there's more than one red or green is intuitively understood - particularly by women ;)


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What easy to conduct (at home) reactions could yield solid carbon?
I am trying to isolate pure carbon in its elemental form. Are there any decomposition reactions or other easy to carry out reactions to yield solid carbon? and in the case that it is isolated, does it start reacting with other elements in the air?

A classic reaction to yield carbon is the dehydration of sucrose (table sugar) with concentrated sulfuric acid: Put some sugar in a beaker, carefully add sulfuric acid and watch a voluminous "snake" of porous carbon (blow up due to the steam released during the reaction) from the beaker.
You'll find lots of videos on this on youtube or other platforms.
Note that this is performed with concentrated sulfuric acid! If you decide that you must do this at home, protect yourself (googles, gloves, lab coat) and keep the sulfuric acid locked away.
But there's a more simple way to yield carbon:
Hold a spoon (with a long handle) into the flame of a candle. The black stuff that deposits on the spoon is carbon. Play with the distance for optimal results.
As far as the reactivity of carbon is concerned: There's no danger in that! Millions of people stockpile carbon in the form of coal at home in their cellars or sheds for heating.
And actually, this points to the easiest and most safe option: Don't perform any reaction at home and grind some charcoal or anthracite coal instead.

If concentrated sulphuric acid is on the table, wouldn't it be possible to get very pure carbon by taking a pencil lead, crushing it into dust and then submerging it in the acid for a while? Not much other than the pure graphite could survive that. Bonus points for being a single allotrope! The candle soot is what I had first thought, but while it is a very simple and convenient procedure, the soot isn't exactly pure carbon.

@NicolauSakerNeto You're probably right. The clay used as a binder won't survive.

@NicolauSakerNeto Purity is a relative term. Doesn't make any sense to make statements what is pure what is not without reference to purity level and what kind of impurities we care. Natural graphite in penciles is not an "absolutely pure" substance either.

Sorry I know this is definitely not the most technical or probably correct term, but in the sucrose and sulfuric acid reaction, why does the carbon product look 'wet' and not very compact as other carbon based materials like charcoal pellets or solid graphite rods.

@AlanZ2223 During the dehydration of sucrose with sulfuric acid, alot of heat is released - enough to convert the water (= reaction product) to steam. As a consequence, the carbon adopts a foam-like structure - the carbon is build around bubbles of water vapour. When the mass cools down, the water condenses and forms a wet film on the carbon surface. In addition, traces of sulfuric acid dragged out of the mix might add to the wet look.

Why don't you just buy activated carbon? Generally it is rather pure and contains no carcinogen or poisonous substances, so it is very safe to handle, unlike most pyrolysis products.

Well I wanted to perform an experiment, just for the whole funs and what not.

You should be able to drive off water from a pure carbohydrate (like sucrose) simply by heating it enough. Think of it as a very simple example of destructive distillation.
If you prefer something a bit more exotic, you could ignite a magnesium strip and drop it onto a block of dry ice. The magnesium will combine with the oxygen, leaving carbon soot. I'm not sure how easy it would be to collect that soot, though, or to separate it from the magnesium oxide co-product.


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Separating metals out of a magnet
I have acquired the crushed magnet out of a motor I was given. If I'm correct that the magnet is $\ce{NdFeB}$ magnet, how could I separate out all the metals into a pure or oxide form? And for hard drive magnets, is there any way to remove the $\ce{Ni}$ coating without having to get nitric acid?  I don't have any proper chemistry equipment at all.

From this paper,  it seems Electrochemical recovery of Neodymium is possible from the Magnet.


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The feasibility of the electrochemical cell.
This is the question extracted from my country high school level' exam.
So it asks to discuss the feasibility of the electrochemical cell to carry out electrolysis of an aqueous solution of sodium chloride at standard condition.
Al(s)/Al(aq)//Cu(aq)/Cu
My question is how I should start to answer this question?

Electrode potentials , electrolytic cell and Google will get you started .

What are the reactions required for electrolysis? What are the possible reactions given the solutions and the electrodes? What are the potentials of the various combinations? Start writing these out and you should see patterns emerging.


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Nomenclature of methyl groups in an alkanol

I'm having trouble with this question since I'm getting (A), when the answer says its (B).
My logic was that there is a methyl group on the first carbon, and a hydroxide functional group on the second carbon, hence the name 1-methyl-2-pentanol.
Could someone please explain why I am wrong?

Don't let the graphical layout of the molecule deceive you!  There are no branches in the carbon chain.  (That is, no carbon atom is bound to more than two carbon atoms.)  The molecule ethane is not called "methylmethane", so calling your molecule a methylpentane derivative would similarly be an error.
With no branches (and no rings and no double bonds etc.), and with six carbon atoms, we must be dealing with a hexane derivative.  Once you know that, you just need to apply IUPAC rules that minimize the numbering of atoms to which substituents are bound.  So answer B makes more sense than C, for that reason.

Alright, I understand now! Thankyou!
So, just to be 100% sure, under what circumstances would the molecule be a pentane derivative with a methyl chain branch?

For simple alkanes with methyl "branches", the numbering gives it away.  You can't have a "1-methyl" substituent on an otherwise unsubstituted alkane, because it would make the chain longer, thus making it a hexane.  So "1-methylpentane" doesn't exist, but 2-methylpentane and 3-methylpentane do.  (But not "4-methylpentane" or "5-methylpentane" because the former is the same as the 2-methyl compound, and the latter thas the same "not really a branch" problem as the "1-methyl".)

Could someone please explain why I am wrong?

Find the longest chain! This determines the base name: hexane.
Find the position of the $\ce{OH}$ substituent. Lower number wins.


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How to make diamonds out of ashes?
Several companies are offering a service to turn ashes of human remains into diamonds.
Does anyone know any details about the chemical aspects of the process?
I have no idea, but suspect that the ashes are composed mostly of phosphates and oxides. Where does the carbon hide? Any carbonate would have decomposed at the creation temperature. Or does the cremation chamber has such a high $\ce{CO2}$ partial pressure that $\ce{CO2}$ from burners is captured by the $\ce{CaO}$?
How is the carbon then extracted?
Also, diamond is normally made out of graphite, which agrees with a statement on the LifeGem website.
Let's assume they can squeeze out a little $\ce{CO2}$ by heating the ashes in an oxygen stream. Then they concentrate the $\ce{CO2}$ on a molecular sieve, use the Sabatier reaction to create $\ce{CH4}$ which they decompose on a hot metal wire. Then they dissolve that in acid and get a little bit of powder rich in $sp^2$ bonds and similar to graphite. Could one make diamonds out of that?
On the other hand, the page mentioned above states that they first extract the carbon and then turn it into graphite. How do you turn a few mg of fine carbon powder into graphite?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. || Well done! This question is really thought through!

According to US published patent application 20030017932 "Cremeting; filtering carbon particles; sublimation; polishing"

The preferred process for collection begins with the oven operator positioning the body in the oven so that the head and chest area are not positioned directly underneath the main burner. This can be accomplished by positioning the body to the left or right side of the main burner, or positioning the body so that the legs and feet are underneath the main burner rather than the head and torso. Positioning the body in this manner assures that carbon will remain in the body's head area. The carbon can then be gathered by hand, or by using a metal shovel or scoop, or the like.

So, in this technique (or proposed technique) the idea is to not let the normal cremation process occur, which would change much of the carbon to CO2.  It seems more like pyrolysis to charcoal.
They go on to say:

Another alternative embodiment for carbon collection involves collecting carbon from pulverized cremated remains. These remains consist mostly of ash, but depending on how the cremation was performed, there may be traces of carbon particles mixed in with the ashes. The remains can be placed in a vacuum induction furnace. The furnace is heated to 2000 degrees centigrade in a vacuum ranging from 30 toir to 500 torr. Chlorine gas is injected into the furnace, and reacts with the impurities to form chlorides. The impurities leave the carbon in the form of chloride gases, and are filtered as they exit the furnace.

For more possible methods read this and other patent applications such as US20040031434.

In another alternative embodiment of the invention, the remains can be cremated conventionally, mixed with additional carbon from another source, and purified as described above. It is contemplated that, using this technique, a gem containing at least some of the original carbon from the cremated remains can be prepared, even if the amount of carbon present in the remains alone is insufficient to make a gemstone of desired size or type.

There is no way to say what actually occurs at a particular company, unless someone has first hand knowledge, which I don't.

That is one creepy patent. 'The resulting synthetic gem quality crystal comprising carbon from a vertebrate has a unique character to it, because it specifically relates to the vertebrate which supplied at least a portion of the carbon for the gem.' I found that LifeGem uses ashes and doesn't depend on this 'special' carbonization process http://www.lifegem.com/secondary/lifegemfaq2006.aspx#previouslycremated

LIFEGEM is a registered trademark of International Research and Recovery Corporation.  The patent application is assigned to the same corporation.  The inventors are CEOs of the corporation.  http://www.google.com/finance?cid=11437443

That makes me wonder if they have a problem getting carbon out of the ashes. The homepage says that normal ashes are ok. Here is a more detailed chemical analysis, unfortunately they didn't look at carbon. http://www.scattering-ashes.co.uk/general/cremation-ashes-chemicalcompostion/

But do they say what percent of the carbon is from the person? If they aren't saying, then it could be only a small fraction.  You may find this report interesting http://www.cremationdiamondreport.com/LifeGems.html

Human cremation is not usually done to analytical completion and the ashes are usually carbon rich. So I suspect that speculating about how to extract that carbon from small amounts is silly.

Let's assume they can squeeze out a little CO2 by heating the ashes in an oxygen stream. Then they concentrate the CO2 on a molecular sieve, use the Sabatier reaction to create CH4 which they decompose on a hot metal wire. Then they dissolve that in acid and get a little bit of powder rich in sp2 bonds and similar to graphite. Could one make diamonds out of that?

You ask good questions about how $\ce{CO2}$ is obtained.  But if there is a way to get $\ce{CO2}$, I doubt methanation is done en route to graphite.  At elevated temperatures, direct reduction of $\ce{CO2}$ to graphite can be accomplished by many metals.
I don't know if it used in the ashes-to-diamonds industry, but in sample preparation for accelerator mass spectrometry, $\ce{CO2}$ is converted to graphite routinely.  Here is one paper from the 1980s, which uses finely divided magnesium at 550 °C to reduce carbon dioxide to elemental carbon.

On the other hand, the page mentioned above states that they first extract the carbon and then turn it into graphite. How do you turn a few mg of fine carbon powder into graphite?

A much later report from Lawrence Livermore National laboratory uses zinc for the same purpose, and shows that their protocol is usable on only a few mg of carbon.  (Note since they are interested in analytical applications, their protocol is extremely rigorous/tedious, but I would guess that for diamond manufacturing applications this level of rigor would not be required.)

Interesting LLNL report, thank you. I wasn't aware that CO2 could be reduced to graphite at such low temperatures. The iron catalyst can then be used as a matrix for the diamond synthesis.

The simple answer is that the ashes of organic material contain large amounts of sodium carbonate, potassium carbonate and calcium carbonate (along with the respective oxides).
Add a few drops of some acid, and you've got the carbon dioxide.

He specifically mentions that carbonates would decompose at those temperatures to make $\ce{CO2}$ and metal oxides. If you're going to say that is incorrect, shouldn't you back that up with evidence/a reference of some sort?

Those carbonates do not decompose significantly until > 1000 K.

See this.

I cite (emphasis from me):  "... can reach temperatures of 2,000 degrees Fahrenheit (1093 degrees Celsius)

Yep. If the can reach 1350K, I don't think regularly reaching > 1000K is unlikely.


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Product of reaction between aluminium chloride and trimethylamine
Can anyone tell the product of the reaction between $\ce{AlCl3}$ (aluminium chloride) and $\ce{N(CH3)3}$ (trimethylamine)?

Have you read about Lewis acids and bases?

Here are a few tips to get you started:

Lewis acids are electron pair acceptors; Lewis bases are electron pair donors.
This looks like a Lewis acid-base reaction. You mention that this is an acid-base reaction but there are no easily accessible protons in either reactant. Aluminum trichloride has no protons; trimethylamine has does have protons but these are $\ce{sp^3~C-H}$ type protons - which are extremely weak. Also, there is no path to stability; deprotonation of one of the methyl groups leaves a localized negative charge on a carbon.

There are other heuristics one can use. For example, note that aluminum trichloride has an electron-deficient aluminum atom (the Al doesn't have an octet). Acceptance of an electron pair will provide it with an octet. This heuristic works fairly well; consider boron trichloride for example. Boron is similarly electron-deficient. And boron trichloride is also a Lewis acid.
Also note that the amine has a lone pair. Keeping in mind that Lewis acid-base reactions are about the movement of electrons (rather than protons) then we might reasonably suspect this to be a Lewis acid-base reaction. Lone pairs generally indicate Lewis basicity, as these are generally available for donation to a Lewis acid. Delocalization of lone pairs can weaken the lone pair's Lewis basicity. Thus, the presence of a lone pair only serves as an indicator of Lewis basicity rather than the strength of Lewis basicity. And of course you can also donate bonding electrons rather than lone pairs (halogenation of ethene would be an example; the $\ce{C=C}$ double bond is broken and the electrons previously involved in that double bond are put into the antibonding orbital of some diatomic halogen)

thanks for your answer. do you then know how to find the VSEPR structure?

ALCL3NH3 Basically nitrogen donates a pair of electrons to form a dative bond, however i cant seem to deduce the central atom to find the VSEPR

@chemstudent you have to determine which atom the nitrogen donates its lone pair electrons to.

in this case aluminium, so do i use ALCL3 to determine VSEPR?

@chemstudent - if you want the structure of the end product, then you use the end product to determine structure.

i see, so in this case it will be a trigonal planar

@chemstudent are you sure the end product is trigonal planar? You can't use AlCl3 to determine the structure of the end product because the end product is not AlCl3.

well the end product is alcl3nh3, but how do i get the central atom?

@chemstudent you have to figure that out using your knowledge of Lewis acids and bases. Hint: the nitrogen lone pair goes into an empty p-orbital on Al in AlCl3.

tetrahedral it is then

@chemstudent - here's an example of how AlCl3 generally acts as a Lewis acid ... apply this generalization to your problem. http://1.bp.blogspot.com/_OdZZiyNKEoU/SeH3Ma_Ul3I/AAAAAAAAAP0/qMIGqHkvrxc/s1600-h/Slide3.jpg

yeah its tetrahedral


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the molecular formula of an unknown hydrocarbon
A student analysed a hydrocarbon X and found it to be containing 43g carbon and 7.2g hydrogen.The relative molecular mass of X was found to be 42.What is the molecular formula of X?

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$43\ \mathrm g$ of carbon for $7.2\ \mathrm g$ hydrogen.
Molar mass of carbon is $12\ \mathrm{g/mol}$ and that of hydrogen is $1\ \mathrm{g/mol}$.
Hence there are $\frac{43\ \mathrm g}{12\ \mathrm{g/mol}} = 3.58\ \mathrm{mol} \approx 3.6\ \mathrm{mol}$ of carbon for every $\frac{7.2\ \mathrm g}{1\ \mathrm{g/mol}}=7.2\ \mathrm {mol}$ of hydrogen.
Hence the empirical formula of the compound is $\ce{CH2}$.
Relative molecular mass of $\ce{CH2}$ is $12+2\times1 = 14$. Given relative molecular mass is $42$.
$\displaystyle\frac{42}{14}=3$
Hence the formula of the compound is $\ce{C3H6}$

Your answer is correct, well structured and well written (using TeX on your first post, good job!), so you have my upvote. However, for future reference, according to site policy we usually don't directly answer homework questions, especially when the asker shows zero effort in solving them and simply copypastes the problem for us to work on. Just something to keep in mind next time.

@NicolauSakerNeto I'll keep that in mind!


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Which causes of quenching increase specifically in electrochemical experiments?
If quenching is seen as a result of running electrochemical experiments on a solution, what kind of processes may be responsible? I'm thinking about UV/Vis spectroscopy, if it makes any difference.
The redox couple would be iodide ions, $\ce{I-/I^3-}$. I understand iodide ions to be quenching agents, so my suspicion would be that their varying concentrations would cause a variation in quenching over the course of the experiment.

I wonder if, when you say UV/Vis spectroscopy, you are also including fluorescence in that. If you are, then the soft iodide / triiodide ions are pretty good reductive quenching agents and they may be responsible for the reduction in fluorescence emission you observe.
Typically, the $\text{I}^-$ will react with the excited state of your dye $\text{D*}$ and quench it by photo-induced electron transfer:
$$\ce{I^{-} + D}^*\ce{\quad ->[\text{photo-induced electron transfer}]\quad I^{\ .} + D^{\ .\ -}\quad ->[\text{back electron transfer}]}\quad \ce{I- + D}$$
As you can see, this process provides a non-radiative pathway for the excited state of the dye to decay back to its ground state. The process is essentially catalyzed by the iodide anion, which is not consumed in the process. The more iodide you have, the more prevalent this process, the more likely it is for the excited state of your dye to go through this process rather than fluorescing, so ultimately the more quenched your dye.
Such a process can be modeled by the Stern - Volmer equation:
$$ \frac{I_0}{I}=1+k_q\tau_0[\text{Q}]$$
Where $I_0$ is the fluorescence intensity in the absence of quencher Q; $I$ the intensity in the presence of a certain concentration [Q] of quencher; $k_q$ is the kinetic constant of the quenching process; $\tau_0$ is the intrinsic lifetime of the fluorophore, i.e. the lifetime you would measure in the same conditions, but in the absence of quencher Q.


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How can decreasing in temperature indicate an endothermic reaction?
I’m really wondering how can decreasing the temperature refer to an endothermic reaction.
I faced this when I was calculating the heat of a reaction Q in a constant-pressure calorimeter.
What I believe is that when something becomes cooler, then it has released some amount of heat.

Endothermic systems are the systems that proceed by gaining heat from the surrounding. Inserting a thermometer into such system, the heat is taken from the thermometer by this system as your thermometer becomes the immediate surrounding.

@GaiSamuelPandakDeng Reactions or processes can be endothermic, but systems can't. The reaction does not gain heat, it gains energy. This energy is mostly from the thermal energy of the solvent, and the energy exchange is called heat transfer. Sometimes heat transfer goes along with temperature changes, and sometimes it does not (e.g. phase transition, infinitely large stirred thermal bath).

Let’s say that you have a beaker of water. You measure the water’s temperature using a thermometer, and you determine that it is 25 °C. Now you add two reactants to the beaker, and mix them together well until they are completely reacted. If the reaction between these two reactants is endothermic, then the chemicals will absorb heat from the water. If you were to now remeasure the water’s temperature, you might find that it was something like 23 °C. This is because the heat that was absorbed by the chemicals was lost by the water.

What if the reaction is endothermic and exergonic? The opposite reaction would be exothermic. So in the equilibrium products are favor. Therefore shouldn't the temperature increase?

In case of endothermic reactions, products possess more energy than substrates. The energy difference (more precisely - enthalpy difference) was taken from surroundings (e.g. solvent), so the surroundings energy is lower. Thus, its temperature, which is a measure of average kinetic energy of molecules, is lower.

By definition, an exothermic reaction gives off heat, and endothermic reaction (e.g. dissolution of ammonium nitrate in water, as in "cool-packs") absorbs heat from the environment in order to proceed.
Your assumption is based on the idea that heat leaves the reactants in order to cool them, which is not the case in an endothermic reaction, but is the case with a heat pump. These are different mechanisms; in the endothermic reaction, heat energy is used to force the reaction to proceed, e.g. a change of state from solid to liquid.

You are absolutely right. And also wrong. You are right, when something becomes cooler, heat is released from it. Your mistake is you are looking for your answer from the wrong perspective, I believe.
When you are dealing with calorimetry and heat of reaction, it can seem contradictory at first. You learn that an exothermic reaction is one in which heat leaves, and an endothermic reaction is one in which heat enters the system. Then you go to determine if a process is endo or exo, and it seems backwards. When measuring, say a chemical reaction in water, you are determining whether the change was exothermic or endothermic. It can seem like the answer is exothermic when heat leaves the water. And it is- for the water. You want to determine the change for the chemical reaction.
So, if water temperature decreases, then heat has left the water. For the water, this would be exothermic. But for the reaction, heat has entered that system. So the reaction is then, of course, endothermic.
Remember, you are determining heat of the system. Not heat of surroundings.
      System = chemicals     Surroundings = Water

What is your "system" in this case?
The reaction that you are performing--your reactants and your products--constitute your system. Since the reaction is endothermic, heat has to enter your system for the reaction to proceed.
But where does this heat come from? It is obviously drawn from the surroundings, and thus you measure a temperature that is now lower than what you started with.

In an endothermic change, temperature is absorbed from surrounding molecules to continue reacting. If these molecules are losing heat, that means their temperature will drop, resulting in a temperature decrease.

If you place a thermometer into a beaker where an exothermic reaction is reacting ( say, sodium hydroxide and water), the temperature will rise. That means that there is heat being released from the reaction?  Isn't the inside of the beaker the location of the reactants themselves and shouldn't that temperature be lower because those molecules are releasing heat?


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How to correctly define anode and cathode?
I am very much confused: in a simple voltaic cell ($\ce{Zn}$, $\ce{Cu}$ in $\ce{H2SO4}$), $\ce{Cu}$ is positively charged and is the anode, where in a Galvanic cell ($\ce{Zn}$ in $\ce{ZnSO4}$ and $\ce{Cu}$ in $\ce{CuSO4}$),
$\ce{Cu}$ is positively charged but is the cathode. How can I define it, where both are electric or voltaic cell?

Cathode is the electrode where the Reduction reaction takes place.
Anode is the electrode where the Oxidation reaction takes place

Voltaic cell is a synonym for a Galvanic cell. In the mentioned system ($\ce{Zn}$, $\ce{Cu}$ in $\ce{H2SO4}$), $\ce{Cu}$ can be charged both positively and negatively. It depends how you connect it to the power source. In a Galvanic/Voltaic cell (as Babounet mentioned):
The cathode is a negatively charged electrode where reduction takes
place, and the anode is a positively charged electrode where oxidation takes place.
Let's analyse the second cell. Without any additional information, I assume that no external power source is connected. This system is known as Daniell cell:

\begin{align}
\ce{Zn | Zn^2+ &|| Cu^2+ | Cu}\\
\ce{Zn - 2e- &-> Zn^2+} &\therefore\text{oxidation} &= \text{anode}\\
\ce{Cu^2+ + 2e- &-> Cu} &\therefore\text{reduction} &= \text{cathode}
\end{align}
The situation is nearly the same for the first cell, but this situation is used and known as Sacrificial Anode. $\ce{Zn}$ acts in the system as an anode and it is dissolved.
You can have two electrodes that are both positively charged such as $\ce{Ag}$ and $\ce{Cu}$. Standard electrode potential of both metals are positive: $\ce{Ag+|Ag (s)} = \pu{+0.80 V}$ and $\ce{Cu^2+ | Cu (s)} = \pu{+0.34 V}$, but if you connect them in a galvanic cell, one will act as cathode ($\ce{Au}$) and the other as anode ($\ce{Cu}$).
As far as I can tell (based on provided informations), the copper electrode will be in both cases positively charged and therefore be a cathode. There must be a mistake in the description or in the statement.

We can take this definition -
Cathode is the electrode which ejects electrons into the site of interest/study.
This holds for both electrolytic and galvanic cells.
In an electrolytic cell, the solution in which the electrodes are immersed is the area of interest and cathode would be that electrode which would eject electrons into the solution.
In a galvanic cell, the half cells (the electrodes dipped in their metal salt solutions) where chemical reactions happen responsible for electric current generation is the area of interest and not the wire. In a typical galvanic cell, the Daniell cell, we can see that Cu electrode would be the cathode as it ejects electrons into the CuSO4 solution reducing the copper(II) ions to copper.
In a cathode ray tube experiment, the CRT would be the area of interest and electrons are ejected from the cathode into the tube and are incident on the glass behind the anode.
The statement that cathode is the electrode where reduction takes place is slightly misleading. This is because oxidation and reduction take place simultaneously(have to). In Daniel cell, Cu electrode is the cathode as Cu(II) ions get reduced to Cu. In electrolysis of molten KCl, the electrode where K(I) ions get reduced to K metal is the cathode. However, note that K(I) ions in the second case come from the solution and the electrodes need not be made of K whereas in the Daniel cell, Cu metal is the electrode itself. In one case we talk about the reduction of electrode while in the other we talk about reduction of something present in the vicinity of electrode. If this is kept in mind, then we could remember cathode as that electrode where reduction takes place.
So, the bottomline is, it is important to know the site of interest/study which is different for a galvanic and an electrolytic cell as the cathode and anode are terms based on that.
Edit: In a galvanic cell, the cathode is positive because it ejects electrons and becomes positive (the electrode being neutral in the beginning and attains an equilibrium). In an electrolytic cell, we are maintaining a potential difference across the electrodes and forcing the reaction to take place and electrons move from negative to positive terminal (being negatively charged, they feel a force opposite to the direction of electric field) and cathode has negative polarity.

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text:
Synthesis of a bicyclo[2.2.2]octane from benzoic acid

I think that it is a Diels Alder reaction, but I don't recognize the start materials!
First I make the tertiary alcohol from carboxylic acid with organo-lithium reagent; after I try to make a Diels Alder reaction, but I don't find the right conjugated diene and dienophile.

Supposed that the synthesis does involve a Diels-Alder reaction, it is nowhere written that it is the last step in the sequence.
Take another look at your product and imagine a $\ce{C=C}$ double bond somewhere. Does it now look more like a Diels-Alder product? The $\ce{C=C}$ can be turned to $\ce{C-C}$ using... (You figure that out).
Work your path back from here and try to identify diene and dienophile.

Don't look until you've figured out the diene ;)

 Convert acid to ester, 2 x $\ce{MeMgBr}$, Diels-Alder reaction with vinylmethyl ether, hydrogenate double bond. You didn't cheat, did you? ;)

The vinyl ether reacts with the aromatic alcohol by way of a stereospecific Diels-Alder?  Isn't a Birch reduction and subsequent isomerization required?

Also, it's probably better to keep the ester intact for an inverse electron demand Diels-Alder with vinyl methyl ether.

@ron Sssssssh, there wouldn't be a diene otherwise ;) But this wasn't your or my homework assignment.

@KlausWarzecha thank you so much for the clue! I swear that I don't look your complete answer. I think that my dienophile is the aromatic ring, so first I reduct it with Birch reduction. But which is my diene?


text:
how to separate gamma-butyrolactone from N-methyl-2-pyrrolidone
Since these two chemicals have different freezing temps (pyrrolidone is $-11~\mathrm{^\circ F}$ [$-24~\mathrm{^\circ C}$] and $\unicode[Times]{x3b3}$-butyrolactone freezes (also called melting) at $-46~\mathrm{^\circ F}$ [$-43~\mathrm{^\circ C}$]), could a person wanting to separate these just immerse them in liquid nitrogen (or something else?) to get the desired chemical to freeze, then filter off the unfrozen butyrolactone? Other routes seem quite difficult for a non chemist. Any objections or other routes? Anything you can add? Even a step by process will be much appreciated.

I do wonder why you are interested in the separation of GHB as a non-chemist.

Can you add info on the relative fraction of each in the mixture as well as which one you want to recover pure? Both of those factors have significant impact on what would be a useful method to apply.

Cud [sic] a person wanting to separate these just emerse [sic] them into liquid nitro [sic] or any [sic] to get the desired chemical to freeze then filter off the unfrozen butyrolacton [sic]?

With liquid nitrogen this would not be possible because liquid nitrogen condenses/boils at -195 °C.  It would freeze both the pyrrolidone and the lactone.

Well. Then say what would a person use to separate these two compounds that would not require distillation and adding of chemicals. Freezing the undesirable must be possible ?

I'm not sure which compound (both?) you need to recover.  If you want to recover just the NMP, then you could probably separate it from the GBL by hydrolyzing the GLB with a strong base, for 4-hydroxybutyrate salts, and then doing a solvent extraction with chloroform or hexanes to separate the NMP from the GBL-derived salts.  Then the NMP can be re-isolated by evaporation of the chloroform or hexanes.

My answer originally said LN2 "freezes" but of course I meant "condenses".  I'm talking about the boiling point not the freezing point so I am editing my answer.

As Curt F. has already pointed out, liquid nitrogen is an overkill and won't help here.
You might want to try one of the following cooling mixtures - values are taken from (http://www.chemieunterricht.de/dc2/tip/08_98.htm):

81 g $\ce{CaCl2*6H2O}$ + 100 g ice : -21,5 °C (this might not be enough)
66 g $\ce{NaBr}$ + 100 g ice : -28 °C
105 g ethanol + 100 g ice : -30 °C

to enrich NMP in ther solid fraction.


text:
Show that there can be 18 electrons in a "g" subshell
I'm learning about quantum numbers right now, and I can't figure out how to calculate the number of electrons in each shell without memorizing it. Is there a formula to find the possible numbers of electrons in the s, p, d, f, and other (theoretical) subshells? If it helps, this problem is put under the "Quantum Numbers, Exclusion Principle" topic.

Simple mnemonic: each orbital, s, p, d and f, has double the consecutive odd numbers of electrons, i.e. 1, 3, 5 & 7 --> 2, 6, 10 & 14.
BTW, from spectroscopy, s, p, d & f refer to splitting of lines, the sharp, principal, diffuse and fine; see Atomic orbital. As Tom Lehrer stated, "These are the only ones of which the news has come to Hahvahd, and there may be many others, but they haven't been discovahd." Although element 121, unbiunium, begins the theoretical g-block...


text:
Why is the vertical part of a titration curve longer for a strong acid by a strong base than for a weak?
As the title says, I would like to understand chemically why this occurs. My theory behind it is that in the former case both are completely dissociated, and so at the equivalence point (and approaching it), everything is controlled increasingly by the autoionization of water, whereas in weak acids there is also a much more sizable contribution of $H^+$ by the acid itself, so the change is not as drastic. If this is correct, I'd like further explanation as to why it is, since I don't quite feel comfortable enough with it intuitively.
This is not a duplicate. I am not asking why the strong acid curve is steep, but rather why the strong acid curve is steep while the weak acid one is not.!
Weak acid and strong acid titration curves with addition of a strong base

Could you provide an example of titration of a strong acid by a strong base? Usually titrations are of weak acids and bases

This is irrelevant. The tendency is the same for all such titrations. However, consider the titration of 0.1 M HCl by 0.2 M NaOH, for instance.

In regards to your edit, is this an example of what you're talking about?

@JohnSnow, yes, it is exactly the example.

Compare (1) the titration of $\ce{NaOH}$ with $\ce{HCl}$ and (2) that of $\ce{NaOH}$ with $\ce{AcOH}$.
At the equivalence point of each titration, solution (1) contains $\ce{Na+}$, $\ce{Cl-}$, and $\ce{H2O}$, with negligible concentration of $\ce{H+}$ and $\ce{OH-}$. In contrast, solution (2) contains $\ce{Na+}$, $\ce{AcO-}$, and $\ce{H2O}$, and in particular $\ce{AcO-}$ will react with water to form a buffer solution of $\ce{AcO-/AcOH}$.
This will be true in general for any weak acid/base titration. The resulting buffer resists changes in pH; hence a weak-acid titration curve is not as steep at the equivalence point as is a strong-acid titration curve.


text:
Why does boron fume in moist air? Is the following accurate?
I initially guessed it was due to the formation of $\ce{H3BO3}$ but am now doubtful. Does it have anything to do with $\ce{HCl}$ formation? If the question referred to a boron halide, I could answer with that. Either the question is wrong, or my guess is. What's accurate?
I'd appreciate any help.
Grade 11 Chem.

Boron does not react with (moist or dry) air at room temperature. Do you mean ‘borane’?

If so, does my answer work?

If you're talking about boranes (simplest example $B_{2}H_{6}$), then the gas produced by reaction with water is hydrogen, and the boron containing product is $B(OH)_{3}$. I suspect most boranes would explode if you exposed them to oxygen though so not sure what experiment you are referring to.


text:
What are the signs for work done by (on?) a gas during isothermal reversible compression?
For the expansion of a gas, isothermally and reversibly, what is the sign taken?
Do I use this equation
$-nRT\ln  (Vf/Vi)$
or this one
$nRT\ln  (Vf/Vi)$ ?
Moreover, when I'm asked about the 'work done', what exactly does the work? The gas itself? The surroundings? I'm confused about this because my text says one thing and the guidebook, another.
Thanks.

The easiest way to remember the sign conventions is to think of the gas particles as "wanting" to get away from each other. To reduce the volume of the gas, you have to do work on the gas - the surroundings are doing the work. In chemistry we always take the point of view of the system. If we do work on the system, we add energy to it, and therefore, the sign of work is positive.
This means the opposite is true as well. When the volume of the system increases, the sign of the work term must be negative - the system is losing energy and doing work on the surroundings.
For your equations, all of this holds for the first one:
$$-nRT\ln  (Vf/Vi)$$
This is because a decrease in volume will give you a fraction in the $\rm{ln}$ term, which leads to a negative sign. The leading negative sign reverses that. The result is that any compression will have a positive work term, and any expansion will have a negative work term.
You can see how the correct sign is a result of the derivation in the wikipedia article.

THANKS!!!
Is the physical or chemical view? Moreover, during the expansion of a gas, is the work done by the gas? If so, is it positive? My textbook tells me to calculate merely 'work done' at constant external pressure and ends up with +16.1 L atm. What does this work?

Work is a transfer of energy: E = q + w. This means that work done on the system is adding energy to the system, so the sign must be positive. During expansion, the system (the gas) is doing work on the surroundings, so the sign should be negative. If they have a positive sign for work during an expansion, then they are taking the opposite convention, or asking from the point of view of the surroundings.

I see, so essentially. In the case given here (this is Chemistry not Physics by the way), they're asking for the work done 'on' the system. The case in topic, to be specific is here, and the reason I ask is in lieu of my prime examination tomorrow.
https://imgur.com/jKGvHRN

I would check with whomever is grading the exam. In the classes I teach, if the question is worded ambiguously, I tell my students to write something like "sign is negative because the system is doing the work" - I don't care if they can guess what I meant in the question, what I care about is whether or not they understand what the signs mean and what the direction of energy transfer is. Your examiner might care about something different.

Shouldn't it be "when the volume of the system increases" instead?

@SohamChowdhury yes - that was a mistake. Thanks for catching that!


text:
Combustion in liquid state
Normally, we think of combustion as occurring only in the gas phase. I think I understand the main reason for this - in the gas phase it's very hard for a bimolecular reaction to produce a single product, which makes it more likely for radical chain reactions to propagate rather than being terminated - but I'm curious about whether it's possible for combustion to occur in the liquid state at all.
For example: suppose I mixed liquid $\ce{CH4}$ with liquid $\ce{O2}$ under high pressure and reasonably low temperature. Would they react immediately? If not, could I 'ignite' the mixture in the same way as I could in the gas phase? (Note that the combustion of $\ce{CH4}$ into $\ce{H2O}$ and $\ce{CO2}$ doesn't change the mole number, so the volume change should be relatively small.) Would the reactions that occur be similar to the reactions that occur during combustion of $\ce{CH4}$ in the gas phase, or would they be completely different?
If this reaction wouldn't resemble combustion, is there some other example of a self-propagating oxidation reaction in liquid or aqueous phase with similar properties to combustion?

Your question is addressed by Explosive Systems Containing Liquid Oxygen: Liquid Oxygen-Liquid Methane Mixtures. J. Chem. Eng. Data, 1959, vol. 4, pp. 127–131.
If liquid methane and liquid oxgygen at 90K are mixed, they are miscible in all proportions.  Mixtures over a wide range of proportions detonated with use of a blasting cap. 100 grams of mixture punched a 10cm hole in a 3/16 inch steel plate.
For comparision, the TNT control only dented the metal plate!
Further tests on the stoichimetric liquid methane/oxgygen mixture found that it was detonated by a .22 caliber copper bullet at 40 foot range, flame and spark.
So nothing happens if the liquids are simply mixed, but a powerful explosion occurs if a stimulus is applied.
Liquid oxygen and liquid methane have also been used in rocket engines.  See Liquid Methane/Liquid Oxygen Injectors for Potential Future Mars Ascent Engines for example.

Nice, thank you. I suppose the explosion is due to the very sudden generation of lots of heat. I was interested in the theoretical case of constant volume (so that the products would also be liquids), but I guess that would be rather hard to achieve in practice!

Perhaps your question is rather what would it take to have a fluidic version of a controlled "burn" rather than an explosion. Interesting! The necessary conditions would be introduction of one liquid ("fuel") into another ("oxidizer") via a nozzle, with a reaction occurring at the interface at a rate sustained by continuous introduction of more of the "fuel" component. Methane and oxygen react too violently - maybe you could go to fluoromethane or difluoromethane and tone it down. Or go to hydrogen peroxide as the oxidizer.
A lot of work has been done on developing extremely compact, lightweight energy sources that can power an engine in extreme environments. For example, liquid lithium and sulfur have been used to power a reciprocating engine.


text:
Strain Energy in di-axial vs di-equatorial
I am dealing with two conformations of cis-1,4-dimethylcyclohexane. One conformation is di-axial, and the other one is di-equatorial. Why does di-axial have higher strain energy. I would make sense if the molecule was 1,3-dimethylcyclohexane because then it would the steric hinderance that would cause higher energy. Is it similar with 1,4-.. conformation? If its not steric repulsion, what factors cause di-axial to have higher strain energy?

Di-axial and di-equatorial in cis-1,4-dimethylcyclohexane in a chair conformation?

Look at your molecule again. Are you sure you are not talking about trans-1,4-dimethylcyclohexane? In the cis-isomer both methyl groups must point in the same direction relative to the ring. That means one must be equatorial and one must be axial.

You can't have both in a chair conformation!

If the methyl groups in 1,4-dimethylcyclohexane adopt a bis-axial orientation, they are trans and they stay trans after ring flip:

If your 1,4-dimethylcyclohexane is cis-substituted the orientation of the methyl groups is axial and equatorial, respectively:

They could be both equatorial in a boat or twisted configuration, though. Currently I am not too sure which is the thermodynamically most stable one.


text:
What is the priority of a phenyl group in a compound such as this?
I assumed this:
http://www.chemspider.com/Chemical-Structure.12491.html?rid=222625d4-811c-4b07-b263-8930fafa831a
to be 1-phenylbut-4-ene
Where is the priority here? No source I know of tells me of where the priority of the phenyl  group lies.
It is wrong. The book states it is 4-PhenylBut-1ene.
What is the priority of the phenyl group?

For preferred IUPAC names, the ring is preferred to the chain.
Therefore, the preferred IUPAC name for the given structure is but-3-en-1-ylbenzene.
However, the context may favour the chain, so that, for example, an unsaturated acyclic structure may be recognized.
When one double bond is present, an unbranched alkene is named by changing the ending to ‘ene’. The double bond is assigned the lower locant possible, which is placed immediately in front of the ending ‘ene’. When several structural features appear, low locants are assigned to ‘ene’ endings before detachable alphabetized prefixes.
Therefore, a correct name for the given structure is 4-phenylbut-1-ene.


text:
Carbocation stability order
I was wondering whether the following order (from here) is correct:

In the first inequality, why did 9 hyperconjugations dominated resonance effect of benzene ring?
In the last inequality doesn't the 2 hyperconjugations of vinylic carbocation overpower the methyl carbocation?

In middle there is:
allyl = secondary

Why is there an equality here when resonance dominates inductive effect?

The vinyl group exerts a -I effect that destabilises the positive charge and makes it more unstable than the methyl carbocation.

@BinaryGeek But what about the HC ?

related http://chemistry.stackexchange.com/questions/10509/instability-of-vinylic-and-arylic-carbocations?rq=1

also http://chemistry.stackexchange.com/questions/19027/inductive-effect-and-hyperconjugation-one-elephant-different-parts?lq=1

I'm not sure about this but I don't think vinylic hydrogens participate in hyperconjugation.

@BinaryGeek I think they do. Do you have some link for stability orders?

@pikachu Sorry, but I don't

@pikachu Vinyl hydrogens can't participate in hyperconjugation because they don't have the correct geometry to allow the orbital interactions.

Also related: https://chemistry.stackexchange.com/q/74943/4945

About the first two, it's an exception to the rule that resonance dominates hyperconjugation.
About the last inequality, the vinyl cation is sp hybridized which means it is highly electronegative, and a positive charge on it will be very unstable.
About the middle inequality, the benzyl carbocation (with five resonance structures) is more stable than allyl carbocation (two resonance structures). You may verify this here.

Would you have a source for the claim "sp hybridized which means it is highly electronegative"?

https://books.google.co.in/books?id=gY-Sxijk_tMC&pg=PA16#v=onepage&q&f=false

You got the middle equality wrong . Look at the highlighted text . I had asked for allyl and secondary . Thanks for the other parts .

What will the stability order of $2^{\circ}$ and $3^{\circ}$ benzylic carbocation in comparison to tertiary carbocation?

As for the 1st two:
Consider the 3 R's as independent(I am just explaining) ,each pushing a part of the electron cloud in its bond towards the central carbon through inductive effect as well as Hyper-conjugation. So in a way each R is reinforcing each other and an overall combined stability is achieved.
However consider  the resonating structures of Benzyl carbocation. If you research a bit deeper , you will find that of all the 4 resonating structures possible, the form that is given ( + on the external carbon) contributes to almost 90% of the original stable structure. This is because, if you take any other resonating structure( say + charge on a carbon inside the ring ), you will find that the structure is quite unstable and very highly tends to resort back to the + charge outside the ring ( after all, breaking a Benzene ring isn't very favorable) . So in a way , the resonating structures do not 'reinforce' each other so "Resonance" does not stabilize the way it is expected to do (as in case of similar resonating structures)
Talking of the last inequality:
The "Hyper-conjugation" that you mentioned wouldn't play the role here as you imagine. You might- have overlooked the fact that vinyl cation is least stable MOSTLY because it has a very high tendency to get converted to another structure, namely the left carbon(in your diagram) with a + charge and the right one with a lone pair. This is because ,then it will have many more hyper-conjugations to get stable. This is what makes vinyl carbocation least stable, it very very rarely ever exists or tries to exist in this particular form.
As for the middle equality,
Yes, in a way your doubt is right. (Since there is an equivalent resonance structure in the left one (as given) and that is a powerful one!!).However the thing you have missed while comparing the 2nd one is the fact that "Inductive" effect plays a 'minor' role in contributing to the stability of the secondary carbocation.The 'Major' role is played by the 7 "Hyper-conjugations" which is no less than a single resonance. Of course 1 hyperconjugation is less than 1 resonance but basic intuition will tell you that 7 hyperconjugations will have a fair chance to cope with 1 resonance.

For discussion of the structure of the vinyl cation see here. Your analysis is completely wrong I'm afraid.

Are my first 2 reasonings also wrong? @bon

They are not wrong, per se, just poorly worded. I do take some issue with this statement though Resonance" does not stabilize the way it is expected to do. The benzyl cation has extensive resonance stabilization which accounts for the fact that it is far more stable than otherwise expected. It's just that tertiary alkyl cations are a little bit more stable.

But ,is a compound having non equivalent resonating structures as stable as another which has stable resonating structures? @bon

What do you mean? Stable is only a relative term. Non equivalent resonance structures can still be stable compared to a different structure.

Am I wrong in saying whatever I have said in the second paragraph except the last part of the last line(on which you raised a question)?


text:
dependence of Intensity on penetration of light
I want to know about a transparent material,In which penetration of light decreases as intensity of light increases.

Nonlinear optics is what you're looking for.

@NicolauSakerNeto I am not sure that it is intentional. For me the question sounds like much more basic, and I think OP should first check Lambert-Beer law http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law

Are you trying to describe photochromism? See http://en.wikipedia.org/wiki/Photochromism.

In general, when light penetrates a material (water for example) its intensity decreases exponentially with increasing depth. This process is known as attenuation or absorption. The severity of attenuation differs with the wavelength of electromagnetic radiation. In the region of visible light, the red part of the spectrum attenuates more rapidly than the shorter-wavelength blue part.
But what you are asking about is a non-linear optical property known under the name: Saturable absorption It is a property of materials where the absorption of light decreases with increasing light intensity. At sufficiently high incident light intensity, atoms in the ground state of a saturable absorber material become excited into an upper energy state at such a rate that there is insufficient time for them to decay back to the ground state before the ground state becomes depleted, and the absorption subsequently saturates. The material becomes transparent at that wavelength. Saturable absorbers are useful in laser cavities.
Another kind  of non-linear materials is known under the name "optical power limiting materials". In these materials the absorption of light increases with intensity such that beyond a certain input intensity the output intensity approaches a constant value. Such a material can be used to limit the amount of optical power entering a system. This can be used to protect expensive or sensitive equipment such as sensors, can be used in protective goggles, or can be used to control noise in laser beams.
For more details, please see: http://en.wikipedia.org/wiki/Saturable_absorption
and http://en.wikipedia.org/wiki/Two-photon_absorption


text:
IUPAC Nomenclature of Benzene+Aliphatic chain

What is this compound's IUPAC name?
1) 2-phenyloctane.
2) octan-2-ylbenzene.
Is using phenyl correct or is taking benzene as the base correct?

I did the following searches to find that if aliphatic chain has more than 6 carbon atoms then it prefers using phenyl : http://m.wolframalpha.com/input/?i=2-phenylpentane&x=0&y=0   &   http://m.wolframalpha.com/input/?i=2-phenylhexane&x=0&y=0 . Also this says so too. So is it correct ?

I cant add more than 2 links , hence the comments . Can someone add it in the question ?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. Let me get this straight: You can't add links?! And, [tag:iupac] is used for the IUPAC org itself, not nomenclature.

@Loong So both of them are correct ?

@MARamezani Not more than two links

@pikachu Yes, both ‘octan-2-ylbenzene’ and ‘2-phenyloctane’ are correct; ‘octan-2-ylbenzene’ is the preferred IUPAC name.

@Loong But this says While compounds like these are usually named by simple benzene type naming (chlorobenzene and bromobenzene), the phenyl group naming is usually applied to benzene rings where a substituent with six or more carbons is attached, such as in the diagram below.

@Loong Please have a look at my previous comment . thanks.

@pikachu That site is not necessarily wrong, since according to current IUPAC recommendations (2013), “the context may favor (…) the chain, so that (…) an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms (…) in the principal chain of the acyclic structure.” However, the approach for the preferred IUPAC name (PIN) prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

@pikachu The old IUPAC rule (from 1979) actually was: “Choice between these methods is made according to the more appropriate of the following principles: (…) (b) treatment of a smaller of structure as a substituent into a larger. (…) In accordance with the principle (b) (…), a hydrocarbon containing a small cyclic nucleus attached to a long chain is generally named as a derivative of the acyclic hydrocarbon; (…)”

@Loong Okay . Thanks !!! So the site goes with the 1979 rule .


text:
Trying to figure out heat of reaction
A question in Thermodynamics of Natural Systems by G. M. Anderson is to find the "standard heat of reaction" of this reaction:
$$\ce{C8H16N2O3(aq) + H2O(l) -> C6H13NO2(aq) + C2H5NO2(aq)}$$
As far as I understand it, it should be a rather simple problem of finding $\Delta_\mathrm r H$.
So it should be:
$\Delta_\mathrm r H=\Delta_\mathrm fH (\ce{C6}) + \Delta_\mathrm fH (\ce{C2}) - \Delta_\mathrm fH (\ce{C8}) - \Delta_\mathrm fH (\ce{H2O})$ (note that I shortened the organic formulae)
According to the data from the same textbook, it's:
$(-632077~\pu{J/mol}-513988~\pu{J/mol})-(-847929~\pu{J/mol}-285830~\pu{J/mol})=-12306~\pu{J/mol}$
However, the answer given is $−89475~\pu{J/mol}$.
Am I missing something? Do I need to add some terms of pressure or volume expansion? Is this not the enthalpy that I'm looking for?

Have you considered estimating the enthalpy change using bond energies? It's more approximate than using standard formation enthalpies, but might be useful to compare against your value and the textbook value. It would require knowing the structures for the organic molecules, however.

@NicolauSakerNeto this is a basic textbook for introductory thermodynamics. I'm certain that using bond energies is not required as it is not even covered in this book. I'm thinking that it may be an error either in the data provided or in the answer given...

It might not be required, but it's a way to check! Do you have the structures or the compound names?

Leucylglycine + water = leucine + glycine

Interesting, a calculation by bond energies assuming Dave has the right structures (Dave, are you a wizard?!) suggests the reaction enthalpy should be $-85\ kJ\ mol^{-1}$. Maybe there is a print error in the enthalpy data.

@baffledu What do you mean by "the answer given is -89745" The book says the answer is -12306 at page 267.

@DavePhD weird - I didn't get that far in the book. I'm using what they have on their website: http://www.cambridge.org/download_file/203255 See chapter 3 Q&A8. So I guess it's just a mistake in that file?

I think it's a mistake in that answer key I was looking a page 267 here: https://www.google.com/?gws_rd=ssl#q=Thermodynamics+of+Natural+Systems&tbm=bks

There is a "readme" note from the author that says:

The answers are not guaranteed, in the sense that not only
  might I make a mistake here and there, but I might have used different sources of
  data, or the data in a database may have changed. I have not been able to check all
  the answers, or to make them all consistent with the data in Appendix B, although I
  believe that most are.

So I think that this problem is an example of one of those situations.
The book itself works out the problem on page 267 and gets the same answer as you: −12306 J/mol.